BOOK Geankoplis Transport Processes and Unit Operations, Third Edition PDF

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Transport Processes and Unit Operations I. ! f ) . \ \r CHRISTIE J. GEANKOPLIS University of Minnesota Transport Processes and Unit Operations THIRD EDITION Prentice-Hall International. Inc. ISBN 0-13-045253-X This edition may be sold only in those countries to which it is consigned by Prentice-Hal...

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Transport Processes and Unit Operations

I. !

f

) \

\r

.

CHRISTIE J. GEANKOPLIS University of Minnesota

Transport Processes and Unit Operations THIRD EDITION

Prentice-Hall International. Inc.

ISBN 0-13-045253-X

This edition may be sold only in those countries to which it is consigned by Prentice-Hall International. It is not to be re-exported and it is not for sale in the U.S.A., Mexico, or Canada.

© 1993, 1983, 1978 by P T R Prentice-Hall, Inc. A Simon & Schuster Company Englewood Cliffs, New Jersey 07632

All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher.

Printed in the United States of America 10

9

ISBN 0-13-045253-X

Prentice-Hall International (UK) Limited, London Prentice-Hall of Australia Ply. Limited, Sydney Prentice-Hall Canada Inc., Toronto Prentice-Hall Hispanoamericana, S.A., Mexico Prentice-Hall of India Private Limited, New Delhi Prentice-Hall of Japan, Inc., Tokyo Simon & Schuster Asia Pte. Ltd., Singapore Editora Prentice-Hall do Brasil, Ltda., Rio de Janeiro Prentice-Hall, Inc., Englewood Cliffs, New Jersey

I

.;

Dedicated to the memory of my beloved mother, Helen, for her love and encouragement

Contents xi

Preface

PART 1 TRANSPORT PROCESSES: MOMENTUM, HEAT, AND MASS Chapter I 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

Classification of Unit Operations and Transport Processes SI System of Basic Units Used in This Text and Other Systems Methods of Expressing Temperatures and Compositions Gas Laws and Vapor Pressure Conservation of Mass arid Material Balances Energy and Heat Units Conservation of Energy and Heat Balances Graphical, Numerical, and Mathematical Methods

Chapter 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11

Principles of Momentum Transfer and Overall Balances

Introduction Fluid Statics General Molecular Transport Equation for Momentum, Heat, and Mass Transfer Viscosity of Fluids Types of Fluid Flow and Reynolds Number OveraII Mass Balance and Continuity Equation Overall Energy Balance Overall Momentum Balance Shell Momentum Balance and Velocity Profile in Laminar Flow Design Equations for Laminar and Turbulent Flow in Pipes Compressible Flow of Gases -

Chapter 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11

Introduction to Engineering Principles and Units

Principles of Momentum Transfer and Applications

Flow Past Immersed Objects and Packed and Fluidized Beds Measurement of Flow of Fluids Pumps and Gas-Moving Equipment Agitation and Mixing of Fluids and Power Requirements Non-Newtonian Fluids Differential Equations of Continuity Differential Equations of Momentum Transfer or Motion Use of Differential Equations of Continuity and Motion Other Methods for Solution of Differential Equations of Motion Boundary-Layer Flow and Turbulence Dimensional Analysis in Momentum Transfer

1

1 3 5 7

9 14 19 23

31 31 32

39 43

47 50 56

69 78 83 lOl 114

114

127 133

140 153 164

170 175

184 190 202

vii

Chapter 4

4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 1.10 4.11 4.12 4.13 4.14 4.15

Introduction and Mechanisms of Heat Transfer Conduction Heat Transfer Conduction Through Solids in Series Steady-State Conduction and Shape Factors Forced Convection Heat Transfer Inside Pipes Heat Transfer Outside Various Geometries in Forced Convection Natural Convection Heat Transfer Boiling and Condensation Heat Exchangers Introduction to Radiation Heat Transfer Advanced Radiation Heat-Transfer Principles Heat Transfer of Non-Newtonian Fluids Special Heat-Transfer Coefficients Dimensional Analysis in Heat Transfer Numerical Methods for Steady-State Conduction in Two Dimensions

Chapter 5

5.1 5.2 5.3 5.4 5.5 5.6 5.7

Principles of Steady-State Heat Transfer

Principles of Unsteady -State Heat Transfer

Derivation of Basic Equation Simplified Case for Systems with Negligible Internal Resistance Unsteady-State Heat Conduction in Various Geometries Numerical Finite-Difference Methods for Unsteady-State Conduction Chilling and Freezing of Food and BiolOgIcal Materials Differential Equation of Energy Change Boundary-Layer Flow and Turbulence in Heat Transfer

Chapter 6

Principles of Mass Transfer

6.1 Introduction to Mass Transfer and Diffusion 6.2 Molecular Diffusion in Gases 6.3 Molecular Diffusion in Liquids 6.4 Molecular Diffusion in Biological Solutions and Gels 6.5 Molecular Diffusion in Solids 6.6 ,_Nu~erical Methods for Steady-State Molecular Diffusion in Two Dimensions

Chapter 7

7.1 7.2 73

7.4 7.5 7.6 7.7 7.8 7.9 viii

Principles of Unsteady -State and Convective Mass Transfer

Unsteady-State Diffusion Convective Mass-Transfer Coefficients Mass-Transfer Coefficients for Various Geometries Mass Transfer to Suspensions of Small Particles Molecular Diffusion Plus Convection and Chemical Reaction Diffusion of Gases in Porous Solids and Capillaries Numerical Methods for Unsteady-State Molecular Diffusion Dimensional Analysis in Mass Transfer Boundary-Layer Flow and Turbulence in Mass Transfer

214 214 220 223 233

236 247 253 259 267 276 281 297 300

308 310

330 330 332

334 350 360

365 370

381 381 385 397 403 408 413

426 426 432 437 450

453 462 468 474

475 Contents

PART 2

UNIT OPERATIONS Chapter 8

8.1 8.2 8.3 8.4 85 8.6 8.7 8.8

Evap H 2 0

(1.5-10)

An accounting of the total moles of O 2 in the fuel gas is as follows: mol 02in fuel gas

(i)27.2(CO)

+ 5.6(C0 2) + 0.5(0 2 )

= 19.7 mol O 2

For all the Hz to be completely burned to H 20, we need from Eq. (1.5-10) mol O 2 for 1 mol H2 or 3.1(t) = 1.55 total mol O 2 , For completely burning the CO fwm Eq. 0.5-9), we need 27.2(~) = 13.6 mol O 2 , Hence, the amount of O 2 we must add is, theoretically, as follows:

i

mol O 2 theoretically needed = 1.55

+ 13.6 -

0.5 (in fuel gas)

1{65 mol O 2 For a 20"/" excess, we add 1.2(14.65), or 17.58 mol O 2 , Since air contains 79 mol % N z • the amount ofN2 added is (79/21)(17.58), or 66.1 mol N z . To calculate the moles in the final flue gas, all theH z gives H 20, or 3.1 mol H 2 0. For CO, 2.0% does not react. Hence, 0.02(27.2), or 0.54, mol CO will be unburned. A total carbon balance is as follows: inlet moles C = 27.2 + 5.6 = 32.8 mol C. In the outlet flue gas, 0.54 mol will be as CO and the remainder of 32.8 - 0.54, or 32.26, mol as CO 2 . For calculating the outlet mol O 2 , we make an overall O 2 balance. O 2 in O 2 out

=

19.7 (in fuel gas)

+ 17.58 (in

air)

=

37.28 mol O 2

(3.1/2) (in H 2 0) + (0.54/2) (in CO) + 32.26 (in CO 2) + free O 2

Equating inlet O 2 to outlet, the free remaining O 2 = 3.2 mol O 2 , For the N2 balance, the outlet = 63.6 (in fuel gas) + 66.1 (in air), or 129.70 mol N 2 . The outlet flue gas contains 3.10 mol H 2 0, 0.54 mol CO, 32.26 mol CO 2 , 3.20 mol O 2 , and 129.7 mol N 2 • In chemical reactions with several reactants, the limiting reactant component is defined as that compound which is present in an amount less than the amount necessary for it to react stoichiometrically with the other reactants. Then the percent completion df a reaction is the amount of this limiting reactant actually converted, divided by the amount originally present, times 100. Sec. 1.5

Conservation of Mass and Material Balances

13

1.6 1.6A

ENERGY AND HEAT UNITS

Joule, Calorie, and Btu

In a manner similar to that used in making material balances on chemical and biological processes, we can also make energy balances on a process. Often a large portion of the energy entering or leaving a system is in the form of heat. Before such energy or heat balances are made, we must understand the various types of energy and heat units. In the Sf system energy is given in joules (1) or kilojoules (kJ). Energy is also expressed in btu (British thermal unit) or cal (calorie). The g calorie (abbreviated cal) is defined as the amount of heat needed to heat 1.0 g water 1.0D C (from 14.5°C to 15.5°C). Also, 1 kcal (kilocalorie) 1000 cal. The btu is defined as the amount of heat needed to raise 1.0 lb water 1°F. Hence, from Appendix A.I, 1 btu

1.6B

252.16 cal

= 1.05506 kJ

(1.6-1)

Heat Capacity

The heat capacity of a substance is defined as the amount of heat necessary to increase the temperature by 1 degree. It can be expressed for I g, lib, 1 g mol, 1 kg mol, or I Ib mol of the substance. For example, a heat capacity is expressed in SI units as J/kg mol· K; in other u!1its as cal/g· dc. caljg mol, DC, kcal/kg mol· DC, btu/lb m • OF, or btu/lb mol· OF. It can be shown that the actual numerical value of a heat capacity is the same in mass units or in molar units. That is,

1.0 caljg' DC

=

1.0 btu/Ibm' OF

(1.6-2)

1.0 caljg mol· °C

=

1.0 btu/lb mol, OF

(1.6-3)

For example, to prove this, suppose that a substance has a heat capacity of 0.8 btu/Ibm' OF. The conversion is made using 1.8°F for PC or 1 K, 252.16 cal for 1 btu, and 453.6 g for 1 Ibm' as follows: cal ') heat capacity ( g 0c

btu )( cal)( 1 )( OF) ( 0.8 Ibm' OF 252.16 btu 453.6 g/lb 1.8 DC m

0.8

cal 0C g'

The heat capacities of gases (sometimes called specific heat) at constant pressure c p are functions of temperature and for engineering purposes can be assumed to be independent of pressure up to several atmospheres. I n most process engineering calculations, one is usually interested in the amount of heat needed to heat a gas from one temperature t I to another at 12 , Since the cp varies with temperature, an integration must be performed or a suitable mean cpm used. These mean values for gases have been obtained for Tl of 298 K or 25 c C (77°F) and various T2 values, and are tabulated in Table 1.6-1 at 101.325 kPa pressure or less as Cpm in kJ/kg mol, K at various values ofT2 in K or 0c. EXAMPLE 1.6-1. Heating ofN z Gas The gas N2 at 1 atm pressure absolute is being heated in a heat exchanger. Calculate the amount of heat needed in J to heat 3.0 g mol N z in the 14

Chap. }

Introduction to Engineering Principles and Units

TABLE

1.6-1.

Mean Molar Heat Capacities o/Gases Between 298 and TK (25 and roC) at 101.325 kPa or Less(SI Units: c p kJ/kg mol- K)

T(KI 298

25

373 473 573 673 773 873 973 1073 1173 1273 1473 1673

100 200 300 400 500 600 700 800 900 1000 1200 1400

H2

N2

co

Air

O2

H 20

CO 2

CH 4

S02

28.86 28.99 29.13 29.18 29.23 29.29 29.35 29.44 29.56 29,63 29,84 30,18 30.51

29.14 29.19 29.29 29,46 29.68 29.97 30.27 30.56 30,85 3U6 31.43 31.97 32.40

29.16' 29.24 29.38 29.60 29.88 30,19 30,52 30.84 31.16 31.49 31.77 32.30 32.73

29.19 29.29 29.40 29.61 29.94 30.25 30.56 30.87 31.18 31.48 31.79

29.38 29.66 30.07 30.53 31.01 31.46 31.89 32.26 32.62 32.97 33.25 33.78 34.19

33.59 33.85 34.24 34.39 35.21 35.75 36.33 36.91 37.53 38.14 38.71 39.88 40.90

37.20 38.73 40.62 42.32 43.80 45.12 46.28 47.32 48.27 49,15 49,91 51.29 52.34

35.8 37.6 40.3 43.1 45.9 48.8 51.4 54.0 56.4 58.8 61.0 64.9.

39.9 41.2 42,9 44.5 45.8 47.0 47.9 48.8 49.6 50.3 50.9 51.9

32.32 3276

Mean Molar Heat Capacities of Gases Between 25 and TOC at 1 atm Pressure or Less (English Units: cp = btu/lb mol· °F) T('C)

Hz

N,

CO

25 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 {700 1800 1900 2000 2100 2200

6.894 6.924 6.957 6.970 6.982 6.995 7.011 7.032 7.060 7.076 7.128 7.169 7.209 7.252 7.288 7,)26 7.386 7.421 7,467 7.505 7.548 7,588 7.624

6.961 6.972 6.996 7.036 7.089 7.159 7.229 7.298 7.369 7.443 7,507 7,574 7.635 7.692 7.738 7.786 7.844 7.879 7.924 7,957 7.994 8,028 8.054

6,965 6.983 7.017 7.070 7.136 7.210 7.289 7.365 7.44) 7.521 7.587 7,653 7.714 7,772 7.818 7,866 7,922 7,958 8.001 8.033 8.069 8.101 8.127

Air

O2

NO

6.972 7,017 7.134 6.996 7.083 7.144 7.021 7.181 7.224 7.073 7.293 7.252 7.152 7.406 7.301 7.225 7.515 7.389 7.299 7,616 7,470 7.374 7.706 7.549 7.447 7.792 7.630 7.520 7.874 7.708 7.593 7.941 7.773 7.660 8,009 7.839 7.719 8.068 7.898 7.778 8,123 7.952 7,824 8.166 7.994 7.873 8.203 8.039 7.929 8,269 8,092 7.965 8.305 8.124 8.010 8.349 8.164 8.043 8.383 8.192 8.081 8.423 8.225 8,115 8.460 8.255 8.144 8.491 8,277

H,O

CO,

8.024 8.084 8.177 8.215 8.409 8.539 8.678 8.816 8.963 9.109 9.246 9.389 9.524 9,66 9,77 9.89 9,95 10.13 10.24 10.34 10,43 10.52 10.61

8.884 9.251 9.701 10.108 10.462 10.776 11.053 11.303 11.53 11.74 11.92 12.10 12,25 12.39 12.50 12.69 12.75 12.70 12.94 13.01 13.10 13.17 13,24

HC! Cl 2 CH.

S02 C2 H. SO, C,H 6

6,96 6.97 6.98 7.00 7.02 7.06 7.10 7.15 7.21 7.27 7.33 7.39 7.45

9.54 9.85 10.25 1D.62 10.94 11.22 11.45 11.66 11.84 12.01 12.15 12.28 12.39

8:12 8.24 8.37 8,48 8.55 8.61 8.66 8.70 8.73 8.77 8.80 8.82 8.94

8.55 8.98 9.62 10.29 10.97 11.65 12.27 12.90 13.48 14.04 14.56 15.04 15.49

10,45 11.35 12.53 13.65 14.67 15.60 16.45 17.22 17.95 18.63 19.23 19.81 20.33

12.11 12.84 13.74 14.54 15.22 15.82 16.33 16.77 17.17 17.52 17.86 18.17 18.44

12.63 13.76 15.27 16.72 18.11 19.39 20.58 21.68 22.72 23.69 24.56 25.40 26.15

Source: O. A, Hougen, K. W. Watson, and R. A. Ragatz, Chemical Process Principles, Par! I, 2nd ed. New York: John Wiley & Sons, Inc" 1954. With pennission.

rollowing temperature ranges: (a) 298-673 K (25·400°C) (b) 298-1123 K (25-850°C) (c) 673-1123 K (400-850°C) Sec. 1.6

Energy and Heal Units

15

Solution:

For case (a), Table 1.6-1 gives c pm values at 1 atm pressure or less and can be used up to several atm pressures. For N z at 673 K,c pm = 29.68 kJ/kg mol· K or 29.68 J/g mol· K. This is the mean heat capacity for the range 298-673 K. heat required = M g mol

(c

pm

g

m~I' K}Tz- TdK

(1.6-4)

Substituting the known values, heat required

= (3.0)(29.68)(673 - 298) = 33390 J

For case (b), the cpm at 1123 K (obtained by linear interpolation between 1073 and 1173 K) is 3l.00 J /g mol· K. heat required

=

3.0(31.00)(1123 - 298)

=

76725 J

For case (c), there is no mean heat capacity for the interval 673-1123 K. However, we can use the heat required to heat the gas from 298 to 673 K in case (a) and subtract it from case (b), which includes the heat to go from 298 to 673 K plus 673 to 1123 K. heat required (673-1123 K)

=

heat required (298-1123 K) - heat required (298-673)

(1.6-5)

Substituting the proper values into Eq. (l.6-5), heat required

=

76725 - 33390

=

43 335 J

On heating a gas mixture, the total heat required is determined by first calculating the heat required for each individual component and then adding the results to obtain the total. The heat capacities of solids and liquids are also functions of temperature and independent of pressure. Data are given in Appendix Ac2, Physical Properties of Water; A.3, Physical Properties of Inorganic and Organic Compounds; and A.4, Physical Properties of Foods and Biological Materials. More data are available in (PI).

EXAMPLE 1.6-2.

Heating o/Milk

Rich cows' milk (4536 kg/h) at 4.4°C is being heated in a heat exchanger to 54.4°C by hot water. How much heat is needed?

Solution:

From Appendix A.4 the average heat capacity of rich cows' milk is 3.85 kJ/kg' K. Temperature rise, Ll T = (54.4 - 4.4 tC = 50 K. heat required

=

(4536 kgjh)(3.85 kJ/kg' K)(1/3600 h/s)(50 K)

=

242.5 kW

The enthalpy, H, of a substance in J/kg represents the sum of the internal energy plus the pressure-volume term. For no reaction and a constant-pressure process with a change in temperature, the heat change as computed from Eq. (1.6-4) is the difference in enthalpy, LlH, of the substance relative to a given temperature or base point. In other units, H = btu/Ibm or cal/g. 1.6C

Latent Heat and Steam Tables

Whenever a substance undergoes a change of phase, relatively large amounts of heat changes are involved at a constant temperature. For example, ice at oDe and 1 atm pressure can absorb 6013.4 kJ/kg mol. This enthalpy change is called the latent heat of fusion. Data for other compounds are available in various handbooks (PI, WI). 16

Chap. 1

Introduction to Engineering Principles and Units

When a liquid phase vaporizes to a vapor phase under its vapor pressure at constant temperature, an amount of heat called the latent heat oj vaporization must be added. Tabulations of latent heats of vaporization are given in various handbooks. For water at 25°e and a pressure of23.75 mm Hg, the latent heat is 44020 kJ/kg mol, and at 25°e and 760 mm Hg, 44045 kJ/kg mol. Hence, the effect of pressure can be neglected in engineering calculations. However, there is a large effect of temperature on the latent heat of water. Also, the effect of pressure on the heat capacity of liquid water is small and can be neglected. Since water is a very common chemical, the thermodynamic properties of it have been compiled in steam tables and are given in Appendix A.2 in SI and in English units.

EXAMPLE 1.6-3. UseoJSteam Tables Find the enthalpy change (i.e., how much heat must be added) for each of the following cases using SI and English units. (a) Heating 1 kg (Ibm) water from 2Llloe (70°F) to 60 e (140"F) at 101.325 kPa (1 atm) pressure. (b) Heating 1 kg (Ibm) water from 21.l1"e (70°F) to 115.6"e (24{rF) and vaporizing at 172.2 kPa (24.97 psi a). (c) Vaporizing 1 kg (Ibm) water at 115.6°e (240°F) and 172.2 kPa (24.97 psia). 0

Solution: For part (a), the effect of pressure on the enthalpy ofliquid water is negligible. From Appendix A.2, H at 21.11°e

88.60 kJ/kg

or

at

Hat 60 0 e

251.13 kJ/kg

or

at 140°F = 107.96 btu/Ibm

change in H = 6.H = 251.13

70"F

=

38.09 btu/Ibm

88.60 = 162.53 kJ/kg

= 107.96 - 38.09 = 69.87 btu/lb m

In part (b), the enthalpy at 115.6°e (240°F) and 172.2 kPa (24.97 psia) of the saturated vapor is 2699.9 kJ/kg or 1160.7 btu/Ibm. change in H

6.H

2699.9

88.60 = 2611.3 kJ/kg

1160.7

38.09 = 1122.6 btu/Ibm

The latent heat of water at 115.6°e (240°F) in part (c) is 2699.9

484.9 = 2215.0 kJ/kg