An Instructor’s Solutions Manual To Accompany
An Instructor’s Solutions Manual to AccompanyPRINCIPLES OF HEAT TRANSFER, 7TH EDITION, SIFRANK KREITHRAJ M. MANGLIKMARK S. BOHNSI EDITION PREPARED BY: SHALIGRAM TIWARI
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INSTRUCTOR'S SOLUTIONS MANUAL TOACCOMPANYPRINCIPLES OFHEAT TRANSFERSEVENTH EDITION, SIFRANK KREITHRAJ M. MANGLIKMARK S. BOHNSI EDITION PREPARED BY:SHALIGRAM TIWARIIndian Institute of Technology Madras
TABLE OF CONTENTSCHAPTERPAGE1 . 12 . 853 . 2314 . 3115 . 4216 . 5137 . 6078 . 6839 . 78110 . 871
67706 00 FM pi-xxiii.qxd5/14/109:32 AMPage ixCONTENTSChapter 1 Basic Modes of Heat Transfer1.11.21.31.41.51.61.71.8The Relation of Heat Transfer to Thermodynamics 3Dimensions and Units 7Heat Conduction 9Convection 17Radiation 21Combined Heat Transfer Systems 23Thermal Insulation 45Heat Transfer and the Law of Energy Conservation 51References 58Problems 58Design Problems 68Chapter 2 Heat n 71The Conduction Equation 71Steady Heat Conduction in Simple Geometries 78Extended Surfaces 95Multidimensional Steady Conduction 105Unsteady or Transient Heat Conduction 116Charts for Transient Heat Conduction 134Closing Remarks 150References 150Problems 151Design Problems 163Chapter 3 Numerical Analysis of Heat Conduction3.13.23.3166Introduction 167One-Dimensional Steady Conduction 168One-Dimensional Unsteady Conduction 180ixCopyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
67706 00 FM pi-xxiii.qxdx5/14/109:32 AMPage xContents3.4*3.5*3.6*3.7Two-Dimensional Steady and Unsteady Conduction 195Cylindrical Coordinates 215Irregular Boundaries 217Closing Remarks 221References 221Problems 222Design Problems 228Chapter 4 Analysis of Convection Heat 4.13*4.14Introduction 231Convection Heat Transfer 231Boundary Layer Fundamentals 233Conservation Equations of Mass, Momentum, and Energy for Laminar Flow Overa Flat Plate 235Dimensionless Boundary Layer Equations and SimilarityParameters 239Evaluation of Convection Heat Transfer Coefficients 243Dimensional Analysis 245Analytic Solution for Laminar Boundary Layer Flow Over a Flat Plate 252Approximate Integral Boundary Layer Analysis 261Analogy Between Momentum and Heat Transfer in Turbulent Flow Overa Flat Surface 267Reynolds Analogy for Turbulent Flow Over Plane Surfaces 273Mixed Boundary Layer 274Special Boundary Conditions and High-Speed Flow 277Closing Remarks 282References 283Problems 284Design Problems 294Chapter 5 Natural Convection5.15.25.35.4*5.55.6*230296Introduction 297Similarity Parameters for Natural Convection 299Empirical Correlation for Various Shapes 308Rotating Cylinders, Disks, and Spheres 322Combined Forced and Natural Convection 325Finned Surfaces 328Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
67706 00 FM pi-xxiii.qxd5/14/109:32 AMPage xiContents5.7Closing Remarks 333References 338Problems 340Design Problems 348Chapter 6 Forced Convection Inside Tubes and Ducts6.16.2*6.36.4*6.56.66.7Introduction 351Analysis of Laminar Forced Convection in a Long Tube 360Correlations for Laminar Forced Convection 370Analogy Between Heat and Momentum Transfer in Turbulent Flow 382Empirical Correlations for Turbulent Forced Convection 386Heat Transfer Enhancement and Electronic-Device Cooling 395Closing Remarks 406References 408Problems 411Design Problems 418Chapter 7 Forced Convection Over Exterior Surfaces7.17.27.3*7.47.5*7.6*7.7420Flow Over Bluff Bodies 421Cylinders, Spheres, and Other Bluff Shapes 422Packed Beds 440Tube Bundles in Cross-Flow 444Finned Tube Bundles in Cross-Flow 458Free Jets 461Closing Remarks 471References 473Problems 475Design Problems 482Chapter 8 Heat n 485Basic Types of Heat Exchangers 485Overall Heat Transfer Coefficient 494Log Mean Temperature Difference 498Heat Exchanger Effectiveness 506Heat Transfer Enhancement 516Microscale Heat Exchangers 524Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.xi
67706 00 FM pi-xxiii.qxdxii5/14/109:32 AMPage xiiContents8.8Closing Remarks 525References 527Problems 529Design Problems 539Chapter 9 Heat Transfer by al Radiation 541Blackbody Radiation 543Radiation Properties 555The Radiation Shape Factor 571Enclosures with Black Surfaces 581Enclosures with Gray Surfaces 585Matrix Inversion 591Radiation Properties of Gases and Vapors 602Radiation Combined with Convection and Conduction 610Closing Remarks 614References 615Problems 616Design Problems 623Chapter 10 Heat Transfer with Phase on to Boiling 625Pool Boiling 625Boiling in Forced Convection 647Condensation 660Condenser Design 670Heat Pipes 672Freezing and Melting 683References 688Problems 691Design Problems 696Appendix 1 The International System of UnitsAppendix 2 Data TablesA3A6Properties of Solids A7Thermodynamic Properties of Liquids A14Heat Transfer Fluids A23Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
Chapter 1PROBLEM 1.1The outer surface of a 0.2m-thick concrete wall is kept at a temperature of –5 C, while theinner surface is kept at 20 C. The thermal conductivity of the concrete is 1.2 W/(m K).Determine the heat loss through a wall 10 m long and 3 m high.GIVEN10 m long, 3 m high, and 0.2 m thick concrete wallThermal conductivity of the concrete (k) 1.2 W/(m K)Temperature of the inner surface (Ti) 20 CTemperature of the outer surface (To) –5 CFINDThe heat loss through the wall (qk)ASSUMPTIONSOne dimensional heat flowThe system has reached steady stateSKETCHL 0.2 mL 10mH 3mqkTi 20 CTo – 5 CSOLUTIONThe rate of heat loss through the wall is given by Equation (1.2)qk AK(ΔT)Lqk (10 m) (3m) (1.2 W/(m K) )(20 C – (–5 C))0.2 mqk 4500 WCOMMENTSSince the inside surface temperature is higher than the outside temperature heat is transferred from theinside of the wall to the outside of the wall.1 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
PROBLEM 1.2The weight of the insulation in a spacecraft may be more important than the spacerequired. Show analytically that the lightest insulation for a plane wall with a specifiedthermal resistance is that insulation which has the smallest product of density timesthermal conductivity.GIVENInsulating a plane wall, the weight of insulation is most significantFINDShow that lightest insulation for a given thermal resistance is that insulation which has the smallestproduct of density (ρ) times thermal conductivity (k)ASSUMPTIONSOne dimensional heat transfer through the wallSteady state conditionsSOLUTIONThe resistance of the wall (Rk), from Equation (1.13) isRk LAkwhereL the thickness of the wallA the area of the wallThe weight of the wall (w) isw ρALSolving this for LL wρASubstituting this expression for L into the equation for the resistanceRk wρ k A2 w ρ k Rk A2Therefore, when the product of ρ k for a given resistance is smallest, the weight is also smallest.COMMENTSSince ρ and k are physical properties of the insulation material they cannot be varied individually.Hence in this type of design different materials must be tried to minimize the weight.PROBLEM 1.3A furnace wall is to be constructed of brick having standard dimensions 22.5 cm 11 cm 7.5 cm. Two kinds of material are available. One has a maximum usable temperatureof 1040 C and a thermal conductivity of 1.7 W/(m K), and the other has a maximumtemperature limit of 870 C and a thermal conductivity of 0.85 W/(m K). The bricks costthe same and can be laid in any manner, but we wish to design the most economical wallfor a furnace with a temperature on the hot side of 1040 C and on the cold side of200 C. If the maximum amount of heat transfer permissible is 950 W/m2 for each squarefoot of area, determine the most economical arrangements for the available bricks.2 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
GIVENFurnace wall made of 22.5 cm 11 cm 7.5 cm bricks of two typesType 1 bricks Maximum useful temperature (T1, max) 1040 C Thermal conductivity (k1) 1.7 W/(m K)Type 2 bricks Maximum useful temperature (T2, max) 870 C Thermal conductivity (k2) 0.85 W/(m K)Bricks cost the sameWall hot side temperature (Thot) 1040 C and wall cold side temperature (Tcold) 200 CMaximum permissible heat transfer (qmax/A) 950 W/m2FINDThe most economical arrangement for the bricksASSUMPTIONSOne-dimensional, steady state heat transfer conditionsConstant thermal conductivitiesThe contact resistance between the bricks is negligibleSKETCHType 1 BricksType 2 BricksTmax 1040 CTmax 200 CT12 870 CSOLUTIONSince the type 1 bricks have a higher thermal conductivity at the same cost as the type 2 bricks, themost economical wall would use as few type 1 bricks as possible. However, there should be thickenough layer of type 1 bricks to keep the type 2 bricks at 870 C or less.For one-dimensional conduction through type 1 bricks (from Equation 1.2)kAqk ΔTLqmaxk 1 (Thot – T12)L1Awhere L1 is the minimum thickness of the type 1 bricks.Solving for L1k1(Thot – T12)L1 qmax A L1 1.7 W/(m K)950 W/m 2(1040 – 870)K 0.3042 m 30.42 cm3 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
This thickness can be achieved by using 4 layers of type 1 bricks using the 7.5 cm dimension.Similarly, for one-dimensional conduction through type 2 bricksL2 L2 k2(T12 – Tcold) qmax A 0.85 W/(m K)950 W/m 2(870 – 200)K 0.6 m 60 cmThis thickness can be achieved with 8 layers of type 2 bricks using the 7.5 cm dimension.Therefore, the most economical wall would be built using 4 layers of type 1 bricks and 8 layers oftype 2 bricks with the three inch dimension of the bricks used as the thickness.PROBLEM 1.4To measure thermal conductivity, two similar 1-cm-thick specimens are placed in anapparatus shown in the accompanying sketch. Electric current is supplied to the6-cm by 6-cm guarded heater, and a wattmeter shows that the power dissipation is 10watts (W). Thermocouples attached to the warmer and to the cooler surfaces showtemperatures of 322 and 300 K, respectively. Calculate the thermal conductivity of thematerial at the mean temperature in W/(m K).GIVENThermal conductivity measurement apparatus with two samples as shownSample thickness (L) 1 cm 0.01 cmArea 6 cm 6 cm 36 cm2 0.0036 m2Power dissipation rate of the heater (qh) 10 WSurface temperatures Thot 322 K Tcold 300 KFINDThe thermal conductivity of the sample at the mean temperature in W/(m K)ASSUMPTIONSOne dimensional, steady state conductionNo heat loss from the edges of the apparatusSKETCHEGuard Ring and InsulationSSimilar SpecimenThotHeaterTcoldWattmeter4 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SOLUTIONBy conservation of energy, the heat loss through the two specimens must equal the power dissipationof the heater. Therefore the heat transfer through one of the specimens is qh/2.For one dimensional, steady state conduction (from Equation (1.3)qk qkAΔT hL2Solving for the thermal conductivityqhLk 2AΔTk (5 W)(0.01m)(0.0036 m 2 ) (322 K 300 K)k 0.63 W/(m K)COMMENTSIn the construction of the apparatus care must be taken to avoid edge losses so all the heat generatedwill be conducted through the two specimens.PROBLEM 1.5To determine the thermal conductivity of a structural material, a large 15 cm-thick slabof the material was subjected to a uniform heat flux of 2500 W/m2, while thermocouplesembedded in the wall 2.5 cm apart were read over a period of time. After the system hadreached equilibrium, an operator recorded the readings of the thermocouples as shownbelow for two different environmental conditions.Distance from the surface (cm)Temperature ( C)Test 1:05101540659713205101595130168208Test 2:From these data, determine an approximate expression for the thermal conductivity as afunction of temperature between 40 and 208 C.GIVENThermal conductivity test on a large, 15 cm slabThermocouples are embedded in the wall, 2.5 cm apartHeat flux (q/A) 2500 W/m2Two equilibrium conditions were recorded (shown in Table above)5 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
FINDAn approximate expression for thermal conductivity as a function of temperature between 40 and208 C.ASSUMPTIONSOne-dimensional conductionSKETCHThermocouplesDistance (cm) 051015SOLUTIONThe thermal conductivity can be calculated for each pair of adjacent thermocouples using the equationfor one-dimensional conductionq kAΔTLSolving for kk q LA ΔTThis will give a thermal conductivity for each pair of adjacent thermocouples which are assigned to theaverage temperature of the pair of thermocouples. As an example, for the first pair of thermocouplesin Test 1, the thermal conductivity (ko) is 5 10 –2 m ko (2500 W/m 2 ) o 5 W/(m K) 65 C 40 o C The average temperature for this pair of thermocouples isTavg 40 65 52.5 C2The average temperature and the thermal conductivity for all other pairs of thermocouples are given inthe table below.n( C)12345652.581114.5112.5149188Thermal Conductivity W/(m K)53.93.573.383.293.1256 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
These points are displayed graphically.5k(W/(mK))4Mean Variation of kwith Temperature3100150200Temperature ( C)We will use the best fit quadratic function to represent the relationship between thermal conductivityand temperaturek (T) a b T c T 2The constants a, b, and c can be found using a least squares fit.Let the experimental thermal conductivity at data point n be designated as kn. A least squares fit of thedata can be obtained as followsThe sum of the squares of the errors isS [kn k (Tn )]2NS kn2 2 a kn N a 2 2ab Tn 2b knTn 2ac Tn2 b2 Tn2– 2c knTn2 2bc Tn3 c 2 Tn4By setting the derivatives of S (with respect
An Instructor’s Solutions Manual to Accompany . copy and download functionality disabled, and accessible solely by your students who have purchased the associated textbook for the Course. You may not sell, licens
INSTRUCTOR SOLUTIONS MANUAL. Instructor’s Manual to accompany Modern Physics, 3rd Edition Kenneth S. Krane Department of Physics Oregon State University 2012 John Wiley & Sons . ii Preface This Instructor’s Manual
Independent Personal Pronouns Personal Pronouns in Hebrew Person, Gender, Number Singular Person, Gender, Number Plural 3ms (he, it) א ִוה 3mp (they) Sֵה ,הַָּ֫ ֵה 3fs (she, it) א O ה 3fp (they) Uֵה , הַָּ֫ ֵה 2ms (you) הָּ תַא2mp (you all) Sֶּ תַא 2fs (you) ְ תַא 2fp (you
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Instructor Candidate Application Revised: January 2018 Instructions: . 2.1 Delivers all core content consistent with AHA published guidelines, Instructor Manual, Lesson Plans, and agenda . Yes Yes with req. No Not observed Reviewer's comments: _ _ American Heart Association Emergency Cardiovascular Care Program : Instructor Monitor Tool : Instructor Monitor Tool Revised: January 2018 .
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