PROCESS HEAT TRANSFER D. Q. KERN
INTERNATIONAL STUDENT EDITION
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PROCESS HEAT TRANSFER D. Q. KERN
INTERNATIONAL STUDENT EDITION
..... .. .. ... .. ... .•• •. ..•.. ··........... ......
·····•••··· ·
PROCESS HEAT TRANSFER
PROCESS HEAT TRANSFER BY
DONALD Q. KERN D. Q. Kern Associates, and Professorial Lecturer in Chemical Engineering Case Institute of Technology
INTERNATIONAL STUDENT EDITION
McGRAWHILL INTERNATIONAL BOOK COMPANY Auckland Bogota Guatemala Hamburg Johannesburg Lisbon London Madrid Mexico New Delhi Panama Paris San Juan Sao Paulo Singapore Sydney Tokyo
PROCESS HEAT TRANSFER
INTERNATIONAL STUDENT EDITION Copyright© 1965 Exclusive rights by McGrawHill Book Company Japan, Ltd. for manufacture and export. This book cannot be reexported from the country to which it is consigned by McGrawHilL 21st printing 1983
Copyright, 1950, by the McGrawHill Book Company, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording. or otherwise, without the prior written permission of the publisher. When ordering this title use ISBN 0070853533
TOSHO PRINTING CO,, LTD., TOKYO, JAPAN'
To my wife
NATALIE W. KERN for her real help
PREFACE It is the object of this text to provide fundamental instruction in heat transfer while employing the methods and language of industry. This treatment of the subject has evolved from a course given at the Polytechnic Institute of Brooklyn over a period of years. The possibilities of collegiate instruction patterned after the requirements of the practicing process engineer were suggested and encouraged by Dr. Donald F. Othmer, Head of the Department of Chemical Engineering. The inclusion of the practical aspects of the subject as an integral part of the pedagogy was intended to serve as a supplement rather than a substitute for a strong foundation in engineering fundamentals. These points of view have been retained throughout the writing of the book. To provide the rounded group of heattransfer tools required in process engineering it has been necessary to present a number of empirical calculation methods which have not previously appeared in the engineering literature. Considerable thought has been given to these methods, and the author has discussed them with numerous engineers before accepting and including them in the text. It has been a further desire that all the calculations appearing in the text shall have been performed by an experienced engineer in a conventional manner. On several occasions the author has enlisted the aid of experienced colleagues, and their assistance is acknowledged in the text. In presenting several of the methods some degree of accuracy has been sacrificed to permit the broader application of fewer methods, and it is hoped that these simplifications will cause neither inconvenience nor criticism. It became apparent in the early stages of writing this book that it could readily become too large for convenient use, and this has affected the plan of the book in several important respects. A portion of the material which is included in conventional texts is rarely if ever applied in the solution of runofthemill engineering problems. Such material, as familiar and accepted as it may be, has been omitted unless it qualified as important fundamental information. Secondly, it was not possible to allocate space for making bibliographic comparisons and evaluations and at the same time present industrial practice. Where no mention has been made of a recent contribution to the literature no slight was intended. Most of the literature references cited cover methods on which the author has obtained additional information from industrial application. vii
Vlll
PREFACE
The author has been influenced in his own professional development by the excellent books of Prof. W. H. McAdams, Dr. Alfred Schack, and others, and it is felt that their influence should be acknowledged separately in addition to their incidence in the text as bibliography. For assistance with the manuscript indebtedness is expressed to Thomas H. Miley, John Blizard, and John A. Jost, former associates at the Foster Wheeler Corporation. For checking the numerical calculations credit is due to Krishnabhai Desai and Narendra R. Bhow, graduate students at the Polytechnic Institute. For suggestions which led to the inclusion or exclusion of certain material thanks are due Norman E. Anderson, Charles Bliss, Dr. John F. Middleton, Edward L. Pfeiffer, Oliver N. Prescott, Everett N. Sieder, Dr. George E. Tait, and to Joseph Meisler for assistance with the proof. The Tubular Exchanger Manufacturers Association has been most generous in granting permission for the reproduction of a number of the graphs contained in its Standard. Thanks are also extended to Richard L. Cawood, President, and Arthur E. Kempler, VicePresident, for their personal assistance and for the cooperation of The Patterson Foundry & Machine Company. DoNALD
NEW YORK, N.Y. April, 1950
Q.
KERN
CONTENTS PREFACE . . . . • . . . . . . . . . . . . . . .
vii
INDEX TO THE PRINCIPAL APPARATUS CALCULATIONS.
xi
CHAPTER
1. 2. 3. 4. 5. 6.
7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21.
Process Heat Transfer. Conduction. Convection. . Radiation . . Temperature . . Counterflow: Doublepipe Exchangers . 12 Parallelcounterflow: ShellandTube Exchangers. Flow Arrangements for Increased Heat Recovery . Gases . . . . . . . . . . . . . . . . Streamline Flow and Free Convection . Calculations for Process Conditions . Condensation of Single Vapors . Condensation of Mixed Vapors . . . Evaporation . . . . . . . . . . . Vaporizers, Evaporators, and Reboilers. . Extended Surfaces . . . . . . . . . . . Directcontact Transfer: Cooling Towers. Batch and Unsteady State Processes. Furnace Calculations. . . . . . . . . . Additional Applications . . . . . . . . The Control of Temperature and Related Process Variables.
1
6 25 62 85 102 127 175 190 201 221 252 313 375 453 512 563 624 674 716 765
APPENDIX OF CALCULATION DATA.
791
AUTHOR INDEX.
847
SUB.Jl!lCT lNDmx.
851
ix
INDEX TO THE PRINCIPAL APPARATUS CALCULATIONS ExcHANGERS
Double"pipe counterflow exchanger (benzenetoluene) . Doublepipe seriesparallel exchanger (lube oilcrude oil) Tubular exchanger (kerosenecrude oil) Tubular exchanger (waterwater) . . . . . . . . Tubular cooler (K.PO, solutionwater). . . . . . Tubular heater, unbaf!led (sugar solutionsteam) Tubular 24 cooler (33.5° API oilwater) . . . . Tubular exchangers in series (acetoneacetic acid) . Tubular gas aftercooler (ammonia gaswater) . . . Tubular gas intercooler (C02water vaporwater). . . Tubular streamline flow heater (crude oilsteam). Tubular free convection heater (kerosenesteam). Core tube heater (gas oilsteam) . . . Tank heater (anilinesteam) . . . . . . Tubular exchanger (straw oilnaphtha.) . Tubular 48 exchanger (lean oilrich oil) . Tubular cooler (NaOH solutionwater) . Tubular heater (alcoholsteam). . . . . Tubular splitflow cooler (flue gaswater) Jacketed vessel (aqueous solutionsteam) Tube coil (aqueous solutionsteam) . Pipe coil cooler (slurrywater) . . . . . Trombone cooler (S02 gaswater). . . . Atmospheric cooler (jacket waterwater). Electric resistance heater . . . . . . .
113
121 151 155 161 167 181 184 193 346 203 207 211
217 231 235 238 241 246 719
723 725 729
736 758
CONDENSERS (TUBULAR)
Condenser, horizontal (propanolwater). . . . . . Condenser, vertical (propanolwater) . . . . . . . Desuperheatercondenser, horizontal (butanewater) Condensersub cooler, vertical (pentaneswater) . . . Condensersub cooler, horizontal (pentaneswater). . 11 Reflux condenser, vertical (carbon disulfidewater) Surface condenser (turbine exhaust steamwater). . Condenser, horizontal (hydrocarbon mixturewater) .. Condenser, horizontal (steam, C02 mixturewater) . . Condenser, horizontal (hydrocarbon mixture, gas, steamwater)
274 277
285 290 295
299 308 331 346 356
EvAPORATORS (T\Jl!ULAR)
381 388
Raw water evaporator. Power plant makeup evaporator
xi
xii
INDEX TO THE PRINCIPAL APPARATUS CALCULATIONS Process multiple effect evaporator. Heat transformer evaporator . . . Salt water distiller . . . . . . . Cane sugar multiple effect evaporator. Paper pulp waste liquor multiple effect evaporator Caustic soda multiple effect forced circulation evaporator Thermocompression cane sugar evaporator. . . . . . .
389 390
393 418 427
437 447
VAPORIZING ExcHANGERS (TUllULA.R)
Vaporizer, forced circulation (butantHltea.m). Kettle reboiler (hydrocarbonssteam) • . . . Thermosyphon reboiler, horizontal (naphthagas oil) Thermosyphon reboiler, vertical (butanesteam)
464 475 482 488
ExTENDED SuRFACEs
Longitudinal fin doublepipe cooler (gas oilwater) Tubular longitudinal £n cooler (oxygenwater). Transverse £n crossflow cooler (airwater).
530 535 556
DIRECT CoNTACT TRANSFER
Cooling tower requirement. Cooling tower guarantee. . Cooling tower rerating . . . Gas cooler (nitrogenwater) Gas cooler, approximate solution (nitrogenwater)
602
605 609
615 620
RADIANT HEATERS
Tube still . . Direct fired vessel.
702 709
CHAPTER 1 PROCESS HEAT TRANSFER Heat Transfer. The science of thermodynamics deals with the quantitative transitions and rearrangements of energy as heat in bodies of matter. Heat transfer is the science which deals with the rates of exchange of heat between hot and cold bodies called the source and receiver. When a pound of water is vaporized or condensed, the energy change in either process is identical. The rates at which either process can be made to progress with an independent source or receiver, however, are inherently very different. Vaporization is generally a much more rapid phenomenon than condensation. Heat Theories. The study of heat transfer would be greatly enhanced by a sound understanding of the nature of heat. Yet this is an advantage which is not readily available to students of heat transfer or thermodynamics because so many manifestations of heat have been discovered that no simple theory covers them all. Laws which may apply to mass transitions may be inapplicable to molecular or atomic transitions, and those which are applicable at low temperatures may not apply at high temperatures, For the purposes of engineering it is necessary to undertake the study with basic information on but a few of the many phenomena. The phases Qf a single substance, solid, liquid, and gaseous, are associatedwith its energy content. In the solid phase the molecules or atoms are close together, giving it rigidity. In the liquid phase sufficient thermal energy is present to extend the distance of adjacentmolecules such that rigidity is lost. In the gas phase the presence of additional thermal energy has resulted in a relatively complete separation of the atoms _or molecules so that they may wander anywhere in a confined space. It is also recognized that, whenever a change of phase occurs outside the critical region; a large amount of energy is involved in the transition. For the same substance in its different phases the various thermal properties have different orders of magnitude. As an example, the specific heat per unit mass is very low for solids, J:righ for liquids, and usually intermediate for gases. Similarly in any body absorbing or losing heat, special consideration must be given whether the change is one of sensible or latent heat or both. Still further, it is also known that a hot source is 1
2
PROCESS HEAT TRANSFER
capable of such great subatomic excitement that it emits energy without any direct contact with the receiver, and this is the underlying principle of radiation. Each type of change exhibits its own peculiarities. Mechanisms of Heat Transfer. There are three distinct ways in which heat may pass from a source to a receiver, although most engineering applications are combinations of two or three. These are conduction, convection, and radiation .. Conduction.· Conduction is the transfer of heat through fixed material such as the stationary wall shown in Fig. 1.1. The direction of heat flow will be at right angles to the wall if the wall surfaces are isothermal Hof face and the body homogeneous and isotropic. Assume that a source of ?empemfllre Direction of ofnetT/ fl< heat exists on the left face of the hofbody wall and a receiver of heat exists on Temp:f*'re the right face~ It has been known ~ ~ld body and later it will be confirmed by ~ derivation that the flow of heat per ~·'~~ ~ hour is proportional to the change x=O x=x DistAncein temperature through the wall and FIG. 1.1. Heat flow through a wall. the area of the wall A. If t is the temperature at any point in the wall and x is the thickness of the wall in the direction of heat flow, the·quantity of heat flow dQ is given by
t
dQ
=
kA ( 
:!)
Btu/hr
(1.1)
The term dtjdx is called the temperature gradient and has a negative sign if the temperature has been assumed higher at the face of the wall where x = 0 and lower at the face where x = X. In other words, the instantaneous quantity of heat transfer is proportional to the area and temperature difference ·dt, which drives the heat through the wall of thickness dx. The proportionality constant k is peculiar to conductive heat transfer and is known as the thermal conductivity. It is evaluated experimentally and is basically defined by Eq. (1.1). The thermal conductivities of solids have a wide range of numerical values depending upon whether the solid is a relatively good conductor of heat such as a metal or a poor conductor such as asbestos. The latter serve as insu:.. lators. Although heat conduction is usually associated with heat transfer through solids, it is also applicable with limitations to gases and liquids. Convection. Convection is the transfer of heat· between relatively hot and cold portions of a fluid. by mixing. Suppose a can of liquid were
PROCESS HEAT TRANSFER
3
placed over a hot flame.. The liquid at the bottom of the can becomes heated and less dense than before owing to its thermal expansion. The liquid adjacent to the bottom is also less dense than the cold upper portion and rises through it, transferring its heat by mixing as it rises. The transfer of heat from the hot liquid at the bottom of the can to the remainder is natural or free convection. If any other agitation occurs, such as that produced by a stirrer, it is forced convection. This type .of heat transfer may be described in an equation which imitates the form of the conduction equation and is given by dQ=hAdt
(1.2)
The proportionality constant h is a term which is influenced by the nature of the fluid and the nature of the agitation and must be evaluated experimentally. It is called the heattransfer coefficient. When Eq. (1.2) is written in integrated form, Q = hA t:.t, it is called Newton's law of cooling. Radiation. Radiation involves the transfer of radiant energy from a source to a receiver. When radiation issues from a source to a receiver, part of the energy is absorbed by the receiver and part reflected by it. Based on the second law of thermodynamics Boltzmann established that the rate at which a source gives off heat is dQ =
UE
dA T4
(1.3)
This is known as the fourthpower law in which Tis the absolute temperature. U is a dimensional constant, but E JS a factor peculiar to l'adiation and is called the emissivity. The emissivity, like the thermal conductivity k or the heattransfer coefficient h, must also be determined experimentally. Process Heat Transfer. Heat transfer has been described as the study of the rates at which heat is exchanged between heat sources and receivers usually treated independently. Process heat transfer deals with the rates of heat exchange as they occur in the heattransfer equipment of the engineering and chemical processes. This approach brings to better focus the importance of the temperature difference between the source and receiver, which is, after all, the driving force whereby the transfer of heat is accomplished. A typical problem of process heat transfer is concerned with the quantities of heats to be transferred, the rates at which they may be transferred because of the natures of the bodies, the driving potential, the extent and arrangement of the surface separating the source and receiver, and the amount of mechanical energy which may be expended to ~acilitate the transfer. Since heat transfer involves an exchange in a system, the loss of heat by the one body will equal the heat absorbed by another within the confines of the same system.
4
PROCESS HEAT TRANSFER
In the chapters which follow studies will first be made of the three individual heattransfer phenomena and later of the way in which their combination with a simultaneous source and receiver influences an apparatus as a whole. A large number of the examples which follow have been selected from closely related processes to permit gradual comparisons. This should not be construed as limiting the broadness of the underlying principles. Many of the illustrations and problems in the sueeeeding chapters refer to liquids derived from petroleum. This is quite reasonable, since petroleum refining is a major industry, petroleum products are an important fuel for the power industry, and petroleum derivatives are the starting point for many syntheses in the chemical industry. Petroleum is a mixture of a great many chemical compounds. Some can be isolated rather readily, and the names of common hydrocarbons present in petroleum may be identified on Fig. 7 in the Appendix. But more frequently there is no need to obtain pure compounds, since the ultimate use of a mixture of reiated compounds will serve as well. Thus lubricating oil is a mixture of several compounds of high molecular weight, all of which are suitable lubricants. Similarly, gasoline which will ultimately be burned will be composed of a number of volatile combustible compounds. Both of these common petroleum products were present in the crude oil when it came from the ground or were formed by subsequent reaction and separated by distillation. When dealt with in a process or marketed as mixtures, these products are called fractions or cuts. They are given common names or denote the refinery operation by which they were produced, and their specific gravities are defined by a scale established by the American Petroleum Institute and termed either degrees API or •API. The • API is related to the specific gravity by (1.4)
Being mixtures of compounds :the petroleum fractions do not boil isothermally like pure liquids but have boiling ranges. At atmospheric pressure the lowest temperature at which a liquid starts to boil is identified as the initial boiling point, IBP, °F. A list of the common petroleum fractions derived from crude oil is given below: Fractions from crude oil Light ends and gases ............. . Gasoline ....................... . Naphtha ...... ·................. . Kerosene ....................... .' Absorption oil .................. . Straw oil ....................... . Distillate .... : .................. . Gas oil ......................... . Lube oil ........................ . Reduced crude .................. . Paraffin wax and jelly ............ . Fuel oil (residue) ................ . Asphalt ........................ .
Approx 0 API
114 75 60 45 40
Approx IBP, °F
200 300 350 450
40 35 28 1830
500
2535
500
550 600
5
PROCESS HEAT TRANSFER
A method of defining the chemical character of ·petroleum and correlating the properties of mixtures was introduced by Watson, Nelson, and Murphy.! They observed that, when a crude oil of uniform distilling behavior is distilled into narrow cuts, the r.atio of the cube root of .the absolute average boiling points to the speeifio gravities of the cuts is a constant or TB~
K=8
(1.5)
where K = characterization factor TB =average boiling point, oR 8 = specific gravity at 60°/60° NOMENCLATURE FOR CHAPTER 1
A h K k Q 8
T Ta t
z, X
11'
Heattransfer surface, ft 2 Individual heattransfer coefficient, Etu/(hr)(ft•)("F) Characterization factor Thermal conductivity, Btu/(hr) (ft•)(°F /ft) Heat fiow, Btu/hr Specific gravity, dimensionless Temperature, 0 R Absolute average boiling temperature, 0 R Temperature in general, °F Distance, ft . A constant, Btuj(hr)(ft"WR«) Emissivity, dimensionless
1 Watson, K. M., E. F. Nelson, and G. B. Murphy, Ind. Eng. Chem., 26, 880 (1933 27, 1460 (1935).
CHAPTER 2 CONDUCTION The Thermal Conductivity. The fundamentals of heat conduction were established over a century ago and are generally attributed to Fourier. In numerous systems involving flow such as heat flow, fluid flow, or electricity flow, it has been observed that the flow quantity is directly proportional to a driving potential and inversely proportional to the resistances applying to the system, or
Fl ow
cc
potential resistance
(2.1)
In a simple hydraulic path the pressure along the path is the driving potential and the roughness of the pipe is the flow resistance. In an electric circuit the simplest applications are expressed by Ohm's law: . The voltage on the circuit is the driving potential, and the difficulty · with which electrons negotiate the wire is the resistance. In heat flow through a wall, flow is effected by a temperature difference between the hot and cold faces. Conversely, from Eq. (2.1) when the two faces of a wall are at different temperatures, a flow of heat and a resistance to heat flow are necessarily present. The conductance is the reciprocal of the resistance to heat flow and Eq. (2.1) may be expressed by Flow
cc
conductance X potential
(2.2)
To makeEq. (2.2) an equality the conductance must be evaluated in such a way that both sides will be dimensionally and numerically correct. Suppose a measured quantity of heat Q' Btu has been transmitted by a wall of unknown size in a measured time interval 0 hr with a measured temperature difference t:.t °F. Rewriting Eq. (2.2)
Q=
~'
= conductance X t:.t
Btujhr
(2.3)
and the conductance has the dimensions of Btu/(hr)(°F). The conductance is a measured property of the entire wall, although it has also been found experimentally that the flow of heat is influenced independently by the thickness and the area of the wall. If it is desired to design a wall to have certain heatflow characteristics, the conductance obtained above 6
7
CONDUCTION
is not useful, being applicable only to the experimental wall. To enable a broader use of experimental information, it has become conventional to report the conductance only when all the dimensions are referred to unit values. When the conductance is reported for a quantity of material 1 ft thick with heatflow area 1 ft 2, time unit 1 hr, and temperature difference 1 °F, it is called the thermal conductivity k. The relationship between the thermal conductivity and the Wafer ovf Wafer in conductance of an entire wall of t~~ thickness L and area A is then given by Chamber · He()ffin1 ~~~~n?~~~~ p~fe A Guard rimy Conductance = k y; ~~
and
A
Q = k L ilt
Auxibixry
(2.4)
henter
where k has the dimensions resulting from the expression QL/A At or Btu/(hr)(ft2 of flow area) (°F of temperature difference)/(ft of wall thickness). 1 Experimental .Determination of k: Nonmetal Solids. An apparatus for the determination of the thermal conductivity of nonmetal solids is shown in Fig. 2.1. It consists of an electrical heating plate, two identical test specimens through which heat Fla. 2.1. Guarded conductivity apparatus. passes, and two water jackets which remove heat. The temperatures at both faces of the specimens and at their sides are measured by thermocouples. A guard ring is provided to assure that all the measured heat input to the plate passes through the specimens with a negligible loss from their sides. The guard ring surrounds the test assembly and consists of an auxiliary heater sandwiched between pieces of the material being tested. While current enters the heating plate, the input to the
*
1 In the metric system it is usual to report the thermal conductivity as cal/(sec)(cm2) (0 0/cm). * An excellent review of experimental methods will be"found in Saba and Srivastava, "Treatise on Heat," The Indian Press, Calcutta, 1935. Later references are Bates, 0. K., Ind. Eng. Chern., 26,432 (1933); 28, 494 (1936); 38, 375 (1941); 37, 195 (1945). Bollimd, J. L. and H. W. Melville, Trans. Faraday Soc, 33, 1316 (1937). Hutchinson, E., Trans. Faraday Soc., 41, 87 (1945).
8
PROCESS HEAT TRANSFER
auxiliary heater is adjusted until no temperature differences exist between the specimens and adjacent points in the guard ring. Observations are made when the heat input and the temperatures on both faces of each specimen remain steady. Since half of the measured electrical heat input to the plate flows through each specimen and the temperature difference and dimensions of the specimen are known, k can be computed directly from Eq. (2.4). Liquids and Gases. There is greater difficulty in determining the conductivities of liquids and gases. If the heat flows through a thick layer of liquid or gas, it causes free convection ·and the conductivity is deceptively high. To reduce convection it is necessary to use very thin films and small temperature differences with attendant errors of measurement. A method applicable to viscous fluids consists of a bare electric wire passing through a horizontal Ouler tube filled with test liquid. The tube is cylinder immersed in a constanttemperature bath. The resistance of the wire is calibrated against its temperature. For a given rate of heat input and for the temperature of FIG. 2.2. Liquid conductivity ap the wire obtained from resistance measure· paratus. (After J. F. D. Smith.) · • can b e ca1cu1ate d ment s th e conduct1v1ty by suitable equations. A more exact method, however, is that of Bridgman and Smith, 1 consisting of a very thin fluid annulus between two copper cylinders immersed in a constanttemperature bath as shown in Fig. 2.2. Heat supplied to the inner cylinder by a resistance wire flows through the film to the outer cylinder, where it is removed by the bath. This apparatus, through the use of a reservoir, assures that the annulus is full pf liquid andis adaptable to gases. The film is U4 in. thick, and the temperature difference is kept very small. Influence of Temperature and Pressure on k. The thermal conductivities of solids are greater than those of liquids, which in tum are greater than those of gases. It is easier to transmit heat through a solid than a liquid and through a liquid than a gas. Some solids, such as metals, have high thermal conductivities and are called conductors. Others have low conductivities and are poor conductors of heat. These are insulators. In experimental determinations of the type described above the thermal conductivity has been assumed independent of the temperature at any point in the test material. The reported values of k are consequently t
Smith, J. F. D., Ind. Eng. Chern., 22, 1246 (1930); Trans. ASME, 68, 719 (1936).
CONDUCTION
9
the averages for the entire specimen, and the error introduced by this assumption can be estimated by an examination of Tables 2 to 5 in the Appendix. The conductivities of solids may either increase or decrease with ·temperature and in some instances may even reverse their rate of change from a decrease to an increase. For the most practical problems there is no need to introduce a correction for. the variation of the thermal conductivity with temperature. However, the variation can usually be expressed by the simple linear equation k = ko
+ "(t
where ko is the conductivity at 0°F and 'Y is a constant denoting the change in the conductivity per degree change in temperature. The conductivities of most liquids decrease with increasing temperature, although water is a notable exception. For all the common gases and vapors there is an increase with increasing temperature. Sutherland1 deduced an equation from the kinetic theory which is applicable to the variation of the conductivity of gases with temperature T )~' ++ckck ( 492
492 k = k~ 2 T
where Ck = Sutherland constant T = absolute temperature of the gas, 0 R ksz = conductivity of the gas at 32°F The influence of pressure on the conductivities of solids and liquids appears to be negligible, and the reported data on. gases are too inexact owing to the effects of free convection and radiation to permit generalization. From the kinetic theory of gases it can be concluded that the influence of pressure should be small except where a very low vacuum is encountered. Contact Resistance. One of the factors which causes error in the deter:mination of the thermal conductivity is the nature of the bond formed between the heat source and the fluid or solid specimen which contacts it and transmits heat. If a solid receives heat by contacting a solid, it is almost impossible to exclude the presence of air or other fluid from the contact. Even when a liquid contacts a metal, the presence of minute pits or surface roughness may permanently trap infinitesimal bubbles of air, and it will be seen presently that these may cause considerable error. Derivation of a General Conduction Equation. In Eqs. (2.1) to (2.4) a picture of heat conduction was obtained from an unqualified observation of the relation between heat flow, potential, and resistance. It is now feasible to develop an equation which will have the broadest applicability 1 Sutherland,
W., PMl. Mag., 36, 507 (1893).
10
PROCESS HEAT TRANSFER
and from which.other equations may be deduced for special applications. Equation (2.4) may be written in ilifferential form dQ'
do
dt
(2.5)
= k dA dx
In this statement k is the only property .of the matter and it is assumed to be independent of the other variables. Referring to Fig. 2.3, an elemental cube of volume dv = dx dy dz receives a differential quantity of heat dQ~ Btu through its left yz face in the time interval dO. Assume all but the left and right yz faces are z insulated. In the same interval the quantity of heat dQ~ leaves at the right face. It is apparent that any of three effects may occur: dQ~ may be greater than dQ~ so that \.~i•dQ~ the elemental volume stores ·heat, ..~~X increasing the average temperature of the cube; dQ~ may be greater y than dQi so that the cube loses X heat; and lastly, dQi and dQ; may FIG. 2.3. Unidirectional heat flow. be equal so that the heat will simply pass through the cube without affecting the storage of heat. Taking either of the first two cases as being more general, a storage or depletion term dQ' can be defined as .the difference between the heat entering and the heat leaving or (2.6)
According to Eq. (2.5) the heat entering on the left face may be given by dQ~ =
d8
The temperature gradient .
!!•
~t
uX
as f(x) only is 
a(afax) · X
Over the
+ dx, if dQ~ > dQi, the total change in the tem
per.ature gradient will be gradient is 
(2.7)
~~ may vary with both time and position in
the cube: The variation of distance dx from x to x
at) ox
kdydz (  
and at x
a(a~ax) dx
or  : ;2 dx.
Then at x the
+ dx the temperature gradient is at ox
()2t
oX 2
dx
11
CONDUCTION dQ~
out of the cube at the right face and in the same form as Eq. (2. 7) is given by ( at ot dx)   . dO = k dy dz .  ox iJx 2 2
dQ'2 .
from which dQ' dO
dQ~
dQ~
(i1 t) dx
(2.8)
2
= de  de = k dy dz
i)x2
(2.9)
The cube will have changed in temperature by dt deg. The change in temperature per unit time will be dtjd6 and over the time interval dO it is given by (dt/dO) dO deg. Since the analysis has been based on an elemental volume, it is now necessary to define a volumetric specific heat, cv, Btu/(ft 3WF) obtained by multiplying the· weight specific heat c Btu/ (lb) (°F) by the density p. To raise the volume dx dy dz by dt d8 OF dO
requires a heat change in the cube of dQ'
dO
at
(2.10)
= cp dx dy dz ao
and combining Eqs. (2.9) and (2.10)
at cp dx dy dz ao
(a t) 2
=
k dy dz ax 2 dx
(2.11)
from which
:~ = ;p(:;2)
(2.12)
which is Fourier's general equation, and the term kjcp is called the thermal diffusivity, since it contains all theproperties involved in the conduction of heat and has the dimensions of ft 2/hr. If the insulation is removed from the cube so that the heat travels along the X, Y, and Z axes, Eq. (2.12) becomes 2 at _ k 2t a2t (2.13) {)(} 
Cp
(a
oX 2
at)
+ iJy 2 + i'Jz 2
When the flow of heat into and out of the cube is constant as in the steady state, t does not vary with time, and dtjdO = 0, .in Eq. (2.12). atjax is a constant and CJ 2tjax 2 = 0. dQ~ = dQ~, and Eq. (2.8) reduces to Eq. (2.5) where dx dy = dA. Substituting dQ for dQ'(dfJ, both terms having
12
PROCESS HEAT TRANSFER
dimensions of Btu/hr, the steadystate equation is dQ
= k dA
dt
(2.14)
dz
Equation (2.14) applies to many of the common engineering problems. Thermal Conductivity from Electricalconductivity Measurements. The relationship between the thermal and electrical conductivities of metals demonstrates an application of Fourier's derivation incorporated in Eq. (2.9) and is a useful method t, t of determining the thermal conductivities of metals. An insulated bar of metal as shown in Fig. 2.4 has its left and right crosssectional faces exposed to different constanttemperature baths at h and t2, respectively. By fastening electric leads to the left and right faces, respectively, a current of I amp may· be passed in the direction indicated, generating heat throughout the Fra. 2.4. Heat flow in a metal. length of the bar. The quantities of heat leaving both ends of the bar in the steady state must be equal to the amount of heat received as electrical energy, l 2R"', where R"' is the resistance in ohms. From Ohm's law
where E 1  E2 is the voltage difference, u the resistivity of the wire in ohmft and K; the reciprocal of the resistivity, is the electrical conductivity. I  KA dE

(2.15)
4
u dz dx R"'=y=KA
{2.16)
Substituting Eqs. (2.15) and (2.16) for I 2R"', dx KA (dE) dx dx (dE) KA2
2 2 dQ J2R"' K A dx
2

(2.17)
But this is the same as the heat transferred by conduction and given by
13
CONDUCTION
Eq. (2.9).
When t1
= t2 and equating (2.9)
and (2.17), (2.18)
But (2.19) Differentiating,
dt = (dE) ~ ~ (d E) dx dx dE + dE dx 2
2
2
2
2
2
(2.20)
If I and A are constant for the bar, then K(dEjdx) is constant. Since K does not vary greatly with t or x, dE/dx is constant, d 2E/dx 2 = 0, and from Eq. (2.18) substituting Eq. (2.20) for d 2t/dx 2
(2.21) (2.22) (2.23) where C1 and C2 are integration constants. Since there are three constants in Eq. (2.23), C1, C2, and k/K, three voltages and three temperatures must be measured along the bar for their evaluation. C1 and 0 2 are determined from the end temperatures and k is obtained from k/K using the more simply determined value of the electrical conductivity K. Flow of Heat through a Wall. Equation (2.14) was obtained from the general equation when the heat flow and temperatures into and out of the two opposite faces of the partially insulated elemental cube dx dy dz were constant. Upon integration of Eq. (2.14) when all of the variables but Q · are independent the steadystate equation is
Q=
kt
At
(2.24)
Given the temperatures existing on the hot and cold .faces of a wall, respectively, the heat flow can be computed through the use of this equation. Since kA/L is the conductance, its reciprocal R is the resistance to heat flow, orR= L/kA (hrWF)/Btu. Example 2.1. Flow of Heat through a Wall. The faces of a 6in. thick wall measuring 12 by 16ft will be maintained at 1500 and 300~F, respectively. The wall is made of kaolin insulating brick. How much heat will escape through the wall? SolutiQn. The average temperature of the wall will be 900°F. From Table 2 in the
14
PROCESS HEAT TRANSFER
Appendix the thermal conduetivity at 932°F is 0.15 Btu/ (hr) (ftS) (°F/ft). tion to 900°F will not change this value appreciably.
Extrapola
kA ..
Q =r; .... Where At = 1500  300 = l200°F A = 16 X 12 = 192ft' L = X2 = 0.5 ft .
192
Q = 0.1~ X O.S X 1200 = 69,200 Btujhr
Flow of Heat through a Composite Wall: Resistances in Series. Equation (2.24) is of interest when a wall consists of several materials placed together in series such as in the construction of a furnace or boiler firebox. Several types of refractory brick are usually employed, since those capable of withstanding the higher inside temperatures are more fragile and expensivethan those required near the outer surface, where the temperatures are considerQ. Q ably lower. Referring to Fig. 2.5, three different refractory materials are. placed together indicated by the. subscripts a, b, and c. For the entire wall Fio. 2.5. Hea.t flow through a composite wall.
Q = ilt
R
(2.25)
The heat flow in Btu per hour through material a. must overcome the resistance R". But in passing through material a the· heat must also pass through materials b and c in series. The heat flow entering at the left face must be equal to the heat flow leaving the right face, since the steady state precludes heat storage. If Ra, Ro, and Ra are unequal, as the result of differing conductivities and thicknesses, the ratio of the temperature difference across each layer to its resistance must be the same as the ratio of the total temperature difference is to the total resistance or (2.26) For any composite system using actual temperatures (2.27)
CONDUCTION
15
Rearranging and substituting,
Q
llt
to  ts
= R = (L./koA)
+ (Lb/k,A) + (L./k.A)
(2.28)
Example 2.2. Flow of Heat through a Composite Wall. The wall of an oven consists of three layers of brick. The inside is built of 8 in. of firebrick, k = 0.68 Btu/ (hr)(ft 1)(°F/ft), surrounded by 4 in. of insulating brick, k = 0.15, and an outside layer of 6 in. of building brick, k = 0.40. The oven operates at 1600°F and it is anticipated that the outer side of the wall can be maintained at 125°F by the circulation of air. How much heat will be lost per square foot of surface and what are the temperatures at the interfaces of the layers? Solution:
For the firebrick, R. = L./k.A = 8/12 X 0.68 X 1 = 0.98 (hr)(°F)/(Btu) Insulating brick, Rb = Lb/kbA = 4/12 X 0.15 X 1 = 2.22 Building brick, R. = L./k.A = 6/12 X 0.40 X 1 = 1.25 R = 4.45 Heat loss/ft 2 of wall, Q = l:l.t/R = (1600  125)/4.45 = 332 Btujhr For the individual layers: At = QR and I:J.t. = I:J.t. = 332 X 0.98 = 325°F I:J.tb = 332 X 2.22 "" 738•F
QR., etc. t, = 1600  325 = 1275°F t2 = 1275  738 = 537•F
Example 2.3. Flow of Heat through a Composite Wall with an Air Gap. To illustrate the poor conductivity of a gas, suppose an air gap of X in. were left between the insulating brick and the firebrick. How much heat would be lost through the wall if the inside and outside temperatures are kept constant? Solution. From Table 5 in the Appendix at 572°F air has a conductivity of 0.0265 Btu/(hr)(ft 2)(°F/ft), and this temperature is close to the range of the problem.
= 0.25/12 X 0.0265 = 0.79 (hT)(°F)/Btu R = 4.45 0.79 = 5.24 Q = 1600  125 = 281 Btu/hr . 5.24 .
Rair
+
It is seen that in a wall 18 in. thick a stagnant air gap only heat loss by 15 per cent.
X in. thick reduces the
Heat Flow through a Pipe Wall. In the passage of heat through a flat wall the area through which the heat flows is constant throughout the entire distance of ~G~~v=L=Iftj the heat flow path. Referring to Fig. 2.6 " showing a unit length of pipe, the area of the heat flow path through the pipe wall increases with the distance of the path from r 1 to r 2• !':;~!·!an~eat flow through The area at any radius r is given by 211'1'1 and if the heat flow~ out of the cylinder the temperature gradient for the
16
PROCESS HEAT TRANSFER
incremental length dr is dt/dr.
Equation (2.14) becomes
2~k( ~~)
q=
Btu/ (hr) (lin ft)
(2.29)
Integrating,
t= When r ·= r;, t
;:k In
r
+ C1
(2.30)
= t;; and when r = r., t = t.; where i and o refer to the inside and outside surfaces, respectively.
q = 211k(t;  t.) 2.3 log r.jr,
Then (2.31)
and if D is the diameter,
Fm. 2. 7.
Cylindrical re
sistances in series.
Referring to Fig. 2.7 where there is a composite cylindrical resistance, (2.32) (2.33)
Adding,
2.3q I Dz  ogt1  ta_ 211ka
D1
+ 2.3q l ogDa 2?rko
D2
(2.34)
Example 2.4. Heat Flow through a Pipe Wall. A glass pipe has an outside diameter of 6.0 in. and an inside diameter of 5.0 in. It will be used to transport a fluid which maintains the inner surface at 200°F. It is expected that the outside of the pipe will be maintained at 175°F. What heat flow will occur? Solution. k = 0.63 Btu/(hr)(ft 2 )(°F/ft) (see Appendix Table 2). q
=
Z,.k(t,  t.)
2.3log D./D;
= 2 X 3.14 X 0.63(200  175) = 538 Btujlin ft 2.3 log 6.0/5.0
If the inside diameter of a cylinder is greater than 0.75 of the outside diameter, the mean of the two may, be used. Then per foot of length llt
q= 
R
=
llt
;,~
La./kaAm
(D2.,.. Dt)/2 1rka(D1 + Dz)/2
(2.35)
where (D2  Dt)/2 is the thickness of ~he pipe. Within the stated limitations of the ratio Dz/D 1, Eq. (2.35) will differ from Eq. (2.34) by
17
CONDUCTION
about 1 per cent. Actually there are 1.57 ft 2 of external surface per linear foot and 1.31 ft 2 of internal surface. The heat loss per square foot is 343 Btu/hr based on the outside surface and 411 Btu/hr based on the inside surface. Heat Loss from a Pipe. In the preceding examples it was assumed that the cold external surface could be maintained at a definite temperature. Without this assumption the examples would have been indeterminate, since both Q and ~t would be unknown and independent in a single equation. In reality the temperature assigned to the outer wall depends not only on the resistances between the hot and cold surfaces but also on the ability of the surrounding colder atmosphere to remove the heat arriving at the outer surface. Consider a  pipe as shown in Fig. 2.8 covered ta (lagged) with rock wool insulation and carrying steam at a temperature t. considerably above that of the atmosphere, ta. The overall temperature difference driving heat out of the pipe is t.  ta. The Fra. 2.8. Heat loss from an insulated pipe. resistances to heat flow taken in order are (1) the resistance of the steam to condense upon and give up heat to the inner pipe surface, a resistance which has been found experimentally to be very small so that t. and are nearly the same; (2) the resistance of the pipe metal, which is very small except for thickwalled conduits so that and t~' are nearly the same; (3) the resistance of the rock wool insulation; and (4) the resistance of the surrounding air to rei:nove heat from the outer surface. The last is appreciable, although the removal of heat is effected by the natural convection of ambient air in addition to the radiation caused by the temperature difference between the outer surface and colder air. The natural convection results from warming air adjacent to the pipe, thereby lowering its density. The warm air rises and is replaced continuously by cold air. The combined effects of natural convection arid radiation cannut be represented by a conventional resistance term R,. == La!KaA, since La is indefinite and the conductance of the air is simultaneously supplemented by the transfer of heat by radiation. Experimentally, a temperature difference may be created between a known outer surface and the air, and the heat passing from the outer surface to the air can be determined from measurements on the fluid flowing in the pipe. Having Q, A, and ~t, the combined resistance of both effects is obtained as the quotient of ~t/Q. The flow of heat from a pipe to ambient air is usually a heat loss, and it is therefore desirable to report the data as a unit conductance term k/L Btu/(hr)(ft1 of
~~t:
t:
18
PROCESS HEAT TRANSFER
external surface)(°F of .temperature difference). The unit conductance is the reciprocal of the unit resistance L/k instead of the reciprocal of the resistance for the entire surface LjkA. In other words, it is the conductance per square foot of heatflow surface rather than the conductance of the total surface. The unit resistance has the dimensions
0 0 FIG. 2.9.
200 300 500 600 TemperOiture difference ct,10), °F
700
Heat transfer by convection and radiation from horizontal pipeS at temperature ·
t1 to air at 70°F.
(hr)(ft 2 )(°F)/Btu. The reciprocal of the unit resistance, ha, has the dimensions of Btu/(hr)(ft 2)(°F) and is sometimes designated the surface coefficient of heat transfer. Figure 2.9 shows the plot of the surface coefficient from pipes of different diameters and surface temperatures to ambient air at 70°F. It is based upon the data of Heilman, 1 which has been substantiated by the later experiments of Bailey and Lyell.' 2 The four resistances in terms of the equations already discussed are Condensation of steam; q=
h.1rD~(t.  t~)
Heilman, R. H., Ind. Eng. Ohem., 16, 445452 (1924). 2 Bailey, A., and N. C. Lyell, Engineering, 14'1:, 6062 (1939). 1
(1.2)
19
CONDUCTION
Pipe wall: 21fkb (1 "} q = 2.3 log D';;n. t, t,
(2.31)
2nkc ( 11 ) q = 2.3log DdD': t.  t 1
(2.31)
Insulation:
Radiation and convection to air: (1.2)
or combining·
t,  ta.
1
= q ( h,1rD~
+ 2nkb 2.3 l D~' + 2.3 I Dt + 1 ) og D~ 2nkc og D:' h,.1rD1
The terms inside the parentheses are the four resistances, and of these the first two can usually be neglected. The equatic;m then reduces to 7r(t.  ta.)
q = 2.3 Dt 2k0 log D~'
1
+ ha.Dt
From the abscissa of Fig. 2. 9 it is seen that h,. depends upon not only the temperature difference but the actual temperatures at the outside of the insulation and of the air. Its reciprocal is also one of the resistances necessary for the calculation of the total temperature difference, and therefore the surface coeffibient ha cannot be computed except by trialanderror methods. Example 2.J. Heat Loss from a Pipe to Air. A 2in. steel pipe (dimensions in Table 11 in the Appendix) carries steam at 300°F. It is lagged with ;!.i in. of rock wool, k = 0.033, and the surrounding air is at 70°F. What will be the heat loss per linear foot? Solution. Assume t1 = l50°F, t,  70 = 80°F, ha = 2.23 Btu/(hr)(ft 2)(°F) .. q
=
3 ·~~~;~  70) 1 . 23 z· X 0.033 log 2.S75 + 2.23 X 3.375/12
= 104.8 Btu/(hr)(lin ft)
Check between t, and t1, since At/R = At,jR,: _ _ 2 X 3.14 X 0.033(300  t1) q  104 ·8 2.3.log 3.375/2.375 t 1 = 123.5°F No check Assume tl
= 125°F; f,
q =
 70
= 55°F, h. = 2.10 Btu/(hr)(ft')(°F).

3 ·i~~~~ 70) 1 = 103.2 Btu/(hr)(lin ft) . 23 2 X 0.033 log 2.375 + 2.10 X S.375/12
20
PROCESS HEAT TRANSFER
Check between t. and h. =
q
103 2 _ 2 X 3.14 X 0.033(300  t,) • 2.3log 3.375/2.375 t1 = 125.s•F Check
The total heat loss q does not appear to vary significantly for the different assumed values of t~o This is because the insulation and not the small surface coefficient affords the major resistance to heat flow. When the variation in q is considerable for different assumed temperatures of t1, it indicates insufficient insulation.
The Maximum Heat Loss through Pipe Insulation. It would seem at first that the thicker the insulation the less the total heat loss. This is always true for flat insulation but not for curved insulation. Consider a pipe with successive layers of cylindrical insulation. AB the thickness of the insulation is increased, the surface area from which heat may be removed by air increases and the total heat loss may also increase if the area increases more rapidly than the resistance.. Referring to Fig. 2.10, the resistance of the insulation per linear foot of pipe is (2.36) FIG. 2.10. dius.
The critical ra
and the resistance of the air per linear foot of pipe, although a function of the surface and air temperatures, is given by 1 Ra = h:Jnrr
(2.37)
The resistance is a minimum and the heat loss a maximum when the derivative of the sum of the resistances R with respect to the radius r is set equal to zero or dR 1 r 1 1 dr = 0 = 21rkb d ln r;: + h,.21r d (2.38) 1 1
r
= 2rrkbr
 h,.21rr 2
At the maximum heat loss r = rc, the critical radius, or r.
kb
= h;.
(2.39)
In other words, the maximum heat loss from a pipe occurs when the critical radius equals the ratio of the thermal conductivity of the insulation to the surface coefficient of heat transfer. The ratio has the dimension of ft. It is desirable to keep the critical radius as small as possible
21
CONDUCTION
so that the application of insulation will result in a reduction and not an increase in the heat loss from a pipe. This is obviously accomplished by using an insulation of small conductivity so that the critical radius is loos than the radius of the pipe, or r. < '1'1. The Optimum Thickness of Insulation. The optimum thickness of insulation is arrived at by a purely economic approach. If a bare pipe were to carry a hot fluid, there would be a certain hourly loss of heat whose value could be determined from the cost of producing the Btu in the plant heatgenerating station. The lower the heat loss the greater the thickness and initial cost of the insu· •lation and the greater the annual fixed charges (maintenance and depreciatio~} ] which must be added to the annual heat .:g loss. The fixed charges on pipe insulation will be about 15 to 20 per cent of the initial installed cost of the insulation. By "'::sc:c: assuming a number of thicknesses of insu IS lation and adding the fixed charges to the ~ value of the heat lost, a minimum cost Thickness of insuiGltion will be obtained and the thickness corresponding to it will be the Qptimum eco. FIG. 2.11. Optimum thickness of insulation. nomic thickness of the insulation. The form of such an analysis is shown in Fig. 2.11. The most difficult part is obtaining reliable initialinstallationcost data, since they vary greatly with plant to plant and with the amount of insulating to be done at a single time. Graphical Solutions of Conduction Problems. Thus far in the treatment of conduction, only those cases have been considered in which the heat input per square foot of surface was uniform. It was also characteristic of these cases that the heat removal per square foot of surface was also uniform. This was likewise true for the cylinder, even though the internal al).d external surfaces were not identical. Some of the common problems of steadystate conduction in solids involve the removal or input of heat where it is not uniform over a surface, and although the solution of such problems by mathematical analysis is often complicated it is possible to obtain close approximations graphically. The method employed here is that of Awbery and Schofield 1 and earlier investigators. Consider the section of a metalsheathed wall, as shown in Fig. 2.12, with hot side ABC at the uniform temperature h. At recurring intervals DF on the cold side DEF at the uniform temperature tz, metal bracing
§
1 Awhery, J. and F. Schofield, P:roc. Intern. Congr. Refrig., 5th Confll'., 3, 591610 (1929).
22
PROCESS HEAT TRANSFER
ribs are attached to the outer sheath and imbedded twothirds into the thickness of the wall. Since the sheath and metal rib both have a high thermal conductivity compared with the wall material itself, the rib and sheath may both be considered to be at very nearly the same temperature. The predominantly horizontal lines indicated on the drawing represent isothermal planes perpendicular to the plane of "the drawing. Consequently there is no heat flow to be considered in the direction perpendicular to the plane of the drawing. B
tz FIG. 2.12.
E
Graphical representation of heat conduction.
Since the drawing is symmetrical about the vertical line BE, consider only the right half of the drawing bounded by BCFE. Assume an arbitrary number of isotherms n, in the direction from B to E so that, if k 2
is constant, t.t = n, l::.t.,.
If k varies with t then k t.t., = ..!_ { k dt.
.
n,Jl
The
greater the assumed number of isotherms the greater the precision of the solution. Next, consider the heat to flow from t1 to metal at t2 through n 1 lanes emanating from BC and forming the network indicated. Now refer to any small portion of any lane, such as abed with length x, mean width y, where y = (ab cd) /2, and unit depth z = 1 perpendicular to the drawing. The steadyheat flow into each lane is Q1• The conduction equation is then Q1 = k(yz) t.t,jx. The temperature difference from one isotherm to the next is naturally the same, and since Q1 is constant for the lane it is evident from the conduction equation that the ratio y/x must also be constant, although y and x may vary. The network of the drawing is constructed such that, for each quadrilateral, y = x. Where x is small it is because the isotherms are crowded together owing to the high heat removal by the rib. The heat flow per lane is then given by
+
k (tl t2).
n.
CONDUCTION
23
The total heat flow from BC thus requires nz = Qn,Jk(tl l2) lanes, where Q is the total heat flow. Figure 2.12 was constructed in this manner starting with six isotherms. Although the individual portions of the network are neither squares nor rectangles their corners are at right angles in accordance with the steadystate principle that the flow of heat is always at right angles to the isotherms comprising the temperature difference. In Fig. 2.12 it is seen that 11 lanes were obtained for each half of a symmetrical section. If the isotherms were undisturbed by the rib, the portions abed would then be squares and the heat entering BC would flow normal to it and 8.3 lanes would be required. The rib is therefore equivalent to increasing the heat removal by 33 per cent. When the ribs are spaced more closely together, the fractional heat removal increases. PROBLEMS 2.1. A furnace is enclosed by walls made (from inside out) of 8 in. of kaolin firebrick, 6 in. of kaolin insulating brick, and 7 in. of fireclay brick. What i!> the hea~ loss per square foot of wall when the inside of the furnac(l is maintained at 2200°F and the outside at 200°F? 2.2. A furnace wall is to consist in series of 7 in. of kaolin firebrick, 6 in. of kaolin insulating brick, and sufficient fireclay brick to reduce the heat loss to 100 Btu/ (hr) (ft•) when the face temperatures are 1500 and 100°F, respectively. What thickness of fireclay brick should be used? If an effective air gap of % in. can be incorporated between the fireclay and insulating brick when erecting the wall without impairing its structural support, what thickness of insulating brick will be required? 2.3. A furnace wall consists of three insulating materials in series. 32 per cent chrome brick, magnesite bricks, and lowgrade refractory bricks (k = 0.5). The magnesite bricks cannot withstand a face temperature above 1500°F, and the lowgrade bricks cannot exceed 600°F. What thickness of the wall will give a heat loss not in excess of 1500 Btu/(hr) (ft 2) when the extreme face temperatures are 2500 and 200°F, respectively? 2.4. A 6in. IPS pipe is covered with three resistances in series consisting from the inside outward of 72 in. of kapok, 1 in. of rock wool, and 72 in. of powdered magnesite applied as a plaster. If the inside surface is maintained at 500°F and the outside at 100°F, what is the'heat loss per square foot of outside pipe surface? 2.6. A 2in. IPS line to a refrigerated process covered with 72 in. of kapok carries 25% NaCl brine at 0°F and at a flow rate of 30,000 lb/hr. The outer surface of the kapok will be maintained at 90°F. What is the equation for the flow of heat? Calculate the heat leakage into the pipe and the temperature rise of the fluid for a 60ft length of pipe. 2.6. A vertical cylindrical kiln 22 ft in diameter is enclosed at the top by a hemispherical dome fabricated from an 8in. layer of interlocking and selfsupporting 32 per cent chrome bricks. Derive an expression for .conduction thTough the dome. When the inside and outside of the hemispherical dome are maintained at 1600 and 300°F, respectively, what is the heat loss per square foot of internal dome surface? How does the total heat loss for the dome compare with the total heat loss for a fiat structurally supported roof of the same material when exposed to the same difference in temperature?
24.
PROCESS HEAT TRANSFER
2. 7: A 4in. steel pipe carrying 450°F steam is lagged with 1 in. of kapok surrounded by 1 in. of powdered magnesite applied as a plaster. The surrounding air is at 70•F. What is the loss of heat from the pipe per linear foot? 2.8. A 3in. IPS main carries steam from the powerhouse to the process plants at a linear velocity of 8000 fpm. The steam is at 300 psi (gage), and the atmosphere is at 70°F. What percentage of the total heat How is the baretube heat loss per 1000 ft of pipe? If the pipe is lagged with half a thickness of kapok and half a thickness of asbestos, what total thickness will reduce the insulated heat loss to 2 per cent of the baretube heat loss? 2.9. In a 6in. steam line at 400°F the unit resistance for the condensation of steam at the inside pipe wall has been found experimentally to be 0.00033 (hr)(ft2)("F)/Btu. The line is lagged with 32 in. of rock rool and 32 in. of asbestos. What is the effect. of including the condensation and metal pipewall resistances in calculating the total heat loss per linear foot to atmospheric air at 7o•F? NOMENCLATURE FOR CHAPTER 2
HeatHow area, .ft 2 Constants of integration Sutherland constant Volumetric specific heat, Btu/ (ft3) (°F) c. Specific heat at constant pressure, Btu/(lb)(°F) c D Diameter, ft E Voltage or electromotive force Surface coefficient of heat transfer, Btu/(hr)(ftt)(°F) h. I Current, amp Electrical conductivity, 1/ohmft K Thermal conductivity, Btu/(hr)(ft 2 )(°F/ft) k Thickness of wall or length of pipe, ft L Number of heatHow lanes nz Number of isotherms n~ Heat How, Btu/hr Q Qz Heat How per lane, Btu/hr Q' Heat, Btu q Heat How, Btu/(hr)(lin ft) Resistance to heat How, (hr)(°F)/Btu R Resistance to electric How, ohms R"' r Radiuth ft Temperature at any point, °F At Temperature difference promoting heat How, °F T Absolute temperature, 0 R Volume, ft3 IJ x, y, z Coordinates of distance, ft Change in thermal conductivity per degree 'Y (J
Time,hr
p
Density, lb/fta Resistivity, ohmft
(}"
CHAPTER 3
CONVECTION Introduction. Heat transfer by convection is due to fluid motion. Cold fluid adjacent to a hot surface receives heat which it imparts to the bulk of the cold fluid by mixing with it. Free or natural convection occurs when the fluid motion is not implemented by mechanical agitation. But ·when the fluid is mechanically agitated, the heat is transferred by forced convection. The mechanical agitation may be supplied by stirring, although in most process applications it is induced by circulating the hot and cold fluids at rapid rates on the opposite sides of pipes or tubes. Free and forcedconvection heat transfer occur at very different speeds, the latter being the more rapid and therefore the more common. Factors which promote high rates for forced convection do not necessarily have the same effect on free convection. It is the purpose of this chapter to establish a general method for obtaining the rates of heat transfer particularly in the presence of forced convection. Film Coefficients. In the flow of heat through a pipe to air it was seen that the passage of heat into the air was not accomplished solely by conduction. Instead, it occurred partly by radiation and partly by free convection. A temperature difference existed between the pipe surface and the average temperature of the air. Since the distance from the pipe surface to the region of average air temperature is indefinite, the resistance cannot be computed from Ra = La/koA., using k for air. Instead the resistance must .. be determined experimentally by appropriately measuring the surface temperature of the pipe, the temperature of the air, and the heat transferred from the pipe as evidenced by the quantity of steam condensed in it. The resistance for the entire surface was then computed from _ !:..ta Ra  Q (hrWF)/Btu If desired, La can also be calculated from this val~e of Ra and would be the length of a fictitious conductiop. film of air equivalent to the combined resistance of conduction, free convection, and radiation. The length of the film is of little significance, although the concept of the fictitious film finds numerous applications. Instead it is preferable to 25
26
PROCESS HEAT TRANSFER
deal directly with the reciprocal of the unit resistance h, which has an experimental origin. Because the use of the unit resistance L/k is so much more common than the use of the total surface resistance L/kA, the letter R will now be used to designate L/k (hr)(ft2WF)/Btu and it will simply be called the resistance. Not all effects other than conduction are necessarily combinations of two effects. Particularly in the case of free or forced convection to liquids and, in fact, to most gases at moderate temperatures and temperature differences the influence of radiation may be 0 neglected and the experimental resistance corresponds to forced or free convection alone as the case may be. Consider a pipe wall with forced convection of different magnitudes on both sides of the pipe as shown in Fig. 3.1. On the to inside, heat is deposited by a hot flowing · liquid, and on the outside, heat is received by ·a cold flowing liquid. Either resistance can be measured independently by obtaining the temperature difference between the pipe surface and the average temperature of the liquid. The heat transfer can be determined FIG. 3.1. Two convection from the sensibleheat change in either fluid coefficients. over the length of the pipe in which the heat transfer occurs. Designating the resistance on the inside by R; and on the outside by R., the inside and outside pipew·all temperatures by tp and tw, and applying an expression for the steady state,
Q = A,(T, R,
tp)
= A.(t,,  t.) R.·
(3.1)
where T, is the temperature of the hot fluid on the inside and t. the temperature of the cold fluid on the outside. Replacing the resistances by their reciprocals h; and h., respectively, (3.2)
The reciprocals of the heattransfer resistances have the dimensions of Btu/(hr)(ft2 )(°F of temperature difference) and are called individual film coefficients or simply film coefficients. Inasmuch as the film coefficient is a measure of the heat flow for unit surface and unit temperature difference, it indicates the rate or speed with which fluids havi11:g a variety of physical properties and under varying degrees of agitation transfer heat. Other factors influence the
27
CONVECTION
film coefficient such as the size of the pipe and whether or not the fluid is considered to be on the inside or outside of the pipe. With so many variables, each having its own degree of influence on the rate of heat transfer (film coefficient), it is fairly understandable why a rational derivation is not available for the direct calculation of the film coefficient. the other hand, it is impractical to run an experiment to determine the coefficient each time heat is to be added or removed from a fluid. Instead it is desirable to study some method of correlation whereby sev1.c~xl eral basic experiments performed with a wide · range of the variables can produce a relation?JJ/77777//77777 ship which will hold for any other combina(or) tions of the variables. The immediate problem is to establish a method of correlation and then apply it to some experimental data. The Viscosity. It is not possible to proceed tldx.1 very far in the study of convection and fluid y flow without defining a property which has h.r///777/7,?)7//7 an important bearing upon both, viscosity. Rd (allowed), as after a period of service, the apparatus no longer delivers a quantity of heat equal to the process requirements and must be cleaned. Numerical values of the dirt or fouling factors for a variety of process services are provided in Appendix Table 12. The tabulated fouling factors are intended to protect the exchanger from delivering less than the required process heat load for a period of about a year to a year and a ID.alf. Actually the purpose of the tabulated fouling factors should be considered from another point of view. In designing a process plant containing many heat exchangers but without alternate or spare pieces of heattransfer equipment, the process must be discontinued and the equipment cleaned as soon as the first exchanger becomes fouled. It is impractical to shut down every time one exchanger or another is fouled, and by using the tabulated fouling factors, it can be arranged so that all the exchangers in the process become dirty at the same time regardless of service. At that time all can be dismantled and cleaned during a single shutdown. The tabulated values may differ from those encountered by experience in particular services. If too frequent cleaning is necessary, a greater value of Rd should be kept in mind for future design. It is to be expected that heattransfer equipment will transfer more heat than the process requirements when newly placed in service and that it will deteriorate through operation, as a result of dirt, until it just fulfills the process requirements. The calculation of the temperatures delivered initially by a clean exchanger whose surface has been designed for U» but which is operating without dirt and which is consequently oversurfaced is not difficult. Referring to Eqs. (5.18) and (5.19) use U a for U and the actual surface of the exchanger A (which is based on U»). This calculation is also useful in checking whether or not a clean exchanger will be able to deliver the process heat requirements when it becomes dirty. Pressure Drop in Pipes and Pipe Annuli. The pressuredrop allowance in an exchanger is the static fluid pressure which may be expended to drive the fluid through the exchanger. The pump selected for the circulation of a process fluid is one which develops sufficient head at the desired capacity to overcome the frictional losses caused by connecting
109
COUNTERFLOW
piping, fittings, control regulators, and the pressure drop in the exchanger itself. To this head must be added the static pressure at the end of the line such as the elevation or pressure oj the ·final receiving vessel. Once a definite pressure drop allowance has been designated for an exchanger as a part of a pumping circuit, it should always be utilized as completely as possible in the exchanger, since it will otherwise be blown off or expanded through a pressure reducer. Since in Eq. (3.44) (nearly, since f varies somewhat with
D.uG)
and in Eq. (6.2) for turbulent flow h;
0:
_G0.8
(nearly)
the best use of available pressure is to increase the mass velocity which also increases hi and lessens the size and cost· of the apparatus. It 'is customary to allow a pressure drop of 5 to 10 psi for an exchanger or battery of exchangers fulfilling a single process service except where the flow is by gravity. For each pumped stream 10 psi is fairly standard. For gravity flow the allowable pressure drop is determined by the elevation. of the storage vessel above the final outlet z in feet of fluid. The feet of fluid may be converted to pounds per square inch by multiplying z by p/144. The pressure drop in pipes can be computed from the Fanning equation [Eq. (3.44)], using an appropriate value of f from Eq. (3.46) or Eq. (3.47b), depending upon the type of flow. For the pressure drop in fluids flowing in annuli, replace D in the Reynolds number by D~ to obtain f. The Fanning equation may then be modified to give !:J.F = 4fG2L
2gp 2 D~
(6.14)
Where several double pipe exchangers are connected in series, annulus to annulus and pipe to pipe as in Fig. 6.5, the length in Eq. (3.44) or (6.14) is the total for the entire path. The pressure drop computed by Eq. (3.44) or (6.14) does not include the pressure drop encountered when the fluid enters or leaves exchangers. For the inner pipes of double pipe exchangers connected in series, the entrance loss is usually negligible, but for annuli it may be significant. The allowance of a pressure drop of one velocity head, V 2 /2g', per hairpin will ordinarily suffice. Suppose water flows in an annulus with a mass velocit,y of 720,000 lb/(hr)(ft 2). Since p = 62.5lb/ft 8 (approximately),
G
V
720,000
= 3600p = 3600 X 62.5 = 3·2 fps.
110
PROCESS HEA1' TRANSFER
The pressure drop per hairpin will be 3.2 2/(2 X 32.2) = 0.159 ft of water or 0.07 psi. Unless the velocity is well above 3 fps, the entrance and exit losses may be neglected. Values of V 2/2g' are plotted directly against the mass velocity for a fluid with a specific gravity of 1.0 in Fig. 27 in the Appendix. The Calculation of a Double Pipe Exchanger. All the equations developed previously will be combined to outline the solution of a double pipe exchanger. The calculation consists simply of computing h. and h;. to obtain U c. Allowing a reasonable fouling resistance, a value of U n is calculated from which the surface can be found with the use of the Fourier equation Q = UnA At. Usually the first problem is to determine which fluid should be placed in the annulus and which in the inner pipe. This is expedited by estab'lishing the relative sizes of the flow areas for both streams. For equal allowable pressure drops· on both the hot and cold streams, the decision rests in the arrangement producing the most nearly equal mass velocities and pressure drops. For the standard arrangements of double pipes the flow areas are given in Table 6.2. TABLE
6.2.
FLow AREAs AND EcrnvALE: = 1.0) or
haDe(~)!" k
k
(\;,..J.l.)o.r D. (c~)Y3 X 1.0 = ho Btu/(hr)(ft WF) k 4
2
}:__
(6.15b)
Overall coefficients:
+
(11) Compute U a = h;oho/(h;o ho), Btu/(hr)(ft2WF). (6.7) (12) Compute UD from 1/UD = 1/Ua +Rd. (6.10) (13) Compute A from Q = U llA ilt which may be translated into length.
If the length should not correspond to an integral number of hair
pins, a change in the dirt factor will result. The recalculated dirt factor should equal or exceed the required dirt factor by using the next larger integral number of hairpins. Calculation of AP. This requires a knowledge of the total length of path satisfying the heattransfer requirements. Inner pipe: (1) For Rep in (6) above obtain f from Eq. (3.46) or (3.47b). (2) t.F'p = 4fG 2L/2gp 2D, ft. AFpp/144 = APp, psi.
(3.45)
Annulus: 4"~~"Cm . D'• = 47r(D2 (1 ') Ob tam
 Dl) DV +
=
(D 2 .D 1) •
(6.4) .
Compute the frictional Reynolds number, Re~ = D~Ga/p.. For Re~ obtain/from Eq. (3.46) or (3.47b). ' (6.14) (2') t.F'a = 4fG2L/2gp 2D~, ft. (3') Entrance and exit· losses, one velocity head per hairpin: AFz = (AF,
~; ft/hairpin
+ AF!)p/144 = APa, psi.
There is an advantage if both fluids are computed side by side, and the use of the outline in this manner will be demonstrated in Example 6.1.
COUNTERFLOW
113
Example 6.1. Double Pipe BenzeneToluene Exchanger. It is desired to heat 9820 lb /hr of cold benzene from 80 to 120•F using hot toluene which is cooled from 160 to 100•F. The specific gravities at 68°F are 0.88 and 0.87, respectively. The other fluid properties will be found in the Appendix. A fouling factor of 0.001 should be provided for each stream, and the allowable pressure drop on each stream is 10.0 psi. A number of 20ft hairpins of.2 by 1~in. IPS pipe are available. How many hairpins are required? Solution: (1) Heat balance:
Benzene, t..v =
80
+ 120 =
1oo•F
2
c = 0.425 Btu/(lb)(°F)
(Fig. 2)
Q = 9820 X 0.425(120  80) = 167,000 Btu/hr Toluene, Tav =
w
160
+ 100 =
130°F c = 0.44 Btu/(lb)(°F) 2 167,000 = 0.44(160  100) = 6330 lb/hr
(Fig. 2)
(2) LMTD, (see the method of Chap. 3):
Diff.
Cold fluid
Hot fluid 160
Higher temp
100
Lower temp
LMTD = At,  Ah · 2.3.log At,jAtt
I
120
40
80
20 20
20 28 8°F 2.3 log 4% 0 = ·
(5.14)
.(8) Caloric temperatures: A check of both streams will show that neither is viscous
at the cold terminal (the viscosities less than 1 centipoise) and the temperature ranges and temperature difference are moderate. The coefficients may accordingly be evaluated from properties at the arithmetic mean, and the value of (p.f,..,.)o.u may be assumed equal to 1.0. Tav = ~(160
+ 100)
= 130°F
lav
= %(120
+ 80) = 100°F
Proceed now to the inner pipe. A check of Table 6.2 indicates that the flow area of the inner pipe is greater than that of the annulus. Place the larger stream, benzene in the inner pipe. Hot fluid: annulus, toluene (4') Flow area, D, = 2.067/12 = 0.1725 ft
Dt = 1.66/12 = 0.138 ft a. = ,.(D~  DD/4
= ,.(0.1725 2  0.138 2)/4 = 0.00826 ft• Equiv diam, D. = (D~  DD /Dt ft . [Eq (6.3)] D, = (0.1725 2  0.1382)/0.138 = 0.0762 ft
Cold fluid: inner pipe, benzene = 1.38/12 = 0.115 ft Flow area, ap = ,.D•/4 = ,. X 0.115 2 /4 = 0.0104 ft' (4) D
114
PROCESS HEAT TRANSFER Hot fl1fid: annulus, toluene
Cold fl·uid: inner pipe, benzene
Mass vel, Ga = W laG (5) Mass vel, GP = w/ap 6330/0.00826 = 767,000 lb/(hr)(ft 2) = 9820/0.0104 = 943,000 lb/(hr)(ft2) At 130°F, p. = 0.41 ep [Fig. 14] (6) At wo•F, p. = 0.50 cp [Fig. 14] 0.41 X 2.42 = 0.99 lb/{ft)(hr) = 0.50 X 2.42 = 1.21lb/(ft)(hr) DG Reynolds no., Re. = D,Ga Reynolds no., Rep = _ P (6') = (6') =
= 0.0762
p.
Jl.
X 767,000/0.99
= 59,000
= 0.115
X 943,000/1.21
= 89,500
in = 167 [Fig. 24) (7) iii = 236 [Fig. 24] (8') At 130°F, c = 0.44 Btn/(lb)(°F) (8) At 100°F, c = 0.425 Btu/0b)(°F) [Fig. 2) [Fig. 2} k = 0.085 Btu/(hr)(ft 2 )("F/ft) [Table 4] k = 0.091 Btu/(hr) (ft 2)(°F /ft) [Table 4] (7')
( ~)l> k
= (0.44 X 0.99)H = 1 725 0.085 .
(~)!i k
(9') h.
= }H ~. (t)%
. H . k (9) h;. = J D k
(;:,.ru
[Eq. (6.15b)]
(c")% (
Jl. )"·" Jl..,
[Eq. (6.15a)] 0.091 = 236 X O.ll 5 X 1.78 X 1.0
0.085
= 167 X 0.0 762 X 1.725 X 1.0
= .323 Btu/(hr)(ft•)(°F)
= (0.425 X 1.2l)li = l 78 0.091 .
= 333 Btu/(hr)(ft2)(°F) (10) Correct h; to the surface at the OD ID
h;. = h; X OD
[Eq. (6.5)]
= 333 X 1. 38 = 276 1.66 Now proceed to the annulus. (11) Clean overall coefficient, Uc: 276 323 Uc = ~ = X = 149 Btu/(hr)(ft2)(°F) h;. + h. 276 + 323 (12) Design overall coefficient, U D:
...!... =...!... + Ra· UD Uc
(6.7)
(6.10)
Rd = 0.002 (required by problem) 1 1 UD = 149 0 ·002 U D = 115 Btu/(hr)(ft2 WF)
+
Summary 323
I
h outside
Ua
149
UD
115
.1~
(13) Required ilurface: Q ='
uDA At
.4 =
uDQ At
167,000 = 50 5 ft• Surface  lUi X 28.8 ·
115
COUNTERFLOW From Table 11 for per foot length.
Ho:J:in. IPS standard pipe there are 0.435 ft 2 of external surface Required length
=:.~a~
= 116lin ft
This may be fulfilled by connecting three 20ft hairpins in series. (14) The surface supplied will actually be 120 X 0.435 = 52.2 ft2.
will accordingly be greater than required.
uD
=
Rd =
The dirt factor The actual design coefficient is
52~tx~. 8 = 111 Btu/(hr)(ft'WFJ
Ua UD 149 111 o VaUD = 149 X 111 = 0.0023 (hr)(ft 2)(F)/Btu
(6.13)
Pressure Drop (1') D; for pressure drop differs from D, (1) For Rep = 89,500 in (6) above for heat transfer. 0.264 D; = (D, _ D 1) [Eq. (6.4) f = 0.0035 (DG/p.)"· 42 [Eq. (3.47b)] = (0.1725  0.138) = 0.0345 ft 0.264 Re' = D;Ga = 0.0035 + 89,500"·4' = 0.0057 • p. s = 0.88, p = 62.5 X 0.88 = 55.0 = 0.0345 X 767,000/0.99 = 26,800 [Table 6)
+
I
= 0.0035
s = 0.87,
p
+ 26°8~~. 4, , =
= 0.0071
(2) t::.F., =
[Eq. (3.47b)) 62.5 X 0.87 = 54.3
A"' t.ll'a
== ==
4fG!L
AD
== 2 •D'
2gp D
4 X 0.0057 X 943,000 2 X 120 8 X 55.0 2 X 0.115 =Uft 8.3 X 55.0 32 .
= 2 X 4.18 X 10
[~k~
(2 ')
4 fG'f,L 2
~P = 144 UP , 4 X 0.0071 X 767,0002 X 120 Allowable t:.Pp 2 X 4.18 X 108 X 54.3' X 0.0345 23.5 ft
=
· psl
== 10.0 psi
·(S') V ·
F1 = 3
G 767,000 3 92f  3600p  3600 X 54.3  . ps
(2V') g,
=3
·X 2 3.92 X 32 _2 = 0. 7 ft 2
+ 0.7)54.3  9 2 . ··144  · psi Allowable t:.P a == 10.0 psi
tJ>  (23.5
A check of U1o and U." gives 161 and 138, respectively, and K, = 0.17. From Fig. 17 for At./Ath = 2%o == 0.5, F.= 0.43, whereas in the solution. above the arithmetic mean temperatures were used. The .arithmetic mean assumes F, = 0.50. However, since the ranges are small for both fluids, the error is too small to be significant. If the ranges of the :fluids or their viscosities were large, the error might be considerable for Fc =· 0.43.
Double Pipe Exchangers in Series.parallel Arrangements. Referring to Example 6.1, it is seen that a calculated pressure drop of 9.2 psi is
116
PROCESS HEA1' TRANSFER
obtained against an allowable pressure drop of 10.0 psi. Suppose, however, that the calculated pressure drop were 15 or 20 psi and exceeded the available head. How then might the heat load be transferred with the available pressure head? One possibility is the use of a bypass so that only threequarters or twothirds of the fluid flows through the exchanger and the remainder through the bypass. This does not provide an ideal solution, since the reduced flow causes several unfavorable changes in the design. (1) The reduced flow through the exchanger reduces the mass velocity Ga and the film coefficient h.. Since both of the coefficients are nearly alike, 323 vs. 276, any sizable reduction in Ga alone decr(lases U c by nearly G~·s. (2) If less liquid circulates through the annulus, it has to be cooled over a longer range than from 160 to W0°F so that, upon T,
.,1'""!
~
Tz }'IG.
0.5.
t2
J
n
tt,
Double pipe exchangers in series.
~
w,c;fz t,,w,c FIG. 6.6.
Seriesparallel arrangement.
mixing with the bypass fluid, the process outlet temperature of W0°F results. As an example, the portion circulating through the annulus might have to be cooled over the range from 160 to 85°F depending upon the percentage bypassed. The outlet temperature of 85°F is closer to the inner pipe inlet of 80°F than originally, and the. new coldterminal difference .ilt1 of only 5°F greatly decreases the LMTD. The two effects, decreased U c and LMTD, increase th~ required number of hairpins greatly even though the heat load is constant. Reversing the location of the streams by placing the benzene in the annulus does not provide a solution in this case, since the benzene stream is larger than the toluene stream. The possibility of reversing the location of the streams should always be examined .first whenever the allowable pressure drop cannot be met. A solution is still possible, however, even when all the above have failed. When .two double pipe exchangers are connected in series, the arrangement is shown in Fig. 6.5. Suppose that the stream which is too large to be accommodated in several exchangers in series is divided in half and each half traverses. but one exchanger through the inner pipes in
117
COUNTERFLOW
Fig. 6.6. Dividing a stream in half while keeping the flow area constant produces about oneeighth of the series pressure drop, since G and L will be half and the product of G2L in Eq. (6.14) will be oneeighth. While the film coefficient will also be reduced, the unfavorable temperature difference of bypassing can be circumvented.. Where there is a substantial unbalance between the weight flow of the two streams because one operates over a long range and the other over a very short range, the large stream may be divided in three, four, or more parallel streams. In larger services each parallel stream may also flow through several exchangers in series in each parallel bank. The term "parallel streams" should not be confused with "parallel flow." The former refers to the division of the ·flow of one fluid, whiie the latter refers to the direction of flow between two fluids. The True Temperature Difference for Seriesparallel Arrangements. The LMTD calculated from T1, T2, t1, and t2 for the series arrangement will not be the same for a seriesparallel arrangement. Half of the pipe fluid enters the upper exchanger II in Fig. 6.6 where the annulus fluid is hot, and half enters the lower exchanger I in which the annulus fluid has already been partially cooled. While exchangers in series do not transfer equal quantities of heat, the seriesparallel relationship is even more adverse, the lower exchanger accounting for relatively less of ·the total heat transfer. If the true temperature difference is called t1t, it will not be id!')ntical with the LMTD for the process conditions although both of the exchangers operate. in counterflow. Consider the two exchangers in Fig. 6.6 designated by I and II. The intermediate temperature is T, and the outlets of the parallel streams are designated by t¥ and t~. Their mixed temperature is t2. For exchanger I, containing half the surface, Q.
= WC(T · T2)
and LMTD1
=
~A
X LMTDr
_ (T  tD  (T2  t1) In (T  tD/CT2  t1)

{6.16) {6.17)
Substituting in Eq. (6.16), UA (T .T2) I T  t~ 2WC  (T  ti)  (T2  t1) n T2  t1 ~arranging,
UA = (T  T2) ln (T tD 2WC (T  T2)  (t~  h) (T2  t1) 1 ITt~ 1  (t~  t1)/(T  T2) n T2  t~
(6.18)
118
PROCESS HEAT TRANSFER
Let
Ri = (T  T2) = we (t~ t1) 2WC UA = ~ln Tt~ 2WC R 1  1 T2  tl
(6.19)
Similarly for exchanger II
UA . Qu = WC(T1  T) = 2 X LMTDn
(6.20)
1
_ (T1  1~ )  (T  t1)
LMTDu  ln (T1  t~1 )/(T  t1) ·
(6.21)
Let
RII = Tl T = ~ t~  l1 2WC UA RII Tl  t~1 2WC = R 11  1 ln T  t1 1
(6.22)
Since c and C were assumed constant,
R1
we = RII = R' = 2WC
(6.23)
Let
Similarly let
R' and S are ratios which recur frequently in obtaining the true temperature difference t:.t from the LMTD. S is the ratio of the cold fluid range to the maximum temperature span, the latter being the difference between both inlet temperatures, T1 and t1. But
M11 1 _
sr =
T  t~ T  t1
= R'SX1 = T 
Tt~
T2  tl
t1 _
T  it
t~  t1 T  it
1SX
= 1  R'S1
and from Eq. (6.19)
R'
181
UA 2WC
= R' 1 ln 1  R'S1
(6.24)
UA 2WC
= R'
R' 1  su 111 1 1 R'Su
(6.25)
and from Eq. (6.22)
119
COUNTERFLOW
and equating Eqs. (6.24) and (6.25), 1BI 1Su 1  R' 8 1 = 1  R'su Therefore
Adding Eqs. (6.24) and (6.25),
UA 2R' 1  SI 2R' T  t! WC = R'  lln 1  R'St = R' ;__ lln T 2  t1 in which T is the only unknown, and since MI
=
6 ·26 )
MI1,
Tt T T T2 Tt tr = T li T 2  2ttT + t1(T1 + T2)  T1T2 Equation (6.27) is a quadratic whose solution is T
(
=0
v'4ti 4h[{Tt + T2)  4TtT2] 2 = it± V(Tt  h)(T2 it)
(6.27)
= 2t1 ±
(6.28)
The minus sign applies when the heating medium is in the pipes. · The plus sign applies when the cooling medium is in the pipes. Substituting for T in Eq. (6.26),
UA We
=
2R' In [(R' l)(T t  h)+ y'(Tt t1)(T2R1  1 R1 y'(Tt tt)(T2 h)
= R;~
1In [ (R'R; 1) (~:: !)~ + ~~]
tt)]
(6.29)
bot is the single value for the entire seriesparallel arrangement; thus
(6.30) (6.31)
It is convenient in this derivation to employ a definition for the true temperature difference in terms of the maximum temperature span
Tt
t1: .6.t
= 'Y(Tt
 h)
Equating (6.31) and (6.32),
we UA (Tt T2) 'Y
=
'Y(Tt tt)
= We(Tt. T2) UA(Tt h)
(6.32)
PROCESS HEAT TRANSFER
120
Since M
=
(T1  T2)j(T1  h), define P'
UA/WC
=
Mh; then
P'
+M =1
M
or
=
(T2  h)/(Tl  h) and
= 1P'
Substituting in Eq. (6.29), UA 2R' WC = R' 1ln
or
[(R'w  1) ( P'1 )l2 + R'1 ]
1 P' = 2 ( R'R' 1) ln· [(R'  1) (P'1)}' + R'1] .~ 'Y
(6.33)
(6.34)
If developed in a generalized manner it Gan be shown that, for one series hot stream and n parallel cold streams, Eq. (6~34) becomes
1 P'
'Y =
(R'  1) (P'1)
11 "
nR' [ 2.3 R'  1 log ~
+ R'1]
(6.35a)
where
For one series cold stream and n parallel hot streams, 1  P" 'Y
= 2.3 1 _:: R" log
[
)l/n + R" ]
( 1 (1  R") P"
(6.35b)
where and Example 6.2. Calculation of the True Temperature Difference. A bank of double pipe exchangers operates with the hot fluid in series from· 300 to 200°F and the cold fluid in six parallel streams from 190 to 220°F, What is the true temperature difference !:J.t'l
P' = T 2 T1 
tt = 200 t1
 190 = 0 091 300  190 . .
Substituting in Eq. (6.35a) and solving, At
I Tl  T. 300  200 R = n(t,  t1) = 6(220  190)
'Y =
= 0 ·558
0.242.
= 0.242(300  190)
= 26.6°1!'
(6.32)
The LMTD would be 33. 7°F, and an error of 27 per cent would he introduced by its use.
Exchangers with a Viscosity Correction, rf>. For heating or cooling fluids, the use of Fig. 24 with an assumed value of (p./p.w) 0 •14 = 1.0 also assumes a negligible deviation of fluid properties from isothermal flow. For nonvi~cous fluids the deviation from isothermal flow during heating
121
COUNTERFLOW
an
or cooling does not introduce appreciable error in the calculation of the heattransfer coefficient. When the pipewall temperature differs appreciably from the caloric temperature of the controlling fluid and the controlling fluid is viscous, the actual value of cp = (f.l/ ji.w) 0 •14 ro:ust be taken jnto account. To include the' correction, tw may be determined by Eq. (5.31) or by (5.32) from uncorrected values of ho/cfJrr. and hio/cPp, which are then corrected accordingly by multiplication by c/Ja and cfJp respectively. The corrected coefficients where cfJ ....= 1.0 are
=(!:)~a
(6.36)
cpp cpp ' = (h··)
(6.37)
ho t!;u
Similarly for two resistances in series .employing the viscosity corrections for deviation from the isothermal the clean overall coefficient is again (6.38) Example 6.3. Double Pipe Lube OilCrude Oil Exchanger. 6,900 lb/hr of a 26°API lube oil must be cooled from 4:50 to 350°Fby 72,500 lb/hr of 34°API midcontinent crude oil. The crude oil.will be heated from 300 to 310°F. A fouling factor of 0.003 should be provided for each stream, and the allowable pressure drop on each stream will be 10 psi. A number 5(1)0
~~
_]XC:
0~0 ~~~ fx'\ (;'\ r;'\
;)
(b)Triomgu!Glr pitch
(c)·Square pitch rotc;ttecl
FIG. 7.3.
<ellTriornqular pitch with cleorninc:Jianes
Common tube layouts for exchangers.
eter and 1 in. OD are most common in heatexchanger design. The data of Table 10 have been arranged in a manner which will be most useful in heattransfer calculations. Tube Pitch. Tube holes cannot be drilled very close together, since too small a width of metal between adjacent tubes structurally weakens the tube sheet. The shortest distance between two adjacent tube holes is the clearance or ligament, and these are now fairly standard. Tubes are laid out on either square or triangular patterns as shown in Fig. 7;3a and b. The advantage of square pitch is that the tubes are accessible for external cleaning and cause a lower pressure drop when fluid flows in the direction indicated in Fig. 7.3a. The tube pitch PT is the shortest centertocenter distance between adjacent tubes. The common pitches forsquare layouts are% in. OD on lin. square pitch and 1 in. OD on 1~in. square pitch. For triangular layouts these are % in. OD on 1X6in. triangular pitch, % in. OD on lin. triangular pitch, and lin. OD on 1~in. triangular pitch. ln Fig. 7.3c the squarepitch layout has been rotated 45°, yet it is essentially the same as in Fig. 7.3a. In Fig. 7.3d a mechanically cleanable modification of triangular pitch is
12 PARALLELCOUNTERFLOW
129
·shown. If the tubes are spread wide enough, it is possible to allow the cleaning lanes indicated. Shells. Shells are fabricated from steel pipe with nominal IPS diameters up to 12 in. as given in Table 11. ,Above 12 and including 24 in. the actual outside diameter and the nominal pipe diameter are the same. The standard wall thickness for shells with inside diameters from 12 to 24 in. inclusive is % ·in., which is satisfactory for shellside operating pres::mres up to 300 psi. Greater wall thicknesses may be obtained for greater pressures. Shells above 24 in. in diameter are fabricated by rolling steel. plate. Stationary Tubesheet Exchangers. The simplest type of exchanger . is the fixed or stationary tubesheet exchanger of which the one shown in Fig. 7.4 is an example. The essential parts are a shell (1), equipped with two nozzles and having tube sheets (2) at both ends, which also serve as flanges for the attachment of the two channels (3) and their respective
FIG. 7.4.
Fixedhead tubular exchanger.
channel covers (4). The tubes are expanded into both tube sheets and are equipped with transverse baffles (5) on the shell side. The calculation of the effective heattransfer surface is frequently based on the distance between the inside faces of the tube sheets instead of the overall tube length. Baffles. It is apparent that higher heattransfer coefficients result when a liquid is maintained in a state of turbulence. To induce turbulence outside the tubes it is customary to employ baffles which cause the liquid to flow. through the shell at right angles to the axes of the tubes. This causes considerable turbulence even when a small quantity of liquid flows through the shell. The centertocenter distance between baffles is called the baj!le pitch or baffle spacing. Since the baffies may be spaced close together or far apart, the mass velocity is not entirely dependent ~'.pon the diameter of the shell. The baffie spacing is usually not greater than. a distance equal to the inside diameter of the shell or closer than a distance equal to onefifth the inside diameter of the shell. The baffles are held securely by means of baffle spacers (6) as shown in Fig. 7.4, which consist of throughbolts screwed into the tube sheet and a number of
130
PROCESS HEAT TRANSFER
smaller lengths of pipe w·hich form shoulders between adjacent baffles. An enlarged detail is shown in Fig. 7.5. There are several types of baffles which are employed in heat exchangers, but by far the most common are segmental baffles as shown 'in Fig. 7 .6. Segmental baffles are drilled plates ·with heights which a.re generally 75 per cent of the inside diameter of the shell. These are known .Shell
w===============~======~~ifknge Tvbe ~ T2, then t2  T 2 is called the temperature cross. It is useful to investigate several typical process temperatures and to note the influence of different approaches and crosses upon the value of
t:tt~·L
1.0

Q9
...
1
1 1  rwo
ftv,1s Jiln

IT~ ~ ..._ 50°Fran!JI. Two fluids wilh ~ 100°Franges"'

.....
r\
as Fr
Limiting praclical Fr 
Q7
~
~ ~\
\
\ \
!\
0.6
\ \ \
I 05 · 100
so
foE
60
40
20
0
20
40
Approach°F *Cross°F>I
FIG. 7.22. Influence of approach temperature on F7' with fluids .having equal ranges in a 12 exchanger.
1.0
 :.:t1l
0.9 
Tw~Fiui~ ~
/00°Fn:rnqes
Flu~ds 1wti'~+
1
!'.(, ..f.OO ana' 20"Fr~tn!f.e&
.......
.....
0.8
'
N
rt\.
.:..r 1\ ~ \
limifinq pracfica/ Fr ~
~
0.6
0.5
1\
I
w
w
w
\ \
00 00 40 0 ~ ~Approach °F ...JCross °F>1 Influence of approach temperature on F7' with fluids having unequal ranges in
Fro. 7.23. a 12 exchanger.
·
For a given service the reduction of F7' below unity in Eq. (7.42) is compensated for by increasing the surface. Thus if the process temperatures are fixed it may be inadvisable to employ a parallel flowcounterflow exchanger as against a counterflow exchanger, since it increases the cost of the equipment beyond the value of its mechanical advantages. In Fig. 7.22 two pairs of fluids each with equal ranges of 100 and 50°F
F7'.
147
12 PARALLELCOUNTERFLOW
are studied. The operating temperatures of the cold fluid are :fixed, while the hotfluid temperatures are variable thereby changing the approach in each case. Note the conditions under which FT shrinks rapidly, particularly the approach at the practical minimum FT = 0.75 and the influence of the relationship between T2 and t2. The calculation of several points is demonstrated. Example 7.2. Calculation of FT for Fluids with Equal Ranges Point: (a) 50° approach (b) Zero approach (c) 20° cross (T,) aso 200 ct.) (T ,) 300 200 (t 2) (T,) 280 200 (t2 ) (T 2) 250 100 (t,) (T.) 200 100 (t,) (T•) 180 100 (t,) 100 100 100 100 100 100 T, T 2 100 R = 1.0 R = ~ = 100  1.0 R = 1.0 S FT
t.  t,
100
= T,  t, = 350 = 0.925 (~g. 18)
100 = 0.40
S = 0.50 FT = 0.80
s=
0.555
FT = 0.64
In Fig. 7.23 are shown the results of the calculation when one fluid has a range :five times as great as the other. Shellside Pressure Drop. The pressure drop through the shell of an exchanger is proportional to the number of times the fluid crosses the bundle between baffles. It is also proportional to the distance across the bundle each time it is crossed. Using a modification of Eq. (3.44) a correlation has been obtained using the product of the distance across the bundle, taken as the inside diameter of the shell in feet D. and the number of times the bundle is crossed N + 1, where N is the number of baffles. If L is the tube length in feet, Number of crosses, N
+1=
tube length, in./baffle space, in: = 12 X L/B
(7.43)
If the tube length is 16'0" and the bafll.es are spaces 18 in. apart, there will be 11 crosses or 10 baffles. There should always be an odd number of crosses if both shell nozzles are on opposite sides of the shell and an even number if both shell nozzles are on the same side of·the shell. With close baffie spacings at convenient intervals such as 6 in. and under, one baffle may be omitted if the number of crosses is not an integer. The equivalent diameter used for calculating the pressure drop is the same as for heat transfer, the additional friction of the shell itself being neglected. The isothermal equation for the pressure drop of a fluid being lreated or cooled and including entrance and exit losses is
l:lP
•
= fG~D.(N + 1) = 2gpD.q,, ·
JG;D8(N + 1) 5.22 X 1010D.sq,,
psf
(7.44)
148
PROCESS HEAT TRANSFER
where 8 is the specific gravity of the fluid. Equation (7.44) gives the pressure drop in pounds per square foot. The common engineering unit is pounds per square inch. To permit the direct solution for t:.P, in psi dimensional shellside friction factors, square foot per square inch, have been plotted in Fig. 29~ To obtain the pressure drop in consistent units by Eq. (7.44) multiply fin Fig. 29 by 144. Tubeside Pressure Drop. Equation (3.44) may be used to obtain the pressure drop in tubes, but it applies principally to an isothermal fluid. Sieder and Tate have correlated friction factors for fluids being heated or cooled in tubes. They are plotted in dimensional form in Fig. 26 and are used in the equation psf
(7.45)
where n is the number of tube passes, L the tube length, and Ln is the total length of path in feet. The deviations are not given, but the curve has been accepted by the Tubular Exchanger Manufacturers Association. In flowing from one pass into the next at the channel and floating head the fluid changes direction abruptly by 180°, although the flow area provided in the channel and floatinghead cover should not be less than the combined flow area of all the tubes in a single pass. The change of direction introduces an additional pressure drop t:.P., called the return loss and accounted for by allowing four velocity heads per pass. The velocity head V 2 /2g' has been plotted in Fig. 27 against the mass velocity for a fluid with a specific gravity of 1, and the return losses for any fluid will be · 4n V2 t:.P = 'psi (7.46) • 8 2g' where V = velocity, fps s = specific gravity g' = acceleration of gravity, ftjsec 2 The total tubeside pressure drop t:.P'r will be psi
(7.47)
The Analysis of Performance in an Existing 12 Exchanger. When all the pertinent equations are used to calculate the suitability of an existing exchanger for given. process conditions, it is known as rating an exchanger. There are three significant points in determining the.suitability of an existing exchanger for a new service. 1. What clean coefficient U c can be "performed" by the two fluids as the result of their flow and individual film coefficients h;o and ho?
12 PARALLELCOUNTERFLOW
149
2. From the heat balance Q = WC(Tt  T2) = wc(t2  tt), known surface A 1 and the true temperature difference for the process temperatures a value of the design or dirty coefficient, U D is obtained. U a must exceed U D sufficiently so that the dirt factor, which is a measure of the excess surface, will pennit operation of the exchanger for a reasonable period of service. 3. The allowable pressure drops for the two streams may not be ·exceeded. "When thes~ are fulfilled, an existing exchanger is suitable for the process conditions for which it has been rated. In starting a calculation the first point which arises is to determine whether the hot or cold fluid should be placed in the shell. There is no fast rule. One stream may be large and the other small, and the baffle spacing may be such that in one instance the shellside flow area a. will be larger. Fortunately any selection can be checked by switching the two streams and seeing which arrangement gives the larger value of U a without exceeding the allowable · pressure drop. Particularly in preparation for later methods there is some advantage, however, in starting calculations with the tube side, and it may be well to establish the habit. The detailed steps in the rating of an exchanger are outlined below. The subscripts s and t are used to distinguish between the shell and tubes, and for the outline the hot fluid has been assumed to be in the shell. By always placing the hot fluid on the left the usual method of computing the LMTD may be retained. The Calculation of an Existing 12 Exchanger. Process conditions required Hot fluid: Tt, T2, W, c, s, p., k, Rd, AP Cold fluid: it, t2, w, c, s, ~' k, Ra, A.P For the exchanger the following data must be known: Shell side
ID Baffle space Passes
Tube side Number and length OD, BWG, and pitch Passes
(1) Heat balance, Q = WC(Tt  T2) = wc(t2  it) (2) True temperature difference At: LMTD R = Ti  T2, S = t2  h I t2  tt Tt  it At = LMTD X F'r (F r from Fig. 18)
(5.14) (7.42)
150
PROCESS HEAT TRANSFEIJ,
(5,28), (5.29)
(3) Caloric temperature Tc and t.: 1
Hot fluid: shell side Cold fluid: tube side (4') Flow area, a. = ID X C'B/144PT, ftt (4) Flow area, a,: [Eq. (7.1)] Flow area per tube a~ from Table 10, in. 1 No. of tubes X flow area/tube a, = No. of passes = Nt a~/144n, ft 2 Eq. (7.48) (6) Mass vel, Gt = w/at, lb/(hr)(ft 1 ) (6') Mass vel, G. = W /a., lb/(hr)(ft2) [Eq. (7.2)] (6') Obtain D. from Fig. 28 or compute (6) Obtain D from Table 10, ft. from Eq. (7.4). Obtain I' at t,, lb/(ft)(hr) = cp X 2.42 Obtain I' at T., lb/(ft)(hr) = cp X 2.42 Ret = DGt!l'
Re.
=
Dll./I'
(7) Obtain iH from Fig. 24. (7') Obtain j H from Fig. 28. (8') At Tc obtain c, Btu/(lbWF) and k, (8) At t. obtain c, Btu/ (lb) (°F) and k, Btu/(hr)(ftt)(°F /ft). Btu/(hr)(ft2)(°F /ft). Compute* (cJ,
= 10.0 psi
(4)
4
4
0 ~3
fli>'I'
[Fig. 27]
[E q.
(7 46)] .
X 0.15 = 2.9 psi
+ fli>;. + 2.9 = 9.2 psi
= tJ>,
= 6.3
Allowable
2Vg~ = 0.15
fli>'I'
[Eq. (7.47)1
= 10.0 psi
It is seen that a dirt factor of 0.00348 will be obtained although only 0.003 will be ~equired to provide reasonable maintenance period. The pressure drops have not been: exceeded and the exchanger will be satisfactory for the service.
154
PROCESS HEAT TRANSFER
Exchangers Using Water. Cooling operations using water in tubular equipment are very common. Despite its abundance, the heattransfer characteristics of water separate it from all other fluids. It is corrosive to steel, particularly when the tubewall temperature is high and dissolveQ. air is present, and many industrial plants use .nonferrous tubes exclusively for heattransfer services involving water. The commonest nonferrous tubes are admiralty, red brass, and copper, although in certain localities there is a preference for Muntz metal, aluminum bronze, and 8.Iuminum. · Since shells are usually fabricated of steel, water is best handled in the tubes. When water flows in the tubes, there is no serious problem of corrosion of the channel or floatinghead cover, since these parts are often made of cast iron or cast steel. Castings are relatively passive to water, and large corrosion allowances above structural requirements can be provided inexpensively by making the castings heavier. Tube sheets may be made of heavy steel plate with a corrosion allowance of about 7§ in. above the required structural thickness or fabricated of naval brass or aluminum without a corrosion allowance. When water travels slowly through a tube, dirt and slime resulting from microorganic action adhere to the tubes which would be carried away if there were greater turbulence. As a standard practice, the use of cooling water at velocities less than 3 fps should be avoided, although in certain localities minimum velocities as high as 4 fps are required for continued operation. Still another factor of considerable importance is the deposi" tion of mineral scale. When water of average mineral and air content is brought to ·a temperature in excess of 120°F, it is found that tube action becomes excessive, and for this reason an outlet water temperature above 120°F should be avoided. Cooling water is rarely abundant or without cost. One of the serious problems facing the chemical and power industries today results from the gradual deficiency of surface and subsurface water in areas of industrial concentration. This can be partially overcome through the use of cooling towers (Chap. 17), which reuse the cooling water and reduce the requirement to only 2 per cent of the amount of water required in oncethrough use. River water may provide part of the solution to a deficiency of ground water, but it is costly and presupposes the proximity of a river. River water must usually be strained by moving screens and pumped considerable distances, and in some localities the water from rivers servic· ing congested industrial areas requires cooling in cooling towers before it can be used. Many sizable municipalities have legislated against the use of public water supplies for large cooling purposes other than for makeup in cooling towers or spraypond systems. Where available, municipal water
1fa PARALLELCOUNTERFLOW
155
may average about 1 cent per 1000 gal., although it has the advantage of being generally available at from 30 to 60 psi pressure which is adequate for most process needs including the pressure drops in heat exchangers. Where a cooling tower is used, the cost of the water is determined by the cost of fresh water, pumping power, fan power, and writeoff on the original investment. The shellside heattransfer curve (Fig. 28) correlates very well for the flow of water across tube bundles. The high thermal conductivity of water results in relatively high film coefficients compared with organic ,fluids. The use of the tubeside curve (Fig. 24), however, gives coefficients which. are generally high. In its place the data of Eagle and Ferguson 1 for water alone are given in Fig. 25 and are recommended whenever water flows in tubes. Since this graph deals only with water, it has been possible to plot film coefficients vs. velocity in feet per second with temperature .as the parameter. The data have been plotted with the %in., 16 BWG tube as the base, and the correction factor obtained from the insert in Fig. 25 should be applied when any other inside diameter is used. In a watertowater exchanger with individual film coefficients ranging from 500 to.1500 for both the shell and tube, the selection of the required dirt factor merits serious judgment. As an example, if film coefficients of 1000 are obtained on the shell and tube sides, the combined resistance is 0.002, or Uc = 500. If a fouling factor of 0.004 is required, the fouling factor becomes the controlling resistance. When the fouling factor is 0.004, U D must be less than 1/0.004 or 250. Whenever high coefficients exist on both sides of the exchanger, the use of an unnecessarily large fouling factor should be avoided. The following heatrecovery problem occurs in powerhouses. Although it involves a moderatesize exchanger, the heat recovery is equivalent to nearly 1500 lb/hr of steam, which represents a sizable economy in the course of a year. Example 7.4. Calculation of a DistilledwaterRawwater Exchanger. 175,000 lb/hr of distilled water enters an exchanger at 93°F and leaves at 85°F. The heat will be transferred to 280,000 lbjhr of raw water coming from supply at 75°F and leaving the exchanger at so•F. A 10 psi pressure drop may be expended on both streams while providing a fouling factor of 0.0005 for distilled water and 0.0015 for raw water when the tube velocity exceeds 6 fps. Available for this service is a 157.4: in. ID exchanger having·l60% in. OD, 18 BWG tubea 16'0" long and laid out on 1X 6 in. triangular pitch. ·The bundle is arranged for two passes, and baffies are spaced 12 in. apart. Will the exchanger be suitable? 1
Eagle, A., and R. M. Ferguson, Pro. Roy. Soc., A127, 540566 (1930).
156
PROCESS HEAT TRANSFER
Solutwn:
Exchanger: Tube side· Null.lber and lf..ngth = 160, 16'0" OD, BWG, pitch =%in., 18 BWG, Passes= 2
Shell aide ID = 15),4 in. BafHe space = 12 in. Passes= 1
IM 8 in. tri.
(1) Heat balance:
Distilled water, Q = 175,000 X 1(93  85) = 1,400,000 Btu/hr Raw water, Q = 280,000 X 1(80  75) = 1,400,000 Bt.u/hr (2) At:
Hot Fluid
Cold Fluid
93
Higher Temp
80
13
85
Lower Temp
75
10
5
3
8
Differences
(5.14)
LMTD = ll.4°F
R
=
8
5=
Diff.
1.6
5 8 = 93 _ 75 = 0.278
(Fig. 18) (7.42) = 0.945 X 11.4 = 10.75°F (3) T, and t,: The average temperatures Ta and t,. of 89 and 77.5°F Will be satisfactory for the short ranges and .p. and op 1 taken as 1.0. Try hot fluid in shell as a trial, since it is the smaller of the two. F'r = 0.945 At
Hot fluid: ahell side, rlistilled water Cold fluid: tube side, raw water (4') a, = ID X C'B/144P7! [Eq. (7.1)] (4) a~ = 0.334 in.• [TablelO] a, = N ,a;/144n [Eq. (7.48)] = 15.25 X 0.1875 X 12/ 144 X 0.9375 = 0.254 ft' = 160 X 0.334/144 X 2 = 0.186 ftl (5') G. = W fa, [Eq. (7.2)] (5) G, = wja, = 175,000/0.254 = 280,000/0.186 = 690,000 lb/(hr)(ft1) = 1,505,0001b/(hr)(ft') Vel, V = G,j3600p = 1,505,000/3600 X 62.5 = 6.70 fps (6) At t. = 77.5°F, ,. = 0.92 X 2.42 (6') At T. = 89"F, p. = 0.81 X 2.42 = 1.96lb/(ft)(hr) = 2.23 lb I (ft )(hr) [Fig. 14] [Fig.14] D, = 0.55/12 = 0.0458 ft [Fig. '28] D = 0.65/12 = 0.054 ft (Re, is for pres. sure drop only) [Table 10] Re, = D,.G,fp. [Eq. (7.31 Re, = DG,j,. = 0.0458 X 690,000/1,96 = 16,200 (7') jg = 73 [Fig. 28] = 0.054 X 1,505,000/2.23 = 36,400 (8') At T. = gg•F, c = 1.0 Btu/(lb)(°F) k = 0.36 Btu/(hr)(ft')(°F /ft) [Table 4] (cp.fk)'li = (1.0 X 1.96/0.36)'li = 1.76
157
12 PARALLELCOUNTERFLOW Cold fluid: tube side, raw water
Hot fluid: shell side, distilled water
(9') h; =
iH
i. (tt'
[Fig. 25] X 1 [Eq. (6.15b)J (9) h; = 1350 X 0.99 = 1335 h;. = h; X ID/OD = 1335 X 0.65/0.75 = 73 X 0.36 X 1.76/0.0458 = 1010 = 1155 [Eq. (6.5)] (10') (11') (12') The small difference in. the average temperatures eliminates the need for a tubewall correction, and .p, = 1. . (13) Clean overall coefficient U o: U0
=
h;ch. + ho
h;o
= 1155 1155
X 1010
+ 1010
= 537 Bt /(hr)(fttJ(OF) U
(6.38)
·when both. film coefficients are high the thermal resistance of the tube metal is not necessarily insignificant as assumed in the derivation of Eq. (6.38). For a steel 1.8 BWG tubeR,. = 0.00017 and for copper R .. = 0.000017. (14) Design overall coefficient UD:
External surface/ft, a" = 0.1963 ft 2 /ft A = 160 X 16'0" X 0.1963 = 502 ft• Q 1,400,000 259 U D = A. = ~= 2.on (4) tiFT = p, + P, = 4.9 2.6 = 7.5 psi
+
Allowable tiFT
[Eq. (7.46)] •
pSI
[Eq. (7.47)]
= 10.0 psi
It, is seen that the overall coefficient for this problem is five times that of the oiltooil exchange of Example 7.3, the principal difference being due to the excellent thermal properties of water. The exchanger is satisfactory for the service.
Optimum Outletwater Temperature. ln using water as the cooliag medium for a given duty it is possible to circulate a large quantity with a small temperature range or a small quantity with a large temperature range. The temperature range of the water naturally affects the LMTD. If a large quantity is used, t2 will be farther from T1 and less surface is required as a result of the larger LMTD. Although this will reduce the original investment and fixed charges, since depreciation and maintenance will also be smaller, the operating cost will be increased owing to the greater quantity of water. It is apparent that there must be an optimum between the two conditions: much water and small surface or little water 11.nd large surface. In the following it is assumed that the line pressure on the water is sufficient to overcome the pressure drop in the exchanger and that the cost of the water is related only to the amount used. It is also assumed that the cooler operates in true counterflow so that !:.t = LMTD. If the approach is small or there is a temperature cross, the derivation below requires an estimate of FT by which the LMTD is multiplied. The total annual cost of the exchanger to the plant will be the sum of the annual cost of water and the fixed charges, which include maintenance and depreciation. If aT is the total annual COSt, aT = (water cost/lb)(lb/hr)(annual hr) + (annual fixed charges/ft2)(ft2) Q = wc(t2  t1) = UA (LMTD) (7.49) Substituting the heatbalance terms in Eq. (7.49), where w = Q/[c(t 2  t 1)] and the surface A = Q/U(LMTD)
a
Qoaw T
= c(ta t1)
al'Q
+ U(LMTD)
12 PARALLELCOUNTERFLOW
159
where 8 = annual operating hours Cw = water cost/lb CF ,;., annual fixed cbarges/ft2 Assuming U is constant
LMTD = At2  AlL ln At2/Ah
Keeping all factors constant except the wateroutlet temperature and consequently At2, CT =
Q8Cw ~;(t2  tz)
+
CFQ U [ T 1  t2  Atz ] In (Tz t2)/Atz
(7.50)
The optimum condition will occur when the total annual cost is a minimum, thus when dCT/dt2 = 0. Differentiating and equating the respective parts, 2 1 ] (7.51) U8Cw(Tz h At1) l Tz t2 [ CFc · t2 tz = n""A4  1  (Tt t2)/At1 Equation (7.51) has been plotted by Colburn and is reproduced in Fig. 7.24. Example '7;5. · Calculation of the Optimum Outletwater Temperature. A viscous .fluid is cooled from 175 to 150°F with water available at 85°F. What is the optimumoutlet water temperature? 175 z = Mz 150  85 = t..tl =.65 It will first be necessary to assume a value of U. Since the material is viscous, assume U = 15. To evaluate the group UBCw/Cpe: (J = 8000 operating hours annually Cw computed at $0.01/1000 gal. = 0.01/8300, dollars/lb For annual charges assume 20 per cent repair and maintenanc.e and 10 per cent depreciit.tion. At a unit cost of $4 per square foot the annual fixed charge is $4 X 0.30 = $1.20
The specific heat of water is taken as 1.0.
UoCw = 15 X 8000 ( 0.01) = 0 _1205 (fpc 1.20 X 1.0 8300 · Tt  Tz = 175  150 = 0.39 t.t~
150  85
From Figure 7.24,
~: = 0.96 &z = T,  t2 = 0.96 X 65 = 62.3°F t: = 175  62.3 = 112.7°F
160
PROCESS HEAT TRANSFER
When the value of U is high or there is a large hotfluid range, the optimum outletwater temperature may be considerably above the upper limit of 120°F. This is not completely correct, since the maintenance cost will probably rise considerably above 20 per cent of the initial cost when the temperature rises above 120°F. Usually information is not available on the increase in maintenance cost with increased wateroutlet 10 8 6
4 J 2
At2 1.0

~ '"~ "' r
t;,•z. fn !G':Ln (2) ..r'  5.22 X l0 10Ds,
[Fig. 26]
+ 170)
= 0209 lb/ft 3 0
= 0.00335
D, = 23.25/12 = 1.94 ft
,
JG~D,(N
+ 1)
(3) G, = 828,000, V 2/2g' = 0.090 [Fig. 27] (3 ) AP. = 5.22 X 1010D.s.p, AP, = (4n/s)(V 2 /2g') [Eq. (7.46)] 0.00162 X 25,400 2 X 1.94 X 8 4 8 1 = ~ X 0.090 = 2. 9 psi = 5.22 X 10 0 X 0.0458 X 0.00335 X 1.0 (4) t,.Pp = f1Po + f>Pr [Eq. (7.47)] = 2.0 psi = 3.6 2.9 = 6.5 psi Allowable 11FT = 10.0 psi Allowable AP, = 2.0 psi
1
+
The ability to meet the allowable pressure drop hinges closely upon the density of the If the gas had been air at the same pressure, the density and pressure drop would have been 0.209 X 2J1 7 = 0.357lb/ft 3 and 1.2 psi. Similarly an exchanger can be used for gases at vacuum pressure only when a very small mass velocity is employed. The latter results in very low transfer rates for vacuum services, values of U D being as low as 2 to 10 Btu/(hr)(ft')("F). gas.
Air Compressor Intercoolers. In the compression of air for utility purposes it is common to subject atmospheric air to four or more stages of compression. The allowable pressure drop in the intercoolers following the initial stages of compression is extremely critical. Assuming that. the compressors operate with compression ratios of roughly 2;!1: 1 or 2~: 1, a pressure drop of 1 psi in the firststage intercooler represents a reduction in the total pressure delivered after the fourth stage of 1 X 2.5 X 2.5 X 2.5 = 13.1 psi and nearly 80 psi after the sixth stage. In addition, the presence of moisture in the inlet air makes it impossible to compute the heat load as a simple sensibleheat change. Suppose saturated air is taken into the compressor at 95°F (during a summer shower), compressed, and then cooled back to 95°F between each stage of compression. The air and water vapor both occupy the same total volume. The compression of a saturated gas raises the dew point above
196
PROCESS HEAT TRANSFER
its initial dew point. Cooling the compressed gas back to its original temperature requires that it be cooled below its dew point. This can occur only if water condenses out of the compressed gas during cooling: The reasoning follows: The original water vapor in the saturated atmos· pheric air was all that could exist in the original volume of air at 95°F. Its total weight can be obtained from the specific volume of steam, in cubic feet per pound, for 95°F as found in Table 7. After compression and cooling back to 95°F the total volume of air is reduced but the specific volume of steam at 95°F is unchanged. The same specific volume can be maintained in the reduced gas volume only if some of the water condenses out of the gas. Example 9.2. Calculation of the Heat Load for an Air Intercooler. 4670 cfm of air saturated at 95°F enters a fourstage adiabatic compressor, having a compression ratio of 2.33:1, at atmospheric pressure. (a) How much heat must be removed in the firststage intercooler? (b) How much heat must be removed in the secondstage intercooler? Solution: (a) Inlet 4670 cfm: Saturation partial pressure of water at 95°F = 0.8153 psi (Table 7) Saturation specific volume of water at 95°F = 404.3 ft 8/lb (Table 7) The air and water both occupy the same volume at their respective partial pressures. Lb waterjhr entering = 4670 X 60/404.3 = 692lb First stage:
Mter 2.33 compression ratio P2 = 14.7 X 2.33 = 34.2 psi TT•) = (~)(rIl/'Y h = 1.40 for air*) ( 1 &bl PI T>abo = (2 33)(1.tll/1.( 460+95 . T2abo = 705°R or 245°F Intercooler:
Final gas volume = 4670 X 60 X 14.7/34.2 = 120,000 ft 1/hr Water remaining in air = 120,000/404.3 = 297 lb/hr Condensation in intercooler = 692  297 = 395lb/hr Specific volume of atmospheric air = (359/29) (555/492) 14.7/(14.7  0.8153) = 14.8 ft'/lb Air in inlet gas = 4670 X 60/14.8 = 18,900 lb/hr Heat load (245 to 95°F):
Sensible heat: Q.;. = 18,900 X 0.25(245  95) = Qwater = 692 X 0.45(245  95) Latent heat: Qter = 395 X 1040.1 * The correction of
'Y
708,000 Btu/hr 46, 700 Btu/hr
411,000 Btu/hr Total 1, 165,700 Btu/hr
for the presence of water vapor is usually omitted.
GASES
197
H condensation had not been accounted for, an error of 33 per cent would have resulted. It should also be noted that over half of the water condenses in the firststage intercooler. (b) Second stage: P• = 34.2 X 2.33 = 79.8 psi. Final gas volume = 4670 X 60 X 14.7/79.8 = 51,500 ft•jhr Pounds of water remaining in air = 51,500/404.3 = 127.5 lbjhr Condensation in intercooler = 297  127.5 = 169.5lb/hr
Heat load (245° to 95°F): Sensible heat: Qair
= 18,900
X 0.25(245  95)
= 708,000 Btujhr
Q,....,. = 297 X 0.44(245  95) = 19,600 Btujhr Latent heat:
Q = 169.5 X 1040.1
= 170,700 Btujhr
898,300 Btu/hr Example 9.3. Calculation of the Dew Point after Compression. The dew point and saturation temperature of the saturated inlet air are the same. After the firststage compression the dew point is raised. What is the dew point when air saturated at 95°F and 14.7 psi is compressed to 34.2 psi?
Solutron: At inlet: Mols air = 18,900/29 = 652 Mols water = 692/18 = 38.4 690.4 After compression: Partial pressure of water vapor = (38.4/690.4)34.2 = 1.90 psi From Table 7 equivalent to 1.90 psi, dew point = 124°F In other words, the gas and water vapor are cooled sensibly from 245 to 124°F in the first stage intercooler and the water vapor will start to condense at 124°F.
The Calculation of Coolers for Wet Gases. The calculation of the heat load and film coefficients in the intercoolers of adiabatic compression systems starting with initially dry gases offers no particular difficulty. The heat load is the sensibleheat requirement to cool the gas back between stages. The film coefficient is that of the dry gas. Coolers which are required to cool wet gases present a number of additional problems. If the wet gas is to be cooled below its dew point, two zones will appear: (1) from the inlet temperature to the dew point in which both the gas and the vapor are cooled sensibly and (2) from the dew point to the outlet temperature, in which the gas and vapor are cooled and part of the vapor condenses. The first zone can be calculated rather simply as a dry gas, but the calculation of the second zone is extremely lengthy. A relatively accurate calculation will be demon
198
PROCESS HEAT TRANSFER
strated as an example of condensation in Chap. 13, where it will be seen that both the condensation and gas film coefficients are closely related. The film coefficient for the mixture varies considerably from the dew point to the outlet temperature as the concentration of the condensable vapor diminishes. It is also seen that, in any wet gas cooling service if the tubewall temperature is below the dew point of the gas, even though the gas is not cooled below its dew point, the tube wall will be wet with condensate. As droplets of condensate fall from the tube, they will reflash into the gas and some fraction may drain from the cooler if the temperature should fall below the dew point. However, the film of liquid on the tube actually introduces a resistance film through which the heat must b~ transferred. If the condensable vapor is water, the resistance can be omitted because of the high conductivity of the film. If it is a vapor whose condensate is a viscous fluid, it may be necessary to calculate the mean resistance of the film from methods given in Chap. 13 based on the properties of the condensate. It is also well to consider that gases which are not particularly corrosive when containing a small concentration of water vapor may be corrosive when dissolved in condensate water at the cold tube wall. Desuperheaters, which are simply gas coolers, frequently operate with part of the surface moist although no actual condensate drains from the system. The performance of commercial intercoolers for permanent gases saturated at atmospheric pressure with water at 100•F or less can be predicted rapidly by empirical rules. These rules are as follows: (1) Calculate the entire heat load ofsensible cooling and condensation as if transferred at the dry gas rate, and (2) use the value of ~t = FT X LMTD obtained from the inlet and outlet temperatures of the gas to and from the cooler and the temperatures of the water. These rules are actually the combination of a safe and an unsafe generalization which tend to offset each other. The combined film coefficient for condensation and gas cooling below the dew point is greater than that given by (1). The true temperature difference is less than that calculated by (2), since the log mean for the portion of the heat load delivered from the dew point to the outlet is less than that calculated by the rule. PROBLEMS
9.1. 3500 cfm of dry nitrogen at 17 psig and 280"F is cooled to 100•F by water with an inlet temperature of s5•F. Available for the service is a 31 in. ID 12 exchanger having 600 % in. OD,, 16 BWG tubes 12'0" long arranged for eight passes on lin. triangular pitch. The baffie spacing is 24 in. center to center. Pressure drops of 2.0 psi for the gas and 10.0 for the water should not be exceeded, and a minimum dirt factor of 0.01 should be provided. Will the cooler work? 9.2. 17,500 lb/hr of oxygen at atmospheric pressure is cooled from 300 to 100•F by water from 85 to 100°. Available for the service is a 31 in. ID 12 exchanger containing 600 % in. OD, 16 BWG tubes 12'0" long arranged for eight passes on lin. triangular pitch. Baffles are spaced 24 in. apart. What are the dirt factor and the pressure drops?
GASES
199
9.3. 5000 cfm of saturated air at 100°F enters the first stage of a compre.ssor having a 2.45:1 compression ratio. The air is at atmospheric pressure. (a) How much heat must be removed after each of the four· stages, assuming a 2.0 psi pressure drop in each intercooler. (b) Available for the firststage intercooler is a 29 in. ID exchanger having 508 tubes %:ODin., 14 BWG by 12'0" long arranged for eight passes on lin. triangular pitch. The ba:ffie spacing is 24 in. Using cooling water with an 85°F inlet, what are the pressure drops and dirt factor? 9.4. For the secondstage intercooler of &le 9.2 in the text the following 12 exchanger is available: 21Xi: in. ID containing 294% in. OD, 14 BWG tubes 12'0" long arranged for eight passes on 1~ 6 in. triangular pitch. Baffies are spaced 20 in. apart. What are the dirt factor and the pressure drops?
NOMENCLATURE FOR CHAPTER 9
Heattransfer surface, ft 2 Flow area, ft' External surface per linear foot, ft Ba:ffie spacing, in. B Clearance between tubes, in. C' Specific heat of cold fluid, Btu/(lb)(°F) c D Inside diameter of tubes, ft D, Equivalent diameter, ft Temperature difference factor, At = FT X LMTD, dimensionless Fr Friction factor, ft2/in.' ! :Mass velocity, lb/(hr)(ft 2) G g' Acceleration of gravity, ft/see 2 Heattransfer coefficient in general, for inside fluid, and for outside h, h;, h. fluids, respectively, Btu/ (hr) (ft2 ) (°F) Value of k; when referred to the tube OD, Btu/(hr)(ft 2){°F) hi.o Factor for heat transfer (hD/k)(cp,jk)Y., dimensionless jn Factor for heat transfer (hjcG)(c!'/k)%, dimensionless ih Thermal conductivity, Btu/(hr)(ft 2) (°F/ft) k Tube length, ft L LMTD Log mean temperature difference, °F Number of baffies N N; Number of tubes Number of tube passes n Tube pitch, in. PT Pressure drop, psi AP APT, APt, APr Total, tube side and return drop, psi p Pressure, psia Temperature group (T,  To)/(t•  t,), dimensionless R Combined dirt factor, (hr)(ft 2)(°F/Btu) Rd Re Reynolds number, dimensionless Temperature group (t2  t1)/(T1  t,), dimensionless s Specific gravity, dimensionless Absolute temperature, "R Average temperature of hot fluid, oF Caloric temperature of hot fluid, °F Inlet and outlet temperature of hot fluid, op Average temperature of cold :fluid, °F
A
a a"
200
PROCESS HEAT TRANSFER
p
Calorie temperature of cold fluid, °F Inlet and outlet temperature of cold fluid, °F True temperature difference in Q = U DA ll.t, °F Clean and design overall coefficient of heat transfer, Btu/(hr)(ft 1)(°F) Velocity, fps Specific volume, ft3jlb Weight flow of hot fluid, lbjhr Weight flow of cold fluid, lbjhr Weight, lb Ratio of specific heats of a gas, dimensionless Viscosity, centipoises X 2.42 = lb/(ft)(hr) Viscosity at the tube wall, centipoises X 2.42 = lb/(ft)(hr) Density, lb/ft:
1>
(p./P.w) O.U
v
w
w
w' "(
p.
Subscripts (except as noted above) 8
Shell side Tube side
CHAPTER 10 STREAMLINE FLOW AND FREE CONVECTION Streamline Flow in the Tubes of Exchangers. From the Fourier equation alone for a single tube Q = wc(t2  t1) = h/rrDL) llt. When the inside tubewall temperature tp is constant, the temperature difference At in streamline flow may be replaced by the arithmetic mean of the hot and cold terminal temperature differences At,.= [(tp  t1) + (tp  t2)]/2. Solving for h;D jk, h;D
T
=
(2 we) kL (tp ;;:
h)
(h tt) (tp  tz)
+
{10.1)
It is interesting to note that the highest outlet temperature attainable in a tube is the constant temperature of the hot tube wall tp. For this case Eq. (10.1) reduces to hJ)
k
=(~we) 7(
kL
(10.2)
No obf3erved average value of h; can exceed that given by Eq. (10.2), and it is a useful tool by which erroneous observations may be rejected. Graetz obtained Eq. (3.27) from purely theoretical considerations on the assfunption of a parabolic distribution of velocities as the fluid flows in a tube in laminar flow. He did not include any corrections for modifications of the parabolic distribution during heating and cooling. Sieder and Tate evaluat€d the equivalent empirical equation [Eq. (3.32)] and obtained Eq. (6.1), which may be credited with correcting for the modifications of the velocity distribution during heating and cooling. Streamline flow in tubes may be construed as a conduction effect, and it is also subject to the simultaneous occurrence of free convection as well. Free convection is only significant in nonviscous fluids. Fluids flow in streamline flow due to three conditions: (1) The fluid is viscous; (2) the fluid is not viscous, but the quantity is small for the flow area provided; and {3) the flow quantity and viscosity are intermediate but combine to give streamline flow. Only when free convection is surpressed owing to the high average viscosity of the liquid, say several centipoises and over, or when the temperature difference is small does Eq. (6.1) give h; within the stated deviation as pure conduction. For cases falling under (2) or (3) above, lihe value of h,from Eq. (6.1) may be 201
202
PROCESS HEAT TRANSFER
conservative, the true value of h, being as much as 300 per cent greater due to the influence of free convection. In the Graetz derivation the value of L in the Graetz number wcjkL or in the ratio D/L is assumed to be the length of the path over which the fluid moves with a conduction temperature gradient at right angles to the long axis of the tube. Naturally, if mixing occurs at any point in the heattransfer tube, the distance from the inlet to the point of mixing must be regarded as the length of the path over which the temperature gradient is effective whether or not it corresponds to the total tube length of the exchanger, nL. Boussinesq 1 has advanced the theory that under ideal conditions streamline flow is not established until the liquids has traveled a length of approximately 15 tube diameters. In multipass heattransfer equipment it is sometimes possible to consider the fluid in the tubes mixed at the end of each pass. Internal mixing is desirable, just as convection is desirable and also because it restricts the length of streamline path to the length of each pass, L. The shorter the unmixed length the greater the value. of h., although it is not always safe to assume that mixing occurs at the end of each tube pass of an exchanger. In modern multjpass exchangers, the flow areas in the floating head and channel are usually designed to be identical with or slightly greater than the flow area of the tubes in each passs. In this way it is possible to eliminate excessive return pressure drops. If no turbulence or mixing is induced at the ends of each pass, nL is the total length of path instead of L, which leads to the calculation of safe values of h; should mixing actually occur. Frequently it will be found that the Reynolds number based on the viscosity at ta = (tt t2)/2 is less than 2100 but near the outlet the Reynolds number based on the viscosity at i2 is greater than 2100. Equation (6.1) is not applicable in the transitional or turbulent flow range. In calculations on mUltipass exchangers, the point at which Re = 2100 must then be obtained by trial and error, and the path beyond it excluded from calculation as streamline flow. If a portion of the tube is in the transitional range, it can best be computed by means of Fig. 24. For a twotube pass exchanger the maximum error in h, calculated from Eq. (6.1) between a mixing or nonmixing assumption between passes is (7f)lS = 1.26 or 26 per cent, since h, a: 1/LlS, For an eighttube pass exchanger the error is (M) ~ = 2.0 or 100 per cent. Ordinarily a decision should not be arrived at without consulting the design of the exchanger and noting whether or not provision for mixing between passes has been included.
+
1
Boussinesq, J., Gompt. rend., 113, 9 (1891).
203
STREAMLINE FLOW AND FREE CONVECTION
Example 10.1. Crude Oil Heater: Streamline Flow. A line carries 16,000 lb/hr of.34°API crude oil. It enters the tubes at 95°F and is heated to 145°F using steam at 250°F.. Consider the fluid mixed between passes. The viscosities of the crude oil are
•F 250 200 150 125 100
P.
cp
1.15 1.7 2.8
3.8 5.2
Available for temporary service is a horizontal 12 exchanger having a 15% in. ID shell with 86 1 in. OD, 16 BWG tubes 12'0" long laid out on 1Uin. triangular pitch. The bundle is arranged for two tube passes, and baflles are spaced 15 in. apart. Since the heater is for temporary use, no dirt factor will be included. Will the exchanger fulfill the requirement? Solution:
Exchanger: SheU side
ID =
15~
in. Baflle space = 15 in. Passes = 1
Tube side Number and length = 86, 12'0" OD, BWG, pitch = 1 in., 16 BWG, U'li:in. tri. Passes = 2
(1) Heat balance:
Crude, Q = 16,000 X 0.485(145  95) = 388,000 Btu/hr Steam, Q = 410 X 945.5 = 388,000 Btu/hr (2) t.t: Hot Fluid
Diff.
250
Higher Temp
145
105
250
Lower Temp
95
155
Differences
50
50
0
at
Cold Fluid
= LMTD = 129°F (true.counterflow)
(5.14)
(3) T. and t.: This fluid is in streamline flow throughout the heater [see (6), p. 204].
For streamline flow the arithmetic mean should be used.
For the first pass
125 = 110.F (approx. ) t• = t, +  t; = 95 + 2 2 On the assumption that the fluids are mixed between passes, each pass m\1St be solved independently. Since only two passes are present in this exchanger, it is simply a matter of assuming the temperature at the end of the first pass. More than half the heat load must be transferred in the first pass; therefore assume t; at' the end of the first pass is 125°F and t;  t, isao•F.
204
PROCESS HEAT TRANSFER
Cold fluid: tube side, crude oil 2 [Table 10] a, = N,al /144n [Eq. (7.48)] = 86 X 0.594/144 X 2 ... 0.177 ft 1 (5) G, = w/a 1 = 16,000/0.177 = 90,400 lb/(hr)(fV) (6) D = 0.87/12 = 0.0725 ft [Table 10] Re, = DG,jp. At t. = 145°F (the outlet temperature) p. = 2.95 X 2.42 = 7.15lb/(ft)(hr) Re, = 0.0725 X 90,400/7.15 = 915 At ta = no•F, . p. = 4.8 X 2.42 = 11.6lb/(ft)(hr) Re, = 0.0725 X 90,400/11.6 = 565
Hot fluid: shell side steam
a; = 0.594 in.
(4)
(7) h.=
1.86~ (~;;~)~ "'' [Eq. (6.1)]
(S) ~ = 0.485 X 11.6 = 72 5 k 0.0775 .
!!. L
(9)
(9') h. = 1500
(10') lp
=
la
+ h; ~ ho (T. 
= 0.0725 = 0 0060
12
'
!!!
= 1.86 X 0.0775 q,, 0.0725 (565 X 72.5 X
0.0060)~
= 12.4
(10) At t, = 249•F,
la)
p.
1500 = 110 + 12.4 1500 (250  110) = 249°F
+
q,,
= 1.20 X 2.42 = 2.9lb/(ft)(hr) =
(;~J.u
 (11.6)"· 2.9 (11) h; =
i
h·
14
= 1.20
q,, = 12.4 X 1.20 = 14.9 Btu/(hr)(ft 2)("F)
At =
t,  la
= 249  110
t; ,t, =
= 139°F
h.A; At we
Internal surface per foot of length = 0.228 ft
A; =
~6
X 12'0" X 0.228 = 117.5 ft 2
X 117.5 X 139 = 31 4 .F t'. _ t1 = 14.9 16,000 X 0.485 •
Assumed value of t;  t, = so.o•F (close enough for check) The oil now enters the second pass at 126.4°F and leaves at 152 instead of 145°F, indica.ting tha.t the heater is oversurfaced if no dirt factor i8 required.
205
STREAMLINE FLOW AND FREE CONVECTION
If four or more passes are present, the calculation is carried out in the same manner as shown above with a new assumed temperature at the end of each pass. If the calculated outlet from the last pass equals or exceeds t2, the heater will operate satisfactorily.
When a dlrt factor is to be provided, obtain the initial outlet temperature t2 delivered by the oversized heater, when freshly placed in service, from the heat balance and Eq. (5.18), using Ua for U, Ua is obtained from UD and Rd by Eq. (6.10) instead of from hio and ho, since in this case it is desired to calculate the value of h;o which will produce the initial value of the outlet temperature. Having calculated t2, solve for each pass until the outlet temperature corresponds to the initial value of tz before dirt accumulates. Free Convection in Tubes. Streamline flow is calculated by equations employing weight flow or mass velocity as one of the variables. However, if a horizontal tube surrounded by condensing steam carries cold
(a)
(b)
FIG. 10.1.
(c)
(d)
Free convection in tubes.
liquid and its flow is suddenly halted, the liquid in the tube continues to heat. According to Eq. (6.1) the film coefficient should be zero if the mass velocity is zero. Where the heat is transferred through movements within the liquid itself without forced circulation, it occurs. by free or natural convection. Some of the factors influencing free convection in liquids can be observed very readily in the laboratory owing to changes in the index of refraction which accompany changes in density. When a fluid is heated in a glass vessel on a hot plate, the convection currents are visible. Liquid at the bottom of the vessel and adjacent to the heat source is heated by conduction. The heat absorbed reduces the density of the bottom layer of liquid so that it rises and the colder liquid tends to settle. In a horizontal tube the process is somewhat more orderly. Starting with a stationary liquid, heat is applied from the outside, raising the temperature of an outer layer of liquid as shown in Fig. 10.1a. The cold central core is heavier than the liquid adjacent to the wall and settles toward the bottom of the tube somewhat as shown in Fig. lO.lb. The rate of settling is retarded by the temperatureviscosity relationship between the hot fluid at the wall and the cooled core of liquid. As the
206
PROCESS HEAT TRANSFER
freeconvection currents are established, they develop and mix with the bulk of liquid presumably as in Fig. lO.lc, and if the tube is large, mixing may be accelerated as· in Fig. lO.ld. The film coefficient for free convection is a function of the inside diameter of the tubeD, density of the liquid p, coefficient of expansion {J, gravitational constant g, thermal conductivity k, viscosity IJ., and lastly the temperature difference At. between the hot tube wall and the bulk of the fluid.
h,
=
j(D, p, fJ, c, g, k,
!L,
At,.)
Solving by dimensional analysis
h.D _ (D 32
C1 k
p g{J p. 2
(CJ.L)b
At..)" 
k
(10.3)
where D 8p2gfJ At.,/!.1. 2 is the Grashof number. Combined Free Convection and Streamline Flow in Horizontal Tubes. Just as there is a transition region and not a single point separating streamline and turbulent flow, there must also be some transition region between free convection to a fluid standing still and streamline flow. At low linear velocities both are undoubtedly operative. Equation (6.1) was correlated from data obtained on small tubes with moderately viscous fluids and under moderate temperature differences such that the Grashof numbers were relatively small. Kern and Othmer 1 investigated this region in horizontal tubes under large temperature differences and tube diameters and evaluated free convection as a correction to the SiederTate equation. Their final equation is
h.Dk (.1:)o. 14 = l.S6 [(DG) (CJ.l.) (Q)]% 2.25(1 + O.OlOGr}>) k L log Re
(lOA)
p.
f.Lw
where Gr., is the Grashof number evaluated from properties taken at the average fluid temperature t., = (t1 t2)j2. hi as ordinarily calculated by Eq. (6.1) can be corrected for free convection by multiplying by
+
if;
= 2.25(1 + O.OlOGr.,%) log Re
(10.5)
Inspection of Eq. (10.4) indicates that the influence of the freeconvection currents are dissipated in the transition and turbulent region. In view of the delicate nature of freeconvection currents this is quite plausible. The two factors which ordinarily influence free convection most are a low viscosity and large temperature difference. Martinelli et al.2 studied Kern, D. Q., and D. F. Othmer, Trans. AIChE, 89, 517555 (1943). Martinelli, R. C., C. J. Southwell, G. Alves, H. L. Craig, E. B. Weinberg, N. E. Lansing, and L. M. K. Boelter, Trans. AIChE, 38, 943 (1942). 1
2
STREAMLINE FLOW AND FREE CONVECTION
207
the influence of free convection in upward and downward flow in vertical tubes. They found a slight increase in the coefficient when heating water in upward flow compared with downward flow. Their final correlation is rather complicated, although they also obtained a correlation of free convection involving the Reynolds number. Example 10.2. Kerosene Heater: Streamline Flow and Free Convection. A line carries 16,000 lb/hr of 40"API light distillate or heavy kerosene with corrosive contaminants. It enters thetubes at 95°F and is heated to l45°F using steam at 250°F. Consider the liquid unmked between passes. The viscosities of the kerosene are •F
250 200 150 125 100
J.l, cp 0.60 0.85 1.30 1. 70 2.10
Available for the service is a horizontal 12 exchanger, having a 15X in. ID shell with 86 1 in. OD, 16 BWG tubes 12'0" long laid out on lXin. triangular pitch. The bundle is arranged for two tube passes, and baffles are spaced 15 in. apart (same as Example 10.1). What is the true dirt factor? Solution:
Exchanger: Shell aide 1D = 15~ in. Baffle space = 15 in. Passes= 1
Tube side Number and length = 86, 12'0" OD, BWG, pitch = 1 in., 16 BWG, 1Uin. tri. Passes= 2
(1) Heat balance: Kerosene, Q = 16,000 X 0.50(145  95) = 400,000 Btujhr Steam, Q = 4230 X 945.5 = 400,000 Btu/hr (2)
t.t: Hot Fluid
Cold Fluid
Diff:
250
Higher Temp
145
105
250
Lower Temp
95
155
Differences
50
50
0
t.t = LMTD = 129"F (true counterflow) (5.14) (8) T. and t.: This fluid is in streamline flow throughout the heater fsee (6), p. 208]. For streamline flow the arithmetic mean is
208
PROCESS HEAT TRANSFER Hot fluid: shell side, steam
Cold fluid: tube side, 40°API kerosene (4) = 0.594 in.• [Table 10] a,= N,a:f144n [Eq. (7.48)] = 86 X 0.594/144 X 2 = 0.177 It• (6) G, = w/at = 16,000/0.177 = 90,400 lb/(hr)(ft') (6) D "" 0.87/12 = 0.0725 ft [Table 10]
ai
Ret= DG,jp. At t, = 145°F, p. = 1.36 X 2.42 = 3.29 lb/(ft)(hr) [Fig. 14] Re, = 0.0725 X 90,400/3.29 = 1990 At t. = 120"F, p.. = 1.75 X 2.42 = 4.23lb/(ft)(hr) Re, = 0.0725 X 90,400/4.23 = 1,550 Streamline flow, fluid unmixed between passes Ln/D = 12 X 2/0.0725 = 331 (7) in = 3.10 [Fig. 24] .(8) At p. = 1.75 cp and 40"API, k(cp../k)~ = 0.24 Btu/(hr)(ft•) (°F /ft) [Fig. 16] (9') Condensation of steam: h.= 1500
= jH ~(~)~ [Eq. (6.15a)] D k h;/ 1 = 3.10 X 0.24/0.0725 = 10.25
(10') t,.:
(10)
+ _h_. (T. t.) [Eq. (5.31)] ho. + ~· 1000 120 + (250  120) 8.9 + 1500
t., = t. =
= 249•F
!!:! q,,
(9)
~:
= ~X ID/OD [Eq. (6.5)] = 10.25 X 0.87/1.0 = 8;91
(11) At t., = 249°F, p.., = 0.60 X 2.42 = 1.45 lb/(ft)(hr)
q,, = (p./p.w)O.U
= (4.23/1.45) 0·u = 1.16
h~: "''
(12) h;. =
[Eq. (6.37)]
= 8.91 X 1.16 = 10.3 Btu/(hr)(fts)(°F) tJ.t. = t,.  t. = 249  120 = 129°F Since the kerosene has a viscosity of only 1.75 cp at the caloric temperature and tJ.t. = 1299F, free convection should be investigated. if! = 2.25(1 O.OlGr.Y.I) [Eq. (10.5) log Re Grashof number, Gr. = D'p 2g{j !J.t.jp.i 8 = 0.80, [Fig. 6] p = 0.8 X 62.5 = 50.0 lb/fta (j _ _! _ 1/PI  1/Pt  °F  (t,  lt)(1/Pov)
+
8~

s:
STREAMLINE FLOW AND FREE CONVECTION Hot fluid: shell side, steam
209
Gold fluid: tube side, 40° API kerosene At 95•F, 81 == 0.810 [Fig. 6] At 145•F, 82 = 0.792 {I = 0.00045lfOF Gr~ = 0.0725 3 X 50.0 2 X 0.00045 X 4.18 X 10 8 X 129/4.239 = 1,300,000 y, = 2.25(1 0.01 X 1,300,000%) = 1 47 · log 1550 Corrected h;. = 10.3 X 1.47 = 15.1
+
(13) Clean overall coefficient Uc:
Uc =
h;:~·ho
=
~::~ ~ ~=
= 14.9 Btu/(hr)(ft 2)(DF)
(6.38)
(14) Design overall coefficient U »:
a" = 0.2618 ft2/lin ft Total surface, A = 86 X 12'0" X 0.2618 = 270 ft• UD =A~t
(Table 10)
= 2:g ~~~9 = 11.5 Btu/(hr)(ft2)(DF) 0
(16) Dirt factor:
Rd
= Uc  UD = 14.9  11.5 = 0.0198 (hr)(ft•)("F)/Btu UcUD
14.9 X 11.5
(6.13)
Summary: Figures in parentheses are uncorrected for free convection Summary 1500
I
h outside
Uc
(10.2)14.9
UD
11.5
I
(10.3) 15.1
Rd Calculated 0.0198
If the correction for free convection were not included, it would have given the impression that the unit would not work, since Uc would be smaller than U D· Corrected, however, it is very safe. If the viscosity were 3 or 4 centipoises and the Reynolds number remained the same because of increased weight flow, the exchanger would not be suitable. The pressure drop may be computed as heretofore except that q,, for pressure drop in streamline flow is (P./1',.) 0•25•
The Use of Core Tubes. Employing the horizontal exchanger of the preceding example, suppose that 50,000 lb/hr of a 28°API gas oil was to be heated from 105 to 130°F using steam at 250°F. Would the exchanger be satisfactory? Solving as before, the value of the viscosity correction tPt is 1.18 but the freeconvection correction 1/1 is negligible, and U c would be Jess than U D· Cores, dummy tubes with one end sealed (by clamping in a vise), may
210
PROCESS HEAT TRANSFER
be placed within the tubes of an exchanger as shown in Fig. 10.2. They constrict the crosssectional area by forming an annulus which replaces the inside diameter of the tube with a smaller equivalent diameter and which increases the mass velocity of the tube fluid. Although the use of a core decreases the effective tube diameter and increases the mass
~.::: . . .
Tube wall
~.:::======:..·=::::::::::~==
FIG. 10.2.
Core detail.
velocity, it does not alter the Reynolds number from its value before introduction of the core. Using a core,
i
a.= (Di Df) where D1 is the outside diameter of the core and D2 is the inside diameter of the tube. Wetted perimeter for heat transfer 1 = 1TD2
n. = e:4 en~ nv = m m 1TD2
D2
w
('1T/4)(D~ = Di  m G=
Ret = D.G fl.
 DV
w ('1T/4)(m  DD
D2fJ.
4w
= 1TD2p,
Without a core the result is the same. DG
Ret
D2
= ;;: = ;;
w (7r/4)m
4w
= 'ITD2f1.
Since the Reynolds number reinains the same, the advantage of the cores is manifested by the smaller value of D, to D in Eq. (6.1). However, the use of cores eliminates the possibility of free convection as correlated by Eq. (10.4), since the cores disrupt the natural~convection currents. Cores are most advantageously employed only when the fluid is viscous and free convection is precluded. Any size of core may be employed, and it is customary to use 18 BWG or lighter heatexchanger tubes for the purpose. If the tube is sealed by clamping in a vise, the core must be of such size that on~half its perimeter is greater than the inside diameter of the tube into which it is to be inserted. If the clamped width is less, the core is secured by welding 1 Unlike the annulus of a double pipe exchanger, the wetted perimeter is the circumference at the inside diameter of the outer tube rather than the outer diameter of the inner tube.
211
STREAMLINE FWW AND FREE CONVECTION
small rods at the clamped ends, which are larger than onehalf the perimeter of the tube. The cores are inserted with the sealed end at the inlet of each pass, so that in multipass exchangers half are inserted at the channel and half at the opposite end of the bundle. Since cores produce considerable inlet and exit turbulence, it is fair to assume that substantial mixing occurs between passes. In such cases the unmixed length of path may be taken as the tube length. The equivalent diameter for pressure drop is smaller than that for heat transfer because the wetted perimeter is the sum of the circumferences of the inside of the tube plus the outside of the core. Ordinarily, if there is a large quantity of liquid flowing but the viscosity is high, the pressure drop will be quite large. If the flow is scanty and the viscosity is small, the pressure drop will usually be negligibly small despite the use of a core. Example 10.3. Gas Oil Heater Using Cores. In an effort to make the exchanger in Example 10.2 suitable for the flow of 50,000 lb fhr of 28°API gas oil, the use of cores will be investigated. The oil will be heated from 105 to 130°F using steam at 250°F. Comparisons will be made with Example 10.2. The correct core cannot always be selected for the first trial, and it is usually necessary to assume several core sizes and carry out the calculation. Viscosities of the oil in the low temperature range are
•F 250 200 150 125 100
cp 2.0 3.1 5.0 6.3 8.2
JJ.,
Solution:
Exchanger: Shell side ID = 15~in. Baffle space = 15 in. Passes = 1
Tube side Number and length = 86, 12'0" OD, BWG, pitch = 1 in., 16 BWG, Passes= 2
1~in.
tri.
Assume that %in., 18 BWG cores are used. (1)
Heat balance: Gas oil, Q = 50,000 X 0.47(130  105) = 587,000 Btu/hr Steam, Q = 6220 X 945.5 = 587,000 Btu/hr
(2) At:
Cold Fluid
Hot Fluid
Diff.
250
Higher Temp
130
120
250
Lower Temp
105
145
Differences
25
25
0
L1t = LMTD = 132.5°F (true counterflow)
(5.14)
212
PROCESS HEAT TRANSFER
(3) T. and t.:
(t, +to) 130 + 105 _ 117 .F t = 2,., 5 2 G
0
Hot fluid: shell side, steam
Cold fluid: tube side, crude oil
(4') a. = ID X C'B/144PT {Eq. (7.1)] (4) = 15.25 X 0.25 X 15/144 X 1.25 = 0.318 ft•
a,'
4
= ... (d:  dD = ~ (0.87~  0.50•)
= 0.40 in.•
(d2 and d, are the annulus diameters, in.) a, = N,a;/144n [Eq. (7.48)] = 86 X 0.40/144 X 2 = 0.119 ft2 (6') G, = Wja. [Eq. (7.2)] (5) G, = w/a 1 = 6220/0.318 = 50,000/0.119 = 19,600 lbj(hr)(ft•) = 420,000 lb/(hr)(fV)
= d~  eli = 0.872
2  0.50 0.87 ~' = 0.013 X 2.42 = 0.0314lb/(ft)(hr) = 0.582 in. [Fig. 15] D, = 0.72/12 = 0.060 ft [Fig. 29) D, = 0.582/12 = 0.0485 ft At t. = 117.s•F, Re. = D,G,jp. (for pressure drop) [Eq. (7.3)) JJ = 6.9 X 2.42 = 16.7lb/(ft)(hr) Re, = D/},jp. = 0.06 X 19,600/0.0314 = 37,400 = 0.0485 X 420,000/16.7 = 1,220 Assume mixing between passes L/ D, "" 12/0.0485 = 247 (7) jH = 3.10 (8) At JJ = 6.9 cp and 2s•API k(cp./k)'l.l = 0.35 Btu/(hr)(ft')("F/ft) [Fig. 16]
(6') At T. = 2so•F,
(6) d.
(9') Condensation of stea.m:
h; • k [Eq. (6.15a)] (9) "it=:JHD, k = 3.10 X 0.35/0.0485 = 22.4
d.
(cp.)Y,
h.= 1500 (10') t.,:
t,. = t.
(10)
+ _ho_ (T. h;. +h.
~:·
 t.) [Eq. (5.31)]
1500 = 117.5 + 19.5 + 1500 (250  117.5)
= 249"F
=
!:
X
~~
[Eq. (6,5)
= Z!.4 X 0.87/1.0 = 19.5 (11) At t,. = 249°F,
= 2.0 X 2.42 = 4.84lb/(ft)(hr) [Fig. 14] t = (JJ./JLw) o.t4 = (16. 7 /4.84) 0·14 = 1.18
J.<w
(12) h;.
!':
= q,, = 19.5 X
[Eq. (6.37)] 1.18
= 23.0 Btuj(hr)(ft2)(°F) (13) Clean overall coefficient Uc: Uc
"'h.~•~·h.
"';::g ~ ~:~
= 22.6 Btu/(hr)(ft 2)(°F)
(6.38)
213
STREAMLINE FLOW AND FREE CONVECTION (1') Design overall coefficient UI>:
A = 270 ftl
UD =}At=
27 gs~~.S = 16.4 Btu/(hr)(ft")(°F)
(16) Dirt factor Rd:
Ra
=
Uc UD UcUv
=
22·6  16·4 22.6 X 16.4
=
0.0172 (hr)(ft')(°F)/Btu
(6.13)
Summary 1500
I
h outside
Uc
22.6
Uv
16.4
119.4
R. Calculated 0.0172 Note that the cores make the heater operable.
Pressure Drop (1') For Re, = 37,400,
! = 0.0016 ftl/in ..
[Fig. 29)
(1) d~ = (d2  d,) = (0.87  0.50)
n:, = o.37/12 = o.o3o9 rt Re~ =
n;p.;"
= 0.0309
(2') No. of crosses, N
= 12 X 12/15
+1
[Eq. (6.4)]
= 0.37 in.
X 420,000/16.7 = 777
f = 0.00066 ft'/in.• = 12L/B
= 10
II
= 13.82 ft3jlb
8
1 = 13.82 X 62.5 = O.OOUa
[Fig. 26]
(2) For streamlineflow pressure drop:
[Eq. (7.43)] q,, = (16.7/4.84) 0 •26 = 1.35 8 = 0.85 (Table 7) f:J.P _ JG:Ln '  5.22 x to••n;,s.,
[Fig. 6]
[Eq. (7.45)]
D. = 15.25/12 = 1.27 ft (S') AP = ~ JG!(D,)(N + 1) '
0.00066 X 420,000 2 X 12 X 2 (3) 5.22 X 10 10 X 0.0309 X 0.85 2 5.22 X 10 10D,sq,, X 1.35 [Eq. (7.52)J [Eq. (7.46) I = 1.5 psi 2 0;0016 X 19,600 X 1.27 1 X 10 f:J.P r = negligible = 2 X 5.22 X 1010 X 0.060 X 0.00116 X 1.0 = 1.4 psi
The choice of a core was satisfactory. A large dirt factor should be used because of the inaccuracies of the calculation. It would also be advisable to investigate the use Qf a. ~ or 74 in. OD core. although either 9f these cores will cause a considerably grea.ter pres2ure drop.
214
PROCESS HEAT TRANSFER
Streamline Flow in Shell. When an exchanger is selected to fulfill process conditions from among existing exchangers, streamline flow in tubes should be avoided if possible. By comparison with turbulent flow, streamline flow requires the use of a larger surface for the delivery of equal heat. Sometimes streamline flow in tubes is unavoidable as when there is a great unbalance between both the sizes of the two streams and the shellside and tubeside flow areas. If the flow area of the shell is greater than that of the tubes and one of the fluid quantities is much larger than the other, it may be imperative in meeting the allowable pressure drop to place the smaller stream in the tubes. Still other instances of streamline flow arise when a liquid is placed in the tubes for corrosion reasons whereas from the standpoint of fluid flow it logically belongs in the shell. Streamline flow in shells, on the other hand, is fortunately a rare problem. It occurs at a Reynolds number well below 10, as plotted in Fig. 28, due to the constantly changing flow area across the long axis of the bundle. The shell is consequently an excellent place in which to locate a scanty stream. The fouling factors of Table 12 have been predicated upon turbulent flow, and for low tube velocities they should be increased perFra. 10.3. Free convection outside haps 50 to 100 per cent to provide additubes. (After Ray.) tiona! protection. Free Convection outside Tubes and Pipes. The mechanism of free convection outside a horizontal cylindrical shape differs greatly from that within it. On the outside of a pipe the convection currents are not restrained as they are within it, and heated fluid is usually free to rise through greater heights of cold fluid thereby increasing the convection. The atmosphere about a pipe has been explored by Ray/ and the isotherms are shown schematically in Fig. 10.3. Cold air from the relatively ambient atmosphere travels in toward the hot tube whence it is heated and rises. Numerous investigators have established the influence of the Grashof and Prandtl numbers on the correlation of free convection. Unfortunately most of the experimental information has been obtained on apparatuses such as single tubes and wires rather than industrial equipment. Accordingly, however, the film coefficient for free convection to 1
Ray, B. B., Proc. Indian Assoc. Cultivation Sci., 6, 9!i (1920).
STREAMLINE FLOW AND FREE CONVECTION
21.5
gases from horizontal cylinders can be represented by
[(DMg~ At) (c'~')] '
0 25
hJ) = a
k, ~} k, where h. is the free convection coefficient and all the properties are evaluated at the fictitious film temperature t1 taken as the mean of the temperature of the heating surface and the bulk temperature of the fluid being heated. Thus
t _tw+ta
,2
(10.6)
For the most part it is difficult to obtain good data for the various types and sizes of equipment used in industry. This is due in part to the interference and complexities of freeconvection heating elements such as banks of tubes and the inability to control an atmosphere of fluid to the extent necessary for good experimental results. Freeconvection correlations from the outside surfaces of different shapes which are of direct engineering value fall largely into two classes: free convection about single tubes or pipes and free convection about vessels and walls. McAdams~ gives an excellent review of the work done heretofore in this field. It is apparent that freeconvection currents are not only influenced by the position of the surface but also by its proximity to other surfaces. A hot horizontal surface sets up currents which differ greatly from those set up by a vertical surface. McAdams has summarized these in a simplified dimensional form 2 for free convection to air.
~ty·25
Horizontal pipes:
he= 0.50 do
Long vertical pipes:
h.
Vertical plates less than 2 ft high:
(Atr· z he = 0.3 At h. = 0.38 Atn2 5 h. = 0.2 At 0·25
Vertical plates more than 2ft high: Horizontal plates: Facing upward: Facing downward:
= 0.4 etr·2· d.
he= 0.28 
0 25 •
(10.7)
(10.8)
2 •
(10.9)
(10.10) (10.11) (10.12)
where At is the temperature difference between hot surface and cold fluid in °F, do is the outside diameter in inches, and z is the height in feet. 1 McAdams, W. H., "Heat Transmission," 2d ed., pp. 237246, McGrawHill Book Company, Inc., New York, 1942. sPerry, J. H., "Chemical Engineers' Handbook," 3rd Ed., p. 474, McGrawHill Book Company, Inc., New York, 1950.
216
PROCESS HEAT TRANSFER
;
Liquid No 0 fhanol so nl'en one 22 SulfuriC Acid 98 '1'o 00 '26 Sulfuric Acid 6!t% 40() ~ \000,000 0 30!t 600,0011 20 Toluene 0 2 Water
. "
a;
J
a:
"'
""0
· OA
~ Q3
::::; Q2
0 0 0 0

I
No 7 9 12 7 I 4 8 6 5 14 II
Gas Air Ammoma Carbon DIOXide Carbon Mcnl))(ide Hydrogen Methane Nitric Oxide Nitroqen Oxygen Sulfur Dioxide Water
20Q: 300,000 100,000 100::1=,60,000 8~ t:30,000
70
~Q:r,~ (\000 30"
6,000
2~
3,000
~
150 170 8
190 210
6
3
p .~~~~~g~~~:isof do Outside diameter of I cvlinderinches
2
kf "The~/,"' conducti~ Btu (hr)(ft2)(°F/
4
10
Mean temperature of fl'lrn°F he Film ;oefficient Btu (hrXftzJ'(•F) At Temperooture difference across film•F
aI
l't ~~~tyft't I atm. ~ecific heQt (d Coeff. ofth'i/c"~' expansion I °F
Cf
f.ut Visco&i~ lb/(hr) ft)
I,000 10:: 600
8
300
4 ~~ 3 60
00
2
30 20
Fla. 10.4. Free convection outside horizontal pipes and tubes. [T. H. Chilton, A. P. Colburn, R. P. Generaux, and H. C. Vernon, TrcmB. ASMB Petroleum Mech. Eng., 66, 5 (1933).]
For single horizontal pipes the dimensionless expression will apply except that a will vary between 0.47 and 0.53 between small and large pipes. Thus
h,D = 0.47 [(D!pjg{J Llt) (CP.t)] 0
k,
p.j
k,
0"25
(10.13)
STREAMLINE FLOW AND FREE CONVECTION
217
Chilton, Colburn, Generaux, and Vernon 1 have developed an alignment chart which gives conservative coefficients for a single pipe but which has been used by the author and others without noticeable error for the calculations of free convection outside banks of tubes. The dimensional equation, plotted in Fig. 10.4 for gases and liquids, is
he= 116 [
e'~'{j) (~)
r
25
(10.14)
where tJ.i is in centipoises. Of the four axes in the alignment chart one is the reference line of values of k'pjc1{}/w which allows its use for fluids other than those whose key numbers are given in Fig. 10.4. The use of the chart for tube bundles .requires that the pipes or tubes not be located too closely to the bottom of the vessel or too closely among themselves so that they interfere with the naturalconvection currents. The tubes should not be located closer than several diameters from the bottom of the vessel, nor should the clearance between tubes be less than a diameter. Notwithstanding the data available, freeconvection design is not very accurate, and reasonable factors of safety such as large fouling factors are recommended. The use of the several types of freeconvection correlations for fluids outside tubes is demonstrated by .the typical problem which follows. Example 10.4. Calculation of a Heating Bundle for an Aniline Storage Tank. A horizontal outdoor storage tank 5'0" ID by 12'011 long and presumed cylindrical is to
r1:=============..:;r1r1 L,,t=,;;.;::...r_=_;;_=;_::;_~=_;=;..;:7.=:;..
~
t
Sfeam Drips Fra. 10.5. Heating a 11uid in a tank by free convection.
be used for aniline storage. at a constant temperature of l00°F even though the prevailing atmospheric temperature should fall to 0°F. The tank is not insulated but is shielded from the wind. Heat will be supplied by means of exhaust steam through a bundle connected through the bottom of the tank as shown in Fig. 10.5 and consisting of 6'0"lengths of lin. IPS pipe. The use of a temperature controller (Chap. 21) to throttle the steam would make the entire operation automatic. How many lengths of pipe should be provided? Solution. The first part of the problem is to determine how much heat will be lost from the tank to the atmosphere. This gives the heat load for which the coil must be designed. I
Chilton, T. H., A. P. Colburn, R. P. Generaux, and H. C. Vernon, Trans. A.S.M.E.,
Pelroleum Meek. Eng., 56, 5 (1933).
218
PRQCESS HEAT TRANSFER
From Eq. (10.7) to (10.12) the convection coefficient to air will be somewhere between h. = 0.2 A.t 0• 26 and h. = 0.3 A.t 0• 26 for the entire tank surface. Since the equation giving the higher value of h. is safer. Convection loss: he = 0.3 ~t 0 · 26
Neglecting the temperature drop through the tank metal A.t = 100  0 = 100°F
= 0.95 Btuj(hr)(ft')(°F) Rad' t' t h _ 0.173 e[(TI.obs/100)4  (T2.oba/100)'] la wn ra e, ' T,,.ba  To.aba he
= 0.3
X 100°· 26
(4.32)
Assuming an emissivity of about 0.8, h,
= 0.173
X
0.~:!·6' 4.6')
= 0.75 Btu/(hr)(fV)(oF)
Combined loss: h. + h, = 0.95 + 0.75 = 1.70 Btu/(hr)(ft')(°F) 2,. X 52 Total tank area =  +,. X 5 X 12 = 227.8 ft' 4 Total ~eat loss, Q = (h. + h,.)A At = 1. 70 X 227.8 X (100  O) = 38,800 Btujhr
This heat must be supplied by the pipe bundle. Assuming exhaust steam to be at 212°F,
From Figure 10.4, h.
U _
~ d.
= 212
 100
tl
== 212
+2 100
1.32
= 85 = 156oF (nearly)
= 48 Btu/(hr)(ft•)(°F). h.oho _ 1500 X 48 _ 46 S B /(h )(f ')(oF) + h.  1500 + 48  · tu r t
c  h;.
(6.38)
Assume a dirt factor of 0.02 (hr)(ft 2)("F)/Btu UD =
Total surface,
g~ ~
:;;; !::! ~: = 24.1 Btu/(hr)(ft'WF) =
A = U ~At
= 38,800/24.1 Area/pipe Number of pipes
X (212  100) = 14.4 ft 2 = 0.344 X 6 = 2.06 ft•
(Table 11)
= 14.4/2.06 = 7
These may be arranged in series or parallel. PROBLEMS 10.1. 9000 lb/hr of 34°API crude oil at 100°F (see Example 10.1 for viscosities)
enter the tubes of a 17~ in. ID 12 exchanger containing 118 tubes 1 in. OD, 16 BWG, 12'0" long arranged for two passes in 1~in. triangular pitch. It is desired to attain an outlet temperature of 150°F using steam at 250°F. What is the actual outlet temperature?
STREAMLINJ.ff FLOW AND FREE CONVEC7'ION
2H1
10.2. 37,000 lb/hr of an organic compound whose properties are closely approximated by the 28°API gM oil in the Appendix is to be heated by steam in the tubes of a clean exchanger because the shell baffles are halfcircle support plates. The ail is to be heated from 100 to 200°F by steam at 325°F. Available for the service is a 27 in. ID ~2 exchanger containing 334 1 in. ODt 16 BWG, 16'0" long tubes arranged for two passes on Uiin. triangular pitch. (a) Determine the dirt factor if the oil is not mixed between passes? (b) What is the actual outlet temperature of the oil if the fluid is mixed between passes? 10.3. 22,000 lb /hr of a flushing oil closely resembling 28°API gas oil is to be heated from 125 to 175°F to improve filtration. Available as a heating medium is another line corresponding to a 35°API distillate at 280°F. The temperature range of the distillate will be only 5°F so that the exchanger is essentially in counterflow. Available for the service is a 21% in. ID 12 exchanger containing 240 % in., 16 BWG tubes 8'0" long. Tubes are arranged for six passes on lin. square pitch. Ba.ffies are spaced 8 in. apart. Because of contaminants the flushing oil will flow in the tubes. Investigate all the possibilities, including the use of cores, to make the exchanger operable. Pressure drops are 10 psi, and a combined dirt factor is 0.015 (not to be confused with 0.0015). 10.4. A vertical cylindrical storage tank 6'0" ID and 8'0" tall is filled to 80 per cent of its height with ethylene glycol at 150°F. The surrounding atmosphere may fall to o•F in the winter, but it is desired to maintain the temperature within the tank by using a rectangular bank of I in. IPS pipes heated with exhaust steam at 225°F How much surface is required? How should the pipes be arranged? NOMENCLATURE FOR CHAPTER 10 A
a" a B
a,, a, D D, D,, D', d
d.
I
G
Gr Gr. g
h, hi, h.
n
Heat transfer surface, ft• External surface per linear foot, ft Flow area, ftt Baffie spacing, in. Constants, dimensionless Inside diameter of tubes or pipes, ft Outside diameter of tubes or pipes, ft Equivalent diameter for heat transfer and pressure drop, ft Inside diameter of tubes, in. Outside diameter of tubes, in. Friction factor, ft; /in. 2 Mass velocity, lb/(hr)(ft2 ) Grll8hof number, dimensionless Grll8hof number at the average or bulk temperature, dimensionless Acceleration of gravity, ft/hr2 Heattransfer coefficient in general, for inside fluid, and for outside fluid, respectively, Btu/(hr)(ft2)(°F) Value of h; when referred to the tube outside diameter Heattransfer coefficient for free convection, Btu/(hr)(ft 2)(°F) Heattransfer coefficient for radiation, Btuj()>t.)(ft2)(°F) Factor for hel).t transfer, dimensionless Thermal conductivity, Btu/ (hr) (ft 2) (°F /ft) Tube length or length of unmixed path, fl:. Number of tube passes
220
PROCESS HEAT TRANSFER
Pressure drop in general, psi tJ> tJ>T, tJ>,, tJ>, ·Total, tube, and return pressure drop, respectively, psi
PT Rd Re, Re' 8
T., T.
t,, t.
t., t. t, t; tp, t., At At.
Uc, UD
"
W,w z a
Tube pitch, in. Combined dirt factor, (hr) (ft2WF) /Btu Reynolds number for heat transfer and pressure drop, dimensionless Specific gravity, dimensionless Average and caloric temperature of hot fluid, °F Inlet and outlet temperatures of cold fluid, °F Average and caloric temperatures of cold fluid, °F Film temperature 3'2 (t., + t.), °F Temperature at end of first pass, °F Inside and outside tube wall temperatures, °F Temperature difference for ·heat transfer, °F Temperature difference between tube wall and average fluid temperature, oF Clean and design overall coefficients, Btu/(hr)(ft 2)(°F) Specific volume, ft3 jlb Weight flow of hot and cold fluid, lb/hr Height, ft Constant Coefficient of thermal expansion, l/°F Viscosity, centipoises X 2.42 = lb/(ft)(hr) Viscosity at tube wall temperature, centipoises X 2.42 = lb/(ft)(hr) Viscosity at film temperature, centipoises Density, lb/fta (p,/JLt»)o.u. For pressure drop in streamline flow, (p,jp,.,) 0• 26 Free convection correction, Eq. (10.5), dimensionless
Subscripts (except as noted above) Evaluated at the film temperature Shell side 8 t Tube side
f
CHAPTER 11 CALCULATIONS FOR PROCESS CONDITIONS Optimum Process Conditions. The experience obtained from the calculation of existing tubular exchangers will presently be applied to cases in which only the process conditions are given. Before undertaking these calculations an investigation will be made to determine whether or not several related pieces of equipment can avail themselves of the overall process temperatures in an optimum manner. This is an economic affair similar to that for the optimum outletwater temperature
_tM:h
gas /n
Fra. 11.1.
Typical process employing heat recovery.
and the optimum use of exhaust steam discussed in Chap. 7. Often an exchanger operates in series with a cooler and a heater as shown in Fig. 11.1. The cooler regulates the final temperature of the hot fluid, and the heater adjusts the final temperature of the cold f!uid 1 to the requirements of the next step in the process. The quantity of heat recovered in the exchanger alone greatly influences its size and cost, since the true temperature difference in the exchanger approaches zero as the heat recovery approaches 100 per cent. On the other hand, whatever 1 This is actually the way to compensate and take advantage of the overperformance of an exchanger when it is clean if it has been designed for a dirty coefficient U0 . Unless equipped with a bypass to reduce the flow through the exchanp;er, the temperature ranges of both fluids will exceed the process conditions. This is offset by throttling the steam and water in the heater and cooler and accordingly reducing the utility costs.
221
222
PROCESS HEAT TRANSFER
heat is not recovered in the exchanger must be removed or added through the use of additional steam in the heater and additional water in the cooler, which raises the initial costs of the two as well as the operating cost of the process. This arrangement suggests the presence of an optimum distribution of temperatures so that the fixed and the operating charges will combine to give a minimum. Among a number of examples of heat recovery available in the chemical and power industries, that of Fig. 11.1 is typical. It shows the flow sheet of a simple vaporrecovery system such as is employed in the stripping of gasoline from natural gas, an absorption and distillation process, although the absorber and distilling column are not a part of the present analysis. Only the pertinent heatflow lines are dash enclosed. The natural gas coming from the earth is laden with gasoline vapors which command a higher price when separated from the natural gas and condensed. The rich gas enters the absorber where it contacts an absorbent, usually a nonviscous oil, in which the gasoline vapors selectively dissolve. The outlet gas of reduced gasolinevapor concentration is lean gas. The absorbent leaves the bottom of the absorber with the dissolved vapors as rich oil. It must next be fed to a distilling column where the gasoline and the oil are separated by steam distillation. The oil leaving the bottom of the distilling column is substantially free of the solute and therefore lean oil. Absorption is favored by low temperatures, while distillation requires higher temperatures so that the gasoline can be vaporized from the oil. The exchanger, heater, and cooler are represented in Fig. 11.1 by E, H, and CR, respectively. The temperatures to and from the absorber and distilling column will be considered fixed by equilibrium conditions whose definitions are beyond the scope of this study. The temperatures of the steam and water will also be fixed. The nucleus of the problem lies in determining the exchanger outlet either of which fixes the other, so that the total temperatures T., or annual cost of the three heattransfer items will be a minimum. If CE, CH, and Cc11 are the annual fixed charges per square foot for the exchanger, heater, and cooler, and if AE, AH, and AcB are their surfaces, the five elements of cost are 1. Fixed charges of exchanger, CEAz 2. Fixed charges of heater, CHAH 3. Cost of steam, Cs, dollars/Btu 4. Fixed charges of cooler, Cc~oR 5. Cost of water, Cw, dollars/Btu The surface of each unit, A = Q/U !J.t, is dependent upon its true tem· perature difference. In each piece of equipment, however, the true temperature difference depends upon either T, or ty. To obtain an
t,,
223
CALCULATIONS FOR PROCESS CONDITIONS
expression for the minimum cost it will be necessary to differentiate the total annual cost with respect to either T., or ly and solve for T., or t, after setting the derivative equal to zero. For simplicity assume that only true counterflow exchangers are employed so that the true temperature difference and LMTD of all units are identical and that all the unit fixed charges C E, C "' CoR are the same and independent of the total number of square feet of each unit. The equation for the total annual cost CT is obtained after eliminating t11 through the use of the dimensionless group R.
Cr
=
CEWC(Tt T.,) ln (1  R)Tt h Us[(l  R)Tt (R  l)T.,] T., it
+
+ C.r;;c ln T.
h;;.
CaRWC(T., T2)
~~: + RT., + C.9wc(t2
+ UcR[(T., t;) (T2 tDJ
l T., t~ n T 2  t~
+
C
wB
WC(
+ RT.,
 t1  RTt T., Tz
)
+ RTz) (11.1)
where Us, U "' and U aR are the overall coefficients, 8 is the total number of annual operating hours, and the other temperatures are indicated in Fig. 11.1. When differentiated with respect to T:tl and set equal to zero, the equation for the optimum value of T., is
CH
0=
1
c.e + Cw8 + UH [R(T:tl Tr) + (T.
tt)] +CaR (ti  l~) l (T., t;) UcR [(T.,  Tz) (ti  ~)] n (T2  tf) +CaR (T., T2) 1 UaR [(T., T2) +(tit~)] (T., t~) CE (Tl  h) 1  UE [(Tt it) + R(T., Tt)l (T., TJ
+
(11.2)
T., can be obtained by a trialanderror calculation. Actually where the disposition of the temperatures justifies obtaining the optimum, the operation will be too large to permit the use of true counterflow in all of the equipment. When the flow pattern in the exchanger alone deviates from true counterflow and At is given by Eq. (7.37), the problem is somewhat more difficult to solve and less direct. However, if FT is placed in the denominator of the last term in Eq. (11.2), a simplified trialanderror solution can be obtained for a system using a 12, 24, etc., exchanger. Equation (11.2) is not particularly useful, however, unless extensive data are available on the installed costs and fixed charges of exchangers, since the cost per square foot of surface also varies with the size of the exchanger. Sieder1 has shown in F'ig. 11.2 how the 1 Sieder,
E. N., Chem. Met. Eng., 46,322325 (1939)
PROCESS HEAT TRANSFER
224
relative cost (per square foot) of surface decreases from small to large exchangers. The fixed charges are usually arrived at as a percentage (30 per cent) of the initial cost per square foot, and this varies so much 2.2 2.0
~1.8 ~1.6
........
1il.4
1\
\ \.
u
~
1.2
r.... ....... i"" t
~1.0 0::
0.8 0.60
600
1000
1500
2000
2500
3000
Surf01ce,sq ft FIG. 11.2.
Cost of tubular surface vs. size of exchanger.
(Sieder, Chemical. Engineerinu.)
with the size of the final exchanger that several successive trials are required to establish the proper range of individual costs. Other obstacles are also encountered in the solution of a general equation for the optimum. If a typical problem is solved for a system with a 12 exchanger and the calculated value ofT= comes out to be less than the practical limit of FT = 0.75  0.80 in the 12 exchanger, the entire calculation, though valid, must be repeated using a different exchanger. The task of obtaining the optimum process tem~ OphmL/m 1'0~ / peratures can be achieved more readily ......... .,_ by graphical means. As the tempera0 u '\. ture T=is varied, there are two opposed costs as follows: As T= is increased, the ~ cost of the utilities increases but the / ~c.?,~r9'e.s initial cost and fixed charges on the ex/ changer decrease. By assuming several values of T., the required sizes of all the FIG. 11.3. Optimum recovery tem equipment can be computed for each perature. assumption. From the overall heat balances the operating costs of the utilities can also be obtained. The total annual costs are then the sum of the two costs as plotted in Fig. 11.3 with the optimum corresponding to the point indicated. The Optimum Exchanger. The factors which are favorable for the attainment of higher film coefficients for fluids in exchangers also increase
/"/
U)
\
+ _. ./fY
17
VAJ
CALCULATIONS FOR PROCESS CONDITIONS
225
their pressure drops. For turbulent flow in tubes as given by Eq. (6.2) the film coefficient varies with G~· 8, whereas the pressure drop in Eq. (7.45) varies with Gl. This means that if the tubeside coefficient is the controlling coefficient and if the design of the exchanger can be altered so as to increase the tubeside mass velocity, the overall coefficients U a, UD also increase and the size of the exchanger can be reduced accordingly. Offsetting this advantage is an increase in the pressure drop and pumping power costs for the greater mass velocity. As in the calculation of the optimum process conditions, there is an optimum exchanger capable of fulfilling process conditions with a minimum annual cost. The achievement of this design, however, requires an exchanger capable of providing the optimum fluidflow velocities on the shell as well as the tube sides. This would frequently entail the use of an odd number of tube passes or an odd tube length, which is inconsistent with industrial practices. McAdams1 has given an excellent resume of the equations and approximations required for establishing the optimum exchanger to which the reader is referred. Rating an Exchanger. In Chaps. 7 through 10 the calculations were carried out on existing exchangers. In the application of the Fourier equation to an existing exchanger, Q was determined from the heat balance; A from the number, outside diameter, and length of the tubes; and D.t from Yr X LMTD, permitting the solution for U n. From the fluidflow conditions, h., h;., U a, and the pressure drops were calculated. The criterion of performance Ra was then obtained from UD and Uc. When there is no available exchanger and only the process conditions are known, calculation may proceed in an orderly way by assuming the existence of an exchanger and testing it as in previous examples for a suitable dirt factor and pressure drops. To prevent the loss of considerable time, rational methods of assuming an exchanger should be developed. By returning to the component parts of the Fourier equation Q = U DA At, the heat load Q is seen to be fixed by the process conditions, while At is obtained by assuming a fluidflow pattern. Thus if it is desired to set up a 12 exchanger, and if the process temperatures give FT > 0.75  0.80, the remaining unknowns are Un and A. The design or dirty coefficient UD is in turn related by a reasonable dirt factor to U c, which reflects the heattransfer characteristics of the two fluids. Previous examples in the text suggest that different numerical film coefficients may be expected within definite ranges in welldesigned exchangers for different classes of fluids. It is also apparent that, except where both coefficients are approximately 1 McAdams,
W. H., "Heat Transmission," 2d ed., pp. 363367, McGrawHill Book
Company, Inc., New York, 1942.
226
PROCESS HEAT TRANSFER
equal, the lower film coefficient determines the range of U c and U D· In the light of this experience if a trial value of U D is assumed and then substituted into the Fourier equation to supplement calculated values of Q and At, it permits the trial calculation of A. To facilitate the use of reasonable trial values of U D approximate overall coefficients have been presented in Appendix Table 8 for common pairs of fluids. When the value of A is combined with tube length and pitch preferences, the tube counts in Appendix Table 9 become a catalogue of all the possible exchanger shells, from which only one will usually best fulfill the process conditions. Having decided which fluid will tentatively Bow in the tubes, the trial number of tube passes can be approximated by a consideration of the quantity of fluid flowing in the tubes and the number of tubes corresponding to the trial value of A. The trial mass velocity should fall somewhere between 750,000 and 1,000,000 lb/(hr)(ft2) for fluids with allowable tubeside pressure drops of 10 psi. If the trial number of tube passes has been assumed incorrectly, a change in the total number of passes changes the total surface which is contained in a given shell, since the number of tubes for a given shell diameter varies with the number of tubepass partitions. The assumed number of tube passes for the trial surface is satisfactory if it gives a value of h, above U D and a pressure drop not exceeding the allowable pressure drop for the fluid. One may next proceed to the shell side by assuming a trial baffie spacing which can be varied, if in error, over a wide range without altering h;, A, or At as computed previously for the tube side. It is always advantageous therefore to compute the tube side first to validate the use of a particular shell. In calculating an exchanger the best exchanger is the smallest one with a standard layout which just fulfills the dirtfactor and pressuredrop requirements. There are only a few limitations to be considered. It is still assumed that there is no advantage in using less than the allowable pressure drop and that, in accordance with Fig. 28, 25 per cent cut segmental baffles will be employed within the minimum and maximum spacing. The extremes of the spacing range are Maximum spacing, B = ID of shell, in. B = ID of shell' or 2 m., . whichever lS . Iarger . . . spacmg, M1mmum 5
(11.3) (11 .4)
These limitations stem from the fact that at wider spacings the flow tends to be axial rather t.han across the bundle and at closer spacings there is excessive leakage between the baffies and the shell. Owing to the convention of placing the inlet and outlet nozzles on opposite sides of the shell, the end baffles may not exactly conform to the chosen spacing
CALCULATIONS FOR PROCESS CONDITIONS
227
for an even number of baffles and an odd number of crosses. When using a close convenient spacing, if an odd number of baffles is indicated, the pressuredrop and heattransfer coefficients can be calculated for the chosen spacing, although one baffle will be omitted by respacing the extreme end baffles. The different combinations of the number of tube passes and the bailie pitch which can be employed in a given shell permit the variation of the mass velocities and film coefficients over broad limits. The number of tube passes can be varied from two to eight and in larger shells to sixteen. As indicated above, the shellside mass velocity can be altered fivefold between the minimum and maximum baffle spacings. It is desirable to bear in mind the latitude this permits in the event that the first trial baffle spacing and tube passes have been wide of fulfilling the process conditions. In 12 exchangers the least performance is obtained with two tube passes and the maximum baffle space. For the tube side in turbulent flow h; ex: G2·s APt ex: GtnL where nL is the total length of path. Going to eight tube passes in the same shell inside diameter the changes incurred are
but
h;(S """"')
= (~)0.8
hi(2  · · )
2
=
~ 1
APt
••...> 82 X 8 X 1 tlP,(2 ....•••) = 22 X 2 X 1
64
=T
or although the heattransfer coefficient can be increased threefold, the pressure drop must be increased 64 times to accomplish it. For streamline flow the expenditure of the larger amount of pumping energy will increase the tubeside coefficient only by h;(8 .......)
= (~)~ = 1.58 2
h;(2pasaea)
1
provided the fluid is in streamline flow in both cases. be represented approximately by hu AP.
The shell side may
ex: G~· 5
+ 1) where N is the number of bailies and N + 1 the number of bundle crosses. ex:
G;(N
The changes for the shell side between minimum and maximum bailie spacings are 05 ho,mia 2.23 ' ho,m•~ = l = 1
(5)
228
PROCESS HEAT TRANSFER
but t:.Pa,min t:.Ps.max
=
5z X 5 P X 1
=
125 1
0ffsetting this, however, is the fact that the shell side yields a higher order of film coefficients for the smaller of the two streams if there is a great difference in their weight flow rates. Through the use of the overall coefficients suggested in Table 8 and a judicious analysis of the summary of the first trial assumptions, it should be possible to obtain the most suitable exchanger on the second trial. Tube Standards. There are numerous advantages in standardizing the outside diameter, gage, and length of the tubes used in a plant. 1.4
)'
;:; 1.3
;t: ,1.2 ",;;
81.1 4>
~1.0
..
~0.9
,_
0.8 0.6
,..........
o.s
v
/
1.0
v
v 1.2
/
Tube OD, in.
1.4
1.6
FIG. 11.4. Cost of tubular surface vs. tube outside diameter. Engineering.)
1.8 (Sieder, Chemical
Standardization reduces the number of sizes and lengths which must be carried in storage for the replacement of tubes which develop leaks. It also reduces the number of installing and cleaning tools required for maintenance. Pitch standards have already been discussed in Chap. 7, but the selection of the tube diameter has an economic aspect. Obviously the smaller the diameter of the tube the smaller the shell required for a given surface, the greater the value of h;, and the smaller the first cost. The nature of the variation in cost, taken from Sieder, is shown in Fig. 11.4. The difference is small between the use of % and 1 in. OD tubes. The cost per square foot rises greatly, however, as the diameters increase above 1 in. OD. Similarly the longer the tubes the smaller the shell diameter for a given surface, and from Fig. 11.5 the cost variation between the use of 12, 16, and 20ft tubes is not very great although 8ft tubes sharply increase the first cost. The lower cost of surface obtained from the smaller outside diameter and greater length tubes is offset by the fact that maintenance and particularly cleaning are costlier with long tubes of small outside diameter. If the tubes are too small, under % in. OD, there are too many to be
229
CALCULATIONS FOR PROCESS CONDITIONS
cleaned, and there is less facility in handling and cleaning the smaller tubes. If the tubes are too long, it is difficult to remove the bundle and plot space must be allocated not only for the exchanger itself but also for the withdrawal of the bundle. Long tubes are also very difficult to replace, especially where the baffles are closely spaced. It is difficult to obtain comparative maintenance data per square foot as a function of tube diameter or length, since few industrial users appear to employ an assortment of tubes or have kept cost data. It may be significant 1.4 c:;1.3
) [Eq. (7.2)] [Eq. (7.3)] (6') Re, = D,G,/JL Obtain D. from Fig. 28 or compute from Eq (7.4). Obtain JL at T. (7') iH from Fig. 28 (8') At T. obtain c Btu/(lb)(°F) and k Btu/(hr)(ft~)("F /ft) (CJL/k)l'S (9') ho == jH
~.
e:)lS q,.
(10!) Tubewall temp,
t,. = t,
(6) Re, = DG,jJL Obtain D from Table 10. Obtain JL at t,.
(7)
iH from Fig.
(8) At
24
t, obtain c Btu/(lb)(°F) and k
Btu/ (hr) (ft 2) (°F /ft) (CJL/k)l'S , k (CJL)% [Eq. (6.15)] (9) h i =JHjj k t
t,.
+ h;.ho+ h. (T. 
From Table 10 a, == N,a;/144n, ft 2 [Eq. (7.48)) ID will be obtained from Table 10. (6) Mass vel, G, = wja, lb/(hr)(ft 2)
{10) h;. = !!:!_ ID
q,,
t,)
q,, OD
(11) Obtain JL" and q,, == (JL/P.w) o.u
[Fig. 24]
[Fig. 24] h = h.q,. 0
q,.
[Eq. (6.5)]
Eq. [5.31]
(11') Obtain JL .. and q,, = (JL/JL,.)o.u
(12') Corrected coefficient,
[Eq. (6.15a)]
(12) Corrected coefficient. [Eq. (6.36)] h·'0  ;;;;h;, "''
[Eq. (6.37)]
Check pressure drop. If unsatisfactory, Check pressure drop. If unsatisfactory, assume a new pass arrangement. assume a new baffle spacing.
CALCULATIONS FOR PROCESS CONDITIONS
231
Pressure Drop (1') For Re. in (6') obtain J (Fig. 29) (1) For Reo in (6) obtain f (2') No. of crosses, N + 1 = 12L!B ( ) tll' fG'fLn 2 1 [Eq. (7.43)] = 5.22 X 101oDsq,1 ilP = JG';D.(N 1) [Eq. (7.44)) • 5.22 X 1010D.sq,. 4n V1 1 3) tll'r e ~ 8 2g'
lFig. 26] [Eq. (7.45)]
+
(4) APT= AP, +APr
[Eq. (7.46)] [Eq. (7.47)]
If both sides are satisfactory for film coefficients and pressure drop, the trial may be concluded. (13) Clean overall coefficient Uc: Uc = h,.h. (6.38) h;.
+ h"
(14) Dirt factor Rd: UD has been obtained in (c) above.
Rd = Uc UD UcUD
(6.13)
The calculation of a number of exchangers for typical sensibleheattransfer conditions are given in this chapter. Each presents a different aspect of design. Together they should provide the perspective necessary for meeting a variety of applications encountered in modern industry. Since the method of approach involves trialanderror calculations, the analyses and comments included in each solution should reduce the time required for subsequent calculations. Example 11.1. Calculation of a Straw OilNaphtha Exchanger. 29,800 lb/hr of a 35•API light oil at 340°F is used to preheat 103,000 lb /hr of 48• API naphtha from 200 to 230°F. The viscosity of the oil is 5.0 centipoises at 100•F and 2.3 centipoises at 210•F. The viscosity of the naphtha is 1.3 centipoises at 100•F and 0.54 centipoises at 2to•F. Pressure drops of 10 psi are allowable. Because the oil may tend to deposit residues, allow a combined dirt factor of 0.005 and use square pitch. Plant practice employs %, in. OD, 16 BWG tubes 16'0" long wherever possible.
Solution: (l) Heat balance: Straw oil, Q = 29,800 X 0.58(340 240) = 1,780,000 Btujhr Naphtha, Q = 103,000 X 0.56(230 200) = 1,730,000 Btujhr (2) ko ~
q,,
=is£. (Wt ·
= 46
Cold fluid: tube side, naphtha ('1) jg = 102 [Fig. 24) (8) For p. = 0.54 cp and 48'API
. [Fig. 16) k(cp.jk)*' == 0.167 Btu/(hr)(ft•WF /ft)
tEq. C6.15b)l (9) h; =
!!:!
X 0.224 = 130 0.0792
"''
js~ (1;t q,,
[Eq. (6.15a)]
= 102 X 0.167/0.0517 = 329
= !!:! X ID = 329 X 0 ·62 = 272 (10'), (11'), (12') Omit the viscosity cor (10) h;, q,, q,, OD 0.75 rection for the trial or q,, = 1.0. [Eq. (6.5)] (11), (12) Omit the viscosity correction h, = ~ = 130 Btu/(hr)(ft2)(°F) for the trial or ¢ 1 = 1.0.
h 1, = h;, = 272 Btu/(hr)(ft•WF)
q,,
Proceed with the pressure .drop calculation.
Pressure Drop (1) For Re, = 31,300, [Fig. 29] f = 0.0002 ft"/in.• [Fig. 6J s = 0. 72
(1') For Re, = 7000,
f
= 0.00225 ft•/in. • s = 0.76
+1
[Fig. 26] [Fig. 6)
JG',Ln
= 12L/B (2) AI', = . X 1010Ds.p, [Eq. (7.45)] 5 22 [Eq. (7.43)] 0.00020 X 792,000 2 X 16 X 2 = 12 X 16/3.5 = 55 10 X 0.0517 X 0.72 = 5.22 X 10 [Fig. 6] B = 0.76 X 1.0 D, = 15.25/12 = 1.27 ft = 2.1 psi (3) and (4) may be omitted for the trial. (S') AP = fG;D,(N + 1) [Eq. (7.44)] Now proceed to the shell side. 10 • 5.22 X 10 D.s.p. 0.00225 X 321,0002 X 1.27 X 55 5.22 X 1010 X 0.0792 X 0.76 X 1.0 = 5.2psi (2') No. of crosses, N
(13) Clean overall coefficient Uo: 272 X 130 _ 8 8 2 B /(h )(ft•)(•F) U0 _ h;,h;.ho + h, _ 272 + 130  · tu r
(6.38)
(14) Dirt factor Rd: U»from (c) is 72.3. Rd = Uc U» = 88·2  ~ 2 · 3 = 0.0025 (hr)(ft2)(°F)/Btu UcU» 88.2 X 72.3
(6.13)
234
PROCESS HEAT TRANSFER
Summary
h outside
130
Uc
88.2
UD
72.3
272
Ra Calculated 0. 0025 Ra Required
0.0050
5.2
Calculated .t.P
2.1
10.0
Allowable .t.P
10.0
Discussion. The first trial is disqualified because of failure to meet the required dirt factor. What conclusions may be drawn so that the next trial will produce the .satisfactory exchanger? Could any advantage be gained by reversing the streams? Obviously the film coefficient for the distillate, which is controlling, would drop considerably if the streams were reversed. Could four passes be used for the tubes? Doubling the number of tube passes would approximately double the mass velocity and give eight times the tub<Jside pressure drop thereby exceeding the allowable M'. All the assumptions above have been reasonable. The exchanger is simply a little too small, or in other words, the value assumed for UD must be reduced. It will be necessary to proceed anew.
Trial 2: Assume U D = 60, two tube passes and the minimum shell baffle space. Proceeding as above and carrying the viscosity correction and pressure drops to completion, the new summary is given using a 1771. in. ID shell with 166 tubes on two passes and a 3.5in. bafHe space. Summary 115.5
h outside
1
Uc
74.8
UD
54.2
213
Ra Calculated 0. 005 Ra Required
0.005
4.7
Calculated .t.P
2.1
10.0
Allowable .t.P
10.0
The final exchanger will be Shell side
ID = 17~in. Baffle space = 3.5 in. Passes= 1
Tube side Number and length = 166, 16'0" OD, BWG, pitch = % in., 16 BWG, lin. square Passes= 2
235
CALCULATIONS FOR PROCESS CONDITIONS
Example 11.2. Calculation of a LeanoilRichoil Exchanger. 84,348 lb/hr of a 35°API lean absorption oil in a process identical with Fig. 11.1 leaves a stripping column to transfer its heat to 86,357 lb/hr of rich oil leaving the absorber at 100°F with a gravity of closely 86°API at 60°F. The range for the lean oil will be from 350 to 160°F, and the outlet temperature of the rich oil will be 295°F. The viscosity of the oil is 2.6 centipoises at 100°F and 1.15 centipoises at 210°F. Pressure drop.s of 10 psi are available, and in accordance with Table 12, a combined dirt factor of 0.004 should be allowed. Plant practice again employs.% in. OD, 16 BWG tubes 16'0" long and laid out on square pitch.
Solution: (1)
Heat balance: Lean oil, Q = 84,438 X 0.56{350  160) = 8,950,000 Btu/hr Rich oil, Q = 86,357 X 0.53(295  100) = 8,950,000 Btu/hr
(2) At:
Hot Fluid
Cold Fluid
Diff.
350
Higher Temp
295
55
160
Lower Temp
100
60
190
Differences
195
5 (5.14)
LMTD = 57.5°F R
= ~~~ = 0.9i5
s
195 = 350  100
= 0.78
12 exchanger, FT = inoperable 24 exchanger, FT =inoperable 36 exchanger, FT = 0.725 48 exchanger, FT = 0.875
(7.18) (Fig. (Fig. (Fig. (Fig.
18) 19) 20) 21)
A 48 exchanger arrangement will be required. This may be met by four 12 exchangers in series or two 24 exchangers in series. The latter will be used. At =
0.875 X 57.5 = 50.3°F
(7.42)
(3) T, and t,: At, = At,,
1.09
K. = 0.32 F, = 0.48 T, = 160 + 0.48 X 190 = 251 °F k = 100 + 0.48 X 195 = 193.5°F
(Fig. 17) (5.28) (5.29)
7'rial: (a) Assume U D = 50: Although both oils and quantities are almost identical, the
temperature range of the cold rich oil and correspondingly greater viscosity will make the rich oil controlling. For this reason, allowable pressure drops being equal, the cold fluid should be placed in the shell. The coefficients will be lower than those of Table 8, since the pressure drops are more difficult to meet in a 48 exchanger and mass velocities must accordingly be kept down. In Example 11.1 with a similar controlling oil the value of Uc was approximately 75 with a minimum baffie spacing which did not completely take advantage of the allowable pressure drop. The
236
PROCESS HEAT TRANSFER
assumption of U D = 50 is a compromise between medium and heavy organics and will probably be high but will help to establish the correct unit on the following trial. A
= _Q_ =
uD l.lt
8,950,000 = 3560 ft• 50 X 50.3
Use two 24 exchangers in series with removable longitudinal baffles. Number of tubes per shell, Nt = 3560/2 X 16'0" X 0.1963 = 567 (Table 10) (b) Assume six tube passes: From previous problems a mass velocity of 700,000 gave satisfactory tubeside pressure drops. Since the number of tube passes in two units will be greater, a maximum of about 450,000 should be employed. Six passes (flow area/tube, = 0.302 in. 2) G = wja 1 = w144njN,a; = 84.438 X 144 X 6/567 X 0.302 = 426,000 lb/(hr)(ft') From the tube counts (Table 9): 567 tubes, six passes, % in. OD on lin. square pitch Nearest count: 580 tubes in a 31 in. ID shell (c) Corrected coefficient U D: a" = 0.1963 ft 2 /lin ft (Table 10) A = 2 X 580 X 16'0" X 0.1963 = 3640 ft2 8,950,000 49 0 U D = A Ql.lt = 3640 X 50.3 = .
a;
Hot fluid: tube sule, lean oil (4) Flow area, = 0.302 in. •
a;
a,:
[Table 10) a 1 = N,a;/144n [Eq. (7.48)) = 580 X 0.302/144 X 6 = 0.203 ft 2
G, = W fa, = 84,438/0.203 = 416,000 lb/(hr)(ft 2)
(6) Mass vel,
(6) At T. = 251 °F, J.£
= 0.88
= 2.13 lb/(ft)(hr) D
= 0.62/12 = 0.0517 ft
X 2.42
[Fig. 14) [Table 10)
Re, = DG,fp.
= 0.0517 X 416,000/2.13 = 10,100
Cold fluid: shell side, rich oil (4') Flow area, a,: Since the quantity of fluid is large, any baffle spacing may be arbitrarily assumed. Assume B = 12 in. a, = ID X C'B/144PT [Eq. (7.1)) = 72(31 X 0.25 X 12/144 X 1.0) = 0.323 ft 2 (24 exchanger) (6') Mass vel, G, = wja, [Eq. (7.2)) = 86,357/0.323 = 267,000 lb/(hr)(ft") (6') At t. = 193.5°F, J.£ = 1.30 X 2.42 = 3.15lb/(ft)(hr) [Fig. 14) D, = 0.95/12 = 0.0792 ft [Fig. 28) Re, = D,G,jp. [Eq. (7.3)) = 0.0792 X 267,000/3.15 = 6,720 (7') jH = 45 [Fig. 28) (8') For p. = 1.30 cp and 35° API k(cp.jk)'h = 0.213 Btu/(hr)(ft 2 WF/ft) [Fig. 16]
(7) jH = 36.5 [Fig. 24] (8) For p. = 0.88 cp and 35°API k(cp.jk)'h = 0.185 Btu/(hr)(ft•WF/ft) [Fig. 16) , k [Eq. (6.15a)) (9') ho = jH .!_ (~)';i .p, [Eq. (6.15)] (9) h; =JHD k. q,, D, k 0 213 0 185 !:.!_ = 36.5 X · = 130 .!:! = 45 X · = 121 "'· 0.0792 "'' 0.0517 ID [Eq. (6.5)] (10') Omit the viscosity correction for the (10) h;. = h; X OD trial, q,, = 1.0. = 130 X 0.62/0.75 = 107 = 121 Btu/(hr)(ft•WF) (11) (12) Omit the viscosity correction for h. the trial, q,1 = 1.0
(C}.£);3
=!!:!
h,. = h,. =
"''
107 Btu/(hr)(ft•WF)
"''
237
CALCULATIONS FOR PROCESS CONDITIONS Pressure Drop (1) For Ret = 10,100,
! 8
= 0.00027 ft'/in.• = 0.77
[Fig. 26] [Fig. 6]
JG";Ln
(2) APe = 5.22 X 1010Dsq,,
(1') For Re, = 6720, = 0.0023 ft• /in.• 8 = 0.79
!
[Eq. (7·45 )] (2') No. of crosses, N
[Fig. 29] [Fig. 6)
+1
12L/B [Eq. (7.42)) = 2 X 2 X 12 X 16/12 = 64 D, = 31/12 = 2.58 ft
0.00027 X 416,000' X 16 X 6 X 2
=
= 5.22 X 10 10 X 0.0517 X 0.77 X 1.0 = 4.3 psi (3') AP = (3) G, = 416,000, V•f2g' = 0.024
jG;D.(N + 1 ) [Eq. (7.44)1 ' 5.22 x 10 10D.sq,. 0.0023 X 267,0002 X 2.58 X 64 = 5.22 X 10 10 X 0.0792 X 0.79 X 1.0 = 8.3 psi
[Fig. 27} AP =4nV• =4X2X6X0024 r 8 2g' 0.77 • = 1.5 psi jEq. (7.46)] APr= AP, + AP11. = 4.3 + 1.5 = 5.8psi [Eq. (7.47) l The pressure drop suggests the possibility of using eight passes but a rapid check shows the pressure drop for eight passes would exceed 10 psi. Now proceed to the shell side. (13) Clean overall coefficient Uc:
Uc =,::~·h.
=
i~~ ~iii
= 56.8 Btu/(hr)(ft•WF)
(6.38)
= 0.0028 (hr)(ft•)(OF)/Btu
(6.13)
(14) Dirt factor R..r, = 5.22 X 10 1 ~D.sq,. [Eq. ( . )] 3 2g' 4 4 0.0019 X 533,0002 X 1.60 X 28 = ~ X 0.18 = 2.9 psi 5.22 X 1010 X 0.06 X 1.115 X 1.0 = 7.0 psi (4) APT = AP, +APr [Eq. (7.47)] = 4.3 + 2.9 = 7.2 psi Now proceed to the shell side. (2') No. of crosses, N
== 12L/B
{Eq. (7.43 )]
(2) AP,
v•
(13) Clean overall coefficient Uo:
U0 
h_;.h.
h;.
+ h.
 972 X 717  413 B /(h )(f ')(•F)  972 + 717 tu r t
(6.38)
(14) Dirt factor Ra: U n from (c) is 242.
Rd =
Uc Un 413242 0 UcUn = 413 X 242 = 0.0017 (hr)(ft 2)(F)/Btu
(6.13)
241
CALCULATIONS FOR PROCESS CONDITIONS SUlllmary h outside
717 Uc
413
UD
242
972
Ra Calculated 0.0017 Ra Required
0.0020
7.0
Calculated AP
7.2
10.0
Allowable AP
10.0
Discua8ion. Adjustment of the baffie space to use the full10 psi will still not permit the exchanger to make the 0.002 dirt factor. The value of U D has been assumed too high. Try the next size shelL Trial2:
Try a 21:!{ in. ID shell with four tube passes and a 6in. baffie·space. sponds to 170 tubes.
720
houtside
Uc
390
UD
200
This corre
840
Rd Calculated 0 . 0024
Rd Required 0.002
9.8
Calculated AP
4.9
10.0
Allowable AP
10.0
The use of six tube passes exceeds the allowable tube side pressure drop. The final exchanger will be Shell side
ID = 21~ in. Baffle space = 6 in. Passes= 1
Tube side Number and length = 170, 16'0" OD, BWG, pitch = 1 in., 14 BWG, Hiin. tri. Passes = 4
Example 11.4. Calculation of an Alcohol Heater. 115,000 lb/hr of absolute alcohol (100 per cent ethyl alcohol, s = 0. 78) is to be heated under pressure from astor
242
PROCESS HEAT TRANSFER
age temperature of 80 to 2oo•F using steam at 225°F. A dirt factor of 0.002 is required along with an allowable alcohol pressure drop of 10 psi. Plant practice is established using 1~in. OD tubes, 14 BWG, 12'0"long. Triangular pitch is satisfactory for clean services. Sol:ution: (1) Heat balance:
Alcohol, Q = 115,000 X 0.72(200  80) = 9,950,000 Btu/hr Steam, Q = 10,350 X 962 = 9,950,000 Btu/hr (2) 11t: LMTD (true counterflow): Hot Fluid
Cold Fluid
Diff.
225
Higher Temp
200
25
225
Lower Temp
80
145
120
120
0
Differences
LMTD = 68:3•F
(5.14)
(3) T. and t.: Use T. and t. because of the low alcohol viscosity.
Trial. (a) Assume U D = 200: From Table 8 values of U D from 200 to 700 may be expected when a dirt factor of 0.001 is employed. Since the dirt factor required is 0.002, the very maximum value of U D would be 500 corresponding to the dirt alone. Q  9,950,000  728 ft2 A  UDAt200 X 68.3a" = 0.2618 ft2 /lin ft (Table 10) 728 Number of tubes, N1 = , , X . = 232 12 0 0 2618 (b) Assume two tube passes: Only one or two passes are required for steam heaters. From the tube counts (Table 9): 232 tubes, two passes, 1 in. OD on 1 :!1:in. triangular pitch Nearest count: 232 tubes in a 23X in. ID shell (c) Corrected coefficient U D: A = 232 X 12'0" X 0.2618 = 728 ft 2 Q 9,950,000 UD =A 11t = 728 X 68.3 = 200
Hot fluid: tube side, steam Cold fluid: shell side, alcohol = 0.546 in. 2 [Table 10] (4') To obtain a mass velocity between Nta;/144n [Eq. (7.48)] 400,000 and 500,000 use a 7in. baffle space. 232 X 0.546/144 X 2 = 0.44 ft• [Eq. (7.1)] a, = ID X C'B/144PT = 23.25 X 0.25 X 7/144 X 1.25 ft' = 0.226
(4) Flow area,
a, =
=
a;
243
CALCULATIONS FOR PROCESS CONDITIONS
Cold fluid: shell side, alcohol (5') G, = wja, [Eq. (7.2)] = 115,000/0.226 = 10,350/0.44 = 23,500 lb/(hr){ft 2) = 508,000 lb/(hr)(ft') (6) At 225°F, (6') At ta = l40°F, p. = 0.60 X 2.42 = 1.45 lb/(ft)(hr) p. = 0.013 X 2.42 = 0.0314 lb/(ft)(hr) [Fig. 15] [Fig. 14] D. = 0.72/12 = 0.06 ft [Fig. 28] J) = 0.834/12 = 0.0695 ft Re, = DG,(p. (for pressure drop only) Re, = D,G,jp. [Eq. (7.3)] = 0.0695 X 23,500/0.0314 = 0.06 X 508,000/1.45 = 21,000 = 52,000 (7') jH = 83 (8') At 140•F, k = 0.085 Btu/(hr)(ft 2 WF/ft) [Table 4] (cp.jk)~~ = (0.72 X 1.45/0.085)~ = 2.31 Hot fluid: tube side,. steam
(6) G, = Wja,
(9) h;.
(9') h. = jH
= 1500 Btu/(hr)(ft2WF)
~
~' (7;t <J>,
[Eq. (6.15b)]
= 83 X 0.085 X 2.31/0.06
(10') q,,
h. =
~ tf>,
= 270
= 1.0, = 270 Btu/(hr)(ft'WF)
Pressure Drop (1) For Re, = 52,000,
(1') For Re, = 21,000, 0.000175 ft2/in. 2 [Fig. 26] f = 0.0018 ft 2/in. 2 From Table 7 the specific volume is ap 8 = 0.78 proximately 21 fVjlb, p = ~l = 0.0477lb/ft3 s = 0.0477/62.5 = 0.00076
I=
[Fig. 29]
= !.
!G!Ln [E . . )] (2') No. of crosses, N + 1 = 12LjB 7 53 2 5.22 X 1010Dsq., q [Eq. (7.43)] 0.000175 X 23,500' X 12 X 2 = 12 X 12/7 = 21 1 = 2 X 5.22 X 10'~ X 0.0695 X 0.00076 D, = 23.25/12 = 1.94 ft X 1.0 = 0.42 psi (3) t..P,: Negligible because of partial con1 • D JG!D,(N + 1) (3 ) ur, 5.22 X 1010D.s¢, [Eq. (7 .44 )] densation at end of first pass. ( ) 1lP 2 1
0.0018 X 508,000 2 X 1.94 X 21 5.22 X 10 10 X 0.06 X 0.78 X 1.0 = 7.8 psi (13) Clean overall coefficient Uc:
Uc
=
h;.h. = l500 X 270 = 229 Btu/(hr)(ft2)(°F) h;. h. 1500 270
+
+
(6.38)
(14) Dirt factor Rd: UD from (c) is 200. _ Uc UD _ 229 200 _ 0 Rd UcUD X  0.000633 (hr)(ft 2)(F)/Btu 200 229
(6.13)
244
PROCESS HEAT TRANSFER
Summary 1500
h outside
Uc
229
UD
200
270
Rd Calculated 0.000633 Rd Required 0.002
0.42
Calculated P
7.8
Neg.
Allowable P
10.0
Discussion. This is clearly an instance in which UD was assumed too high. It is now a question of how much too high. With the aid of the summary it is apparent that in a larger shell a clean overall coefficient of about 200 may be expected. To permit a dirt factor of 0.002 the new U D should be 1
1
1
uD = uc + Rd = 200 + 0.002 UD = 143 Trial2: A
=
_g_
= 9,950,000 = 1020 ft2
143 X 68.3 1020 = 325 , , X _ 12 0 0 2618 Nearest count: 334 tubes in a 27 in. ID shell (Table 9). The same bafHe pitch should be retained, since the pressure drop increases with the inside diameter. Summary UDAt
No. of tubes =
1500
I
h outside
Uc
214
UD
138.5
250
Rd Calculated 0. 0025 Rd Required
0.002
0.23
Calculated AP
7.1
Neg.
Allowable AP
10.0
If a 25in. exchanger had been used, the dirt factor would be less than 0.002 and a 6in. baf!le space would give a pressure drop exceeding 10 psi. The final exchanger is
CALCULA1'IONS FOR PROCESS CONDITIONS Shell side
ID = 27 in. Baffle space = 7 in. Passes = 1
245
Tube side Number and length = 334, 12'0" OD, BWG, pitch = 1 in., 14 BWG, 1;liin. tri. Passes = 2
Split Flow. Sometimes it is not possible to meet the pressuredrop requirements in a 12 or 24 exchanger. Instances will occur when (1) the true temperature difference or U» is very great and a small exchanger is indicated for the quantity of heat to be transferred, (2) one fluid stream has a very small temperature range compared with the other, or (3) the allowable pressure drop is small. In gases and vapors the last is the more critical because of the low density of the gas or vapor. In liquids an excellent example of (2) is found in the quenching of steel, where it is customary to cool a large volume of circulated quench oil over a small range. It is also characteristic of a number of nearconstant temperature operations such as the removal of heat from exothermic reactions by continuous recirculation of the reacting fluids through an external 12 cooler. T,
~T; ::J
g., +1
a. E ~ tl
0 l?w. ll.G.
Tz Splitflow exchanger.
Fw. 11. 7. flow.
L/2
Length
L
Temperature relations in split
The failure to meet the allowable pressure drop by the conventional methods in a 12 exchanger should be construed as an indication that fluid flow and not heat transfer is the controlling factor. Reducing the length of the tubes and increasing the diameter of the shell provides one means of reducing the pressure drop, but other means are available. By locating the shell inlet nozzle at the center instead of at the end of the shell and using two outlet nozzles as shown in Fig. 11.6, the shellside pressure drop will be but oneeighth as great as in a conventional 12 exchanger of the same shell diameter. The reduction is due to halving the mass velocity and halving the length of the shellside path. This type of flow is known as split flow. As seen in Fig. 11.7 the temperature relations in a splitflow exchanger are not in true counterflow or identical with a 12 exchanger, being dis
246
PROCESS HEAT TRANSFER
continuous at the shell midpoint. A direct solution of the equation' for the true temperature difference is somewhat tedious, since the values of t~' and t;' are related to the actual temperature differences and heat transferred in both parts of the exchanger on either side of the shell inlet. If T2 > t2, it is satisfactory to obtain At by multiplying the J,MTD by the value of FT obtained for a 12 exchanger. For services in which there is a temperature cross the actual splitflow equation should be used. Another type of flow giving even lower pressure drops is divided flow, which is usually· reserved for lowpressure gases, condensers, and reboilers. It will be discussed in Chap. 12. Example 11.6. Calculation of a Flue Gas Cooler. 10,500 cfm of flue gas (mol. wt. = 30) at 2 psig and 250°F is to be cooled to 125°F with an allowable pressure drop of 1.0 psi. Cooling will be effected by water from 80 to 100°F and with an allowable pressure drop of 10 psi. An overall dirt factor of 0.005 should be provided with a reasonable minimum water velocity. Plant practice uses 1 in. OD, 14 BWG tubes on square pitch for all services, and because it is sometimes difficult to meet the pressure drop in gas coolers, no tube length is specified.
Solution: (1) Heat balance: Gas: 10,500 cfm of flue gas at 250°F 10,500 X 60 X 30 Total gas = 359 X (711/492) X (14.7/16.7) = 41 •300 lb/hr Gas, Q = 41,300 X 0.25(250  125) = 1,290,000 Btu/hr Water, Q = 64,500 X 1(100  80) = 1,290,000 Btu/hr (2) At:
Hot Fluid
Cold Fluid
Diff.
250
Higher Temp
100
150
125
Lower Temp
80
45
125
Differences
20
105
1 Unpublished notes, D. Q. Kern and C. L. Carpenter. The equation in terms of terminal temperatures for a splitflow exchanger with two tube passes is At _ (T 1  T,)>../2
~(T,2 31 · og
n
T,)2J\n e(T•
+1
T,)
G t)/2M (7\ T,)(A + 2) + t, + t, _ 4Tt
+ 2(n 
(T, _ T,)(>. _ Z) + t, + t, _ 4Tt + 2(n  1)T,
n+l
where n = QB!QA
' "=
v'4F+1 R
and QBIQ"' = ratio of heat transfer in each half
n+l
1)T,
CALCULATIONS FOR PROCESS CONDITIONS
247
LMTD = 87.4°F
(5.14) 20 (Fig. 18) 0 118 s = 250  80 = · A conventionall2 exchanger will be satisfactory. llt = 0.935 X 87.4 = 81.6°F (7.42} (3) T, and t,: The average temperatures T. and t, will be satisfactory because of the small variations in the individual viscosities. R
= ~; = 6.25
Trial: (a) Assume U » = 15: From the examples of Chap. 9, at atmospheric pressure and 2 psi allowable pressure drop a coefficient of about 20 might be anticipated. Since
the allowable pressure drop in this example is only 1.0 psi, the trial value of U n must be reduced accordingly. Assume 12'0" tubes to increase the shell eross section. A
= _!L_ U » t.t
= 1,290,000 = 1055 ft2
15 X 81.6 a" = 0.2618 ft 2 /lin ft
1055 , , X _ = 336 12 0 0 2618 (b) Assume eight tube passes: Because of the low design coefficient gas exchangers are large for the amount of cooling medium required. From the tube counts (Table 9): 336 tubes, eight passes, 1 in. OD on 1;!4in. ~quare pitch Nearest count: 358 tubes in a 31 in. ID shell (c) Corrected coefficient U»: A = 358 X 12'0" X 0.2618 = 1125 fti 1,290,000 14 0 UD = AQD.t = 1125 X 81.6 = . When solved in a manner identical with the preceding examples and using the smallest integral number of bundle crosses (five) corresponding to a 28.8in. spacing the summary is Summary Number of tubes =
24.0
I
h outside
Uc
22.7
Un
14.0
I
392
Rn Calculated 0. 027 RJ Required
0.005
5.2
I Calculated D.P
1 0
Allowable AP
1.0
I 10.0
Diacusllion. The exchanger selected as a solution to the requirements combines two conditions which have not been met previously: The dirt factor is considerably greater than neceBSary, a:nd the pressure drop is five times greater than the allowable. Had 8ft tubes been used in place of the 12ft tubes for U D = 15, the shell inside diameter would have been 37 in. The baffles could have been spaced 32 in. apart to provide three bundle crosses, but the resulting pressure drop would be 1.7 psi. This would be unsatisfactory, since gases require large inlet connections and the flow distribution on
248
PROCESS HEAT TRANSFER
the first and third bundle crosses would be poor and the conditions of allowable pressure drop would still not be met. The solution is found in a splitflow arrangement.
Trial 2. Split flrno: (a) Assume U D = 15. Referring to the summary of the first trial it is evident that, if the pressure drop is to be met, the mass velocity must be reduced so that the new gas film coefficient will be considerably below the value of 24.0 obtained for ordinary flow. (b) Alssume 12 tube passes. The low water coefficient of 392 corresponds to a velocity of only 1. 7 fps which is extremely low for corrosion and dirt even where good water may be employed. Since the size of the shell will not be altered appreciably, having a large inside diameter, it is justifiable that 12 passes be employed. Fewer tube passes would be needed if tube cores were inserted_in the tubes. These may be calculated in the manner of Example 10.3. When using more than 8 passes in large shells, the tube count for 8 passes should be reduced by 5 per cent for 12 passes and 10 per cent for 16 passes. For smaller shells it is advisable to avoid the use of 12 and 16 passes. Using the same shell as in Trial1 for 12 passes, the new tube count will be 358 X 0.95 = 340 tubes. (c) Corrected coefficient UD: A = 340 X 12'0" X 0.2618 = 1070 ft 2 Q 1,240,000 14 8 U D = AAi = 1070 X 81.6 = . Hot fluid: shell side, flue gas (4') Flow area,
Cold fluid: tube side, water (4) Flow area,
a;
= 0.546 in. 1 a. = ID X C'B/144PT [Eq. (7.1)] There must be an odd number of crosses in a, = N,a;/144n = 340 X 0.546/144 X 12 each half of the shell and the largest spacing is 31 in. 72 in./31 in. = 2 crosses; say 3 crosses (odd) . 12 X 12 . Actua1 spacmg = 2X3 = 24 m.
[Table 10] =
0.107 ft'
a. = 31 X 0.25 X 24/144 X 1.25 = 1.03 ft 2 (6') Mass vel, split flow: (6) G, = w/a 1 G. = UW fa, [Eq. (7.2)] = 64,500/0.107 = 602,000 lb/(hr)(ft')  }2 X 41,300/1.03 = 20,000 lb/(hr)(ft2) Vel, V = G,/3600p = 602,000/3600 X 62.5 = 2.68 fps (6) At ta = go•F, (6') At T. = 187.5•F, p. = 0.0206 X 2.42 = 0.050 lb/(ft)(hr) p. = 0.81 X 2.42 = 1.96 lb/(ft)(hr) [Fig. 15] [Fig. 141 D. = 0.99/12 = 0.0825 [Fig. 28] D = 0.834/12 = 0.0695 ft Re, = D.G,jp. [Eq. (7.3)] Re, = DG,jp. = 0.0825 X 20,000/0.05 = 33,000 = 0.0695 X 602,000/1.96 = 21,300 (Re 1 is for pressure drop only) (7') jn = 105 !Fig. 28] (8') At 187.5•F, k = 0.015 Btu/(hr)(ft2)(°F/ft) (Table 5) (cp/k)» = (0.25 X 0.050/0.015)» = 0.94
CALCULATIONS FOR PROCESS CONDITIONS Cold fluid: tube side, water
Hot fluid: shell side, flue gas (9') ko = jH ;,
~
(1;t .p,
(Eq. (6.15)]
= 105 X 0.015 X 0.94/0.0825
"''
= 17.9
(10') (11') (12') q,,
h, =
~ q,,
249
(9) h; = 710 X 0.94 = 667 (10) h;, = h; X ID/OD
= 1.0
[Fig. 25]
= 667 X 0.83/1.0
= 17.9 Btu/(hr)(ft•WFJ
= 557 Btu/(hr)(ft 2)("F)
Pressure Drop (1) For Re, = 21,300, {Fig. 26] !Fig. 29] f = 0.00012 ft 2 /in. • (2') No. of crosses, N + 1 = 3 [Eq. (7.43)] AP _ fGfLn 2 [Eq. (7.45)] ( ) '  5.22 X. 10 10Ds.p, D, == 31/12 = 2.58 ft 8 = 0.0012 _ 0.00022 X 602,000 2 X 12 X 12  5.22 X 1010 X 0.0695 X 1.0 X 1.0 = 3.1 psi (3') AP = f(f;D,(N 1) [Eq. (7.44)] (3) AP, = 4n/s(V2 /2g') [Fig. 27] ' 5.22 X 1010D,.sq,, = (4 X 12/1)0.052 = 2.5 psi 0.00167 X 20,0002 X 2.58 X 3 = 5.22 X 10 10 X 0.0825 X 0.0012 X 1.0 = 1.0 psi (4) APT = AP, + AP, [Eq. (7.47)] = 3.1 + 2.5 = 5.6 psi
(1') For Re, = 33,000, f = 0.0017 ft 2 /in. 2
+
[13) Clean overall coefficien~ U c:
Uc = h;~'+\. = ~~; ~ i;:~ = 17.3 Btu/(hr)(ft )(°F) 2
(6.38)
(14) Dirt factor Ra: U D from (c) = 14.8
Ra =
U~;U~D
=
i;:: ~ i!::
= 0.0098 (hr)(ft•)("F)/Btu
Summary 17.9
h outside
Uo
17.3
UD
14.8
557
Ra Calculated 0.0098 Ra Required 0.005 1.0
Calculated AP
5.6
1.0
Allowable AP
10.0
(6.13)
250
PROCESS HEA1' TRANSFER
A 16pass uni.t would also be suitable but is not warranted.
Shell side ID = 31 in. Baffie space = 24 in. Passes = split flow
The final exchanger is
Tube side Number and length = 340, 12'0" OD, BWG, pitch = 1 in., 14 BWG, I7iin. square Passes= 12
PROBLEMS For the following process conditions determine the size and arrangement of exchanger to fulfill the conditions allowing pressure drops of 10 psi each stream and a combined dirt factor of 0.004. Employ 12 exchangers wherever possible. 11.1. 60,000 lb/hr of 42° API kerosene is cooled from 400 to 225°F by heating 35°API distillate from 100 to 2t, •F Cold and hot terminal temperature diff.erence, •F Clean and design overall coefficient of heat transfer, Btu/(hr)(ft')(°F) Overall coefficient of heat transfer for cooler, exchanger, and heater, respectively, Btu/(hr)(fV){°F) Velocity, fps Weight flow of hot fluid, lb/hr Weight flow of. cold fluid, lb/hr Villcosity ratio, (p./,...,)o.u Viscosity, centipoises X 2.42 = lb/(ft)(hr) VIScosity at tubewall temperature, centipoises X 2.42 = lb I (ft,)(hr) Annual operating hours
Subscripts (except as noted above)
Shell Tube
CHAPTER 12 CONDENSATION OF SINGLE VAPORS
Introduction. A fluid may exist as a gas, vapor, or liquid. The change from liquid to vapor is vaporization, and the change from vapor to liquid is condensation. The quantities of heat involved in the vaporization or condensation of a pound of fluid are identical. For a pure fluid compound at a given pressure the change from liquid to vapor or vapor to liquid occurs at but one temperature which is the saturation or equilibrium temperature. Since vaporliquid heattransfer changes usually occur at constant or nearly constant pressure in industry, the vaporization or condensation of a single compound normally occurs isothermally. When a vapor is removed upon formation from further contact with a liquid, the addition of heat to the vapor causes superheat, during which it behaves like a gas. If a mixture of vapors instead of a pure vapor is condensed at constant pressure, the change does not take place isothermally in most instances. The general treatment of vapor mixtures differs in certain respects from single compounds and will be studied in the next chapter with the aid of the phase rule of J. Willard Gibbs. Condensation occurs at very different rates of heat transfer by either of the two distinct physical mechanisms, which will be discussed pres· ently, dropwise or filmwise condensation. The condensing film coefficient is influenced by the texture of the surface on which condensation occurs and also by whether the condensing surface is mounted vertically or horizontally. In spite of these apparent complications condensation, like streamline flow, lends itself to direct mathematical study. Dropwise and Filmwise Condensation. When a saturated pure vapor comes into contact with a cold surface such as a tube, it condenses and may form liquid droplets on the surface of the tube. These droplets may not exhibit an affinity for the surface and instead of coating the tube fall from it, leaving bare metal on which successive droplets of condensate may form. When condensation occurs by this mechanism, it is called dropwise condensation. Usually, however, a distinct film may appear as the vapor condenses and coats the tube. Additional vapor is then required to condense into the liquid film rather than form directly on the bare surface. This is film or filmwise condensation. The two mechanisms are distinct and independent of the quantity of vapor condensing per square foot of surface. Filmwise condensation is therefore 252
CONDENSATION OF SINGLE VAPORS
253
not a transition from dropwise condensation because of the rapidity at which condensate forms on the tube. Due to the resistance of the condensate film to the heat passing through it the heattransfer coefficients for dropwise condensation are four to eight times those for film wise condensation. Steam is the only pure vapor known to condense in a dropwise manner, and special conditions are required for its occurrence. These are described by Drew, Nagle, and Smith1 and principally result from the presence of dirt on the surface or the use of a contaminant which adheres to the surface. Materials have been identified by Nagle 2 which promote the dropwise condensation of steam although these also introduce an impurity into the steam. Dropwise condensation also occurs when several materials condense simultaneously as a mixture and where the condensate mixture is not miscible, as in the case of a hydrocarbon and steam. However, during various periods in the normal operation of a steam condenser the mechanism may initially be filmwise condensation, shift to dropwise condensation, and at some later time revert to film condensation. Because of the lack of control it is not customary in calculations to take advantage of the high coefficients which have been obtained in dropwisecondensation experiments. This chapter consequently deals with the calculation of condensers for various conditions and is based solely upon filmcondensation heattransfer coefficients. It is fortunate that the phenomenon of film condensation lends itself to mathematical analysis, and the nature of condensation on a cold surface may be considered one of selfdiffusion. The saturation pressure of vapor in the vapor body is greater than the saturation pressure of the cold condensate in contact with the cold surface. This pressure difference provides the potential for driving vapor out of the vapor body at a great rate. Compared with the small resistance to heat transfer by diffusion from the vapor into the condensate, the film of condensate on the cold tube wall contributes the controlling resistance. It is the slowness with which the heat of condensation passes through this film that determines the condensing coefficient. The ultimate form of an equation for the condensing coefficient may be obtained from dimensional analysis where the average condensing coefficient h is a function of the properties of the condensate film, k, p, g, 11, and L, tlt, and:\, the last being the latent heat of vaporization. Nusselt theoretically derived the relationships for the mechanism. of film condensation, and the results he obtained are in excellent agreement with experiments. Process Applications. In chemical industry it is a common practice to separate a liquid mixture by distilling off the compounds which have t I
Drew, T. B., W. M. Nagle, and W. Q. Smith, Trans. AIChE, 81, 605621 (1935) Nagle, W. M., U.S. Patent 1,995,361.
254
PROCESS HEAT TRANSFER
lower boiling points in the pure condition from those having higher boiling points. In a solution· of several compounds each exerts a partial pressure and the most volatile cannot be boiled off from the rest without carrying some of the heavier or higher boiling compounds along with it. The proportion of heavier compounds carried off when a solution starts to boil is less than existed in the original solution before boiling commenced. If the vapor coming off initially is condensed, it has a lower boiling point than the original solution, indicating the increase in the proportion of the more volatile compounds. By successively boiling off
Boftoms
Distilling column with auxiliaries.
only part of a liquid mixture, condensing the vapor formed, and boiling off only a part of the condensate, it is possible to obtain a nearly pure quantity of the most volatile compound by numerous repetitions of the procedure. Thus the separation by distillation is accomplished by partial vaporization and subsequent condensation. In distillation it is customary to obtain a number of partial vaporizations and condensations by directly contacting a vapor and a liquid cooling medium in a continuous distilling column. The bubble cap distilling column shown in Fig. 12.1 is representative of modern practice and derives its name from a series of inverted slotted .caps which are placed over vapor risers on each plate of the column. Vapor from below a plate enters the risers and is broken into bubbles as it passes through the slotted bottoms of the bubble caps and thence through the layer of liquid
CONDENSATION OF SINGLE VAPORS
Z55
maintained by the downcomer on each plate. The feed, which is usually a liquid, is a mixture of niore and less volatile compounds and enters the distilling column at the feed plate where the volatile compounds are partially vaporized by the rising vapors as the feed travels across the plate. The remainder of the liquid on the plate is less volatile than the feed and overflows to the plate below through the downcomer. The boiling points of the liquids on each of the lower plates is consequently higher. To vaporize a portion of the feed, vapor from below the feed plate must exchange heat with the liquid on the feed plate, thereby driving the more volatile compounds to the plate above the feed. By supplying heat at the bottom of the column where the increased concentration of the least volatile compounds represents the highest boiling temperature in the system, a thermal gradient is established plate by plate between the bottom of the column and the top. Heat supplied at the bottom by vaporization in a reboiler is transmitted to the top of the column platebyplate due to the temperature diffelences corresponding to the differences in boiling points between plates. Continuous distillation requires the presence of liquid at all times on the plates, so that vapors of the less volatile compounds in the feed may be condensed and carried downward. To accomplish this, some volatile liquid from the condenser, which represents one plate above the top plate and which is therefore colder, is introduced onto the top plate and flows downward in the column. The volatile liquid which is poured back into the column from the condenser is the reflux. The quantity of volatile components removed from the system at the top and having the same composition as the reflux is called the distillate or overhead product. The heavier compounds removed at the bottom are variously called waste or residue or, if they are of value, bottoms product. The quantitative aspects of the heat balance are treated in Chap. 14. It is the condensing temperature in the condenser which determines the operating pressure of the distilling column, since the saturation temperature of a vapor varies wit.h its pressure. The overhead product must condense in the condenser at a temperature sufficiently high so that its latent heat can be removed by cooling water. The size of the condenser is dependent upon the difference between the condensing temperature and the range of the cooling water. If the condensing temperature is very close to the cooling water range at atmospheric pressure, the distillation pressure must be elevated to permit the attainment of a larger .At.
In the power industry the term surface conden8er is reserved for tubular equipment which condenses steam from the exhaust of turbines and engines. Since a turbine is primarily designed to obtain mechanical work
256
PROCESS HEA1' 'l'RANSFER
from heat, the max:imum conversion is obtained in the turbine by maintaining a low turbinedischarge temperature. If the turbine were to discharge to the atmosphere, the lowest attainable steam temperature would be 212°F, but if the steam were to discharge into a condenser under vacuum, it would be possible to operate at discharge temperatures of 75°F and lower and to convert the enthalpy difference from 212 to 75°F into useful work. Condensation on SurfacesNusselt's Theory. In condensation on a vertical surface a film of condensate is formed as shown in Fig. 12.2 and further condensation and heat transfer to the surface occurs by conduction through the film which is assumed to be in laminar flow downward. The thickness of this film greatly influences the t,__....,_ocY rate of condensation, since the heat accompanying the remova1 of vapors from the vapor phase encounters the condensate film as a resistance which may be qul.te large. The thickness of the film is a function of the velocity of drainage which varies with the deviation of the surface from a vertical position. For a vertical surface the thickness of the film cumulatively increases from top to bottom. For this reason the condensing coefficient for a vapor condensing on a vertic.al surface decreases from top to bottom, x',':/J and for the attainment of a large condensing coefficient the height of the surface should not be v~r.y gr~at. The v~locity o! drainage for FIG. 12.2. The vertical con• a1so a funcdensate film. equal quantities of cond ensate IS tion of the viscosity of the condensate: The lower the viscosity the thinner the film. For all liquids the viscosity decreases as the temperature increases, and the condensing coefficient consequently inereases with the condensate temperature. The derivations given in. this chapter through Eq. (12.34) are those of Nusselt. 1 The following assumpti()ns are involved: 1. The heat delivered by the vapor is latent heat only. 2. The drainage of the condensate filmJrom the surface is by laminar flow only, and the heat is transferred through the film by conduction. 3. The thickness of the film at any point is a function of the mean velocity of flow and of the amount of condensate passing at that point. 4. The velocity of the individual layers of the film is a function of the relation between frictional shearing force and the weight of the film (see Chap. 3). 1
Nusselt, W., Z. Ver. deut. lng., 60, 541 (1916).
CONDENSATION OF SINGLE VAPORS
257
5. The quantity of condensate is proportional to the quantity of heat transferred, which is in turn related to the thickness of the film and of the temperature difference between the vapor and the surface. 6. The condensate film is so thin that the temperature gradient through it is linear. 7. The physical properties of the condensate are taken at the mean film temperature. 8. The surface is assumed to be relatively smooth and clean. 9. The temperature of the surface of the solid is constant. 10. The curvature of the film is neglected. Condensation. Vertical Surfaces. In Fig. 12.2 the rate at which heat passes from the vapor through the liquid condensate film and into the cooling surface per unit area is given by
~=
k(t'y; t)
= AW' = h(t'
 t)
(12.1)
where }. is the latent heat of vaporization, W' the pounds of condensate formed per hour per square foot, andy' is the thiclmess of the condensate film at the generalized point in the figure whose coordinates are x', y'. The other symbols have their conventional meaning. The rate at which the vapor condenses is then given by W' = k(t'  t) ).y'
(12.2)
The liquid flows downward over the vertical surface with a velocity u varying from zero at the tubefilm interface and increasing outward to the condensatevapor interface. The velocity also increases vertically as the condensate flows downward. Consider a small cube of unit depth dz = 1, defined by dx dy 1 in the moving condensate illm of Fig. 12.2. On the side near the cold vertical surface there is a tangential force acting upward and tending to support the cube. On the side away from the cold vertical surface there is a tangential force acting downward due to the more rapid movement of the liquid downward as the distance from the surface is increased. If the resultant force upward through the cube is designated by 7', then the respective forces are r  dr/2 and 7' + dr/2. The differential tangential force must be offset by gravity acting down. pdxdyl
=
On unit area, dx dz = 1
(r ;~di) (r + ;;~) = dr
(12.3)
258
PROCESS HEAT TRANSFER
From the basic definition of viscosity in Chap. 3 the tangential stress is defined by Eq. (3.3), using (lbforce)(hr)/ft 2 as the dimensions for the viscosity. Since it is customary to use the dimensions (lbmass)/(ft)(hr) for the dimensions, Eq. (3.3) becomes (12.4) (12.5)
p
Take pfJ.' as constant. d2u _ pg dy2 = J.l.
u = 
(12.6)
pgyz 2p,
+ ely + c2'
(12;7)
The constants Ct and C2 must now be evaluated. Since the liquid adheres to the wall, u must equal zero at y = 0, making C2 equal zero. z
/
w' X (b) (a> FIG. 12.3. Vertical condensate film flow.
At the outer boundary of the film (condensatevapor interface) there is no tangential stress and from T = p, dtt/dy when y = y',
(dtt) dy
= u'
u =
0= 
pgy'
+ c1
J.l.
(12.8)
~ (yy' ~)
At a distance x from the top of the condensing surface the average velocity downward u is given by
·a = .I:_ {v' ·u dy = pg y'' y'
)o
3p,
(12.9)
CONDENSATION OF SINGLE VAPORS
259
When the value of x from the top of a vertical wall is taken to be unity as shown in Fig. 12.3a, the quantity of downward flow across a horizontal plane of area ly' of the condensate is ly'up
At x + dx there is a gain in the amount of downward flow of condensate as shown in Fig. 12.3b. Using the value for iZ from Eq. (12.9), multiplying by py', and differentiating with respect to x to obtain the increase from x to x + dx, •
d(puy')
)
2
= d ( ~: y'' = P ~Y
,.
d.y'
(12.10)
And this increase must come from condensation out of the vapor and into the condensate film d(puy')
= W'1
dx
where W' is the condensate flow out of the vapor and "normal" to the falling condensate layer per unit area as in Fig. 12.3b. From Eq. (12.2), however, W' was defined in terms of the heat transfer as W' = k(t' t) 'Ay'
Substituting for W' in Eq. (12.10) the value from Eq. (12.2), k 2 ,. 
'Ay'
(t' t) dx
= P gy
dy'
~
2 'A ,. (t'  t) dx = UL.JL dy' kp.
For a limited range set t'  t, When y'
= 0, x =
p, ).., p.,
(12.11) (12.12)
and k constant and integrate.
0
y''
~;; =
y'
= [4~tk p2]..g
(t'  t)x (t' 
(12.13)
t)x]H
(12.14)
The heattransfer coefficient across the condensate layer at the distance the origin per unit of interfacial area is given from Eq. (12.1) by
~:from
h = Q.,/A.., = ~ .• t'  t y' Substitute y' from Eq. (12.14): k 3p2Ag [ h. = 4JL(t'  t)
]J4
(12.15)
1 xl4
(12.16)
260
PROCESS HEA'l' TRANSFER
The total heat through the condensate layer from 0 to x is Q., ["'
Q"' == }o
["' [
h.,(t'  t) dx = }o
(k3: 2},_g) = 4~• 3
J4
k3 2J\ ]1.4 d 4JJ.(/ g t) (t'  t) x~
[(t'  t)x]H
(12.17)
If the average coefficient between the two points is ii,
, (kjp}A.g)>' [(t' 
4~•
h
=
(Q.,).,=L (t' t)L
fi
=
0.943 (k}p}A.g)J4
t)L]~
)J.j
= 3 ~~+:(t'':ct)'"""L(12.18)
fl.!L t1t1
where k" Ph and JJ./ are evaluated at the film temperature the film temperature is t1 = ~(t' + t) = ~(Tv + iw) and f1t1 = t1 
t1
and where (12.19)
iw
In the above, as in the derivation which follows, the stress caused by the passage of the saturated vapor over the condensatevapor interface has been neglected. It can be included, although it is not of practical consequence. The variation of the film thickness and local heattransfer coefficient are shown in Fig. 12.4. The shapes of the curves follow the thickness and consequently the resistance of the condensate film. Inclined Surfaces. Consider a cube making the angle a as shown in Fig. 12.5. The gravity component acting in a plane parallel to the surface is p sin a, and Eq. (12.3) becomes p
sin a dy dx 1 =
(r 
On unit area dx dz = 1 p
dr dy)  (r dy 2
dr sin a=  dy
+ dy dr dy) 2
= dr
(12.20) (12.21)
Equation (12.6) becomes gp •
sma J.l.
Equation (12.7) becomes
(12.22)
(12.23)
At the start of condensation on the tube where y = 0 and there is no velocity along the tube, u = 0 and c2 = 0 gpy' . cl =sma J1.
CONDENSATION OF SINGLE VAPORS 0
1""'
261
L...~"
'V J
1\
1/
0.5
\ \ I.0
II
.....
I  f.~
Fm. 12.5.
I l:1:> I l~ (IJ
x,ft
Film on an inclined surface
~t:: :,__
I. 51 18
,__ f~ ,__ f~ ~~ ,__ 1.... !S
,__ ff
l(
~ ~
~
,.. ,__ ,__ 1
~
l;:
2. 01 ~~
2. 5
3.0
0
400
800
1200
1600
woo
Coefficient ,8tuj(hr)( ft1 J(°F)
0.8 1.6 2.4 3.2 4.0x 10~ Film thickness,ft FIG. 12.4. Vertical film thickness and condensing coefficients. for a descending film. 0
FIG. 12.6. tube.
Condensate film on a horizontal
(After Nwmlt.)
and Eq. (12.9) becomes u = gpy
,.
3M
and Eq. (12.18) becomes
ii = 0.943
(k
sin a
3 2'
1P1" 0
•
sm a
(12.24) )~
MIL llt,
(12.25)
Consider a cube of unit length at radius The mass flow of vapor into the condensate film through the area r dcx and with a film Honzontal TUbular Surfaces.
r making an angle a with the vertical as seen in Fig. 12.6.
262
PROCESS HEAT TRANSFER
thickness y' is. given by the conduction equation.
= k(t' 
W'
t)r da
t..y' The condensation must give rise to an increase in the inclined falling film. For a differential amount of condensation the increase through the condensing area r dais d(puy') and for Eq. (12.10) 2
d(puy')
= P g d(y'' sin a) = W' dx 3JJ.
Substituting for lV', Eq. (12.11) becomes p 2g
k(t'  t)r da _ )..y'
 3J.I.
d( ,, . ) y sm a
3p.k(t'  t)r da _ 'd( ,, . ) y ysma 2 ~ p g"
Let m
m da
=
3J.I.k(t'  t)r '~o'
p2gf..
= y'd(y'' sin a)
Differentiating m da
+ y'' cos ada) + y'' cos a da
= y'(3y'' sin a dy' = 3y'' sin a dy'
(12.26)
In Eq. (12.26) the term 3y'' dy' appears, but d(y'') = 4y'' dy' and
3y'' dy'
= % dy''.
Rearranging Eq. (12.26) and substituting,
3 da =  sin a dy'' 4m
Let y''/m = f
4
=
,.
+ '!L cos a da m
z da =%sin adz+ z cos ada 3 . dz+ zcosa 1 = 0 sma4 da
(12.27)
(12.28)
Equation (12.28) is a linear differential equation whose solution is
z When a
= 0, Ca = 0,
+ (3 f
= Sin
z=
4
a
•.
1 ~3sm~a .
sinli ada
J
+ c3)
sinli ada
(12.29)
(12.30)
CONDENSATION OF SINGLE VAPORS
263
The value of this integral for different values of a may be determined by graphical methods. From the substitution in Eq. (12.27) y' = if;ml4 = ..p [3f.Lk(t'  t)r]l1: p2g}..
(12.31)
As shown in Eq. (12.15), h., = kjy'. The thickness of the film actually decreases slightly as a increases from 0 to 5°, and then it increases steadily and breaks into drops. The localheat transfer coefficient at any point is then (12.32)
The average heattransfer coefficient h,, of the segment between angles is
a 1 and a2
(12.33)
Employing graphical methods as before, where D. is the outside diameter of the tube, the average heattransfer coefficients are found to be
From 0 to 180° which is onehalf the tube, the other half being sym. metrical,
ii = 0.725 (
k}p}Ag )Y1 11tDa fltt
(12.34:)
The variation of the film thickness and heattransfer coefficient for steam on a horizontal tube is shown in Fig. 12.7. As in the preceding case it is governed by the resistance of the condensate film to conduction. Development of Equations for Calculations. McAdams 1 found from the correlation of the data of several investigations that observed condensing coefficients for steam on vertical tubes were 75 per cent greater than the theoretical coefficients calculated by Eq. (12.18). The values calculated from Eq. (12.18) agree, however, for a condensate in streamline flow with the values calculated from Eq. (6.1) for ordinary streamline flow. When 1l. liquid descends vertically on the outside of a tube, it is certainly 1
McAdams, op. cit., p. 264.
264
PROOESS HEAT TRANSFER
F:w. 12.7. Plot of the heattransfer coefficient and film thickness of water on a horizontal tube. (After Nmaelt.)
in streamline flow at the top of the tube, where the accumulation of condensate is small. If a relatively large amount of vapor is condensed on the tube, it is possible that at some point below the top the film will change to turbulent flow. This may be estimated from the diameter and length of the tube, the viscosity of the condensate, and the quanAr tity being condensed. Referring to the tube as shown in Fig. 12.8, the crosshatched area outside the tube represents condensate film as seen at any point looking down. This is similar to the flow in the annulus of a double pipe F1a. 12.8. Vertical de exchanger except that the outer surface of the scending :film. film is not formed by a concentric pipe. In the case of the double pipe exchanger the equivalent diameter was taken as four times the hydraulic radius.
CONDENSATION OF SINGLE VAPORS
265
Then _ X _ D • 4r,. 4
free flow area
wetted perimeter
and
DIJ Re=Jl.
For vertical tubes let A 1 = crosssectional area (shaded) P = wetted perimeter per tube D.= 4 X AJ!P Letting the loading per tube be w' = W /Nt, where Nt is the number of tubes, G = w'jA 1 lb/(hr)(ft 2) (12.35) Re = D.Gjp. = (4Ai/P)(w'/A,)jp. = 4w'fp.P Calling the condensate loading per linear foot G',
w'
lb/(hr)(lin ft)
G' = p'
(12.36)
Eq. (12.35) becomes Re
= 4G' (J.
The total heat load is given by Q = }..w'. Q AW1 h =A !J.t1 = PL!lt1
i\ G'
=Lilt,
(12.37)
fi)l4 G'
(12.38)
Substituting in Eq. (12.18),
h=
0.943
(kMu P.!
Multiplying the right term by (4p./4p.)l4,
;i%. = 0 943 ( 4k}pjg fl_)l4 . p.J 4G'
h ( pj )~ kjpjg
=
1.47 (4G' )li ,.,.,
(12.39)
For horizontal tubes Eq. (12.39) becomes
h ( p.J
k}p}.g
)J.3 = 1.51 (40!J.J")~
(12.40)
where the loading for a single horizontal tube is
w
G" = LN:
(12.41)
266
PROCESS HEAT TRANSFER
Using the corresponding loading as given by either Eq. (12.36) or (12.41) as the case may be, Eqs. (12.39) and (12.40) may both be represented by
);i = 1.5 (4G')~ _1.5 (4G")li 
h ( p.} 
kjp]g
JI.J
(12.42)
jJ.f
Equations {12.39) and (12.40) were obtained for condensation on single tubes. In a verticaltube bundle the presence of one or more tubes does not alter the assumptions on which t.he derivation was predicated. However, on horizontal tubes in tube bundles it has been found that the splashing of the condensate as it drips over successive rows of tubes causes G" to be more nearly inversely proportional toNi~" rather than N 1 so that it is preferable to use a fictitious value for horizontal tubes G"
W = LN,%
lb/(hr)(lin ft)
(12.43)
Figure 12.9 is a line chart of solutions of Eq. (12.42) prepared for convenience. Its use requires that the film be in streamline flow corresponding to an average Reynolds number of about 1800 to 2100 for the flow gradient assumed by the condensate. For steam at atmospheric pressure Eq. (12.42) reduces to the equations given by McAdams: 1 For horizontal tubes ~12.44a)
and for vertical tubes 
h
4000
= LV. ilt,>>
(12.44b)
where ilt1 ranges from 10 to 150°F. It is frequently desirable to apply Eqs. (12.39), (12.40), and (12.42) to the calculation of condensers which are modifications of the 12 exchanger with condensation in the shell. Such condensers have baflled tube bundles. The baffles do not affect the condensing film coefficients in horizontal condensers, since the coefficients are independent of the vapor mass velocity, but they do influence the accumulation of condensate on the tubes of vertical condensers. Moreover, in condensers with multipass tubes the tubewall temperature is different at every point in each pass, whereas the surface temperature was assumed constant in the derivations. A correction for the latter cannot be accounted for in the calculations except by the treatment of small surface increments of each pass individually. The error introduced by using the mean tubewall temperature as being effective over all the surface is apparently too small to justify the lengthier calculation. Since the baffie holes are ordinarily 1
Ibid., p. 270.
1.0
ConclensinCJ coefficient ii ~he 100 1000
10
~r~c~},;;J
10,000 ~
~
~__.t;:;
~~~
~
~
~
~~
::
~~
~
v
2 § ~
VERTICAL lVBES G~Wft'NtD
_o.l v•so·
~~~~~~~~~+t~~~~~&4~~~~~
HORIZONTAL 7VBES
G"~WILAf*
7ubeQO.,ff, ff f~~=t~~~~~~~~~;:;f;f1~Eflf§~~ 0L =• Tube length,D=II), ff 0
Nt= No. of lubes in hund!e ~:;t::l~~~~~~~~f;;;::J;;;:P.f::tj,!.l~;....4 W= Condensafe massflowJb/nr_ ;..+
v
..lb/hr
Tripleeffect process evaporator.
from a bleed point or from the boiler directly. The selection of the number of effects is closely allied to the relationship among the fixed charges and the steam operating cost. M1.1ltipleeffect evaporators with parallel feed need not have all effects simultaneously in operation and can therefore be adjusted if the demand for distilled water varies. Evaporators for this type of service are generally of medium size, about 500 to 2000 ft 2 per shell. A ·typical tripleeffect process evaporator is shown in Fig. 14.12, where 83,205 lb/hr of steam at 35 psia (20 lb · gage) and saturated is split to mix directly with the cold feed at 70°F and for vaporization in the evaporator. This process employs several of the elements discussed under makeup evaporators. The final product is 222,015 lb/hr of distilled treated water. In order to obtain a maximum quantity of vaporization from a given quantity of steam initially near atmospheric pressure, temperature differences are maintained from the first to the last effect by operating Vvith the last effect under vacuum. This is accompllshed ·by using a steam jet vacuum pump on the last effect. Thus the shell side of the first effect in Fig. 14.12 operates at 18.9 psia or 225°F, the second at 8.6 psia or 186°F, and the third at
390
PROCESS HEAT TRANSFER
2.5 psia or 134°F. This establishes differences .ih the effects. of 34, 39, and 52°F. The operating pressures for each of the effects is determined by trial and error, so that all three effects will have the same surface as calculated by A = Q/UD At. This procedure will be demonstrated in detail in the treatment of a chemical evaporator. If the vacuum were not applied, the maximum available temperature difference over the three effects would be from 259 to 212°F or 47 instead of 125°F as shown in the flow sheet. The principle of vacuum evaporation is extensively used in chemical evaporation. Since the vapor from the last effect is at a low temperature, it has little value in the preheating of the feed, its temperature in this case being 134°F. The evaporatorcondenser, therefore, operates with coolingtower water instead of feed water. It will be noted that, although
40£800/h 'hr 210'F
Evap feedP'""P Heattransformer evaporator.
the heat from the lasteffect vapor is rejected, the blowdow'n heat losses of such a system are reduced considerably by being at a low heat level. The steam required for the vacuum pump must also be taken into account in computing the efficiencies of vacuum processes. 3. Heattransformer Evaporators. The heattransformer evaporator is a singleeffect system of one or more shells in parallel receiving steam from the exhaust of a highpressure turbine or highpressure engine. Flow sheets are shown in Figs. 14.13 and 14.14. The. purpose of this type of evaporator is to condense steam from a highpressure boiler which has passed through a highpressure turbine and into the evaporator. The condensate is then returned directly to the highpressure boiler by a pressure booster, thereby keeping the highpressure circuit closed and continuously supplied with high:pressure boiler water and steam. Obviously highpressure boiler and turbine installations are favorably affected by this circuit. By the condensation of the exhaust steam from the highpressure turbine or engine the heat transfer in the evaporator is used to produce large quantities of process steam, all or a large part of which is never returned to the evaporator system. If the condensate is ·not
391
EVAPORATION
returned, it is because it may be difficult to collect or the vapor may be consumed in a chemical or heating process or it may be continuously contaminated. This type of evaporator is relatively large, having been built in a single unit in sizes up to 11,000 ft2 of surface and capable of producing 150,000 to 200,000 lb/hr of steam. Single evaporators of this size are 10 to 11ft ;aoooJrw £
lilrbine .., JOQ()()()/b/hr ~ ';::. ,_ , VI 1250#ga ) I ~ _J.§.JOO.f.bf!!:. _ 750"F 0 c::.~ 2JS.J#qa 40rF
J:K
~ y
/Jesuperhi!IY Inlet
fnfl:rcondenser: fat/pipe
ffof well
·
.,
Condensing
Water Outlet
Fro.. 14.17. Twostage ejector with jet intercondenser, serving a barometric condenser. (Footer Wheeler Corporation.) 
:b'ro. 14.18a. Twostage ejector with jet inter and aftercondensers. (Footer Wheeler Corporation.)
barometric condenser with air ejectors, an example of which is shown in Fig. 14.17. It is operated by the two steam jet ejectors. A single ejector connected to a condenser is capable of maintaining a vacuum of about 26.5 in. Hg abs and can be made with several jets
EVAPORATION
395
replacing the singlemotive steam nozzle. This provides a more uniform distribution of steam in the mixing zone. When a vacuum of 26.5 to 29.3 in. is desired, it can be accomplished by means of a twostage ejector as shown in Fig. 14.18a and b.· Forhigher vacuums the use of a threestage ejector is required. The twostage detail in Fig. 14.18a is the same as that employed in Fig. 14.17. The condenser in Fig. 14.17 is a barometric condenser equipped with a water inlet and distribution plates, so that the water entering the inlet cascades or sprays over the incoming fliqh
,l===========~~=[~~~pressure steam
Orain FIG. 14.18b. Company.)
Twostage ejector with surface inter and aftercondensers.
(The Lummus
stream from the evaporator or precondenser and removes a large part of the steam from the process as condensate. The remaining air with less steam flows overhead into the first stage ejector. After compression in the first stage the partial pressure of the steam will have been increased and most of the remaining vapor can be condensed by another direct contact with cooling water. Again referring to Fig. 14.17, in order to remove the water and condensate from the assembly without losirig vacuum it is necessary that a leg of liquid be maintained with a hydrostatic head zp equal to the difference between the vacuum and atmospheric pressure, where z is the height and p the density. In this manner the upper surface of the liquid in the tail pipe is at a pressure corresponding to the vacuum and the liquid at the bottom of the tail pipe is at atmospheric pressure due to the weight of the hydrostatic head. Thus liquid· under vacuum continu
396
PROCESS HEAT TRANSFER
ously enters the tail pipe, and liquid at atmospheric pressure continually leaves by way of the hot well at the bottom of the tail pipe. Atmospheric pressure correaponds to a hydrostatic head of 34ft of water, and complete vacuum corresponds to zero hydrostatic head. To maintain a process at substantially complete vacuum requires that a leg of 34 ft of water be maintained between the barometric condenser and the hot well. If a vacuum of less than 29.92 in. Hg is to be maintained by the ejectors but a leg of 34 ft equivalent to 29.92 in. Hg has been provided, it means merely that the height of liquid in the tail pipe will automatically drop
Water
In
.1 To fail pipe and /!of well
J. To fail ptpe
and /!of we//
(a)COUNTERFLOW TYPE Fra. 14.19.
(b)PARALLEL FLOW TYPE
Barometric condensers.
to provide only the necessary hydrostatic difference between the operating vacuum of the ejector and the atmospheric pressure. Barometric condensers are of two types, counterflow and parallel flow as shown in Fig. 14.19a and b. In counterliow types the temperature of the water at the liquid level may approach the vapor temperature more. closely than in parallelflow types. Counterflow types are preferable where water is at a premium or it is difficult to have the vapor enter from the top. If a pump is used to remove the tail liquid instead of a total barometric height,· whatever head is supplied by the pump can be deducted from the total barometric height and the assembly is known as a lowlevel condenser. The quantity of water required in the barometric condenser can be computed from 
Q
Gpm  500( T a  . t W  ta )
(14.4)
EVAPORATION
397
where T. = saturation temperature of the vapor, °F tw = temperature of the water, °F ta = degrees of approach, to T., °F In counterflow barometric condensers ta is taken as 5°F. The twostage ejector assembly in Fig. 14.18a will produce the same results as that of Fig. 14.18b. It differs only in that the condensation after each stage is accomplished by means of tubular surface instead of the direct contact of the cooling water with the steam or vapor mixture. The tubular intercondenser and aftercondenser are combined in a single shell, since the total surface required is usually quite small. Surface condenlilation is mandatory where the vacuum exhaust cannot be mixed with the cooling water for reasons of corrosion or chemical reaction.
Aftercondem;er Drain Trap
Fro. 14.20. Singlestage ejector with surface aftercondenser serving a surface condenser. (Foster Wheeler Corporation.)
In the operation of a steam turbine or engine the surface condenser discussed in Chap. 12 actually serves as the precondenser for the maintenance of vacuum across the turbine. The surface condenser yields not only condensate but also a mixture of air saturated with water vapor, which must be continuously removed. Failure to remove the air causes an increase in the pressure and temperature of the condenser, as well as a noncondensable air blanket which reduces the overall heattransfer coefficient. The air removal can be accomplished as shown in Fig. 14.20 with a surface intercondenser and a removal pump. The hotwell pump is provided for lowlevel installation. A counterpart using a lowlevel barometric condenser can also be used. For the design and selection of ejectors Jackson 1 has presented a comprehensive discussion. I Jackson, D. H., Chem. Eng. Proyress, 44, 347352 (1948).
398
PROCESS HEAT TRANSFER
CHEMICAL EVAPORATION
Comparison between Powerplant and Chemical Evaporation. The purpose of the majority of powerplant evaporators is the separation of pure water fl·om raw or treated water. The impurities are continuously withdrawn from the system as blowdown. In chemical industry the manufacture of heavy chemicals such as caustic soda, table salt, and sugar starts with dilute aqueous solutions from which large quantities of water must be removed before final crystallization can take place in suitable equipment. In the powerplant evaporator the· unevaporated
Feed,wF,c
.t
Stec:~m
WsAS
WFwl
WF·W,Wt
Drips w2
wrw,·wtwa
(OI)·FORWARD FEED
( b1 BAC~WARD FEED Fw. 14.21. feed.
Quadrupleeffect chemical evaporator arranged for forward and backward
portion of the feed is the residue, whereas in the chemical evaporator it is the product. This leads to the first of several differences between powerplant and chemical evaporation. These are as follows: Absence of Blowdown. Chemical evaporators do not operate with blowdown, and instead of liquid being fed in parallel to each body it is usually fed to multipleeffect systems in series. Common methods of feeding are shown in Fig. 14.21a and b. The feed to the first effect is partially. evaporated in it and partially in each of the succeeding effects. When the liquid feed flows in the same direction as the vapor, it is forward feed, and
EVAJ>ORATION
399
when fed in the reverse direction, it is backward feed. From the standpoint of effectively using the temperature potentials forward feed is preferable. If the liquid is very viscous, there is an advantage to the use of backward feed, since the temperature of the first effect is always the greatest and the corresponding viscosity will be less. The advantages and disadvantages of both will be discussed later. The absence of blowdown enables greater heat recovery in a chemical evaporator. Boiling Point Rise (BPR). Although chemical evaporators are capable of high heat efficiencies, they are incapable under certain conditions of a high utilization of temperature potentials and consequently require greater surfaces. This is due to the fact that a concentrated aqueous solution undergoes a boiling point rise above the saturation temperature corresponding to pure water at the same pressure. Suppose steam enters the tubes or chest of a chemical evaporator at 45 psia and is to evaporate water from a caustic soda solution. The steam temperature is 274°F. If pure water is evaporated at 18 in. Hg, the temperature .of the vapor formed woUld be 169°F. But because of the dissolved salt the liquor boils at 246°F at 18 in. Hg instead of at 169°F. The temperature difference across the heattransfer surface is only 274  246 = 28°F, and the difference of 246  169 = 77°F represents lost potential which cannot be attained owing to the presence of the dissolved material. The difference between i;he temperature of heating vapor and the saturation temperature corresponding to the pressure of the evaporating vapor is the apparent temperature drop (Ll.t)a, or 274  169 = 105°F in the example above. Heattransfer coefficients, which are reported on a basis of Q 1 UD  A(At)a
are apparent overaU coefficients. If coefficients are based on the temperature difference across the heating surface between the heating vapor and the evaporating liquid, as in most cases; U » = Q/A At, where At = 28°F in the example above. If solutions have boiling point rises above about 5°F, the latent heat of vaporization of steam from the solution differs from that obtained from the steam tables (Table 7) at the saturation pressure of the vapor. The latent heat of vaporization for steam from a . solution can be computed either from Duhring's relationship or from the equation of Othmer. 1 According to Duhring's rule, (14.5) I
Othmer, D. F., Ind. Eng. Chem., 32, 841856 (1940).
400
PROCESS HEAT TRANSFER
where A.. = latent heat of I lb of pure water from solution at the temperature t and pressure p. Xw = latent heat of l.lb of pure water at the temperature t:V but at p., the same pressure as t t~, t:r = boiling points of the solution and water at the same pressure, p., OR ilt~jilt:V = rate of change of the two boilingpoint curves over the same pressure range According to Othmer's method and based on the ClausiusClapeyron equation, x. d log p. (14.6) A.w = d log Pw where p. and pw are the respective absolute vapor pressures of the solution and pure water over an identical range of temperatures. The BPR ca,n be computed but only for dilut~ solutions which are relatively ideal. For real solutions the data on boilingpoint elevation must be obtained experimentally by measuring the vaporpressu;re curve for a given concentration at two different temperatures. Additional determinations can be made for other concentrations if more tha.n a single effect is used. Fruid Properties. In the powerplant evaporator the watersoftening process is modified in different localities so that the evaporator feed composition causes a minimum of foaming and other operating difficulties. In the chemical evaporator the residue, a concentrated solution, is the desired product, and usually no adjustment can be made to the solution to prevent foaming or to eliminate the deposition of scale. This must be taken into consideration in the design of the equipment. Furthermore, concentrated solutions, as discussed in Chap. 7, produce liquors of high viscosity. Particularly since boiling is a combination of vaporization and free convection the overall coefficient of heat transfer is a function of both the concentration and the temperature at which evaporation occurs. The influence of viscosity may be so great that a negligible Grashof group, D 3p2g(3 ilt/J.1. 2, results for evaporators operating with natural circulation. Under these circumstances the liquor cannot be relied upon to circulate very rapidly about the heating element and it is necessary to use forced circulation instead of natural circulation as presumed heretofore. CHEMICAL EVAPORATORS
Chemical evaporators fall into two classes: natural circulation and forced circulation. N atural;.circulation evaporators are used singly or in multiple effect for the simpler evaporation requirements. Forced
EVAPORATION
401
circulation evaporators are used for viscous, salting, and scaleforming solutions. Naturalcirculation evaporators fall into four main classes: 1. 2. 3. 4.
Hori:rontal tube Calandria vertical tube Basket vertical tube Long tube vertical
The discussion of evaporator. design in this chapter deals only with those which are designed on a basis of flux and accepted overall coefficients. Those employing film coefficients are treated in the next chapter. Horizontaltube Evaporators. Horizontaltube evaporators as shown in Fig. 14.22 are the oldest type of chemical evaporator. Although they once enjoyed widespread use, they have given way to other types. They consist of a round. or square shell and a horizontal tube bundle which is usually square. They do not take very good advantage ofthe thermal currents induced by heating and therefore. are not so acceptable as thetypeswhichhavereplaced them. The horizontal evaporator is the only distinct type of chemical evaporator employing steam in the tubes. The principal advantages of horizontal evaporators lie in the relatively small headroom they require and the !'Lbility to arrange the bundle so that air.brought ii::t with the steam FIG. 14.22. Horizontaltube evaporator. (Swenson Evaporator Company.) does not collect and blanket useful surface. The horizontal evaporator is least satisfactoryJor fluids which form scale or deposit salt, the deposit being on the outside of the tube, and it is therefore used only for relatively simple problems of concentration rather than for the preparation of a liquid for ultimate crystallization. It is well suited to processes in which the final product is a liquor instead of a solid, such as industrial sugar sirups, where the large volume of liquid stored in the evaporator can permit a close adjustment of the final density by changing the holdup in the evaporator. The tube length is determined by the size of the evaporator body itself. Because evaporation occurs on the outside of the tubes, eliminating a scale problem inside the tubes, the horizontaltube evaporator uses smaller tubes than any other type, from % to lX in. OD. Calandriatype Evaporators. The calandria evaporator is shown in Fig. 14.23. It consists of a short verticaltube bundle, usually not more
402
PROCESS HEAT TRANSFER
than 6'0" high, set between two fixed tube sheets which are bolted to the shell flanges. The steam flows outside the tubes in the steam chest, and there is a ·large circular space for downtake in the center of the bundle whereby the cooler liquid circulates back to the bottom of the tubes. The flow area of the downtake is from onehalf the flow area of the tubes to an area equal to it. Tubes are large, up to 3 in. OD, to Sfeom reduce the pressure drop and permit rapid circulation and are installed in tube sheets with ferruletype packing. The layout of a typical calandria is shown in Fig. 14.24. One of the problems is to bafHe the steam chest so that there is a FIG. 14.23. Calandriatype evaporator. (Swe7U!on Evaporator Company.) relatively uniform tube coverage.

Steam in/ef
f)ownfr;;rke
Fro. 14.24. Typical calandria baffiing. Arrows indicate direction of steam flow. areas indicate location of noncondensable bleed points.
Shaded
403
EVAPORATION
Anotheris to provide adequate bleed points so that no pockets of noncondensable gas develop. The condensate is removed at any convenient point. The space above the liq\lid level on the tube sheet serves primarily to disengage the liquid, which is carried along by the vapor. A common accessory of evaporators is a catchall which is installed on the vapor line for the purpose of removing entrained liquid and returning it to the bulk of the fluid. Two typical catchalls are shown in Fig. 14.25a and b. They operate on the principle of centrifugally removing the liquid droplets. Calandria evaporators are so common· they are often referred to .as standard evaporators. Since scaling occurs inside the tubes, it is possible to use the standard evaporator for more rigorous services than the hori_vapor out
i
~~~ lJ:G. 14.25a.
Catchall with bottom outlet.
l;'IG.
14.251t.
Catchall with top outlet.
zontal tube evaporator and, in addition, a propeller can be installed in the dished or conical bottom to increase the rate of circulation. Baskettype Evaporators. A baskettype evaporator is shown in Fig. 14.26. It is similar to a calandria evaporator except that it has a removable bundle which can be cleaned quite readily. The bundle is supported on internal brackets, and the downtake occurs between the bundle and the shell instead of in a central downtake. Because the tube sheets hang freely, the problem of differential expansion between the tubes and the steam chest shell is not important. This type frequently is designed with a conical bottom and may also have a propeller installed to increase circulation. As a result of these mechanical advantages the basket evaporator can be used for liquors which have a tendency to scale, although they are not recommended for liquids with high viscosities or great rates of scaling. The selection of a basket or calandria evaporator usually follows the established policies of the different industries in which they are used after years of experience, or modifications suggested by manufacturers. Some manufacturers have preferences for the one type
404
PROCESS HEAT TRANSFER
in a certain application, whereas another will prefer the second type for the same service. Longtube Vertical Evaporators. A longtube vertical evaporator is shown in Fig. 14.27. It consists of a long tubular heating element designed for the passage of the liquor through the tubes but once by natural circulation. The steam enters through a vapor belt as dis

Feed
FIG. 14.26. Baskettype evaporator. (Swenson Evaporator Company.)
FIG. 14.27. Longtube vertical evaporator. (General American Transportation Company.)
cussed in Chap. 12, and the bundle is baffied so that there is a free movement of steam, condensate, and noncondensable downward. The upper tube sheet is free, and just above it there is a vapor deflector to reduce the entrainment.· This type of evaporator is not especially adapted to scaling or salting liquors, but it is excellent for the handling of foamy, frothy liquors. The velocity of the vapor issuing from the tubes is greater than in the shorttube vertical types. Tubes are usually 17.4: to 2 in. OD and from 12 to 24ft in length. When arranged for recirculation the apparatus will be as shown in Fig. 14.28. In this type, disengagement
EVAPORATION
405
occurs outside the evaporator body. The calculation of this type of evaporator for aqueous solutions with known physical and thermal properties will be discussed in Chap. 15. Forcedcirculati,on Evaporators. Forcedcirculation evaporators are made in a variety of arrangements as shown in Figs. 14.29 through 14.31. Forcedcirculation evaporators may not be so economical to operate as naturalcirculation evaporators, but they are necessary where the concen
FIG. 14.28. Longtube recirculation evaporator. (General American Transportation Company.)
FIG. 14.29. ·Forcedcirculationtypo evaporator with inlride vertical heating element. (Swenson Evaporator Company.)
tration problem involves a solution with poor flow, scale, and thermal characteristics. Since the Grashof group varies inversely with the square of the viscosity, there is a limit to the viscosities of solutions which will naturally recirculate. With very viscous materials there is no alternative but to use this type of evaporator. Also, where there is a tendency to
406
PROCESS HEAT TRANSFER
form scale or deposit salts, the high velocities obtainable by the use of circulating pumps are the only means of preventing the formation of excessive deposits. Forcedcirculation evaporators are well adapted to a close control of flow, particularly when a long time of contact may be injurious to the chemical in solution. Tubes for forcedcirculation evaporators are smaller than in naturalcirculation types, usually not exceeding 2 in. OD.
14.30. Forcedcirculation evaporator with vertical external element. American Transportation Corporation.)
FIG,
(General
In Fig. 14.27 the steam enters the bundle outside the evaporator body and contacts the tubes at the top of the bundle by means of the annular space provided. A deflector plate is installed above the upper tube sheet, and the circulating pump is installed at ground level. In Fig. 14.30 the same effect is produced by means of a vertical external bundle, which simplifies construction to a degree but which is not so compact. Figure 14.31 is a variation with a horizontal bundle which is particularly adaptable where the headroom is low.
EVAPORATION
407
Effect of Hydrostatic Head. Consider a pure 'fluid with a boiling surface somewhat above the top of a bundle of horizontal tubes. The boiling point is regarded as being set by the pressure at the liquidvapor interface. If there is a great layer of liquid above the tube bundle, it will exert a hydrostatic pressure upon the liquid in contact with the tube surface. The added pressure upon the liquid raises the boiling temperature
FIG. 14.31. Forcedcirc.ulation evaporator with horizontal external element. American Tramportation Corporation.)
(General
at the heattransfer surface above that necessary to produce vapors of the saturation temperature corresponding to the pressure at the liquidvapor interface. The effect of hydrostatic head, as in the case of BPR, reduces the useful temperature difference effective upon the heattransfer surface. Since evaporators operate on fixed apparent temperature differences, the size of the heattransfer surface must be increased accordingly for the presence of a hydrostatic head. The influence of temperature and pressure on theoverall coefficient is essentially that shown in Fig. 14.7.
408
PROCESS HEAT TRANSFER
At high temperatures the liquid is less viscous and is more favorably suited to evaporation. The effect of hydrostatic head may be estimated from
(14.7) where f:..th = hydrostatic elevation of the boiling point, °F TR = solution boiling temperature, 0 R v = specific volume of water vapor at TR ft3/lb X. = latent heat of vaporization corresponding to the saturation pressure, Btu/lb f:..p = hydrostatic head, ft Usually f:..p may be taken as corresponding to onehalf the indicated liquid level. It is apparent that the influence of hydrostatic pressure will be greater as the vacuum upon the system is increased, since v varies considerably with the pressure whereas X. varies. but little. For' all evaporators operating with natural circulation a ;loss in available capacity due to hydrostatic head cannot be avoided but the loss can be reduced by maintaining the lowest liquid levels consistent with the efficient operation of the equipment. If the froth is regulated to 10 in. above the upper tube sheet of vertical heating elements, good operating control can usually be ~ffected. The achievement of good operating control is facilitated by the choice of the tube diameter and length and in general by designing for a high fluid velocity into the disengaging space. Problems of arrangement for natural circulation will be treated in Chap. 15. Multipleeffect Chemical Evaporation. In the study of parallelfeed multipleeffect powerplant evaporators it was shown that in a tripleeffect evaporator 1 lb of steam evaporated approximately 2.25 lb of water. The use of parallel feed is by no means the most economical and is used in chemical evaporation only when the feed solution is nearly saturated to begin with and evaporation is intended for the purpose of supersaturation. In chemical evaporation it is customary to employ forward feed, backward feed, or a modification of both, known as mixed feed. Returning to Fig. 14.21 there a;re certain opposing advantages and disadvantages resulting from the use of either forward .or backward feed. In forward feed if the feed liquor is at a higher temperature than the saturation temperature of the first effect, some evaporation will occur automatically as vapor flashing. Since a vacuum is usually maintained on the last effect, the liquor. flows by itself from effect to effect and a.
EVAPORATION
409
liquor removal pump is required only on the last effect. Similarly, since the saturation temperature of the boiling solution in each effect is lower than the temperature of the effect preceding it, there is flashing or "free" evaporation in each succeeding effect, which reduces the overall steam requirement. In ~n evaporator the boiling film is the controlling resistance and the numerical value of the overall coefficient decreases with concentration because the viscosity is increasing. In forward feed the concentrated liquor is in the last effect, and obviously that effect has the lowest overall coefficient, since the liquor is most concentrated there and at the same time coldest. When backward feed is employed, it overcomes the objection of having the most concentrated liquor in the coldest effect. Here the dilute liquor enters the last and coldest effect and leaves concentrated in the first effect, which is at the highest temperature. In this feed, arrangement liquor must be heated in each effect as compared with solution flashing in each effect in forward feed. Furthermore, the feed must be pumped from effect to effect, which means that the number of places for air leakage, such as the pump and flanges, increases the maintenance and power cost. The temperature relations in backward feed usually offset these disadvantages in part, since the system is in counterflow and the economy of steam is greatest under these conditions. If the feed liquor to a backwardfeed evaporator is initially hot, its introduction into the last effect is wasteful, since the vapors which flash off in the last effect are lost to the condenser. In forward feed not only would these vapors flash but in each succeeding effect they would reevaporate additional water. The problem of the direction of feed to be employed is, as in most alternate problems of heat transfer, an economic one. Backward feed may or may not lead to smaller surface requirements, depending upon the extent of concentration and the viscosity of the desired final solution. The steam cost will be less for backward feed if the feed is cold and less for forward feed if the feed liquor is at approximately the operating temperature of the first effect or higher. The computation of problems by both methods will readily establish the most favorable operating relationship. THE CALCULATION OF CHEMICAL EVAPORATORS BY JOSEPH MEISLERl
Referring to Fig. 14.21a the surface and steam requirements for multipleeffect chemical evaporation can be computed by imposing a heat balance across each effect individually and a material balance 1
The Air Reduction Company, Inc., and the Polytechnic Institute of Brooklyn.
410
PROCESS HEAT TRANSFER
over the whole system. for a quadruple effect: CF
The following nomenclature will be employed
= specific heat of feed, = temperature of feed,
Btu/(lb)( 0 F) °F
tF WF = feed, lb/hr T s = saturation temperature of steam to first effect, °F Ws = steam to first effect, lb/hr WI4 = total water removed by evaporation, lb/hr cl, c2, Cs, C4 = specific heat of liquor in effects 1 to 4, Btu/ (lb) eF) tl, t2, ta, t4, = boiling points of liquor in effects 1 to 4; °F WI, w2, wa, w4 = water removed in effects 1 to 4, lb/hr
Assume that there are no chemical heat effects as a result of the concentration (i.e., negative heats of solution) and that there is no BPR. Forward feed
Heat balance on first effect: W.sXs
+ WFCF(tF 
h)
= w1>.1
(14.8)
Heat balance on second effect: wr>.I
+ (wF 
WI)ci(h  t2) = w2>.2
(14.9)
Heat balance on third effect: w2X2
+ (WF 
WI  w2)c2(t2  ta)
= wa>.s
(14.10)
=
(14.11)
Heat balance on fourth effect: wa>.a
+ (wF ....,. W1 
W2 :.._
Wa)ca(ta 
t4)
W4A4
Material balance: (14.12)
The surface requirements will be A _ 1
A2

=
Aa= A4
=
Q
_
Ws>.s
UDAt U1(Ts h)
w1Xr U2(h  t2) w2>.2 Ua(t2  ta)
WaAa U4(ts  t4)
(14.13)
411
EVAPORA'l'ION
Let (14.14)
where U 1, U 2, U 3, and U 4 are the design overall coefficients in the respective effects. From the material balance and the heat balances there are five equations and five unknowns: W s, w1, w2, wa, and W4. These may be solved for simultaneously. Backward feed. Referring to Figure 14.21b: Heat balance on fourth effect: (14.15)
Heat balance on third effect: W2A2
+ (WF 
W4)c4(ta 
t4)
= WaAa
(14.16)
Heat balance on second effect: W1X1
+ (WF 
W4 
Wa)ca(t2  ta) = W2A2
(14.17)
Heat balance on first effect: WsXs
+ (wF W4 
Wa. w2)c2Ch  t2)
= w1X1
(14.18)
Material balance: (14.19)
The surface relations will be the same as before, since it is practical to impose the restriction that the surface in each of the bodies be identical. It has also been shown by experience that under this condition the pressure differences between effects will be approximately equal. If steam enters the first effect of a quadrupleeffect evaporator at atmospheric pressure and the last effect is at 26 in. Hg vacuum corresponding to 1.95 psia, the pressure difference between the steam and the first effect and from effect to effect will be (14.7  1.95)/5. This will enable selection of the saturation pressures in the individual effects. Since the heattransfer coefficients will be different in the individual effects, it may be found that the surfaces defined by Eqs. (14.13) and (14.14) are unequal. This means that owing to the inequality in the overall coefficient in the different effects the t:..t across each effect does not correspond to the assumption of an equal division of the total pressure differential. This will be particularly true ·when the overall coefficients in the different effects differ greatly or when there is considerable flashing in the first effect. To equalize the surface in each body the temperature differences in the individual effects can be adjusted so that a larger temperature
412
PROCESS HEAT TRANSFER
difference will be employed in the effect having the lowest heat transfer coefficient, the heat loads in all effects remaining nearly equal. Multipleeffect evaporators may be designed for minimum· surface or minimum initial cost. These cases have been treated by Bonilla. 1 The design of multipleeffect evaporators for optimum conditions, however, is more the exception in industry than the rule, the trend being in the dl.rection of standardization. Example 14.2. Calculation of a Tripleeffect Forwardfeed Evaporator. It is desired to concentrate 50,000 lbjhr of a chemical solution at l00°F and 10.0 per cent solids to a product which contains 50 per cent solids. Steam is available at 12 psig, and the last effect of a tripleeffect evaporator with equal heattransfer surfaces in each effect will be assumed to operate at a vacuum of 26.0 in. Hg referred to a 30in. barometer. Water is available at 85°F for use in a barometric condenser. Assume a negligible BPR, an average specific heat of 1.0 in all effects, the condensate from each effect leaves at its saturation temperature and that there are negligible radiation losses. Calculate: (a) Steam consumption, (b) heating'surface required for each body, (c) condenser water requirement. The accepted overall· coefficient~ of heat transfer for the different effects will be U1 = 600, U2 = 250, and U~ = 125 Btu/ (hr)(ft2)("F),
Solution: Total feed Wp = 50,000 lb)hr Total solids in feed = 0.10 X 50,000 = 5000 lbjhr 5000 Total product = . = 10,000 lbjhr 0 50 Total evaporation, w 1_, = 50,000  10,000 = 40,000 lb)hr Cp = 1.0. 'l.'he balances applying to this problem are First effect: W s}.s + WF(tF  t,) = w,x, Second effect: wzXz + (wp  wr)(tr  t2) = w,x. Third effect: w,}.. + (wp  W1  wo)(t.  to) = w,}., Material: wz + w 2 + ws == W1• tp = 1oo•F · Ts at 12 psig = 244°F T, at 26 in. Hg (1.95 psia) ,; 125°F Total temperature difference = l19°F When a forwardfeed multiple~effect evaporator employs equal surfaces in each effect, as noted above, experience indicates that the differences in the pressures between effects will be nearly equal. This will rarely be entirely true, but it forms an excellent starting point for the calculation of the pressures in the effects. Any discrepancies can be adjusted later. Average pressure difference = 25.70  1.95 = 8 .25 ps1·;effect 3 1
Bonilla, C. F., Trans.AIChE, 41, 529537 (1945).
413
EVAPORATION BREAKUP OF THE TOTAL PRESSURE DIFFERENCE
Pressure, psia
Steam chest, 1st effect ...... 26.70 Steam chest, 2d effect ....... 18.45 Steam chest, 3d effect ....... 10.20 (20. 7 in. Hg) 1. 95 (26 in. Hg) Vapor to condenser ••....•..
AP,
psi
.... 8.25 8.25 8.25
Steam or vapor,
OF
Ts = 244 tt =.224 t. = 1,!)4
ta = 125
"A, Btu/lb
AS= 949
961 981 As= 1022
il.t =
x.
=
949Ws + 50,000(100  224) = 961w 1 96lw1 + (50,000  Wt)(224 ...;.. 194) = 981w 2 98lw• + (50,000  Wt wo)(194 125) = 1022w 3 w, + Wo + wa = 40,000 Solving simultaneously, W! = 12,400 w. = 18,300. w. = 14,800 Wt8 = Wt + Wo + Ws = 40,000 Ws = 19,100
A,= A. =
Aa =
Ws;>..s = 19,100 X 949 = 1510 ft• U,(Ts  t,) 600 X 20 WtAl
U o(t,  t.) WoAo Ua(to  ta)
= 12,400 X 961 = 1590
250 X 80
= 13,300 ·
X 981 125 X 69
= 1510
(Use 1600 ft 2 /effect)
Heat to condenser = Wsil.s = 14,300 X 1022 = 14,710,000 Btujhr Water requixement = 14,710,000/(120  85) = 420,000 lbfhr or 420,000/500 = 840 gpm Economy, lb evaporationflb steam = 40,0oo/19,100 = 2.09lbflb Remarkll. During operation the equal pressuredrop distribu.tion may not maintain itself. This will occur if there is undue scaling in one of the effects, if a body is gas bound, or if liquor levels are not properly maintained. Another factor may be the withdrawal of a large quantity of steam from one of the effects as a source of lowpressure heating steam. Any deviation from an equal pressuredrop distribution does not mean that the entire multipleeffect assembly will fail to·operate but instead that the unit will assume a new pressure distribution and operate with a reduced capacity and steam economy.
Nonalgebra.ic Solution of Evaporators. It will presently be shown that the use of· a simultaneous ~ebraic solution such as that above can acarcely be applied advantageously to more complicate. 1 + (wF  wo)(t,  ta} = W•A• First effect: W8A8 + (wF  wa  w,)(t,  t,) = w,Jit
415
EVAPORATION Material: W1
+ Wz + Wa .= Wl3
981w. + 50,000(100  125) = 1022w 3 961wl (50,000  wa)(125  194) = 981w• 949Ws + (50,000  W3  w.)(194  224) = 96lw 1 W1 + W2 + Wa = 40,000 W1 = 15,950 w. = 12,900 w. = 11,150 W1•. = Ui! + W2 + w 8 = 40,000 Ws = 16,950
+
A1
=
16,950 X 949 = 2010 ft2 400X20
A 2 = 15,950 X 961 = 2040 ft2 250 X 30 A 3 = 12,900 X 981 = 1050 ft2 175 X 69 Since the heating surfaces for the three effects are far from equal, the temperature differences employed in the effects must therefore be modified to meet the conditions of the problem. For the first trial the average surface was (2010
+ 2040 + 1050)
=
1700 ft~
3
With a better distribution of temperatures and pressures, however, less than 3 X 1700 = 5100 ft2 of surface may be expected, since the At in both of the first two effects will be improved at the expense of pruy the last effect. Recalculation:
Assume an average surface of 1500 ft 2/effect, and find the temperature differences which will provide these surfaces. Assume First effect: Ts it = 28°F, At = 2%s X 2010 = 1450 ft 2 Second effect: t1  t 2 = 41 °F, A2 = a%1 X 2040 = 1490 ft 2 Third effect: t2  t 8 = 50°F, A. = 6%o X 1050 = 1440 ft 2 Ts  ta = l19°F. The new temperaturepressure distribution is Pressure, psia Steam chest, 1st effect ........... Steam chest, 2d effect ........... Steam chest, 3d effect ........... Vapor to condenser .............
Steam or vapor, °F
26.7 Ts = 244 16.0 tl = 216 16.4in. Hg t. = 175 26.0in. Hg t. = 125
X, Btujlb
949 968 992 1022
PROCESS HEAT TRANSFER
416 Solving again for
w,,_ w,, wa,
WI=
W:
and W s,
15,450
= 13,200
Wa = 11,350 Ws = 16,850
A 1 = 1450 A,= 1470 Aa = 1490
(Use 1500 ft 2/effect)
Heat to condenser = 11,350 X 1022 = 11,600,000 Btu/hr Water requirement = 11,600,000/(120  85) = 332,000 lb/hr
= 332,000/500 = 664 gpm
Economy, lb evaporation/lb steam = 40,000/16,850 = 2.37lb/lb COMPARISON OF FORWARD AND BACKWARD FEED
Total steam, lb /hr .............. Cooling water, gpm ............. Total surface, ft' ...............
Forward
Backward
19,100 840 4,800
16,850 664 4,500
'l:he operating conditions for both forward and backward feed are shown in Fig. 14.32a and b. · Substantiating the simple reasoning, backward feed is more effective thermally than forward feed. Omitted, however, is the maintenance and investment on backwardfeed pumps between each effect and the problems of air leakage and flow control. All of these are considerably greater in backward feed.
Optimum Number of Effects. The greater the number of effects the larger the amount of evaporation per pound of steam admitted to the first effect. The operating costs will be less the larger thenumber of effects. This is offset, however, by the increased first cost of the apparatus and increased maintenance charges for cleaning and replacement, both of which enter as fixed charges. Supervisory labor will be the same for the operation of any number of effects. The cost of condenser water must also be included, and it, too, will decrease the greater the number of effects employed. The optimum number of effects may be obtained by computing the process requirements with two, three, four, or up to six or eight effects and determini:Q.g the fixed charges and operating cost resulting from each arrangement. When the total cost is plotted against the number of effects, a minimum will occur corresponding to the optimum number of effects. Actually, however, the number of effects in the various heavy chemical industries are fairly standardized. For example, table salt is concentrated in four effects in which parallel feed of liquor is used, caustic soda with two or three effects and backward feed, and sugar with five or six effects and forward feed. Except when introducing an entirely new chemical process it will rarely be necessary to carry out a complete economic analysis.
EVAPORATION
417
Bleed Steam. In certain industries and particularly in the production of sugar, there is a great need for lowpressure steam, say 10 to 15 psia, for the large amount of liquor preheating required throughout the plant. For the preheating it is found advantageous to use some of the steam from the first or subsequent effects for various preheating services such as in the decolorization of sugar sirup. Since any vapor formed in the first and later effects will have already been used one or more times for
w"=sqooo J1:S=I~!OO
(a)
FIG. 14.32a.
Example 14.
Forward feed.
(b)
FIG. 14.32b.
Example 14.
Backward feed.
evaporation, the cost per Btu of the vapor is cheaper than that of fresh steam. Steam can be bled economically from one or .more effects for these additional services while reducing the overall heating cost of the plant. To account for this in the heatbalance equations [Eqs. (14.8) through 14.18)] one need only include the term WB for the pounds per hour of bleed steam in the effect in which it occurs. The quantity of bleed steam must also be introduced into the material balance of Eqs. (14.11) and (14.18). To prevent the introduction of one or more
418
PROCESS HEAT TRANSFER
unknowns than there are simultaneous equations, W B should not be entered as another unknown but as a definite number of pounds of bleed steam or as a percentage of the total evaporation if the calculation is for an existing installation. The Solution of Industrial Problems. 1 The preceding examples have served to introduce the elementary methods of calculation and the principles of multipleeffect evaporation. Actually, industrial problems are rarely so simple. Instead, the evaporator system must be integrated with the operation of the entire manufacturing process, and this complicates the calculation greatly. The element of experience is essential to the completion of a calculation within a reasonable period of time. In the remainder of this chapter several of the most widely met commercial problems will be analyzed. These are sugar concentration, wasteliquor evaporation in the paper pulp industry, and the production of caustic soda. The analyses involved in their solution should be readily adaptable to the majority of other problems. A fourth process, distillery waste concentration, has been omitted in favor of evaporation by therinocompression. The methods indicated may be employed either for the design of a new evaporator or for the calculation of performance in an existing evaporator. THE CONCENTRATION OF CANESUGAR LIQUORSFORWARD FEED
Description of the Process. It is the practice in canesugar production to strain the juice containing the sugar after it has been pressed from the sugar cane and clarified chemically. In the initial stage of clarification, called defecation, large amounts of colloids and inorganic and organic salts are removed. This is accomplished by the addition of lime to the juice at 200°F, forming a heavy slime which is removed by settling, decantation, and filtration. The clear solution is then fed to the multipleeffect evaporator for concentration. Example 14.4. An evaporator installation is to have a capacity for concentrating 229,000 lbjhr of 13°Brix (degrees Brix is the per cent by weight of sugar in the solution) to 6o•Brix, at which concentration it will be decolorized. This quantity of sugar solution results from milling 2300 tons of sugar cane per 20hr day. The raw juices will be heated from 82 to 212°F by the use of exhaust vapors bled from the first and second effects. Steam will be available to the first effect at 30 psig. Solution. The arrangement of equipment is shown in Fig. 14.33 and is justified by the fact that bled vapor is a cheaper preheating medium than fresh steam. In this analysis it is arbitrarily assumed that 37,500 lbjhr of 15 psig vapor is bled from the first effect for use in the vacuum pans in the crystallizing section of the plant. The 1 Readers who are not directly interested in chemical evaporation problems may omit the remainder of this chapter without affecting the context of later chapters.
Skam i!!'ecfoJ.
...a::
.J:
~~
\1~·~
"16 .,
Z;~'~ '1)"5
! t>;J
~ I
.
~r
r '
L1...~
drt:ri'ns
•,
Condensate
ff
I
Condensafe+ft\ boiler house
~
* *
. .........H.1,..'w..Jw.... . f l ,..w,..1 'f t t ! ;.~~~~~~~~~~~'========':.~J_~ ____________ J ! Condensate t
! i :~'
!
1 I
~
~
auf
1
~Condens01fe
L:,..l:,.
out
L1([uor cl!scharqe
"19.600 lb/hc oo•ex FIG. 14.33,
Sixbodr sugar evaporator with preheaters.
fl>. .....
c:o
420
PROCESS HEAT TRANSFER
BPR and specific heats of sugar solutions are given in Fig. 14.34. A flow diagram such as that in Fig. 14.33 is usually arrived at from experience or from preliminary l!tudies of total surface requirements as in Table 14.1. TABLE
14.1.
AVERAGE EvAPORATIOl'l PER SQuARE FooT HEATil'lG SURFACE FOR
SUGAR EvAPORATORS
Water evaporated, lb I (hr) (jt•) 1416 68
Effects
1 2 3 4
56
45 34
5
In the matter of choosing the number of effects from experience, it is the praetice to use a straight quadruple effect if little or no vapor bleeding is required and to add a "preevaporator" as' a first effoot if excessive bleeding of low pressure vapor are :required for preheating feed, etc., or if relatively lowpressure live steam is. bled for other plant processes, i.e., vacuum pans. 1 The forwardfeed arrangement is common in the sugar industry, since stronger \uices or. sirups are sensitive to high temperatures. This procedure places a severe handicap on the last effect of a multipleeffect evaporator where the most viscous juice
10
J
v 10
.............
zo
/
30
/
v
v
.....
1.0
...........
fg 0.9
"'
~0.8 .., ~0.1
~ .. ~
Complete vetpor
00
bype~ss
~
g ~
CQ
~
::..:
!oil
~
..,.
~
Prod.
N
~~ flash /'' \li '
....,
60%
g 0.680
..r;:
~
10%
·g 0.640 0.
V)
0.60075
100
125
150
115
200
Temperc:~ture,
225 °F
250
215
300
(a) FIG. 14.40a.
S_pecific heats of highconcentration caustic soda solutions.
FiG. 14.40b.
Specific heats of lowconcentration caustic soda solutions.
(Columbia
Alkali Corporation.)
noncondensables from the system eventually reach the condenser through a system of vent piping, and there they are removed by a steam jet ejector n. Example 14•• 6. Electrolytic cell liquor containing 8.75% NaOH, 16.60% NaCl, and 74.65% water at a temperature of 120°F is to be concentrated to a product containing 50% NaOH, 2.75% NaCI, and 47.25% water at a 'temperature of 192°F. The
438
PROOESS HEAT TRANSFER
capacity of the unit is 25 tons per ~hour day based on 100% caustic soda. Physical properties are given in Figs. 14.SS to 14.41. Solution. Table 14.7 is a material balance. It will be observed that the calculations are conveniently based on 1 ton NaOH production per hour, and therefore the 'numerals in Fig. 14.37 are obtained by using a factor of 1.25 greater than those shown on Table 14.7. From the material balance and from the temperaturepressure distribution in Table 14.8 the heat balance (Table 14.9) can be worked out rather easily for this evaporator.
Percent NGIOH Fw. 14.41.
Relative heat content of caustic scda solutions.
(Columbia Alkali Cor
poration.)
In Table 14.8 the total temperature difference between the saturated live steam and 27 in. Hg vacuum is 274  115 = 159°F. From Fig. 14.38 the estimated BPR of the solutions in the first and second effects is 77 and 26°F, respectively, corresponding to concentrations appearing in Table 14.7. Thus, the effective temperature difference is 159  (77 + 26) = 56°F. The latter is arbitrarily divided evenly between the two effects. In addition to the two effects of the evaporator; an additional amount of evaporation is obtained in the forcedcirculation flash tank. The latte.r operates at 27 in. Hg and receives the concentrated slurry from the first .effect and cools it from 246 to 192°F. The last colW:nn of Table 14.8 summarizes the thermal and material quantities for the flash tank.
439
EVAPORATION TAllLE
BGBis: 1 ton/hr NaOH
14.7.
CAUSTIC EvAPORATOR MATERIAL BALANCE
CeU l;g..or a.t 8.75% NaOH 16.6% NaCI 74.65% H,o Total oellliquor
Waeh a.t SO'F
120'F = 2000 = 3800 = 17050  22s5o
25% NaCI 340 75% H,o = 1020 Total wash = Til6ii NaOH
NaCl
H,o, Lb
%
Lb
Total, Lb
Lb
             ·         ·Overall operation: Cellliquor.................. .... .... ....... 8.75 2000 16.60 3800 17,050 22,850 Wash .......•.............................. _._._._._. _._._ .._ 25.00 ~~~ Totalin.................................... . .. .. 2000 .. .. . 4140 18,070 24,210 Product.................................... 50.00 2000 2. 75 110 1,890 4,000 Remo?ed ................................... .....~::::; 4030 16,180 20,210 Weakliquor settler oper.ation:a From 2d eflect.............................. .. .. . 6000 • • • • • 5505 20,775 32,280 From 50% settler ........................... 50.00 2500 ....!.!.!.!..:.~~~ Total in.. .. . .. • .. . • .. .. . .. . . . . . . . . . . . . . . . . . .. .. . 8500 . .. .. 6895 23,136 38,780 Solids ............... : .. .. .. .. .. .. .. .. .. .. .. 24.62 8500 8.22 2835 23,135 34,470 Bait pptd ...•.•............................ ..... ~4000       
Ce;,_t;!u~e~~'ll~~~~~ettler.....................
.. . .. 4000 5395 10,905 20,300 Wash .. , ....•• , ............................ _._._._ .._ _,_,_,_,_~~~~ Totalin.................................... . .. .. 4000 5735 11,925 21,660 Solids ...................................... 23.05 4000 9.84 1705 11,925 17,630 Salt pptd, .....•...•....................... ::::; ....::::: 4030 ~   
2d effect opera.tion:c:
From centrifuge ............................. Cellliquor ................ : ................ Evaporator feed. . . . . . . . . . . . . . . . . . . . . . . . . . . . Evaporation................................ Product .................................... Solids ...................................... Salt pptd .................................. let effect operation:• From weak liquor settler .................... Evaporation.. .. .. .. .. .. .. . .. .. . .. . .. .. .. .. . Product .................................... Solids ...................................... Salt pptd .......•............•.......... , .. Flashtank operation:• From lot effect............................. Evaporation.. . .. .. .. .. .. .. . .. .. .. .. .. .. .. .. Product ...................... ·.............. Solids ....... : .............................. Balt.pptd .................................. From·fiash tank............................... 50%out ................................... ,. Underllow .................................... Saltpptd.................................... SolidS ........................................ • Solution temperature ~approximate) • Solution temperature approximate) • Solution temperature approximate) tl Solution temperature approxims.te) • Solution temperature approximate)
..,. 
23.05 8.75 14.82 .... . .....19.96
4000 P.84 1705 11,!125 17,630 2000 16.60 3800 17,050 22,850 6000 13. 60 "5505 28, 97 5 ~ . . .. . .. .. ... . 8,200 8,200 6000    5505 20,775 32.280 6000 11.22 3370 20,775 30,045 ~ ....::::; 2135 ~   24.62 4500 8.22 1500 12,230 18,230 .. . .. .. .. • .. .. 7, 537 7, 537 .....45oll 1500 ~ 10,693 47.05 4500 3.84 357 4,693 9,560 ::::; ~::::; "''l3a ~ ~ · . .. .. 4500 1500 4,693 10,693 .. . .. • .. • • .. .. .. .. 443 443 .... . 4500 1500 4';250 10, 250 50.00 4500 2.75 248 4,250 8,998 ::::: ~ ~ 1252 ...... 1';252 .. .. . 4500 . .. .. 1500 4,250 10,250 50.00 2000 2.75 110 1,890 4,000 ..... 2500  ....."'i390 2,360 6,250 ..... .... ...;. 1252 ...... 1,252 50.00 2500 ~ ~r.aao {',998
150°F. 145°F. l41°F. 246°F. 19ZOF.
It was noted that in a forcedcirculation evaporator the surface is not a direct function of the temperature difference between the boiling solution and the saturated vapor temperature in the steam chest. The surface in a forcedcirculation evaporator is determined by an economic balance between the size of the evaporator and the rata of circulation or, what amounts to the same thing, the power requirements of th&
pumps.
440
PROCESS HEAT TRANSFER TABLE 14.8.
CAUSTIC EvAPORATOR
SUMMARY*
Effect Item
Flash tank
I 1. 2. 3. 4. 5. 6. 7. 8. .9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 18. 20.
II
Steam pressure, psi ................. 30 Steam temperature, °F .............. 169 274 tJ.t, °F ............................. 28 28 Liquor temperature, °F ............. 141 192 246 BPR, °F .......................... 26 77 77 Vapor temperature, °F .............. 115 115 169 X, Btu/lb .......................... 1,027 1,027 997 13,367 Feed, lb/hr* ....................... 22,788 50,602 Product, lb/hr* .................... 12,813 13,367 40,352 Evaporation, lb/hr* ................ 10,250 554 9,421 Heat flow, Btujhr* ................. 11,890,000 11,020,000 U D, Btu/ (hr) (ft 2 WF) ............... 700 A, ftt ............................. 683 683 Tubes, OD, in. and BWG ........... 1, 16 1, 16 Tube length, ft ..................... 7 7 No. tubes ......................... 432 432 Circulating pump, gpm ............. 3200 at 20ft 3200 at 20ft 167 at 45ft Apparent efficiency, % .............. 64 54 BHP ............................. 35 8.2 38 Motor, hp ......................... 40 40 10.0
* Correspond to actualftows in Fig. 14.37. A method generally used to solve for surface requirements is indicated below. In designing for optimum surface it is customary to use about 8 fps velocity through the tubes of the heater. Furthermore, in this illustration tubes will be 1 in. OD, 16 BWG,. solid nickel, 7'0" long arranged on a twopass layout. The heaters must be designed for sufficient surface to transfer enough heat from the vapor or steam side to satisfy the heat quantities under 11 of Table 14.8. For effect I. 0.8w(t~  246) = 11,840,000 = UDA tJ.t •. That is, by circulating w lb /hr of liquid through the tubes and raising the temperature of the latter to t~, a small temperature differential above the saturatedsolution boiling temperature in the flash chamber, a gentle flash will occur with not too rapid a crystal growth. By assuming a· given value t~, both w and tJ.t are determined. Furthermore, by designing for a given velocity of 8 fps, the value of U c, the clean overall coefficient, is determined from the thermal characteristics of the solution. This is illustrated below: G
= V(a
X 62.5 X 3600)
=8
X 1.5 X 62.5 X 3600
= 2,700,000, lb/(hr)(ft2)
For this mass velocity a tubefilm coefficient approximately equal to 1375 can be obtained from the methods employed in Chap. 7. Combining the latter with a steam film coefficient of approximately 1500 will give a value for Ua or U D equal to 700. Setting up the following table:
441
EVAPORATION
t., •F
w, lb/hr
251 252 252.5 253
2,970,000 2,480,000 2,290,000 2,120,000
flt
Ua
A, ft 2
a,, flow area.
670 680 686 695
0.87 0.88 0.89 0.90
  25.4 25.0 24.7 24.5
700 700 700 700
Goa~.
per pass, ft 2
Uoo.tc
3,420,000 2,820,000 2,570,000 2,520,000
700
Thus the gain per minute for circulation is 3200, and from pressuredrop calculations .in tubes, the static head, etc., a total dynamic head of 20 ft is found to be required. TABLE 14.9.
CAUSTIC EvAPORATOR HEAT BALANCE Basis = 1 ton/hr NaOH
Effect 1. a. Heat in steam = 10,500 X 930 X 0.974
b. Heating liquor = 18,230 X (246  150) X 0.83 ............... c. Resultant heat.: ................... ; d. Heat of concentrate .................. e. Heat of vapors ...................... 2. a. Heat of vapors .............. : . ...... b. Condensate flash = 10,500(274  169) c. Heat flow .......................... d. Heating liquor = 40,480(141  130)0.92 ............ e. Heat of vapors ...................... Flash tank: a. Liquor flash = 10,693(246  192)0.79). .b. Heat of concentrate .................. c. Heat of vapors ...................... Total evaporation ...................... Condenser: Heat of flash tank vapors (above no•F). Heat of vapors 2 (above l10°F) .......... Total heat in (above 110°F) ............. Water flt = (110  90) = 20• ...........
Btu/hr
Evaporation, lb/hr
9,500,000 1,470,000 8,030,000 300,000 7,730,000/997 7,730,000 1,090,000 8,820,000 410,000 8,410,000/1027 460,000 10,000 450' 000/1027
.....
':,
......
7,750
8,200
443 16,393
452,000 8,450,000 8,902,000/20 = 445,000 lb = 890 gpm/(ton)(hr) ofNa.OH
For effect II. 0.92w(t,.  141) = 11,020,000 = U DA At. For a velocity of 8 fps G = 8 X 1.35 X 62.5 X 3600 = 2,430,000 lb/(hr){ft'). Using thermal characteristics for this solution gives an approximate U D = 700 Btu/(hr)(ftt)("F).
442
PROCESS HEAT TRANSFER
As for effect I:
t., °F
w, lb/hr
t:.t
Uc
A, ft2
a,, flow area per pass, ft2
Goa~.
Uoa1o
2,790,000 2,430,000
700
146 146.5
2,400,000 2,160,000
25.4 25.2
700 700
620 683.
0.80 0.89
From the design coefficient above a smaller surface could have been used. However, since it is desirable to duplicate both the surface and pump of the first effect, the larger surface is chosen. THERMOCOMPRESSION
The principle of thermocompression is continuously finding wider application in industry. Consider a singleeffect evaporator. It is fed with steam and generates vapors which have nearly as much heat content as was originally present in the steam. These vapors are condensed with water as a convenient means of removal, but this is a severe waste of both Btu and water. Were it not for the fact that the temperature of the generated vapors is lower than the steam, it would be possible to circulate the vapor back into the steam chest and evaporate continuously without supplying additional steam. But apart from the heat balance, a temperature difference must exist between the steam and the generated vapors or no heat would have been transferred originally. If the vapors from the evaporator were compressed to the saturation pressure of FIG. 14.42. Thermocompression evaporator. (Buffalo Foundry & Machine the steam, however, the temperature Company.) of the vapors could be raised to that of the original steam. The cost of supplying the necessary amount of compression is usually small compared with the value of the latent heat in the vapors. The practice of recompressing a vapor to increase its temperature and permit its reuse is thermocompression. Where fuel is costly, the compression may be accomplished with a centrifugal compressor as in the Kleinschmidt still for the production of distilled water. Where steam is available at a pressure higher than that required in the evaporator, the
443
EVAPORATION
recompression can be effected in a steam jet booster. The latter operates on the principle of an ejector and is used to compress vapors instead of a noncondensable gas. A thermocompression evaporator of the type using a booster ejector is shown in Fig. 14.42. The pressure at which steam is discharged from a thermocompressor depends upon the pressure and proportion at which the live steam and the vapors are supplied. It will be apparent from the analysis below that the higher the desired discharge pressure the greater the percentage of live steam required. Table 14.10 compares the proportions of live steam to entrained vapors for steamchest pressures of 18 to· 26 psig when live steam is available to the jet at 150 psig and the entrained vapors are taken off at 15 psig. TABLE
14.10.
THERMOCOMPRESSOR EcoNOMY
Steamchest pressure, psig ................. 18 Entrainment ratio, lb entrained vapor jib live steam ................................. 3.33 Amount of live steam saved, % ............. 66
20
22
24
2 67
1.43 59
52
26
1.08
87
46
Table 14.10 shows substantial savings of live steam for fixed quantities of lowpressure steam discharged from the compressor. Figure 14.43 represents the variation of the reciprocal of the entrainment ratio with steamchest pressure for entrained vapors at different pressures. Derivation of charts similar to Fig. 14.43 can be made by analyzing the thermodynamic conditions imposed on the jet thermocompressor. Referring to Fig. 14.44, highpressure steam is caused to expand in the nozzle a fro:rn which it issues with a high velocity·into the mixing space b where it transfers some of its momentum to the vapor sucked in. In the diffuser section d, which is the reverse of a nozzle, the mixed vapors are compressed to the steamchest pressure pa. The work of compression results from the conversion of kinetic energy of the highvelocity mixture into pressure head. Thus, vapor at low pressure Pz is entrained and compressed to a higher pressure Pa at the expense of energy in the motive steam. Using the treatment of Kalustian,l let H 1 = enthalpy of steam at P1, Btu/lb H 2 = enthalpy of steam after isentropic expansion in nozzle to pressure p2, Btu/lb H2' = enthalpy of steam after actual expansion in nozzle to pressure P2, Btu/lb e1 = efficiency of the nozzle 1
Kalustian, P., Refrig. Eng., 28, 188193 (1934).
444
PROCESS HEAT TRANSFER Entrained vapor pressure, ps1q
.. .
2& 242220 18 16 14 12 10 ! I' ,' ' '' : ' i
8
Entrained vapor pressure, psiq
.'
6
60 55 50 45 40
35
30
25
20
1/ I 1/ lL. L ./ / / VI5 28 v v / v Vz ~ I I / / / / // ./ / I0 26 v 'I Jl I v / 'I !/ v v / 48 / I / / / / / 24 !J 'I /5 v / / L/ / / / '/ v / v v / /0 rL ILL 22 .,44 v / / J/ ~40 II .,zo I/, '/ / / I / v v / .,.,o ·;;; I 'I v 'I v / / 1/ / I v / ""18 o) '~32 36 I I v / v / / v 'I / / / 516 / 'I ::l I / v I / / 1/ / / ~ 14 8,28 I I v I 1/ v / / I I / ~24 ~12 / vv II I 'I I / (520 ~ .., 10 1/ I / v ;5 8 / / 1/ v 16 (b)100 LB GAGE (al·90 LB GAGE I I / 12 I !/ / 6 OPERATING PRESSURE OPERATING PRESSURE I. I I / 4 8 i
~
v
Q.
0.
1/)
zI
0v 0
v
v
4
I 2 3 4 Lb live steamflb entr01ined vapor
0 0
5
2 4 6 8 Lb live steam/lb entr01ined vapor
Entrained vapor pressure, psig 60 555045 40 35 30 25
20
15
Entre~~Jnecl
10
10
v01por pressure,psitJ
60 5550454035 30 25 20
15
1/. I I / / / v / /5 56 Ill It 1/ / 56 v v v / / r; I / J / 1/ I / I 52 52 'l I I v / / / / 'l 'I I I I 1/ 0 48 48 v I; I 1/ 1/ / / v I I / '/ 44 'l 'I 44 / I v 1/ ... 'lj v j If I / / "' ~40 '[40 fL 1/ / /j v / IV ~ 36 l i 36 1/ I I I / 'I I I I / / ~~ 32 / I ~ 32 J !J / I 1/ / '/ I I ~28 ;v !zs v, / I J ""ls 24 / ~24 'I
5
10
/
/ 0
/ /
:J
..r::
.1§ 20 0
lj
I
vI
I I vv 'I 12 / I 8
£
;520 16
16
4 0
I o
If 1/
4
6
8
FIG. 14.43.
(d)· 150 LB GA OPERATING PRESSURE
'I II 8
4 2
10
/
/, / I
12 /
(c)125 LB GA OPERATING PRESSURE
Lb live steam/lb entr111ined VCIIPOr
I
I
I 0
o
2
4
6
8
10
Lb live steam/lb entr111ined vapor
Thermocompressor steam entrainment.
(Schutte &: Koerting Company.)
EVAPORATION
445
Then _ H 1  H 2' _ actual work of expansion e1  H1 H2 theoretical work of expansion
( 1 ~· 21 )
Let H a = enthalpy of the mixture at start of compression in the diffuser section at p2, Btu/lb H4 = enthalpy of the mixture after isentropic compression from p 2 to the discharge pressure pa, Btujlb e2 = efficiency of compression in the diffuser Then
H4 Ha
e2 = ;.,::..=.actual work of compression
Let M 2 = entrained vapor at pressure P2 lb M 1 = motive highpressure steam, at pressure p1, lb The actual work required for compressing the mixture, of vapors in the diffuser is given by (M 1 M 2) (H 4  H a)/e2. The actual work obtain
+
FIG. 14.44.
Ejector.
able from the expansion, of the highpressure steam is less than the theoretical value (H 1  H 2) because of friction in the nozzle, which is taken into account by the nozzle efficiency e1. There is a further loss in available energy in the transfer of momentum from the highvelocity jet to the relatively slowmoving entrained vapor. If ea is the efficiency of momentum transfer, the net available work from the jet is M1e1ea(H1  H2).
By equating the net available work obtained from the jet to the actual work required for compressing the mixture in the diffuser, one obtains Entrainment ratio,
=
~: = [ ~~: ~:~ e1e2ea 
1],
lb vapor entrained/lb steam
(14.22)
Equation (14.22) must be solved by trial and error, since Hs and H4 are functions of x 3, the quality of the steam at p 3 which is in itself a func
446
PROCESS HEAT TRANSFER
tion of M 2 and M 10 If one assumes no heat and no enthalpy change due to the mixing a balance on the mixing process gives
+ M2) = X2"M1 + x,M2 X2"Ml + X4M2 Xa = M1 + M2
(14o23)
Xa(M1
(14o24)
where X2" is the quality of motive steam after expansion to P2 and after it has lost its kinetic energy in the entraining process, X4 is the quality of the entrained steam, and x 2" is related to X2', the quality after expansion but before the entrainment step, by the equation (14025) where A is the latent heat of vaporization of the fluid at P2o to x 2, the quality after isentropic e;x:pansion, by
X2'
is related
(1  e1)(H1  H2) ;, (x2'  x2)A As an illustration, take the case where P1 = 150 psig, P2 = 9 psig, H 1 = 1196 Btu/lb (from Table 7), H2 = 1050 Btu/lb (after isentropic expansion from P1 to p2). and x2 = Oo885o It will be assumed that the values of e1, e2, ea are, respectively, Oo98, Oo95, and.Oo85o H1  H2' = Oo98(1196  1050) = 143 and therefore H2' = 10530
From Table 7 A = 954 and from Eqo (14o22)
954(x2'  x2) = (1  Oo98)(1196  1050) and therefore X2' = Oo89o
From Eqo (14o25)
(1  Oo85) (143) = 954(x2"  Oo89) and therefore
X2"
= Oo91.
Assuming xa = Oo95, Ha = 1159(0o95) = 1100o
Since the entrained steam at P2 is assumed saturated, X4 = 1 and therefore H, = 1164; hence from Eqo (14022)
~: =
1 6:: (0079)  1
=
1.80 1 = Oo8
As a check on the assumed value of xa, from Eqo (14o24) above Xa
= X2"
+ x,(M2/M1) + M 2/M 1
1
= Oo91
1
+ 0080 =
+ Oo80
1.71 = 0 95 1.8 °
(which checks the assumed value) 0 Since the entrainment ratio M 2/M 1 is a function of the thermal (expansion storage is hot, returned directly to storage. Now suppose that, for greater cleanliness and the attainment of a higher coefficient, the liquid is speeded through the vaporizer at a faster flow rate by means of a larger pump. It may be desired that liquid enter the vaporizer ·at a rate of 40,000 instead of 10,000 lb/hr suggested by the process. Since only 8000 lb /hr of vapor will be produced, only 8000 lb/hr of fresh liquid will continue to come from storage. The difference, 32,000 lb/hr, will have to be supplied by recirculation again and again of unvaporized liquid through the vaporizer. For 40,000 lb/hr entering the vaporizer, only 8000 lb/hr, or 20 per cent will actually be vaporized, and the higher velocity over the heattransfer surface and the low percenta~e vaporized will permit longer operation without excessive fouling.
456
PROCESS HEAT TRANSFER
The advantage of recirculation can be computed economically. The high recirculating rate will increase the power cost but decrease the size of the equipment and the maintenance cost. The vaporizer may also be connected with a disengaging drum without the use of a recirculating pump. This scheme is natural circulation and is shown in Fig. 15.2. It requires that the disengaging drum be elevated above the vaporizer. R~circulation is effected by the hydrostatic head difference between the column of liquid of height Z1 and the column of mixed vapor and liquid of height Za. The head loss in the vaporizer itself due to frictional pressure drop corresponds to z2. The hydrostatic head difference between zs and Z1 is available to cause liquid to circulate at
Coldt>eed
~I
Steam
I
:
j
: : I
:
__t
FIG. 15.2.
Vaporizing process with natural circulation.
such a velocity that it produces a pressure drop z~ in the vaporizer equal to the hydrostatic difference between zs and Z1. The cold feed is usually fed high on the return pipe so that the leg z1 will have as great a density and hydrostatic pressure Z1P as possible. If the pressure of the feed is greater than the operating pressure of the system, the feed will have to join the return pipe at a lower point unless a jet is used, so that there will be a sufficient liquid head above the junction to prevent feed from backing up directly into the drum. Piping the feed directly to the disengaging drum usually makes control of the operating variables more difficult. If there is a sufficiently great pressure difference between the feed and operating pressure, the feed can be used as the motive fluid in an ejector to increase the recirculation of liquid through the vaporizer. This is
VAPORIZERS, EVAPORATORS, AND REBOILERS
457
inefficient if the fresh liquid is compressed purposely to serve as the motive fluid, but if the fresh ·liquid is available under pressure for other reasons, the ejector permits recovery of some of the pressure head. The advantages of forced circulation or natural circulation are in part economic and in part dictated by space. The forcedcirculation arrangement requires the use of a pump with its continuous operating cost and fixed charges. As with forcedcirculation evaporators, the rate of feed recirculation can be controlled very closely. If the installation is small, the use of a pump is preferable. If a naturalcirculation arrangement is used, pump and stuffingbox problems are eliminated but considerably more headroom must be provided and recirculation rates cannot be controlled so readily. Naturalcirculation steam generators are frequently laid out in accordance with Fig. 15.2 but employing a vertical
Fro. 15.3.
Forcedcirculation reboiler arrangement.
11 exchanger with vaporization in the tubes. Water is especially adaptable to naturalcirculation arrangements, since the density differences between the liquid and vapor at a given temperature are very large. Reboiler Arrangements. When reboilers · are used, the space at the bottom of the column between the liquid level and the bottom plate is employed for disengagement and the bubble caps serve as separators. A typical arrangement for a forcedcirculation reboiler is shown in Fig. 15.3. This type is called a pumpthrough reb oiler. All the liquid on the bottom plate, frequently called trapout to distinguish it from bottom product, is carried by the downcomer below the liquid level of the column. The liquid can be recirculated through the reboiler as many times as is economically feasible, so that the percentage vaporized per circulation is kept low while the bottom product is withdrawn from a separate connection. In general, forcedcirculation or pumpthrough reboilers are used only on small installations or those in which the bottoms liquid is so viscous and
458
PROCESS HEAT TRANSFER
t,he pressure drop through the piping and reboiler so high that natural circulation is impeded. By far the greater number of large reboiler installations employ natural circulation. This can be achieved in either of two simple ways as showu in Fig. 15.4a and b. In Fig. 15.4a all the liquid on the bottom plate is circulated directly to the reboiler, whence it is partially vaporized. The unvaporized portion, on being disengaged under the bottom plate, is withdrawn as bottom product. In Fig. 15.4b the liquid passes through the downcomer below the liquid level of the column as in forced circulation. The bottom liquid is free to recirculate through the reboiler as many times as the hydrostatic pressure difference between Zt and zs will permit. Because there is no opportunity for recirculation in the arrange
Bottoms CalONCETHROUGH REBOILER FIG. 15.4.
Bottoms (b),RECIRCULATING REBOILER
Naturalcirculation reboiler arrangements.
ment in Fig. 15.4a, it is called a oncethrough reboiler arrangement. Figure 15.4b is referred to as a recirculating reboiler. Classification of Vaporizing Exchangers. There is a greater hazard in the design of vaporizing exchangers than any other type of heat exchanger. For this reason it is convenient to set up a classification based on the method of calculation as employed for each distinct type of service. Each of the common classes below is distinguishable by some difference in calculation. A. Forcedcirculation vaporizing exchangers 1. Vaporization in the shell a. Vaporizer or pumpthrough reboiler with isothermal boiling b. Vaporizer or pumpthrough reboiler with boiling range c. Forcedcirculation evaporator or aqueoussolution reboiler 2. Vaporization in the tubes a. Vaporizer or pumpthrough reboiler with or without boiling range b. Forcedcirculation evaporator or aqueous solution reboiler
YAPORIZERS, EVAPORATORS, AND REBOILERS
459
B. Naturalcirculation vaporizing exchangers 1. Vaporization in the shell
a. Kettle reboiler b. Chiller c. Bundleincolumn reboiler d. Horizontal thermosyphon reboiler 2. Vaporization in the tubes a. Vertical thermosyphon reboiler b. Longtube vertical evaporator
Heatflux and Temperaturedifference Limitations. It may be assumed that process conditions will always be established which provide for the vaporization of only part of the liquid fed to a vaporizer. When vaporizing liquids from pools, extremely high maximum fluxes have been obtained. For water a maximum flux is reported at 400,000 Btu/(hr)(ft 2) and for organics from 70,000 to 125,000 Btu/(hr)(ft 2), although these have been obtained only in laboratoryscale apparatuses with perfectly clean surfaces. It i::; again pointed out that the maximum flux occurs at the critical temperature difference and is a limitation of the maximum coefficient ·which may be attained. Beyond the critical temperature difference both the coefficient and the flux decrease, the decrease being due to the formation of a layer of gas on the tubes. It is the phenomenon of vapor blanketing which poses the principal difficulty in the design and operation of vaporizing exchangers. Fluxes of magnitudes as high as those above are of little practical value in design. It will be recognized that vaporization in a 12 exchanger, occurring as it does without· continuous disengagement, is very different from vaporization out of liquid pools. By restricting disengagement in the 12 exchanger, the possibility of vapor blanketing is increased greatly, so that it is necessary also to restrict the flux to an allowable value safely out of the range in which blanketing might occur. The flux is defined by QjA or uD f1t but not by h~(f1t)~, where hv is the vaporization coefficient .!tnd (f1t),. is the temperature difference between the tube wall and the boiling temperature. h~(f1t)w is the flux based on the clean surface Ac, while Q/A is the flux based on the actual surface A, and A is greater than Ac in a vaporizer designed with a dirt factor. However, it is customary to restrict both Q/A and hv to safe. maxima, the two also serving to prevent the presence of too great a temperature difference (f1t)w. The following restrictions will be observed throughout this chapter: I. Flux a. The maximum allowable flux for forced circulation vaporizers· and reboilers vaporizing organics is 20,000 Btu/(hr)(ft 2) and for natural circulation 12,000 Btu/ (hr) (ft2).
460
PROCESS HEAT TRANSFER
b. The maximum allowable flux for the vaporization of water or aqueous solutions of low concentration using forced or natural circulation is 30,000 Btu/(hr)(ft 2 ). II. Film coefficient a. The maximum allowable vaporizing film coefficient for the forced or naturalcirculation vaporization of organics is 300 Btu/(hr)(ft 2 )(°F). b. The maximum vaporizing film coefficient for the forced or naturalcirculation vaporization of water and aqueous solutions of low concentration is 1000 Btu/ (hr )(ft 2) (°F).
Relationship between Maximum Flux and Maximum Film Coefficient. The objects of the limitations above are the elimination of all vapor blanketing possibilities. Suppose it is desired partially to vaporize an organic compound boiling at 200°F in a forcedcirculation vaporizer using steam at a temperature of 400°F so that At = 200°F and the flow is such that a vaporizing coefficient of 300 Btu/(hr)(ft2WF) can be obtained. If the condensing steam coefficient is 1500, U a = 250, and if R 4 = 0.003, UD = 142.. The flux will be 142 X 200 = 28,400 Btu/(hr)(ft 2), which exceeds. limitation Ia. Since Q/A or U D At may not exceed 20,000, any change to permit compliance with Ia amounts to an increase in the total surface for vaporization. If the original steam and vapor temperatures are retained, the new overall coefficient U D will be 20,000/200 = 100 Btu/(hr)(ft2)(°F). The temperature difference (At)w may be greater than the critical temperature difference, since it does not occur at the maximum attainable flux and under these conditions the critical temperature difference can be exceeded within limits without the danger of vapor blanketing. There is no advantage to the use of very high temperature differences, however, since at the maximum allowable flux any increase in At must be offset by a decrease in the allowable value of U D· Only when UD is naturally small may the use of a high At be partially justified. The test of whether or not a vaporizer exceeds the allowable flux is always determined by dividing the total vaporizing heat load by the total available surface for vaporization. By the same token the maximum value of UD which may be anticipated is given by UD = (Q/A)(1/At) regardless of the dirt factor which results. When setting the temperature of the heating medium, it is seen that the use of a large At and corresponding (At)w also requires decreasing U D which in turn gives a large value of Rd = 1/Un  1/Ua. The large dirt factor is not essential to the continued operation of the vaporizer from the standpoint of dirt but only as a preventive against vapor blanketing. Accordingly, when the temperature of the heating medium may be selected independently such as by setting the pressure in the case of steam, it need not be set at a value of At greater than that which gives a U D corresponding to the desired dirt factor.
VAPORIZERS, EVAPORATORS, AND REBOILERS
461
FORCEDCIRCULATION VAPORIZING EXCHANGERS
1. v.A.PORIZATION IN THE SHELL a. Vaporizer or Pumpthrough Reboiler with Isothermal Boiling. The calculations employed in the solution of this type of vaporizer are common to the many simple vaporizing problems found in a plant whether or not connected with a distilling column. If a liquid is substantially pure or a constantboiling mixture, it will boil isothermally. This usually applies to the bottom liquid of a distilling column separating a binary mixture into relatively pure compounds. For utility boiling operations, such as the vaporization of a cold liquid coming from storage, the liquid may not be at its boiling point and may require preheating to the boiling point. Since the shell of a forcedcirculation vaporizer is essentially the same as any other 12 exchanger, the preheating can be done in the same shell as the vaporization. If the period of performance of a vaporizer is to be measured by a single overall dirt factor, it is necessary to divide the shell surface into two successive zones, one for preheating and one for vaporization, in much the same manner employed in condensersubcoolers. The true temperature difference is the weighted temperature difference for the two zones, and the clean coefficient is the weighted clean coefficient as given by Eqs. (12.50) and (12.51). If steam is the heating medium, only two tube passes are required and these need not be equally divided, since the return pass carries considerably less vapor than the first pass. If a hot stream such as gas oil is the heating medium, there is a problem in determining the true temperature difference in each zone. If the approach between the heatingmedium outlet temperature and the vapor outlet temperature is not too small, the true temperature difference can be approximated by considering the temperature fall iri each zone proportional to the heat removed from the heating medium. The method of using zones has also been discussed in Chap. 12 along with condensersubcoolers. Film Coefficients. Where there is a preheating zone, it can be computed by the use of Fig. ~8 like any other heater with the cold fluid in the shell. The isothermal boiling film coefficient is also obtained by the use of Fig. 28 based on the premise that the heat must first be absorbed by the liquid by forced convection before passing into the vapor bubbles and that the liquid heating coefficient is the controlling coefficient in the sequence. Pressure Drop. The shellside pressure drop for the vaporizing zone is computed by introducing the average specific gravity into the denominator of Eq. (7.44). If the vaporizing liquid boils isothermally at t. and receives heat from a medhim having a range T 1  T2, two possibilities may be considered:
462
PROCESS HEAT TRANSFER
Case I: The vapor and heating medium may be in counterflow. Case II: The vapor and heating medium may be in parallel flow. CASE I: VAPORS AND BEATING MEDIUM IN COUNTERFLOW. Referring to Fig. 15.5, if W is the weight flow of the heating medium, 0 its specific heat, and Tis the temperature of the heating medium at any tube length x, (15.1)
WO dT = Ua" dx(T  t,)
dx
is the surface. Intewhere a" grating T with respect to x, ln (T  t,) = At
X=
wa
= ln (T2
T t, In T2  t.
x=O
+ 01
(15.2)
 t,)
(15.3)
0, Ot
X X=L FIG. 15.5. Temperatures during vaporization of an isothermal fluid by a nonisothermal heating medium.
Ua"~
Ua"x
= wa
= 0.115 X 2.42 == 0.278lb/(ft)(hr) [Fig. 14] D = 0.87/12 ,;, 0.0725 ft (Re, is for pressure drop only) D. = 0.99/12 = 0.0825 ft [Fig. 28] Re, = DG,fp. Re. = D.G.jp. = 0.0725 X 31,100/0.0363 = 62,000 = 0.0825 X 233,000/0.278 = 69,200 (f') jH = 159 [Fig. 28] (8') At 172°F(1WAPI) k(cp.jk)'r3 = 0.12 Btu/(hr)(ft 2)("F/ft) [Fig. 16] , = 1.0 (6) At Ts = 338°F,
(9') h.
(9) h;. for condensing steam
= 1500 Btu/(hr)(ft'WF)
=
(~)'r3 D. k
jH .!£_
[Eq. (6.15b)j
= 159 X 0.12/0.0825 = 231 Btu/(hr)(ft'WF)
Clean overall coefficient for preheating U,: U "
=
h;.h. = 1500 X 231 = 200 Bt j(h )(ft•)(oF) h;. h. 1500 231 u r
+
+
(6.38)
Clean surface required for preheating A,:
q, A :P  U,(t.t),
13 400 • 200  67 ' 0 ft•
Vaporization: (6') At 235°F, p. = 0.10 X 2.42
= 0.242 lb/(ft)(hr)
[Fig. 14] Re. = 0.0825 X 233,000/0.242 = 79,500 ('1') jH = l'[Q (Fig. 28} (8') At 235°F k(cp.fk)Y. = 0.115 Btuj(hr)(fV)(°F /ft) q,, = 1.0 (Fig. 16) (9') h. = jH
(9)
h~o
for condensing steam = 1500
~. (?;) 14
[Eq. (6.15)]
= 170 .X 0.115/0.082.') = 237
Clean overall coefficient for vaporization U .:
U = •
h;.h. = 1500 X 237 = 205 + h. 1500 + 287
h;.
Clean surface required for vaporization A.:
A, = _q_, = 21,100 = 103 ftl U,(At). 205 Total clean surf~ce Ac:
Ac =A,+ A.= 67.0
+ 103 =
170ft•
(6.38)
467
VAPORIZERS EVAPORATORS, AND REBOILERS (13) Weighted clean overall coefficient Ua:
Uo = 2:UA = 13,400
Aa
+ 21,100
=
170
203
(12.50)
(14) Design overall coefficient: Surface/lin ft of tube = 0.2618 Total surface.= 76 X 16'0" X 0.2618 = 318 ft 2 Q 4,290,000 Un =A At= 318 X 124.5 = 108·5 Check of maximum flux:
(Table 10)
A total of 170 ft2 are required of which 103 are to be used for vaporization. For the total surface required 318 ft 2 will be provided. It can be assumed, then, that the surface provided for vaporization is 1o~ 10
X 318 = 193 ft2
The flux is Q/A = 2,170,000/193 = 10,700 Btu/(hr)(ft 2).
(Satisfactory)
(15) Dirt factor:
Rd =
u'U;u~»
=
~~; ~ i~~:~ = 0.0043
(hr)(ft•)("F)/Btu
(6.13)
Pressure Drop (1) For Ret = 62,000,
= 0.000165 ft2/in.• [Fig. 26] From Table 7, specific vol of steam at 115 psia = 3.88 ft 3 fib 1 s = 3.88 X 62.5 = 0.00413 1 jG•Ln (2) t:.P, = 2 5.22 X 10 10Da.p, [Eq. (7.45)] 1
PreheaJ: (1') Re, = 69,200, f = 0.00145 ft 2/in. 2
f
=zx
0.000165 X 31,1002 X 16 X 2 5.22 X 10 10 X0.0725 X 0.00413 X 1 = 0.16 psi
[Fig. 29] (2') Length of preheat zone
L, = LA,/Aa
= 16 X 67.0/170 = 6.3 ft (3') No. of crosses, N + 1 = 12L,/B [Eq. (7.43)] = 12 X 6.3/5 = 15 [Fig. 6] 8 = 0.50 D, = 15.25/12 = 1.27 ft
+
(4') AP = _ JG!D,(N 1) • 5.22 X 1010D,sq.., [Eq. (7.44)] _ 0.00145 X 233,000 2 X 1.27 X 15  5.22 X 101° X 0.0825 X 0.50 X 1.0 = 0.70 psi Vaporization: (1 1) Re, = 79,500, f = 0.00142 ft 2 /in. 2 (2') Length of vaporization zone L, = 16  6.3 = 9.7 ft (S') No. of crosses, N + 1 = 12LfB [Eq. (7.43)] = 9.7 X 1% = 23 Mol. wt. = 58.1
468
PROCESS HEAT TRANSFER
Pressure Drop D ensi"ty,
P
8outlet liquid Poutlot liquid 8c;.u tlet mix
58.1 . X 14.7/300 = 2.34 lb /ft3 = 0.43 [Fig. 6] = 0.43 X 62.5 = 26.9 lb/ft' = 359 X
69
%9 2
=
24,700/62.5 0 046 19,750/2.34 +4950/26.9 = · 8inlet = 0.50 Smeau = (0.50 0.046)/2 = 0.28 AP _ 0.00142 X 233,0002 X 1.27 X 23 '  5.22 X 1010 X 0.0825 X 0.28 X 1.0 = 1.9 psi AP, (total) = 0.7 + 1.9 ~ 2.6 psi
+
Summary
h outside
1500
Ua
203
UD
108.5
23%37
Ra Calculated 0.0043 RaRequired 0.16
Calculated llP ·
2.6
Neg
Allowable llP
5.0
b. Vaporizer or Pumpthrough Reboiler with Boiling Range. If a liquid undergoing vaporization is a mixture of a number of miscible compounds, it does not boil isothermally. Instead it has an initial boiling temperature (bubble point) and a final boiling temperature (dew point) at which the last bit of liquid is vaporized. When the mixture starts to boil at its bubble point, the more volatile compounds are driven out of the solution at a greater rate and as the volatile compounds enter the vapor phase the boiling temperature of the residual liquid rises. This means that throughout the vaporizer there is a temperature range over which boiling occurs, and the greater the percentage of the total liquid vaporized the more nearly the range extends from the bubble point to the dew point of the inlet liqUid. Because of the boiling range, sensible as well as latent heat must be absorbed simultaneously by the liquid as it proceeds through the vapor
VAPORIZERS, EVAPORATORS, AND REBOILERS
469
izer so that it will possess a range of boiling temperatures. Furthermore the sensible heat is absorbed over the same surface as the heat for vaporization in contrast to the preheaterisothermal vaporizer in which the two occur in separate zones. The calculation of the boiling coefficient in this case, however, is the same as for a preheatervaporizer as calculated in Example 15.2. Here the heat from the tube wall is first absorbed by the liquid as sensible heat before it is transformed into vaporization. Since the rate of heat transfer from a hot liquid into an incipient vapor bubble is very great, it may be assumed that the sensibleheattransfer coefficient as calculated from Fig. 28 for either direct vaporization or simultaneous sensibleheat transfer is the limiting resistance. The coefficient for the combined sensibleheat transfer and vaporization is calculated as if the entire vaporizing heat load were transferred as sensible heat to the liquid over its boiling range in the vaporizer. The true temperature difference may be taken as the LMTD if the heating medium is isothermal. This assumes that the heat transferred is proportional to the change in temperature, namely, that onehalf of the total load is delivered while the temperature rises onehalf of the total temperature range for vaporization. If the bulk of a mixture consists of closely related compounds with some ·more or less volatile compounds, the assumption that the heat and temperature proportions are equal may cause considerable error. The true temperature difference can be obtained by graphical integration as treated in 'Example 13.3. Film Coefficients. The sensibleheattransfer coefficient should be regarded as the boiling coefficient when applying the restrictions of allowable flux and allowable coefficient even though it is computed from Fig. 28. When a liquid has a boiling range, the average flux, Q/A may be less than 20,000 but because of the variation in temperature difference, UD At 1 at the greater terminal temperature difference may exceed 20,000. Discrepancies of this sort may cause bumping or erratic vaporization if the initial compounds to vaporize are very volatile compared with the bulk and tend to separate from the liquid too readily. A check of the inlet flux can prevent this difficulty. If the inlet flux does not exceed, say, 25,000 Btu/(hr)(ft 2), there is no need to penalize the entire design by providing excess surface simply because of an excessive flux over the first few per cent of the vaporizing surface. Pressure Drop. The pressure drop is computed in the same manner as for isothermal boiling, using the Reynolds number based· on inlet conditions and a gravity which is the mean of inlet and outlet gravities. It is also quite possible that a fluid with a boiling range may enter a vaporizer below its bubble point. In such cases the surface is again divided into two consecutive zones, one for pure preheating and the other for vapori
470
PROCESS HEAT TRANSFER
zation of a mixture with a boiling range. The weighted coefficients and temperature difference may be obtained as before through the use of Eqs. (12.50) and (12.51). c. Forced ..circulation Evaporator or Aqueoussolution Reboiler. As seen in Chap. 14, the shells of 1,2 exchangers are not used in forcedcirculationevaporator processes, since the properties of water are excellent for naturalcirculation equipment. However, a 12 exchanger can easily serve as a forcedcirculation evaporator. In distillation processes such as the distillation of an acetonewater or alcoholwater solution the bottom product is nearly pure water. It may be advantageous in smaller operations of th:is nature to use a pumpthroqgh reboiler in preference to natural circulation, since the losses in the connecting piping may be unduly high and the use of larger connecting pipes does not assure smooth operation. The aqueoussolution reboiler can be calculated in the same manner as the pumpthrough reboiler, with or without a boiling range, except that the allowable flux and film coefficient is greater. This type of equipment is usually designed with the dirt factor as the controlling resistance. The applicability of a method of computing watervaporization rates is therefore of value only at low mass velocities. Since water vapor has a very low vapor density, low mass velocities must be employed whenever the allowable pressure drop is small. Film coefficients for boiling water and aqueous solutions can be obtained by the use of Fig. 28, although these will be about 25 per cent lower than those which have been obtained experimentally. Where data on the physical properties of aqueous solutions are lacking, they can be approximated by the methods of Chap. 7. If the mass velocity is very low, the value of the coefficient so obtained may be multiplied by 1.25 and the overall value of Ua should rarely exceed 600 Btu/(hr)(ft 2)(°F).
2. VAPORIZATION IN THE TUBES a. Pumpthrough Vaporizer or Reboiler with or without Boiling Range. The coefficients for vaporization with or without a boiling range can be obtained from Fig. 24 for organic liquids. The number of tube passes may be as gr¢at in horizontal exchangers as the pressure drop will allow. If the number of tubes in the final passes are greater than the number in the initial passes, it is possible to obtain a reduced pressure drop. The shell side,· when employing steam, may be laid out on triangular pitch, since cleaning will be infrequent and the shell side can be cleaned by boiling out. The pressure drop can be computed using Eq. {7.45) with a Reynolds number based on the inlet properties and a ·specific gravity which is the mean between inlet and outlet. The tube fluid should flow upward.
VAPORIZERS, EVAPORATORS, AND REBOILERS
471
b. Forcedcirculation Evaporator or Aqueoussolution Reboiler. The tubeside data for the evaporation of water and aqueous solutions can also be obtained from Fig. 24. The boiling coefficients will be about 25 per cent greater than the computed values, and at low mass velocities the coefficient can be multiplied by a correction factor of 1.25. This class also includes the forcedcirculation evaporators similar to 11 exchangers. The calculation of the vertical iongtube evaporator will be treated as a naturalcirculation vaporizer. NATURALCIRCULATION VAPORIZING EXCHANGERS
1. VAPOitiZATION
IN THE SHELL
a. Kettle Reboiler. The kettle reboiler is shown in Fig. 15.6. It is a modification of the powerplant evaporator. The relation of the bundle to the shell is seen better from an end elevation. Another form of the kettle reboiler employing a tube sheet which covers the entire shell is
FIG. 15.6.
FIG. lli. 7.
Kettletype reboiler.
(Paiterson Foundry and Machine Oo.)
Kettletype reboiler with integral tube sheet.
(Paiterson Foundry and Machine
Co.)
shown in Fig. 15.7. In this type the bundle is not circular but conforms to the shell as seen from an end elevation. The method of connecting this type of reboiler to the distilling column is shown in Fig.l5.8. Kettle reboilers are fitted with a weir to ensure that the liquid level in the reboiler is maintained and that the tube surface is not exposed. Since only about 80 per cent of the liquid bottoms entering at the inlet are vaporized, provision must be made for the removal of the bottoms product which is on the discharge side of the weir. There are a number of
472
PROCESS HEAT TRANSFER
arbitrary rules on the volume required above the liquid level for disengagement and the maximum number of pounds per hour which may be vaporized from the liquid surface. If the top row of tubes is not higher than 60 per cent of the shell diameter, adequate disengaging space will
Bottom producf FIG. 15.8.
Kettle reboiler arrangement.
be provided when a liquid level covering the top row of tubes is ensured by the weir. b. Chiller. The chiller is shown in Fig. 15.9. It is 1,1. typical kettle reboiler, except for the weir, and the bundle has tubes to a height of about 60 per cent of the diameter. The vapor space above is for disengagement
FIG. 15.9.
Chiller.
(Patterson Foundry and Machine Co.)
of the vapor from the liquid. Chillers are used in refrigeration processes of the vaporcompression type as shown in Fig. 15.10. The refrigeration cycle begins at the point a, where liquid refrigerant at a temperature higher than that of the condenser water and at high pressure passes to a constantenthalpy throttle valve where its pressure is reduced. The pressure and temperature of the liquid on the downstream side of the valve are naturally less than on the highpressure side. The expansion is adiabatic, and some of the liquid flashes into vapor cooling the remainder of the refrigerant on the lowpressure side at b. If the cold refrigerant is to be circulated directly to a refrigerator, the saturati'< = 0.118 Btuj(br}(ft 2)(°F/ft) (9) h; = jH
i (~)M
1
[Eq. (6.15d)]
(5.28) Cold fluid: shell side, gasoline
(9') Assume h. = 300 for trial.
h1jq,, = 220 X 0.118/0.0694 = 374 (10)
!;: = ~
X
g~
[Eq. (6.9)] (10') t'" = t,
= 374 X 0.834/1.0 = 311 The correction for (;,;) o.u is negligible.
+ h;.h; h. (T,
 t,) [Eq. (5.31)]
311
·= 400 + 311 + 300 (517 (.{·in. circular bundle 16'0" long. The column required for Liquid ljrzy§/__ 28,100 lb/hr of the vapor at 200 psig has a diameter of less than 3 ft. If the bundle is inserted in the bottom of such a column, many short tubes will be required and the height of the bottom of the column must be increased so as to maintain the same holdup space. Another obvious disadvantage lies in the size of the flanged connection which must be welded to the side of the column to accommodate the larger b"undle. Internal supports are also required to keep Fw. 15.12. Bundlein the weight of the bundle from forming a cantilever column reboiler. with the column connection flange. These difficulties are usually surmountable when the column diameter is greater than 6 ft, but overall experience favors the use of external reboilers in preference to the small savings realized by eliminating the shell. The calculations for the bundleincolumn reboiler are identical with those for the kettle reboiler, using coefficients from Fig. 15.11.
t
FIG. 15.13.
Horizontal thermosyphon reboiler.
(Patterson Foundry and Machine Co.)
d. Horizontal Thermosyphon Reboiler. This is perhaps the commonest type of reboiler. Figure 15.13 shows a horizontal thermosyphon reboiler. It consists of centrally located inlet and outlet nozzles, a vertical circular support plate between the nozzles, and a horizontal longitudinal bafHe. Horizontal thermosyphons operate on the principle of divided flow as first outlined under condensers in Fig. 12.17, with halves of the entering fluid flowing away from each other below the longitudinal bafHe and toward each other above it. Disengagement takes place in the column, and the reboiler may be connected by the
VAPORIZERS, EVAPORATORS, AND REBOILERS
479
arrangement of Fig. 15.4a or b. In Fig. 15.4a, as stated previously, all the liquid from the bottom plate is led directly to the reboiler. The rate of feed to the reboiler is the hourly trapout rate, which passes through the reboiler but once. In Fig. 15.4b the reboiler is connected to the bottoms, which are free to recirculate at a rate such that the fric'tional pressure drop in the reboiler and other resistances of the circuit just balance the hydrostatic head difference between the liquid and liquidvapor legs. The hydrostatic head available in the latter arrangement, however, is less than in the oncethrough arrangement although a greater head is required for recirculation. The head is provided by raising the bottom liquid level in the column or by raising the column itself. Occasionally the reboiler may be set in a well, but this practice finds little favor in modern plants. Film Coefficients in Horizontal Reboilers. The coefficients used for thermosyphons are substantially the same as those employed for kettl~ reboilers and are given by Fig. 15.11. When there is a boiling range, it is imperative that the overall clean coefficient be weighted for sensible and latentheat loads individually, although the procedure differs from the weighting of successive .zories; since both sensible heating and boiling occur in the same temperature range. This problem was not encountered in forcedcirculation reboilers and vaporizers because the rates of boiling and sensibleheat transfer are ordinarily very nearly identical. However, in a shell without forced convection the rate .of sensibleheat transfer by free convection is usually less than onesixth the naturalcirculation boiling rate. In natural circulation, however, where both sensibleheat transfer and boiling occur over the same surface, the free convection coefficient is undoubtedly modified by the bubble movements which far exceed the agitation derived from ordinary free convection currents. To account for this modification, the sensible portion of the heat load is assumed to be transferred by ordinary free convection and the boiling portion is assumed to be transferred as naturalcirculation vaporization, Although the flow is not counterflow, it usually does not deviate greatly from it, because one or both of the fluids is substantially isothermal. If steam is the heating medium, the counterflow temperature difference applies directly. If a liquid is the heating medium rather than a vapor, the counterflow temperature difference applies only if· the range on the material being vaporized is small and the appro~ch between the heating medium and vaporizing medium inlet temperatures is appreciable. If FTfor a 12 exchanger exceeds 0.90 an insignificant error may be anticipated from the use of the 12 parallel flowcounterflow temperature difference.
480
PROCESS HEAT TRANSFER
Since the temperature differences for sensible heating and vaporization are the same, there is no weighted temperature difference. But the sensible heat q. is transferred with a free convection coefficient of h., and the latent heat qv is transferred with the considerably higher coefficient h.. To permit obtaining a single dirt factor, as the index of performance or maintenance of the reboiler, the weighted coefficient may be obtained as follows: From q = hA At, A,(At). = ~
h.
A.(At)v
The weighted coefficient is then h =
=~ h.
Q q,jk, + q./ku Since neither h. nor h, is influenced by the velocity through the reboiler, it will be of no consequence in the calculation whether the reboiler has been connected for oncethrough or recirculation operation. Pressure Drop. There is an obvious need in the recirculation arrangement to keep the pressure drop through the thermosyphon as small as possible. When studying condensers it was observed that the greater the pressure drop through the condenser the higher the condenser must be el~vated above the column to permit the gravity return of condensate. The effect of pressure drop on the elevation of the bottom tower liquid level above the reboiler is even more critical. The greater the pressure drop through the reboiler the higher the entire column and auxiliaries must be elevated above the ground level to produce sufficient hydrostatic head to overcome the pressure drop. Pressure drops of about 0.25 psi are generally allowed for the reboiler and attendant losses. If the column is small in diameter or height, a pressure drop as high as 0.50 psi may be allowed, but concessions of this nature to the design of the reboiler are rare. For a reboiler vaporizing a small fraction of the liquid entering it, the required elevation is greater, since the return leg to the column contains more liquid than vapor and the difference in density of the streams to and from the reboiler is small. While half ba:(lles may occasionally be used to increase the turbulence in the shell, the tubes are usually prevented from sagging by the vertical support plate between the inlet and outlet nozzles and additional quartercircle support plates. The liquid entering the horizontal thermosyphon flows onehalf of the tube length on the underside of the longitudinal baffie and onehalf the tube length on the upper side, so that all the liquid travels the entire tube length but in each instance with a massvelocity based on onehalf the total flow. The length of path of each parallel stream equals
VAPORIZERS, EVAPORATORS, AND REBOILERS
481
the tube length, and it is sufficiently accurate to treat the pressure drop the same as for a shell without baffles and with axial flow as in Example 7.8. The diameter of a horizontal thermosyphon reboiler is greater than that which corresponds to the tube count of a conventionall2 exchanger because of the free space which must be provided in the upper half to allow the light mixture of vapor and liquid readily to reach the outlet nozzle. If the surface of a 25 in. ID layout sufficed for heat transfer, the tubes would be relocated in a 27 in. ID shell while retaining the tube pitch so as to leave a free vapor flow channel at the top of the shell and a lesser entrance channel at the bottom. The equivalent diameter is computed· directly by Eq. (7.3) from the wetted perimeter ofthe tubes, half the shell, and ·the width of the longi
FIG. 15.14. Horizontal thermosyphon with double nozzles. Machine Co.)
(Patterson Foundry and
tudinal baffle. The flow area is the difference between half a circle and the number of tubes in the upper or lower shell pass. In the absence of the actual tube layout these may be assumed to be equal. The Reynolds number is computed from the inlet liquid viscosity and the equivalent diameter. The pressure drop is based on the mean specific gravity between inlet and outlet, using a friction factor obtained from Fig. 26 for the tube side. When there is a single inlet nozzle on the shell, it is customary not to use a tube length more than five times the shell diameter. Long, thin reboilers do not thermosyphon well. When a long, thin reboiler is indicated, it is usually equipped with two inlet nozzles as shown in Fig. 15.14 with a mass velocity based on onefourth the flow rate in onehalf the flow area. The following table will serve as a guide to wellproportioned horizontal thermosyphons: Shell ID, in. 1217.J.a 19.li:29 31 and ..over
Tube length 8'0" 12'0" 16'0"
When using a recirculating arrangement with a horizontal thermosyphon, recirculation can be computed approximately as the rate at
482
PROCESS HEAT TRANSFER
which the pressure drop through the reboiler equals the hydrostatic difference of Zt and zs in Fig. 15.4b although the rate does not affect the film coefficient. Instead of the recirculation rate it is preferable in reboilera to specify the recirculation ratio, which is defined as the pounds per hour of liquid leaving the reboiler compared with the pounds per hour of vapor alone. This should not be confused with the conventional definition of recirculation rates, which is defined as the ratio of the total hourly throughput to the hourly requirement. In a recirculating reboiler the temperature range is not identical with that of a oncethrough arrangement. If the liquid recirculates, only a small amount of vapor is formed in each circulation and the vaporization occurs over a smaller temperature range, although the outlet temperatures in both arrangements are identical. In the recirculating arrangement the temperature difference is somewhat smaller. Usually the reduction in temperature difference is not significant unless the range of the heating medium is very close to that of the vaporization. A recirculation ratio four times or greater than the hourly oncethrough vapor rate is considered favorable from a cleanliness standpoint. The method of calculating recirculation ratios will be discussed in connection with the vertical thermosyphon reboiler, where a higher order of recirculation is usually obtained. Example 16.4. Calculation of a Oncethrough Horizontal Thermosyphon Reboiler. 38,500 lb/hr of 60°API naphtha in a oncethrough arrangement is to enter a horizontal thermosyphon reboiler and produce 29,000 lbjhr of vapor in the temperature range from 315 to 335°F at an operating pressure of 5.0 psig. Heat will be supplied by 28° API gas oil with a range from 525 to 400"F. Available for the service is a 21% in. ID reboiler containing 116 1 in. OD, 14 BWG tubes 12'0" long laid out on 1%in. square pitch. The bundle has a support plate above the single inlet nozzle and is arranged for eight passes. What are the dirt factor and the pressure drops? Solution: Reboiler: Shell side ID = 21}i in. Support plates = X circles Passes = divided
Tube side Number and length = 116, 12'0" OD, BWG, pitch = 1 ,in., 14 BWG, 1%in. square Passes = 8
(1) Heat balance:
Enthalpy of liquid at 315°F and 19.7 psia = 238 Btu/lb Enthalpy of liquid at 335°F and 19.7 psia = 252 Btu/lb Enthalpy of vapor at 335°F and 19.7 psia = 378 Btu/lb Naphtha q. = 29,000(378  252) = 3,650,000 q. = 38,500(252  238) = 540,000 Q = ~ 4,190,000 Btu/hr Gas oil, Q = 51,000 X 0.66(525  400) = 4,190,000
(Fig. 11)
VAPORIZERS, EVAPORATORS, AND REBOILERS (2)
483
~t:
Hot Fluid
Cold Fluid
525
Higher Temp
335
190
400
Lower Temp
315
85
125
Differences
20
105
LMTD = 131°F 125 R = 20 = 6.25 FT = 0.97 At = FT X LMTD
20
s = 525 = 0.97
315 = 0.095
X 131
= 127°F
(3) T,:
~t. = 85 = 0447 ~th 190 • K. =.0.42 F, = 0.41 T, = 400 + 0.41(525..,.. 400) = 451°F t, = 815 + 0.41(335  815) = 823°F
(Fig. 17)
(5.28) (5.29)
Cold fluid: shell side, naphtha Hot fluid: tube side, gas oil [Table 10] Assume weighted h. = 200 h;. from (10) = 286 a, = N 1a:/l44n = 116 X 0.546/144 X 8 = 0.055 ft' t., = t, h;. h,. h. (T •  t,) [Eq. (5.31)] (6) G, = Wja, = 51,000/0.055 286 = 323 286 200 (451  323) = 928,000 lb/(hr)(ft•) (6) At T, = 451°F, = 382°F p. = 0.45 X 2.42 = 1.09lb/(ft)(hr) (At)., = 382  323 = 59°F [Fig. 14] D = 0.834/12 = 0.0695 ft [Table 10] From Fig. 15.11, h. = > 300, use 300 Re, = DG,jp. h.= 60 = 0.0695 X 928,000/1.09 = 59,200 q./h. = 3,650,000/300 = 12,150 (7) ja = 168 [Fig. 24] q./h. = 540,000/60 = 9,000 (8) At T. = 451 °F (28°API) [Fig. 16] 21,150 k(cp.jk)ri = 0.142 Btu/(hr)(ft 2 )(°F/ft) (9) h; = (jak/D)(c,../k)Y.q,, [Eq. (6.15a)] !!:£ = 168 X 0 ·142 = 343 "'' 0.0695 q,, "" 1.0 [Eq. (6.6)] (10') h. = 4,190,000/21,150 (10) h;. = h; X ID/OD = 198 Btuj(hr)(ft'WF) = 348 X 0.834/1.0 = 286 Btu/(hr)(ft•WF) Checks 200 assumed for h•. (4) a~ = 0.546 in. 2
+
+
+
+
(13) Overall clean coefficient Ua:
Ua
=
h;.h. h;, + h,
= 286 286
198
X = 116 Btu/(hr)(ft )("F) + 198 2
(6.38)
484
PROCESS HEAT TRANSFER
(14) Design overall coefficient UD: Surface per linear foot = 0.2618· Total surface = 116 X 12'0" X 0.2618 = 364 ft 1 uD = AQt.t = 364 4,100,000 90 7 X 127 = .
(Table 10)
Cheek of maximum flux (based on total transfer through the surface):
~ = 4 • 1 ;~oo = 11,500 vs. 12,000 allowable (15) Dirt factor Ra: Ra
= Ua
UD = 11690.7
UaUD
116 X 90.7
= 00024 ·
(6.13)
Pressure Drop 0.000168 ft2/in. 2 (1') D~ = 4 X flow area/frictional wetted [Fig. 26] perhpeter [Fig. 6] Assume half of tubes above and half of 8 = 0.73 tubes below longitudinal baffle.. (2) t.P _ !G~Ln Flow area = U shell cross section  72 '  5.22 X 1010Ds.p, tube cross section 2 0.000168 X 928,000 X 12 X 8 = 522 X 10 10 X 0.0695 X 0.73 X 1 = (21.25 2  1.0 X 116) = 132 in.2 = 5.3 psi (3) G, = 928,000 V"/21 = 0.11 [Fig. 27] a, = 137{ 4 4 = 0.917 ftl .,.. X 21.25 . n V2 4 X 8 . + ,..2 X 1 t.Pr = 4 8 2g' = 0 .73 X 0.11 = 4.8 ps1 W e tted perrmeter = 2 X 116 + 21.25 = 236.7 in. [Eq. (7.46)1 = 4 X 132/236.7 = 2.23 in. (4) t.Pr = t.P, + t.P, [Eq. (7.47)] [Eq. (7.3)) = 5.3 + 4.8 10.1 psi D; = 2.23/12 = 0.186 ft G, = (w/2)/a, = 3138,500 X 0.917 = 21,000 lb/(hr)(ft') For 60°API at 315 use data in Fig. 14 for 56°API gasoline as an approximation. ,. = 0.18 X 2.42 = 0.435lb/(ft)(hr) (1) Re,
= 59,200, f
=
§
=
d;
Re, = D;G,ft< = 0.186 X 21,000/0.435 = 8950
f = 0.00028 ftO/in.'
[Fig. 26] From Fig. 13.14, mol. wt. = 142 142 D enst'ty, P = 359 X 79 %92 X 14.7/19.7 = 0.337 lb /ft 1 souttetliquld at 335°F = 0.61 Poutlet liquid "" 0.61 X 62.5 = 38.1lb/ft• 38,500/62.5 Soutlettnix = 29,000/0.337 + 9500/38.1 0.071 8!nlet at 315°F = 0.625 SsT "" ~(0,625 + 0,071) = 0.35
VAPORIZERS, EVAPORATORS, AND REBOILER'S
485
Pressure Drop
AP 
JG! (Lt.tal) [Eq. (7.45)] •  5.22 X 1010D:s.p. 1 0.00028 X 21,000 X 12 5.22 X 101D X 0.186 X 0.35 X 1.0 = 0.0004psi
Summary h outside
286
Uc
116
UD
Ra
I
SO%o
90.7 Caic.ulated 0.0024
Ra Required 10.1
Calculated r.P
Neg
10.0
Allowable IJ.P
0.25
The dirt factor is somewhat low for continued service. gas oil line is insignificant.
The high pressure drop on the
When a reboiler is overdesigned, it may operate by breathing. As liquid enters the reboiler, it may be completely vaporized very quickly because of the overdesign. New liquid replaces it and cools the surface down. The new liquid remains in the rebojler momentarily and is heated and completely vaporized also, so that intermittent bursts of vapor issue from the reboiler outlet instead of a smooth continuous ;flow of vapor and liquid mixture. This can be overcome by reducing the pressure on the steam if steam is the heating medium or by placing an orifice on the shell outlet flange so as to cause an increased pressure drop on the vapor. Horizontal Thermosyphons with Baffles. Horizontal thermosyphons are occasionally designed with vertically cut baffies such as were discussed in connection with the 24 exchanger. The baffies do not affect the boiling film, but they do affect the sensibleheating coefficient, increasing it beyond the freeconvection value. If the baffies are the usual25 per cent, vertically cut, segmental baffies arranged for sidetoside flow, the sensible film coefficient can be computed from shellside data as given in Fig. 28 and as discussed under the 24 exchanger. If the baffies are cut 50 per cent corresponding to quartercircle support plates, the coefficient is treated on the basis of axial flow as above, using the equivalent diameter as calculated in Example 7.8 and the tubeside data of Fig. 24.
486
PROCESS HEAT TRANSFER
2. VAPORIZA.TION
IN THE TUBES
The members of this class are vertical units operating with a relatively large hydrostatic head and a low pressure drop. For this reason the vaporization usually occurs in the tubes of a onetubepass exchanger, which permits a greater recirculation rate than is common to horizontal units with vaporization in the shell. The three main classes of equipment employing this arrangement are the Sfeam vertical longtube evaporator, the t vertical thermosyphon reboiler, and the unfired steam generator. The
11 rtotdtnq head
exclumr~er ;I
Fro. 15.15. Naturalcirculation steam generator. This is similar to Fig. 14.27 but for foaming and scaling provisions.
Fro. 15.16. Vertical thermosyphon reboiler connected to tower.
steam generator is shown in Fig. 15.15, and it is similar to the vertical longtube evaporator in Fig. 14.27. The vertical thermosyphon reboiler is shown in Fig. 15.16, and it will presently be treated in detail to demonstrate the calculation common to naturalcirculation units. Recirculation Ratios. The recirculation ratio is attained when the· sum of the resistances in the vaporization circuit is equal to the hydrostatic driving force on the vaporizing fluid. Referring to the vertical thermosyphon in Fig. 15.16, there are five principal resistances: 1. Frictional pressure drop through the inlet piping 2. Frictional pressure drop through the reboiler (z2) 3. Expansion or acceleration loss due to vaporization in the reboiler
487
VAPORIZERS, EVAPORATORS, AND REBOILERS
4. Static pressure of a column· of mixed liquid and vapor (za) in the reboiler 5. Frictional pressure drop through the outlet piping Expansion Loss Due to Vaporization. This is taken as two velocity heads based on the mean of the inlet and outlet specific gravities.
GZ
psi
API=144gpa..
(15.14)
Particularly where the recirculation ratio and the operating pressure are great, the difference in the densities between the inlet and outlet are not very large and the expansion loss is negligible. Weight of a Column of Mixed Liquid and Vapor. This is difficult to evaluate if precision is required, since the expansion of the vapor is a function of the recirculation ratio, average specific volume of the vapor, coefficient of expansion of the liquid, etc. For nearly all practical cases it may be assumed that the variation of the specific gravity is linear between the inlet and the outlet. If v is the specific volume at any height x iii the vertical tube of Fig. 15.17 whose total length is L and whose inlet and outlet specific volumes are
. + (vo.
v = v.
 v,)x L
(15.15)
If the weight of the column of mixture is m, the change in weight with height is dm, and if a is the crosssection flow area, dm
=!!: dx
(15.16)
v
FIG. 16.17. Volume change in a single tube
If the static pressure of the column of liquid and vapor is designated by ZaPav and the crosssection area a is unity, 23
Pav
=
!V f Ldx
=
dx
Vi
+ (Vo 
V;)x/L
psi
(15.17)
Integrating and dividing by 144 to obtain the head per square inch Za Pav
144
=
2.3£
1
Vo
144(vo  v;) og V,
psi
(15.18)
Rational solutions for the recirculation ratio can be established by taking all the heads iii the circuit into account as functions of the mass velocity G, and upon solution for G the recirculation rate can be obtained directly. Because the gravity of the reboiier outlet mixture also varies
488
PROCESS HEAT TRANSFER
with the recirculation ratio, the expression becomes complex and it is simpler to solve by trial and error. If the height of an existing reboiler is given, the recirculation ratio can be computed. If the recirculation ratio is given, the required head Z1p1 may be computed. a. Vertical Thermosyphon Reboiler. The vertical thermosyphon is usually a conventional 11 exchanger with the channel end up. The upper tube sheet is placed close to the liquid level of the bottoms in the distilling column. Since the reboiler can be set close to the column, the frictional loss in the inlet and outlet piping is usually negligible. The recirculation ratio is determined from the difference between the hydrostatic head in the distilling column corresponding to the tube length of the reboiler and the weight of the vaporliquid mixture. Recirculation ratios exceeding 4:1 are usually employed. Example 16.6. Calculation of a Vertical Thennosyphon Reboiler. A vertical thermosyphon reboiler is to provide 40,800 lb /hr of vapor which is almost pure butane. In an arrangement identical with Fig. 15.16 the column operates at a pressure of 275 psig corresponding to a nearly isothermal boiling point of 228°F. Heat will be supplied by steam at 125 psig. A recirculation ratio of 4: 1 or greater should be employed. What is the optimum exchanger to fulfill this requirement? %in. OD, 16 BWG tubes on lin. triangular pitch will be used.
Solution: (1) Heat balance:
Enthalpy of liquid at 228°F and 290 psia = 241 Btujlb Enthalpy of vapor at 228°F and 290 psia = 338 Btujlb Butane, Q = 40,800(338  241) = 3,960,000 Btujhr Steam, Q = 4570 X 868 = 3,960,000 Btujhr (2) At: Isothermal boiling t.t = LMTD = 125"F (8) Tc and tc: Both streams are isuthermal.
(Fig, 9)
(Table 7) {5.14)
Trial1 (see Chap. 11 for method of approach): (a) Whe.n establishing reboiler surface the first trial should always be taken for the maximum allowable flux A = __!L = 3,960,000 = 330 ft 2 Q/A 12,000
Assume 16'0" long tubes. These will reduce the shell diameter and provide the cheapest reboiler. However, it will also require the greatest elevation of the column. Number of tubes = 330/16'0" X 0.1963 = 105 (Table 10) (b) Since this will be a 11 exchanger, only one tube pass From the tube counts: 105 tubes, 1 pass, %:in. OD, lin. triangular pitch. Nearest count: 109 tubes in a 13X in. ID shell (c) Corrected coefficient U D! A = 109 X 16'0" X 0.1963 = 342 ft 2 3,960,000 92 5 U D = 342 X 125 = .
489
VAPORIZERS, EVAPORA.TORS, AND REBOILERS Recirculation ratio: Assume 4: 1 recirculation ratio.
Stat1c · pressure of reb oiler 1eg, Za Pav = 2.3L 144 144 (v. _ Vapor density, l'vaPC>r
IIJtauid
p,
= 359 X
= 2 _~ 7
68
%::
!1;)
Vo 1og U.
ps1•
(15.18)
3 X 14 _ 71290 = 2.27lb/ft
= 0.44 ft 3/lb
= !li = 62,5 ~ 0.4S = 0.0372 ft'/lb
Weight flow of recirculated liquid = 4 X 40,800 Total volume out of reboiler:
(Fig. 6)
= 163,200 1bjhr
Liquid, 163,200 X Q.0372 = 6,100 ft 8 Vapor, 40,800 X 0.44 = 17,950 ft 8 Total == 24,050 ft 8 Vo
z,
=
(163,2~~,~~0,800)
= 0.1175 ftaj1b
2.3 X 16 X I 0.1175 60 • P ressure 0 f 1eg, 144 = 144(0.1175  0.0372) og 0.0372 = 1· psi PaT
Frictional resistance: Flow area:
a; = a,= N,144n
0.302 109 X   = 0.229 ft•. 144
(Table 10)
a, = a, ~ = 163,200 + 4o,soo = 891,ooo lb/(hr)(ft•) 0.229 At 228°F, p. = 0.10 X 2.42 = 0.242lb/(ft)(hr)
R6
'
J.'
2
0.242
'
f = 0.000127 ft2/in. 2 meon = (0,43 + 1/0il75 X 62.5) = _ 8 0 285 AP,
= 5.22
=
(Fig. 26)
. 0.000127 X 891,000 2 X 16 = _ si 2 09 p 5.22 X 101 • X 0.0517 X 0.285 X 1.0 1.60 2.09 = 8.69 psi
X 10 10Daq,,
Total resistance
(Fig. 14)
i~ = o.os17 ft = DG = 0.0517 X 891,000 = 190 OOO
D =
0
( . ) 7 45
+
Driving force, ~: = 16 X 0.43 X 62.5/144 = 2.98 psi no check The resistances are greater than the hydrostatic head can provide; hence the recirculation ratio will be less than 4: 1. Of the resistances the frictional pressure drop may be reduced by the square of the mass velocity if the tubes are made shorter. The other alternative is to raise the liquid level in the column above the upper tube sheet. Trial2: Assume 12'0" tubes and 4:1 recirculation ratio: (a) Number of tubes "" 330 ft2 = 330/12'0" X 0.1963 = 140 (b) from the tube counts: 140 tubes, one pass, %in. OD, lin. triangular pitch
Nearest count: 151 tubes in a 15H in. ID shell
490
PROCESS HEAT TRANSFER
(c) Corrected coefficient U D :
A
= 151 X 12'0" X 0.1963 = 356ft'
uD
= 3,960,000/356 X 125 = 89.0
Recirculation ratio: Assume 4: 1 recirculation ratio v; = 0.0372 as before v. = 0.1175 • fl Z3P&v 2.3 X 12 l 0.1175 Stat lC pressure 0 eg, 144 = 144(0.1175  0.0372) og 0.0372
=
.
1.20 ps!
Frictional resistance: 0 302 · a 1 = 151 X. 144 = 0.316 ft• G,
2
0
= ~3~ = 645,000 lb/(hr)(ft•) 645,000
Re, = 0.0517 X 0. 242 = 138,000 f = 0.000135 ft 2 /in.• 0.000135 X 645,000 2 X 12
M>,
= 5.22 X 1010 X 0.0517 X 0.285 X 1.0
.
= 0 ·88 psi
= 1.20 + 0.88 = 2.08 psi ~t~ = 12 X 0.43 X ~~ = 2.24 psi
Total resistance Driving force,
Since the driving force is slightly greater than the resistances, a recirculation ratio better than 4:1 is assured. With a mass velocity of 645,000 lb/(hr)(ft 2 ) equivalent to an wet velocity (V = G,j3600p) of 645,000/3600 X 62.5 X 0.43 = 6.7 fps the butane boiling coefficient may be computed as for forced circulation. Hot fluid: shell side, steam
(9') Condensing steam h. = 1500 Btu/(hr)(WWF)
Cold fluid: tube side, butane (4), (5), (6) Re, = 138,000 (7) jH = 330 [Fig. 24] (8) k(cp.jk)'A = 0.115 Btu/(hr)(ft')(°F/ft) (9) he = (jnk/D)(cp.jk)~ = 330 X 0.115/0.0517 = 735
This exceeds the maximum. Use 300. (10) h., = h. X ID/OD = 300 X 0.62/0.75 = 248 Btu/(hr)(ft)("F) (13) Clean overall coefficient Uc:
Uc
=
h;.h. h;. + h.
= 1500 X·248 1500
+ 248
= 213 Btu/(hr)(fts)("F)
(6.38)
(14) Dirt factor Ra: U D has been obtained above.
R = Ua  U D = 213  89 = 0 0065 (hr)(ft')(oF)/Bt d UaUD 213 X 89 · u
(6.13)
Pressure Drop: The pressure drop through the reboiler has been computed, 0.88 psi.
The head elevation z1 will be 12ft .. The pressure drop on the shell using halfcircle support plates is negligible.
491
VAPORIZERS, EVAPORATORS, AND tlEBOILERS Summary 1500
I
h outside
Uo
213
UD
89
I
248
R" Calculated 0.0065 R" Required 0.0040.006
Neg
Calculated tJ'
0.88
Neg
Allowable tJ'
0.88
The large dirt factor must be retained because of the flux requirements. This is clearly an instance in which the high temperature of the steam yields no advantage. If the steam temperature were lower, a higher value of U D could be used and the surface would remain the same. The final reboiler will be Shell side ID = 157.. Viscosity at the caloric temperature, lb/(ft)(hr) I" K
K.
VAPORIZERS, EVAPORATORS, AND REBOILERS p,.
p
q/
Viscosity at the tube wall temperature, lb/(ft)(hr) Density, lb/ft• Viscosity ratio, (p./J.<w)o.u
Subscripts (except as noted above) B D F
l m
m+1 n n i 1 p R
Bottoms Distillate Feed Liquid Plate below the feed in a distilling column Plate below the m plate Plate above the feed Plate below the n plate Bottom plate Reflux Shell Tube Vapor
511
CHAPTER 16 EXTENDED SURFACES 1
Introduction. When additional metal pieces are attached to ordinary heattransfer surfaces such as pipes or tubes, they extend the surface available for heat transfer. While the extended surface increases the total transmission of heat, its influence as surface is treated differently. from simple conduction and convection. Consider a conventional double pipe exchanger whose cross section is shown in Fig. 16.1a. Assume that hot fluid flows in the annulus and cold
(01)
FIG. 16.1.
(b) Ordinary and finned tube.
fluid in the inner pipe, both in turbulent flow, and that the effective temperatures over the cross section are Tc and t., respectively. The heat transferred can be computed from the inner pipe surface, the annulus coefficient, and the temperature difference T.  tw,' where tw is the temperature of the outer surface of the inner pipe. Next assume that strips of metal are welded to the inner pipe as shown in Fig. 16.lb. Since the metal strips are attached· to the cold tube wall, they serve to transfer additional heat from the hot fluid to the inner pipe. The total surface available for heat transfer no longer corresponds to the outer circumfer ence of the inner pipe but is increased by the additional surface on the sides of the. strips. If the metal strips do not reduce the conventional 1 Comprehensive mathematical treatments of extended surfaces are. given by Carslaw, H. 8., and J. C. Jaeger, "Conduction of Heat in Solids," Clarendon Press, Oxford, 1947. Jakob, M., "Heat Transfer," John W'tley & &ns, Inc., New York, 1949.
512
EXTENDED SURFACES
513
annulus heattransfer coefficient by appreciably changing the fluidflow pattern, more heat will be transferred from the annulus fluid to the pipe fluid. Pieces which are employed to extend the heattransfer surfaces are known as fins. It will be shown in the case of pipes and tubes, however, that eacl_! square foot of extended surface is less effective than a square foot of unextended surface. Referring again to Fig. 16.1b, there is a temperature difference T.  t1 between the annulus fluid and the fin, and the heat which flows into the fin will be conducted by it to the inner
FIG. 16.2. Some commercial forms of extended surface. a. Longitudinai fins. (GTiscomRussell Co.) b. Transverse fins. (GTiscomRussell Co.) c. Discontinuous fins. (Babcock and Wilcox Co.) d. Pegs or stud$. (Babcock and Wiloox Co.) e. Spines. (Thermek Corporation.) (Gardner, Tramactiom of the ASME.)
pipe. For heat to be conducted to the pipe, t1 must be greater than the pipewall temperature t,.. Then T.  t1 is less than T.  t... Since the effective temperature difference between the fluid and fin is less than that between the fluid and pipe, it results in less heat transfer per square foot of fin than for the pipe. Furthermore, the temperature difference between the fluid and the fin changes continuously from the outer extremity to the base of the fin owing to the rate at which heat enters the fin by convection and the rate at which it is transferred to its base by conduction.
514
PROCJESB HEAT TRANSFER
It will be found that two fundamental heattransfer principles enter into the various relationships in fills; (1) to determine from the geometry and conductivity of the fin the nature of the temperature variation and (2) to determine the heattransfer coefficient for the combination of fin and unextended surface. In the case of the double pipe exchanger, for example, the fins suppress the spiral eddies a;bout the annulu!!!, which in turn reduces the convection coefficientfor the annulus below its conventional value as determined from Eq. (6.2). Classification of Extended Surfaces. Fins of a number of industrial types are shown in Fig. 16.2. Pipes and tubes with longitudinal fins are marketed by several manufacturers and consist of long metal strips or channels attached to the outside (a) (b) of the pipe. The strips are attached PEENED WELDED either by grooving and peening the tube FIG. 16.3. Fin attachment. as in Fig. 16.3a or by welding continuously along the base. When channels are attached, they are integrally welded to the tube. as in Fig. 16.3b. Longitudinal fins of this type are commonly used in double pipe exchangers or in unbaffied shellandtube exchangers when the flow proceeds along the axis of the tube. Longitudinal fins are most commonly employed in problems involving ljvb Groove 
rorhof
. n"mf
{01)
HELICAL FINS
(b)·WELDED DISCTYPE FINS
(c)·SHRUNK ON
DISCTYPE FINS FIG. 16.4.
Transverse fins.
gases and viscous liquids or when the smallness of one of a pair of heattransfer streams causes steamline flow. Tran8Verse fins are made in a variety of types and are employed primarily for the cooling and heating of gases in crossjiow. The helical fins in Fig. 16.4a are classified as transverse fins and are attached in a variety of ways such as by grooving and peening, expanding the tube metal itself to form the fin, or welding ribbon to the tube continuously. Disctype
EXTENDED SURFACES
515
fins are also transverse fins and are usually welded to the tube or shrunken to it as shown in Fig. 16.4b and c. In order to shrink a fin onto a tube a disc, with inside diameter slightly hiss than the outside diameter of ths tube, is heated until its inside diameter exceeds the outside diameter of the tube. It is slipped onto the tube, and upon cooling, the disc shrinks to the tube and forms a bond with it. Another variation of the shrunkon fin in Fig. 16.4c employs a hollow ring in its hub into which a hot metal ring is driven. Other types of transverse fins are known as discontinuous fins, and several shapes such as the star fin are shown in Fig. 16.5. Spine or pegtype fins employ cones, pyramids, or cylinders which extend from the pipe surface so that they are usable for either longitudinal flow or crossflow. Each type of finned tube has its own characteristics and effectiveness for the transfer of heat between the fin and the fluid inside the tube, and the remainder of this chapter deals with the derivation of the relationships and the appli(a I· STARTYPE FIN cations of the commonest types. Perhaps the principal future uses will be in the field of atomic energy for the recovery of controlled heat of fission, in the reversing exchanger for commercial oxygen plants, in jet propulsion, and in gasturbine cycles. LONGITUDINAL FINS
Derivation of the Fin Efficiency. The simplest fin from the etandpoint of manufacture as (bl·MDDIFIED STAR FIN well as mathematics is the longitudinal fin of FrG. 16.5. Discontinuous uniform thickness. For the derivation of its fins. characteristics it is necessary to impose the limitations and assumptions given by Murray 1 and later by Gardner. 2 1. The heat flow and temperature distribution throughout the fin is independent of time; i.e., the heat flow is steady. 2. The fin material is homogeneous and isotropic. 3. There is no heat source in the fin itself. 4. The heat flow to or .from the fin surface at any point is directly proportional to the temperature difference between the surface at that point and the surrounding fluid. 5. The thermal conductivity of the fin is constant. 1
·2
Murray, W. M., J. Applied Mechanics, 5, A7880 (1938). Gardner, K. A., Trans. ASME, 67, 621632 (1945).
516
PROCESS HEAT TRANSFER
6. The heattransfer coefficient is the same over the entire fin surface. 7. The temperature of the fluid surrounding the fin is uniform. 8. The temperature of the base of the fin is uniform. 9. The fin thickness is so small compared with its height that temperature gradients across the width of the fin may be neglected. 10. The heat transferred through the outermost edge of the fin is negligible compared with that passing into the fin through its sides. 11. The joint between the fin and the tube is assumed to offer no bond resistance. At any cross section as in Fig. 16.6 let T. be the constant temperature of the hot fluid everywhere surrounding the fin and let t be the tempera
FIG. 16.6.
Derivation of the longitudinal fin efficiency.
ture at any point in the fin and variable. Let e be the temperature difference driving heat from the fluid to the fin at any point in its cross section. Then (16.1) e = T. t If l is the height of the fin varying from 0 to b,
(16.2) The heat within the fin which passes through its cross section by conduction is
de Q = ka~ df
(16.3)
EXTENDED SURFACES
517
where a., is the crosssectional area of the fin. This is equal to the heat which passed into the fin through its sides from l = 0 down to the shaded cross section. If P is the perimeter of the fin, the area of the sides is P dl and the film coefficient from liquid to fin side, whether on .fin surface or tube surface, is h1• or
(16.4)
Differentiating Eq. (16.3) with respect to l (16.5) Equations (16.4) and (16.5) are equalities. (16.6) Rearranging, (16.7) The direct solution of this equation is (16.8)
Let m
= (h1P)~ ka.,
The general solution is (16.9) At l
=
0 (16.10)
where the subscript refers to the outer edge of the .fin. If no heat enters the fin at the extreme end, as qualified by assumption 10, dB/dl = 0 when e = o·and Ct C2 = 0.
~·
(16.11)
e eml + eml e. = 2
(16.12)
e =e. cosh ml
(16.13)
c1 Equation (16.8) becomes or in general terms
=
c2
=
518
PROCESS HEAT TRANSFER
At the base of the fin where l = b
e, = e. cosh mb
(16.14)
where the subscript refers to the base of the fin. Thus an expression has been obtained for the temperature difference between the constant fluid temperature and variable fin temperature in terms of the length of the fin. It is now necessary to obtain an expression for Q in terms of l. From Eq. (16.4) by differentiation with respect to the fin height l, (16.15)
Substituting in Eq. (16.3), (16.16) (16.17) As before, the solution is
(16.18)
Atl = 0 c~
+ c; = o dQ
and
di
~~ =
h1Pe. =
c~
c~
=0 mC~ c~
c~ = h1Pe.
=

mC~ =
(16.19)
= _ h,Pe.
2m
Q = h1Pe.
0
2m eml _
2m
h1Pe. e'"l 2m
(16.20)
In terms of hyperbolic functions h1Pe • . h ml Q = sm m
(16.21)
1Pe • . h b Qb= h~sin m
(16.22)
The ratio of heat load Qb to the temperature difference
e. at the base is
Q. h1Pe. sinh mb eb = me. cosh mb
(16.23)
hp Q .2 ==  1 tanh mb
(16.24)
or
eb
m
519
EXTENDED SURFAOJiJS
Define 1w, as the value of h, to the fin surface alone when referred to the base area of the fin at l = b. Calling the ratio 1w,jh1 = n the efficiency 1 of the fin, the average value of the heattransfer coefficient at the base of the fin is given by Fourier's equation: (16,25) The fin efficiency hb/h1 may be defined by Eqs. (16.24) and (16.25). 51
=
!!:!!.
h1
=
Qb/ebbP mQb 1 e~. tanh mb
tanh mb mb
(16.26)
Equation (16.26) applies only to the fin and not to the bare portion of the tube between the fins. To give the total heat removed by a finned tube, the heat flowing into the fin with a coefficient h1 must ultimately be combined with that flowing into the bare tube at the tube outside diameter. For this it is necessary to establish some reference surface at which the coefficient on the different parts can be reduced to the same heat flux. In an ordinary exchanger ho is referred to the tube outside diameter. Because there is no simple reference surface on the outside of finned tubes it is convenient to use the inside diameter of the tube as the reference surface at which local coefficients are corrected to the same heat flux. By definition h1 is the coefficient to all the outside surface, whether it be the fin or bare tube. Beneath the bases of the fins there is naturally a greater heat flux than on the bare tube surface between fins, since the heat flowing through the bases of the fins is greater per unit of tube area. It may also be expected that part of the heat passing from the bases of the fins is conducted into the tube metal, so that the temperature difference between the annulus fluid and bare tube is not strictly constant. It is not usually necessary to correct for this effect, since the bare tube area so affected by the increased flux near the basef? of the fins is ordinarily small compared with the total bare tube area. At the inside diameter of the tube, however, the heat from both. the fins and the bare tube surface is assumed to have attained a uniform flux. The total heat removed from the annulus liquid and arriving at the tube inside diameter is a composite of the heat transferred by the fins to the tube outside diameter and that transferred directly to the bare tube surface. These niay be combined by means of the weighted effi~ 1 Another term which is used in the literature is the "fin effectiveness," which is the efficiency multiplied by the ratio of fin eurfaee to fin base area.
520
PROCESS HEAT TRANSFER
ciency 0'. If the heat transferred through the bare tube surface at the tube outside diameter is designated by Q., then (16.27)
where A. is the bare tube surface at the outside diameter exclusive of the area beneath the bases of the fins. If there are N 1 fins on the tube, bPN1 is all of the fin surface. The total heat transfer at the outside diameter is given by
+ Q. = hbbPN,e. + h1A.e. = (h•bPN1A. + h,bPN,A.) e.
Q=
Qb
A.
=
bPNt
(~•. + b;Iv) QPN A.e.
(16.28)
1
Substituting Eq. (16.26) to eliminate hb
Q = (bPN1 tan!tb
+ A.) h18•
(16.29)
Calling h1o the composite value of h1 to both the fin and bare tube surfaces when referred to the outside diameter of the tube, the weighted efficiency is by .definition O' = h1./h1 . Combining Eqs. (16.24) and (16.29) 0,
= h1• = ( h1
bPN, tanh mb +A.) mb bPN1 +A.
(16.30)
But as in Eq. (6.2) the values of the coefficients vary inversely with the heatflow area. If h1i is the value of the composite coefficient h1• referred to the tube inside diameter,
hli hfo
=
bPN1
+A.
(16.31)
A;
Substituting in Eq. (16.31) h . = (bPN tanh mb I•
I
mb
+ A •)
h1
A,
(16.32)
From Eq. (16.32) (16.33)
or simply h1,
=
(OAt +A.) },
(16.34)
EXTENDED SURF ACES
521
Thus an equation has been. obtained which gives directly the heattransfer coefficient on the inside of an extended tube which is the equivalent of a value of h1 on the outside surface of the tube. By substituting the physical and geometrical factors for ·a given tube and fin arrangement a weighted efficiency curve can be developed relating h, to h,, based on the inside tube surface. Bonilla 1 has presented a graphical solut]on of Eq. (16.34) for the rectangular fin of uniform thickness. The method of derivation employed here is applicable to other types of 1ongitudinal fins as well, although it is ,rather awkward for the longitudinal fin of triangular cross section. A table of the hyperbolic tangents is given ]n Table 16.1. TABLE
16.1.
HYPERBOLIC TANGENTS
mb
Tanh mb
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 2.0 3:0 4.0
0.0000 0.0997 0.1974 0.2913 0.380 0.462 0.537 0.604 0.664 0.716 0.762 0.801 0.834 0.862 0.885 0.905 0.964 0.995 0.999
Tanh mb !Til)
1.000 0.997 0.987 0.971 0.950 0.924 0.895 0.863 0.830 0.795 0.762 0.728 0.695 0.663 0.632 0.603 0.482 0.333 0.250
Example 16.1. Calculation of the Fin Efficiency and a Weighted Efficiency Curve. To illustrate the use of Eq. (16.34) a weighted fin efficiency curve will be developed for one of the common types of finned tubes used in double pipe exchangers. A double pipe exchanger employs a 1~ in. OD, 13 BWG steel inner tube to which are attached 20 fins 2{) BWG, % in. high. Through suitable bushings the inner pipe is inserted into a 3in. IPS outer pipe. Determine (a) the efficiency of the fins for various values of h~, (b) the weighted efficiency curve. l
Bonilla, C. F., Ind. Eng. Chem., 40, 10981101 (1948).
522
PROCESS HEAT TRANSFER
Solutzon.: Total outside surface, A: Fin surface, At = (20 X 0.75 X 12 X 2)/144 = 2.50 ft 2 /lin ft Bare tube surface, A. = (3.14 X 1.25  20 X 0.035)17{ 44 = 0.268 ft 2 /lin ft A 1 + A. = 2.50 + 0.268 = 2. 77 ft2 /lin ft Total inside surface, A; = (3.14 X 1.06 X 12)/144 = 0.277 ft•/lin ft (a) Fin efficiencies:
h,
5.24h,~
mb
tanh mb
0 =tanh mb mb
4 16 36 100 400 625 900
10.48 20.96 31.44 52.4 104.8 131 157
0.655 1.31 1.97 3.28
0.5717 0.864 0.962 0.995 1.00 1.00 1.00
0.872 0.660 0.498 0.303 0.152 0.122 0.102
6.55 8.19 9.82
The fin efficiencies are given in the last column. Note that, when the outside coefficient ht is 4.0, both sides of the fui contribute heattransfer surface which is 87.2 per cent as effective as bare tube surface. When ht = 100, the surface is only 30.3 per cent as effective, and when ht = 900 (water or steam), the extended surface is only 10.2 per cent as effective. This is because at the higher film coefficient the fin metal more nearly attains. the temperature of the hot fluid and the temperature difference between the fluid and fin is correspondingly less for given values of T. and t•• (b) The weighted efficiency curve directly gives the value of h1 ; obtained at the tube inside diameter when the coefficient to the fin and the bare tube is h1 . h,.
ht 4 16 36 100 400 625 900
X X X X X X X X
=
(flAt+
(OA1 (0.872 X 2.50 (0.660 X 2.50 (0.498 X 2.50 (0.303 X 2.50 (0.152 X 2.50 (0.122 X 2;50 (0.102 X 2.50
A.)!~
+ A.) / A; + 0.268) I 0.277 + 0.268) I 0.277 + 0.268) I 0,277 + 0.268) I 0.277 + 0.26'8) I 0.277 + 0.268) I 0.277 + 0.268) I 0.277
(16.34)
= = =
= = =
=
h1; 35.4 110.8 193.5 370 935 1295 1700
These values of h1 and h1; are plotted in Fig. 16.7. If the film coefficient to the fin is large, as with condensation or the cooling or heating of water in the annulus, there i~ little advantage to be gained from the use of fins particularly when a dirt ·factor must be included which may be the controlling resistance. Actually, a fin of a desired metal
523
EXTENDED SURFACES
can be designed to give a high efficiency for any reasonable value of ht, but as h1 increases, the size and cost of the fin also increases. Greater efficiencies can also be obtained by using fins which are not uniformly thick such as are treated later on. 2000
Ill
1500
I I
:1
1114 in. 0 D tube x 20 fins x 20 BWGlC34 in.hiqhx 3 in. IPS shell
erooo
..,...v
~y
1/
..8::J 800
[/
:600 :;S 500 .f! 400
7
I/
.s.
t .30.0
~ Q)
';;_200 .c:
i: 150
/
*~
~ 100 0 0
v
1/
v~'
v
80
"' 60 =r 50 .s:;'+. 40 30
/
f7
f 3
4 5 6 8 10 15 20 30 40 60 80 100 150 200 300 500 h; heat tr~:~nsfer coefficient to fin, Btu/
(16.40)
k;;
8. From h; and obtain U v•, the corrected overall design coefficient based on the inside tube diameter. 1
1
1
=+Uv; h;; h;
(16.41)
9. Obtain the fiuli: for the surface actually employed.
~;
= Ur;;At
(16.42)
:tO. Impos_e the flux from 9 on the individual resistances and determine whether the summation of the first two individual differences is the same as the assumed value
530
PROCESS HEAT TRANSFER
of T.  .tr.. and if the sum of all differences equals the true temperature difference. In the same order as before, the temperature drops a.re a. Annulus :film: At1 =
9/.41 h,,
{16.43a)
b. Annulus dirt, corrected to the inside diameter: (16.43b)
c. Fin and tube metal: .(16.43c) d. Tubeside dirt: (16.43d)
e. Tubeside film:: (16.43e)
When the tube fluid is a gas, water, or a similar nonviscous fluid, the tubeside viscosity correction can be omitted and rp(J can be taken as 1.0. The metal of which the fin is fabricated greatly affects the efficiency, particularly since k for steel is 26 and for copper 220 Btu/(hr)(ft2WF/ft). The Calculation of a Double Pipe Finned Exchanger. The method employed here close1y follows that of the outline in Chap. 6. The true temperature difference for a double pipe exchanger is the LMTD calculated for true counterflow or a value computed from Eq. (6.35) for seriesparallel flow. The effective tube length does not include the return bend between legs of a hairpin or any of the unfinned portion of the inner tube, The inside of the tube is used as the reference surface. Most manufacturers prefer to. use the outside as the reference surface, since it has a larger numerical value. The solutions of problems can be converted to manufacturers' data by multiplying the inside surface by the ratio (A 1 + Ao)/A, and dividing the overall coefficient U»>by this ratio. Tube lengths of 12, 15, 20, and 24 ft are considered reasonable for extendedsurface hairpins. Large tube lengths are permissible, since the fins on the inner pipe rest snugly on the outer pipe and there is no sagging. As explained previously, fouling factors cannot be combined as in ordinary double pipe exchangers because they are effective on widely different surfaces and each must be treated separately, Example 16.3. Calculation of a Double Pipe Extendedsurface Gas Oil .Cooler. It is desired to cooll8,000 lb/hr of 28°API gas oil from 250 to 200°F in double.Jiipe exchangers consisting of 3'in. IPS shells with 1,%in. IPS inner pipes on which are mounted 24 fins% in. high by 0.035 in. (20 BWG) wide. Water from 80 to 120°F will serve as the cooling medium. Pressure drops of 10.0 psi are allowo.ble on both streams, and fouling factors of 0.002 for the gas oil and 0.003 for the water are required, How many 20ft hairpins will be required?
EXTENDED SURFACES
531
Solution: (1) Heat balance: Gas oil, Q = 18,000 X 0.53(250  200) = 477,000 Btu/hr (Fig. 4) Water, Q = 11,950 X 1.0(120  80) = 477,000 Btu/hr (2) Ll.t: Assume true counterflow for the first trial. Unless the allowable pressure drop on either stream is exceeded, it will not be necessary to consider seriesparallel arrangements calculated by Eq. (6.35). Ll.t = LMTD = 124°F (5.14) (3) Caloric temperatures T, and t.: 120 Ll.t, = 130 = 0 .92 }'T 0 .27 F c = 0 .47 (Fig. 17) Ll.th ~. = T.
t.
= 200
+ 0.47(250
= 80 + 0.47(120
 200) = 224°F  80) = 99°F
(5.28) (5.29)
Hot fluid: annulus, gas oil Cold fluid: inner pipe, water (4') 3 in. IPS, ID == 3.068 in. [Table 11] (4) D = 1.61/12 = 0.134ft [Table 11] 1~ in. IPS, OD = 1.90 in. [Table 11] a, = .. D 2/4 = 1r X 0.134•/4 = 0.0142 ft• Fin cross section, 20 BWG, 72 in. high = 0.035 X 0.5 = 0.0175 in. 2 [Table 10]
aa =
(""4"
X 3.068 2

"" X 4
1. 90 2

24
X 0.0175)
in. 2
= 4.13 = 4.13/144 = 0.0287 ft2 Wetted perimeter* = (,. X 1.90  24 X 0.035 24 X 2 X 0.5) = 29.13 in. d, = 4 X 4.13/29.13 = 0.57 in. D. = 0.57/12 = 0.0475 ft (6'') Ga = W /a. = 18,000/0.0287 = 628,000 lb/(hr) (ft2)
+
(5) G, = wja, = 11,950/0.0142 = 842,000 lb/(hr)(ft 2 ) V = G,j3600p = 842,000(3600 X 62.5 = 3.75 fps (6) At t, = 99°F, (6') At Tc = 224°F, I' = 0.72 X 2.42 = 1.74lb/(ft)(hr) p = 2.50 X 2.42 = 6.05lb/(ft)(hr) [Fig. 14] [Fig. 14] Re, = DG,jp(Re,isforpressuredroponly) Re. = D.G./p = 0.134 X 842,000/1.74 = 65,000 = 0.0475 X 628,000/6.05 = 4930 (7') it = 18.4 [Fig. 16.10) (8') At T. = 224°F k(cp/k)'tB = 0.25 Btu/(hr)(ft 2)("F /ft) [Fig. 16]
(9') h1 = j, ~. (
t)rs q,.
[Eq~
(6.15)] (9) h; = 970 X 0.82 = 795 Btu/(hr)(ft 2 J(OF)
[Fig. 25J 0 25 = 96.7 · "'" 0.0475 *In the derivation, the outermost edge of the fins was assumed to have zero heat
!!:£
= 18.4 X
transfer.
532
PROCESS HEAT TRANSFER
Calculation oft,,. (numbers refer to the outline procedure): (1) Assume T.  t1 ,. = 40°F t,,. = 224  40 = 184°F (2) At 184°F, Ww = 3.5 cp
(Fig. 14)
= (....!:..)b,14 = (2.5)0.1&
= 0.95 ,.,,. 3.5 (3) h1 = 96.7 X 0.95 = 91.8 Btu/(hr)(ft')(°F) r/>o
(4) R.w
1
h'
= 0.002, R1 = 9{.8
= 0.0109 (hr)(ft')(°F)/Btu
+ 0.0109
= 0.002
(16.37)
I
h;
= 77.5 (5) h~; = 255
,
(Fig. 16.9)
, A,+ A.
(6) h" = h1 ~ = 77.5 X 5.76 = 447 1 1 = 0.00169 255 447 (7) Assume rj>, = 1.0 for cooling water: Rd; = 0.003 R; = fr795 = 0.00126 1 h; = 0.00126 + 0.003 Rmotal =
1
1
(16.38) and (Fig. 16.9)
h;; .:... h'j,
(16.39)
=
(16.40)
h:
= 235 1 1 (S) UDi = 255
1
+ 235
(16,41)
Um = 122 (9) To obtain the true flux the heat load must be divided by the actual heattransfer surface. For a 1}2in. IPS pipe there are 0.422 ft 2 /lin foot (Table 11)
Trial: Ai =
uDiQ At
47 000 2 = 122 •X 124 = 31 . 5 ft
Length of pipe required =
;!:
= 74.8lin ft 2 Use two 20ft hairpins = 80 lin ft A; = 80 X 0.422 == 33.8 ft•
J,
=
4
;;3~800
= 14,100 Btu/(hr)(ft')
(~.) / h';.
(a) Annulus .film At1 = (b) Annulus dirt, At.w
=
1
~~f
= {~.) Rdo AI~ A.
(c) Fin and tube metal,
At....;~a~
(d) Tubeside dirt,
l!Jd; =
(e) Tubeside film,
At,=
=
(JJ
(JJ
31.6°
= 14,100 X
Rmetal
~~r
5~::
4.9 T,  t1 .. == 36.5°
= 14,100 X 0.00169 "" 23.8
Rd; = 14,100 X 0.003
(J,)/"• = •
0
42.3 11.1 120.3°
533
EXTENDED SURFACES
Assumed T. t1,. = 40.0° Calculated T.  t1,. = 36.5° LMTD = 124° The difference of 40  36.5 = 3.5°F will not materially change the value of (p./w..)•·u. Note: The temperature drop of 120.3°F corresponds to a fouling factor based on 74.8lin ft while the flux corresponds to 80 lin ft. The.correetion appears at the end of the solution. Pre1sure Drop
+'IT' X
3.07) (1) Re, = 65,000 (pipe) = 0.43 in. f = 0.000192 ft 2/in.2
(1') d; = 4 .X 4.13/(29.13
D;
&: f 8
!Fig. 26]
= 0.43/12 = 0.0359 ft = D:GB/,. 0.0359 X 628,000/6.05 = 3730 = 0.00036 ftl/in. 2 fFig. 16.10] = 0.82 [Fig. 6]
(2')
=
J(/fLn
fG!Ln
(2 ) AP, = 5.22 X l0 10Ds¢t
APe. = 5.22 X 1010D:s¢.
[Eq. (7.45)1 0.00036 X 628,000 2 X 80 5;22 X 1010 X {).0359 X 0.82 X 0.95 = 7.9psi
[Eq. (7 .45)
0.000192 X 842,000 2 X 80 5.22 X 101• X 0.134 X 1.0 X 1.0 1.4 psi
Summary Dirt Factor
0.002
h inside
255
Um
0.003 235
122
7.9
AP
1.4
Allowable AP
10.0
Calculated
10.0
AdjustTiulnl of the fouling factor. U Di = 122 was based on 74.8 lin ft. Based on 80 lin ft U n; = 114 and the difference is excess dirt factor equal to 0.00057. This !',an readily· be added to the tube side to give a dirt factor of 0.00357, or it can be added to the annulus by taking 0.00057 X 5:76 = 0.00328. The total annulus dirt factor for the latter will then be 0.002 + 0.00328 = 0.00528. For the new value of h~
1 ~ = 0.00528
1
+ 91.8
(16.37)
h~ = 62
h;, = 220 U ._ Do 
h~,h~
(Fig. 16.9) _ 220 X 235 _ 114 220 + 235 
h;; + h: 
(16.41)
534
PROCESS HEAT TRANSFER
The corrected summary is 0.0053
Dirt Factor
220
h inside
Un;
0.003 235

114
7.9
Calculated AP
1.4
10.0
Allowable AP
10.0
Remarks: There is no need to readjust the calculation oft,,., since only item (b) will be affected and this is usually insignificant in its contribution to the value of A!. Thus the corrected value is Atdo
= 14,100 X 5~~~ = 13.0° 0
3
This gives a computed At of 128.4° vs. 124.0°1 but this difference does not justify a new trial value of t1.,. The need for readjusting the fouling factor is of particular value when ch~cking the performance of an existing hairpin or battery of hairpins for a new service. As reported by a manufacturer the total surface would be 33.8 X 5.76 = 194 ft 2• The reported overall coefficient would be 114/5.76 = 19.8 Btu/(hr)(ft 2)(°F).
Extendedsurface Shellandtube Exchangers. The use of extended surfaces in double pipe exchangers permits the transfer of a great deal of heat in a compact unit. The same advantages may be obtained from the use of longitudinally finned tubes in shellandtube arrangements equivalent to the 11, 12 or 24 exchanger. Because they are relatively uncleanable, extendedsurface tubes are usually laid out on triangular pitch and are never spaced so closely that the .fins of adjacent tubes intermesh. To prevent sagging and the possibility of tube vibration which might result from intermeshing, each tube in a bundle is supported individually. This cannot be done with conventional support plates because they introduce a certain amount of flow across the bundle which cannot be accomplished very well with longitudinal finned tubes. Support is accomplished, however, by welding or shrinking small circumferential rings about each tube which enclose the fins but at different points along the length of each tube. The rings prevent any tubes from interme5hing and at the same time afford a positiye elimination of vibration damage. At several points along its length the entire bundle is then bound with circumferential bands which keep all of the finned tubes firmly pressed against the rings of adjacent tubes. Longitudinal fin exchangers are relatively expensive and, since they are
535
EXTENDED SURFAGES
not cleanable, can be used only for fluids which ordina.rily have very low film coefficients and which are clean or form dirt that can be boiled out. This makes them ideal for gases at low pressure where the density is low and the allowable pressure drop is accordingly small. The prototype of this exchanger is the 12 exchanger without bafiles as calculated in Example 7 .8. Longitudinal fin shellandtube exchangers are computed in the same manner as double pipe extendedsurface exchangers using the same efficiency curves for identical tubes. Only the equivalent diameters for heat transfer and pressure drop differ. These are computed in the conventional manner for the entire shell using four times the hydraulic radius as discussed in Chap. 6 and demonstrated in Example 7.8. Example 16.4:. Calculation of a Longitudinal Fin Shellandtube Exchanger. 30,000 lb/hr of oxygen at 3 psig pressure and 250°F is to be cooled to 100°Fusing water from 80 to 120°F, The maximum allowable pressure drop for the gas is 2.0 psi and for the water 10.0 psi. Fouling factors of not less than 0.0030 should be provided for each stream. Available for the service is a 19;\i in. ID 12 exchanger equipped with 70 16'0" tubes each with 20 fins J1 in. high of 20 BWG (0.035 in.) steel. The tubes are 1 in. OD, 12 BWG and are laid out on 2in. triangular pitch for four passes. Will the exchanger fulfill the service? What is the final gas side dirt factor? Solution:
Exchanger: Shell side Tube side ID = 19_!,4 in, Number and length = 70, 16'0", 20 fins, 20 BWG, in. Baflle Space = ring supports OD, BWG, pitch = 1 in., 12 BWG, 2in. tri. Passes= 4 Passes = 1 (1) Heat balance:
Oxygen at 17.7 psia, Q = 30,000 X 0.225(250 100) = 1,010,000 Btu/hr Water, Q = 50,500 X 1(100  80) = 1,010,000 Btu/hr (2) At:
Hot Fluid
Cold Fluid
250
Higher Temp
100
150
100
Lower Temp
80
20
150
Difference
.20
130
= 64.6°F R = ~O = 7.5
(5.14)
LMTD
20
s = 250 
At = 0.8i X 64.6 = 56.2°F
80
= 0.1175
(7.18)
Fr
= 0.87
(Fig. 18)
536
PROCESS HEAT TRANSFER
(3) T. and t.: The average temperatures of 175 and gooF will be adequate. Hot fluid: shell side, oxygen (4') a, =
11"
70
4
X 1g,252
(i X 1 + 2
= 211.5 in. 2
Cold fluid: tube side, water
a;
(4) = 0.47g in.• [Table 10] a, = N,a;/144n [Eq. (7.48)] 20 X 0.035 X 0.5) . = 70 X 0.47g/144 X 4 = 0.0582 ft• D = 0.782/12 = 0.0652 ft
= 211.5/144 = 1.47 ft 2 'Vetted perimeter = 70(.,. X 1  20 X 0.035 + 20 X 2 X 0.5) = 1570 in. d, = 4 X 211.5/1570 = 0.54in. [Eq. (6.4)] D, = 0.54/12 = 0.045 ft (6') G, = W fa, (6) Gt = w/at = 30,000/1.47 = 50,500/0.0582 = 20,400 lb/(hr)(ft 2) = 868,000 lb/(hr)(ft 2) V = Gt/3600p = 868,000/3600 X 62.5 = 3.86 fps (6') At T 4 = 175°F, JL = 0.0225 X 2.42 (6) At .t4 = gooF, = 0.0545lb/(ft)(hr) [Fig. 15) p. = 0.80 X 2.42 = l.g4Ib/(ft)(hr) Re, = D,G,jp.' [Fig. 14) = 0.045 X 20,400/0.0545 = 16,850 Re 1 = DG,fp. (for' pressure drop only) = 0.0652 X 868,000/l.g4 = 2g,100 ('1') jH = 5g,5 [Fig. 16.10a] (8') A1l175°F, c ,. 0.225; k = 0.0175 (cp./k)Y. = (0.225 X 0.0545/0.0175)'1& = 0.8g q,, = 1.0 (for gases) (9') ht
=it;, (7;)y.
q,,
[Eq. (6.15)] (9) h; = g4o X O.g6 == go3 [Fig. 25] Rdi = 0.003, hd; ~ 1/0.003 = 333 = 5g,5 X 0.0175 X 0.8g/0.045 hd;h; 333 X go3 = 20.5 h·I = · = =:::::::::=• hdi + h; 333 + go3 Rdo = 0.003, hdo = 1/0.003 = 333 = 243 Btu/(hr)(ft 2)(°F) [Eq. (16.40)] h' = hdoht = 333 X 20.5 = 1g. 3 I hdo + ht 333 + 20.5 [Eq. (16.37)) = 142 Btu/(hr)(ft•WF) [Fig. 16.g)
h;,
Overall design coefficient based on inside of tube U Di! U . D•
=
= 142 X243 = 8g.6 Btu/(hr)(ft2)(0F) h;;h;,h; + h; 142 + 243
(16.41)
Actual overaii coefficient based on inside of tube: Internal surface per lin ft = 0.2048 ft A; = 70 X 0.2048 X 16'0" = 230 ft 2 . 1,010,000 78 2 U Di = A;Qt:..t = 230 X 56.2 = .
(Table 10)
EXTENDED SURFACES
537
Adjustment of the fouling factor:
(u .1 )
Excess fouling factor =
D'aotu•l
1  U .= Ds.
Adding to the outside fouling factor: A,
1:
1 ~ I 2 0.~
1  89 6 = 0.00165 •
Ao = 9.27
+ 9.27 X 00165 = 0.0183 1 h1 ,aotual = 0.0183 + . = 14.9 20 5 Ra. = 0.003 I
h~;
= 113
Check of actual overall coefficient:
U

Dl 
243 h;,h:  113 X  77 3 h;; + h~  113 + 243 •
Check vs. 78.2
Pressure Drop
+1!' X
(1') d: = 4 X 211.5/(1570
= 0.52 in.
19.25) (1) For Re, = 29,100 [Eq. (6.5)] f = 0.00021 ft 2 /in.•
[Fig. 26]
D~ = 0.52/12 = 0.0433 ft Re; = D;G,jp.
f
= 0.0433 X 20,400/0.0545 = 16,200 = 0.00025 ft 2/in.• [Fig. 16.10b)]
Mol. wt. oxygen = 32 32 P = 359 X 6 3%92 X 14.7/17.7 = 0.083lb/ft3 8 = 0.083/62.5 = 0.00133 (2') AP JG!Ln •  5.22 X 1010D:sq,,
fG.Ln (2 ) AP, = 5.22 X l0 10Dsq,,
[Eq. (7.45)1
[Eq. (7.45)] 0.00021 X 868,000 2 X 16 X 4 0.00025 X 20,400' X 16 5.22 X 10 10 X 0.0652 X 1.0 X 1.0 = 5.22 X 10 10 X 0.0433 X 0.00133 X 1.0 = 3.0 psi = 0.6 psi Summary 0.018 113 UD;
Dirt Factor h inside
0.003 243
78.2
0.6
Calculated AP
3.0
2.0
Allowable AP
10.0
The large shellside dirt factor might suggest that the exchanger is consideral;>ly oversized. However, the number of finned tubes which can be fitted on 2in. triangular
538
PROCESS HEAT TRANSFER
pitch into a 17~in. shell is only 54. This gives a required value of U Di of 102 which is slightly greater than ean be performed in a 17Uin. exchanger.
TRANSVERSE FINS
Derivation of the Fin Efficiency. ·The different types of transverse finned pipes and tubes have been discussed at the beginning of the chapter. Expressions for their efficiencies are somewhat more difficult to derive than for longitudinal fins, since even t~e transverse fin of uniform  yb cross section does not reduce to the simple 1 ....2 differential equations of the longitud.inal fin of + I uniform cross sectio)l. The derivations given ~ here are those of Gardner, 1 which are ingenious ~ e because they develop a general expression that is applicable on modification to all the types of manufactured fins including the longitudinal fins. For the derivation of a general case, a transverse fin of varying cross section will be considered. The fluid surrounding the fin is again assumed to be hotter than the fin itself, and the flow of heat is. from the outer fluid to the fin. The same assumptions apply as before. Referring to Fig. 16.12, let e == T.  t wherP Tc is the constant hotfluid temperature FIG. 16.12. Derivation of the and t is the metal temperature at any point in transversefin efficiency. the fin. The heat entering the two sides of the fin between 22' and 11' is dependent upon the surface between the two radii r. and r 1 • Thus the total surface between 22' and 00' is a function of r (16.44) dQ = htedA
r
I
The heat which enters the fin between 22' and 11' flows toward its base through the cross section of the fin at 11'. Applying the Fourier equation de
Q = ka" dr
(16.45)
where k is the thermal conductivity and a., is the crosssectional area of the fin which in this case also varies with r Differentiating Eq. (16.45), _ dQ
dr
'Gardner, op. cit.
= !:._ (ka de) dr
"' dr
=
2
ka d 6 "' dr 2
+ k da,. de dr
dr
{16.46)
539
EXTENDED SURFACES
Equating Eqs. (16.44) and (16.46) 2
de + dr 2
(!a, daz) de + (!!L dA) e dr dr ka, dr
=
0
(16.47)
This secondorder differential equation is somewhat more difficult to evaluate than Eq. (16;6), which had simple roots, and can be solved by Bessel junctions. A number of secon.dorder differential equations arise in the solution of engineering problems which can be solved by a, power series. The solution .can be written as the sum of two arbitrary functions and two arbitrary constants as suggested by Eq. (16.9). In a general form the Bessel equ~ti.on is (16.48)
where n is a constant. Many types of functions which are independent solutions of the Bessel equation have been developed'; and their properties are tabulated. 1 Douglass has provided the following solution for Eq. (16.48) into which Eq. (16.47) may be transformed by multiplying by r 2• d2e r 2 dr 2
+ [(1  2m)r 2