solving problem in food engineering pdf

Academia.edu no longer supports Internet Explorer.

To browse Academia.edu and the wider internet faster and more securely, please take a few seconds to  upgrade your browser .

Enter the email address you signed up with and we'll email you a reset link.

• We're Hiring!
• Help Center

Solving Problems in Food Engineering

except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights.

Related Papers

Mathivazhan Pandian

I Made Agus Putrawan

Adriana Lopes

মোমিনূর রশীদ রনি

Trần Minh Nhật

Shwan Kamiran

Miriam VasFlow

Antonio Lozano

Phillipe Gabriel

thermodinamics

anza khawaja , Waki Ahmed

Loading Preview

Sorry, preview is currently unavailable. You can download the paper by clicking the button above.

RELATED PAPERS

Richard Adio

Guruh Guruh

Nitin Prajapat

pedro badillo

Yara Rivera

Sitha Nurfitriani

chemical enginnering

Cecília Lenzi

Kamran Javed

Minh Anh Ngô

Ahmad Dzulfikri Nurhan

Andrés Romero

Ayan Saha Roy

Yuri G Melliza

RELATED TOPICS

•   We're Hiring!
•   Help Center
• Find new research papers in:
• Health Sciences
• Earth Sciences
• Cognitive Science
• Mathematics
• Computer Science
• Academia ©2023

3,000 Solved Problems in Physics (Schaum's Solved Problems ) (Schaum's Solved Problems Series)

2500 Solved Problems in Fluid Mechanics and Hydraulics (Schaum's Solved Problems )

101 Activities For Teaching Creativity And Problem Solving

Food Engineering Handbook, Two Volume Set: Food Engineering Handbook: Food Process Engineering

The Art and Craft of Problem Solving

Problem Solving : Best Strategies to Decision Making, Critical Thinking and Positive Thinking

3000 Solved Problems in Calculus

Learning to Solve Problems : A Handbook for Designing Problem - Solving Learning Environments

The Art Of Thinking In Systems: Improve Your Logic, Think More Critically, And Use Proven Systems To Solve Your Problems - Strategic Planning For Everyday Life

Animal, Vegetable, Miracle: A Year of Food Life

536 Puzzles and Curious Problems

Editor's Picks

Most Popular

Academic & Education

Business & Career

Children & Youth

Environment

Fiction & Literature

Health & Fitness

Personal Growth

Politics & Laws

Science & Research

• © 2008

Solving Problems in Food Engineering

• Stavros Yanniotis 0

Department of Food Science and Technology, Agricultural University of Athens, Athens, Greece

You can also search for this author in PubMed   Google Scholar

Introduction to food engineering problems for those who have very little to no background in engineering

Supplemental text that covers the basics of food engineering problem solving

A progressive degree of difficulty in the questions

Part of the book series: Food Engineering Series (FSES)

79k Accesses

3 Citations

• Table of contents

About this book

Authors and affiliations, bibliographic information, buying options.

• ISBN: 978-0-387-73514-6
• Instant PDF download
• Readable on all devices
• Own it forever
• Exclusive offer for individuals only
• Tax calculation will be finalised during checkout
• ISBN: 978-0-387-73513-9
• Dispatched in 3 to 5 business days
• Free shipping worldwide Shipping restrictions may apply, check to see if you are impacted .

This is a preview of subscription content, access via your institution .

Table of contents (18 chapters)

Front matter, conversion of units.

Stavros Yanniotis

Use of Steam Tables

Mass balance, energy balance, heat transfer by conduction, heat transfer by convection, heat transfer by radiation, unsteady state heat transfer, mass transfer by diffusion, mass transfer by convection, unsteady state mass transfer, pasteurization and sterilization, cooling and freezing, evaporation, psychrometrics, back matter.

Solving Problems in Food Engineering is a step by step workbook intended to enhance students' understanding of complicated concepts and to help them practice solving food engineering problems. The book covers problems in fluid flow, heat transfer, mass transfer, and the most common unit operations that have applications in food processing, such as thermal processing, cooling and freezing, evaporation, psychrometrics and drying. Included are theoretical questions in the form of true or false, solved problems, semi-solved problems, and problems solved using a computer. The semi-solved problems guide students through the solution. Some of the problems progress from elementary level increasing difficulty so that the book is useful to food science as well as food engineering students.

Aus den Rezensionen:

"… Das Buch ist mit Sicherheit eine Bereicherung für alle, die mit dem Vermitteln von Verständnis für lebensmitteltechnische Berechnungen in ihrer ganzen Bandbreite befasst sind. Auf der dem Buch beiliegenden CD sind einige komplexere Beispiele in Microsoft Excel zu finden - wenn auch mit etwas gewöhnungsbedürftiger Farbgestaltung ..." (Prof. Dr. Harald Rohm, in: Lebensmitteltechnik, 2008, Vol. 40, Issue 6, S. 65)

• Lebensmittel
• Lebensmitteltechnik
• food engineering
• food processing
• food science

Book Title : Solving Problems in Food Engineering

Authors : Stavros Yanniotis

Series Title : Food Engineering Series

DOI : https://doi.org/10.1007/978-0-387-73514-6

Publisher : Springer New York, NY

eBook Packages : Chemistry and Materials Science , Chemistry and Material Science (R0)

Copyright Information : Springer-Verlag New York 2008

Softcover ISBN : 978-0-387-73513-9 Published: 10 December 2007

eBook ISBN : 978-0-387-73514-6 Published: 03 December 2007

Series ISSN : 1571-0297

Edition Number : 1

Number of Pages : XI, 302

Number of Illustrations : 86 b/w illustrations

Topics : Food Science

• Most Popular
• Explore all categories

solving problems in food engineering

• Solving problems in food engineering

Author: profjorge-silva

Post on 02-Jun-2015

3.029 views

449 download

food engineering students

Food science students, similar problems, solutions of problems, solution of problems, solved computer problems, preface food engineering, mass transfer problems.

Embed Size (px) 344 x 292 429 x 357 514 x 422 599 x 487

• 1. Solving Problems in Food Engineering

2. Stavros Yanniotis, Ph.D. Author Solving Problems in Food Engineering 3. Stavros Yanniotis, Ph.D. Department of Food Science and Technology Agricultural University of Athens Athens, Greece ISBN: 978-0-387-73513-9 eISBN: 978-0-387-73514-6 Library of Congress Control Number: 2007939831 # 2008 Springer Science+Business Media, LLC All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, LLC., 233 Spring Street, New York, NY10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed on acid-free paper 9 8 7 6 5 4 3 2 1 springer.com 4. Tell me and I will listen, Show me and I will understand Involve me and I will learn Ancient Chinese Proverb 5. Preface Food engineering is usually a difficult discipline for food science students because they are more used to qualitative rather than to quantitative descrip- tions of food processing operations. Food engineering requires understanding of the basic principles of fluid flow, heat transfer, and mass transfer phenomena and application of these principles to unit operations which are frequently used in food processing, e.g., evaporation, drying, thermal processing, cooling and freezing, etc. The most difficult part of a course in food engineering is often considered the solution of problems. This book is intended to be a step-by-step workbook that will help the students to practice solving food engineering problems. It presumes that the students have already studied the theory of each subject from their textbook. The book deals with problems in fluid flow, heat transfer, mass transfer, and the most common unit operations that find applications in food processing, i.e., thermal processing, cooling and freezing, evaporation, psychometrics, and drying. The book includes 1) theoretical questions in the form true or false which will help the students quickly review the subject that follows (the answers to these questions are given in the Appendix); 2) solved problems; 3) semi- solved problems; and 4) problems solved using a computer. With the semi- solved problems the students are guided through the solution. The main steps are given, but the students will have to fill in the blank points. With this technique, food science students can practice on and solve relatively difficult food engineering problems. Some of the problems are elementary, but problems of increasing difficulty follow, so that the book will be useful to food science students and even to food engineering students. A CD is supplied with the book which contains solutions of problems that require the use of a computer, e.g., transient heat and mass transfer problems, simulation of a multiple effect evaporator, freezing of a 2-D solid, drying, and others. The objectives for including solved computer problems are 1) to give the students the opportunity to run such programs and see the effect of operating and design variables on the process; and 2) to encourage the students to use computers to solve food engineering problems. Since all the programs in this CD are open code programs, the students can see all the equations and the logic behind the calculations. They are encouraged to see how the programs work vii 6. and try to write their own programs for similar problems. Since food science students feel more comfortable with spreadsheet programs than with program- ming languages, which engineering students are more familiar with, all the problems that need a computer have EXCEL1 spreadsheet solutions. I introduce the idea of a digital SWITCH to start and stop the programs when the problem is solved by iteration. With the digital SWITCH, we can stop and restart each program at will. When the SWITCH is turned off the program is not running, so that we can change the values of the input variables. Every time we restart the program by turning the SWITCH on, all calculations start from the beginning. Thus it is easy to change the initial values of the input variables and study the effect of processing and design parameters. In the effort to make things as simple as possible, some of the spreadsheet programs may not operate on some sets of parameters. In such cases, it may be necessary to restart the program with a different set of parameters. I am grateful to Dr H. Schwartzberg, who read the manuscripts and made helpful suggestions. I will also be grateful to readers who may have useful suggestions, or who point out errors or omissions which obviously have slipped from my attention at this point. Athens Stavros Yanniotis May 2007 viii Preface 7. Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii 1. Conversion of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Examples Exercises 2. Use of Steam Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Review Questions Examples Exercises 3. Mass Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Review Questions Examples Exercises 4. Energy Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Theory Review Questions Examples Exercises 5. Fluid Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Review Questions Examples Exercises 6. Pumps. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Theory Review Questions Examples Exercises ix 8. 7. Heat Transfer By Conduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Theory Review Questions Examples Exercises 8. Heat Transfer By Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 Theory Review Questions Examples Exercises 9. Heat Transfer By Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Review Questions Examples Exercises 10. Unsteady State Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Theory Review Questions Examples Exercises 11. Mass Transfer By Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 Theory Review Questions Examples Exercises 12. Mass Transfer By Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 Theory Review Questions Examples Exercises 13. Unsteady State Mass Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 Theory Review Questions Examples Exercises 14. Pasteurization and Sterilization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 Review Questions Examples Exercises x Contents 9. 15. Cooling and Freezing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 Review Questions Examples Exercises 16. Evaporation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 Review Questions Examples Exercises 17. Psychrometrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 Review Questions Examples Exercises 18. Drying . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 Review Questions Examples Exercises References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 Appendix: Answers to Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . 275 Moody diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280 Gurney-Lurie charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 Heisler charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 Pressure-Enthalpy chart for HFC 134a . . . . . . . . . . . . . . . . . . . . . . . 285 Pressure-Enthalpy chart for HFC 404a . . . . . . . . . . . . . . . . . . . . . . . 286 Psychrometric chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288 Roots of d tand=Bi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290 Roots of dJ1(d)-Bi Jo(d)=0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 Roots of d cotd=1-Bi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 Error function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 Index. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 Contents xi 10. Chapter 1 Conversion of Units Table 1.1 Basic units Time Length Mass Force Temperature SI s m kg K, 0 C CGS s cm g K, 0 C US Engineering s ft lbm lbf 0 R, 0 F Table 1.2 Derived units SI US Engineering Force N (1 N = 1 kg m/s2 ) Energy J (1 J = 1 kg m2 /s2 ) Btu Power W (1 W = 1 J/s) HP, PS Area m2 ft2 Volume m3 (1m3 = 1000 l) ft3 Density kg/m3 lbm/ft3 Velocity m/s ft/s Pressure Pa (1 Pa = 1 N/m2 ) bar (1 bar = 105 Pa) torr (1 torr = 1 mmHg) atm (1 atm = 101325 Pa) psi=lbf/in2 Table 1.3 Conversion factors 1 ft = 12 in = 0.3048 m 0 F = 321.8* 0 C 1 in = 2.54 cm 0 C = (0 F-32)/1.8 1 US gallon = 3.7854 l 0 R = 460 0 F 1 lbm = 0.4536 kg K = 273.15 0 C 1 lbf = 4.4482 N 1 psi = 6894.76 Pa 0 C = 0 F/1.8 1 HP =745.7 W 0 C = K 1 Btu = 1055.06 J = 0.25216 kcal 0 F = 0 R 1kWh = 3600 kJ S. Yanniotis, Solving Problems in Food Engineering. Springer 2008 1 11. Examples Example 1.1 Convert 100 Btu/h ft2o F to kW/m2o C Solution 100 Btu h ft2 8F 100 Btu h ft2 8F 1055:06 J 1 Btu 1 kJ 1000 J 1 h 3600 s 1ft2 0:3048 m 2 1:8 8 F 18C 1 kW 1 kJ=s 0:5678 kW m2 8C Example 1.2 Convert 100 lb mol/h ft2 to kg mol/s m2 Solution 100 lb mol h ft2 100 lbmol h ft2 0:4536 kg mol lb mol 1 h 3600 s 1 ft2 0:3048 m 2 0:1356 kg mol s m2 Example 1.3 Convert 0.5 lbf s/ft2 to Pas Solution 0:5 lbf s ft2 0:5 lbf s ft2 4:4482 N lbf 1 ft2 0:3048 m 2 1 Pa 1 N=m2 23:94 Pa s Exercises Exercise 1.1 Make the following conversions: 1) 10 ft lbf/lbm to J/kg, 2) 0.5 Btu/lbm o F to J/kgo C, 3) 32.174 lbmft/lbfs2 to kgm/ Ns2, 4) 1000 lbmft /s2 to N, 5) 10 kcal/min ft o F to W/mK, 6) 30 psia to atm, 7) 0.002 kg/ms to lbmft s, 8) 5 lb mol/h ft2 mol frac to kg mol/s m2 mol frac, 9) 1.987 Btu/lbmol o R to cal/gmol K, 10) 10.731 ft3 lbf/in2 lbmol o R to J/kgmol K 2 1 Conversion of Units 12. Solution 1) 10 ft lbf lbm 10 ft lbf lbm ::::::::::::::m ft :::::::::::::::N 1 lbf ::::::::::::::lbm 0:4536 kg :::::::::::::J m N 29:89 J kg 2) 0:5 Btu lbm8F 0:5 Btu lbm8F 1055:06 J :::::::::: ::::::::::::: ::::::::::::: 1:88F 18C 2094:4 J kg8C 3) 32:174 lbm ft lbf s2 32:174 lbm ft lbf s2 ::::::::::::::: :::::::::::::::lbm ::::::::::::::::::m 1 ft ::::::::::::::: 4:4482 N 1 kg m N s2 4) 1000 lbm ft s2 1000 lbm ft s2 0:4536 kg :::::::::::::::: :::::::::::::::: 1 ft 1 N 1 kg m=s2 138:3 N 5) 10 kcal min ft oF 10 kcal min ft oF 1055:06 J 0:252 kcal ::::::::: min 60 s ::::::::::::ft :::::::::::m :::::::::::8F ::::::::::K :::::::::W ::::::::::J=s 4121 W m K 6) 30 psia 30 lbf in2 ::::::::::::::::in2 :::::::::::::::::m2 ::::::::::::::::::N ::::::::::::::::::lbf :::::::::::::::::Pa ::::::::::::::N=m2 :::::::::::::::::atm :::::::::::::::::Pa 2:04 atm 7) 0:002 kg m s 0:002 kg m s ::::::::::::::lbm :::::::::::::::kg :::::::::::::::m ::::::::::::::::::ft 0:0013 lbm ft s 8) 5 lb mol h ft2 mol frac 5 lb mol h ft2 mol frac ::::::::::::::::kg mol ::::::::::::::: lb mol :::::::::::::::::h ::::::::::::::::::s :::::::::::::::::ft2 :::::::::::::::::::m2 6:78 103 kg mol s m2 mol frac 9) 1:987 Btu lb mol 8R 1:987 Btu lb mol 8R ::::::::::::::cal ::::::::::::::Btu ::::::::::::::::lb mol ::::::::::::::::g mol ::::::::::::::8R ::::::::::::K 1:987 cal g mol K 10) 10:731 ft3 lbf in2 lb mol8R 10:731 ft3 lbf in2 lb mol8R ::::::::::::::::m3 :::::::::::::::::::ft3 ::::::::::::::::N ::::::::::::::::::lbf ::::::::::::::::in2 :::::::::::::::::::m2 :::::::::::::lb mol :::::::::::::kg mol 1:88R K 8314 J kg mol K Exercises 3 13. Exercise 1.2 Make the following conversions: 251o F to o C (Ans. 121.7 o C) 500o R to K (Ans. 277.6 K) 0.04 lbm/in3 to kg/m3 (Ans. 1107.2 kg/m3 ) 12000 Btu/h to W (Ans. 3516.9 W ) 32.174 ft/s2 to m/s2 (Ans. 9.807 m/s2 ) 0.01 ft2 /h to m2 /s (Ans. 2.58x10-7 m2 /s) 0.8 cal/go C to J/kgK (Ans. 3347.3 J/kgK) 20000 kg m/s2 m2 to psi (Ans. 2.9 psi) 0.3 Btu/lbm o F to J/kgK (Ans. 1256 J/kgK) 1000 ft3 /(h ft2 psi/ft) to cm3 /(s cm2 Pa/cm) (Ans. 0.0374 cm3 /(s cm2 Pa/cm ) 4 1 Conversion of Units 14. Chapter 2 Use of Steam Tables Review Questions Which of the following statements are true and which are false? 1. The heat content of liquid water is sensible heat. 2. The enthalpy change accompanying the phase change of liquid water at constant temperature is the latent heat. 3. Saturated steam is at equilibrium with liquid water at the same temperature. 4. Absolute values of enthalpy are known from thermodynamic tables, but for convenience the enthalpy values in steam tables are relative values. 5. The enthalpy of liquid water at 273.16 K in equilibrium with its vapor has been arbitrarily defined as a datum for the calculation of enthalpy values in the steam tables. 6. The latent heat of vaporization of water is higher than the enthalpy of saturated steam. 7. The enthalpy of saturated steam includes the sensible heat of liquid water. 8. The enthalpy of superheated steam includes the sensible heat of vapor. 9. Condensation of superheated steam is possible only after the steam has lost its sensible heat. 10. The latent heat of vaporization of water increases with temperature. 11. The boiling point of water at certain pressure can be determined from steam tables. 12. Specific volume of saturated steam increases with pressure. 13. The enthalpy of liquid water is greatly affected by pressure. 14. The latent heat of vaporization at a certain pressure is equal to the latent heat of condensation at the same pressure. 15. When steam is condensing, it gives off its latent heat of vaporization. 16. The main reason steam is used as a heating medium is its high latent heat value. 17. About 5.4 times more energy is needed to evaporate 1 kg of water at 100 8C than to heat 1 kg of water from 0 8C to 100 8C. 18. The latent heat of vaporization becomes zero at the critical point. 19. Superheated steam is preferred to saturated steam as a heating medium in the food industry. S. Yanniotis, Solving Problems in Food Engineering. Springer 2008 5 15. 20. Steam in the food industry is usually produced in water in tube boilers. 21. Water boils at 08C when the absolute pressure is 611.3 Pa 22. Water boils at 1008C when the absolute pressure is 101325 Pa. 23. Steam quality expresses the fraction or percentage of vapor phase to liquid phase of a vapor-liquid mixture. 24. A Steam quality of 70% means that 70% of the vapor-liquid mixture is in the liquid phase (liquid droplets) and 30% in the vapor phase. 25. The quality of superheated steam is always 100%. Examples Example 2.1 From the steam tables: Find the enthalpy of liquid water at 50 8C, 100 8C, and 120 8C. Find the enthalpy of saturated steam at 50 8C, 100 8C, and 120 8C. Find the latent heat of vaporization at 50 8C, 100 8C, and 120 8C. Solution Step 1 From the column of the steam tables that gives the enthalpy of liquid water read: Hat508C 209:33kJ=kg Hat1008C 419:04kJ=kg Hat1208C 503:71kJ=kg Step 2 From the column of the steam tables that gives the enthalpy of saturated steam read: Hat508C 2592:1kJ=kg Hat1008C 2676:1kJ=kg Hat1208C 2706:3kJ=kg Step 3 Calculate the latent heat of vaporization as the difference between the enthalpy of saturated steam and the enthapy of liquid water. Latent heat at 508C 2592:1 209:33 2382:77kJ=kg Latent heat at 1008C 2676:1 419:09 2257:06kJ=kg Latent heat at 1208C 2706:3 503:71 2202:59kJ=kg 6 2 Use of Steam Tables 16. Example 2.2 Find the enthalpy of superheated steam with pressure 150 kPa and temperature 150 8C. Solution Step 1 Find the enthalpy from the steam tables for superheated steam: H steam 2772:6kJ=kg Step 2 Alternatively find an approximate value from: H steam H saturated cp vapor T Tsaturation 2693:4 1:909 150 111:3 2767:3 kJ=kg Example 2.3 If the enthalpy of saturated steam at 50 8C and 55 8C is 2592.1 kJ/kg and 2600.9 kJ/kg respectively, find the enthalpy at 53 8C. Solution Find the enthalpy at 53 8C by interpolation between the values for 50 8C and 558C given in steam tables, assuming that the enthalpy in this range changes linearly: H 2592:1 53 50 55 50 2600:9 2592:1 2597:4 kJ=kg Exercises Exercise 2.1 Find the boiling temperature of a juice that is boiling at an absolute pressure of 31.19 Pa. Assume that the boiling point elevation is negligible. Solution From the steam tables, find the saturation temperature at water vapor pressure equal to 31.19 kPa as T = ...................8C. Therefore the boiling temperature will be ....................... Exercises 7 17. Exercise 2.2 A food product is heated by saturated steam at 100 8C. If the condensate exits at 90 8C, how much heat is given off per kg steam? Solution Step 1 Find the the enthalpy of steam and condensate from steam tables: H steam::::::::::::::::::::::::::::::::::kJ=kg; H condensate::::::::::::::::::::::::::::::::::kJ=kg: Step 2 Calculate the heat given off: H ::::::::::::::::::::::::: ::::::::::::::::::::::::::: 2299:2 kJ=kg Exercise 2.3 Find the enthalpy of steam at 169.06 kPa pressure if its quality is 90%. Solution Step 1 Find the enthalpy of saturated steam at 169.06 kPa from the steam tables: H steam ::::::::::::::::::::::::::::::::::::::::::::::::::: Step 2 Find the enthalpy of liquid water at the corresponding temperature from the steam tables: H liquid :::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 3 Calculate the enthalpy of the given steam: H xsHs 1 xs HL :::::::::::::::::: :::::::::::::::::::: :::::::::::::::::: ::::::::::::::::: 2477:3 kJ=kg Exercise 2.4 Find the vapor pressure of water at 72 8C if the vapor pressure at 70 8C and 75 8C is 31.19 kPa and 38.58 kPa respectively. (Hint: Use linear interpolation.) 8 2 Use of Steam Tables 18. Exercise 2.5 The pressure in an autoclave is 232 kPa, while the temperature in the vapor phase is 1208C. What do you conclude from these values? Solution The saturation temperature at the pressure of the autoclave should be ........................... Since the actual temperature in the autoclave is lower than the saturation temperature at 232 kPa, the partial pressure of water vapor in the autoclave is less than 232 kPa. Therefore air is present in the autoclave. Exercise 2.6 Lettuce is being cooled by evaporative cooling in a vacuum cooler. If the absolute pressure in the vacuum cooler is 934.9 Pa, determine the final tem- perature of the lettuce. (Hint: Find the saturation temperature from steam tables.) Exercises 9 19. Chapter 3 Mass Balance Review Questions Which of the following statements are true and which are false? 1. The mass balance is based on the law of conservation of mass. 2. Mass balance may refer to total mass balance or component mass balance. 3. Control volume is a region in space surrounded by a control surface through which the fluid flows. 4. Only streams that cross the control surface take part in the mass balance. 5. At steady state, mass is accumulated in the control volume. 6. In a component mass balance, the component generation term has the same sign as the output streams. 7. It is helpful to write a mass balance on a component that goes through the process without any change. 8. Generation or depletion terms are included in a component mass balance if the component undergoes chemical reaction. 9. The degrees of freedom of a system is equal to the difference between the number of unknown variables and the number of independent equations. 10. In a properly specified problem of mass balance, the degrees of freedom must not be equal to zero. Examples Example 3.1 How much dry sugar must be added in 100 kg of aqueous sugar solution in order to increase its concentration from 20% to 50%? S. Yanniotis, Solving Problems in Food Engineering. Springer 2008 11 20. Solution Step 1 Draw the process diagram: S1 S2 S3 20% 100% 50% MIXING 100 kg Step 2 State your assumptions: l dry sugar is composed of 100% sugar. Step 3 Write the total and component mass balances in the envelope around the process: i) Overall mass balance 100 S2 S3 (3:1) ii) Soluble solids mass balance 0:20 100 S2 0:50 S3 (3:2) Solving eqns (3.1) and (3.2) simultaneously, find S2=60 kg and S3=160 kg. Therefore 60 kg of dry sugar per 100 kg of feed must be added to increase its concentration from 20% to 50%. Example 3.2 Fresh orange juice with 12% soluble solids content is concentrated to 60% in a multiple effect evaporator. To improve the quality of the final product the concentrated juice is mixed with an amount of fresh juice (cut back) so that the concentration of the mixture is 42%. Calculate how much water per hour must be evaporated in the evaporator, how much fresh juice per hour must be added back and how much final product will be produced if the inlet feed flow rate is 10000 kg/h fresh juice. Assume steady state. 12 3 Mass Balance 21. Solution Step 1 Draw the process diagram: 10000 kg/h X Y F 60% 12% 42%12% W MIXINGEVAPORATION III Step 2 Write the total and component mass balances in envelopes I and II: i) Overall mass balance in envelope I 10000 W X (3:3) ii) Soluble solids mass balance in envelope I 0:12 10000 0:60 X (3:4) iii) Overall mass balance in envelope II X F Y (3:5) iv) Soluble solids mass balance in envelope II 0:60 X 0:12 F 0:42 Y (3:6) From eqn (3.4) find X=2000 kg/h. Substituting X in eqn (3.3) and find W=8000 kg/h. Solve eqns (iii) and (iv) simultaneously and Substitute X in eqn (3.3) and find=1200 kg/h and Y=3200 kg/h. Therefore 8000 kg/h of water will be evaporated, 1200 kg/h of fresh juice will be added back and 3200 kg/h of concentrated orange juice with 42% soluble solids will be produced. Exercise 3.3 1000 kg/h of a fruit juice with 10% solids is freeze-concentrated to 40% solids. The dilute juice is fed to a freezer where the ice crystals are formed Examples 13 22. and then the slush is separated in a centrifugal separator into ice crystals and concentrated juice. An amount of 500 kg/h of liquid is recycled from the separator to the freezer. Calculate the amount of ice that is removed in the separator and the amount of concentrated juice produced. Assume steady state. Solution Step 1 Draw the process diagram: 1000 kg/h Ice J 40%10% SEPARATIONFREEZING I Step 2 Write the total and component mass balances in the envelope around the process: i) Overall mass balance 1000 I J (3:7) ii) Soluble solids mass balance 0:10 1000 0:40 J (3:8) From eqn (3.8) find J=250 kg/h and then from eqn (3.7) find I=750 kg/h. Comment: Notice that the recycle stream does not affect the result. Only the streams that cut the envelope take part in the mass balance. Exercises Exercise 3.1 How many kg/h of sugar syrup with 10% sugar must be fed to an evaporator to produce 10000 kg/h of sugar syrup with 65% sugar? 14 3 Mass Balance 23. Solution Step 1 Draw the process diagram: 10000 kg/hX 65%10% W EVAPORATION Step 2 State your assumptions: .......................................................................................................................... Step 3 Write the mass balance for sugar on the envelope around the process: 0:10 X :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 4 Solve the above equation and find X=............................................................... kg/h Exercise 3.2 How much water must be added to 200 kg of concentrated orange juice with 65% solids to produce orange juice with 12% solids Solution Step 1 Draw the process diagram: 200 kg W J 65% 12% MIXING Water Exercises 15 24. Step 2 Write the mass balance for solids on the envelope around the process: .............................................................................. Solve the above equation and find J=........................................ kg Exercise 3.3 Milk with 3.8% fat and 8.1% fat-free solids (FFS) is used for the production of canned concentrated milk. The process includes separation of the cream in a centrifuge and concentration of the partially defatted milk in an evaporator. If the cream that is produced in the centrifuge contains 55% water, 40% fat, and 5% fat-free solids, calculate how much milk is necessary in order to produce a can of concentrated milk that contains 410 g milk with 7.8% fat and 18.1% fat-free solids. How much cream and how much water must be removed in the centrifuge and the evaporator respectively? Assume steady state. Solution Step 1 Draw the process diagram: X 410 g FFS 8.1% EVAPORATIONCENTRIFUGATION Fat 3.8% FFS 18.1% Fat 7.8% WaterCream 5% FFS C W55%W 40% F Step 2 Write the total and component mass balances in the envelope around the process: i) Overall mass balance :::::::::::::::::::::: :::::::::::::::::: W :::::::::::::::::::::: (3:9) ii) Fat-free solids mass balance :::::::::::::::::::::::::::::: 0:05 C :::::::::::::::::::::::::::: (3:10) 16 3 Mass Balance 25. iii) Fat mass balance 0:038 X ::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::: (3:11) Solve eqns (3.9), (3.10) and (3.11) simultaneously and find X=.................. g, C= ............................... g and W=............................... g. Exercise 3.4 According to some indications, crystallization of honey is avoided if the ratio of glucose to water is equal to 1.70. Given the composition of two honeys, find the proportions in which they have to be mixed so that the ratio of glucose to water in the blend is 1.7. What will be the composition of the blend? Honey H1: glucose 35%, fructose 33%, sucrose 6%, water 16%. Honey H2: glucose 27%, fructose 37%, sucrose 7%, water 19%. Solution Step 1 Draw the process diagram: Glucose 27% Fructose 37% Sucrose 7% Water 19% MIXING H2 Glucose 35% Fructose 33% Sucrose 6% Water 16% H1 Hb Step 2 Select 1000 kg of blend as a basis for calculation (Hb=1000 kg). Step 3 Write the total and component mass balances in the envelope around the process: i) Overall mass balance :::::::::::::::::::::::::: :::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::: (3:12) ii) Glucose mass balance :::::::::::::::::::::::::::: :::::::::::::::::::::::::::: :::::::::::::::::::::::::::: (3:13) Exercises 17 26. iii) Fructose mass balance :::::::::::::::::::::::::::: :::::::::::::::::::::::::::: :::::::::::::::::::::::::::: (3:14) iv) Sucrose mass balance :::::::::::::::::::::::::::: :::::::::::::::::::::::::::: :::::::::::::::::::::::::::: (3:15) v) Water mass balance :::::::::::::::::::::::::::: :::::::::::::::::::::::::::: :::::::::::::::::::::::::::: (3:16) vi) Ratio of glucose to water in the blend G=W 1:70 (3:17) Solve eqns (3.12) to (3.17) simultaneously and find: H1 .............................................................. kg H2 .............................................................. kg H1/H2 ............................................................ The composition of the blend will be: glucose .......................................................... fructose ......................................................... sucrose .......................................................... water ............................................................. Exercise 3.5 How much glucose syrup with 20% concentration has to be mixed with 100 kg glucose syrup with 40% concentration so that the mixture will have 36% glucose? Exercise 3.6 How many kg of saturated sugar solution at 70 8C can be prepared from 100 kg of sucrose? If the solution is cooled from 70 8C to 20 8C, how many kg of sugar will be crystallized? Assume that the solubility of sucrose as a function of temperature (in 8C) is given by the equation: % sucrose 63.2 0.146T 0.0006T2 . 18 3 Mass Balance 27. Exercise 3.7 Find the ratio of milk with 3.8% fat to milk with 0.5% fat that have to be mixed in order to produce a blend with 3.5% fat. Exercise 3.8 For the production of marmalade, the fruits are mixed with sugar and pectin and the mixture is boiled to about 65% solids concentration. Find the amount of fruits, sugar, and pectin that must be used for the production of 1000 kg marmalade, if the solids content of the fruits is 10%, the ratio of sugar to fruit in the recipe is 56:44, and the ratio of sugar to pectin is 100. Exercise 3.9 For the production of olive oil, the olives are washed, crushed, malaxated, and separated into oil, water. and solids by centrifugation as in the following flow chart. Find the flow rate in the exit streams given that: a) the composition of the olives is 20% oil, 35% water, and 45% solids; b) the composition of the discharged solids stream in the decanter is 50% solids and 50% water; c) 90% of the oil is taken out in the first disc centrifuge; and d) the ratio of olives to water added in the decanter is equal to 1. 2000 kg/h olives WASHER HAMMER MILL MALAXATOR DECANTER DISC CENTRIFUGE DISC CENTRIFUGE water oil oil solids water S O2W O1 Exercises 19 28. Chapter 4 Energy Balance Theory The overall energy balance equation for a system with one inlet (point 1) and one outlet (point 2) is: H1 vml 2 2 z1g _m1 H2 vm2 2 2 z2g _m2 q Ws d mE dt The overall energy balance equation for a system at steady state with more than two streams can be written as: X H v2 m 2 zg _m ! q Ws where H = enthalpy, J/kg vm = average velocity, m/s= correction coefficient (for a circular pipe= 1/2 for laminar flow,% 1 for turbulent flow) z = relative height from a reference plane, m m = mass of the system, kg _m = mass flow rate, kg/s q = heat transferred across the boundary to or from the system (positive if heat flows to the system), W Ws = shaft work done by or to the system (positive if work is done by the system), W E = total energy per unit mass of fluid in the system, J/kg t = time, s In most of the cases, the overall energy balance ends up as an enthalpy balance because the terms of kinetic and potential energy are negligible com- pared to the enthalpy term, the system is assumed adiabatic (Q 0), and there is no shaft work (Ws 0). Then: X _mH 0 S. Yanniotis, Solving Problems in Food Engineering. Springer 2008 21 29. Review Questions Which of the following statements are true and which are false? 1. The energy in a system can be categorize as internal energy, potential energy, and kinetic energy. 2. A fluid stream carries internal energy, potential energy, and kinetic energy. 3. A fluid stream entering or exiting a control volume is doing PV work. 4. The internal energy and the PV work of a stream of fluid make up the enthalpy of the stream. 5. Heat and shaft work may be transferred through the control surface to or from the control volume. 6. Heat transferred from the control volume to the surroundings is considered positive by convention. 7. For an adiabatic process, the heat transferred to the system is zero. 8. Shaft work supplied to the system is considered positive by convention. 9. The shaft work supplied by a pump in a system is considered negative. 10. If energy is not accumulated in or depleted from the system, the system is at steady state. Examples Example 4.1 1000 kg/h of milk is heated in a heat exchanger from 458C to 728C. Water is used as the heating medium. It enters the heat exchanger at 908C and leaves at 758C. Calculate the mass flow rate of the heating medium, if the heat losses to the environment are equal to 1 kW. The heat capacity of water is given equal to 4.2 kJ/kg8C and that of milk 3.9 kJ/kg8C. Solution Step 1 Draw the process diagram: 1000 kg/h milk q HEAT EXCHANGER 45C 75C 72C 90C milk waterwater 22 4 Energy Balance 30. Step 2 State your assumptions: l The terms of kinetic and potential energy in the energy balance equation are negligible. l A pump is not included in the system (Ws 0). l The heat capacity of the liquid streams does not change significantly with temperature. l The system is at steady state. Step 3 Write the energy balance equation: Rate of energy input _mw in Hw in _mm in Hm in Rate of energy output _mw out Hw out _mm out Hm out q (with subscript w for water and m for milk). At steady state rate of energy input rate of energy output or _mw in Hw in _mm in Hm in _mw out Hw out _mm out Hm out q (4:1) Step 4 Calculate the known terms of eqn (4.1) i) The enthalpy of the water stream is: Input: Hw in cpT 4:2 90 378 kJ=kg Output: Hw out cpT 4:2 75 315 kJ=kg ii) The enthalpy of the milk stream is: Input: Hm in cpT 3:9 45 175:5 kJ=kg Output: Hm out cpT 3:9 72 280:8 kJ=kg Step 5 Substitute the above values in eqn (4.1), taking into account that: _mw in _mw out _mw and _mm in _mm out _mw 378 1000 175:5 _mw 315 1000 280:8 1 3600 Examples 23 31. Step 6 Solve for _mw _mw 1728:6 kg=h Example 4.2 A dilute solution is subjected to flash distillation. The solution is heated in a heat exchanger and then flashes in a vacuum vessel. If heat at a rate of 270000 kJ/h is transferred to the solution in the heat exchanger, calculate: a) the temperature of the solution at the exit of the heat exchanger, and b) the amount of overhead vapor and residual liquid leaving the vacuum vessel. The following data are given: Flow rate and temperature of the solution at the inlet of the heat exchanger is 1000 kg/h and 508C, heat capacity of the solution is 3.8 kJ/kg8C, and absolute pressure in the vacuum vessel is 70.14 kPa. Solution Step 1 Draw the process diagram: q HEAT EXCHANGER II I VACUUM VESSEL mL, TL, HL III TFi mFo mv, Hv TFo HFo mFi HFi Step 2 State your assumptions: l The terms of kinetic and potential energy in the energy balance equation are negligible. l A pump is not included in the system (Ws 0). l The heat losses to the environment are negligible. l The heat capacities of the liquid streams do not change significantly with temperature and concentration. l The system is at steady state. Step 3 Write the energy balance equation in envelope II: _mFiHFi q _mFoHFo (4:2) or 24 4 Energy Balance 32. _mFicpFTFi q _mFocpFTFo (4:3) Substitute known values: 1000 3:8 50 270000 1000 3:8 TFo (4:4) Solve for TFo: TFo 121 o C Step 4 Write the mass and energy balance equations in envelope I: i) Overall mass balance: _mFi _mV _mL (4:5) ii) Energy balance: _mFiHFi q _mVHV _mLHL (4:6) or _mFicpFTFi q _mVHV _mLcpLTL (4:7) Step 5 Calculate mv using equations (4.5), (4.6) and (4.7): i) From eqn (4.5): _mL _mFi _mV (4:8) ii) Substitute eqn (4.8) in (4.7): _mFicpFTFi q _mVHV _mFi _mV cpLTL (4:9) iii) Find the saturation temperature and the enthalpy of saturated vapor at 70.14 kPa from the steam tables: TL=908C V=2660 kJ/kg iv) Substitute numerical values in eqn (4.9): 1000 3:8 50 270000 _mV 2660 1000 _mV 3:8 90 Examples 25 33. v) Solve for _mV _mv 50:9 kg=h Step 6 Alternatively, an energy balance in envelope III can be used instead of envelope I: i) Write the energy balance equation: _mFocpFTFo _mVHV _mLcpLTL (4:10) ii) Combine eqns (4.5) and (4.10) and substitute numerical values: 1000 3:8 121 _mV 2660 1000 _mV 3:8 90 iii) Solve for _mV _mv 50:9 kg=h Exercises Exercise 4.1 How much saturated steam with 120.8 kPa pressure is required to heat 1000 g/h of juice from 58C to 958C? Assume that the heat capacity of the juice is 4 kJ/ kg8C. Solution Step 1 Draw the process diagram: HEAT EXCHANGER 5C 95C juice condensate 120.8kPa steam 120.8kPa mji = 1000 kg/h juice mjo ms ms 26 4 Energy Balance 34. Step 2 Write the energy balance equation: _mjiHji _msHs ::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::: or _mjicpjTji _msHs ::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::: Step 3 Substitute numerical values in the above equation. (Find the enthalpy of saturated steam and water [condensate] from steam tables): :::::::::::::::::::::::::::: ::::::::::::::::::::::::::: ::::::::::::::::::::::::: :::::::::::::::::::::::: Step 4 Solve for _ms _ms ::::::::::::::::::::::::::::::::::::::kg=h Exercise 4.2 How much saturated steam with 120.8 kPa pressure is required to concentrate 1000 kg/h of juice from 12% to 20% solids at 958C? Assume that the heat capacity of juice is 4 kJ/kg8C. Solution Step 1 Draw the process diagram: EVAPORATOR water vapor 95o C 95o C juice condensate 120.8kPa steam 120.8kPa mji = 1000 kg/h juice mjo mv ms ms Exercises 27 35. Step 2 Write the overall mass balance equation on the juice side: 1000 _mV _mjo Step 3 Write the solids mass balance equation: 0:12 1000 ::::::::::::::: _mjo Solve for _mjo and _mV _mjo :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::kg=h _mV :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::kg=h Step 4 i) Write the enthalpy balance equation: _mjicpjTji _msHs ::::::::::::::::::::::: ::::::::::::::::::::::::: ::::::::::::::::::: ii) From steam tables, find the enthalpy of water vapor at 958C, of saturated steam at 120.8 kPa, and of water (condensate) at 120.8 kPa. iii) Substitute numerical values in the above equation: :::::::::::::::::: ::::::::::::::::: :::::::::::::::::: ::::::::::::::::: ::::::::::::::::: iv) Solve for _ms _ms ::::::::::::::::::::::::::::::::::::::::kg=h Exercise 4.3 2000 kg/h of milk is sterilized in a steam infusion sterilizer. The milk is heated to 1458C by introducing it into the steam infusion chamber H and then is cooled quickly by flashing in the flash vessel F. The vapor that flashes off in the vessel F is condensed in the condenser C by direct contact of the vapor with cooling water. To avoid dilution of the milk, the pressure in the vessel F must be such that the rate at which vapor flashes off in the vessel F is equal to the steam that is added in the vessel H. Calculate the cooling water flow rate in the condenser that will give the required pressure in the flash vessel. The following data are 28 4 Energy Balance 36. given: The temperature of the milk at the inlet of H is 408C, the temperature of the cooling water at the inlet of the condenser is 208C, the steam introduced into the chamber H is saturated at 475.8 kPa pressure, and the heat capacity of the milk is 3.8 kJ/kg8C at the inlet of the infusion chamber and 4 kJ/kg8C at the exit of the infusion chamber. Solution Step 1 Draw the process diagram: III III milk steam H milk vapor F cooling water C mmi, Hmi mv, Hv mwi, Hwi mwo, Hwommo, Hmo mms, Hms ms, Hs Step 2 State your assumptions: l The terms of kinetic and potential energy in the energy balance equation are negligible. l A pump is not included in the system (Ws 0). l The heat losses to the environment are negligible. l The water vapor pressure of the milk is equal to that of water at the same temperature. l The water vapor pressure in the condenser is equal to the water vapor pressure in the flash vessel. l The system is at steady state. Step 3 Write the mass and energy balance equations in envelope I: i) Energy balance in envelope I: _mmiHmi _msHs _mmsHms ii) Overall mass balance in envelope I: :::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::: ::::::::::::::::::::::::::::: Exercises 29 37. iii) Substitute numerical values and combine the last two equations: :::::::::::::::::::::::::::::::::: 2746:5 _ms ::::::::::::::::::::::::::::::::: iv) Solve for _ms _ms ::::::::::::::::::::::::::::::::::::::::::::::::kg=h Step 4 i) Write the energy balance in envelope II: ::::::::::::::::::::::::: ::::::::::::::::::::::::::: :::::::::::::::::::::::::::: ii) Substitute values taking into account that _ms _mv, in order to avoid dilution of the milk: ::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::: or 1389158 7600 T 395:1 HV iii) Solve the last equation by trial and error to find the value of T that will give a value of HV in agreement with steam tables. T = . . .. . .. . .. . .. . .. . .. . ...8C. Step 5 Write the overall mass balance and energy balance in envelope III: i) Overall mass balance: :::::::::::::::::::::::::::: ::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::: ii) Energy balance: :::::::::::::::::::::::::::: ::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::: iii) Substitute numerical values in the last equation and solve for mwi. The temperature of the water at the exit of the condenser must be equal to :::::::::::::::::::: C, because the water vapor pressure in the condenser was assumed equal to that in the flash vessel F. _mwi :::::::::::::::::::::::::::::::::::::::kg=h 30 4 Energy Balance 38. Exercise 4.4 Find the amount of saturated steam at 270.1 kPa required to heat 100 kg of cans from 508C to 1218C, if the heat capacity of the cans is 3.5 kJ/kg8C. Exercise 4.5 One ice cube at 108C weighing 30g is added to a glass containing 200ml of water at 208C. Calculate the final water temperature when the ice cube melts completely. Assume that 3 kJ of heat are transferred from the glass to the water during the melting of the ice? Use the following values: the latent heat of fusion of the ice is 334 kJ/kg, the heat capacity of the ice is 1.93 kJ/kg8C, and the heat capacity of the water is 4.18 kJ/kg8C. Exercise 4.6 For quick preparation of a cup of hot chocolate in a cafeteria, cocoa powder and sugar are added in a cup of water and the solution is heated by direct steam injection. If the initial temperature of all the ingredients is 158C, the final temperature is 958C, the mass of the solution is 150g initially, and the heat capacity of the solution is 3.8 kJ/kg8C, calculate how much saturated steam at 1108C will be used. State your assumptions. Exercise 4.7 Calculate the maximum temperature to which a liquid food can be preheated by direct steam injection if the initial temperature and the initial solids concentra- tion of the food are 208C and 33% respectively, and the final solids concentra- tion must not be less than 30%. How much saturated steam at 121 kPa pressure will be used? Assume that the heat capacity of the food is 3.0 kJ/kg8C initially and 3.1 kJ/kg8C after the steam injection. Exercises 31 39. Chapter 5 Fluid Flow Review Questions Which of the following statements are true and which are false? 1. The Reynolds number represents the ratio of the inertia forces to viscous forces. 2. If the Reynolds number in a straight circular pipe is less than 2100, the flow is laminar. 3. The velocity at which the flow changes from laminar to turbulent is called critical velocity. 4. The Reynolds number in non-Newtonian fluids is called the Generalized Reynolds number 5. The velocity profile of Newtonian fluids in laminar flow inside a circular pipe is parabolic. 6. The velocity profile of Newtonian fluids in laminar flow is flatter than in turbulent flow. 7. The maximum velocity of Newtonian fluids in laminar flow inside a circular pipe is twice the bulk average velocity. 8. The average velocity of Newtonian fluids in turbulent flow inside a circular pipe is around 80% of the maximum velocity. 9. The maximum velocity of pseudoplastic fluids in laminar flow inside a circular pipe is more than twice the bulk average velocity. 10. The Hagen-Poiseuille equation gives the pressure drop as a function of the average velocity for turbulent flow in a horizontal pipe. 11. The pressure drop in laminar flow is proportional to the volumetric flow rate. 12. The pressure drop in turbulent flow is approximately proportional to the 7/4 power of the volumetric flow rate. 13. In a fluid flowing in contact with a solid surface, the region close to the solid surface where the fluid velocity is affected by the solid surface is called boundary layer. 14. The velocity gradients and the shear stresses are larger in the region outside the boundary layer than in the boundary layer. 15. Boundary layer thickness is defined as the distance from the solid surface where the velocity reaches 99% of the free stream velocity. S. Yanniotis, Solving Problems in Food Engineering. Springer 2008 33 40. 16. The viscosity of a liquid can be calculated if the pressure drop of the liquid flowing in a horizontal pipe in laminar flow is known. 17. The viscosity of non-Newtonian liquids is independent of the shear rate. 18. The flow behavior index in pseudoplastic liquids is less than one. 19. In liquids that follow the power-law equation, the relationship between average velocity and maximum velocity is independent of the flow behavior index. 20. The apparent viscosity of a pseudoplastic liquid flowing in a pipe decreases as the flow rate increases. Examples Example 5.1 Saturated steam at 1508 C is flowing in a steel pipe of 2 in nominal diameter, schedule No. 80. If the average velocity of the steam is 10 m/s, calculate the mass flow rate of the steam. Solution Step 1 Find the inside diameter for a 2 in pipe schedule No. 80 from table: D 4:925cm Step 2 Calculate the inside cross-sectional area of the pipe: A pD2 4 p0:04925m2 4 0:001905 m2 Step 3 Calculate the volumetric flow rate: Q A vaver: 0:001905m2 10m=s 0:01905m3 =s Step 4 Find the specific volume of saturated steam at 150 8C from the steam tables: v = 0.3928 m3 /kg Step 5 Calculate the mass flow rate: _m Q v 0:01905 m3 =s :0:3928 m3=kg 0:0485 kg=s 34 5 Fluid Flow 41. Example 5.2 A 50% sucrose solution at 20 8 C is flowing in a pipe with 0.0475 m inside diameter and 10 m length at a rate of 3 m3 /h. Find: a) the mean velocity, b) the maximum velocity, and c) the pressure drop of the sucrose solution. The viscosity and the density of the sucrose solution at 20 8 C are 15.43 cp and 1232 kg/m3 respectively. Solution Step 1 Calculate the cross-section area of the pipe: A pD2 4 p0:0475m2 4 1:77 103 m2 Step 2 Calculate the mean velocity of the liquid: vm Q A 8:33 104 m3 =s 1:77 103m2 0:471 m=s Step 3 Calculate the Reynolds number: Re Dvp m 0:0475m0:471m=s1232kg=m3 15:43 103kg=ms 1786 Since Re2100, the flow is laminar and vmax 2vm 2 0:471m=s 0:942m=s Step 4 Calculate the pressure drop using the Hagen-Poiseuille equation: P 32vmL D2 320:471m=s0:01543Pas10m 0:0475m2 1030:Pa Exercises Exercise 5.1 Calculate the Reynolds number for water flowing at 5 m3 /h in a tube with 2 in inside diameter if the viscosity and density of water are 1 cp and 0.998 g/ml respectively. At what flow rate does the flow becomes laminar? Exercises 35 42. Solution Step 1 Convert the units to SI: Q 5 m3 =h :::::::::::::::::::::::::::::::::m3 =s D 2 in ::::::::::::::::::::::::::::::::::m m 1 cp :::::::::::::::::::::::::::::::::: kg ms r 0:998 g=ml ::::::::::::::::::::::::::::::::kg=m3 Step 2 Calculate the cross-section area of the pipe: A ::::::::::::::::::::::::::::::::::::::::m2 Step 3 Calculate the mean velocity of the liquid: vm :::::::::::::::::::::::::::::::::::m=s Step 4 Calculate the Reynolds number: Re ::::::::::::::::::::::::::::::::::::::::: Step 5 For the flow to be laminar, Re must be less than or equal to 2100. i) Calculate the velocity from the Reynolds number: 2100 0:0508mvm998Kg=m3 0:001kg=ms Solve for vm: vm ::::::::::::::::::::::::::::::::::::::m=s ii) Calculate the flow rate for the flow to be laminar using vmfound above: Q Avm :::::::::::::::::::::::::::::m3 =s or::::::::::::::m3 =h 36 5 Fluid Flow 43. Exercise 5.2 Calculate the Reynolds number for applesauce flowing at 5 m3 /h in a tube with 2 in inside diameter if the consistency index is 13 Pa s0.3 , the flow behavior index is 0.3, and the density is 1100 kg/m3 . Solution Step 1 Convert the units to SI: Q 5 m3 =h :::::::::::::::::::::::::::::::::::m3 =s D 2 in ::::::::::::::::::::::::::::::::::::::::::::m Step 2 Calculate the cross-section area of the pipe: A ::::::::::::::::::::::::::::::::::::::::::::m2 Step 3 Calculate the mean velocity of the liquid: vm ::::::::::::::::::::::::::::::::::::::::m=s Step 4 Calculate the Reynolds number: Since n 6 1, the fluid is non-Newtonian. The Generalised Reynolds number will be: ReG 23n n 3n 1n Dn v2n m r k ::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Exercise 5.3 Olive oil is flowing in a horizontal tube with 0.0475 m inside diameter. Calculate the mean velocity if the pressure drop per meter of pipe is 1000 Pa. The viscosity of olive oil is 80 cp and its density is 919 kg/m3 . Solution Step 1 Assume the flow is laminar and calculate the mean velocity using the Hagen- Poiseuille equation: vm D2 P 32Lm ::::::::::::::::::::::::::::::::::::::::::::::m=s Exercises 37 44. Step 2 Verify that the flow is laminar: Re :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Exercise 5.4 Honey at 1 liter/min is flowing in a capillary-tube viscometer 2cm in diameter and 50 cm long. If the pressure drop is 40 kPa, determine its viscosity. Solution The viscosity can be calculated using the Hagen-Poiseuille equation. Step 1 Find the mean velocity: vm Q :::::: :::::::::::::::::::::::m=s Step 2 Calculate the viscosity: m D2 P 32Lvm :::::::::::::::::::::::::Pas Exercise 5.5 Tomato concentrate is in laminar flow in a pipe with 0.0475 m inside diameter and 10 m length at a rate of 3 m3 /h. Find: a) the mean velocity, b) the maximum velocity, and c) the pressure drop of the tomato concentrate. The consistency index and the flow behavior index are K = 18.7 Pas0.4 and n = 0.4 respectively. Compare the pressure drop for the sucrose solution of Example 5.2 with the pressure drop of tomato concentrate. Solution Step 1 Calculate the cross-section area of the pipe: A pD2 4 :::::::::::::::::m2 Step 2 Calculate the mean velocity of the liquid: vm Q A ::::::::::::::::::::::m=s 38 5 Fluid Flow 45. Step 3 Use the relationship between mean and maximum velocity for a power-law non- Newtonian fluid in laminar flow to calculate vmax: vmax vm 3n 1 n 1 Therefore vmax :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::m=s Step 4 Find the relationship between mean velocity and pressure drop P in laminar flow for non-Newtonian fluids: vm P 2kL1=n n 3n 1 R n1 n where K is consistency index (Pasn ), n is flow behaviour index, L is pipe length (m), and R is pipe diameter (m). Step 5 Solve for the pressure drop, substitute values, and find P: P ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::Pa Step 6 Compare the above pressure drop with the pressure drop calculated for the sucrose solution. Exercise 5.6 Develop a spreadsheet program to find and plot the velocity distribution as a function of pipe radius for the sucrose solution of Example 5.2 and for the tomato concentrate of Exercise 5.5. Compare the results. Solution Step 1 Find the equations for the velocity distribution in laminar flow in a circular pipe i) for a Newtonian fluid: vr PR2 4mL 1 r R2 Exercises 39 46. ii) for a non-Newtonian fluid: vr P 2KL1=n n n 1 Rn1=n 1 r Rn1=n ! Step 2 Calculate the velocity for various values of radius using the above equations. Step 3 Plot the results. You must end up with the following figure for the sucrose solution and the tomato concentrate: 0.03 0.02 0.01 0.00 0.01 0.02 0.03 Velocity, m/s Radius,m sucrose solution tomato concentrate 0 0.2 0.4 0.6 0.8 1 Step 4 Compare the results. Exercise 5.7 Develop a spreadsheet program to find and plot the velocity distribution as a function of pipe radius for a dilatant liquid with n = 2 and K = 18.7 Pas2 flowing at a rate of 3 m3 /h in a pipe with 0.0475 m inside diameter and 10 m length. 40 5 Fluid Flow 47. Chapter 6 Pumps Theory The mechanical energy balance equation is used to calculate the required power for a pump. The mechanical energy balance equation for a system with one inlet (point 1) and one outlet (point 2) is: vm2 2 vm1 2 2 P2 P1 r z2 z1 g F ws (6:1) where vm = average velocity, m/s a = kinetic energy correction coefficient (for a circular pipe 1=2 for laminar flow,% 1 for turbulent flow) P = pressure, Pa r = density, kg/m3 z = relative height from a reference plane, m g = acceleration of gravity, 9.81 m/s2 F = friction losses per unit mass of fluid, J/kg ws = work supplied by the pump per unit mass of fluid, J/kg The available Net Positive Suction Head (NPSHa) is: NPSHa P pv rg z1 Fs g (6:2) where P = pressure in the suction tank, Pa pv = vapor pressure of liquid in the pump, Pa z1 =distance of the pump from the liquid level in the suction tank, m (z1 positive if the pump is below the liquid level in the tank, z1 negative if the pump is above the liquid level in the tank) Fs = friction losses in the suction line, J/kg S. Yanniotis, Solving Problems in Food Engineering. Springer 2008 41 48. Review Questions Which of the following statements are true and which are false? 1. Mechanical energy includes kinetic energy, potential energy, shaft work, and the flow work term of enthalpy. 2. Mechanical energy cannot be completely converted to work. 3. The fluid pressure drop due to friction in a straight pipe is proportional to the velocity of the fluid. 4. The pressure drop due to skin friction in a pipe can be calculated from the Fanning equation. 5. The friction factor f in laminar flow depends on the Reynolds number and the surface roughness of the pipe. 6. The friction factor f in turbulent flow can be obtained from the Moody chart. 7. In turbulent flow, the higher the surface roughness of the pipe the higher the influence of the Reynolds number on the friction factor f. 8. A sudden change of the fluid velocity in direction or magnitude causes friction losses. 9. Equation 6.1 gives the energy added to a fluid by a pump. 10. The energy added to a fluid by a pump is often called the developed head of the pump and is expressed in m. 11. The required power for a pump is independent of the liquid flow rate. 12. The brake power of a pump depends on the efficiency of the pump. 13. If the pressure in the suction of a pump becomes equal to the vapor pressure of the liquid, cavitation occurs. 14. Under cavitation conditions, boiling of the liquid takes place in the pump. 15. The difference between the sum of the velocity head and the pressure head in the suction of the pump and the vapor pressure of the liquid is called available net positive suction head (NPSH). 16. To avoid cavitation, the available NPSH must be greater than the required NPSH provided by the pump manufacturer. 17. The higher the temperature of the liquid, the lower the available NPSH. 18. It is impossible to pump a liquid at its boiling point unless the pump is below the liquid level in the suction tank. 19. Centrifugal pumps are usually self-primed pumps. 20. Positive displacement pumps are usually self-primed pumps. 21. Positive displacement pumps develop higher discharge pressures than cen- trifugal pumps. 22. The discharge line of a positive displacement pump can be closed without damaging the pump. 23. The discharge line of a centrifugal pump can be completely closed without damaging the pump. 24. The flow rate in a positive displacement pump is usually adjusted by varying the speed of the pump. 42 6 Pumps 49. 25. The flow rate in a positive displacement pump decreases significantly as the head increases. 26. Centrifugal pumps are used as metering pumps. 27. Liquid ring pumps are usually used as vacuum pumps. 28. The capacity of a centrifugal pump is proportional to the rotational speed of the impeller. 29. The head developed by a centrifugal pump is proportional to the speed of the impeller. 30. The power consumed by a centrifugal pump is proportional to the cube of the speed of the impeller. Examples Example 6.1 A liquid food at 50 8C is being pumped at a rate of 3 m3 /h from a tank A, where the absolute pressure is 12350 Pa, to a tank B, where the absolute pressure is 101325 Pa, through a sanitary pipe 1.5 in nominal diameter with 4:6 105 m surface roughness . The pump is 1 m below the liquid level in tank A and the discharge in tank B is 3.3 m above the pump. If the length of the pipe in the suction line is 2 m, the discharge line 10 m, and there are one 90 8 elbow in the suction line, two 90 8 elbows in the discharge line, and one globe valve in the discharge line, calculate the power required, the developed head, and the avail- able Net Positive Suction Head (NPSH). Which of the three pumps that have the characteristic curves given below could be used for this pumping job? The viscosity and the density of the liquid are 0.003 mPas and 1033 kg/m3 respec- tively. The efficiency of the pump is 65%. Assume that the level in tank A is constant. Solution Step 1 Draw the process diagram: Wsz1 z2 V2, P2 V1, P1 x x 1 2 Level of reference A B Examples 43 50. Step 2 Calculate the mean velocity in the pipe: i) Calculate the mass flow rate, _m: _m Qr 3 m3 h 1033 kg m3 1 h 3600 s 0:861 kg=s ii) Find the inside pipe diameter: The inside pipe diameter of 1.5 in nominal diameter pipe is 1.402 in D 1:402 in 0:0254 m in 0:03561 m iii) Calculate the cross-section area of the pipe, A: A pD2 4 p 0:03561 m 2 4 9:959 104 m2 iv) Calculate the mean velocity in the pipe, v: v Q A 3 m3 =h = 3600 s=h 9:959 104 m2 0:837 m=s Step 3 Calculate the Reynolds number: Re Dvr m 0:03561 m 0:837 m=s 1033 kg=m3 0:003 kg=ms 10263 Step 4 Select two points, points 1 and 2, with known v, P, and z values to which to apply the mechanical energy balance equation. The pump must be between points 1 and 2. Step 5 Calculate the frictional losses in the straight sections of the pipe, the elbows, and the valves that are between points 1 and 2: i. Find the friction factor, f, for straight pipes. The friction factor f can be found from the Moody diagram (see Fig A.1 in the Appendix). If roughness e 0:000046m, the relative roughness is: e D 0:000046 m 0:03561 m 0:0013 From the Moody diagram for Re 10263 and e=D 0:0013, read f 0:008. 44 6 Pumps 51. Alternatively, f can be calculated by an empirical relationship e.g., the Colebrook equation: 1 f p 4log10 e=D 3:7 1:255 Re f p 4log10 0:0013 3:7 1:255 10263 f p Solving the above equation by trial and error, find f 0:0082. ii. Find the equivalent length of a 908 standard elbow: Le/D = 32, Equivalent length of straight pipe for 3 elbows: Le 3 32D 332 0:03561 3:42 m iii. Find the equivalent length of the globe valve: Le/D = 300: Le 300D 300 0:03561 10:68 m iv. Use the above results to calculate the frictional losses in the straight pipe sections, the elbows, and the valve: hs 4f v2 2D L 4 0:0082 0:8372 2 0:03561 m2 =s2 m 12 3:42 10:68 m 8:42 m2 s2 8:42 J kg Units equivalence: m2 s2 m2 N s2N m mN s2N mJ s2N mJ s2kgm=s2 J kg Step 6 Calculate the frictional losses in the sudden contraction (entrance from the tank to the pipeline) from: hc 0:55 1 A2 A12 v2 2 2 A2=A1 0 since A1 ) A2. Also, 1 because the flow is turbulent. Therefore, hc 0:55 1 A2 A12 v2 2 2 0:55 1 0:8372 2 1 m2 s2 0:19 m2 s2 0:19 J kg Step 7 Calculate the total frictional losses: F hs hc 8:42 0:19 8:16J=kg Examples 45 52. Step 8 Apply the Mechanical Energy Balance Equation between points 1 and 2 in the diagram. Since the liquid level in the tank is constant, v1 = 0: ws v2 2 v2 1 2 P2 P1 z2 z1 g F 0:8372 0 2 1 m2 s2 101325 12350 1033 Pa kg=m3 3:3 1 m 9:81 m s2 8:61 J kg 117:7 J kg Units equivalence: Pa kg=m3 Pa m3 kg N m2 m3 kg Nm kg J kg Step 9 Calculate the required power: W ws _m 117:7 J kg 0:861 kg s 101:33 J s 101:33 W Since the efficiency of the pump is 65%, the required power (brake power) will be: Wa W Z 101:33 0:65 155:9 W Step 10 Calculate the developed head Hm: Hm ws g 117:7 9:81 J=kg m=s2 12:0 J N 12:0 N m N 12:0 m Step 11 Calculate the available Net Positive Suction Head (NPSHa) using eqn (6.2): i. The total pressure in the suction tank is P = 12350 Pa. ii. The vapor pressure of liquid in the suction is: pv = 12349 Pa (assump- tion: the vapor pressure of the liquid food at 50 8C is the same as that of pure water at the same temperature). 46 6 Pumps 53. iii. The frictional losses in the suction line are: a) Frictional losses in the straight pipe section and the elbow of the suction line: The straight pipe section of the suction line is 2 m. The equivalent straight pipe length of one 908 standard elbow for Le/ D=32, as found in step 5, is Le 132D 32 0:03561 1:14 m Therefore, hss 4f v2 2D L 4 0:0082 0:8372 2 0:03561 2 1:14 1:01 J kg b) Frictional losses in the entrance from the tank to the pipeline: hc 0:19J=kg as calculated in step 6 c) Total losses in the suction line: Fs hss hc 1:01 0:19 1:2 J=kg iv) Substitute values in eqn (6.2) and calculate NPSHa : NPSHa 12350 12349 1033 9:81 1 1:20 9:81 0:88 m Step 12 Select the pump: Find the volumetric flow rate for each one of the pumps A, B, and C for a developed head of 12 m. Find also the required NPSH at the corresponding flow rate: l Pump A: gives 1.6 m3 /h and requires 0.20 m NPSH. Therefore, it does not give the required flow rate of 3 m3 /h when the developed head is 12 m. l Pump B: gives 3.1 m3 /h and requires 1.05 m NPSH. Therefore, it gives the required flow rate of 3 m3 /h, but requires more NPSH than the available of 0.81 m. If used, it will cavitate. l Pump C: gives 3.2 m3 /h and requires 0.45 m NPSH. Therefore, it gives the required flow rate of 3 m3 /h and requires less NPSH than the available of 0.81 m. Therefore, Pump C is suitable for this pumping job. Examples 47 54. Q, m3 /h NPSH,m Developedhead,m PUMP A 0 0 4 8 12 16 20 0 0.5 1 1.5 2 2.5 1 2 3 4 5 NPSH,m Developedhead,m PUMP B 0 4 8 12 16 20 0 0.5 1 1.5 2 2.5 Q, m3 /h 0 1 2 3 4 5 NPSH,m PUMP C 0 4 8 12 16 20 0 0.5 1 1.5 2 2.5 Q, m3 /h 0 1 2 3 4 5 Developedhead,m Exercises Exercise 6.1 Water at 208C is flowing in a horizontal pipe 10 m long with 2 in inside diameter. Calculate the pressure drop in the pipe due to friction for a flow rate of 10 m3 /h. Solution Step 1 Calculate the mean velocity in the pipe: i) Calculate the cross section area of the pipe: A :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::m2 48 6 Pumps 55. ii) Calculate the mean velocity in the pipe, v: v ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: m=s Step 2 Calculate the Reynolds number (find density and viscosity of water from a table with physical properties of water): Re ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 3 Calculate the pressure drop from the Fanning equation (since the flow is turbulent): i) Find the friction factor f from the Moody diagram or from Colebrook equation: f :::::::::::::::::::::::::::::::: ii) Calculate the pressure drop: P 4f L D v2 2 :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::Pa Exercise 6.2 You have available a 550 W pump with 70% efficiency. Is it possible to use this pump to transfer 10 m3 /h of a liquid through a 4.7 cm inside diameter pipe, from one open tank to another, if the liquid is discharged at a point 10 m above the liquid level in the suction tank and the total friction losses are 50 J/kg? The density and the viscosity of the liquid are 1050 kg/m3 and 2 cp respectively. Solution Step 1 Draw the process diagram. Step 2 State your assumptions. ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 3 Select points 1 and 2. Step 4 Calculate the Reynolds number. i) Calculate the cross section area of the pipe, A: A ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::m2 Exercises 49 56. ii) Calculate the mean velocity in the pipe, v: v :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: m=s iii) Calculate the Reynolds number: Re :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Since the flow is turbulent, a = ........................... Step 5 Apply the Mechanical Energy Balance Equation between points 1 and 2. ws :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 149 J kg Step 6 Calculate the power. Ws ws _m :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: W For a 70% pump efficiency, the required power (brake power) will be: WsR :::::::::::::::::: ::::::::::::::: 622 W Since the required power is higher than the available 550 W, the pump is not suitable for this pumping job. Exercise 6.3 A power-law fluid with consistency index K = 0.223 Pa s0.59 , flow behavior index n = 0.59, and density r = 1200 kg/m3 is pumped through a sanitary pipe having an inside diameter of 0.0475 m at a rate of 5 m3 /h from a tank A to a tank B. The level of the liquid in tank A is 2 m below the pump, while the discharge point is 4 m above the pump. The suction line is 3 m long with one 90 8 elbow, while the discharge line is 6 m long with two 90 8 elbows. Calculate the developed head and the discharge pressure of the pump. It is given that the pump is a self-priming pump. Solution Step 1 Draw the process diagram. 50 6 Pumps 57. Ws x 2 Level of reference x 1 A B 3 v3, P3 v2, P2 v1, P1 z1 z2 Step 2 Calculate the Reynolds number. i) Calculate the mass flow rate, _m: _m Qr ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: kg=s ii) Calculate the cross-section area of the pipe, A: A ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::m2 iii) Calculate the mean velocity in the pipe, v : v :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: m=s iv) Calculate the Generalized Reynolds number: ReG Dn v2n r 8n1 3n1 4n n K 0:0475 m 0:59 :::::::::::::::: m=s 20:59 :::::::::::::::::: kg=m3 ::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::: Step 2 Calculate the frictional losses in the straight pipe sections, the elbows, and the valve. i) Find the friction factor f. The friction factor can be calculated from f = 16/ReG in laminar flow or the empirical relationships of Dodge and Metzner (Ref. 1) in turbulent flow. Exercises 51 58. 1 f p 4 n0:75 log10 ReG f 1n=2 0:4 n1:2 :::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ii) Find the equivalent length of a 90 8 standard elbow: Le=D 32 Equivalent length of straight pipe for 3 elbows: Le ::::::::::: m iii) Find the equivalent length of the globe valve: Le=D 300 Equivalent length of straight pipe for 1 globe valve Le ::::: m iv) Calculate the frictional losses in the straight pipe sections, the elbows, and the valve: hs ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: J kg Step 3 Calculate the frictional losses in the sudden contraction.Since the flow is laminar, the kinetic energy correction coefficient is: a 2n 1 5n 3 3 3n 1 2 ::::::::::::::::::::::::::::::::::::::::::::::: hc :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: J kg Step 4 Calculate the total frictional losses. Ft ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: J kg Step 5 Apply the Mechanical Energy Balance Equation between points 1 and 2 in the diagram. Since the liquid level in the tank is constant, v1=0. 52 6 Pumps 59. ws ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::: J kg Step 6 Calculate the required power. Ws ws _m ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: W For a 70% pump efficiency, the required power (brake power) will be: Wa Ws Z ::::::::::::::::::::::::::::::::::::::::::: W Step 7 Calculate the developed head Hm. Hm ws g ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: m Step 8 Calculate the discharge pressure.Apply the Mechanical Energy Balance Equa- tion (MEBE) between points 2 and 3 with v2 v3, z3 0, and ws 0: P3 r P2 r z2 g Fd where Fd = the friction losses in the discharge line. P3 :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::Pa Exercise 6.4 Study the spreadsheet program given in Pump.xls to get familiar with the way the program works. Modify the spreadsheet program given in Pump.xls to solve Example 6.1: a) if a heat exchanger, which gives a pressure drop of 50 kPa, is included in the discharge line; Exercises 53 60. b) if the pump is 1 m above the liquid level in the suction tank (is this pumping possible?); and c) If the required NPSH is 1 m and the absolute pressure in tank A is 101325 kPa, how many meters below the pump could the suction level be without having cavitation problems? If the pump was pumping water at 218C from a well, how many meters below the pump could the suction level be without having cavitation problems? Exercise 6.5 Study the spreadsheet program given in PumpQ.xls to get familiar with the way the program works. Run the program and see how the friction losses, the required power, and the developed head increase with increasing flow rate. Modify the spreadsheet so that instead of variable flow rate, it has variable inside pipe diameter with constant flow rate. 54 6 Pumps 61. Chapter 7 Heat Transfer by Conduction Theory As with all transport phenomena, the rate of the transferred quantity is propor- tional to the driving force and inversely proportional to the resistance. For heat transfer by conduction, the driving force is the temperature difference T and the resistance R x=kA, where x is the wall thickness, k is the thermal conductivity and A is the surface area perpendicular to the direction of transfer. For a cylindrical wall, A is equal to the logarithmic mean surface area, while for a spherical wall, A is equal to the geometric mean surface area. Thus: Rate of heat transfer For a single wall q T R For a composite wall q T P R Resistance to heat transfer Resistance, R Surface area, A Plane wall x=kA A Cylindrical wall r=kALM ALM A1 A2 ln A1=A2 Spherical wall r=kAG AG A1A2 p Review Questions Which of the following statements are true and which are false? 1. Heat is conducted in solids, liquids, and gases by the transfer of the energy of motion from one more energetic molecule to an adjacent less energetic one. S. Yanniotis, Solving Problems in Food Engineering. Springer 2008 55 62. 2. Fouriers law is the basic relationship for heat transfer by conduction. 3. Resistance to heat transfer is proportional to thermal conductivity. 4. Air has low thermal conductivity. 5. Metals have higher thermal conductivity than non-metals. 6. Ice has a thermal conductivity much higher than water. 7. The thermal conductivity of gases is higher than the thermal conductivity of solids. 8. Thermal conductivity is a weak function of temperature. 9. In all cases, thermal conductivity varies with temperature gradient. 10. At steady state, the rate of heat transfer is always zero. 11. At steady state, the temperature at various points in a system does not change with time 12. At steady state, the temperature at various points in a system may change with position. 13. The temperature gradient is positive. 14. For the same heat transfer rate, the slope of the temperature gradientin insulating materials is smaller than in non-insulating materials. 15. The temperature distribution in a plane wall varies linearly with distance in the wall if there is no heat generation in the wall and the thermal conduc- tivity is constant. 16. The temperature distribution in a cylindrical wall varies logarithmically with the distance in the wall if there is no heat generation in the wall and the thermal conductivity is constant. 17. The arithmetic mean area differs from the logarithmic mean area by less than 1.4% if A2/A11.5. 18. In a composite wall at steady state, the heat transfer rate in each layer depends on the thermal conductivity of the layer. 19. The temperature drop in a plane wall is inversely proportional to the resistance. 20. The slope of the temperature gradient in each layer of a composite plane wall depends on the thermal conductivity of the layer. Examples Example 7.1 Calculate the rate of heat transfer through a glass window with 3 m2 surface area and 5 mm thickness if the temperature on the two sides of the glass is 14 8C and 15 8C respectively and the thermal conductivity of the glass is 0.7 W/m 8C. The system is at steady state. Solution 56 7 Heat Transfer by Conduction 63. Step 1 Draw the process diagram: T1 T2 x Step 2 Calculate the resistance of the glass to heat transfer: R x kA 0:005 m 0:7 W=m8C 3 m2 0:00238 8C=W Step 3 Calculate the rate of heat transferred: q T1 T2 R 15 14 8C 0:00238 8C=W 420 W Example 7.2 Hot water is transferred through a stainless steel pipe of 0.04 m inside diameter and 5 m length. The inside wall temperature is 90 8C, the outside surface temperature is 88 8C, the thermal conductivity of stainless steel is 16 W/m 8C, Examples 57 64. and the wall thickness is 2 mm. Calculate the heat losses if the system is at steady state. Solution Step 1 Draw the process diagram: r2 r1 A1 A2 T1 T2 Step 2 Calculate the logarithmic mean area of the wall: i) A1 2p r1L 2p 0:02 m 5 m 0:6283 m2 ii) A2 2p r2L 2p 0:022 m 5 m 0:6912 m2 iii) ALM A1 A2 lnA1 A2 0:6283 0:6912 ln 0:6283 0:6912 0:6592 m2 Step 3 Calculate the resistance of the metal wall to heat transfer: R r2 r1 kmALM 0:002 m 16 W=m 8C 0:6592m2 0:00019 8C=W Step 4 Calculate the rate of heat transfer: q T R T1 T2 R 90 88 8C 0:00019 8C=W 10526 W Example 7.3 The wall of an oven consists of two metal sheets with insulation in between. The temperature of the inner wall surface is 200 8C and that of the outer surface is 50 8C. The thickness of each metal sheet is 2 mm, the thickness of the insulation 58 7 Heat Transfer by Conduction 65. is 5 cm, and the thermal conductivity is 16 W/m 8C and 0.055 W/m 8C respec- tively. Calculate the total resistance of the wall to heat transfer and the heat transfer losses through the wall per m2 of wall area. Solution Step 1 Draw the process diagram: T1 T2 T3 T4 x3x2x1 Step 2 State your assumptions. The system is at steady state. Step 3 Calculate the resistance to heat transfer: i) Inner metal wall R1 x1 k1A 0:002 m 16 W=m8C 1 m2 0:00013 8C=W ii) Insulation: R2 x2 k2A 0:05 m 0:055 W=m8C 1 m2 0:90909 8C=W iii) Outer metal wall: R3 x3 k3A 0:002 m 16 W=m8C 1 m2 0:00013 8C=W Examples 59 66. iv) Total resistance: R R1 R2 R3 0:00013 0:90909 0:00013 0:909358C=W Step 4 Calculate the heat transfer through the wall: q T P R T1 T4 P R 200 50 8C 0:90935 8C=W 165 W Comments: 1) The main resistance to heat transfer (99.97%) is in the insulation layer. 2) The slope of the temperature gradient is steeper in the layer where the resistance is higher. Exercises Exercise 7.1 If an insulation of 2 cm thickness with thermal conductivity equal to 0.02 W/ m 8C is wrapped around the pipe of Example 7.2 so that the outside surface temperature of the insulation is 35 8C, while the inside wall temperature is still 90 8C, what would be the heat loss? What will be the outside surface metal wall temperature T2? Solution Step 1 Draw the process diagram: r2 r1 A1 A2 T1 T2 r3 A3 T3 60 7 Heat Transfer by Conduction 67. Step 2 Calculate the logarithmic mean area of the insulation: i A3 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: m2 ii ALMi :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: m2 Step 3 Calculate the resistance of the insulation layer to heat transfer: R ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 8C=W Step 4 Calculate the total resistance: X R ::::::::::::::::::::::::::::::::::::::::::::::::: Step 5 Calculate the rate of heat transfer: q T P R :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: W Step 6 Calculate temperature T2. The temperature drop is proportional to the resistance: T1 T2 T1 T3 Rw P R where Rw is the metal wall resistance. Therefore, T2::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C Exercise 7.2 The wall of a refrigerator of 4 m2 surface area consists of two metal sheets with insulation in between. The temperature of the inner wall surface is 5 8C and that Exercises 61 68. of the outer surface is 20 8C. The thermal conductivity of the metal wall is 16 W/ m 8C and that of the insulation is 0.017 W/m 8C. If the thickness of each metal sheet is 2 mm, calculate the thickness of the insulation that is required so that the heat transferred to the refrigerator through the wall is 10 W/m2 . Solution Step 1 Draw the process diagram: T1 T2 T3 T4 x3x2x1 Step 2 State your assumptions .................................................................................................................... Step 3 Calculate the resistance to heat transfer: i) Inner metal wall: R1 ::::::::::::::::::::::::::::::::::::::::::::::::::::::: 8C=W ii) Insulation: R2 :::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 8C=W iii) Outer metal wall: R3 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 8C=W iv) Total resistance: X R :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C=W 62 7 Heat Transfer by Conduction 69. Step 4 Calculate the thickness of insulation:Since X R T q :::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::: 8C=W the thickness of the insulation will be x2 :::::::::::::::::::::::::::::::::::::::::::::::::::::::::: m Exercise 7.3 A layer of fat 5 mm thick underneath the skin covers a part of a human body. If the temperature of the inner surface of the fat layer is 36.6 8C and the body loses heat at a rate of 200 W/m2 , what will be the temperature at the surface of the skin? Assume that the thermal conductivity of fat is 0.2 W/m 8C. Solution Step 1 State your assumptions: .................................................................................................................. Step 2 Calculate the resistance to heat transfer: R ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C=W Step 3 Calculate the surface temperature: Since q T R the surface temperature of the skin will be T2 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C Exercise 7.4 A composite plane wall consists of two layers A and B. The thermal conductiv- ity of layers A and B are 0.02 W/m 8C and 15 W/m 8C respectively. If 100 W/m2 are transferred through the wall at steady state, calculate the temperature gradient in the two layers. Exercises 63 70. Solution Step 1 State your assumptions: ...................................................................................................................... Step 2 From Fouriers law: dT dx ::::::::::::::::::::::::::::::::::::::::::::::::: Therefore, dT dx A :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C=m and dT dx B ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::8C=m Exercise 7.5 Find an analytical expression to calculate the heat flux in a plane wall if the thermal conductivity varies with temperature according to the equation k=koaT. Solution Step 1 Use the expression for k in Fouriers law: q A k dT dx koaT dT dx Step 2 Separate the variables and integrate from x1 to x2 and T1 to T2: Exercise 7.6 Develop a spreadsheet program to plot the temperature distribution in a plane wall, a cylindrical wall, and a spherical wall of 4 cm thickness if the outer surface of the wall is at 20 8C and the inner surface is at 200 8C. The inner wall radius for the cylinder and the sphere is 0.1 m, the outer wall radius is 0.14 m, and the thermal conductivity of the wall is 0.02 W/m 8C. Assume steady state. 64 7 Heat Transfer by Conduction 71. Hint: The relationships that give the temperature variation as a function of distance in a wall are: i) Plane wall: T T1 x x1 kA q T1 x x2 x1 T ii) Cylindrical wall: T T1 ln r=r1 2pkL q T1 ln r=r1 ln r2=r1 T iii) Spherical wall: T T1 1 r1 1 r 4pk q T1 1 r1 r 1 r1 r2 T Exercise 7.7 Develop a spreadsheet program to plot the temperature distribution in a plane wall of 4 cm thickness and inner wall surface temperature of 200 8C, at steady state, when a) the thermal conductivity of the wall in W/m 8C varies according to the equation k = 0.03250.0004T and 250 W/m2 of heat are transferred through the wall; and b) the thermal conductivity is constant and equal to the average value for the temperature range of the wall calculated in case a. Calculate the heat transfer rate in case b and compare it to the heat transfer rate of case a. Hint: Calculate the temperature as a function of distance in the wall using the expression developed in Exercise 7.5. Exercises 65 72. Chapter 8 Heat Transfer by Convection Theory For heat transfer by convection, the driving force is the temperature difference T, and the resistance R is equal to 1=hA, where h is the heat transfer coefficient (W/m2 8C ) and A is the surface area (m2 ) perpendicular to the direction of transfer. Heat transfer rate equation Resistance, R Convection only q T R 1=hA Convection combined with conduction q TP R Plane wall P R 1 hiA Pn j1 xj kjA 1 hoA Cylindrical wall P R 1 hiAi Pn j1 rj kjALM 1 hoAo where q = heat transfer rate, W R = resistance to heat transfer, 8C /W T = temperature difference, 8C h= heat transfer coefficient, W/m2 8C hi= inside surface heat transfer coefficient, W/m2 8C ho= outside surface heat transfer coefficient, W/m2 8C A= heat transfer surface area, m2 Ai= inside heat transfer surface area, m2 Ao= outside heat transfer surface area, m2 ALM = logarithmic mean of Ai and Ao , m2 xj and rj = thickness of layer j, m kj = thermal conductivity of layer j, W/m 8C S. Yanniotis, Solving Problems in Food Engineering. Springer 2008 67 73. The heat transfer coefficient is calculated from relationships of the form: Nu f Re; Pr or Nu f Gr; Pr where Nu = Nusselt number Re = Reynolds number Gr = Grashof number Pr = Prandtl number To calculate the heat transfer coefficient: 1. Determine if the flow is natural or forced (free or forced convection). 2. Identify the geometry of the system. 3. Determine if the flow is laminar or turbulent (calculate the Reynolds number). 4. Select the appropriate relationship Nu f Re; Pr 5. Calculate Nu and solve for h. Review Questions Which of the following statements are true and which are false ? 1. The rate of heat transfer by convection is calculated using Newtons law of cooling. 2. The heat transfer coefficient depends on the physical properties of the fluid, the flow regime, and the geometry of the system. 3. The units of the heat transfer coefficient are W/m 8C. 4. The overall heat transfer coefficient has the same units as the local heat transfer coefficient. 5. The resistance to heat transfer by convection is proportional to the heat transfer coefficient. 6. The heat transfer coefficient in gases is usually higher than in liquids. 7. The heat transfer coefficient is lower in viscous fluids than in water. 8. The heat transfer coefficient in forced convection is higher than in natural convection. 9. The heat transfer coefficient in nucleate boiling is higher than in film boiling. 10. The heat transfer coefficient in dropwise condensation is higher than in film condensation. 11. The movement of a fluid in natural convection results from the differences in the density of the fluid. 12. Fouling increases the overall heat transfer coefficient. 68 8 Heat Transfer by Convection 74. 13. Liquid velocities higher than 1 m/s are usually used to reduce fouling. 14. A thermal boundary layer develops on a fluid flowing on a solid surface when the temperature of the fluid is different from the temperature of the solid surface. 15. Temperature gradients exist in the thermal boundary layer. 16. The Prandtl number represents the ratio of thermal diffusivity to momen- tum diffusivity. 17. The Prandtl number relates the thickness of the hydrodynamic boundary layer to the thickness of the thermal boundary layer. 18. The Grashof number represents the ratio of buoyancy forces to viscous forces. 19. The Grashof number in natural convection plays the role of the Reynolds number in forced convection. 20. The fluid properties at the film temperature are used in calculating the heat transfer coefficient outside various geometries. 21. In heat exchangers, counterflow gives a lower driving force than parallel flow. 22. The logarithmic mean temperature difference is used in heat exchangers as the driving force for heat transfer. 23. The arithmetic mean of T1 and T2 differs from their logarithmic mean by more than 1.4% if T1=T21:5. 24. In the case of multiple-pass heat exchangers, the logarithmic mean tem- perature difference must be multiplied by a correction factor. 25. In a plate heat exchanger, the surfaces of the plates have special patterns to increase turbulence. 26. Plate heat exchangers are suitable for viscous fluids. 27. Because the distance between the plates in a plate heat exchanger is small, liquids containing particulates may clog the heat exchanger. 28. A shell and tube heat exchanger cannot be used in high pressure applications. 29. Scraped-surface heat exchangers can handle viscous fluids. 30. Fins are used on the outside surface of a heat exchanger pipe when the heat transfer coefficient on the outside surface of the pipe is higher than the heat transfer coefficient inside the pipe. Examples Example 8.1 Water flows in a pipe of 0.0475 m inside diameter at a velocity of 1.5 m/s. Calculate the heat transfer coefficient if the temperature of the water is 60 8C and 40 8C at the inlet and the outlet of the pipe respectively, and the inside wall temperature of the pipe is 35 8C. Examples 69 75. Solution Step 1 Draw the process diagram: 60C 40C q q Step 2 Find the physical properties of the water. The physical properties must be calculated at the average water temperature: Tm 60 40 2 50 8C Thus, r50 988kg=m3 m50 0:549cp m35 0:723cp cp50 4183J=kg8C k50 0:639W=m8C Step 3 Calculate the Reynolds number: Re Dvr m 0:0475 m 1:5 m=s 988 kg=m3 0:000549 kg=ms 128224 Step 4 Identify the regime of heat transfer: l forced convection l flow inside a cylindrical pipe l turbulent flow 70 8 Heat Transfer by Convection 76. Step 5 Select the most suitable equation of Nu fRe; Pr : Nu 0:023Re0:8 Pr0:33 m mw0:14 Step 6 Calculate the Prandtl number: Pr cpm k 4183 J=kg 8C 0:000549 kg=ms 0:639 W=m 8C 3:59 Step 7 Substitute the values of the Reynolds and Prandtl numbers and calculate the Nusselt number: Nu hD k 0:023 1282240:8 3:590:33 0:549 0:7230:14 411:7 Step 8 Calculate h: h Nu k D 411:7 0:639 W=m 8C 0:0475 m 5538 W=m2 8C Example 8.2 Sucrose syrup flows in a pipe of 0.023 m inside diameter at a rate of 40 lt/min, while steam is condensing on the outside surface of the pipe. The syrup is heated from 50 to 70 8C, while the inside wall temperature is at 80 8C. Calculate 1) the heat transfer coefficient and 2) the required length of the pipe. Solution 50C 70C q q Examples 71 77. 1) Calculation of the heat transfer coefficient: Step 1 Find the physical properties of sucrose syrup at the average syrup temperature: Tm 50 70 2 60 8C Thus, r60 1200kgm3 m60 3:8cp m80 2:3cp cp60 3120J=kg 8C k60 0:46W=m 8C Step 2 Calculate the Reynolds number: The mean velocity is: v Q A 40 lt=min 103 m3 =lt 1 min=60 s p 0:0232 m2 =4 0:000667 m3 =s 0:000415 m2 1:607 m=s Therefore, Re Dvr m 0:023 m 1:607m=s 1200 kg=m3 0:0038 kg=ms 11672 Step 3 Identify the regime of heat transfer: l forced convection l flow inside a cylindrical pipe l turbulent flow Step 4 Select the most suitable equation of Nu f Re; Pr : Nu 0:023Re0:8 Pr0:33 m mw0:14 72 8 Heat Transfer by Convection 78. Step 5 Calculate the Prandtl number: Pr cpm k 3120 J=kg 8C 0:0038 kg=ms 0:460 W=m 8C 25:8 Step 6 Substitute the values of the Reynolds and Prandtl numbers and calculate the Nusselt number: Nu hD k 0:023 116720:8 25:80:33 3:8 2:30:14 129:4 Step 7 Calculate h: h Nu k D 129:4 0:460 W=m 8C 0:023 m 2588 W=m2 8C 2) Calculation of the required pipe length: Step 1 Calculate the heat transferred to the liquid using an enthalpy balance. i) Write the enthalpy balance: Hin q Hout or _mcpTin q _mcpTout or q _mcp Tout Tin (8:1) Tin Tout q q Examples 73 79. ii) Calculate the mass flow rate _m (assume that the density of the syrup at 508C does not differ significantly from that at 608C): _m Qr 40 lt min 1 m3 1000 lt 1 min 60 s 1200 kg m3 0:8 kg s iii) Substitute values into eqn (8.1) and calculate q: q _mc

Solving problems at work - employment.govt.nz › ... › 17ac6cbb3e › solving-problem… · Solving Problems at Work Introduction Problems can often happen in workplaces and common

Solving problems, solving genetic problems, template for manuals - hestan · 30 food safety guidelines 31 special functions 33ven care and cleaning o 39 solving baking problems 40 solving operational problems 41arts / service, aix solving problems, workshop problems solving, solving physics problems with multiple representations · solving physics problems with multiple representations ... solving physics problems with multiple representations ... early, solving kinetics problems, introductory physics: problems solving › faculty › dgaranin › introductory...introductory physics: problems solving d. a. garanin 30 april 2020 introduction solving problems, solving life's problems -, problems solving principles, solving interoperability problems, esp solving-problems, michigan’s problem-solving courts solving problems saving ... · michigan’s problem-solving courts solving problems saving lives michigan supreme court, solving number problems us5235 solving number problems, solving problems 2013, solving problems creatively, solving word problems, problems we're solving, percent problems solving simple percent problems, solving genetics problems, solving nmr problems, solving manufacturing problems, solving percents problems, solving city problems, solving stoichiometry problems, 6-6 solving percent problems - gvsd / overviec2).pdf · 1 6-6 solving percent problems ... solving percent problems using equations ... example e solving percent problems using equations, solving programming problems, solving genetic problems, solving spectroscopy problems, solving flood problems, solving big problems, template for manuals - hestan · 32 food safety guidelines 33 special features menu 35ven care and cleaning o 41 solving baking problems 42 solving operational problems 43arts / service, solving sentence problems, wicked problems: problems worth solving.

• Email: [email protected]

Solving Problems in Food Engineering

• Author / Uploaded
• Carlos Andres

Solving Problems in Food Engineering http://avibert.blogspot.com Stavros Yanniotis, Ph.D. Author Solving Problems in

Views 9,486 Downloads 7,227 File size 5MB

Report DMCA / Copyright

Recommend Stories

Solving Engineering Problems

Contents Solving Engineering Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-1 Introduction

80 5 3MB Read more

Problems solving in accounting

PROBLEMS: Part I 1) On December 31, 2014, the bookkeeper of Graduating Company provided the following information: Accou

237 43 95KB Read more

Solving Dynamics Problems in Maple

Solving Dynamics Problems in Maple Brian D. Harper Mechanical Engineering The Ohio State University A supplement to acc

8,749 1,945 3MB Read more

Solving Dynamics Problems in MATLAB

Solving Dynamics Problems in MATLAB Brian D. Harper Mechanical Engineering The Ohio State University A supplement to ac

3,460 640 2MB Read more

Vibration Problems in Engineering

Vibration Problems in Engineering, 2008, 476 pages, S. Timoshenko, 1443731676, 9781443731676, Lightning Source Incorpora

134 2 112KB Read more

Solving Control Engineering Problems With Matlab

84 39 37MB Read more

Transport Phenomena in Food Engineering

62 0 375KB Read more

Solving Typical FE Problems

engineering economyFull description

71 6 480KB Read more

Solving Lever Problems: Lesson

16 LESSON Solving Lever Problems One of the oldest machines known to humans is the lever. The principles of the lever

36 0 317KB Read more

Citation preview

Solving Problems in Food Engineering http://avibert.blogspot.com Stavros Yanniotis, Ph.D. Author Solving Problems in Food Engineering http://avibert.blogspot.com Stavros Yanniotis, Ph.D. Department of Food Science and Technology Agricultural University of Athens Athens, Greece ISBN: 978-0-387-73513-9 eISBN: 978-0-387-73514-6 Library of Congress Control Number: 2007939831 # 2008 Springer Science+Business Media, LLC All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, LLC., 233 Spring Street, New York, NY10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed on acid-free paper 9 8 7 6 5 4 3 2 1 springer.com ‘‘Tell me and I will listen, Show me and I will understand Involve me and I will learn’’ Ancient Chinese Proverb Preface Food engineering is usually a difficult discipline for food science students because they are more used to qualitative rather than to quantitative descriptions of food processing operations. Food engineering requires understanding of the basic principles of fluid flow, heat transfer, and mass transfer phenomena and application of these principles to unit operations which are frequently used in food processing, e.g., evaporation, drying, thermal processing, cooling and freezing, etc. The most difficult part of a course in food engineering is often considered the solution of problems. This book is intended to be a step-by-step workbook that will help the students to practice solving food engineering problems. It presumes that the students have already studied the theory of each subject from their textbook. The book deals with problems in fluid flow, heat transfer, mass transfer, and the most common unit operations that find applications in food processing, i.e., thermal processing, cooling and freezing, evaporation, psychometrics, and drying. The book includes 1) theoretical questions in the form ‘‘true’’ or ‘‘false’’ which will help the students quickly review the subject that follows (the answers to these questions are given in the Appendix); 2) solved problems; 3) semisolved problems; and 4) problems solved using a computer. With the semisolved problems the students are guided through the solution. The main steps are given, but the students will have to fill in the blank points. With this technique, food science students can practice on and solve relatively difficult food engineering problems. Some of the problems are elementary, but problems of increasing difficulty follow, so that the book will be useful to food science students and even to food engineering students. A CD is supplied with the book which contains solutions of problems that require the use of a computer, e.g., transient heat and mass transfer problems, simulation of a multiple effect evaporator, freezing of a 2-D solid, drying, and others. The objectives for including solved computer problems are 1) to give the students the opportunity to run such programs and see the effect of operating and design variables on the process; and 2) to encourage the students to use computers to solve food engineering problems. Since all the programs in this CD are open code programs, the students can see all the equations and the logic behind the calculations. They are encouraged to see how the programs work vii viii Preface and try to write their own programs for similar problems. Since food science students feel more comfortable with spreadsheet programs than with programming languages, which engineering students are more familiar with, all the problems that need a computer have EXCEL1 spreadsheet solutions. I introduce the idea of a digital SWITCH to start and stop the programs when the problem is solved by iteration. With the digital SWITCH, we can stop and restart each program at will. When the SWITCH is turned off the program is not running, so that we can change the values of the input variables. Every time we restart the program by turning the SWITCH on, all calculations start from the beginning. Thus it is easy to change the initial values of the input variables and study the effect of processing and design parameters. In the effort to make things as simple as possible, some of the spreadsheet programs may not operate on some sets of parameters. In such cases, it may be necessary to restart the program with a different set of parameters. I am grateful to Dr H. Schwartzberg, who read the manuscripts and made helpful suggestions. I will also be grateful to readers who may have useful suggestions, or who point out errors or omissions which obviously have slipped from my attention at this point. Athens May 2007 Stavros Yanniotis Contents http://avibert.blogspot.com Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii 1. Conversion of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Examples Exercises 1 2. Use of Steam Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Review Questions Examples Exercises 5 3. Mass Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Review Questions Examples Exercises 4. Energy Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Theory Review Questions Examples Exercises 5. Fluid Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Review Questions Examples Exercises 6. Pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Theory Review Questions Examples Exercises ix x Contents 7. Heat Transfer By Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Theory Review Questions Examples Exercises 55 8. Heat Transfer By Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Theory Review Questions Examples Exercises 67 Heat Transfer By Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Review Questions Examples Exercises 95 9. 10. Unsteady State Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Theory Review Questions Examples Exercises 11. Mass Transfer By Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 Theory Review Questions Examples Exercises 12. Mass Transfer By Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 Theory Review Questions Examples Exercises 13. Unsteady State Mass Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 Theory Review Questions Examples Exercises 14. Pasteurization and Sterilization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 Review Questions Examples Exercises Contents xi 15. Cooling and Freezing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 Review Questions Examples Exercises 16. Evaporation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 Review Questions Examples Exercises 17. Psychrometrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 Review Questions Examples Exercises 18. Drying . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 Review Questions Examples Exercises References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 Appendix: Answers to Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . Moody diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Gurney-Lurie charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Heisler charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Pressure-Enthalpy chart for HFC 134a . . . . . . . . . . . . . . . . . . . . . . . Pressure-Enthalpy chart for HFC 404a . . . . . . . . . . . . . . . . . . . . . . . Psychrometric chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Roots of d tand=Bi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Roots of dJ1(d)-Bi Jo(d)=0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Roots of d cotd=1-Bi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Error function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 280 281 284 285 286 287 288 290 291 292 293 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 Chapter 1 Conversion of Units http://avibert.blogspot.com Table 1.1 Basic units SI CGS US Engineering Time Length Mass Force Temperature s s s m cm ft kg g lbm – – lbf K, 0C K, 0C 0 R, 0F Table 1.2 Derived units SI Force Energy Power Area Volume Density Velocity Pressure US Engineering 2 N (1 N = 1 kg m/s ) J (1 J = 1 kg m2/s2 ) W (1 W = 1 J/s) m2 m3 (1m3 = 1000 l) kg/m3 m/s Pa (1 Pa = 1 N/m2) bar (1 bar = 105 Pa) torr (1 torr = 1 mmHg) atm (1 atm = 101325 Pa) Table 1.3 Conversion factors 1 ft = 12 in = 0.3048 m 1 in = 2.54 cm 1 US gallon = 3.7854 l 1 lbm = 0.4536 kg 1 lbf = 4.4482 N 1 psi = 6894.76 Pa 1 HP =745.7 W 1 Btu = 1055.06 J = 0.25216 kcal 1kWh = 3600 kJ S. Yanniotis, Solving Problems in Food Engineering. Ó Springer 2008 – Btu HP, PS ft2 ft3 lbm/ft3 ft/s psi=lbf/in2 0 F = 32þ1.8* 0C C = (0F-32)/1.8 0 R = 460 þ 0F K = 273.15 þ 0C 0
0C =
0F/1.8
0C =
K
0F =
0R 1 2 1 Conversion of Units Examples Example 1.1 Convert 100 Btu/h ft2oF to kW/m2oC Solution 100 Btu Btu 1055:06 J 1 kJ 1h 1ft2   ¼100  h ft2 8 F h ft2 8 F 1 Btu 1000 J 3600 s ð0:3048 mÞ2  1:8 8 F 1 kW kW ¼ 0:5678 2 8  8 1 C 1 kJ=s m C Example 1.2 Convert 100 lb mol/h ft2 to kg mol/s m2 Solution 100 lb mol lbmol 0:4536 kg mol 1 h 1 ft2 kg mol   ¼ 100  ¼ 0:1356 h ft2 h ft2 lb mol 3600 s ð0:3048 mÞ2 s m2 Example 1.3 Convert 0.5 lbf s/ft2 to Pas Solution 0:5 lbf s lbf s 4:4482 N 1 ft2 1 Pa ¼ 23:94 Pa s ¼ 0:5   2 ð1 N=m2 Þ ft2 ft2 lbf ð0:3048 mÞ Exercises Exercise 1.1 Make the following conversions: 1) 10 ft lbf/lbm to J/kg, 2) 0.5 Btu/lbmoF to J/kgoC, 3) 32.174 lbmft/lbfs2 to kgm/ Ns2, 4) 1000 lbmft /s2 to N, 5) 10 kcal/min ft oF to W/mK, 6) 30 psia to atm, 7) 0.002 kg/ms to lbmft s, 8) 5 lb mol/h ft2mol frac to kg mol/s m2 mol frac, 9) 1.987 Btu/lbmol oR to cal/gmol K, 10) 10.731 ft3lbf/in2lbmol oR to J/kgmol K Exercises 3 Solution ft lbf ft lbf ::::::::::::::m :::::::::::::::N ::::::::::::::lbm :::::::::::::J   ¼ 10   ft 1 lbf mN lbm lbm 0:4536 kg J ¼ 29:89 kg Btu Btu 1055:06 J ::::::::::::: 1:88F J   ¼ 2094:4 ¼ 0:5  2) 0:5 lbm 8F lbm 8F :::::::::: ::::::::::::: 18C kg8C 1) 10 lbm ft lbm ft ::::::::::::::: ::::::::::::::::::m :::::::::::::::  ¼ 32:174   2 lbf s lbf s2 :::::::::::::::lbm 1 ft 4:4482 N kg m ¼1 N s2 lbm ft lbm ft 0:4536 kg :::::::::::::::: 1N   4) 1000 2 ¼ 1000 2  ¼ 138:3 N s s :::::::::::::::: 1 ft 1 kg m=s2 3) 32:174 kcal kcal 1055:06 J ::::::::: min ::::::::::::ft :::::::::::8F    ¼ 10  min ft o F min ft o F 0:252 kcal 60 s :::::::::::m ::::::::::K :::::::::W W ¼ 4121  ::::::::::J=s mK lbf ::::::::::::::::in2 ::::::::::::::::::N :::::::::::::::::Pa 6) 30 psia ¼ 30 2    in :::::::::::::::::m2 ::::::::::::::::::lbf ::::::::::::::N=m2 :::::::::::::::::atm ¼ 2:04 atm  :::::::::::::::::Pa kg kg ::::::::::::::lbm :::::::::::::::m lbm ¼ 0:002  ¼ 0:0013 7) 0:002  ms m s :::::::::::::::kg ::::::::::::::::::ft ft s 5) 10 8) 5 lb mol lb mol ::::::::::::::::kg mol :::::::::::::::::h  ¼5  h ft2 mol frac h ft2 mol frac ::::::::::::::: lb mol ::::::::::::::::::s :::::::::::::::::ft2 kg mol  ¼ 6:78  103 s m2 mol frac :::::::::::::::::::m2 9) 1:987 Btu Btu ::::::::::::::cal ::::::::::::::::lb mol ¼ 1:987   ¼ lb mol 8R lb mol 8R ::::::::::::::Btu ::::::::::::::::g mol ::::::::::::::8R cal ¼ 1:987 ::::::::::::K g mol K ft3 lbf ft3 lbf ::::::::::::::::m3 ::::::::::::::::N ¼ 10:731 2  10) 10:731 2  in lb mol8R in lb mol8R :::::::::::::::::::ft3 ::::::::::::::::::lbf   ::::::::::::::::in2 :::::::::::::lb mol 1:88R   K :::::::::::::::::::m2 :::::::::::::kg mol ¼ 8314 J kg mol K 4 1 Conversion of Units Exercise 1.2 Make the following conversions: 251oF to oC (Ans. 121.7 oC) 0.01 ft2/h to m2/s (Ans. 2.58x10-7 m2/s) 500oR to K (Ans. 277.6 K) 0.8 cal/goC to J/kgK (Ans. 3347.3 J/kgK) 0.04 lbm/in3 to kg/m3 (Ans. 1107.2 kg/m3) 20000 kg m/s2 m2 to psi (Ans. 2.9 psi) 12000 Btu/h to W (Ans. 3516.9 W ) 0.3 Btu/lbmoF to J/kgK (Ans. 1256 J/kgK) 32.174 ft/s2 to m/s2 (Ans. 9.807 m/s2 ) 1000 ft3/(h ft2 psi/ft) to cm3/(s cm2 Pa/cm) (Ans. 0.0374 cm3/(s cm2 Pa/cm ) Chapter 2 Use of Steam Tables http://avibert.blogspot.com Review Questions Which of the following statements are true and which are false? 1. The heat content of liquid water is sensible heat. 2. The enthalpy change accompanying the phase change of liquid water at constant temperature is the latent heat. 3. Saturated steam is at equilibrium with liquid water at the same temperature. 4. Absolute values of enthalpy are known from thermodynamic tables, but for convenience the enthalpy values in steam tables are relative values. 5. The enthalpy of liquid water at 273.16 K in equilibrium with its vapor has been arbitrarily defined as a datum for the calculation of enthalpy values in the steam tables. 6. The latent heat of vaporization of water is higher than the enthalpy of saturated steam. 7. The enthalpy of saturated steam includes the sensible heat of liquid water. 8. The enthalpy of superheated steam includes the sensible heat of vapor. 9. Condensation of superheated steam is possible only after the steam has lost its sensible heat. 10. The latent heat of vaporization of water increases with temperature. 11. The boiling point of water at certain pressure can be determined from steam tables. 12. Specific volume of saturated steam increases with pressure. 13. The enthalpy of liquid water is greatly affected by pressure. 14. The latent heat of vaporization at a certain pressure is equal to the latent heat of condensation at the same pressure. 15. When steam is condensing, it gives off its latent heat of vaporization. 16. The main reason steam is used as a heating medium is its high latent heat value. 17. About 5.4 times more energy is needed to evaporate 1 kg of water at 100 8C than to heat 1 kg of water from 0 8C to 100 8C. 18. The latent heat of vaporization becomes zero at the critical point. 19. Superheated steam is preferred to saturated steam as a heating medium in the food industry. S. Yanniotis, Solving Problems in Food Engineering. Ó Springer 2008 5 6 2 Use of Steam Tables 20. 21. 22. 23. Steam in the food industry is usually produced in ‘‘water in tube’’ boilers. Water boils at 08C when the absolute pressure is 611.3 Pa Water boils at 1008C when the absolute pressure is 101325 Pa. Steam quality expresses the fraction or percentage of vapor phase to liquid phase of a vapor-liquid mixture. 24. A Steam quality of 70% means that 70% of the vapor-liquid mixture is in the liquid phase (liquid droplets) and 30% in the vapor phase. 25. The quality of superheated steam is always 100%. Examples Example 2.1 From the steam tables: Find the enthalpy of liquid water at 50 8C, 100 8C, and 120 8C. Find the enthalpy of saturated steam at 50 8C, 100 8C, and 120 8C. Find the latent heat of vaporization at 50 8C, 100 8C, and 120 8C. Solution Step 1 From the column of the steam tables that gives the enthalpy of liquid water read: Hat508C ¼ 209:33kJ=kg Hat1008C ¼ 419:04kJ=kg Hat1208C ¼ 503:71kJ=kg Step 2 From the column of the steam tables that gives the enthalpy of saturated steam read: Hat508C ¼ 2592:1kJ=kg Hat1008C ¼ 2676:1kJ=kg Hat1208C ¼ 2706:3kJ=kg Step 3 Calculate the latent heat of vaporization as the difference between the enthalpy of saturated steam and the enthapy of liquid water. Latent heat at 508C ¼ 2592:1  209:33 ¼ 2382:77kJ=kg Latent heat at 1008C ¼ 2676:1  419:09 ¼ 2257:06kJ=kg Latent heat at 1208C ¼ 2706:3  503:71 ¼ 2202:59kJ=kg Exercises 7 Example 2.2 Find the enthalpy of superheated steam with pressure 150 kPa and temperature 150 8C. Solution Step 1 Find the enthalpy from the steam tables for superheated steam: H steam ¼ 2772:6kJ=kg Step 2 Alternatively find an approximate value from: H steam ¼ H saturated þ cp vapor ðT  T saturation Þ ¼ 2693:4 þ 1:909  ð150  111:3Þ ¼ 2767:3 kJ=kg Example 2.3 If the enthalpy of saturated steam at 50 8C and 55 8C is 2592.1 kJ/kg and 2600.9 kJ/kg respectively, find the enthalpy at 53 8C. Solution Find the enthalpy at 53 8C by interpolation between the values for 50 8C and 558C given in steam tables, assuming that the enthalpy in this range changes linearly: H ¼ 2592:1 þ 53  50 ð2600:9  2592:1Þ ¼ 2597:4 kJ=kg 55  50 Exercises Exercise 2.1 Find the boiling temperature of a juice that is boiling at an absolute pressure of 31.19 Pa. Assume that the boiling point elevation is negligible. Solution From the steam tables, find the saturation temperature at water vapor pressure equal to 31.19 kPa as T = ...................8C. Therefore the boiling temperature will be ....................... 8 2 Use of Steam Tables Exercise 2.2 A food product is heated by saturated steam at 100 8C. If the condensate exits at 90 8C, how much heat is given off per kg steam? Solution Step 1 Find the the enthalpy of steam and condensate from steam tables: H steam ¼::::::::::::::::::::::::::::::::::kJ=kg; H condensate ¼::::::::::::::::::::::::::::::::::kJ=kg: Step 2 Calculate the heat given off: H ¼ :::::::::::::::::::::::::  ::::::::::::::::::::::::::: ¼ 2299:2 kJ=kg Exercise 2.3 Find the enthalpy of steam at 169.06 kPa pressure if its quality is 90%. Solution Step 1 Find the enthalpy of saturated steam at 169.06 kPa from the steam tables: H steam ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::: Step 2 Find the enthalpy of liquid water at the corresponding temperature from the steam tables: H liquid ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 3 Calculate the enthalpy of the given steam: H ¼ xs Hs þ ð1  xs ÞHL ¼ ¼ ::::::::::::::::::  :::::::::::::::::::: þ ::::::::::::::::::  ::::::::::::::::: ¼ 2477:3 kJ=kg Exercise 2.4 Find the vapor pressure of water at 72 8C if the vapor pressure at 70 8C and 75 8C is 31.19 kPa and 38.58 kPa respectively. (Hint: Use linear interpolation.) Exercises 9 Exercise 2.5 The pressure in an autoclave is 232 kPa, while the temperature in the vapor phase is 1208C. What do you conclude from these values? Solution The saturation temperature at the pressure of the autoclave should be ........................... Since the actual temperature in the autoclave is lower than the saturation temperature at 232 kPa, the partial pressure of water vapor in the autoclave is less than 232 kPa. Therefore air is present in the autoclave. Exercise 2.6 Lettuce is being cooled by evaporative cooling in a vacuum cooler. If the absolute pressure in the vacuum cooler is 934.9 Pa, determine the final temperature of the lettuce. (Hint: Find the saturation temperature from steam tables.) Chapter 3 Mass Balance http://avibert.blogspot.com Review Questions Which of the following statements are true and which are false? 1. The mass balance is based on the law of conservation of mass. 2. Mass balance may refer to total mass balance or component mass balance. 3. Control volume is a region in space surrounded by a control surface through which the fluid flows. 4. Only streams that cross the control surface take part in the mass balance. 5. At steady state, mass is accumulated in the control volume. 6. In a component mass balance, the component generation term has the same sign as the output streams. 7. It is helpful to write a mass balance on a component that goes through the process without any change. 8. Generation or depletion terms are included in a component mass balance if the component undergoes chemical reaction. 9. The degrees of freedom of a system is equal to the difference between the number of unknown variables and the number of independent equations. 10. In a properly specified problem of mass balance, the degrees of freedom must not be equal to zero. Examples Example 3.1 How much dry sugar must be added in 100 kg of aqueous sugar solution in order to increase its concentration from 20% to 50%? S. Yanniotis, Solving Problems in Food Engineering. Ó Springer 2008 11 12 3 Mass Balance Solution Step 1 Draw the process diagram: 100% S1 100 kg S2 MIXING 20% S3 50% Step 2 State your assumptions: l dry sugar is composed of 100% sugar. Step 3 Write the total and component mass balances in the envelope around the process: i) Overall mass balance 100 þ S2 ¼ S3 (3:1) ii) Soluble solids mass balance 0:20  100 þ S2 ¼ 0:50  S3 (3:2) Solving eqns (3.1) and (3.2) simultaneously, find S2=60 kg and S3=160 kg. Therefore 60 kg of dry sugar per 100 kg of feed must be added to increase its concentration from 20% to 50%. Example 3.2 Fresh orange juice with 12% soluble solids content is concentrated to 60% in a multiple effect evaporator. To improve the quality of the final product the concentrated juice is mixed with an amount of fresh juice (cut back) so that the concentration of the mixture is 42%. Calculate how much water per hour must be evaporated in the evaporator, how much fresh juice per hour must be added back and how much final product will be produced if the inlet feed flow rate is 10000 kg/h fresh juice. Assume steady state. Examples 13 Solution Step 1 Draw the process diagram: W I 10000 kg/h 12% II X EVAPORATION 60% MIXING Y 42% 12% F Step 2 Write the total and component mass balances in envelopes I and II: i) Overall mass balance in envelope I 10000 ¼ W þ X (3:3) ii) Soluble solids mass balance in envelope I 0:12  10000 ¼ 0:60  X (3:4) iii) Overall mass balance in envelope II XþF¼Y (3:5) iv) Soluble solids mass balance in envelope II 0:60  X þ 0:12  F ¼ 0:42  Y (3:6) From eqn (3.4) find X=2000 kg/h. Substituting X in eqn (3.3) and find W=8000 kg/h. Solve eqns (iii) and (iv) simultaneously and Substitute X in eqn (3.3) and find=1200 kg/h and Y=3200 kg/h. Therefore 8000 kg/h of water will be evaporated, 1200 kg/h of fresh juice will be added back and 3200 kg/h of concentrated orange juice with 42% soluble solids will be produced. Exercise 3.3 1000 kg/h of a fruit juice with 10% solids is freeze-concentrated to 40% solids. The dilute juice is fed to a freezer where the ice crystals are formed 14 3 Mass Balance and then the slush is separated in a centrifugal separator into ice crystals and concentrated juice. An amount of 500 kg/h of liquid is recycled from the separator to the freezer. Calculate the amount of ice that is removed in the separator and the amount of concentrated juice produced. Assume steady state. Solution Step 1 Draw the process diagram: Ice I 1000 kg/h 10% FREEZING SEPARATION J 40% Step 2 Write the total and component mass balances in the envelope around the process: i) Overall mass balance 1000 ¼ I þ J (3:7) ii) Soluble solids mass balance 0:10  1000 ¼ 0:40  J (3:8) From eqn (3.8) find J=250 kg/h and then from eqn (3.7) find I=750 kg/h. Comment: Notice that the recycle stream does not affect the result. Only the streams that cut the envelope take part in the mass balance. Exercises Exercise 3.1 How many kg/h of sugar syrup with 10% sugar must be fed to an evaporator to produce 10000 kg/h of sugar syrup with 65% sugar? Exercises 15 Solution Step 1 Draw the process diagram: W X 10% 10000 kg/h EVAPORATION 65% Step 2 State your assumptions: .......................................................................................................................... Step 3 Write the mass balance for sugar on the envelope around the process: 0:10  X ¼:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 4 Solve the above equation and find X= ............................................................. .. kg/h Exercise 3.2 How much water must be added to 200 kg of concentrated orange juice with 65% solids to produce orange juice with 12% solids Solution Step 1 Draw the process diagram: Water W 200 kg 65% MIXING J 12% 16 3 Mass Balance Step 2 Write the mass balance for solids on the envelope around the process: .......................................¼....................................... Solve the above equation and find J=........................................ kg Exercise 3.3 Milk with 3.8% fat and 8.1% fat-free solids (FFS) is used for the production of canned concentrated milk. The process includes separation of the cream in a centrifuge and concentration of the partially defatted milk in an evaporator. If the cream that is produced in the centrifuge contains 55% water, 40% fat, and 5% fat-free solids, calculate how much milk is necessary in order to produce a can of concentrated milk that contains 410 g milk with 7.8% fat and 18.1% fat-free solids. How much cream and how much water must be removed in the centrifuge and the evaporator respectively? Assume steady state. Solution Step 1 Draw the process diagram: Cream Water C 55% W 40% F 5% FFS W X FFS 8.1% Fat 3.8% CENTRIFUGATION EVAPORATION 410 g FFS 18.1% Fat 7.8% Step 2 Write the total and component mass balances in the envelope around the process: i) Overall mass balance :::::::::::::::::::::: ¼ :::::::::::::::::: þ W þ :::::::::::::::::::::: (3:9) ii) Fat-free solids mass balance :::::::::::::::::::::::::::::: ¼ 0:05  C þ :::::::::::::::::::::::::::: (3:10) Exercises 17 iii) Fat mass balance 0:038  X ¼::::::::::::::::::::::::::::::::: þ ::::::::::::::::::::::::::::::::: (3:11) Solve eqns (3.9), (3.10) and (3.11) simultaneously and find X=.................. g, C= ............................... g and W= ............................... g. Exercise 3.4 According to some indications, crystallization of honey is avoided if the ratio of glucose to water is equal to 1.70. Given the composition of two honeys, find the proportions in which they have to be mixed so that the ratio of glucose to water in the blend is 1.7. What will be the composition of the blend? Honey H1: glucose 35%, fructose 33%, sucrose 6%, water 16%. Honey H2: glucose 27%, fructose 37%, sucrose 7%, water 19%. Solution Step 1 Draw the process diagram: H2 H1 Glucose 27% Fructose 37% Sucrose 7% Water 19% MIXING Hb Glucose 35% Fructose 33% Sucrose 6% Water 16% Step 2 Select 1000 kg of blend as a basis for calculation (Hb=1000 kg). Step 3 Write the total and component mass balances in the envelope around the process: i) Overall mass balance :::::::::::::::::::::::::: þ :::::::::::::::::::::::::::: ¼ ::::::::::::::::::::::::::::::::: (3:12) ii) Glucose mass balance :::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::: ¼ :::::::::::::::::::::::::::: (3:13) 18 3 Mass Balance iii) Fructose mass balance :::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::: ¼ :::::::::::::::::::::::::::: (3:14) iv) Sucrose mass balance :::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::: ¼ :::::::::::::::::::::::::::: (3:15) v) Water mass balance :::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::: ¼ :::::::::::::::::::::::::::: (3:16) vi) Ratio of glucose to water in the blend G=W ¼ 1:70 (3:17) Solve eqns (3.12) to (3.17) simultaneously and find: H1 ¼ .............................................................. kg H2 ¼ .............................................................. kg H1/H2 ¼ ............................................................ The composition of the blend will be: glucose ¼ .......................................................... fructose ¼ ......................................................... sucrose ¼ .......................................................... water ¼ ............................................................. Exercise 3.5 How much glucose syrup with 20% concentration has to be mixed with 100 kg glucose syrup with 40% concentration so that the mixture will have 36% glucose? Exercise 3.6 How many kg of saturated sugar solution at 70 8C can be prepared from 100 kg of sucrose? If the solution is cooled from 70 8C to 20 8C, how many kg of sugar will be crystallized? Assume that the solubility of sucrose as a function of temperature (in 8C) is given by the equation: % sucrose ¼ 63.2 þ 0.146T þ 0.0006T2. Exercises 19 Exercise 3.7 Find the ratio of milk with 3.8% fat to milk with 0.5% fat that have to be mixed in order to produce a blend with 3.5% fat. Exercise 3.8 For the production of marmalade, the fruits are mixed with sugar and pectin and the mixture is boiled to about 65% solids concentration. Find the amount of fruits, sugar, and pectin that must be used for the production of 1000 kg marmalade, if the solids content of the fruits is 10%, the ratio of sugar to fruit in the recipe is 56:44, and the ratio of sugar to pectin is 100. Exercise 3.9 For the production of olive oil, the olives are washed, crushed, malaxated, and separated into oil, water. and solids by centrifugation as in the following flow chart. Find the flow rate in the exit streams given that: a) the composition of the olives is 20% oil, 35% water, and 45% solids; b) the composition of the discharged solids stream in the decanter is 50% solids and 50% water; c) 90% of the oil is taken out in the first disc centrifuge; and d) the ratio of olives to water added in the decanter is equal to 1. water olives WASHER 2000 kg/h HAMMER MILL W water MALAXATOR DECANTER DISC CENTRIFUGE DISC CENTRIFUGE O1 O2 oil oil solids S Chapter 4 Energy Balance http://avibert.blogspot.com Theory The overall energy balance equation for a system with one inlet (point 1) and one outlet (point 2) is:     vml 2 v 2 dðmEÞ _ 1  H2 þ m2 þ z2 g m _ 2 þ q  Ws ¼ þ z1 g m H1 þ dt 2 2 The overall energy balance equation for a system at steady state with more than two streams can be written as:   X  v2 _ ¼ q  Ws H þ m þ zg m 2 where H = enthalpy, J/kg vm = average velocity, m/s  = correction coefficient (for a circular pipe  = 1/2 for laminar flow,   1 for turbulent flow) z = relative height from a reference plane, m m = mass of the system, kg _ = mass flow rate, kg/s m q = heat transferred across the boundary to or from the system (positive if heat flows to the system), W Ws = shaft work done by or to the system (positive if work is done by the system), W E = total energy per unit mass of fluid in the system, J/kg t = time, s In most of the cases, the overall energy balance ends up as an enthalpy balance because the terms of kinetic and potential energy are negligible compared to the enthalpy term, the system is assumed adiabatic (Q ¼ 0), and there is no shaft work (Ws ¼ 0). Then: X _ ¼0 mH S. Yanniotis, Solving Problems in Food Engineering. Ó Springer 2008 21 22 4 Energy Balance Review Questions Which of the following statements are true and which are false? 1. The energy in a system can be categorize as internal energy, potential energy, and kinetic energy. 2. A fluid stream carries internal energy, potential energy, and kinetic energy. 3. A fluid stream entering or exiting a control volume is doing PV work. 4. The internal energy and the PV work of a stream of fluid make up the enthalpy of the stream. 5. Heat and shaft work may be transferred through the control surface to or from the control volume. 6. Heat transferred from the control volume to the surroundings is considered positive by convention. 7. For an adiabatic process, the heat transferred to the system is zero. 8. Shaft work supplied to the system is considered positive by convention. 9. The shaft work supplied by a pump in a system is considered negative. 10. If energy is not accumulated in or depleted from the system, the system is at steady state. Examples Example 4.1 1000 kg/h of milk is heated in a heat exchanger from 458C to 728C. Water is used as the heating medium. It enters the heat exchanger at 908C and leaves at 758C. Calculate the mass flow rate of the heating medium, if the heat losses to the environment are equal to 1 kW. The heat capacity of water is given equal to 4.2 kJ/kg8C and that of milk 3.9 kJ/kg8C. Solution Step 1 Draw the process diagram: q milk water 1000 kg/h 45°C 75°C HEAT EXCHANGER 72°C 90°C milk water Examples 23 Step 2 State your assumptions: l l l l The terms of kinetic and potential energy in the energy balance equation are negligible. A pump is not included in the system (Ws ¼ 0). The heat capacity of the liquid streams does not change significantly with temperature. The system is at steady state. Step 3 Write the energy balance equation: _ w in  Hw Rate of energy input ¼ m in _m þm in  Hm in _ w out  Hw out þ m _ m out  Hm out þ q Rate of energy output ¼ m (with subscript ‘‘w’’ for water and ‘‘m’’ for milk). At steady state rate of energy input ¼ rate of energy output or _ w in  Hw in þ m _ m in  Hm in ¼ m _ w out  Hw out þ m _ m out  Hm out þ q m Step 4 Calculate the known terms of eqn (4.1) i) The enthalpy of the water stream is: Input: Hw in Output: Hw ¼ cp T ¼ 4:2  90 ¼ 378 kJ=kg out ¼ cp T ¼ 4:2  75 ¼ 315 kJ=kg ii) The enthalpy of the milk stream is: Input: Hm in Output: Hm ¼ cp T ¼ 3:9  45 ¼ 175:5 kJ=kg out ¼ cp T ¼ 3:9  72 ¼ 280:8 kJ=kg Step 5 Substitute the above values in eqn (4.1), taking into account that: _ w in ¼ m _ w out ¼ m _ w and m _m m in _ m out ¼m _ w  378 þ 1000  175:5 ¼ m _ w  315 þ 1000  280:8 þ 1  3600 m (4:1) 24 4 Energy Balance Step 6 _w Solve for m _ w ¼ 1728:6 kg=h m Example 4.2 A dilute solution is subjected to flash distillation. The solution is heated in a heat exchanger and then flashes in a vacuum vessel. If heat at a rate of 270000 kJ/h is transferred to the solution in the heat exchanger, calculate: a) the temperature of the solution at the exit of the heat exchanger, and b) the amount of overhead vapor and residual liquid leaving the vacuum vessel. The following data are given: Flow rate and temperature of the solution at the inlet of the heat exchanger is 1000 kg/h and 508C, heat capacity of the solution is 3.8 kJ/kg8C, and absolute pressure in the vacuum vessel is 70.14 kPa. Solution Step 1 Draw the process diagram: mv, Hv I II mFi TFi HFi HEAT EXCHANGER III mFo VACUUM TFo VESSEL HFo q mL, TL, HL Step 2 State your assumptions: l l l l l The terms of kinetic and potential energy in the energy balance equation are negligible. A pump is not included in the system (Ws ¼ 0). The heat losses to the environment are negligible. The heat capacities of the liquid streams do not change significantly with temperature and concentration. The system is at steady state. Step 3 Write the energy balance equation in envelope II: _ Fi HFi þ q ¼ m _ Fo HFo m or (4:2) Examples 25 _ Fi cpF TFi þ q ¼ m _ Fo cpF TFo m (4:3) Substitute known values: 1000  3:8  50 þ 270000 ¼ 1000  3:8  TFo (4:4) Solve for TFo: TFo ¼ 121 o C Step 4 Write the mass and energy balance equations in envelope I: i) Overall mass balance: _ Fi ¼ m _Vþm _L m (4:5) _ Fi HFi þ q ¼ m _ V HV þ m _ L HL m (4:6) _ Fi cpF TFi þ q ¼ m _ V HV þ m _ L cpL TL m (4:7) ii) Energy balance: or Step 5 Calculate mv using equations (4.5), (4.6) and (4.7): i) From eqn (4.5): _L¼m _ Fi  m _V m (4:8) ii) Substitute eqn (4.8) in (4.7): _ Fi cpF TFi þ q ¼ m _ V H V þ ðm _ Fi m _ V ÞcpL TL m (4:9) iii) Find the saturation temperature and the enthalpy of saturated vapor at 70.14 kPa from the steam tables: TL=908C V=2660 kJ/kg iv) Substitute numerical values in eqn (4.9): _ V  2660 þ ð1000  m _ V Þ  3:8  90 1000  3:8  50 þ 270000 ¼ m 26 4 Energy Balance _V v) Solve for m _ v ¼ 50:9 kg=h m Step 6 Alternatively, an energy balance in envelope III can be used instead of envelope I: i) Write the energy balance equation: _ Fo cpF TFo ¼ m _ V HV þ m _ L cpL TL m (4:10) ii) Combine eqns (4.5) and (4.10) and substitute numerical values: _ V Þ  3:8  90 _ V  2660 þ ð1000  m 1000  3:8  121 ¼ m _V iii) Solve for m _ v ¼ 50:9 kg=h m Exercises Exercise 4.1 How much saturated steam with 120.8 kPa pressure is required to heat 1000 g/h of juice from 58C to 958C? Assume that the heat capacity of the juice is 4 kJ/ kg8C. Solution Step 1 Draw the process diagram: mji = 1000 kg/h juice 5°C ms steam 120.8kPa mjo HEAT EXCHANGER juice 95°C ms condensate 120.8kPa Exercises 27 Step 2 Write the energy balance equation: _ ji Hji þ m _ s Hs ¼ ::::::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::::::: m or _ ji cpj Tji þ m _ s Hs ¼ ::::::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::::::: m Step 3 Substitute numerical values in the above equation. (Find the enthalpy of saturated steam and water [condensate] from steam tables): :::::::::::::::::::::::::::: þ ::::::::::::::::::::::::::: ¼ ::::::::::::::::::::::::: þ :::::::::::::::::::::::: Step 4 _s Solve for m _ s ¼ ::::::::::::::::::::::::::::::::::::::kg=h m Exercise 4.2 How much saturated steam with 120.8 kPa pressure is required to concentrate 1000 kg/h of juice from 12% to 20% solids at 958C? Assume that the heat capacity of juice is 4 kJ/kg8C. Solution Step 1 Draw the process diagram: water vapor mv mji = 1000 kg/h juice o 95 C ms steam 120.8kPa EVAPORATOR mjo 95oC juice ms condensate 120.8kPa 28 4 Energy Balance Step 2 Write the overall mass balance equation on the juice side: _Vþm _ jo 1000 ¼ m Step 3 Write the solids mass balance equation: _ jo 0:12  1000 ¼ :::::::::::::::  m _ jo and m _V Solve for m _ jo ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::kg=h m _ V ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::kg=h m Step 4 i) Write the enthalpy balance equation: _ ji cpj Tji þ m _ s Hs ¼ ::::::::::::::::::::::: þ ::::::::::::::::::::::::: ¼ ::::::::::::::::::: m ii) From steam tables, find the enthalpy of water vapor at 958C, of saturated steam at 120.8 kPa, and of water (condensate) at 120.8 kPa. iii) Substitute numerical values in the above equation: :::::::::::::::::: þ ::::::::::::::::: ¼ :::::::::::::::::: þ ::::::::::::::::: þ ::::::::::::::::: _s iv) Solve for m _ s ¼ ::::::::::::::::::::::::::::::::::::::::kg=h m Exercise 4.3 2000 kg/h of milk is sterilized in a steam infusion sterilizer. The milk is heated to 1458C by introducing it into the steam infusion chamber H and then is cooled quickly by flashing in the flash vessel F. The vapor that flashes off in the vessel F is condensed in the condenser C by direct contact of the vapor with cooling water. To avoid dilution of the milk, the pressure in the vessel F must be such that the rate at which vapor flashes off in the vessel F is equal to the steam that is added in the vessel H. Calculate the cooling water flow rate in the condenser that will give the required pressure in the flash vessel. The following data are Exercises 29 given: The temperature of the milk at the inlet of H is 408C, the temperature of the cooling water at the inlet of the condenser is 208C, the steam introduced into the chamber H is saturated at 475.8 kPa pressure, and the heat capacity of the milk is 3.8 kJ/kg8C at the inlet of the infusion chamber and 4 kJ/kg8C at the exit of the infusion chamber. Solution Step 1 Draw the process diagram: milk mmi, Hmi I steam cooling water mwi, Hwi III II vapor mv, Hv ms, Hs F H C mms, Hms mmo, Hmo milk mwo, Hwo Step 2 State your assumptions: l l l l l l The terms of kinetic and potential energy in the energy balance equation are negligible. A pump is not included in the system (Ws ¼ 0). The heat losses to the environment are negligible. The water vapor pressure of the milk is equal to that of water at the same temperature. The water vapor pressure in the condenser is equal to the water vapor pressure in the flash vessel. The system is at steady state. Step 3 Write the mass and energy balance equations in envelope I: i) Energy balance in envelope I: _ mi Hmi þ m _ s Hs ¼ m _ ms Hms m ii) Overall mass balance in envelope I: :::::::::::::::::::::::::::::::: þ ::::::::::::::::::::::::::: ¼ ::::::::::::::::::::::::::::: http://avibert.blogspot.com 30 4 Energy Balance iii) Substitute numerical values and combine the last two equations: _ s ¼ ::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::: þ 2746:5 m _s iv) Solve for m _ s ¼ ::::::::::::::::::::::::::::::::::::::::::::::::kg=h m Step 4 i) Write the energy balance in envelope II: ::::::::::::::::::::::::: ¼ ::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::: _s¼m _ v , in order to avoid ii) Substitute values taking into account that m dilution of the milk: ::::::::::::::::::::::::::::::::: ¼ :::::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::::::: or 1389158  7600  T ¼ 395:1  HV iii) Solve the last equation by trial and error to find the value of T that will give a value of HV in agreement with steam tables. T = . . .. . .. . .. . .. . .. . .. . ...8C. Step 5 Write the overall mass balance and energy balance in envelope III: i) Overall mass balance: :::::::::::::::::::::::::::: þ ::::::::::::::::::::::::::: ¼ ::::::::::::::::::::::::::::::: ii) Energy balance: :::::::::::::::::::::::::::: þ ::::::::::::::::::::::::::: ¼ ::::::::::::::::::::::::::::::: iii) Substitute numerical values in the last equation and solve for mwi. The temperature of the water at the exit of the condenser must be equal to :::::::::::::::::::: C, because the water vapor pressure in the condenser was assumed equal to that in the flash vessel F. _ wi ¼ :::::::::::::::::::::::::::::::::::::::kg=h m Exercises 31 Exercise 4.4 Find the amount of saturated steam at 270.1 kPa required to heat 100 kg of cans from 508C to 1218C, if the heat capacity of the cans is 3.5 kJ/kg8C. Exercise 4.5 One ice cube at 108C weighing 30g is added to a glass containing 200ml of water at 208C. Calculate the final water temperature when the ice cube melts completely. Assume that 3 kJ of heat are transferred from the glass to the water during the melting of the ice? Use the following values: the latent heat of fusion of the ice is 334 kJ/kg, the heat capacity of the ice is 1.93 kJ/kg8C, and the heat capacity of the water is 4.18 kJ/kg8C. Exercise 4.6 For quick preparation of a cup of hot chocolate in a cafeteria, cocoa powder and sugar are added in a cup of water and the solution is heated by direct steam injection. If the initial temperature of all the ingredients is 158C, the final temperature is 958C, the mass of the solution is 150g initially, and the heat capacity of the solution is 3.8 kJ/kg8C, calculate how much saturated steam at 1108C will be used. State your assumptions. Exercise 4.7 Calculate the maximum temperature to which a liquid food can be preheated by direct steam injection if the initial temperature and the initial solids concentration of the food are 208C and 33% respectively, and the final solids concentration must not be less than 30%. How much saturated steam at 121 kPa pressure will be used? Assume that the heat capacity of the food is 3.0 kJ/kg8C initially and 3.1 kJ/kg8C after the steam injection. Chapter 5 Fluid Flow http://avibert.blogspot.com Review Questions Which of the following statements are true and which are false? 1. The Reynolds number represents the ratio of the inertia forces to viscous forces. 2. If the Reynolds number in a straight circular pipe is less than 2100, the flow is laminar. 3. The velocity at which the flow changes from laminar to turbulent is called critical velocity. 4. The Reynolds number in non-Newtonian fluids is called the Generalized Reynolds number 5. The velocity profile of Newtonian fluids in laminar flow inside a circular pipe is parabolic. 6. The velocity profile of Newtonian fluids in laminar flow is flatter than in turbulent flow. 7. The maximum velocity of Newtonian fluids in laminar flow inside a circular pipe is twice the bulk average velocity. 8. The average velocity of Newtonian fluids in turbulent flow inside a circular pipe is around 80% of the maximum velocity. 9. The maximum velocity of pseudoplastic fluids in laminar flow inside a circular pipe is more than twice the bulk average velocity. 10. The Hagen-Poiseuille equation gives the pressure drop as a function of the average velocity for turbulent flow in a horizontal pipe. 11. The pressure drop in laminar flow is proportional to the volumetric flow rate. 12. The pressure drop in turbulent flow is approximately proportional to the 7/4 power of the volumetric flow rate. 13. In a fluid flowing in contact with a solid surface, the region close to the solid surface where the fluid velocity is affected by the solid surface is called boundary layer. 14. The velocity gradients and the shear stresses are larger in the region outside the boundary layer than in the boundary layer. 15. Boundary layer thickness is defined as the distance from the solid surface where the velocity reaches 99% of the free stream velocity. S. Yanniotis, Solving Problems in Food Engineering. Ó Springer 2008 33 34 5 Fluid Flow 16. The viscosity of a liquid can be calculated if the pressure drop of the liquid flowing in a horizontal pipe in laminar flow is known. 17. The viscosity of non-Newtonian liquids is independent of the shear rate. 18. The flow behavior index in pseudoplastic liquids is less than one. 19. In liquids that follow the power-law equation, the relationship between average velocity and maximum velocity is independent of the flow behavior index. 20. The apparent viscosity of a pseudoplastic liquid flowing in a pipe decreases as the flow rate increases. Examples Example 5.1 Saturated steam at 1508 C is flowing in a steel pipe of 2 in nominal diameter, schedule No. 80. If the average velocity of the steam is 10 m/s, calculate the mass flow rate of the steam. Solution Step 1 Find the inside diameter for a 2 in pipe schedule No. 80 from table: D ¼ 4:925cm Step 2 Calculate the inside cross-sectional area of the pipe: A¼ pD2 pð0:04925mÞ2 ¼ ¼ 0:001905 m2 4 4 Step 3 Calculate the volumetric flow rate: Q ¼A  vaver: ¼ ð0:001905m2 Þð10m=sÞ ¼ 0:01905m3 =s Step 4 Find the specific volume of saturated steam at 150 8C from the steam tables: v = 0.3928 m3/kg Step 5 Calculate the mass flow rate: _ ¼ m Q 0:01905 m3 =s ¼ ¼ 0:0485 kg=s v :0:3928 m3 =kg Exercises 35 Example 5.2 A 50% sucrose solution at 20 8 C is flowing in a pipe with 0.0475 m inside diameter and 10 m length at a rate of 3 m3/h. Find: a) the mean velocity, b) the maximum velocity, and c) the pressure drop of the sucrose solution. The viscosity and the density of the sucrose solution at 20 8 C are 15.43 cp and 1232 kg/m3respectively. Solution Step 1 Calculate the cross-section area of the pipe: A¼ pD2 pð0:0475mÞ2 ¼ ¼ 1:77  103 m2 4 4 Step 2 Calculate the mean velocity of the liquid: vm ¼ Q 8:33  104 m3 =s ¼ ¼ 0:471 m=s A 1:77  103 m2 Step 3 Calculate the Reynolds number: Re ¼ Dvp ð0:0475mÞð0:471m=sð1232kg=m3 Þ ¼ ¼ 1786 m 15:43  103 kg=ms Since Re 40) 1) Local Temperature a) Rectangular coordinates !   1 Te T 4X ð1Þn ð2n þ 1Þpx ð2n þ 1Þ2 p2 exp  cos Fo ¼ Te To p n¼0 2n þ 1 2L 4 b) Cylindrical coordinates 1   Te  T 2X J0 ðrdn Þn exp R2 d2n Fo ¼ Te  To R n¼1 dn J1 ðRdn Þ with dn being roots of the Bessel function Jo(Rdn) = 0. The first five roots of Jo(x) are: 2.4048, 5.5201, 8.6537, 11.7915 and 14.9309 (see Table A.1). Consequently, d1 ¼ 2:4048=R; d2 ¼ 5:5201=R; d3 ¼ 8:6537=R; d4 ¼ 11:7915=R and d5 ¼ 14:9309=R c) Spherical coordinates 1 Te  T 2R X ð1Þn npr  2 2  exp n p Fo sin ¼ Te  To pr n¼1 n R 2) Mean Temperature a) Rectangular coordinates ! 1 Te  Tm 8X 1 ð2n þ 1Þ2 p2 Fo ¼ ¼ 2 exp  Te  To p n¼0 ð2n þ 1Þ2 4   8 1 1 ¼ 2 expð  2:47FoÞ þ expð  22:2FoÞ þ expð  61:7FoÞ þ ::: p 9 25 b) Cylindrical coordinates 1   Te  Tm 4X p2 ¼ 2 exp R2 d2n Fo ¼ Te  To p n¼1 R2 d2n 4 ¼ 2 1:7066expð  5:783FoÞ þ 0:324expð  30:5FoÞ p  þ 0:132expð  74:9FoÞ þ ::: c) Spherical coordinates 1   Te  Tm 6X 1 ¼ exp n2 p2 Fo ¼ Te  To p2 n¼1 n2   6 1 1 ¼ 2 expð  p2 FoÞ þ expð  4p2 FoÞ þ expð  9p2 FoÞ þ ::: p 4 9 where 2 A = heat transfer surface area,  m 2 a = thermal diffusivity, m s Bi = Biot number ð¼ hR=kÞ cp = heat capacity, J=kg 8C erf = the error function Theory 103 Table 10.2 Solutions of the unsteady state heat transfer equation for uniform initial temperature and both resistances, surface and internal, significant ð0:15Bi540Þ 1) Local Temperature a) Rectangular coordinates   1 X with   cos Lx dn Te  T 2Bi exp d2n Fo ¼ 2 2 dn roots of: d tand ¼ Bi Te  To d þ Bi þ Bi cosðdn Þ n¼1 n b) Cylindrical coordinates   1 X   Te  T 2 Bi Jo Rr dn exp d2n Fo ¼ 2 2 Te  To n¼1 dn þ Bi Jo ðdn Þ c) Spherical coordinates   1   sin Rr dn Te  T RX 2 Bi exp d2n Fo ¼ 2 2 Te  To r n¼1 dn þ Bi  Bi sinðdn Þ with dn roots of: d J1 ðdÞ ¼ Bi Jo ðdÞ with dn roots of: d cotd ¼ 1  Bi For the center of the sphere (r=0) 1 X   Te  T 2 Bi dn exp d2n Fo ¼ 2 2 Te  To n¼1 dn þ Bi  Bi sinðdn Þ 2) Mean Temperature a) Rectangular coordinates 1   Te  Tm X 2Bi2 ¼ exp d2n Fo 2 2 2 Te  To n¼1 ðdn þ Bi þ BiÞdn with dn roots of: d tand ¼ Bi b) Cylindrical coordinates 1   Te  Tm X 4 Bi2 ¼  2  2 exp d2n Fo 2 Te  To n¼1 dn þ Bi dn with dn roots of: d J1 ðdÞ ¼ Bi Jo ðdÞ c) Spherical coordinates 1   Te  Tm X 6 Bi2  2  2 exp d2n Fo ¼ 2 Te  To n¼1 dn þ Bi  Bi dn with dn roots of: d cotd ¼ 1  Bi Table 10.3 Unsteady state heat transfer with negligible internal resistance ð0:1 mcp Te  To ln hA Te  T erfc = the complementary  function defined as erfc = 1 - erf  error Fo = Fourier number ¼ at L2 h = heat transfer coefficient, W m2 8C Jo = Bessel function of the first kind of order zero J1 = Bessel function of the first  kind of order one k = thermal conductivity, W m2 C L = half thickness of the plate, m m = mass, kg 104 10 Unsteady State Heat Transfer Table 10.4 Solutions of the unsteady state heat transfer equation for a semi-infinite body with uniform initial temperature 1) Constant surface temperature   Te  T x ¼ erf pffiffiffiffiffi Te  To 2 at 2) Convection at the surface pffiffiffiffiffi      Te  T x hx h2 at x h at þ 2 erfc pffiffiffiffiffi þ ¼ erf pffiffiffiffiffi þ exp Te  To k k 2 at 2 at k where R = radius of cylinder or sphere, m r = distance of the point from the centerline of the cylinder or center of the sphere, m x = distance of the point from the center plane of the plate, or from the surface in the case of semi-infinite body, m t = time, s Te = equilibrium temperature (temperature of the environment or surface temperature) To = initial temperature T = temperature at time t and point x Tm = mean temperature To calculate the temperature as a function of time and position in a solid: 1. Identify the geometry of the system. Determine if the solid can be considered as plate, infinite cylinder, or sphere. 2. Determine if the surface temperature is constant. If not, calculate the Biot number and decide the relative importance of internal and external resistance to heat transfer. 3. Select the appropriate equation. 4. Calculate the Fourier number. 5. Find the temperature by applying the selected equation (if F0 > 0:2, calculate the temperature either using only the first term of the series solution or using Heisler or Gurnie-Lurie charts). REVIEW QUESTIONS Which of the following statements are true and which are false? 1. If the temperature at any given point of a body changes with time, unsteady state heat transfer occurs. 2. Thermal diffusivity is a measure of the ability of a material to transfer thermal energy by conduction compared to the ability of the material to store thermal energy. Examples 105 3. Materials with high thermal diffusivity will need more time to reach equilibrium with their surroundings. 4. The Biot number (Bi) expresses the relative importance of the thermal resistance of a body to that of the convection resistance at its surface. 5. If Bi, the external resistance is negligible. 6. If Bi > 40, the surface temperature may be assumed to be equal to the temperature of the surroundings. 7. If Bi, the internal resistance is significant. 8. If Bi Bi, the temperature of the body may be assumed to be uniform. 9. Problems with Bi are treated with the lumped capacitance method. 10. The Fourier number (Fo) has dimensions of time. 11. The Heisler charts give the temperature at the center of an infinite slab, infinite cylinder, and sphere when FO > 0:2. 12. The Gurney-Lurie charts give the temperature at any point of an infinite slab, infinite cylinder, and sphere when FO > 0:2. 13. Thermal penetration depth is defined as the distance from the surface at which the temperature has changed by 10% of the initial temperature difference. 14. Until the thermal penetration depth becomes equal to the thickness of a finite body heated from one side, the body can be treated as a semi-infinite body. 15. A cylinder of finite length can be treated as an infinitely long cylinder if the two ends of the cylinder are insulated or if its length is at least 10 times its diameter. Examples Example 10.1 A steak 2 cm thick is put on a hot metallic plate. The surface of the steak in contact with the hot surface immediately attains a temperature of 120oC and retains this temperature. Calculate the temperature 1.1 cm from the hot surface of the steak after 15 min, if the initial temperature of the meat is 5oC and the  7 2 thermal diffusivity of the meat is 1:4  10 m s. Solution Step 1 Draw the process diagram: L x 106 10 Unsteady State Heat Transfer Step 2 Identify the geometry of the system. The shape of the steak is a plate heated from one side. Step 3 Examine the surface temperature. The surface temperature is constant. Step 4 Select the appropriate equation. The equation for a plate with uniform initial temperature and constant surface temperature is (Table 10.1): !   1 Te  T 4X ð1Þn ð2n þ 1Þpx ð2n þ 1Þ2 p2 cos Fo (10:1) ¼ exp  Te  To p n¼0 2n þ 1 2L 4 Step 5 Calculate the Fourier number. Since the steak is heated only from one side, the characteristic dimension is the thickness of the steak, not the half thickness. Therefore: Fo ¼ t ð1:4  107 m2 =sÞð15  60Þ ¼ ¼ 0:315 L2 0:022 m2 Step 6 Calculate the temperature Since FO > 0:2, only the first term of the sum in eqn (10.1) can be used without appreciable error. Alternatively, the solution can be read directly from a Gurney-Lurie chart (see for example Ref. 3), which gives the dimensionless ratio ðTe  TÞ=ðTe  To Þ vs. the Fourier number. The temperature below is calculated both ways: i) From the equation: !   1 Te  T 4X ð1Þn ð2n þ 1Þpx ð2n þ 1Þ2 p2 cos Fo ¼ exp  Te  To p n¼0 2n þ 1 2L 4 !   4 ð1Þ0 ð2  0 þ 1Þpx ð2  0 þ 1Þ2 p2  cos Fo exp  p2  0 þ1 2L 4     2  px   p2  4 4 p  0:009 p ¼ cos exp  Fo ¼ cos exp  0:315 p 2L p 2  0:02 4 4 ¼ 0:445 Calculate the temperature as: Te  T 120  T ¼ 0:445 ! T ¼ 68:8 C ¼ Te  To 120  5 Examples 107 ii) From the Gurnie-Lurie chart: a) Find the value of Fo ¼ 0:315 on the x-axis of the Gurney-Lurie chart. b) Find the curve with k=hL ¼ 0 (constant surface temperature or Bi ! 1) and x=L ¼ 0:009=0:02 ¼ 0:45. c) Read the dimensionless temperature on the y-axis as ðTe  TÞ=ðTe  To Þ ¼ 0:45. d) Calculate the temperature as Te  T 120  T ¼ 0:45 ! T ¼ 68:3 C ¼ Te  To 120  5 (Te–T)/(Te–To) 1.0 k/hL = 0 x/L = 0.45 0.1 0.0 0.0 0.2 0.4 0.6 0.8 Fo Comment: The reading from the chart can only be approximate Example 10.2 A hot dog 1.5 cm in diameter and 16 cm in length with 58C initial temperature is immersed in boiling water. Calculate a) the temperature 3 mm under the surface after 2 min, b) the temperature at the center after 2 min, c) the time necessary to reach 818C at the center, d) the average temperature of the hot dog after 2 min, and e) the heat that will be transferred to the hot dog during the 2 min of boiling.  Assume that the thermal conductivity is 0:5W=m 8C, the density is 1050kg m3 , the heat capacity is 3:35kJ=kg 8C, and the heat transfer coefficient at the surface  of the hot dog is 3000W m2 8C. Solution Step 1 Draw the process diagram: r D R L 108 10 Unsteady State Heat Transfer Step 2 Identify the geometry of the system. The hot dog has a cylindrical shape with L=D equal to: L 0:16 m ¼ ¼ 10:7 D 0:015 m Since L=D > 10, the hot dog can be treated as an infinite cylinder. The contribution of heat transferred through the bases of the cylinder can be neglected. Step 3 Examine the surface temperature. Since the surface temperature is unknown, the Biot number must be calculated: Bi ¼ hR ¼ k   3000 W=m2 o C ð0:0075 mÞ 0:5 W=mo C ¼ 45 Since Bi > 40 the external resistance to heat transfer is negligible. Therefore, it can be assumed that the surface temperature will immediately reach the temperature of the environment, 100oC. Step 4 Select the appropriate equation to use.The solution for constant surface temperature will be applied (Table 10.1): 1   Te  T 2X J0 ðrdn Þn exp R2 d2n Fo ¼ Te  To R n¼1 J1 ðRdn Þ (10:2) Step 5 Calculate the Fourier number. Calculate the thermal diffusivity first: a¼ k 0:5 W=mo C  ¼ 1:42  107 m2 =s ¼ rcp 1050 kg=m3 ð3350 J=kgo CÞ Then: Fo ¼ t ð1:42  107 m2 =sÞð120 sÞ ¼ ¼ 0:30 R2 0:00752 m2 Since Fo > 0:2, only the first term of the series solution can be used without appreciable error. Alternatively, the solution can be read directly from a chart Examples 109 such as Gurney-Lurie or Heisler (see Fig A.3 and A.6 in the Appendix), which gives the dimensionless ratio ðTe  TÞ=ðTe  To Þ vs. the Fourier number. 1) Calculate the temperature 3 mm under the surface. i) From eqn (10.2) above: Step 1 Find the values of dn. Since Fo > 0:2 the first term of the series solution is enough. However, two terms will be used in this problem for demonstration purposes. Since x ¼ Rdn and R = 0.0075, the values of dn will be (see Table A.1 in the Appendix): x dn 2.4048 5.5201 320.6 736.0 Step 2 Substitute these values into eqn (10.2) and calculate T: 1   Te  T 2X J0 ðrdn Þn exp R2 d2n Fo ¼ Te  To R n¼1 dn J1 ðRdn Þ    2 Jo ð0:0045  320:6Þ  expð  2:40482 ð0:30ÞÞ 0:0075 320:6  J1 ð2:4048Þ    Jo ð0:0045  736Þ þ expð  5:52012 ð0:30ÞÞ 736  J1 ð5:5201Þ  Jo ð1:4427Þ ¼ 266:67 expð1:7349Þ 320:6  J1 ð2:4048Þ  Jo ð3:312Þ þ expð9:141Þ 736  J1 ð5:5201Þ  0:5436 ¼ 266:67 expð1:7349Þ 320:6  0:5192  0:3469 þ expð9:141Þ 736  ð0:3403Þ   ¼ 266:67 5:76  104 þ 1:48  107 ¼ 0:154 or Te  T 100  T ¼ 0:154 ! T ¼ 85:48C ¼ Te  To 100  5 110 10 Unsteady State Heat Transfer   Comment: Notice that the 2nd term 1:48  107 is much smaller than the first  term 5:76  104 and could have been neglected. ii) From the Gurney-Lurie chart for an infinite cylinder (see Fig A.3 in the Appendix): Step 1 Calculate the ratio r=R which gives the position of the point (r is the distance of the point of interest from the axis of the cylinder): r 0:0045 m ¼ ¼ 0:6 R 0:0075 m Step 2 Select the group of curves on the Gurney-Lurie chart that will be used based on the k=hR value. Since Bi > 40, the curves with k=hR ¼ 0 on the Gurney-Lurie chart can be used. Step 3 On the Gurney-Lurie chart: l l l l Find the value of Fo ¼ 0:3 on the x-axis Find the curve with k=hR ¼ 0 and r=R ¼ 0:6 Read the dimensionless temperature on ðTe  TÞ=ðTe  To Þ ¼ 0:15 Calculate the temperature as Te  T 100  T ¼ 0:15 ! T ¼ 85:8 C ¼ Te  To 100  5 (Te-T)/(Te-To) 1.00 k/hR = 0 r/R = 0.6 0.10 0.01 0.0 0.2 0.4 0.6 Fo 2) Calculate the temperature at the center. i) From the equation: As above but with r = 0 : 0.8 1.0 the y-axis as Examples 111 1   Te  T 2X J0 ðrdn Þn ¼ exp R2 d2n Fo Te  To R n¼1 dn J1 ðRdn Þ      2 Jo ð0  320:6Þ exp  2:40482 ð0:30Þ  0:0075 320:6  J1 ð2:4048Þ    Jo ð0  736Þ 2 expð  5:5201 ð0:30ÞÞ þ 736  J1 ð5:5201Þ   J o ð 0Þ J o ð 0Þ ¼ 266:67 expð1:7349Þ þ expð9:141Þ 320:6  J1 ð2:4048Þ 736  J1 ð5:5201Þ   1 1 expð1:7349Þ þ expð9:141Þ ¼ 266:67 320:6  0:5192 736  ð0:3403Þ   ¼ 266:67 1:06  103  4:279  107 ¼ 0:283 or Te  T 100  T ¼ 0:283 ! T ¼ 73:1 C ¼ Te  To 100  5 Comment: Notice that the 2nd term (4:279  107 ) is much smaller than the first term (1:06  103 ) and could have been neglected. ii) From the Heisler chart for an infinite cylinder (Fig A.6): On the Heisler chart: l l l Find the value of Fo ¼ 0:3 in the x-axis. Find the curve with k=hR ¼ 0. Read the dimensionless temperature on the y-axis as ðTe  TÞ= ðTe  To Þ ¼ 0:27 Calculate the temperature as Te  T 100  T ¼ 0:27 ! T ¼ 74:4 C ¼ Te  To 100  5 1.00 k/hR = 0 (Te-T)/(Te-To) l 0.10 0.01 0.0 0.2 0.4 0.6 Fo 0.8 1.0 112 10 Unsteady State Heat Transfer 3) Calculate the time necessary to reach 81oC at the center: Calculate the dimensionless temperature Te  T 100  81 ¼ 0:2 ¼ Te  To 100  5 On the Heisler chart: l l l l Find on the y-axis the value ðTe  TÞ=ðTe  To Þ ¼ 0:2 Find the curve k=hR ¼ 0. Read the Fourier number on the x-axis as Fo ¼ 0:36. Find the time from the Fourier number as: t ¼ Fo R2 0:00752 m2 ¼ 0:36 ¼ 143 s  1:42  107 m2 =s (Te-T)/(Te-To) 1.00 k/hR = 0 0.10 0.01 0.0 0.2 0.4 0.6 0.8 1.0 Fo 4) Calculate the average temperature. Step 1 Write the equation for the mean temperature of an infinite cylinder with Bi > 40 ðTable 10:1Þ: 1   Te  Tm 4X p2 ¼ 2 exp R2 d2n Fo p n¼1 R2 d2n Te  To  4 ¼ 2 1:7066 expð  5:783FoÞ þ 0:324 expð  30:5FoÞ p  þ 0:132 expð  74:9FoÞ þ ::: Step 2 Substitute values in the above equation and solve for Tm (since Fo > 0:2, the first term only is enough; in this example, three terms will be used for demonstration purposes): Examples 113 Te  Tm 4 ¼ 2 1:7066 expð5:783  0:3Þ þ 0:324 expð30:5  0:3Þ Te  To p  þ 0:132 expð74:9  0:3Þ þ ::::: ¼  4 0:3011 þ 3:44  105 þ 2:3  1011 þ ::: ¼ 0:122 2 p and Tm ¼ Te  0:122ðTe  To Þ ¼ 100  0:122ð100  5Þ ¼ 88:4 8C Comment: Notice that the 2nd and 3rd terms of the sum are negligible compared to the 1st term and they could have been omitted. 5) Calculate the heat transferred to the solid in 2 min. Because qt mcp ðTm  To Þ Tm  To Te  Tm ¼ ¼ ¼1 mcp ðTe  To Þ qe Te  To Te  To the heat transferred to the hot dog in 2 min will be:   Te  Tm qt ¼ mcp ðTe  To Þ 1  Te  To   Te  Tm ¼ ðVrÞcp ðTe  To Þ 1  Te  To   2   ! 2 p 0:015 m kg J 3350 ¼ ð100  5 o CÞð1  0:122Þ ð0:16 mÞ 1050 3 m kg 8C 4 ¼ 8296 J Example 10.3 12mm  16mm  14mm rectangular fruit pieces are immersed in syrup. Calculate the temperature at the center of a piece of fruit after 5 min if the initial temperature of the fruit piece is 208C, the syrup temperature is 1008C, the heat  transfer coefficient at the surface of the fruit piece is 83W m2 8C, and the physical properties of the fruit piece are k ¼ 0:5 W=m8C, cp ¼ 3:8kJ=kg8C, and r ¼ 900kg m3 . Assume the fruit piece does not exchange matter with the syrup. Solution Step 1 Draw the process diagram: 114 10 Unsteady State Heat Transfer y z Lz x Ly Lx The temperature at the center is affected by the heat transferred from the three directions x, y, and z. The contribution from each direction has to be calculated separately, and the combined effect will be calculated at the end. 1) Heat transferred in the x-direction: Step 1 Select the appropriate equation to use. i) Calculate the Biot number for the x-direction: Bix ¼ hLx ¼ k   83 W=m2 o C ð0:006 mÞ 0:5 W=m8C ¼1 ii) Since 0:1 5 Bi 5 40 both external and internal resistances are important. Therefore, the solution will be (Table 10.2):   Xd 1 cos X n   Te  Tx 2Bix Lx exp d2n Fox ¼ 2 Te  To n¼1 dn þ Bix þ Bix cosðdn Þ Step 2 Calculate the Fourier number Fox for the x-direction: i) Calculate the thermal diffusivity: a¼ k 0:5 W=m 8C  ¼ ¼ 1:46  107 m2 =s 3 rcp 900 kg=m ð3800 J=kg 8CÞ Examples 115 ii) Calculate the Fourier number: t ð1:46  107 m2 =sÞð300 sÞ ¼ ¼ 1:217 L2x 0:0062 m2 Fox ¼ Since Fox > 0:2, only the first term of the sum in the above equation can be used without appreciable error. Alternatively, the solution can be read directly from the Heisler chart. Step 3 Calculate the temperature: i) From the equation: The first root of the equation d tand ¼ Bi for Bi = 1 is: d1 ¼ 0:8603 (see Table A.2 in the Appendix).   cos Lx dn   Te  Tx 2Bix x exp d2n Fox ¼  2 2 Te  To dn þ Bix þ Bix cosðdn Þ   0  0:8603 cos   21 0:006 exp 0:86032  1:217 ¼ 2 2 cos ð 0:8603 Þ 0:8603 þ 1 þ 1 ¼ 0:455 ii) From the Heisler chart for an infinite slab: l l Find the value of Fo = 1.22 on the x-axis. Find the curve with k=hL ¼ 1: Read the dimensionless temperature on the y-axis as: Te  Tx ¼ 0:46 Te  To 1.0 (Te–T)/(Te–To) l k/hLx = 1 0.1 0.0 0.0 0.5 1.0 1.5 Fo 2.0 2.5 3.0 116 10 Unsteady State Heat Transfer 2) Heat transferred in the y-direction: Step 1 Select the appropriate equation to use. i) Calculate the Biot number for the y-direction: Biy ¼ hLy ¼ k   83 W=m2 o C ð0:008 mÞ 0:5 W=mo C ¼ 1:328 ii) Since 0:1 1 Te  Ty X 2Biy ¼ 2 Te  To n¼1 dn þ Biy þ Biy  y dn   Ly exp d2n Foy cosðdn Þ cos Step 2 Calculate the Fourier number Foy for the y-direction: Foy ¼ t ð1:46  107 m2 =sÞð300 sÞ ¼ ¼ 0:684 L2y 0:0082 m2 Since Foy > 0:2, only the first term of the sum in the above equation can be used without appreciable error. Alternatively, the solution can be read directly from the Heisler chart. Step 3 Calculate the temperature: i) From the equation: The first root of the equation d tand ¼ Bi for Bi =1.328 is: d1 ¼ 0:9447 (see Table A.2).   y dn   Ly Te  Ty 2Biy exp d2n Foy  Te  To d2n þ Bi2y þ Biy cosðdn Þ   0  0:9447 cos 2  1:328 0:008 ¼ cosð0:9447Þ 0:94472 þ 1:3282 þ 1:328   exp 0:94472  0:684 ¼ 0:618 cos Examples 117 ii) From the Heisler chart for an infinite slab (Fig. A.5): l l l Find the value of Fo = 0.68 on the x-axis. Find the curve with k=hL ¼ 0:75 (interpolate between curves 0.7 and 0.8). Read the dimensionless temperature on the y-axis as: Te  Ty ¼ 0:61 Te  To 1.0 (Te-T)/(Te-To) k/hLy = 0.75 0.1 0.0 0.0 0.5 1.0 1.5 Fo 2.0 2.5 3.0 3) Heat transferred in the z-direction: Step 1 Select the appropriate equation to use. i) Calculate the Biot number for the z-direction:  Biz ¼ hLz ¼ k  83 W=m2 o C ð0:007 mÞ 0:5 W=mo C ¼ 1:162 ii) Since 0:1  z cos dn 1   Te  Tz X 2Biz Lz exp d2n Foz ¼ 2 Te  To n¼1 dn þ Biz þ Biz cosðdn Þ Step 2 Calculate the Fourier number Foz for the z-direction: Foz ¼ t ð1:46  107 m2 =sÞð300 sÞ ¼ ¼ 0:894 L2z 0:0072 m2 118 10 Unsteady State Heat Transfer Since Foz > 0:2, only the first term of the sum in the above equation can be used without appreciable error. Alternatively, the solution can be read directly from the Heisler chart. Step 3 Calculate the temperature: i) From the equation: The first root of the equation d tand ¼ Bi for Bi =1.162 is: d1 ¼ 0:9017(see Table A.2):   z cos dn   Te  Tz 2Biz Lz exp d2n Foz ¼  2 2 Te  To dn þ Biz þ Biz cosðdn Þ   0  0:9017 cos 2  1:162 0:007 ¼ cosð0:9017Þ 0:90172 þ 1:1622 þ 1:162   exp 0:90172  0:894 ¼ 0:545 ii) From the Heisler chart for an infinite slab (Fig. A.5): l l Find the value of Fo = 0.89 on the x-axis. Find the curve with k=hL ¼ 0:86 (interpolate between curves 0.8 and 1.0). Read the dimensionless temperature on the y-axis as: Te  Tz ¼ 0:55 Te  To 1.0 (Te–T)/(Te–To) l k/hLx = 1 0.1 0.0 0.0 0.5 1.0 1.5 Fo 2.0 2.5 3.0 Examples 119 4) The combined effect of heat transferred in the x, y, and z directions is: Te  Txyz ¼ Te  To  Te  Tx Te  To  Te  Ty Te  To  Te  Tz Te  To  ¼ 0:455  0:618  0:545 ¼ 0:153 and the temperature is: 100  Txyz ¼ 0:153 ! Txyz ¼ 87:8 8C 100  20 Example 10.4 Calculate how long it will take for the temperature on the non-heated surface of the steak of Example 10.1 to increase by 1 % of the initial temperature difference. Solution Step 1 Identify the geometry of the system. As long as the temperature change on the cold surface of the steak is less than 1% of the initial temperature difference, the steak can be treated as a semiinfinite body. Step 2 Examine the surface temperature. The surface temperature is constant at 120oC. Step 3 Select the appropriate equation. The equation for a semi-infinite body with uniform initial temperature and constant surface temperature is (Table 10.4):   Te  T x ¼ erf pffiffiffiffiffi Te  To 2 t with T  To ¼ 0:01 Te  To (since the accomplished temperature change is 1% ) 120 10 Unsteady State Heat Transfer or Te  T T  To ¼1 ¼ 1  0:01 ¼ 0:99 Te  To Te  To Therefore   x 0:99 ¼ erf pffiffiffiffiffi 2 t Step 4  pffiffiffiffiffi Find the erf of the argument x 2 t from an erf table (Table A.5). The argument has to be equal to 1.82 for the error function to be equal to 0.99. Therefore x pffiffiffiffiffi ¼ 1:82 2 t Step 5 Solve for t:    x 2 0:02 m 2 3:64 ¼ 215:6 s t ¼ 3:64 ¼  1:4  107 m2 =s Comment: The distance x where the temperature has increased by 1% of the initial temperature difference is called the ‘‘thermal penetration depth.’’ As long as the thermal penetration depth in a finite body is less than the thickness of the body (or half the thickness of the body, in the case of heating from both sides), the body can be treated as a semi-infinite body. Example 10.5 Concentrated milk is sterilized in a can with 7.5 cm diameter and 9.5 cm height. Calculate the time required to heat the milk from 458C to 1158C given that: a) the can is in such a motion during heating that the milk temperature is uniform inside the can, b) the can contains 410 g of milk, c) the overall  heat transfer coefficient between the heating medium and the milk is 300W m2 8C, d) the heating medium temperature is 1308C, and e) the mean heat capacity of the milk over the temperature range of the heating process is 3650J=kg8C. Exercises 121 Solution Since the milk temperature is uniform inside the can during heating, the internal resistance can be considered negligible. The solution for Bi t¼ mcp Te  To ln ¼ hA Te  T ð0:410 kgÞ ð3650 J=kgo CÞ 130  45 ¼     ln 2 130  115 300 W=m2 8C p ð0:075 mÞð0:095 mÞ þ 2 p 0:075 m2 4 ¼ 277 s Exercises Exercise 10.1 Peas are blanched by immersion in hot water at 908C. Calculate the temperature at the center of a pea after 3 min if the diameter of the pea is 8 mm, the initial temperature of the is 208C, and the heat transfer coefficient at the surface of  pea 2 the pea is 100W m 8C. Assume that the physical properties of the pea are  r ¼ 1050kg m3 , cp ¼ 3:7kJ=kg8C, k ¼ 0:5W=m8C. Solution Step 1 State your assumptions. The pea can be assumed to be a sphere. Step 2 Select the equation for a sphere. i) Calculate the Biot number: Bi ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::: Since 0:1 122 10 Unsteady State Heat Transfer Step 3 Calculate the Fourier number: ¼ :::::::::::::: ¼ ::::::::::::::: ::::::::::::::::::::::::::::: and Fo ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::: Since Fo > 0:2, only the first term of the sum in the above equation can be used without appreciable error. Alternatively, the solution can be read directly from the Heisler chart. Step 4 Calculate the temperature: i) From the above equation: Substitute values in the above equation (d1 ¼ 1:432) and calculate the temperature as 90  T  :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 90  20 T ¼ ::::::::::::::::::::::::::::::::::::::::::8C ii) Also find the temperature from the Heisler chart for a sphere: l l Find the value of Fo on the x-axis. Calculate k=hR and find the corresponding curve. k ¼ ::::::::::::::::::::::::::::::::::: hR l Read the dimensionless temperature on the y-axis as: Te  T ¼ :::::::::::::::::::::::::: Te  To l Calculate the temperature as: ::::::::::::::::::::  T ::::::::::::::::::::::::::::::::::: ! T ¼ :::::::::::::::::::::::: ¼ ::::::: 123 (Te-T)/(Te-To) Exercises Fo Exercise 10.2 A 211x300 can containing a meat product is heated in a retort. The initial o temperature  2 of the can is 50 C, the heat transfer coefficient at the surface is the meat product are: 3000W m 8C, and the physical properties of  k ¼ 0:5W=m8C, cp ¼ 2:85kJ=m8C, and r ¼ 1100kg m3 . Calculate the temperature at the geometric center of the can after 30 min if the steam temperature in the retort is 130oC. Solution Step 1 Determine the shape of the object. The shape of the can is a finite cylinder. The temperature at the center is affected by the heat transferred from the cylindrical surface as well as from the flat bases of the cylinder. The contribution from each direction has to be calculated separately and the combined effect will be calculated at the end. The dimensions of the can are: Diameter ¼ 2 Height ¼ 3 11 inches ¼ 0:06826 m; 16 0 inches ¼ 0:0762 m 16 1) Calculate the heat transferred in the radial direction. Step 1 Select the appropriate equation: i) Calculate the Biot number for the r-direction and determine the relative significance of external and internal resistances: Bir ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 124 10 Unsteady State Heat Transfer ii) Select the equation: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 2 Calculate the Fourier number For for the radial direction: For ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Is For > 0:2? If yes, only the first term of the sum in the above equation can be used without appreciable error. Alternatively, the solution can be read directly from the Heisler chart. Step 3 Calculate the temperature: i) From the equation: Find the values of dn. As in Example 10.2, substitute values in the selected equation and calculate: Te  Tr ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Te  To ii) From the Heisler chart: On the x-axis of the Heisler chart for a infinite cylinder: l l Find the value of For. Calculate k=hR and find the corresponding curve. Read the dimensionless temperature on the y-axis as: Te  Tr ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Te  To (Te-T)/(Te-To) l Fo Exercises 125 2) Calculate the heat transferred in the x-direction. Step 1 Select the appropriate equation: i) Calculate the Biot number for the x-direction and determine the relative significance of external and internal resistances: Bix ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ii) Select the equation: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 2 Calculate the Fourier number Fox for the x-direction: Fox ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Is Fox > 0:2? If yes, only the first term of the sum in the above equation can be used without appreciable error. Alternatively, the solution can be read directly from the Heisler chart. Step 3 Calculate the temperature: i) From the equation: Substitute values and calculate: Te  Tx ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Te  To ii) From the Heisler chart: On the x-axis of the Heisler chart for an infinite slab: l l l Find the value of Fox. Calculate k=hLx and find the corresponding curve. Read the dimensionless temperature on the y-axis as: Te  Tx ¼ :::::::::::::::::::::::::: Te  To 10 Unsteady State Heat Transfer (Te-T)/(Te-To) 126 Fo 3) Calculate the combined effect of the heat transferred through the cylindrical surface and the heat transferred through the flat bases of the can: Te  Trx ¼ Te  To  Te  Tr Te  To  Te  Tx Te  To  ¼ ::::::::::::::::::::::::::::::::::::::::::::::::: Calculate the temperature as: 130  Trx ¼::::::::::::::::::::::::: ! Trx ¼ ::::::::::::::::::::8C ::::::::::::::::::::::::::::::: Exercise 10.3 For the removal of field heat from fruits, hydrocooling is usually used. Calculate the time required to reduce the center temperature of a spherical fruit from 25oC to 5oC by immersion in cold water  of 1oC if the diameter of the fruit is 6 cm, the heat transfer coefficient is 1000W m2 C, and the physical properties of the fruit are: k ¼ 0:4W=m C, r ¼ 900kg m3 , cp ¼ 3:5kJ=kg C. Solution Solve the problem using the Heisler chart for a sphere: l Calculate the value of the dimensionless temperature: Te  T ¼ ::::::::::::::::::::::::::::::::::: Te  To l l Find the value of the dimensionless temperature on the y-axis of the Heisler chart: Calculate the value of k=hR and find the corresponding curve. k ¼ ::::::::::::::::::::::::::::::: hR Exercises l l 127 Read the Fourier number on the x-axis (check if Fo > 0:2 so that the Heisler chart is valid). Calculate the time from: FoR2 ¼ :::::::::::::::::::::::::::::::  (Te-T)/(Te-To) t¼ Fo Exercise 10.4 A root crop is in the ground 5 cm from the surface when suddenly the air temperature drops to 108C. If the freezing point of the crop is -18C, calculate if the temperature of the soil at this depth will drop below the freezing point after 24 h. The initial temperature of  the soil is 128C, the heat transfer coefficient at the surface of the ground is 10W m2 8C, and the physical properties of the soil are k ¼ 0:5W=m 8C, r ¼ 2000kg m3 and cp ¼ 1850J=kg 8C. Neglect latent heat effects. Solution Step 1 Draw the process diagram: Te = –10°C q h = 10W/m2°C x T0 = 12°C 128 10 Unsteady State Heat Transfer Step 2 Select the appropriate equation to use.The ground can be treated as a semiinfinite body. Therefore, the equation for unsteady state in a semi-infinite body with convection at the surface (since the surface temperature is not constant) will be used (Table 10.4): pffiffiffiffiffi      Te  T x hx h2 t x h t þ 2 ¼ erf pffiffiffiffiffi þ exp erfc pffiffiffiffiffi þ Te  To k k 2 t 2 t k (10:3) Step 3 Calculate the thermal diffusivity: ::::::::::::: ¼ ¼ ::::::::::::::::: m2 =s :::::::::::::::::::::::::::: Step 4 Calculate: x pffiffiffiffiffi ¼ 2 t 0:05 m ¼ ::::::::::: :::::::::::::::::::::::::::::::::::::::::::: and   2o pffiffiffiffiffi 10 W=m C  ::::::::::::::::::::::::::::::::::: h t ¼ ::::::::::: ¼ k ::::::::::::::::::::::::::::::::::::::::::::    pffiffiffiffiffi Step 5 Find the value of erf pxffiffiffiffiffi and erfc pxffiffiffiffiffi þ h t from Table A.5 taking k 2 t 2 t into account that erfc ¼ 1  erf. Step 6 Substitute values in eqn (10.3) above and calculate: Te  T ¼ :::::::::::::::::::::::::: Te  To ! T ¼ :::::::::::::::::::::8C Exercises 129 Exercise 10.5 Calculate the temperature of 200 litres of glucose syrup heated in an agitated jacketed vessel for 30 min if the initial temperature of the syrup is 308C, the temperature of the steam in the jacket is 1208C, the density and the heat capacity of the syrup are 1230kg m3 and 3:35kJ=kg 8C respectively, the overall heat  transfer coefficient between the heating medium and the syrup is 200W m2 8C, and the inside surface area of the vessel in contact with the syrup is 1.5 m2. Solution Step 1 Select the appropriate equation to use.Since the syrup is agitated, its temperature can be assumed to be uniform. Therefore the equation to use is: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 2 Calculate the mass of the syrup: m ¼ Vr ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::: Step 3 Apply the selected equation: Te  T ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Te  To ! T ¼ :::::::::::::::::::::::8C 130 10 Unsteady State Heat Transfer Exercise 10.6 A thermocouple with 4 mm diameter and 2 cm length is immersed in milk which is at 608C. How long it will take for the thermocouple to reach a temperature of 59.5oC if the heat transfer coefficient at the surface of the thermocouple is 200W m2 8C, the initial temperature of the thermocouple is 25oC, its mass is 1 g, its heat capacity is 0:461kJ=kg C, and its thermal conductivity is 15W=m 8C. If the thermocouple were exposed to air of 60oC instead of milk, how long would it take to reach 59.5oC? Assume that the heat transfer coefficient in this case is 20W m2 8C. Solution Step 1 Select the appropriate equation: i) Calculate the Biot number: Bi ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ii) Determine the relative importance of internal and external resistance: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: iii) Select and apply the appropriate equation: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 2  For the case of air, repeat the calculations with h ¼ 20W m2 C. Exercise 10.7 Solve Example 10.1 using the spreadsheet program Heat Transfer-Negligible Surface Resistance.xls. Also solve the problem for the case in which the thickness of the steak is 3 cm and for the case in which the steak is heated from both sides. Solution Lx x Exercises 131 Step 1 On the spreadsheet Heat Transfer-Negligible Surface Resistance.xls, go to the sheet ‘‘slab. ’’Turn the ‘‘SWITCH’’ OFF (set the value in cell G1 equal to 0 and press ENTER). Step 2 Insert the parameter values in the yellow cells using: half thickness in the x-direction Lx = 0.02 m (since the steak is heated from one side, Lx is equal to the thickness); half thickness in the y-direction and z-direction Ly = 1 m, Lz = 1 m (since the steak is considered as an infinite plate, use big numbers on Ly and Lz to simulate the fact that heat transfer is significant only in the x-direction); distance of the point from the coldest plane in the x-direction x ¼ 0:02  0:011 ¼ 0:009m; distance in the z and y directions y = 0 and z = 0 (since the steak is considered as an infinite plate); initial temperature To ¼ 5 C; surface temperature Te ¼ 120 C. Also insert the given values for the thermal diffusivity. Step 3 Set the value of the time step to 15  60 ¼ 900 s in cell F4. Step 4 Turn the ‘‘SWITCH’’ ON (set the value in cell G1 equal to 1 and press ENTER). Step 5 Read the value for the temperature of the point that lies 9 mm below the surface in cell K20 (green cell). Step 6 Read the value for the mean temperature in cell K40 (green cell). Step 7 To see the effect of steak thickness, set Lx = 0.03. Repeat steps 2 to 6. Step 8 To see how the temperature of the point changes with time: a) Turn the ‘‘SWITCH’’ OFF (set the value in cell G1 equal to 0 and press ENTER). b) Set the value of the time step in cell F4 equal to 10 or some other small value. c) Turn the ‘‘SWITCH’’ ON (set the value in cell G1 equal to 1 and press ENTER). d) Iterate by pressing F9 until the temperature in cell K20 reaches the value you want. 132 10 Unsteady State Heat Transfer Step 9 To see how much faster the temperature at the same point would have increased if the steak were heated from both sides, i.e., between the hot plates of a toaster: x Lx a) Turn the ‘‘SWITCH’’ OFF (set the value in cell G1 equal to 0 and press ENTER). b) Set the value of half thickness in the x-direction to Lx = 0.01 m. c) Set the value of the coordinates of the point to x = 0.001 m, y = 0, z = 0. d) Set the value of the time step equal to 10. e) Turn the ‘‘SWITCH’’ ON (set the value in cell G1 equal to 1 and press ENTER). f) Iterate by pressing F9 until the same temperatures as in steps 5 and 6 are reached. g) Read the time in cell F3. Exercise 10.8 Solve Example 10.2 using the spreadsheet program Heat Transfer-Negligible Surface Resistance.xls. Solution r D Lx x Step 1 On the spreadsheet Heat Transfer-Negligible Surface Resistance.xls, go to the sheet ‘‘cylinder.’’ Turn the ‘‘SWITCH’’ OFF (set the value in cell G1 equal to 0 and press ENTER). Step 2 Insert the parameter values in the yellow cells using: diameter of the cylinder D = 0.015 m; half of the height of the cylinder Lx = 0.08 m; distance of the point from the center line in the direction of radius r = 0.0045 m Exercises 133 (point 3 mm from the surface); distance of the point from the center of the cylinder in the x direction x = 0; initial temperature To ¼ 5 C; surface temperature Te ¼ 100 C. Also insert the given values for the physical properties and the heat transfer coefficient. Step 3 Read the Biot number for the r and x directions in cells B42 and C42 Step 4 Set the value of the time step to 120 in cell F4. Step 5 Turn the ‘‘SWITCH’’ ON (set the value in cell G1 equal to 1 and press ENTER). Step 6 Read the value for the temperature of the point that lies 3 mm below the surface in cell K20 (green cell). Step 7 Turn the ‘‘SWITCH’’ OFF (set the value in cell G1 equal to 0 and press ENTER). Step 8 Set the value of r = 0 and x = 0 (coordinates of the center of the cylinder). Step 9 Turn the ‘‘SWITCH’’ ON (set the value in cell G1 equal to 1 and press ENTER). Step 10 Read the value for the temperature of the center of the hot dog in cell K20 (green cell). Step 11 Turn the ‘‘SWITCH’’ OFF (set the value in cell G1 equal to 0 and press ENTER). Step 12 Set the value of the time step to 0.5 in cell F4. Step 13 Turn the ‘‘SWITCH’’ ON (set the value in cell G1 equal to 1 and press ENTER). Step 14 Iterate by pressing F9 until the temperature in cell K20 is equal to 81oC. Read the value of the time in cell F3. Step 15 Turn the ‘‘SWITCH’’ OFF (set the value in cell G1 equal to 0 and press ENTER). 134 10 Unsteady State Heat Transfer Step 16 Set the value of the time step to 120 in cell F4. Step 17 Turn the ‘‘SWITCH’’ ON (set the value in cell G1 equal to 1 and press ENTER). Step 18 Read the value for the mean temperature in cell K40 (green cell). Step 19 Read the value for the heat transferred in cell K43. Exercise 10.9 Solve Example 10.3 using the spreadsheet Heat  Transfer-Internal and External Resistance.xls . Run the program for h ¼ 50W m2 8C and h ¼ 200W m2 8C to see the effect of the heat transfer coefficient on the time-temperature relationship. Solution y z Lz Ly x Lx Step 1 On the spreadsheet Heat Transfer-Internal and External Resistance.xls go to the sheet ‘‘slab’’. Turn the ‘‘SWITCH’’ OFF (set the value in cell G1 equal to 0 and press ENTER). Step 2 Insert the parameter values in the yellow cells using: Lx = 0.006 m; Ly = 0.008 m; Lz = 0.007 m; distance of the point from the geometric center x = 0, y = 0, z= 0; initial temperature To = 208C; temperature of the environment Te = 1008C. Exercises 135 Also insert the given values for the density, the heat capacity, the thermal conductivity, and the heat transfer coefficient. Step 3 Read the Biot numbers for the x, y, and z direction in cells B37, C37 and D37. Are they between 0.1 and 40? Step 4 Set the value of the time step to 5  60 ¼ 300 in cell F4. Step 5 Turn the ‘‘SWITCH’’ ON (set the value in cell G1 equal to 1 and press ENTER). Step 6 Read the temperature for the center in cell K15 (green cell). Step 7 Read the mass average temperature in cell K30 (green cell). Step 8 To see how the temperature at the selected point changes with time turn the ‘‘SWITCH’’ OFF (set the value in cell G1 equal to 0 and press ENTER). Step 9 Set the value of the time step equal to 10 (or some other small value) in cell F4. Step 10 Turn the ‘‘SWITCH’’ ON (set the value in cell G1 equal to 1 and press ENTER). Step 11 Iterate (by pressing F9) until the value of time in cell F3 (grey cell) reaches 300. Step 12 See the plot for the center temperature and mass average temperature vs. time and the dimensionless temperature in the x-direction vs. Fourier number for the x-direction in the diagrams. Step 13 Change the value of h and the corresponding values of di. Run the program and see the temperature at the center after 5 min. Step 14 Make the necessary changes in the spreadsheet to plot the dimensionless nonaccomplished temperature change for the y and z directions. Exercise 10.10 For the removal of field heat from certain fruits, hydrocooling, forced air cooling and room cooling have been proposed. Calculate and plot the temperature change at the center of the fruit vs. the time for the three methods. Assume 136 10 Unsteady State Heat Transfer that the fruit is spherical; its initial temperature is 25oC; the cold water and the air temperature are 1oC; the diameter of the fruit is 6 cm;the heat transfer coefficient is 1000W m2 C in the case of hydrocooling, 20W m2 8C in the case of forced air cooling, and 5W m2 C in the case of room cooling; and the physical properties of the fruit are: k ¼ 0:4W=m 8C, r ¼ 900kg m3 , cp ¼ 3:5kJ=kg 8C. Solution D r Step 1 Calculate the Biot number in all cases. Step 2 Select the spreadsheet Heat Transfer-Negligible Surface Resistance.xls or Heat Transfer- Internal and External Resistance.xls depending on the value of Biot number. Step 3 Follow the instructions and run the program for a sphere. Step 4 Run the program using a small time increment, e.g., 10 s. Exercise 10.11 The product development department in a food plant is interested in knowing what shape they should give to a new product for a faster temperature change at the center of the product from 20oC to 90oC. Should the shape be a cube 50mm  50mm  50mm, a cylinder with diameter 62.035 mm and length 41.36 mm, or a sphere with diameter 62.035 mm? All these shapes will have Exercises 137 the same volume and the same mass. The heating is steam at 100oC; the  medium 2 heat transfer coefficient in all cases is 2000W m C; and the density, the heat capacity, and the thermal conductivity of the product are 900kg m3 , 3200J=kg C, and 0:5W=m C respectively. Determine which shape will give a faster temperature change at the center. Solution Step 1 Calculate the Biot number in all cases. Step 2 Select the spreadsheet Heat Transfer-Negligible Surface Resistance.xls or Heat Transfer- Internal and External Resistance.xls depending on the value of the Biot number. Step 3 Follow the instructions and run the program for a cube, a cylinder, and a sphere. Step 4 Find the time necessary in each case for Tlocal to reach 90oC. Exercise 10.12 Understand how the spreadsheet program Heat Transfer-Negligible Surface Resistance.xls calculates and plots the temperature change of a sphere with time. Modify the program to plot the temperature distribution with the radius of the sphere for a certain time. Exercise 10.13 Understand how the spreadsheet program Heat Transfer-Internal and External Resistance.xls calculates and plots the  temperature change of a cylinder with time. Modify the program to plot qt qe vs:Fo0:5 for a certain Biot number value for an infinite cylinder. Exercise 10.14 a) Solve Exercise 10.4 using the spreadsheet program Semi-Infinite body 1.xls. b) Plot the temperature profile for the depth 0-5 cm after 12 h and 24 h. c) What is the minimum depth at which the temperature does not change after 12 h? d) Observe how the temperature changes with time at the depth of 1 cm, 3 cm, and 5 cm for the first 24 hours. http://avibert.blogspot.com 138 10 Unsteady State Heat Transfer Solution Step 1 In the sheet ‘‘T vs. x’’, follow the instructions, run the program for Time ¼ 24  3600 ¼ 86400 s and read the temperature in cell F16 when the distance in cell F3 is equal to 0.05 m. Observe the temperature profile in the plot. Step 2 Set the time equal to 12  3600 ¼ 43200 s in cell B29, follow the instructions, run the program, and observe the temperature profile in the plot. Step 3 Continue running the program until the temperature in cell F16 is equal to 12oC. Read the depth in cell F3. Step 4 In the sheet ‘‘T vs. t’’, follow the instructions, run the program for Distance x = 0.01 m until Time = 86400 s in cell F3. Observe the temperature profile in the plot. Rerun the program for x = 0.03 m and x = 0.05 m following the instructions. Exercise 10.15 Use the spreadsheet program Semi-Infinite body 2.xls, to study how the temperature in a semi-infinite body changes with time when a) the temperature at the surface is constant, and b) the temperaturepatffiffiffiffiffithe  surface is not constant but is affected by convection. Find the value of h t k for which the solution for both cases is approximately the same. Solution Step 1 In the sheet ‘‘Convection at thepsurface’’, follow the instructions and run the ffiffiffiffiffi program for various values of h t k in cell B41. Step 2 Read the temperature in cell H16. Step 3 For the case in which the temperature at the surface is constant, run the sheet ‘‘Constant surface temperature.’’ Step 4 Observe the temperature change in the Temperature vs. Time Observe the  pffiffiffiffiffiplot. difference between the two solutions for low values of h t k. Which is the correct one? Step 5 TTo vs. pxffiffiffiffiffi . Observe the plot of accomplished temperature change T e To 2 t Compare it with the respective plot in your textbook, e.g., Ref 3, 4, 5. Exercises 139 Exercise 10.16 Using the spreadsheet program Semi-Infinite body 3.xls, calculate the time necessary for the thermal penetration depth to reach the non-heated surface of the steak of Example 10.1. If the temperature of the hotplate changes to 150oC, does the time necessary for the thermal penetration depth to reach the non-heated surface of the steak change? Is the temperature at x = 0.02 m the same as in the case of 120oC at the time the thermal penetration depth is equal to the thickness of the steak? Solution Step 1 In the sheet ‘‘Calculations,’’ follow the instructions and insert the parameter  values. For the heat transfer coefficient use a high value, e.g., 3000W m2 C in cell B39 (assume a negligible conduct resistance). Step 2 Run the program until Time in cell F4 remains constant. Read the required time in cell G16. Step 3 Observe the temperature change in the Temperature vs. Distance plot. Step 4 Use a hotplate temperature in cell B31 of 150oC. Rerun the program until Time in cell G16 remains constant. Read the required time in cell G16. Step 5 Read the temperature in cell G18 when x = 0.02 in cell G17. Compare this value with the corresponding temperature when the hot plate temperature is 120oC. Exercise 10.17 In the ‘‘hot-break’’ process, whole tomatoes are heated with live steam to inactivate the pectin methylesterase enzyme before crushing and extraction, thereby yielding high viscosity products. Calculate the time necessary to heat the center of a tomato 6 cm in diameter to 50oC using 100oC steam if the initial temperature of the tomato is 20oC and the density, the  heat capacity, and the thermal conductivity for tomatoes are 1130kg m3 , 3:98kJ=kg C, and 0:5W=m C respectively. Find the time necessary for the mass average temperature to reach a value of 82 oC. Solution Use the spreadsheet Heat Transfer-Negligible Surface Resistance.xls to solve the problem. Write the necessary assumptions. 140 10 Unsteady State Heat Transfer Exercise 10.18 A potato 5 cm in diameter is immersed in boiling water. Calculate the required time for the temperature at the center of the potato to reach 59oC, the onset of starch gelatinization, if the initial temperature of the potato is 21oC and the density, the heat capacity and the thermal conductivity for potatoes are 1120kg m3 , 3:7kJ=kg 8C and 0:55W m 8C respectively. Find the mass average temperature of the potato when the center temperature is 59oC. Study the effect of diameter on the required time. Is it proportional to D or to D2? Solution Use the spreadsheet Heat Transfer-Negligible Surface Resistance.xls to solve the problem. Write the necessary assumptions. Chapter 11 Mass Transfer by Diffusion http://avibert.blogspot.com Theory As with heat transfer, the rate of the transferred quantity in mass transfer is proportional to the driving force and inversely proportional to the resistance. For mass transfer by diffusion, the driving force is the chemical potential difference; but concentration (mol/m3), mole fraction, or pressure differences are usually used. Thus, the mass transfer rate is: ni ¼ driving force
ci ¼ resistance R For gases: Driving force Equimolar Counterdiffusion
pi Diffusion of i through stagnant j
pi Resistance, R
z RG T Di j A
z pj M RG T Di j A P For liquids: Driving force Resistance, R Equimolar Counterdiffusion
xi
z 1 Di j A cm Diffusion of i through stagnant j
xi
z xj M Di j A c m S. Yanniotis, Solving Problems in Food Engineering. Ó Springer 2008 141 142 11 Mass Transfer by Diffusion For solids: Mass transfer rate
ci ni ¼ R
ci ni ¼ P R For a single wall For a composite wall Resistance, R Flat wall Cylindrical wall Spherical wall
z DA
r DALM
r DAG Surface area, A A A1 A2 ln A1 =A2 pffiffiffiffiffiffiffiffiffiffiffiffi AG ¼ A1 A2 ALM ¼ Flat packaging films: Resistance, R For gases For water vapor
z PM A
z PM WV  A 22414 Surface area, A A A where A = surface area perpendicular to the direction of transfer, m2 ALM = logarithmic mean of surface area A1 and A2, m2 AG = geometric mean of surface area A1 and A2, m2 ci = concentration of i, mol/m3 cm = mean concentration of i and j, mol/m3 D = diffusion coefficient, m2/s Dij = diffusion coefficient of i in j, m2/s ni= mass transfer flux, rate, mol/s P = total pressure, Pa pi = partial pressure of i, Pa pjM = logarithmic mean pressure difference of j, Pa PM = gas permeability, cm3 cm/s cm2 atm PM WV = water vapor permeability, g cm/s cm2 Pa RG = ideal gas constant, m3 Pa/mol K T = temperature, K Review Questions 143 xi = mol fraction of i xjM = logarithmic mean mol fraction of j
z and
r = wall thickness, m Review Questions Which of the following statements are true and which are false? 1. Fick’s 1st law refers to mass transfer by diffusion at steady state. 2. Molecular diffusion is a phenomenon analogous to heat transfer by conduction. 3. The resistance to mass transfer increases linearly with diffusivity. 4. Mass diffusivity has the same units as thermal diffusivity and kinematic viscosity. 5. Diffusivity in gasses is about 10000 times higher than in liquids. 6. Diffusivity does not vary with temperature. 7. Diffusivity in solids may vary with concentration. 8. The driving force for mass transfer by molecular diffusion is the difference in chemical potential. 9. There is bulk flow in equimolar counterdiffusion. 10. In equimolar counterdiffusion Ni=Nj, where Ni and Nj are the fluxes of gas i and gas j with respect to a fixed position. 11. There is no bulk flow in the case of diffusion of gas i through stagnant nondiffusing gas j. 12. For the same driving force, the flux Ni in equimolar counterdiffusion is smaller than Ni in diffusion of i through stagnant nondiffusing j due to the bulk motion of i. 13. When concentrations are dilute, the bulk flow may be negligible. 14. Permeability refers to the diffusion of a gas in a solid and is used extensively in calculating mass transfer in packaging materials. 15. Permeability is equal to the product of the diffusion coefficient and the solubility of the gas in the solid. 16. The difference in partial pressure inside and outside the packaging material is being used as the driving force for mass transfer calculations. 17. Permeability decreases as the temperature increases. 18. Layers of different materials may be combined in laminates to give a composite material with good barrier properties for water vapor, gasses, and light. 19. Polyethylene is a good water vapor barrier and serves as an adhesive to the next layer. 20. Aluminum foil is a good gas barrier 144 11 Mass Transfer by Diffusion Examples Example 11.1 Water vapor is diffusing through a stagnant film of air at 308C towards a cold surface at 68C, where it is condensed. Calculate the water vapor diffusion flux if the pressure is 1 atm, the water vapor pressure 10 mm from the cold surface is 3 kPa, and the water vapor diffusion coefficient in the air is 0.26 cm2/s. Solution Step 1 Draw the process diagram: P =1 atm Z =10 mm pA1 = 3 kPa pB1 pA2 pB2 6°C air molecule water molecule Step 2 State your assumptions: l l The system is at steady state. There are no eddies. Step 3 Select the appropriate equation to calculate the water vapor diffusion flux. Water molecules diffuse towards the cold plate; air molecules diffuse in the opposite direction. There is a bulk movement towards the cold plate to keep the system at constant pressure. Therefore the equation to use is: Nw ¼ nw pw1 pw2 pw1 pw2 Daw P ¼ ¼ ðp p Þ ¼ p
z a M RG T
z pa M w1 w2 A RA Daw A P RG T A (with subscript ‘‘w’’ for water vapor and ‘‘a’’ for air). Examples 145 Step 4 Find the values of partial pressure to use in the above equation: i) The water vapor partial pressure at the interphase at 68C is pw2 ¼ 0:935 kPaðfrom steam tablesÞ: ii) The partial pressure of air is pa1 ¼ 101325  3000 ¼ 98325 Pa and pa2 ¼ 101325  935 ¼ 100390 Pa pa M ¼ pa1 pa2 98325  100390 ¼ 99353:9 Pa p ¼ 98325 ln a1 ln pa2 100390 Comment: The arithmetic mean instead of the log mean could have been used with very little error, since pa1 and pa2 values differ from one another by a small percentage. Step 5 Substitute values and calculate the water vapor diffusion flux: Daw P ðp  pw2 Þ¼ R T z paM w1   0:26x104 m2 =s ð101325 PaÞ ð3000  935ÞPa ¼ ¼ ð8314:34 m3 Pa=kmol KÞ ð303 KÞ ð0:01 mÞ ð99353:9 PaÞ Nw ¼ ¼ 2:17x106 kmol=s m2 Example 11.2 A food product is sealed in a flexible laminated package with 0.1 m2 surface area that is made of a polyethylene film layer 0.1 mm thick and a polyamide film layer 0.1 mm thick. The package is stored at 218C and 75% relative humidity. Calculate the transfer rate of oxygen and water 146 11 Mass Transfer by Diffusion vapor through the film at steady state if the partial pressure of O2 inside the package is 0.01 atm, that outside the package is 0.21 atm, and the water activity of the product inside the package is 0.3. The permeability (PM) of polyethylene and polyamide to O2 are 22801011 and 51011 cm3/ (s cm2 atm /cm) respectively, and the water vapor transmission rate (WVTR) for these materials measured at 37.88C using 90% RH water vapor source and 0% RH desiccant sink are 61011 and 371011 g/(s cm2 /cm) respectively. Solution Step 1 State your assumptions: l l The convective resistance to mass transfer on the two sides of the package are negligible compared to the diffusion resistance of the film. The WVTR at 218C does not differ appreciably from that at 37.88C. Step 2 Calculate the diffusion rate of oxygen: i) Select the equation to use:
p nO2 ¼ P O2 R ii) Substitute values and calculate the diffusion rate: X R ¼ R polyethylene þ R polyamid ¼   22414
z polyeth: 22414
z polyam: 22414
z polyeth:
z polyam: þ ¼ þ ¼ A A PM polyeth: A PM polyam: PM polyeth: PM polyam:   22414 cm3 =mol ð0:01 cmÞ   ¼ 11 ð1000 cm2 Þ 2280  10 cm3 cm=s cm2 atm ! ð0:01 cmÞ  ¼ þ 5  1011 cm3 cm=s cm2 atm ¼ ¼ 9:83  106 þ 4:48  109 ¼ 4:49  109 atm s=mol
p 0:21  0:01 atm n O2 ¼ P O2 ¼ ¼ 4:45  1011 mol=s R 4:49  109 atm s=mol Examples 147 Step 3 Calculate the diffusion rate of water vapor. i) Select the equation to use:
p nw ¼ P w R with pw the water vapor difference between the inside and the outside of the package and R¼
z PM WV A ii) Find the water vapor pressure inside and outside the package. The water vapor pressure pw at 218C is 2487 Pa (from steam tables). The water vapor partial pressure outside the package for 75% relative humidity is: pw o ¼ 75 2487 ¼ 1865:3 Pa 100 The water vapor partial pressure inside the package for water activity 0.3 is: pw i ¼ 0:3 x 2487 ¼ 746:1 Pa iii) Calculate the water vapor permeability. Water vapor permeability for polyethylene can be calculated from the Water Vapor Transmission Rate as: WVTRpolyeth: 6  1011 g cm=s cm2 ¼
p 0:90  6586  0 Pa ¼ 1:01  1014 g cm=s cm2 Pa PM WV polyethylene ¼ Similarly for polyamide: WVTRpolyeth: 37  1011 g cm=s cm2 ¼
P 0:90  6586  0 Pa ¼ 6:24  1014 g cm=s cm2 Pa PM WV polyamid ¼ 148 11 Mass Transfer by Diffusion iv) Calculate the total resistance to water vapor transfer for the laminate: X R ¼ Rpolyet h y l e ne þ Rpolyamid ¼ ¼ 
zpolyeth: PM WV polyeth: A þ
zpolyamid ¼ PM WV polyamid A 0:01 cm  1:01  1014 g cm=s cm2 Pa ð1000 cm2 Þ þ  0:01 cm  ¼ 6:24  1014 g cm=s cm2 Pa ð1000 cm2 Þ ¼ 9:90  108 þ 1:60  108 ¼ 1:15  109 Pa s=g v) Calculate the water vapor transfer rate:
p 1865:3  746:1 Pa ¼ 9:73  107 g=s nw ¼ P w ¼ R 1:15  109 Pa s =g Comment: Notice that the polyethylene film layer contributes the main resistance to water vapor transfer (86.1%), while polyamide contributes the main resistance to oxygen transfer (99.8%). Exercises Exercise 11.1 Water evaporates from the flat surface of the leaf of a vegetable and is diffusing away through a stagnant layer of air. The total pressure is 101325 Pa and the temperature is 248C. Calculate the evaporation rate in g/s under the following conditions: water activity (aw) of the leaf surface is 0.98, partial water vapor pressure 5 mm away from the surface of the leaf is 2100 Pa, surface area of the leaf is 50 cm2. The diffusion coefficient of water vapor in air is 2.6105 m2/s. Solution Step 1 State your assumptions: The time interval the calculations are based on is small so that the water vapor pressure at the surface of the leaf is constant and the system is at steady state. Exercises 149 Step 2 Select the equation to use: Since water vapor diffuses through a stagnant layer of air, the equation to be used is: Nw ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::: Step 3 Find the water vapor pressure pw at 248C from the steam tables: .................... Calculate the partial water vapor pressure pw1 at the surface of the leaf as: pw1 ¼ aw : pw ¼ ::::::::::::::::::::::::::::::::::::: Step 4 Calculate the partial pressure of air: pa1 ¼ P  pw1 ¼ ::::::::::::::::::::::::::::::::::::::::::::::: pa2 ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: paM ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 5 Calculate the mass transfer flux: Nw ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::kmol=s m2 Step 6 Calculate the evaporation rate: nw ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::g=s Exercise 11.2 An aroma compound is encapsulated in a nonporous spherical particle made of a homogenous biopolymer film. Calculate the aroma release rate if the particle diameter is 1 mm, the film thickness is 0.1 mm, the concentration of the aroma compound is 0.1 g/cm3 and 0.01g/cm3 on the inside and outside surface of the particle respectively, and the diffusion coefficient of the aroma compound in the film is 11012 m2/s. 150 11 Mass Transfer by Diffusion Solution Step 1 Draw the process diagram: r1 C2 C1 r2 A1 A2 Step 2 State your assumptions: l l The aroma release rate is diffusion-controlled. The release rate is constant (steady state). Step 3 Calculate the resistance to mass transfer in a spherical wall. i) Calculate the geometric mean area of A1 and A2: A1 ¼ pD2 ¼ ::::::::::::::::::::::::::::::::::::::::::::::::: A2 ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: AG ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ii) Calculate the resistance: R¼
r ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: DAG Step 4 Calculate the diffusion rate through the wall of the sphere: n ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Exercises 151 Exercise 11.3 The solubility and the diffusivity of O2 in polyvinyl chloride are 3.84x104 cm3 O2 at STP/cmHg.cm3 solid and 1.18x1012 m2/s respectively. Calculate the permeability of polyvinyl chloride to O2. Solution The permeability is the product of solubility and diffusivity. Therefore: PM ¼ S:D ¼:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: cm3 cm s cm2 Pa Exercise 11.4 A supplier of frozen hamburger patties is thinking of packaging them in 0.1 mm thick polypropylene film. The permeability of polypropylene to O2 and water vapor is 1x108 cm3/(s cm2 atm /cm) and 8x1015 g/(s cm2 Pa/cm) respectively. The partial pressure of O2 is 0 atm and that of water vapor is 0.12 Pa inside the package, while outside the package the partial pressure of O2 is 0.21 atm and that of water vapor is 0.04 Pa. Calculate how much oxygen and how much water vapor will pass through the polypropylene film of 600 cm2 surface area in one week. Solution Step 1 Sate your assumptions: l l The convective resistance to mass transfer on the two sides of the package is negligible compared to the diffusion resistance of the film. .............................................................................................................. Step 2 Calculate the transfer rate of O2. i) Calculate the resistance: R ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ii) Calculate the transfer rate of oxygen: nO2 ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::mol=s Step 3 Calculate the total amount of O2 that will diffuse into the package in one week: mO2 ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: g 152 11 Mass Transfer by Diffusion Step 4 Calculate the transfer rate of water vapor. i) Calculate the resistance: R ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ii) Calculate the transfer rate of water vapor: nw ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::g=s Step 5 Calculate the total amount of water vapor that will diffuse out of the package in one week: mw ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: g Exercise 11.5 Meat is packaged in a plastic disk covered with PVC film. Calculate the required thickness of the film so that 1104 mol of oxygen enters the package through the film in 24 hours. The permeability of PVC to oxygen is 4561011 cm3/ (s cm2 atm /cm), the surface area of the film is 300 cm2, and the partial pressure of oxygen inside and outside the package is 0 atm and 0.21 atm respectively. Solution Step 1 State your assumptions: .......................................................................................................................... Step 2 Calculate the mass transfer rate in mol/s: nO2 ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 3 Calculate the required resistance: R ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 4 Calculate the required thickness of the film
z ¼ PM A R ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 22414 Exercises 153 Exercise 11.6 Potato chips are packed in a plastic bag in a nitrogen atmosphere. Calculate the amount of N2 that will diffuse out of the plastic bag in 3 months if the thickness of the film is 0.05 mm, the absolute pressure of N2 in the bag is 1.05 atm, the surface area of the bag is 1000 cm2, and the permeability of the plastic film to N2 is 1.51010 cm3/s cm2 atm/cm. Solution Step 1 State your assumptions: .......................................................................................................................... Step 2 Calculate the resistance: R ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 3 Calculate the transfer rate of N2: nN2 ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::mol=s Step 4 Calculate the total amount of N2 that will diffuse out of the package in three months: mN2 ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: g Verify that the pressure in the bag will not change appreciably due to N2 diffusion. Hint: Use the ideal gas law. Chapter 12 Mass Transfer by Convection http://avibert.blogspot.com Theory For mass transfer by convection, the driving force is the concentration difference
c expressed in mol/m3 or mol fraction or pressure, and the resistance R is equal to 1=kc A, where kc mass transfer coefficient and A surface area (m2) perpendicular to the direction of transfer. The mass transfer coefficient is calculated from relations of the form: Sh ¼ fðRe; ScÞ where kc L Dij Lvr Re = Reynolds number, Re ¼ m m Sc = Schmidt number, Sc ¼ rDij Sh = Sherwood number, Sh ¼ Dij = diffusion coefficient of component i in component j, m2/s kc = mass transfer coefficient, m/s or mol/s m2 mol fraction or mol/s m2 Pa L = characteristic dimension, m v = fluid velocity, m/s m = fluid viscosity, Pa s r = fluid density, kg/m3 To calculate the mass transfer coefficient: 1. Determine if the flow is natural or forced (free or forced convection). 2. Identify the geometry of the system. 3. Determine if the flow is laminar or turbulent (calculate the Reynolds number). S. Yanniotis, Solving Problems in Food Engineering. Ó Springer 2008 155 156 12 Mass Transfer by Convection 4. Select the appropriate relationship Sh ¼ fðRe; ScÞ. 5. Calculate Sh and solve for kc. Review Questions Which of the following statements are true and which are false? 1. Mass transfer by convection is analogous to heat transfer by conduction. 2. The mass transfer coefficient may be expressed in different units which depend on the units that are used in the driving force. 3. The mass transfer coefficient depends on the physical properties of the fluid, the flow regime, and the geometry of the system. 4. The resistance to mass transfer by convection is proportional to the mass transfer coefficient. 5. A concentration boundary layer develops on a fluid flowing on a solid surface when the concentration of species ‘‘i’’ in the fluid is different from its concentration on the solid surface. 6. Concentration gradients exist in the concentration boundary layer. 7. The Schmidt number represents the ratio of mass diffusivity to momentum diffusivity. 8. The Lewis number represents the ratio of thermal diffusivity to mass diffusivity. 9. The Schmidt number relates the thickness of the hydrodynamic boundary layer to the thickness of the concentration boundary layer. 10. For dilute solutions the mass transfer coefficient for equimolar counterdiffusion is equal to the mass transfer coefficient for i through stagnant j. Examples Example 12.1 In a fruit packaging house, oranges are washed on a perforated belt conveyor by water sprays and dried in a stream of high speed air at room temperature before waxing. Calculate the mass transfer coefficient on the surface of an orange at 208C if the air velocity is 10 m/s, the orange has a spherical shape with a diameter of 7 cm and the diffusivity of water vapor in air is 2.5  10-5 m2/s. Examples 157 Solution Step 1 Draw the process diagram: 10 m/s 20°C Step 2 State your assumptions: l l The moisture in the air is low so that the physical properties of air can be used: The oranges are apart from each other so that air flows around each one. Step 3 Find the physical properties of air at 208C: r ¼ 1:207g=m3 m ¼ 1:82  105 kg=ms Step 4 Calculate the Reynolds number:   3 Dvr ð0:07 mÞ ð10 m=sÞ 1:207 kg=m Re ¼ ¼ ¼ 46423 m 1:82  105 kg=ms Step 5 Calculate the Schmidt number: Sc ¼ m 1:82  105 kg=ms  ¼
¼ 0:602 rDij 1:207 kg=m3 2:5  105 m2 =s Step 6 Select a suitable equation of Sh ¼ fðRe; ScÞ for flow around a single sphere: 158 12 Mass Transfer by Convection Sh ¼ 2 þ 0:552 Re0:53 Sc1=3 Step 7 Calculate the Sherwood number: Sh ¼ kc D ¼ 2 þ 0:552 ð46423Þ0:53 ð0:602Þ1=3 ¼ 140:6 Dij Step 8 Calculate kc from the Sherwood number: kc ¼ Sh Dij 2:5  105 m2 =s ¼ 0:050 m=s ¼ 140:6 0:07 m D Exercises Exercise 12.1 Instant coffee is dried in a spray dryer. Calculate the mass transfer coefficient on the surface of a coffee droplet that falls through the air in the spray dryer at the initial stages of drying if the diameter of the droplet is 0.5 mm, the relative velocity between the air and the droplet is 50 m/s, and the air temperature is 1808C. Assume that the droplet surface temperature is at 608C. Solution Step 1 State your assumptions: l l The moisture in the air is low so that the physical properties of air can be used. The droplet has a spherical shape with constant diameter. Step 2 Find the physical properties of air at the film temperature (the average temperature between droplet surface and bulk fluid): r ¼:::::::::::::::::::::::::::::::::::::::::::::::::::::::kg=m3 m ¼:::::::::::::::::::::::::::::::::::::::::::::::::::::::kg=ms Dij ¼:::::::::::::::::::::::::::::::::::::::::::::::::::::::m2 =s Exercises 159 The diffusion coefficient of water vapor in air at 258C and 1 atm is 2.6x105m2/s. Estimate the diffusion coefficient at 1208C using the relation Dij / T1: 5 where T is degrees Kelvin. Step 3 Calculate the Reynolds number: Re ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 4 Calculate the Schmidt number: Sc ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 5 Select a suitable equation of Sh ¼ fðRe; ScÞ for flow around a single sphere: Sh ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 6 Calculate the Sherwood number: Sh ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 7 Calculate kc from the Sherwood number: kc ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Exercise 12.2 Calculate the mass transfer coefficient for oxygen transfer from an air bubble rising in a non-agitated fermentation broth if the bubble diameter is 0.5 mm, the diffusion coefficient of O2 in the broth is 2x109 m2/s, the density and the viscosity of the broth are 1200 kg/m3 and 0.002 kg/m s, and the density of the gas is 1.1 kg/m3. Solution Step 1 State your assumptions:The bubble has a spherical shape with constant diameter. Step 2 Select a suitable equation for mass transfer to small bubbles: Sh ¼ 2 þ 0:31 Gr1=3 Sc1=3 160 12 Mass Transfer by Convection Step 3 Calculate the Grashof number: Gr ¼ D3b r Dr g ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::: m2 Step 4 Calculate the Schmidt number: Sc ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 5 Calculate the Sherwood number: Sh ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 6 Calculate kc from the Sherwood number: kc ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Exercise 12.3 Several correlations are available in the literature for calculating mass transfer coefficients for different geometries. Three such correlations for flow past single spheres are: Sh ¼ 2 þ 0:60Re1=2 Sc1=2 Sh ¼ 2 þ 0:552Re053 Sc1=2   Sh ¼ 2þ 0:40Re1=2 þ 0:06Re2=3 Sc0:4 Re570000 15Re548000 3:55Re57600 Develop a spreadsheet program to evaluate and plot Sh vs. Re from these correlations for 15Re5100 and for Sc ¼ 0:6 and Sc ¼ 2. Repeat the calculation for 1005Re 50000. Compare the results. Exercise 12.4 Air flows over the surface of a solid tray 0.50 m x 0. 50 m filled with a vegetable cut into small cubes that completely cover the tray. Calculate the Exercises 161 rate of water vapor transfer from the surface of the tray to the air stream if the mass transfer coefficient is 0.03 kg/s m2 Pa. The water vapor pressure is 12000 Pa at the surface of the vegetable and 5000 Pa at the bulk of the air stream. Solution Step 1 Draw the process diagram: pa = 5000 Pa nw kG ps = 12000 Pa Step 2 State your assumptions: The vegetable surface is covered with unbound moisture. Step 3 Calculate the mass transfer rate: nw ¼ kG Aðps  pa Þ ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::: Exercise 12.5 Air passes through a fixed bed of corn of 6 m3 where the mass transfer coefficient is 0.05 kg/s m2 Pa and the surface area per unit volume of the bed is 60 m2/m3. The water vapor pressure at the surface of the corn is 5000 Pa at the bottom of the bed and 5500 Pa at the top. Water vapor pressure for the bulk air is 1800 Pa at the inlet and 2500 Pa at the outlet of the bed. Calculate the rate of water vapor transfer from the corn to the air. 162 12 Mass Transfer by Convection Solution Step 1 Draw the process diagram: Pa2 Pa1 Step 2 State your assumptions:The calculations will be based on steady state conditions. Step 3 Select the equation to use: n ¼ kG A
P i) Calculate the mean driving force Palong the bed. The logarithmic mean driving force should be used in this case:
P ¼ ðps1  pa1 Þ  ðps2  pa2 Þ ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::: p  pa1 ln s1 ps2  pa2 ii) Calculate the total external surface area of the solid particles: A ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::: iii) Calculate the mass transfer rate: n ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::: Chapter 13 Unsteady State Mass Transfer http://avibert.blogspot.com Theory The unidirectional unsteady state mass transfer equation in the x-direction is given by Fick’s 2nd law: @C @2C ¼ Dij 2 @t @x The solutions of the unsteady state mass transfer equation are the same with the solutions of the unsteady state heat transfer equation given in Tables 10.1, 10.2, and 10.4, with the following correspondence between heat and mass transfer variables: where Bim = Biot number for mass transfer CFi = species concentration in the fluid at the interface, mol/m3, kg/m3 Table 13.1 Relations between heat and mass transfer parameters Heat transfer Mass transfer Te  T Te  To Cs  C Cs  Co Fo ¼ at2 L Fom ¼ Bi ¼ hL k Bim ¼ mDkc L ij pxffiffiffiffiffi 2 at 2 pffiffiffiffiffi h at k pffiffiffiffiffiffiffiffi m kc Dij t Dij Dij t L2 pxffiffiffiffiffiffiffiffi Dij t S. Yanniotis, Solving Problems in Food Engineering. Ó Springer 2008 163 164 13 Unsteady State Mass Transfer CFL = species concentration in the bulk of the fluid, mol/m3, kg/m3 Cs = species concentration in the medium at the surface of the solid, mol/m3, kg/m3 Co = initial concentration of the species in the solid, mol/m3, kg/m3 C = concentration of the species in the solid at time t and point x, mol/m3, kg/m3 Dij = mass diffusivity, m2/s Fom = Fourier number for mass transfer kc = mass transfer coefficient, m/s L = characteristic dimension (half thickness of the plate, radius of a cylinder or sphere), m m = equilibrium distribution coefficient x = distance of the point from the center plane of the plate, or from the surface in the case of semi-infinite body, m t = time, s To calculate the concentration as a function of time and position in a solid we follow the same procedure as for calculating the temperature as a function of time and position in a solid: 1. Identify the geometry of the system. Determine if the solid can be considered as plate, infinite cylinder, or sphere. 2. Determine if the surface concentration is constant. If not, calculate the Biot number and decide the relative importance of internal and external resistance to mass transfer. 3. Select the appropriate equation given for unsteady state heat transfer (Tables 10.1, 10.2, 10.4). 4. Calculate the Fourier number. 5. Find the concentration by applying the selected equation (if Fo > 0:2, calculate the temperature either using only the first term of the series solution or using Heisler or Gurnie-Lurie charts). Review Questions Which of the following statements are true and which are false? 1. If the concentration at any given point of a body changes with time, unsteady state mass transfer occurs. 2. Fick’s 2nd law is used in unsteady state mass transfer problems. 3. Gurnie-Lurie and Heisler charts are valid for unsteady state mass transfer problems. 4. The Heisler chart can be used to find the concentration at any point in a body if the Fourier and Biot numbers are known. 5. Gurnie-Lurie charts are used when Fo Examples 165 6. The distribution coefficient of the diffusing component between fluid and solid is used in unsteady state mass transfer problems. 7. The required time for a certain change in concentration due to diffusion is proportional to the square of the thickness of the body. 8. If only one surface of an infinite slab is exposed to the diffusing substance, half the thickness of the slab is used as characteristic dimension. 9. Diameter is used as the characteristic dimension of a sphere for the solution of unsteady state mass transfer problems. 10. The solutions of an infinite cylinder and an infinite slab are combined to find the concentration distribution in a cylinder of finite length. Examples Example 13.1 A food item 2  20  20 cm with 200 kg water/m3 moisture content is exposed to a high speed hot air stream in a dryer. The surface of the food immediately attains moisture equal to 50 kg water/m3. This surface condition is maintained  thereafter. If the diffusivity of moisture in the food is 1:3  109 m2 s, calculate: a) the moisture content at the center after 5 hours, b) how much time is required to bring the moisture content at the center to a value of 100 kg water/m3, and c) the mass average moisture content at this time. Solution Step 1 Draw the process diagram: Lx x Step 2 State your assumptions: l l l l Moisture is initaially uniform throughout the body. Moisture transfer in the solid is diffusion controlled. The volume of the solid is constant. The diffusivity is constant. Step 3 Identify the geometry of the system. The shape of the food can be considered as an infinite slab since the thickness is much smaller than the other two dimensions. 166 13 Unsteady State Mass Transfer Step 4 i) Select the appropriate equation to use. Since the surface concentration is constant, the equation is (see Tables 10.1 and 13.1). !   1 Cs  C 4X ð1Þn ð2n þ 1Þpx ð2n þ 1Þ2 p2 cos Fox ¼ exp  Cs  Co p n¼0 2n þ 1 2L 4 (13:1) ii) Calculate the Fourier number   1:3  109 m2 =s ð5  3600 sÞ Dt Fox ¼ 2 ¼ ¼ 0:234 Lx ð0:01 mÞ2 Since Fox > 0:2 the first term of the series in eqn (13.1) is enough. Alternatively, the Heisler chart can be used. Step 5 i) Find the concentration at the center using eqn (3.1) above: !   1 Cs  C 4X ð1Þn ð2n þ 1Þpx ð2n þ 1Þ2 p2 cos Fox ¼ exp  Cs  Co p n¼0 2n þ 1 2L 4 4 ð1Þ0 ð2  0 þ 1Þ  p  0 ð2  0 þ 1Þ2 p2 exp  ¼ cos  0:234 p ð 2  0 þ 1Þ 2  0:01 4 ! ¼ 0:715 and C ¼ Cs  0:715ðCs  Co Þ ¼ 50  0:715 ð50  200Þ ¼ 157 kg water=m3 ii) Find the concentration at the center using the Heisler chart for an infinite slab (Fig A.5): l l l Find the value of FO ¼ 0:234 on the x-axis. Find the curve with 1=Bi ¼ 0. Read the dimensionless concentration on the y-axis as: Cs  C ¼ 0:7 Cs  Co C ¼ Cs  0:7ðCs  Co Þ ¼ 50  0:7 ð50  200Þ ¼ 155 kg water=m3 Examples 167 (Cs-C)/(Cs-Co) 1.0 Bi-1 = 0 0.1 0.0 0.0 0.2 0.4 0.6 0.8 Fo Step 6 Find the required time for the moisture content to drop to 100 kg water/m3 at the center. The time will be found from the Heisler chart: Calculate the dimensionless concentration Cs  C 50  100 ¼ 0:33 ¼ Cs  Co 50  200 On the Heisler chart: l l l l Find on the y-axis the value ðCs  CÞ=ðCs  C0 Þ ¼ 0:33. Find the curve 1=Bi ¼ 0. Read the Fourier number on the x-axis as Fo ¼ 0:54. Find the time from the Fourier number as: L2 t ¼ Fo Dx ¼ 0:54 ij 0:012 m2 ¼ 41538 s 1:3  109 m2 =s 1.0 (Cs-C)/(Cs-Co) l Bi-1 = 0 0.1 0.0 0.0 0.2 0.4 0.6 Fo 0.8 168 13 Unsteady State Mass Transfer Step 7 Calculate the mass average moisture content. i) Find the equation for the mean concentration of an infinite slab with Bi > 40 in Table 10.1: ! 1 Cs  Cm 8X 1 ð2n þ 1Þ2 p2 Fo ¼ exp  Cs  Co p2 n¼0 ð2n þ 1Þ2 4 ii) Substitute the values in the above equation and solve for Cm (since Fo > 0:2 the first term only is enough). ! 1 Cs  Cm 8X 1 ð2n þ 1Þ2 p2 Fo ¼ ¼ exp  Cs  Co p2 n¼0 ð2n þ 1Þ2 4 8 1 ð2  0 þ 1Þ2 p2  0:54 ¼ 2 exp  p ð2  0 þ 1Þ2 4 ! ¼ 0:214 and Cm ¼ Cs  0:214ðCs  Co Þ ¼ 50  0:214 ð50  200Þ ¼ 82:1 kg water=m3 Exercises Exercise 13.1 Apple slices 4 mm thick are exposed to sulphur fumes in a sulphuring house before drying. If the Biot number for mass transfer is higher than 40 and the surface concentration of SO2 is 0.1 mol SO2/m3, calculate the mass average SO2 concentration in the slices after 1 h. Assume zero initial concentration of SO2 in the apple and a diffusion coefficient of SO2 in the apple of 1  109 m2 s. Solution SO2 Step 1 State your assumptions: Exercises l l 169 Mass transfer through the sides of the slice is negligible (the solution for an infinite slab can be applied). Only the top of the slice is exposed to sulphur fumes. Step 2 Calculate the Fourier number. Since the food is exposed to the fumes from one side the characteristic dimension is equal to the thickness of the slice. Therefore: Fo ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 3 Find the equation for the mean concentration of a plate with Bi > 40 in Table 10.1. Substitute the values in the equation and solve for Cm: Cs  Cm ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Cs  Co and Cm ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Exercise 13.2 http://avibert.blogspot.com Calculate how long it will take for the concentration on the non-exposed surface of the apple slice of Exercise 13.1 to increase by 1% of the initial concentration difference. Solution Step 1 Identify the geometry of the system. As long as the concentration change on the non-exposed surface of the apple slice is less than 1% of the initial concentration difference, the slice can be treated as a semi-infinite body. Step 2 Select the appropriate equation from Table 10.4. Since the surface concentration is constant, the equation for a semi-infinite body with uniform initial concentration and constant surface concentration will be used:   Cs  C x ¼ erf pffiffiffiffiffiffi Cs  Co 2 Dt Step 3 The mass penetration depth is given by (as in heat transfer): x pffiffiffiffiffiffi ¼ ::::::::::::::::::::::::::::::::: 2 Dt 170 13 Unsteady State Mass Transfer Step 4 Solve for t, substitute values, and calculate t: t ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Comment: This result shows that for the first ...................... the body can be treated as a semi-infinite body. Exercise 13.3 A piece of feta cheese 15cm  15cm  10cm is immersed in NaCl brine. Calculate the NaCl concentration in the center of the piece and the mass average concentration after 10 days ifthe concentration at the surface immediately attains a value of 130kgNaCl m3 and remains constant during the brining process, the cheese did not contain any salt at the  beginning, and the diffusion 10 2 coefficient of NaCl in the cheese is 3  10 m s. Solution Step 1 Draw the process diagram: y z Lz Ly x Lx Step 2 State your assumptions: l l NaCl transfer in the cheese is diffusion controlled. The volume of the cheese remains constant. Exercises 171 Step 3 Define the shape of the object. The cheese is a rectangular slab. The concentration in the center is affected by the mass transferred from the three directions x, y, and z. The contribution from each direction has to be calculated separately and the combined effect will be calculated at the end. Step 4 Select the appropriate equation to use. Since the concentration at the surface is constant, external resistance is considered negligible. Therefore, the solution for the concentration at the center will be: !   1 Cs  C 4X ð1Þn ð2n þ 1Þpx ð2n þ 1Þ2 p2 cos Fo ¼ exp  Cs  Co p n¼0 2n þ 1 2L 4 Step 5 Calculate mass transfer in x, y, and z directions. 1) Mass transferred in the x-direction: i) Calculate the Fourier number Fox for the x-direction: Fox ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Since FOx the Heisler chart cannot be used. More than one term of the sum in the above equation has to be used. Three terms will be enough for this problem. ii) Calculate the dimensionless concentration: !   1 Cs  Cx 4 X ð1Þn ð2n þ 1Þ p x ð2n þ 1Þ2 p2 cos Fox  ¼ exp  2Lx Cs  Co p n¼0 2n þ 1 4 2  3   ð 2  0 þ 1Þ p  0 ð1Þ0 ð2  0 þ 1Þ2 p2 :::: 7 exp  6 2  0 þ 1 cos 4 2  0:075 7 6 7 6 7 6 þ:::::::::::::::::::::::::::::::::::::::: 7 46 7¼ 6 ¼ 6 7 p6 7 7 6 7 6 ::::::::::::::::::::::::::::::::::::::::::::::::::: 5 4 þ:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ¼ :::::::::::::::::::::::::::::::::::::: 172 13 Unsteady State Mass Transfer 2) Mass transferred in the y-direction: i) Calculate the Fourier number Foy for the y-direction: Foy ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ii) Calculate the dimensionless concentration: !   1 Cs  Cy 4 X ð1Þn ð2n þ 1Þ p y ð2n þ 1Þ2 p2 cos Foy  ¼ exp  2Ly Cs  Co p n¼0 2n þ 1 4 3 2 ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 7 6 7 6 þ::::::::::::::::::::::::::::::::::::::::::::::::: 7 6 7 6 7 6 46 7 ¼ 6 7¼ 7 p6 ::::::::::::::::::::::::::::::::::::::::::::::::::: 7 6 7 6 6 þ:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 7 5 4 ¼ :::::::::::::::::::::::::::::::::::::: 3) Mass transferred in the z-direction: i) Calculate the Fourier number Foz for the z-direction: Foz ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ii) Calculate the dimensionless concentration. !   1 Cs  Cz 4 X ð1Þn ð2n þ 1Þ p z ð2n þ 1Þ2 p2 cos Foz  ¼ exp  2Lz Cs  Co p n¼0 2n þ 1 4 3 2 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 7 6 7 6 þ::::::::::::::::::::::::::::::::::::::::::::::::: 7 6 7 6 7 6 46 7 ¼ 6 7¼ 7 6 p ::::::::::::::::::::::::::::::::::::::::::::::::::: 7 6 7 6 6 þ:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 7 5 4 ¼ :::::::::::::::::::::::::::::::::::::: Exercises 173 4) The combined effect of mass transferred in the x, y, and z directions is: Cs  Cxyz ¼ Cs  Co  Cs  Cx Cs  Co  Cs  Cy Cs  Co  Cs  Cz Cs  Co  ¼ :::::::::::::::::::::::::::::::::::::::::::::: Calculate the concentration at the center as: ::::::::::::::::::  Cxyz ¼ :::::::::::::::::::::::: ! Cxyz ¼  Step 6 The solution for the mass average concentration is: ! 1 Cs  Cm 8X 1 ð2n þ 1Þ2 p2 Fo ¼ exp  Cs  Co p2 n¼0 ð2n þ 1Þ2 4 1) Mass transferred in the x-direction: ! 1 Cs  Cmx 8 X 1 ð2n þ 1Þ2 p2 Fox ¼ ¼ exp  p n¼0 ð2n þ 1Þ2 Cs  Co 4 3 2 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 7 6 7 6 þ::::::::::::::::::::::::::::::::::::::::: 7 6 7 6 7 6 86 7 ¼ 6 7¼ 7 p6 ::::::::::::::::::::::::::::::::::::::::::::::::::: 7 6 7 6 7 6 þ:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 5 4 ¼ :::::::::::::::::::::::::::::::::::::: 174 13 Unsteady State Mass Transfer 2) Mass transferred in the y-direction: ! 1 Cs  Cmy 8 X 1 ð2n þ 1Þ2 p2 Foy  ¼ exp  p n¼0 ð2n þ 1Þ2 Cs  Co 4 3 2 :::::::::::::::::::::::::::::::::::::::: þ ::::::::::::::::::::::::::::::::::::::::::::: 7 6 7 6 7 6 7 6 8 7¼ ::::::::::::::::::::::::::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::::::::: ¼ 6 7 6 p6 7 7 6 5 4 ¼ :::::::::::::::::::::::::::::::::::::: 3) Mass transferred in the z-direction: ! 1 Cs  Cmz 8 X 1 ð2n þ 1Þ2 p2 Foz  ¼ exp  p n¼0 ð2n þ 1Þ2 Cs  Co 4 3 2 ::::::::::::::::::::::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::::::::::::: 7 6 7 86 7 6 ¼ 6 7¼ p 6 :::::::::::::::::::::::::::::::::::::::::::::: þ :::::::::::::::::::::::::::::::::::::::: 7 5 4 ¼ :::::::::::::::::::::::::::::::::::::: 4) The combined effect of mass transferred in the x, y, and z directions is:     Cs  Cm xyz Cs  Cm x Cs  Cm y Cs  Cm z ¼ Cs  Co Cs  Co Cs  Co Cs  Co ¼ ::::::::::::::::::::::::::::::::::::::: Calculate the mass average concentration as: ::::::::::::::::::  Cm xyz ¼ :::::::::::::::::::::::: ! Cmxyz ¼ :::::::::::::::::::::::::::::::::: Exercise 13.4 To remove the excess salt from salt-stock cucumbers in a pickle factory, the cucumbers are immersed in several changes of fresh water. Calculate the Exercises 175 average NaCl concentration in a cucumber after 5 hours of immersion in fresh water if the cucumber has 2 cm diameter and 6 cm length, the initial NaCl 3 content is 100  kg NaCl/m , and the diffusivity of NaCl in the cucumber is 9 2 1  10 m s. Assume that the change of water is continuous so that the salt concentration in the water is zero. Solution Step 1 Draw the process diagram: r D L x Step 2 State your assumptions: l The external resistance to mass transfer is negligible so that ................. ............................................................................................................... l Diffusion through the ends of the pickle is negligible ................................................................................................................ l Step 3 Identify the geometry of the system. The pickle will be treated as an infinitely long cylinder: Step 4 Select the equation to use. The solution for the mean concentration for a cylinder with constant surface concentration will be applied for the cylindrical surface, that is: 1   Cs  Cm 4X p2 ¼ 2 exp R2 d2n For ¼ 2 2 p n¼1 R dn Cs  Co 4 ¼ 2 1:7066expð  5:783For Þ þ 0:324expð  30:5For Þ p þ 0:132expð  74:9For Þ þ ::: Step 5 Calculate the Fourier number: For ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::: 176 13 Unsteady State Mass Transfer Since For 40:2, more than one term of the sum in the above equation will be used. Two terms are enough for this case. Verify that even the contribution of the 2nd term is small. Step 6 Substitute the values in the above equation and solve for Cm: Cs  Cm 4  ð1:7066expð  5:783For Þ þ 0:324expð  30:5For ÞÞ ¼ Cs  Co p2 ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ¼ ¼ ::::::::::::::::::::::::: and Cm ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::: . Exercise 13.5 The moisture content of the soil in a field is 200 kg water/m3, when suddenly dry air starts blowing. Calculate the moisture content of the soil 5 cm below the surface after 48 hours if the diffusion coefficient of water in the soil is 1:3  108 m2 s and the equilibrium moisture content of the soil in contact with the air is 100 kg water/m3. Assume that the surface immediately equilibrates with the air. Solution Step 1 Draw the process diagram: Cs = 100 kg/m3 Bi > 40 x Co = 100 kg/m3 Exercises 177 Step 2 State your assumptions: l l Initially, the moisture content in the soil is uniform. The moisture transfer in the soil is diffusion controlled. Step 3 Identify the shape of the object.Ground can be treated as a semi-infinite body. Step 4 Select the equation to use.The equation for unsteady state in a semi-infinite body with constant surface concentration will be used:   Cs  C x ¼ erf pffiffiffiffiffiffi Cs  Co 2 Dt i) Calculate: x pffiffiffiffiffiffi ¼ 2 Dt ¼ ::::::::::: ::::::::::::::::::::::::::::::::::::::::::::   x ii) Find the value of erf pffiffiffiffiffi from Table A.5 2 t iii) Substitute values in the original equation and calculate: Cs  C ¼ :::::::::::::::::::::::::::::::::::::::::::::: Cs  Co ! C ¼ ::::::::::::::::::::::::::::::::::::::: . Exercise 13.6 Solve Example 13.1 using the spreadsheet program Mass Transfer-Negligible Surface Resistance.xls. Solution Step 1 On the spreadsheet Mass Transfer-Negligible Surface Resistance.xls, go to the sheet ‘‘slab.’’ Turn the ‘‘SWITCH’’ OFF (set the value in cell G1 equal to 0 and press ENTER). Step 2 Insert the parameter values in the yellow cells using: half thickness in the x-direction Lx ¼ 0:01m (since the slice is exposed to the air from both sides, 178 13 Unsteady State Mass Transfer Lx is taken equal to half thickness), half thickness in the y-direction Ly ¼ 0:10m, half thickness in the z-direction Lz ¼ 0:10m, x ¼ 0, y ¼ 0 and z ¼ 0 (since the concentration will content  be calculated at the center), initial moisture  Co ¼ 200kgwater m3 , surface concentration Cs ¼ 50kgwater m3 . Also insert the given value for the diffusivity. Step 3 Set the value of the time step to 120 s (or some other value) in cell F4. Step 4 Turn the ‘‘SWITCH’’ ON (set the value in cell G1 equal to 1 and press ENTER). Step 5 Iterate by pressing F9 until the moisture in cell K20 reaches the value 100. Step 6 Read the mass average moisture content in cell K40. Observe if the average moisture is affected by the mass transferred from the y and z directions. Step 7 To see how the mass average concentration would have decreased if the slice was exposed to the air from one side only: Lx x i) Turn the ‘‘SWITCH’’ OFF (set the value in cell G1 equal to 0 and press ENTER). ii) Set the value of half thickness in the x-direction to Lx ¼ 0:02m. iii) Set the value of the coordinates of the point to x ¼ 0,y ¼ 0 and z ¼ 0. vi) Turn the ‘‘SWITCH’’ ON (set the value in cell G1 equal to 1 and press ENTER). v) Iterate by pressing F9 until the mass average moisture in cell K40 reaches the same value as in 6 above. vi) Read the time in cell F3. Compare the two time values. Exercise 13.7 Small identical cubes 2cm  2cm  2cm of a cheese-like product were drysalted (covered with dry NaCl crystals). From time to time, one cube was Exercises 179 taken out and analyzed for NaCl and moisture content. Calculate the diffusion coefficient of NaCl in the cheese if the following average NaCl concentration vs. time data were obtained. Time, h kg NaCl/m3 0 2 6 10 24 30 48 60 3 122 163 194 206 215 221 221 Solution Step 1 State your assumptions: .......................................................................................................................... Step 2 On the spreadsheet Mass Transfer-Negligible Surface Resistance.xls, go to the sheet ‘‘slab.’’ Turn the ‘‘SWITCH’’ OFF (set the value in cell G1 equal to 0 and press ENTER). Step 3 Insert the parameter values in the yellow cells using: half thickness in the x-direction Lx = 0.01 m, half thickness in the y-direction Lx ¼ 0:01m, half thickness in the z-direction Lz ¼ 0:01m, x ¼ 0, y ¼ 0 and z ¼ 0, initial NaCl   content Co ¼ 3kgNaCl m3 , surface concentration Cs ¼ 221kgNaCl m3 . Step 4 Insert a first trial value for the diffusivity. Step 5 Set the value of the time step to 1000 s (or some other value) in cell F4. Step 6 Turn the ‘‘SWITCH’’ ON (set the value in cell G1 equal to 1 and press ENTER). Step 7 Iterate by pressing F9 until the mass average NaCl concentration in cell K40 reaches the value of 221 (final experimental concentration). 180 13 Unsteady State Mass Transfer Step 8 Calculate the mean % deviation between the experimental values and the theoretically predicted values of a mass average concentration of NaCl from: N Cexp  Ctheor 100 X Deviation % ¼ N 1 Cexp Step 9 Turn the ‘‘SWITCH’’ OFF (set the value in cell G1 equal to 0 and press ENTER). Step 10 Insert a second trial value for the diffusivity in cell B38. Step 11 Turn the ‘‘SWITCH’’ ON (set the value in cell G1 equal to 1 and press ENTER) Step 12 Iterate by pressing F9 until the mass average NaCl concentration in cell K40 reaches the value 221. Step 13 Calculate the mean % deviation between the experimental values and the theoretically predicted values of a mass average concentration of NaCl as above. Step 14 Repeat steps 9 to 13 until a minimum deviation % is reached. Exercise 13.8 Solve Exercises 13.1, 13.2, 13.3, and 13.4 using the spreadsheet Mass TransferNegligible Surface Resistance.xls. Chapter 14 Pasteurization and Sterilization http://avibert.blogspot.com Review Questions Which of the following statements are true and which are false? 1. The decimal reduction time D is the heating time in min at a certain temperature required for the number of viable microbes to be reduced to 10% of the original number. 2. The z value is the temperature increase required for a ten-fold decrease in D. 3. Thermal death time is the heating time required to give commercial sterility. 4. Thermal death time does not depend on the initial microbial load. 5. The D value does not depend on the initial microbial load. 6. The D value of a microorganism is independent of the food item. 7. The D value of a microbe is a measure of the thermal resistance of the microbe. 8. If the number of microbes in a process has to be reduced from an initial load of 106 to a final 104, the required thermal death time will be 10D. 9. If the number of microbes in a canned product is reduced from 103 to 104, it means that 1 can in 100000 may be spoiled. 10. As the process temperature increases, the thermal death time increases. 11. A 10D process is usually applied as a minimum heat treatment for Clostridium botulinum. 12. Typical z values are 5.5 8C for vegetative cells, 10 8C for spores, and 22 8C for nutrients. 13. A D121.1 = 0.21 min is usually assumed for Clostridium botulinum. 14. The accepted risk for Clostridium botulinum is 1012. 15. The accepted spoilage probability for mesophilic spoilage microorganisms is usually 105. 16. The accepted spoilage propability for thermophilic spoilage microorganisms upon incubation after processing is usually 102. 17. The slowest heating point of a can filled with a liquid food is the geometric center of the can. S. Yanniotis, Solving Problems in Food Engineering. Ó Springer 2008 181 182 14 Pasteurization and Sterilization 18. The worst case scenario for calculating the lethal effect in a holding tube for a Newtonian liquid is to assume that the residence time is half the mean residence time. 19. In calculating the lethal effect in a solid particulate flowing in a two-phase flow in a holding tube, the convective heat transfer resistance at the surface of the particulate can be neglected. 20. The Ball Formula method is an alternative method to the general method for calculating the lethal effect of a sterilization process. 21. For the purpose of thermal processing, foods are divided into low acid foods (pH > 4:5) and high acid foods (pH

Report "Solving Problems in Food Engineering"

Solving Problems In Food Engineering Springer 2008 0387735135 Pdf

• Registration

Solving Problems in Food Engineering

• Marion Norris
• 7 years ago

--> -->