Solution Manual For Heat and Mass Transfer: Fundamentals and Applications, 6th Edition

2-1 Solutions Manual for Heat and Mass Transfer: Fundamentals & Applications 6th Edition Yunus A. Çengel, Afshin J. Ghajar McGraw-Hill Education, 2020 Chapter 2 HEAT CONDUCTION EQUATION PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGraw-Hill Education and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill Education: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill Education. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-2 Introduction 2-1C The term steady implies no change with time at any point within the medium while transient implies variation with time or time dependence. Therefore, the temperature or heat flux remains unchanged with time during steady heat transfer through a medium at any location although both quantities may vary from one location to another. During transient heat transfer, the temperature and heat flux may vary with time as well as location. Heat transfer is one-dimensional if it occurs primarily in one direction. It is two-dimensional if heat tranfer in the third dimension is negligible. 2-2C Heat transfer is a vector quantity since it has direction as well as magnitude. Therefore, we must specify both direction and magnitude in order to describe heat transfer completely at a point. Temperature, on the other hand, is a scalar quantity. 2-3C Yes, the heat flux vector at a point P on an isothermal surface of a medium has to be perpendicular to the surface at that point. 2-4C Isotropic materials have the same properties in all directions, and we do not need to be concerned about the variation of properties with direction for such materials. The properties of anisotropic materials such as the fibrous or composite materials, however, may change with direction. 2-5C In heat conduction analysis, the conversion of electrical, chemical, or nuclear energy into heat (or thermal) energy in solids is called heat generation. 2-6C The phrase “thermal energy generation” is equivalent to “heat generation,” and they are used interchangeably. They imply the conversion of some other form of energy into thermal energy. The phrase “energy generation,” however, is vague since the form of energy generated is not clear. 2-7C The heat transfer process from the kitchen air to the refrigerated space is transient in nature since the thermal conditions in the kitchen and the refrigerator, in general, change with time. However, we would analyze this problem as a steady heat transfer problem under the worst anticipated conditions such as the lowest thermostat setting for the refrigerated space, and the anticipated highest temperature in the kitchen (the so-called design conditions). If the compressor is large enough to keep the refrigerated space at the desired temperature setting under the presumed worst conditions, then it is large enough to do so under all conditions by cycling on and off. Heat transfer into the refrigerated space is three-dimensional in nature since heat will be entering through all six sides of the refrigerator. However, heat transfer through any wall or floor takes place in the direction normal to the surface, and thus it can be analyzed as being one-dimensional. Therefore, this problem can be simplified greatly by considering the heat transfer to be onedimensional at each of the four sides as well as the top and bottom sections, and then by adding the calculated values of heat transfer at each surface. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-3 2-8C Heat transfer through the walls, door, and the top and bottom sections of an oven is transient in nature since the thermal conditions in the kitchen and the oven, in general, change with time. However, we would analyze this problem as a steady heat transfer problem under the worst anticipated conditions such as the highest temperature setting for the oven, and the anticipated lowest temperature in the kitchen (the so called “design” conditions). If the heating element of the oven is large enough to keep the oven at the desired temperature setting under the presumed worst conditions, then it is large enough to do so under all conditions by cycling on and off. Heat transfer from the oven is three-dimensional in nature since heat will be entering through all six sides of the oven. However, heat transfer through any wall or floor takes place in the direction normal to the surface, and thus it can be analyzed as being one-dimensional. Therefore, this problem can be simplified greatly by considering the heat transfer as being one- dimensional at each of the four sides as well as the top and bottom sections, and then by adding the calculated values of heat transfers at each surface. 2-9C Heat transfer to a potato in an oven can be modeled as one-dimensional since temperature differences (and thus heat transfer) will exist in the radial direction only because of symmetry about the center point. This would be a transient heat transfer process since the temperature at any point within the potato will change with time during cooking. Also, we would use the spherical coordinate system to solve this problem since the entire outer surface of a spherical body can be described by a constant value of the radius in spherical coordinates. We would place the origin at the center of the potato. 2-10C Assuming the egg to be round, heat transfer to an egg in boiling water can be modeled as one-dimensional since temperature differences (and thus heat transfer) will primarily exist in the radial direction only because of symmetry about the center point. This would be a transient heat transfer process since the temperature at any point within the egg will change with time during cooking. Also, we would use the spherical coordinate system to solve this problem since the entire outer surface of a spherical body can be described by a constant value of the radius in spherical coordinates. We would place the origin at the center of the egg. 2-11C Heat transfer to a hot dog can be modeled as two-dimensional since temperature differences (and thus heat transfer) will exist in the radial and axial directions (but there will be symmetry about the center line and no heat transfer in the azimuthal direction. This would be a transient heat transfer process since the temperature at any point within the hot dog will change with time during cooking. Also, we would use the cylindrical coordinate system to solve this problem since a cylinder is best described in cylindrical coordinates. Also, we would place the origin somewhere on the center line, possibly at the center of the hot dog. Heat transfer in a very long hot dog could be considered to be one-dimensional in preliminary calculations. 2-12C Heat transfer to a roast beef in an oven would be transient since the temperature at any point within the roast will change with time during cooking. Also, by approximating the roast as a spherical object, this heat transfer process can be modeled as one-dimensional since temperature differences (and thus heat transfer) will primarily exist in the radial direction because of symmetry about the center point. 2-13C Heat loss from a hot water tank in a house to the surrounding medium can be considered to be a steady heat transfer problem. Also, it can be considered to be two-dimensional since temperature differences (and thus heat transfer) will exist in the radial and axial directions (but there will be symmetry about the center line and no heat transfer in the azimuthal direction.) Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-4 2-14C Heat transfer to a canned drink can be modeled as two-dimensional since temperature differences (and thus heat transfer) will exist in the radial and axial directions (but there will be symmetry about the center line and no heat transfer in the azimuthal direction. This would be a transient heat transfer process since the temperature at any point within the drink will change with time during heating. Also, we would use the cylindrical coordinate system to solve this problem since a cylinder is best described in cylindrical coordinates. Also, we would place the origin somewhere on the center line, possibly at the center of the bottom surface. 2-15 A certain thermopile used for heat flux meters is considered. The minimum heat flux this meter can detect is to be determined. Assumptions 1 Steady operating conditions exist. Properties The thermal conductivity of kapton is given to be 0.345 W/mK. Analysis The minimum heat flux can be determined from q = k T L = (0.345 W/m  C) 0.1C = 17.3 W/m2 0.002 m 2-16 Diameter, length, and mass of stainless steel rod is given. The rod is insulated on its exterior surface except for the ends. Temperature distribution in the rod is also given. The heat flux along the rod is to be determined. Assumptions 1 Heat conduction is steady and one-dimensional in the x-direction. 2 Thermal conductivity is constant. Analysis The heat flux can be found from Fourier’s law ?̇ = −? ?? ?? Table A-3 gives values for the thermal conductivity of stainless steels, however we are not told which type of stainless steel the rod is made of, and the thermal conductivity varies between them. We do know the mass of the rod, and can use this to calculate its density:  = ? ? 0.221 kg = = = 7895 kg/m3 ? 2 [(0.018 m)2 (0.11 m)]/4 ??? ( )? 4 From Table A-3, with 7900 kg/m3,it appears that the material is AISI 304 stainless steel. The temperature of the rod from the given equation for temperature distribution varies from 310 K at x=0 to290 K at x = L = 110 mm. Evaluating the thermal conductivity at the average temperature of 300 K, from Table A-3, k = 14.9 W/mK. Thus, ?̇ = −? ?? ?? = −? (− 20K ? ) = −14.9 W/mK(− 20K 0.11 m ) = 2709 W/?? DiscussionIf the temperature of the rod varies significantly along its length, the thermal conductivity will vary along the rod as much or more than the variation in thermal conductivities between the different stainless steels. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-5 2-17 The rate of heat generation per unit volume in a stainless steel plate is given. The heat flux on the surface of the plate is to be determined. Assumptions Heat is generated uniformly in steel plate. e Analysis We consider a unit surface area of 1 m2. The total rate of heat generation in this section of the plate is L E = e V = e ( A  L) = (5 10 6 W/m 3 )(1 m 2 )(0.03 m) = 1.5 10 5 W gen gen plate gen Noting that this heat will be dissipated from both sides of the plate, the heat flux on either surface of the plate becomes E gen 1.5  10 5 W q = = = 75,000 W/m 2 = 75 kW/m 2 Aplate 2 1 m 2 2-18 The rate of heat generation per unit volume in the uranium rods is given. The total rate of heat generation in each rod is to be determined. Assumptions Heat is generated uniformly in the uranium rods. g = 7107 W/m3 D = 5 cm L=1m Analysis The total rate of heat generation in the rod is determined by multiplying the rate of heat generation per unit volume by the volume of the rod E = e V = e (D 2 / 4) L = (7  10 7 W/m 3 )[ (0.05 m) 2 / 4](1 m) = 1.374 10 5 W = 137 kW gen gen rod gen 2-19 The variation of the absorption of solar energy in a solar pond with depth is given. A relation for the total rate of heat generation in a water layer at the top of the pond is to be determined. Assumptions Absorption of solar radiation by water is modeled as heat generation. Analysis The total rate of heat generation in a water layer of surface area A and thickness L at the top of the pond is determined by integration to be E gen = V egen dV =  L x =0 e0 e −bx ( Adx) = Ae0 e −bx −b L = 0 Ae 0 (1 − e −bL ) b Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-6 2-20E The power consumed by the resistance wire of an iron is given. The heat generation and the heat flux are to be determined. Assumptions Heat is generated uniformly in the resistance wire. q = 1000 W D = 0.08 in L = 15 in Analysis A 1000 W iron will convert electrical energy into heat in the wire at a rate of 1000 W. Therefore, the rate of heat generation in a resistance wire is simply equal to the power rating of a resistance heater. Then the rate of heat generation in the wire per unit volume is determined by dividing the total rate of heat generation by the volume of the wire to be E gen E gen 1000 W  3.412 Btu/h  7 3 egen = = =   = 7.820  10 Btu/h  ft V wire (D 2 / 4) L [ (0.08 / 12 ft) 2 / 4](15 / 12 ft)  1W  Similarly, heat flux on the outer surface of the wire as a result of this heat generation is determined by dividing the total rate of heat generation by the surface area of the wire to be E gen E gen 1000 W  3.412 Btu/h  5 2 q = = =   = 1.303  10 Btu/h  ft Awire DL  (0.08 / 12 ft)(15 / 12 ft)  1W  Discussion Note that heat generation is expressed per unit volume in Btu/hft3 whereas heat flux is expressed per unit surface area in Btu/hft2. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-7 Heat Conduction Equation 2-21C The one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and heat  2 T egen 1 T + = generation is . Here T is the temperature, x is the space variable, egen is the heat generation per unit k α t x 2 volume, k is the thermal conductivity,  is the thermal diffusivity, and t is the time. 2-22C The one-dimensional transient heat conduction equation for a long cylinder with constant thermal conductivity and 1   T  egen 1 T = heat generation is . Here T is the temperature, r is the space variable, g is the heat generation per r + r r  r  k  t unit volume, k is the thermal conductivity,  is the thermal diffusivity, and t is the time. 2-23 We consider a thin element of thickness x in a large plane wall (see Fig. 2-12 in the text). The density of the wall is , the specific heat is c, and the area of the wall normal to the direction of heat transfer is A. In the absence of any heat generation, an energy balance on this thin element of thickness x during a small time interval t can be expressed as E element Q x − Q x + x = t where Eelement = Et + t − Et = mc (Tt + t − Tt ) = cAx(Tt + t − Tt ) Substituting, T − Tt Q x − Q x + x = cAx t + t t Dividing by Ax gives T − Tt 1 Q x + x − Q x − = c t + t A x t Taking the limit as x → 0 and t → 0 yields 1   T  T  kA  = ρc A x  x  t since from the definition of the derivative and Fourier’s law of heat conduction, Q − Q x Q   T  lim x + x = =  − kA  x →0 x x x  x  Noting that the area A of a plane wall is constant, the one-dimensional transient heat conduction equation in a plane wall with constant thermal conductivity k becomes  2T x 2 = 1 T α t where the property  = k / c is the thermal diffusivity of the material. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-8 2-24 We consider a thin cylindrical shell element of thickness r in a long cylinder (see Fig. 2-14 in the text). The density of the cylinder is , the specific heat is c, and the length is L. The area of the cylinder normal to the direction of heat transfer at any location is A = 2rL where r is the value of the radius at that location. Note that the heat transfer area A depends on r in this case, and thus it varies with location. An energy balance on this thin cylindrical shell element of thickness r during a small time interval t can be expressed as E element Q r − Q r + r + E element = t where Eelement = Et + t − Et = mc (Tt + t − Tt ) = cAr (Tt + t − Tt ) E element = egenV element = egen Ar Substituting, T − Tt Q r − Q r + r + e gen Ar = cAr t + t t where A = 2rL . Dividing the equation above by Ar gives − T − Tt 1 Q r + r − Q r + egen = c t + t A r t Taking the limit as r → 0 and t → 0 yields 1   T  T  kA  + e gen = c A r  r  t since, from the definition of the derivative and Fourier’s law of heat conduction, Q − Q r Q   T  lim r + r = =  − kA  r →0 r r r  r  Noting that the heat transfer area in this case is A = 2rL and the thermal conductivity is constant, the one-dimensional transient heat conduction equation in a cylinder becomes 1   T  1 T r  + egen = r r  r   t where  = k / c is the thermal diffusivity of the material. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-9 2-25 We consider a thin spherical shell element of thickness r in a sphere (see Fig. 2-16 in the text).. The density of the sphere is , the specific heat is c, and the length is L. The area of the sphere normal to the direction of heat transfer at any location is A = 4r 2 where r is the value of the radius at that location. Note that the heat transfer area A depends on r in this case, and thus it varies with location. When there is no heat generation, an energy balance on this thin spherical shell element of thickness r during a small time interval t can be expressed as E element Q r − Q r + r = t where Eelement = Et + t − Et = mc (Tt + t − Tt ) = cAr (Tt + t − Tt ) Substituting, T −T Q r − Q r + r = cAr t + t t t where A = 4r 2 . Dividing the equation above by Ar gives − T − Tt 1 Q r + r − Q r = c t + t A r t Taking the limit as r → 0 and t → 0 yields 1   T  T  kA  = ρc A r  r  t since, from the definition of the derivative and Fourier’s law of heat conduction, Q − Q r Q   T  lim r + r = =  − kA  r →0 r r r  r  Noting that the heat transfer area in this case is A = 4r 2 and the thermal conductivity k is constant, the one-dimensional transient heat conduction equation in a sphere becomes 1   2 T  1 T r = r  α t r 2 r  where  = k / c is the thermal diffusivity of the material. 2-26 For a medium in which the heat conduction equation is given in its simplest by  2T x 2 = 1 T :  t (a) Heat transfer is transient, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is constant. 2-27 For a medium in which the heat conduction equation is given by  2T  2T x y + 2 2 = 1 T :  t (a) Heat transfer is transient, (b) it is two-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is constant. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-10 2-28 For a medium in which the heat conduction equation is given in its simplest by 1 d  dT   rk  + e gen = 0 : r dr  dr  (a) Heat transfer is steady, (b) it is one-dimensional, (c) there is heat generation, and (d) the thermal conductivity is variable. 2-29 For a medium in which the heat conduction equation is given by 1   T    T   kr  + k  + e gen = 0 : r r  r  z  z  (a) Heat transfer is steady, (b) it is two-dimensional, (c) there is heat generation, and (d) the thermal conductivity is variable. 2-30 For a medium in which the heat conduction equation is given in its simplest by r d 2T dr 2 +2 dT =0: dr (a) Heat transfer is steady, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is constant. 2-31 For a medium in which the heat conduction equation is given by 1   2 T  1 T r = r  α t r 2 r  (a) Heat transfer is transient, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is constant. 2-32 For a medium in which the heat conduction equation is given by 1   2 T  1  2T 1 T r + =   r  r 2 sin 2   2  t r 2 r  (a) Heat transfer is transient, (b) it is two-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is constant. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-11 2-33 We consider a small rectangular element of length x, width y, and height z = 1 (similar to the one in Fig. 2-20). The density of the body is  and the specific heat is c. Noting that heat conduction is two-dimensional and assuming no heat generation, an energy balance on this element during a small time interval t can be expressed as Rate of heat   Rate of heat conduction  Rate of change of         at the surfaces at  conduction at the  −   =  the energy content   surfaces at x and y   x + x and y + y   of the element        E element Q x + Q y − Q x + x − Q y + y = t or Noting that the volume of the element is V element = xyz = xy 1 , the change in the energy content of the element can be expressed as Eelement = Et + t − Et = mc (Tt + t − Tt ) = cxy(Tt + t − Tt ) Substituting, T − Tt Q x + Q y − Q x + x − Q y + y = cxy t + t t Dividing by xy gives −   T − Tt 1 Q x + x − Q x 1 Q y + y − Q y − = c t + t y x x y t Taking the thermal conductivity k to be constant and noting that the heat transfer surface areas of the element for heat conduction in the x and y directions are Ax = y 1 and A y = x 1, respectively, and taking the limit as x, y, and t → 0 yields  2T  2T x y + 2 2 = 1 T α t since, from the definition of the derivative and Fourier’s law of heat conduction, 1 Q x + x − Q x 1 Q x 1   T    T   2T lim = =  − kyz  = − k  = −k 2 x →0 yz x yz x yz x  x  x  x  x   1 Q y + y − Q y 1 Q y 1   T    T   2T  − kxz  = −  k  = −k lim = = y →0 xz y xz y xz y  y  y  y  y 2 Here the property  = k /  c is the thermal diffusivity of the material. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-12 2-34 We consider a thin ring shaped volume element of width z and thickness r in a cylinder. The density of the cylinder is  and the specific heat is c. In general, an energy balance on this ring element during a small time interval t can be expressed as E element (Q r − Q r + r ) + (Q z − Q z + z ) = t z But the change in the energy content of the element can be expressed as Eelement = Et + t − Et = mc (Tt + t − Tt ) = c(2rr )z (Tt + t − Tt ) Substituting, rr r+r T − Tt (Q r − Q r + r ) + (Q z − Q z + z ) = c(2rr )z t + t t Dividing the equation above by (2rr )z gives − T − Tt 1 Q r + r − Q r 1 Q z + z − Q z − = c t + t 2rz r 2rr z t Noting that the heat transfer surface areas of the element for heat conduction in the r and z directions are Ar = 2rz and Az = 2rr , respectively, and taking the limit as r, z and t → 0 yields 1   T  1   T    T  T k  + k  kr +  = c r r  r  r 2     z  z  t since, from the definition of the derivative and Fourier’s law of heat conduction, 1 Q r + r − Q r 1 Q 1   T  1   T  lim = =  − k (2rz ) =−  kr  r →0 2rz r 2rz r 2rz r  r  r r  r  1 Q z + z − Q z 1 Qz 1   T    T  lim = =  − k (2rr )  = − k  z →0 2rr z 2rr z 2rr z  z  z  z  For the case of constant thermal conductivity the equation above reduces to 1   T   2T 1 T = r + r r  r  z 2  t where  = k /  c is the thermal diffusivity of the material. For the case of steady heat conduction with no heat generation it reduces to 1   T   2T =0 r + r r  r  z 2 Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-13 2-35 Consider a thin disk element of thickness z and diameter D in a long cylinder. The density of the cylinder is , the specific heat is c, and the area of the cylinder normal to the direction of heat transfer is A = D 2 / 4 , which is constant. An energy balance on this thin element of thickness z during a small time interval t can be expressed as  Rate of heat   Rate of heat   Rate of heat   Rate of change of           conduction at  −  conduction at the  +  generation inside  =  the energy content   the surface at z   surface at z + z   the element   of the element          or, E element Q z − Q z + z + E element = t But the change in the energy content of the element and the rate of heat generation within the element can be expressed as Eelement = Et +t − Et = mc (Tt + t − Tt ) = cAz(Tt + t − Tt ) and E element = egenV element = egen Az Substituting, T − Tt Q z − Q z + z + egen Az = cAz t + t t Dividing by Az gives T − Tt 1 Q z + z − Q z − + egen = c t + t A z t Taking the limit as z → 0 and t → 0 yields 1   T  T  kA  + e gen = c A z  z  t since, from the definition of the derivative and Fourier’s law of heat conduction, Q − Q z Q   T  lim z + z = =  − kA  z →0 z z z  z  Noting that the area A and the thermal conductivity k are constant, the one-dimensional transient heat conduction equation in the axial direction in a long cylinder becomes  2T z 2 + e gen k = 1 T  t where the property  = k /  c is the thermal diffusivity of the material. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-14 Boundary and Initial Conditions; Formulation of Heat Conduction Problems 2-36C The mathematical expressions of the thermal conditions at the boundaries are called the boundary conditions. To describe a heat transfer problem completely, two boundary conditions must be given for each direction of the coordinate system along which heat transfer is significant. Therefore, we need to specify four boundary conditions for two-dimensional problems. 2-37C The mathematical expression for the temperature distribution of the medium initially is called the initial condition. We need only one initial condition for a heat conduction problem regardless of the dimension since the conduction equation is first order in time (it involves the first derivative of temperature with respect to time). Therefore, we need only 1 initial condition for a two-dimensional problem. 2-38C A heat transfer problem that is symmetric about a plane, line, or point is said to have thermal symmetry about that plane, line, or point. The thermal symmetry boundary condition is a mathematical expression of this thermal symmetry. It is equivalent to insulation or zero heat flux boundary condition, and is expressed at a point x0 as T ( x0 , t ) / x = 0 . 2-39C The boundary condition at a perfectly insulated surface (at x = 0, for example) can be expressed as −k T (0, t ) =0 x or T (0, t ) = 0 which indicates zero heat flux. x 2-40C Yes, the temperature profile in a medium must be perpendicular to an insulated surface since the slope T / x = 0 at that surface. 2-41C We try to avoid the radiation boundary condition in heat transfer analysis because it is a non-linear expression that causes mathematical difficulties while solving the problem; often making it impossible to obtain analytical solutions. 2-42 Heat conduction through the bottom section of an aluminum pan that is used to cook stew on top of an electric range is considered. Assuming variable thermal conductivity and one-dimensional heat transfer, the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem is to be obtained for steady operation. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be variable. 3 There is no heat generation in the medium. 4 The top surface at x = L is subjected to specified temperature and the bottom surface at x = 0 is subjected to uniform heat flux. Analysis The heat flux at the bottom of the pan is E gen Q 0.90  (900 W) q s = s = = = 31,831 W/m 2 2 2 As D / 4  (0.18 m) / 4 Then the differential equation and the boundary conditions for this heat conduction problem can be expressed as d  dT  k =0 dx  dx  −k dT (0) = q s = 31,831 W/m 2 dx T ( L) = T L = 108 C Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-15 2-43 Heat conduction through the bottom section of a steel pan that is used to boil water on top of an electric range is considered. Assuming constant thermal conductivity and one-dimensional heat transfer, the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem is to be obtained for steady operation. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant. 3 There is no heat generation in the medium. 4 The top surface at x = L is subjected to convection and the bottom surface at x = 0 is subjected to uniform heat flux. Analysis The heat flux at the bottom of the pan is E gen Q 0.85  (1250 W) q s = s = = = 33,820 W/m 2 2 As D / 4  (0.20 m) 2 / 4 Then the differential equation and the boundary conditions for this heat conduction problem can be expressed as d 2T =0 dx2 dT (0) −k = q s = 33,280 W/m 2 dx dT ( L) −k = h[T ( L) − T ] dx 2-44 The outer surface of the East wall of a house exchanges heat with both convection and radiation., while the interior surface is subjected to convection only. Assuming the heat transfer through the wall to be steady and one-dimensional, the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem is to be obtained. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant. 3 There is no heat generation in the medium. 4 The outer surface at x = L is subjected to convection and radiation while the inner Tsky surface at x = 0 is subjected to convection only. Analysis Expressing all the temperatures in Kelvin, the differential equation and the boundary conditions for this heat conduction problem can be expressed as 2 d T =0 dx2 −k dT (0) = h1[T1 − T (0)] dx −k dT ( L) 4 = h1 [T ( L) − T 2 ] +  2 T ( L) 4 − Tsky dx   T1 h1 T2 h2 L x Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-16 2-45 Heat is generated in a long wire of radius ro covered with a plastic insulation layer at a constant rate of egen . The heat flux boundary condition at the interface (radius ro) in terms of the heat generated is to be expressed. The total heat generated in the wire and the heat flux at the interface are E = e V = e (r 2 L) gen q s = gen wire Q s = A E gen A gen = o egen (ro2 L) (2ro ) L = egen ro D 2 L egen Assuming steady one-dimensional conduction in the radial direction, the heat flux boundary condition can be expressed as −k dT (ro ) egen ro = dr 2 2-46 A long pipe of inner radius r1, outer radius r2, and thermal conductivity k is considered. The outer surface of the pipe is subjected to convection to a medium at T with a heat transfer coefficient of h. Assuming steady onedimensional conduction in the radial direction, the convection boundary condition on the outer surface of the pipe can be expressed as −k h, T r1 r2 dT (r2 ) = h[T (r2 ) − T ] dr 2-47E A 2-kW resistance heater wire is used for space heating. Assuming constant thermal conductivity and onedimensional heat transfer, the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem is to be obtained for steady operation. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant. 3 Heat is generated uniformly in the wire. 2 kW Analysis The heat flux at the surface of the wire is D = 0.12 in E gen Q 2000 W q s = s = = = 353 .7 W/in 2 L = 15 in As 2ro L 2 (0.06 in)(15 in) Noting that there is thermal symmetry about the center line and there is uniform heat flux at the outer surface, the differential equation and the boundary conditions for this heat conduction problem can be expressed as 1 d  dT  egen =0 r + r dr  dr  k dT (0) =0 dr dT (ro ) −k = q s = 353 .7 W/in 2 dr Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-17 2-48 Water flows through a pipe whose outer surface is wrapped with a thin electric heater that consumes 300 W per m length of the pipe. The exposed surface of the heater is heavily insulated so that the entire heat generated in the heater is transferred to the pipe. Heat is transferred from the inner surface of the pipe to the water by convection. Assuming constant thermal conductivity and one-dimensional heat transfer, the mathematical formulation (the differential equation and the boundary conditions) of the heat conduction in the pipe is to be obtained for steady operation. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant. 3 There is no heat generation in the medium. 4 The outer surface at r = r2 is subjected to uniform heat flux and the inner surface at r = r1 is subjected to convection. Analysis The heat flux at the outer surface of the pipe is Q Q s 300 W q s = s = = = 734 .6 W/m 2 As 2r2 L 2 (0.065 cm)(1 m) Noting that there is thermal symmetry about the center line and there is uniform heat flux at the outer surface, the differential equation and the boundary conditions for this heat conduction problem can be expressed as Q = 300 W h T d  dT  r =0 dr  dr  r1 r2 dT (r1 ) = h[T (ri ) − T ] = 85[T (ri ) − 70 ] dr dT (r2 ) k = q s = 734 .6 W/m 2 dr k 2-49 A spherical container of inner radius r1 , outer radius r2 , and thermal conductivity k is given. The boundary condition on the inner surface of the container for steady one-dimensional conduction is to be expressed for the following cases: (a) Specified temperature of 50C: T (r1 ) = 50C (b) Specified heat flux of 30 W/m2 towards the center: k r1 dT (r1 ) = 30 W/m 2 dr (c) Convection to a medium at T with a heat transfer coefficient of h: k 2-50 A spherical shell of inner radius r1, outer radius r2, and thermal conductivity k is considered. The outer surface of the shell is subjected to radiation to surrounding surfaces at Tsurr . Assuming no convection and steady one-dimensional conduction in the radial direction, the radiation boundary condition on the outer surface of the shell can be expressed as −k  dT (r2 ) 4 =  T (r2 ) 4 − Tsurr dr r2 dT (r1 ) = h[T (r1 ) − T ] dr  k r1 r2 Tsurr  Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-18 2-51 A spherical container consists of two spherical layers A and B that are at perfect contact. The radius of the interface is ro. Assuming transient onedimensional conduction in the radial direction, the boundary conditions at the interface can be expressed as ro T A (ro , t ) = TB (ro , t ) and −kA T A (ro , t ) T (r , t ) = −k B B o r r 2-52 A spherical metal ball that is heated in an oven to a temperature of Ti throughout is dropped into a large body of water at T where it is cooled by convection. Assuming constant thermal conductivity and transient one-dimensional heat transfer, the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem is to be obtained. Assumptions 1 Heat transfer is given to be transient and one-dimensional. 2 Thermal conductivity is given to be constant. 3 There is no heat generation in the medium. 4 The outer surface at r = r0 is subjected to convection. Analysis Noting that there is thermal symmetry about the midpoint and convection at the outer surface, the differential equation and the boundary conditions for this heat conduction problem can be expressed as 1   2 T  1 T r = r   t r 2 r  T (0, t ) =0 r T (ro , t ) −k = h[T (ro ) − T ] r T (r ,0) = Ti k T h r2 Ti 2-53 A spherical metal ball that is heated in an oven to a temperature of Ti throughout is allowed to cool in ambient air at T by convection and radiation. Assuming constant thermal conductivity and transient one-dimensional heat transfer, the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem is to be obtained. Assumptions 1 Heat transfer is given to be transient and one-dimensional. 2 Thermal conductivity is given to be variable. 3 There is no heat generation in the medium. 4 The outer surface at r = ro is subjected to convection and radiation. Analysis Noting that there is thermal symmetry about the midpoint and convection and radiation at the outer surface and expressing all temperatures in Rankine, the differential equation and the boundary conditions for this heat conduction problem can be expressed as  1   2 T  T  kr  = c 2 r r  t r  T (0, t ) =0 r T (ro , t ) 4 −k = h[T (ro ) − T ] + [T (ro ) 4 − Tsurr ] r T (r ,0) = Ti Tsurr k r2 T h Ti Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-19 Solution of Steady One-Dimensional Heat Conduction Problems 2-54C Yes, the temperature in a plane wall with constant thermal conductivity and no heat generation will vary linearly during steady one-dimensional heat conduction even when the wall loses heat by radiation from its surfaces. This is because the steady heat conduction equation in a plane wall is d 2T / dx 2 = 0 whose solution is T ( x) = C1 x + C 2 regardless of the boundary conditions. The solution function represents a straight line whose slope is C1. 2-55C Yes, this claim is reasonable since in the absence of any heat generation the rate of heat transfer through a plain wall in steady operation must be constant. But the value of this constant must be zero since one side of the wall is perfectly insulated. Therefore, there can be no temperature difference between different parts of the wall; that is, the temperature in a plane wall must be uniform in steady operation. 2-56C Yes, this claim is reasonable since no heat is entering the cylinder and thus there can be no heat transfer from the cylinder in steady operation. This condition will be satisfied only when there are no temperature differences within the cylinder and the outer surface temperature of the cylinder is the equal to the temperature of the surrounding medium. 2-57C Yes, in the case of constant thermal conductivity and no heat generation, the temperature in a solid cylindrical rod whose ends are maintained at constant but different temperatures while the side surface is perfectly insulated will vary linearly during steady one-dimensional heat conduction. This is because the steady heat conduction equation in this case is d 2T / dx 2 = 0 whose solution is T ( x) = C1 x + C 2 which represents a straight line whose slope is C1. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-20 2-58 A large plane wall is subjected to specified heat flux and temperature on the left surface and no conditions on the right surface. The mathematical formulation, the variation of temperature in the plate, and the right surface temperature are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since the wall is large relative to its thickness, and the thermal conditions on both sides of the wall are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation in the wall. Properties The thermal conductivity is given to be k =2.5 W/m°C. Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, the mathematical formulation of this problem can be expressed as d 2T =0 dx2 and −k k q=700 W/m2 T1=80°C dT(0) = q 0 = 700 W/m2 dx T (0) = T1 = 80C L=0.3 m (b) Integrating the differential equation twice with respect to x yields dT dx x = C1 T ( x) = C1x + C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give q0 k Heat flux at x = 0: − kC1 = q0 → C1 = − Temperature at x = 0: T (0) = C1  0 + C2 = T1 → C2 = T1 Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T ( x) = − q 0 700 W/m 2 x + T1 = − x + 80 C = −280 x + 80 k 2.5 W/m  C (c) The temperature at x = L (the right surface of the wall) is T (L) = −280  (0.3 m) + 80 = -4C Note that the right surface temperature is lower as expected. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-21 2-59 The base plate of a household iron is subjected to specified heat flux on the left surface and to specified temperature on the right surface. The mathematical formulation, the variation of temperature in the plate, and the inner surface temperature are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since the surface area of the base plate is large relative to its thickness, and the thermal conditions on both sides of the plate are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation in the plate. 4 Heat loss through the upper part of the iron is negligible. Properties The thermal conductivity is given to be k = 20 W/m°C. Analysis (a) Noting that the upper part of the iron is well insulated and thus the entire heat generated in the resistance wires is transferred to the base plate, the heat flux through the inner surface is determined to be Q 800 W q 0 = 0 = = 50,000 W/m 2 − 4 2 Abase 160 10 m Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, the mathematical formulation of this problem can be expressed as and d 2T =0 dx2 dT (0) −k = q 0 = 50,000 W/m 2 dx T ( L) = T2 = 85C (b) Integrating the differential equation twice with respect to x yields dT = C1 dx Q =800 W A=160 cm2 k T2 = 85°C L=0.6 cm T ( x) = C1x + C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give q x = 0: − kC1 = q 0 → C1 = − 0 k q L x = L: T ( L) = C1 L + C 2 = T2 → C 2 = T2 − C1 L → C 2 = T2 + 0 k x Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be q 0 q L q ( L − x) x + T2 + 0 = 0 + T2 k k k (50 ,000 W/m 2 )( 0.006 − x)m = + 85 C 20 W/m  C = 2500 (0.006 − x) + 85 T ( x) = − (c) The temperature at x = 0 (the inner surface of the plate) is T (0) = 2500 (0.006 − 0) + 85 = 100C Note that the inner surface temperature is higher than the exposed surface temperature, as expected. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-22 2-60 A concrete slab with embedded heating cable to provide 1200 W/m 2 of heat to melt snow. Formulate the temperature profile in the concrete slab. Determine the slab thickness so that the temperature difference between the heater surface and the slab surface does not exceed 21°C, as recommended in ASHRAE Handbook to minimize thermal stress. Assumptions1 Heat transfer is steady. 2 One dimensional heat conduction through the concrete slab. 3 The bottom surface at x = 0 is subjected to uniform heat flux from the heating cable. 4 The upper surface at x = L is at a constant temperature of 0°C from the snow melt. 5 There is no heat generation in the concrete slab. 6 Thermal properties are constant. Properties The thermal conductivity of concrete is given as 1.4 W/m·K. Analysis Taking the direction normal to the surface of the concrete slab to be the x direction with x = 0 at the bottom surface (the surface that is in contact with the heater surface), the differential equation for heat conduction can be expressed as ?2? =0 ?? 2 Integrating the differential equation twice with respect to x yields ?? = ?1 ?? ?(?) = ?1 ? + ?2 whereC1 and C2 are arbitrary constants. Applying the boundary conditions yields ? = 0: −? ??(0) = ?̇ 0 = −??1 ?? → ?1 = − ?̇ 0 ? ?̇ 0 ? ? Substituting C1 and C2 into the general solution, the temperature profile in the concrete slab is determined to be ?̇ ? (? − ?) + ?? ?(?) = ? Note that at x = L, T(L) = T2 = 0°C, where the snow melt occurs. The concrete slab thickness such that the temperature difference between the heater surface (T1) and the slab surface (T2) does not exceed 21°C is ? 1.4 W/mK [21 K] = 0.0245 m = ??. ? mm ? = [?1 − ?2 ] = ?̇ 0 1200 W/m2 DiscussionAs the concrete slab thickness increases, the temperature difference between the heater surface and the slab surface will increase. So, 24.5 mm is the maximum thickness for the concrete slab to comply with the recommendation by the 2015 ASHRAE Handbook—HVAC Applications, Chapter 51, for T1 – T2 ≤ 21°C. ? = ?: ?(?) = ?2 = ?1 ? + ?2 → ?2 = ?? − ?1 ? = ?2 + Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-23 2-61 A large plane wall is subjected to specified temperature on the left surface and convection on the right surface. The mathematical formulation, the variation of temperature, and the rate of heat transfer are to be determined for steady onedimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. Properties The thermal conductivity is given to be k = 2.3 W/m°C. Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, the mathematical formulation of this problem can be expressed as d 2T dx 2 =0 and k T (0) = T1 = 90C dT ( L) = h[T ( L) − T ] dx (b) Integrating the differential equation twice with respect to x yields dT = C1 dx −k T1=90°C A=30 m2 L=0.4 m T ( x) = C1x + C2 T =25°C h=24 W/m2.°C x where C1 and C2 are arbitrary constants. Applying the boundary conditions give x = 0: T (0) = C1  0 + C 2 → C 2 = T1 x = L: − kC1 = h[(C1 L + C 2 ) − T ] → C1 = − h(C 2 − T ) h(T1 − T ) → C1 = − k + hL k + hL Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be T ( x) = − =− h(T1 − T ) x + T1 k + hL (24 W/m 2  C)(90 − 25 )C x + 90 C (2.3 W/m  C) + (24 W/m 2  C)( 0.4 m) = 90 − 131 .1x (c) The rate of heat conduction through the wall is h(T − T ) dT Q wall = −kA = −kAC1 = kA 1  dx k + hL (24 W/m 2  C)(90 − 25 )C = (2.3 W/m  C)(30 m 2 ) (2.3 W/m  C) + (24 W/m 2  C)( 0.4 m) = 9045 W Note that under steady conditions the rate of heat conduction through a plain wall is constant. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-24 2-62 A large plane wall is subjected to convection on the inner and outer surfaces. The mathematical formulation, the variation of temperature, and the temperatures at the inner and outer surfaces to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. Properties The thermal conductivity is given to be k = 0.77 W/mK. Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the inner surface, the mathematical formulation of this problem can be expressed as d 2T =0 dx2 The boundary conditions for this problem are: dT (0) h1[T1 − T (0)] = −k dx −k dT ( L) = h2[T ( L) − T 2 ] dx k h2 T2 h1 T1 (b) Integrating the differential equation twice with respect to x yields dT = C1 dx T ( x) = C1 x + C2 L where C1 and C2 are arbitrary constants. Applying the boundary conditions give x = 0: h1[T1 − (C1  0 + C2 )] = −kC1 x = L: −kC1 = h2 [(C1 L + C2 ) − T2 ] Substituting the given values, the above boundary condition equations can be written as 5(27 − C 2 ) = −0.77C1 −0.77C1 = (12)(0.2C1 + C2 − 8) Solving these equations simultaneously give C1 = −45.45 C 2 = 20 Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be T ( x) = 20 − 45.45x (c) The temperatures at the inner and outer surfaces are T (0) = 20 − 45.45  0 = 20C T ( L) = 20 − 45.45  0.2 = 10.9C Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-25 2-63 In this example, the concepts of Prevention through Design (PtD) are applied in conjunction with the solution of steady one-dimensional heat conduction problem. The top surface of the plate is cooled by convection, and temperature at the bottom surface is measured by an IR thermometer. The variation of temperature in the metal plate and the convection heat transfer coefficient necessary to keep the top surface below 47°C are to be determined. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation in the plate. 4 The bottom surface at x = 0 is at constant temperature while the top surface at x = L is subjected to convection. Properties The thermal conductivity of the metal plate is given to be k = 13.5 W/m∙K. Analysis Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the lower surface, the mathematical formulation can be expressed as d 2T =0 dx 2 with boundary conditions T ( 0 ) = T0 dT ( L ) = h [ T ( L ) − T ] dx Integrating the differential equation twice with respect to x yields −k dT = C1 dx T ( x ) = C1 x + C2 whereC1 and C2 are arbitrary constants. Applying the first boundary condition yields T ( 0 ) = C1  0 + C2 → C2 = T0 The application of the second boundary condition gives dT ( L ) = h [ T ( L ) − T ] → −kC1 = h( C1 L + C2 − T ) dx Solving for C2 yields −k C1 = h( T − C2 ) T −T =  0 k + hL (k / h)+ L Now substituting C1 and C2 into the general solution and the variation of temperature is T ( x) = T − T0 x + T0 ( k / h) + L The minimum convection heat transfer coefficient necessary to maintain the top surface below 47°C can be determined from the variation of temperature: T ( L ) = TL = T − T0 L + T0 (k / h)+ L Solving for h gives h= k TL − T0  13 .5 W/m  K  (47 − 60) C = = 413 W/m 2  K  L T − TL  0.025 m  (30 − 47) C Discussion To keep the top surface of the metal plate below 47°C, the convection heat transfer coefficient should be greater than 413 W/m2∙K. A convection heat transfer coefficient value of 413 W/m2∙K is very high for forced convection of gases. The typical values for forced convection of gases are 25–250 W/m2∙K (see Table 1-5 in Chapter 1). To protect workers from thermal burn, appropriate apparel should be worn when operating in an area where hot surfaces are present. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-26 2-64 A series of ASME SA-193 carbon steel bolts of 1 cm thread length are bolted on the upper surface of a metal plate. The upper surface is exposed to convection with the ambient air. The bottom surface is subjected to a uniform heat flux. Formulate the temperature profile in the metal plate, and determine the location in the plate where the temperature begins to exceed 260°C. The compliance of the SA-193 bolts with the ASME Boiler and Pressure Vessel Code (ASME BPVC.IV2015, HF-300) is to be determined. Assumptions1 Heat transfer is steady. 2 One dimensional heat conduction through the metal plate. 3 The bottom surface at x = 0 is subjected to uniform heat flux while the upper surface at x = L is at uniform temperature. 4 There is no heat generation in the plate. 5 Thermal properties are constant. Properties The thermal conductivity of the metal plate is given as 15 W/m·K. Analysis The uniform heat flux on the bottom plate surface (x = 0) is equal to the heat flux transferred by convection on the upper surface (x = L): ?̇ 0 ?̇ 0 = ℎ[?(?) − ?∞ ] → ?(?) = + ?∞ ℎ Taking the direction normal to the surface of the plate to be the x direction with x = 0 at the bottom surface, the differential equation for heat conduction can be expressed as ?2? =0 ?? 2 Integrating the differential equation twice with respect to x yields ?? = ?1 ?? ?(?) = ?1 ? + ?2 whereC1 and C2 are arbitrary constants. Applying the boundary conditions yields ??(0) ?̇ 0 ? = 0: −? = ?̇ 0 = −??1 → ?1 = − ?? ? ?̇ 0 ?̇ 0 ? = ?: ?(?) = ?? = ?1 ? + ?2 → ?2 = −?1 ? + ?? = ? + + ?∞ ? ℎ Substituting C1 and C2 into the general solution, the temperature profile in the metal plate is determined to be ?̇ ? ?̇ ? ?̇ ? ?̇ ? ?̇ ? (? − ?) + ?(?) = − ? + ? + + ?∞ = + ?∞ ? ? ? ? ? Using the temperature profile of the metal plate, the location where the temperature begins to exceed 260°C is ? ? 15 W/mK 15 W/mK (30 − 260)K ?(?) = 260℃: ? = ? + + [?∞ − ?(?)] = 0.05 m + + 2 ℎ ?̇ 0 10 W/m K 2250 W/m2 = 0.0167 m = ?. ?? cm Discussion Since the thread length of the SA-193 bolts is 1 cm; the bolt tips would only reach the location x = 0.04 cm in the metal plate. This is where the bolts are subjected to the highest temperature in the metal plate. At this location, the temperature in the plate can be determined from the temperature profile: 2250 W/m2 2250 W/m2 (0.05 − 0.04)m + ?(0.04) = + 30C = 256.5℃ 15 W/mK 10 W/m2 K Therefore, the SA-193 bolts would comply with the ASME Boiler and Pressure Vessel Code (ASME BPVC.IV-2015, HF300). Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-27 2-65 A plane wall is subjected to uniform heat flux on the left surface, while the right surface is subjected to convection and radiation heat transfer. The variation of temperature in the wall and the left surface temperature are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Temperatures on both sides of the wall are uniform. 3 Thermal conductivity is constant. 4 There is no heat generation in the wall. 5 The surrounding temperature T∞ = Tsurr = 25°C. Properties Emissivity and thermal conductivity are given to be 0.70 and 25 W/m∙K, respectively. Analysis(a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, the mathematical formulation can be expressed as d 2T =0 dx 2 Integrating the differential equation twice with respect to x yields dT = C1 dx T ( x) = C1x + C2 whereC1 and C2 are arbitrary constants. Applying the boundary conditions give q dT (0) x = 0: − k = q0 = −kC1 → C1 = − 0 dx k q x = L : T ( L) = TL = C1L + C2 → C2 = −C1L + TL = 0 L + TL k Substituting C1 and C2 into the general solution, the variation of temperature is determined to be q q q T ( x) = − 0 x + 0 L + TL → T ( x) = 0 ( L − x) + TL k k k (b) The uniform heat flux subjected on the left surface is equal to the sum of heat fluxes transferred by convection and radiation on the right surface: 4 q0 = h(TL − T ) +  (TL4 − Tsurr ) q0 = ( 15 W/m2  K )( 225 − 25 ) K + ( 0.70 )( 5.67  10 −8 W/m2  K 4 )[(225 + 273) 4 − (25 + 273) 4 ] K 4 q0 = 5130 W/m2 (c) The temperature at x = 0 (the left surface of the wall) is q 0 5130 W/m2 ( L − 0) + TL = (0.50 m) + 225 C = 327.6 C k 25 W/m  K Discussion As expected, the left surface temperature is higher than the right surface temperature. The absence of radiative boundary condition may lower the resistance to heat transfer at the right surface of the wall resulting in a temperature drop on the left wall surface by about 40°C. T (0) = Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-28 2-66 A flat-plate solar collector is used to heat water. The top surface (x = 0) is subjected to convection, radiation, and incident solar radiation. The variation of temperature in the solar absorber and the net heat flux absorbed by the solar collector are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation in the plate. 4 The top surface at x = 0 is subjected to convection, radiation, and incident solar radiation. Properties The absorber surface has an absorptivity of 0.9 and an emissivity of 0.9. Analysis Taking the direction normal to the surface of the plate to be the x direction with x = 0 at the top surface, the mathematical formulation can be expressed as d 2T =0 dx 2 Integrating the differential equation twice with respect to x yields dT = C1 dx T ( x) = C1x + C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give q dT (0) → x = 0: −k = q0 = −kC1 C1 = − 0 dx k x = 0: T (0) = T0 = C2 Substituting C1 and C2 into the general solution, the variation of temperature is determined to be q T ( x) = − 0 x + T0 k At the top surface (x = 0), the net heat flux absorbed by the solar collector is 4 q0 = qsolar −  (T04 − Tsurr ) − h(T0 − T ) q0 = (0.9)(500W/m2 ) − (0.9)(5.6710−8 W/m2  K4 )[(35+ 273)4 − (0 + 273)4 )] K4 − (5 W/m2  K)(35 − 25) K q0 = 224 W/m2 Discussion The absorber plate is generally very thin. Thus, the temperature difference between the top and bottom surface temperatures of the plate is miniscule. The net heat flux absorbed by the solar collector increases with the increase in the ambient and surrounding temperatures and thus the use of solar collectors is justified in hot climatic conditions. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-29 2-67 A 20-mm thick draw batch furnace front is subjected to uniform heat flux on the inside surface, while the outside surface is subjected to convection and radiation heat transfer. The inside surface temperature of the furnace front is to be determined. Assumptions 1 Heat conduction is steady. 2 One dimensional heat conduction across the furnace front thickness. 3 Thermal properties are constant. 4 Inside and outside surface temperatures are constant. Properties Emissivity and thermal conductivity are given to be 0.30 and 25 W/m ∙ K, respectively Analysis The uniform heat flux subjected on the inside surface is equal to the sum of heat fluxes transferred by convection and radiation on the outside surface: 4 q 0 = h(TL − T ) +  (TL4 − Tsurr ) 5000 W/m2 = (10 W/m2  K)[TL − (20 + 273 )] K + (0.30)(5.67  10 −8 W/m2  K 4 )[TL4 − (20 + 273 ) 4 ] K 4 Copy the following line and paste on a blank EES screen to solve the above equation: 5000=10*(T_L-(20+273))+0.30*5.67e-8*(T_L^4-(20+273)^4) Solving by EES software, the outside surface temperature of the furnace front is TL = 594 K For steady heat conduction, the Fourier’s law of heat conduction can be expressed as dT dx Knowing that the heat flux and thermal conductivity are constant, integrating the differential equation once with respect to x yields q T ( x) = − 0 x + C1 k Applying the boundary condition gives q q → C1 = 0 L + TL x = L : T ( L) = TL = − 0 L + C1 k k Substituting C1 into the general solution, the variation of temperature in the furnace front is determined to be q 0 = −k q 0 ( L − x ) + TL k The inside surface temperature of the furnace front is T ( x) = T (0) = T0 = q 0 5000 W/m 2 L + TL = (0.020 m) + 594 K = 598 K k 25 W/m  K Discussion By insulating the furnace front, heat loss from the outer surface can be reduced. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-30 2-68E A large plate is subjected to convection, radiation, and specified temperature on the top surface and no conditions on the bottom surface. The mathematical formulation, the variation of temperature in the plate, and the bottom surface temperature are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since the plate is large relative to its thickness, and the thermal conditions on both sides of the plate are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation in the plate. Properties The thermal conductivity and emissivity are given to be k =7.2 Tsky Btu/hft°F and  = 0.7. 75°F x Analysis (a) Taking the direction normal to the surface of the plate to be T  the x direction with x = 0 at the bottom surface, and the mathematical h formulation of this problem can be expressed as and d 2T L =0 dx2 dT ( L) 4 4 −k = h[T ( L) − T ] +  [T ( L) 4 − Tsky ] = h[T2 − T ] +  [(T2 + 460 ) 4 − Tsky ] dx T ( L) = T2 = 75F (b) Integrating the differential equation twice with respect to x yields dT = C1 dx T ( x) = C1x + C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give Convection at x = L: Temperature at x = L: 4 − kC1 = h[T2 − T ] +  [(T2 + 460 ) 4 − Tsky ] 4 → C1 = −{h[T2 − T ] +  [(T2 + 460 ) 4 − Tsky ]} / k T ( L) = C1  L + C 2 = T2 → C 2 = T2 − C1 L Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T ( x) = C1x + (T2 − C1L) = T2 − ( L − x)C1 = T2 + 4 h[T2 − T ] +  [(T2 + 460 ) 4 − Tsky ] ( L − x) k (12 Btu/h  ft 2  F)( 75 − 90 )F + 0.7(0.1714  10 -8 Btu/h  ft 2  R 4 )[( 535 R ) 4 − (480 R) 4 ] = 75 F + (4 / 12 − x) ft 7.2 Btu/h  ft  F = 75 − 20 .2(1 / 3 − x) (c) The temperature at x = 0 (the bottom surface of the plate) is T (0) = 75 − 20.2  (1 / 3 − 0) = 68.3F Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-31 2-69 A series of ASTM B21 naval brass bolts are bolted on the upper surface of a plate. The upper surface is exposed to convection with air and radiation with the surrounding surface. Formulate the temperature profile in the plate, and determine if the bolts comply with the ASME Code for Process Piping. Assumptions1 Heat transfer is steady. 2 One dimensional heat conduction through the plate. 3 The bottom surface at x = 0 is well-insulated while the upper surface at x = L is subjected to convection and radiation. 4 There is no heat generation in the plate. 5 Thermal properties are constant. Properties The emissivity of the plate and bolts is given as 0.3. Analysis Taking the direction normal to the surface of the plate to be the x direction with x = 0 at the bottom surface, the differential equation for heat conduction can be expressed as ?2? =0 ?? 2 Integrating the differential equation twice with respect to x yields ?? = ?1 ?? ?(?) = ?1 ? + ?2 whereC1 and C2 are arbitrary constants. Applying the well- insulated surface boundary condition at the bottom surface (x = 0) yields ??(0) ? = 0: −? = −??1 = 0 → ?1 = 0 ?? Since C1 = 0, the temperature profile in the plate is constant ?(?) = ?2 At the upper surface (x = L) we have ??(?) 4 ] ? = ?: −? = −??1 = ℎ[?(?) − ?∞ ] + ??[?(?)4 − ?surr =0 ?? ? = ?: ?(?) = ?(?) = ?2 Solving for C2 (note that absolute temperatures are used for the calculation), 4 ] ℎ[?2 − ?∞ ] + ??[?24 − ?surr =0 W W (2) ( 2 ) (?2 − 323)(K) + (0.3)(5.67 × 10−8 ) ( 2 4 ) (?24 − 4734 )(K 4 ) = 0 m K m K ?(?) = ?2 = 437 K = ???℃ Discussion The temperature in the plate exceeds the maximum use temperature by 15°C. The use of the ASTM B21 bolts under these conditions does not comply with the ASME Code for Process Piping (ASME B31.3-2014). One way to reduce the plate temperature is by increasing the convection heat transfer coefficient with forced convection. If the convection heat transfer coefficient were to increase to higher than 3.15 W/m 2·K, then the plate temperature would reduce to below 149°C: 4 ?surr − ?(?)4 ℎ > ?? ?(?) − ?∞ W 4734 − 4224 3 (K ) > (0.3)(5.67 × 10−8 ) ( 2 4 ) m K 422 − 323 > 3.15 W/m2 ∙K Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-32 2-70 Chilled water flows in a pipe that is well insulated from outside. The mathematical formulation and the variation of temperature in the pipe are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe. Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as d  dT  r =0 dr  dr  and −k dT (r1 ) = h[T f − T (r1 )] dr dT (r2 ) =0 dr Insulated Water Tf r2 r1 (b) Integrating the differential equation once with respect to r gives r L dT = C1 dr Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, dT C1 = dr r T (r ) = C1 ln r + C 2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give r = r2: r = r1: C1 = 0 → C1 = 0 r2 −k C1 = h[T f − (C1 ln r1 + C 2 )] r1 0 = h(T f − C 2 ) → C 2 = T f Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T (r ) = T f This result is not surprising since steady operating conditions exist. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-33 2-71E A steam pipe is subjected to convection on the inner surface and to specified temperature on the outer surface. The mathematical formulation, the variation of temperature in the pipe, and the rate of heat loss are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe. Properties The thermal conductivity is given to be k = 7.2 Btu/hft°F. Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as d  dT  r =0 dr  dr  and −k T =160F dT (r1 ) = h[T − T (r1 )] dr Steam 250F h=12.5 T (r2 ) = T2 = 160 F (b) Integrating the differential equation once with respect to r gives L = 30 ft dT r = C1 dr Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, dT C1 = dr r T (r ) = C1 ln r + C 2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give C − k 1 = h[T − (C1 ln r1 + C 2 )] r = r1: r1 r = r2: T (r2 ) = C1 ln r2 + C 2 = T2 Solving for C1 and C2 simultaneously gives T − T T − T C1 = 2 and C 2 = T2 − C1 ln r2 = T2 − 2 ln r2 r r k k ln 2 + ln 2 + r1 hr1 r1 hr1 Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T − T r T (r ) = C1 ln r + T2 − C1 ln r2 = C1 (ln r − ln r2 ) + T2 = 2 ln + T2 r2 k r2 ln + r1 hr1 = (160 − 250 )F r r ln + 160 F = −24 .74 ln + 160 F 2.4 7.2 Btu/h  ft  F 2.4 in 2.4 in ln + 2 (12 .5 Btu/h  ft 2  F)( 2 / 12 ft ) (c) The rate of heat conduction through the pipe is C T2 − T dT Q = −kA = −k (2rL ) 1 = −2Lk r k dr r ln 2 + r1 hr1 = −2 (30 ft)( 7.2 Btu/h  ft  F) ln (160 − 250 )F 7.2 Btu/h  ft  F 2.4 + 2 (12 .5 Btu/h  ft 2  F)( 2 / 12 ft ) = 33,600 Btu/h Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-34 2-72 The convection heat transfer coefficient between the surface of a pipe carrying superheated vapor and the surrounding air is to be determined. Assumptions 1 Heat conduction is steady and one-dimensional and there is thermal symmetry about the centerline. 2 Thermal properties are constant. 3 There is no heat generation in the pipe. 4 Heat transfer by radiation is negligible. Properties The constant pressure specific heat of vapor is given to be 2190 J/kg ∙ °C and the pipe thermal conductivity is 17 W/m ∙ °C. Analysis The inner and outer radii of the pipe are r1 = 0.05 m / 2 = 0.025 m r2 = 0.025 m + 0.006 m = 0.031 m The rate of heat loss from the vapor in the pipe can be determined from Q = m c (T − T ) = (0.3 kg/s)( 2190 J/kg  C) (7) C = 4599 W loss p in out For steady one-dimensional heat conduction in cylindrical coordinates, the heat conduction equation can be expressed as d  dT  r =0 dr  dr  and −k Q dT (r1 ) Q loss = = loss dr A 2 r1 L T (r1 ) = 120 C (heat flux at the inner pipe surface) (inner pipe surface temperature) Integrating the differential equation once with respect to r gives dT C1 = dr r Integrating with respect to r again gives T (r ) = C1 ln r + C2 where C1 and C 2 are arbitrary constants. Applying the boundary conditions gives r = r1 : dT (r1 ) C 1 Q loss =− = 1 dr k 2 r1 L r1 1 Q loss ln r1 + C 2 2 kL → C1 = − 1 Q loss 2 kL 1 Q loss ln r1 + T (r1 ) 2 kL Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be r = r1 : T (r1 ) = − → C2 = 1 Q loss 1 Q loss ln r + ln r1 + T (r1 ) 2 kL 2 kL 1 Q loss =− ln( r / r1 ) + T (r1 ) 2 kL T (r ) = − The outer pipe surface temperature is Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-35 1 Q loss ln( r2 / r1 ) + T (r1 ) 2 kL 1 4599 W  0.031  =− ln   + 120 C 2 (17 W/m  C)(10 m)  0.025  = 119 .1 C From Newton’s law of cooling, the rate of heat loss at the outer pipe surface by convection is Q = h(2 r L)T (r ) − T  T (r2 ) = − loss 2 2  Rearranging and the convection heat transfer coefficient is determined to be Q loss 4599 W h= = = 25.1 W/m 2  C 2 r2 L[T (r2 ) − T ] 2 (0.031 m)(10 m)(119 .1 − 25) C Discussion If the pipe wall is thicker, the temperature difference between the inner and outer pipe surfaces will be greater. If the pipe has very high thermal conductivity or the pipe wall thickness is very small, then the temperature difference between the inner and outer pipe surfaces may be negligible. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-36 2-73 A subsea pipeline is transporting liquid hydrocarbon. The temperature variation in the pipeline wall, the inner surface temperature of the pipeline, the mathematical expression for the rate of heat loss from the liquid hydrocarbon, and the heat flux through the outer pipeline surface are to be determined. Assumptions 1 Heat conduction is steady and one-dimensional and there is thermal symmetry about the centerline. 2 Thermal properties are constant. 3 There is no heat generation in the pipeline. Properties The pipeline thermal conductivity is given to be 60 W/m ∙ °C. Analysis The inner and outer radii of the pipeline are r1 = 0.5 m / 2 = 0.25 m r2 = 0.25 m + 0.008 m = 0.258 m (a) For steady one-dimensional heat conduction in cylindrical coordinates, the heat conduction equation can be expressed as d  dT  r =0 dr  dr  and −k dT (r1 ) = h1[T,1 − T (r1 )] dr (convection at the inner pipeline surface) dT (r2 ) = h2 [T (r2 ) − T, 2 ] (convection at the outer pipeline surface) dr Integrating the differential equation once with respect to r gives −k dT C1 = dr r Integrating with respect to r again gives T (r ) = C1 ln r + C2 where C1 and C 2 are arbitrary constants. Applying the boundary conditions gives r = r1 : −k dT(r1 ) C = −k 1 = h1 (T,1 − C1 ln r1 − C 2 ) dr r1 r = r2 : −k dT (r2 ) C = −k 1 = h2 (C1 ln r2 + C 2 − T,2 ) dr r2 C1 and C 2 can be expressed explicitly as C1 = − T,1 − T, 2 k /(r1h1 ) + ln( r2 / r1 ) + k /(r2 h2 ) C 2 = T,1 − T,1 − T,2  k   − ln r1  k /(r1h1 ) + ln( r2 / r1 ) + k /(r2 h2 )  r1h1  Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be T (r ) = − T,1 − T,2  k  + ln( r / r1 ) + T,1  k /(r1h1 ) + ln( r2 / r1 ) + k /(r2 h2 )  r1h1  Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-37 (b) The inner surface temperature of the pipeline is T,1 − T, 2  k  + ln( r1 / r1 ) + T,1  k /(r1h1 ) + ln( r2 / r1 ) + k /(r2 h2 )  r1h1    60 W/m  C (70 − 5) C  2  (0.25 m)( 250 W/m  C)  =− + 70 C 60 W/m  C 60 W/m  C  0.258  + ln  + (0.25 m)( 250 W/m 2  C)  0.25  (0.258 m)(150 W/m 2  C) = 45.5 C (c) The mathematical expression for the rate of heat loss through the pipeline can be determined from Fourier’s law to be T (r1 ) = − dT Q loss = −kA dr dT (r2 ) = −2LkC1 dr T,1 − T, 2 = ln( r2 / r1 ) 1 1 + + 2 r1 Lh1 2Lk 2 r2 Lh 2 = −k (2 r2 L) (d) Again from Fourier’s law, the heat flux through the outer pipeline surface is q 2 = −k = = dT (r2 ) C dT = −k = −k 1 dr dr r2 T,1 − T, 2 k k /(r1h1 ) + ln( r2 / r1 ) + k /(r2 h2 ) r2 (70 − 5) C  60 W/m  C    60 W/m  C 60 W/m  C  0.258   0.258 m  + ln +   (0.25 m)( 250 W/m 2  C)  0.25  (0.258 m)(150 W/m 2  C) = 5947 W/m 2 Discussion Knowledge of the inner pipeline surface temperature can be used to control wax deposition blockages in the pipeline. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-38 2-74 Liquid ethanol is being transported in a pipe where the outer surface is subjected to heat flux. Convection heat transfer occurs on the inner surface of the pipe. The variation of temperature in the pipe wall and the inner and outer surface temperatures are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation in the wall. 4 The inner surface at r = r1 is subjected to convection while the outer surface at r = r2 is subjected to uniform heat flux. Properties Thermal conductivity is given to be 15 W/m∙K. Analysis For one-dimensional heat transfer in the radial r direction, the differential equation for heat conduction in cylindrical coordinate can be expressed as d  dT  r =0 dr  dr  Integrating the differential equation twice with respect to r yields r dT = C1 dr or dT C1 = dr r T (r ) = C1 ln r + C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give r = r2 : −k dT (r2 ) C = − q s = − k 1 dr r2 → r C1 = q s 2 k r = r1 : −k dT (r1 ) = h[T − T (r1 )] dr → k C1 = h (C1 ln r1 + C2 − T ) r1 Solving for C2 gives k 1  C2 = C1 − ln r1  + T h r  1  Substituting C1 and C2 into the general solution, the variation of temperature is determined to be k 1  T (r ) = C1 ln r + C2 = C1 ln r + C1 − ln r1  + T  h r1  → r k 1 r T (r ) = q s 2  + ln  + T k  h r1 r1  The temperature at r = r1 = 0.015 m (the inner surface of the pipe) is T ( r1 ) = q s r2 1000 W/m2  0.018 m  + T =   + 10C = 34.0C h r1 50 W/m2  K  0.015 m  T (r1) = 34.0C The temperature at r = r2 = 0.018 m (the outer surface of the pipe) is r k 1 r  0.018 m  15 W/m  K  1 0.018   T (r2 ) = q s 2  + ln 2  + T = (1000 W/m2 ) + ln   + 10 C k  h r1 r1  15 W/m  K  50 W/m2  K  0.015 m 0.015  T (r2 ) = 34.2C Both the inner and outer surfaces of the pipe are at higher temperatures than the flashpoint of ethanol (16.6°C). Discussion The outer surface of the pipe should be wrapped with protective insulation to keep the heat input from heating the ethanol inside the pipe. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-39 2-75 Liquid water flows in a tube with the inner surface lined with PVDC lining. The tube outer surface is subjected to a known uniform heat flux. The tube inner diameter, the tube wall thickness, the water temperature, and the convection heat transfer coefficient are known. Formulate the temperature profile in the tube wall, and determine if the PVDC lining is in compliance with the ASME Code for Process Piping. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation in the tube wall. 4 The inner surface at r = r1 is subjected to convection while the outer surface at r = r2 is subjected to uniform heat flux. 5 The PVDC lining is very thin and the temperature gradient in the lining is negligible. Properties Thermal conductivity of the tube wall is given to be 15 W/m∙K. AnalysisFor one-dimensional heat transfer in the radial r direction, the differential equation for heat conduction in cylindrical coordinate can be expressed as ? ?? (? ) = 0 ?? ?? Integrating the differential equation twice with respect to r yields ?? ?? ?1 ? = ?1 or = ?? ?? ? ?(?) = ?1 ln ? + ?2 whereC1 and C2 are arbitrary constants. Applying the boundary conditions yields ??(?1 ) ?1 ? = ?1 : −? = ℎ[?∞ − ?(?1 )] → ? = ℎ(?1 ln ?1 + ?2 − ?∞ ) ?? ?1 ??(?2 ) ?1 ?2 ? = ?2 : −? = −? = −?̇ ? → ?1 = ?̇ ? ?? ?2 ? Solving for C2 yields 1? ?2 = ?1 ( − ln ?1 ) + ?∞ ?1 ℎ Substituting C1 and C2 into the general solution, the temperature profile in the tube wall is determined to be 1? ?? ? ?? ?(?) = ?1 ln ? + ?2 = ?1 ln ? + ?1 ( − ln ?1 ) + ?∞ → ?(?) = ?̇ ? (?? + ) + ?∞ ?1 ℎ ? ?? ?? ? At the tube inner surface (r = r1 = D1/2 = 0.012 m), which is lined with PVDC lining, the temperature is W (15 ) ?2 ?1 1 ? W (0.017 m) 1 m ∙ K ) + 20C ?(?1 ) = ?̇ ? (ln + ) + ?∞ = (2300 2 ) ( × ? ?1 ?1 ℎ m (15 W ) 0.012 m 50 W m∙K m2 ∙ K = ??. ?℃ > 79℃ where and r2 = r1 + wall thickness = 0.012 m + 0.005 m = 0.017 m. Discussion The tube inner surface temperature at r = r1 =0.012 m is more than 6°C higher than the recommended maximum temperature of 79°C by the ASME Code for Process Piping (ASME B31.3-2014, A323) for PVDC. Therefore, having the tube inner surface lined with PVDC lining is not in compliance with the code. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-40 2-76 Liquid water flows in a tube with the inner surface lined with PTFE lining. The tube inner surface is subjected to convection with water, the tube outer surface is subjected to convection with superheated steam. Formulate the temperature profile in the tube wall, and determine if the PTFE lining is in compliance with the ASME Code for Process Piping. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation in the tube wall. 4 The inner surface at r = r1 is subjected to convection with water. 5 The outer surface at r = r2 is subjected to convection with superheated steam. 6 The PTFE lining is very thin and the temperature gradient in the lining is negligible. Properties Thermal conductivity of the tube wall is given to be 15 W/m∙K. AnalysisFor one-dimensional heat transfer in the radial r direction, the differential equation for heat conduction in cylindrical coordinate can be expressed as ? ?? (? ) = 0 ?? ?? Integrating the differential equation twice with respect to r yields ? ?? ?? ?1 = ?1 or = ?? ?? ? ?(?) = ?1 ln ? + ?2 whereC1 and C2 are arbitrary constants. Applying the boundary conditions yields ??(?1 ) ?1 = −? = ℎ1 [?∞,1 − ?(?1 )] ?? ?1 ??(?2 ) ?1 −? = −? = ℎ2 [?(?2 ) − ?∞,2 ] ?? ?2 ? = ?1 : −? ? = ?2 : C1 and C2 can be expressed as ?1 = − ?∞,1 − ?∞,2 ? ? ? + ln 2 + ?1 ℎ1 ?1 ?2 ℎ2 ?∞,1 − ?∞,2 ? ( − ln ?1 ) ? ? ? ?1 ℎ1 + ln 2 + ?1 ℎ1 ?1 ?2 ℎ2 Substituting C1 and C2 into the general solution, the temperature profile in the tube wall is determined to be ?2 = ?∞,1 − ?(?) = − ?∞,? − ?∞,? ? ? ( + ?? ) + ?∞,? ? ?? ? ?? ?? ? ? + ?? + ?? ?? ?? ?? ?? At the tube inner surface (r = r1 = 0.012 m), which is lined with PTFE lining, the temperature is Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-41 ?(?1 ) = − ?(?1 ) = − ?∞,1 − ?∞,2 ? ?1 ( + ln ) + ?∞,1 ? ? ? ?1 ℎ1 ?1 + ln 2 + ?1 ℎ1 ?1 ?2 ℎ2 (50 − 600)C 15 W mK [ ] + 50C W W W (0.012 m) (50 2 ) 15 0.017 m mK mK m K W + ln 0.012 m + W (0.012 m) (50 2 ) (0.017 m) (30 2 ) m K m K 15 = ???℃ > 260℃ whereand r2 = r1 + wall thickness = 0.012 m + 0.005 m = 0.017 m. Discussion The tube inner surface temperature at r = r1 =0.012 m is 41°C higher than the recommended maximum temperature of 260°C by the ASME Code for Process Piping (ASME B31.3-2014, A323) for PTFE. Therefore, having the tube inner surface lined with PTFE lining is not in compliance with the code. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-42 2-77 A spherical container is subjected to uniform heat flux on the inner surface, while the outer surface maintains a constant temperature. The variation of temperature in the container wall and the inner surface temperature are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Temperatures on both surfaces are uniform. 3 Thermal conductivity is constant. 4 There is no heat generation in the wall. 5 The inner surface at r = r1 is subjected to uniform heat flux while the outer surface at r = r2 is at constant temperature T2. Properties Thermal conductivity is given to be k = 1.5 W/m∙K. Analysis For one-dimensional heat transfer in the radial direction, the differential equation for heat conduction in spherical coordinate can be expressed as d  2 dT  r =0 dr  dr  Integrating the differential equation twice with respect to r yields r2 dT = C1 dr or dT C1 = dr r 2 C1 + C2 r where C1 and C2 are arbitrary constants. Applying the boundary conditions give T (r ) = − dT (r1 ) C = q1 = −k 21 dr r1 → r2 C1 = −q1 1 k C1 + C2 r2 → C2 = T2 + r = r1 : −k r = r2 : T (r2 ) = T2 = − C1 q r 2 = T2 − 1 1 r2 k r2 Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T (r ) = q1 r12 q r 2 + T2 − 1 1 k r k r2 → r2  1 1  T ( r ) = q1 1  −  + T2 k  r r2  The temperature at r = r1 = 1 m (the inner surface of the container) is r2  1 1 T (r1 ) = T1 = q1 1  −  + T2 k  r r2  T1 = (7000 W/m2 ) (1 m) 2 1   1 −   + 25 C = 247C (1.5 W/m  K)  1 m 1.05 m  Discussion As expected the inner surface temperature is higher than the outer surface temperature. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-43 2-78 A spherical shell is subjected to uniform heat flux on the inner surface, while the outer surface is subjected to convection heat transfer. The variation of temperature in the shell wall and the outer surface temperature are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation in the wall. 4 The inner surface at r = r1 is subjected to uniform heat flux while the outer surface at r = r2 is subjected to convection. Analysis For one-dimensional heat transfer in the radial r direction, the differential equation for heat conduction in spherical coordinate can be expressed as d  2 dT  r =0 dr  dr  Integrating the differential equation twice with respect to r yields r2 dT = C1 dr or dT C1 = dr r 2 C1 + C2 r where C1 and C2 are arbitrary constants. Applying the boundary conditions give T (r ) = − r = r1 : −k dT (r1 ) C = q1 = −k 21 dr r1 r = r2 : −k dT (r2 ) = h[T (r2 ) − T ] dr → → r2 C1 = −q1 1 k −k  C  C1 = h − 1 + C2 − T  2 r2  r2  Solving for C2 gives r 2  k 1 1  C2 = q1 1  − + T k  h r22 r2  Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T (r ) = − C1 r2 1 r 2  k 1 1  + C2 = q1 1 + q1 1  − + T r k r k  h r22 r2  → r2  1 k 1 1  T ( r ) = q1 1  + −  + T k  r h r22 r2  The temperature at r = r2 (the outer surface of the shell) can be expressed as 2 q  r  T (r2 ) = 1  1  + T h  r2  Discussion Increasing the convection heat transfer coefficient h would decrease the outer surface temperature T(r2). Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-44 2-79 A spherical container is subjected to specified temperature on the inner surface and convection on the outer surface. The mathematical formulation, the variation of temperature, and the rate of heat transfer are to be determined for steady onedimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal symmetry about the midpoint. 2 Thermal conductivity is constant. 3 There is no heat generation. Properties The thermal conductivity is given to be k = 30 W/m°C. Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as d  2 dT  r =0 dr  dr  and T1 k T (r1 ) = T1 = 0C r1 r2 T h dT (r2 ) = h[T (r2 ) − T ] dr (b) Integrating the differential equation once with respect to r gives dT r2 = C1 dr Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, dT C1 = dr r 2 −k C1 + C2 r where C1 and C2 are arbitrary constants. Applying the boundary conditions give C T (r1 ) = − 1 + C 2 = T1 r = r1: r1 T (r ) = − r = r2: −k  C  C1 = h − 1 + C2 − T  2 r2  r2  Solving for C1 and C2 simultaneously gives r (T − T ) C T1 − T r2 C1 = 2 1  and C 2 = T1 + 1 = T1 + r r k k r1 r1 1− 2 − 1− 2 − r1 hr2 r1 hr2 Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T (r ) = − =  1 1 C1 C + T1 + 1 = C1  −  + T1 = r r1  r1 r  T1 − T  r2 r2   − +T r2 k  r1 r  1 1− − r1 hr2 (0 − 25)C  2.1 2.1  −   + 0C = 29 .63(1.05 − 2.1 / r ) 2.1 30 W/m  C 2 r   1− − 2 (18 W/m 2  C)( 2.1 m) (c) The rate of heat conduction through the wall is dT C r (T − T ) Q = −kA = −k (4r 2 ) 21 = −4kC1 = −4k 2 1  r k dr r 1− 2 − r1 hr2 = −4 (30 W/m  C) (2.1 m) (0 − 25 )C = 23,460 W 2.1 30 W/m  C 1− − 2 (18 W/m2  C)( 2.1 m) Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-45 2-80 A spherical container is used for storing chemicals undergoing exothermic reaction that provides a uniform heat flux to its inner surface. The outer surface is subjected to convection heat transfer. The variation of temperature in the container wall and the inner and outer surface temperatures are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation in the wall. 4 The inner surface at r = r1 is subjected to uniform heat flux while the outer surface at r = r2 is subjected to convection. Properties Thermal conductivity is given to be 15 W/m∙K. Analysis For one-dimensional heat transfer in the radial r direction, the differential equation for heat conduction in spherical coordinate can be expressed as d  2 dT  r =0 dr  dr  Integrating the differential equation twice with respect to r yields dT C1 dT = r2 = C1 or dr r 2 dr C T (r ) = − 1 + C2 r where C1 and C2 are arbitrary constants. Applying the boundary conditions give r = r1 : −k dT (r1 ) C = q1 = −k 21 dr r1 r = r2 : −k dT (r2 ) = h[T (r2 ) − T ] dr → → r2 C1 = −q1 1 k −k  C  C1 = h − 1 + C2 − T  2 r r2  2  Solving for C2 gives r 2  k 1 1  C2 = q1 1  − + T k  h r22 r2  Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T (r ) = − C1 r2 1 r 2  k 1 1  + C2 = q1 1 + q1 1  − + T r k r k  h r22 r2  → r 2  k 1 1 1  T (r ) = q1 1  + − + T k  h r22 r r2  The temperature at r = r1 = 0.5 m (the inner surface of the container) is r 2  k 1 1 1  T (r1 ) = q1 1  + − + T k  h r22 r1 r2  T (r1 ) = (60000 W/m2 )  (0.5 m) 2  15 W/m  K  1 1 1 + −    + 23 C (15 W/m  K)  1000 W/m2  K  (0.55 m)2 (0.5 m) (0.55 m)  T (r1 ) = 254.4C The temperature at r = r2 = 0.55 m (the outer surface of the container) is 2 T (r2 ) = 2 q1  r1  60000 W/m 2  0.5 m    + T = + 23C = 72.6C   h  r2  1000 W/m 2  K  0.55 m  The outer surface temperature of the container is above the safe temperature of 50°C. Discussion To prevent thermal burn, the container’s outer surface should be covered with insulation. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-46 2-81 A spherical container is subjected to uniform heat flux on the outer surface and specified temperature on the inner surface. The mathematical formulation, the variation of temperature in the pipe, and the outer surface temperature, and the maximum rate of hot water supply are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal symmetry about the mid point. 2 Thermal conductivity is constant. 3 There is no heat generation in the container. Properties The thermal conductivity is given to be k = 1.5 W/m°C. The specific heat of water at the average temperature of (100+20)/2 = 60C is 4.185 kJ/kgC (Table A-9). Analysis (a) Noting that the 90% of the 500 W generated by the strip heater is transferred to the container, the heat flux through the outer surface is determined to be Q Q 0.90  500 W q s = s = s 2 = = 213 .0 W/m 2 2 A2 4r2 4 (0.41 m) Noting that heat transfer is one-dimensional in the radial r direction and heat flux is in the negative r direction, the mathematical formulation of this problem can be expressed as d  2 dT  r =0 dr  dr  and Insulation T (r1 ) = T1 = 100 C k T1 k dT (r2 ) = q s dr Heater r1 r2 r (b) Integrating the differential equation once with respect to r gives r2 dT = C1 dr Dividing both sides of the equation above by r2 and then integrating, dT C1 = dr r 2 C1 + C2 r where C1 and C2 are arbitrary constants. Applying the boundary conditions give T (r ) = − r = r2: k C1 = q s → C1 = 2 r2 r = r1: T (r1 ) = T1 = − q s r22 k q r 2 C1 C + C 2 → C 2 = T1 + 1 = T1 + s 2 r1 r1 kr1 Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T (r ) = −  1 1  1 1  q r 2 C1 C C + C 2 = − 1 + T1 + 1 = T1 +  − C1 = T1 +  −  s 2 r r r1  r1 r   r1 r  k 1  (213 W/m 2 )( 0.41 m) 2 1  1  = 100 C +  −  = 100 + 23 .87  2.5 −  1.5 W/m  C r  0.40 m r   (c) The outer surface temperature is determined by direct substitution to be  1 1   Outer surface (r = r2): T (r2 ) = 100 + 23 .87  2.5 −  = 100 + 23 .87  2.5 −  = 101.5C r2  0.41    Noting that the maximum rate of heat supply to the water is 0.9  500 W = 450 W, water can be heated from 20 to 100C at a rate of Q 0.450 kJ/s Q = m c p T → m = = = 0.00134 kg/s = 4.84 kg/h c p T (4.185 kJ/kg  C)(100 − 20) C Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-47 Heat Generation in a Solid 2-82C Heat generation in a solid is simply conversion of some form of energy into sensible heat energy. Some examples of heat generations are resistance heating in wires, exothermic chemical reactions in a solid, and nuclear reactions in nuclear fuel rods. 2-83C No. Heat generation in a solid is simply the conversion of some form of energy into sensible heat energy. For example resistance heating in wires is conversion of electrical energy to heat. 2-84C The cylinder will have a higher center temperature since the cylinder has less surface area to lose heat from per unit volume than the sphere. 2-85C The rate of heat generation inside an iron becomes equal to the rate of heat loss from the iron when steady operating conditions are reached and the temperature of the iron stabilizes. 2-86C No, it is not possible since the highest temperature in the plate will occur at its center, and heat cannot flow “uphill.” 2-87 Heat is generated uniformly in a large brass plate. One side of the plate is insulated while the other side is subjected to convection. The location and values of the highest and the lowest temperatures in the plate are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since the plate is large relative to its thickness, and there is thermal symmetry about the center plane 3 Thermal conductivity is constant. 4 Heat generation is uniform. Properties The thermal conductivity is given to be k =111 W/m°C. Analysis This insulated plate whose thickness is L is equivalent to one-half of an uninsulated plate whose thickness is 2L since the midplane of the uninsulated plate can be treated as insulated surface. The highest temperature will occur at the insulated surface while the lowest temperature will occur at the surface which is exposed to the environment. Note that L in the following relations is the full thickness of the given plate since the insulated side represents the center surface of a plate whose thickness is doubled. The desired values are determined directly from T s = T + To = T s + e gen L h egen L2 2k = 25 C + (2  10 5 W/m 3 )( 0.05 m) = 252 .3C + 44 W/m 2  C k egen Insulated L=5 cm T =25°C h=44 W/m2.°C = 252.3C (2  10 5 W/m 3 )( 0.05 m) 2 = 254.6C 2(111 W/m  C) Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-48 2-88 Prob. 2-87 is reconsidered. The effect of the heat transfer coefficient on the highest and lowest temperatures in the plate is to be investigated. Analysis The problem is solved using EES, and the solution is given below. “GIVEN” L=0.05 [m] k=111 [W/m-C] g_dot=2E5 [W/m^3] T_infinity=25 [C] h=44 [W/m^2-C] “ANALYSIS” T_min=T_infinity+(g_dot*L)/h T_max=T_min+(g_dot*L^2)/(2*k) Tmin [C] 525 425 358.3 310.7 275 247.2 225 206.8 191.7 178.8 167.9 158.3 150 142.6 136.1 130.3 125 550 Tmax [C] 527.3 427.3 360.6 313 277.3 249.5 227.3 209.1 193.9 181.1 170.1 160.6 152.3 144.9 138.4 132.5 127.3 500 450 Tmin [C] h [W/m2.C] 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 400 350 300 250 200 150 100 20 30 40 50 60 70 80 90 100 2 h [W/m -C] 550 500 450 Tmax [C] 400 350 300 250 200 150 100 20 30 40 50 60 70 80 90 100 2 h [W/m -C] Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-49 2-89 Both sides of a large stainless steel plate in which heat is generated uniformly are exposed to convection with the environment. The location and values of the highest and the lowest temperatures in the plate are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since the plate is large relative to its thickness, and there is thermal symmetry about the center plane 3 Thermal conductivity is constant. 4 Heat generation is uniform. Properties The thermal conductivity is given to be k =15.1 W/m°C. Analysis The lowest temperature will occur at surfaces of plate while the highest temperature will occur at the midplane. Their values are determined directly from T s = T + To = Ts + e gen L h egen L2 2k = 30 C + (5  10 5 W/m 3 )( 0.015 m) = 155 C + 60 W/m  C 2 = 155C k egen T =30°C h=60 W/m2°C T =30°C h=60 W/m2.°C 2L=3 cm (5  10 5 W/m 3 )( 0.015 m) 2 = 158.7C 2(15 .1 W/m  C) Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-50 2-90 Heat is generated in a large plane wall whose one side is insulated while the other side is subjected to convection. The mathematical formulation, the variation of temperature in the wall, the relation for the surface temperature, and the relation for the maximum temperature rise in the plate are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since the wall is large relative to its thickness. 3 Thermal conductivity is constant. 4 Heat generation is uniform. Analysis (a) Noting that heat transfer is steady and one-dimensional in x direction, the mathematical formulation of this problem can be expressed as and d 2T e gen dx k + 2 =0 dT (0) = 0 (insulated surface at x = 0) dx −k k egen dT ( L) = h[T ( L) − T ] dx T h Insulated (b) Rearranging the differential equation and integrating, d 2T egen dx k =− 2 egen dT =− x + C1 dx k → L x Integrating one more time, T ( x) = − egen x 2 (1) + C1 x + C 2 2k Applying the boundary conditions: −egen dT (0) B.C. at x = 0: =0→ (0) + C1 = 0 → C1 = 0 dx k  − egen L2   − e gen  − k  L  = h + C 2 − T   2k   k    B. C. at x = L: e gen L = C2 = Dividing by h: egen L h + − he gen L2 2k egen L2 2k − hT + C 2 → C 2 = e gen L + hegen L2 2k + hT + T Substituting the C1 and C2 relations into Eq. (1) and rearranging give T ( x) = − egen x 2 egen L T ( L) = e gen L2 2k e gen L h + egen L h + egen L2 + T = egen ( L2 − x 2 ) + egen L + T 2k h 2k 2k h which is the desired solution for the temperature distribution in the wall as a function of x. (c) The temperatures at two surfaces and the temperature difference between these surfaces are T (0) = + + T + T Tmax = T (0) − T ( L) = egen L2 2k Discussion These relations are obtained without using differential equations in the text (see Eqs. 2-67 and 2-73). Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-51 2-91E Heat is generated in a large plane wall whose one side is insulated while the other side is maintained at a specified temperature. The mathematical formulation, the variation of temperature in the wall, and the highest temperature in the wall are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional, and there is thermal symmetry about the center plane. 3 Thermal conductivity is constant. 4 Heat generation varies with location in the x direction. Properties The thermal conductivity is given to be k = 5 Btu/h·ft·ºF. Analysis (a) Noting that heat transfer is steady and one-dimensional in x direction, the mathematical formulation of this problem can be expressed as d 2 T e gen ( x) + =0 k dx 2 where egen = ax 2 d 2T dx 2 =− e gen ( x) k =− a 2 x k The boundary conditions for this problem are: T (0) = T0 (specified surface temperature at x = 0) dT ( L) = 0 (insulated surface at x = L) dx (b) Rearranging the differential equation and integrating, d 2T a dT 1a 3 = − x2 → =− x + C1 k dx 3k dx 2 Integrating one more time, 1 a 4 x + C1 x + C2 12 k Applying the boundary conditions: B.C. at x = 0: T (0) = T0 = C 2 T ( x) = − (1) dT ( L) 1a 3 aL3 =− L + C1 = 0 → C1 = dx 3k 3k Substituting the C1 and C2 relations into Eq. (1) and rearranging gives B.C. at x = L: 1 a 4 aL3 (2) x + x + T0 12 k 3k (c) The highest (maximum) temperature occurs at the insulate surface (x = L) and is determined by substituting the given quantities into Eq. (2), the result is T ( x) = − 1 a 4 aL3 aL4 L + L + T0 = + T0 12 k 3k 4k (1200 Btu/h  ft 5 ) (1 ft 4 ) = + 700 F 4 (5 Btu/h  ft  F) = 760 F T ( L) = Tmax = − Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-52 2-92 Heat is generated in a large plane wall whose one side is insulated while the other side is maintained at a specified temperature. The mathematical formulation, the variation of temperature in the wall, and the temperature of the insulated surface are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since the wall is large relative to its thickness, and there is thermal symmetry about the center plane. 3 Thermal conductivity is constant. 4 Heat generation varies with location in the x direction. Properties The thermal conductivity is given to be k = 30 W/m°C. Analysis (a) Noting that heat transfer is steady and one-dimensional in x direction, the mathematical formulation of this problem can be expressed as d 2T e gen ( x) dx k + 2 k egen =0 where egen = e0 e −0.5 x / L and dT (0) = 0 (insulated surface at x = 0) dx and e 0 = 8106 W/m3 Insulated T ( L) = T2 = 30C (specified surface temperature) T2 =30°C L x (b) Rearranging the differential equation and integrating, d 2T dx =− 2 e0 −0.5 x / L e → k e e −0.5 x / L dT =− 0 + C1 → dx k − 0.5 / L dT 2e0 L −0.5 x / L = e + C1 dx k Integrating one more time, T ( x) = 2e0 L e −0.5 x / L 4e L2 + C1 x + C 2 → T ( x) = − 0 e −0.5 x / L + C1 x + C 2 k − 0.5 / L k (1) Applying the boundary conditions: 2e L 2e L dT (0) 2e0 L −0.50 / L = e + C1 → 0 = 0 + C1 → C1 = − 0 B.C. at x = 0: dx k k k B. C. at x = L: T ( L ) = T2 = − 4e0 L2 −0.5 L / L 4e L2 2e L2 e + C1 L + C 2 → C 2 = T2 + 0 e −0.5 + 0 k k k Substituting the C1 and C2 relations into Eq. (1) and rearranging give T ( x ) = T2 + e0 L2 [4(e −0.5 − e −0.5 x / L ) + 2(1 − x / L)] k which is the desired solution for the temperature distribution in the wall as a function of x. (c) The temperature at the insulate surface (x = 0) is determined by substituting the known quantities to be e0 L2 [4(e −0.5 − e 0 ) + (2 − 0 / L)] k (8  10 6 W/m 3 )( 0.05 m) 2 = 30 C + [4(e −0.5 − 1) + (2 − 0)] (30 W/m  C) = 314C Therefore, there is a temperature difference of almost 300°C between the two sides of the plate. T (0) = T2 + Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-53 2-93 Prob. 2-92 is reconsidered. The heat generation as a function of the distance is to be plotted. Analysis The problem is solved using EES, and the solution is given below. “GIVEN” L=0.05 [m] T_s=30 [C] k=30 [W/m-C] e_dot_0=8E6 [W/m^3] “ANALYSIS” e_dot=e_dot_0*exp((-0.5*x)/L) “Heat generation as a function of x” e [W/m3] 8.000E+06 7.610E+06 7.239E+06 6.886E+06 6.550E+06 6.230E+06 5.927E+06 5.638E+06 5.363E+06 5.101E+06 4.852E+06 8.000×106 7.500×106 7.000×106 e [W/m3] x [m] 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 6.500×106 6.000×106 5.500×106 5.000×106 4.500×106 0 0.01 0.02 0.03 0.04 0.05 x [m] 2-94 A nuclear fuel rod with a specified surface temperature is used as the fuel in a nuclear reactor. The center temperature of the rod is to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no change in the axial direction. 220°C 3 Thermal conductivity is constant. 4 Heat generation in the rod is uniform. Properties The thermal conductivity is given to be k = 29.5 W/m°C. Analysis The center temperature of the rod is determined from To = Ts + egen ro2 4k = 220 C + (4 10 7 W/m 3 )( 0.005 m) 2 = 228C 4(29 .5 W/m. C) egen Uranium rod Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-54 2-95E Heat is generated uniformly in a resistance heater wire. The temperature difference between the center and the surface of the wire is to be determined. Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat r generation in the heater is uniform. Ts Properties The thermal conductivity is given to be k = 5.8 Btu/hft°F. Analysis The resistance heater converts electric energy into heat at a rate of 3 kW. The rate of heat generation per unit length of the wire is E gen E gen (3  3412 .14 Btu/h) e gen = = 2 = = 2.933  10 8 Btu/h  ft 3 V wire ro L  (0.04 / 12 ft) 2 (1 ft) Then the temperature difference between the centerline and the surface becomes Tmax = egen ro2 4k = ro 0 Heater (2.933  10 8 Btu/h  ft 3 )( 0.04 / 12 ft) 2 = 140.5F 4(5.8 Btu/h  ft  F) 2-96 A 2-kW resistance heater wire with a specified surface temperature is used to boil water. The center temperature of the wire is to be determined. Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the heater is uniform. 110C Properties The thermal conductivity is given to be k = 20 W/m°C. Analysis The resistance heater converts electric energy into heat at a rate of 2 kW. The rate of heat generation per unit volume of the wire is r E gen E gen 2000 W 8 3 egen = = = = 1.768  10 W/m D V wire ro2 L  (0.002 m) 2 (0.9 m) The center temperature of the wire is then determined from Eq. 2-71 to be To = T s + egen ro2 4k = 110 C + (1.768 10 8 W/m 3 )( 0.002 m) 2 = 118.8C 4(20 W/m. C) Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-55 2-97 Heat is generated in a long solid cylinder with a specified surface temperature. The variation of temperature in the cylinder is given by 2 egen ro2   r    T (r ) = 1 −    + Ts k   ro     (a) Heat conduction is steady since there is no time t variable involved. (b) Heat conduction is a one-dimensional. (c) Using Eq. (1), the heat flux on the surface of the cylinder at r = ro is determined from its definition to be 80C k egen  e gen ro2  2r  dT (ro ) −  q s = −k = −k  dr  k  ro2  r = r0 r D  e gen ro2  2r   − o  = 2e gen ro = 2(35 W/cm 3 )( 4 cm) = 280 W/cm 2 = −k   k  ro2  2-98 Prob. 2-97 is reconsidered. The temperature as a function of the radius is to be plotted. Analysis The problem is solved using EES, and the solution is given below. “GIVEN” r_0=0.04 [m] k=25 [W/m-C] e_dot_gen=35E+6 [W/m^3] T_s=80 [C] “ANALYSIS” T=(e_dot_gen*r_0^2)/k*(1-(r/r_0)^2)+T_s “Variation of temperature” T [C] 2320 2292 2209 2071 1878 1629 1324 964.9 550.1 80 2500 2000 1500 T [C] r [m] 0 0.004444 0.008889 0.01333 0.01778 0.02222 0.02667 0.03111 0.03556 0.04 1000 500 0 0 0.01 0.02 0.03 0.04 r [m] Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-56 2-99 A cylindrical nuclear fuel rod is cooled by water flowing through its encased concentric tube. The average temperature of the cooling water is to be determined. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat generation in the fuel rod is uniform. Properties The thermal conductivity is given to be 30 W/m ∙ °C. Analysis The rate of heat transfer by convection at the fuel rod surface is equal to that of the concentric tube surface: h1 As ,1 (Ts ,rod − T ) = h2 As , 2 (T − Ts , tube ) h1 (2 r1 L)(Ts ,rod − T ) = h2 (2 r2 L)(T − Ts ,tube ) Ts,rod = h2 r2 (T − Ts,tube ) + T h1r1 (a) The average temperature of the cooling water can be determined by applying Eq. 2-68: egen r1 Ts ,rod = T + (b) 2h1 Substituting Eq. (a) into Eq. (b) and solving for the average temperature of the cooling water gives egen r1 h2 r2 (T − Ts, tube ) + T = T + h1 r1 2h1 r egen r1 T = 1 + Ts , tube r2 2h2 = 0.005 m  (50  10 6 W/m 3 )( 0.005 m)    + 40 C 0.010 m  2(2000 W/m 2  C)  = 71.3 C Discussion The given information is not sufficient for one to determine the fuel rod surface temperature. The convection heat transfer coefficient for the fuel rod surface (h1) or the centerline temperature of the fuel rod (T0) is needed to determine the fuel rod surface temperature. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-57 2-100 The heat generation and the maximum temperature rise in a solid stainless steel wire. Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the heater is uniform. Properties The thermal conductivity is given to be k = 14 W/mK. Analysis (a) The heat generation per unit volume of the wire is E gen,electric I 2 Re egen = = Vwire  ro2 L With electrical resistance defined as Re = L () A where  = electrical resistivity (m), L = wire length (m), A = wire cross-sectional area D2/4 (m2) Combining equations for e gen and Re, we have e gen = e gen = I 2 I 2 A (D 2 / 4) = 2 = 2 16 I 2   2D4 16 (120A) 2 (45  10 -8   m)  2 (0.001 m) 4 = 1.05  1010 W/m3 (b) The maximum temperature rise in the solid stainless steel wire is obtained from To − Ts = Tmax,cylinder = Tmax = 2 e gen ro 4k (W/m3) (1.05  10 10 W / m 3 )(0.0005 m)2 = 47ºC 4 (14 W / m  K) Discussion The maximum temperature rise in the wire can be reduced by increasing the convective heat transfer coefficient and thus reducing the surface temperature. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-58 2-101 A long homogeneous resistance heater wire with specified surface temperature is used to heat the air. The temperature of the wire 3.5 mm from the center is to be determined in steady operation. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the wire is uniform. Properties The thermal conductivity is given to be k = 8 W/m°C. Analysis Noting that heat transfer is steady and one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as 1 d  dT  egen =0 r + r dr  dr  k and r 180°C ro egen T (ro ) = Ts = 180C (specified surface temperature) dT (0) = 0 (thermal symmetry about the centerline) dr Resistance wire Multiplying both sides of the differential equation by r and rearranging gives egen d  dT  r r =− dr  dr  k Integrating with respect to r gives egen r 2 dT r =− + C1 dr k 2 (a) It is convenient at this point to apply the boundary condition at the center since it is related to the first derivative of the temperature. It yields B.C. at r = 0: egen dT (0) =−  0 + C1 dr 2k 0 → C1 = 0 Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating, egen dT =− r dr 2k and T (r ) = − egen 4k r 2 + C2 (b) Applying the other boundary condition at r = ro , B. C. at r = ro : Ts = − egen 4k ro2 + C 2 → C 2 = Ts + egen 4k ro2 Substituting this C2 relation into Eq. (b) and rearranging give T (r ) = Ts + egen 4k (ro2 − r 2 ) which is the desired solution for the temperature distribution in the wire as a function of r. The temperature 3.5 mm from the center line (r = 0.0035 m) is determined by substituting the known quantities to be T (0.0035 m) = Ts + egen 4k (ro2 − r 2 ) = 180 C + 5 10 7 W/m3 [(0.005 m) 2 − (0.0035 m) 2 ] = 200C 4  (8 W/ m  C) Thus the temperature at that location will be about 20°C above the temperature of the outer surface of the wire. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-59 2-102 A long homogeneous resistance heater wire with specified convection conditions at the surface is used to boil water. The mathematical formulation, the variation of temperature in the wire, and the temperature at the centerline of the wire are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the wire is uniform. Properties The thermal conductivity is given to be k = 15.2 W/mK. Analysis Noting that heat transfer is steady and one-dimensional in r T the radial r direction, the mathematical formulation of this problem Water h can be expressed as ro 1 d  dT  egen =0 r + r dr  dr  k 0 and −k dT (ro ) = h[T (ro ) − T ] (convection at the outer surface) dr Heater dT (0) = 0 (thermal symmetry about the centerline) dr Multiplying both sides of the differential equation by r and rearranging gives egen d  dT  r r =− dr  dr  k Integrating with respect to r gives egen r 2 dT (a) r =− + C1 dr k 2 It is convenient at this point to apply the second boundary condition since it is related to the first derivative of the temperature by replacing all occurrences of r and dT/dr in the equation above by zero. It yields egen dT (0) B.C. at r = 0: 0 =−  0 + C1 → C1 = 0 dr 2k Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating, egen dT =− r dr 2k egen 2 and (b) T (r ) = − r + C2 4k Applying the second boundary condition at r = ro , B. C. at r = ro : k egen ro 2k egen ro egen 2  egen 2  = h − ro + C 2 − T  → C 2 = T + + ro 2h 4k  4k  Substituting this C 2 relation into Eq. (b) and rearranging give T (r ) = T + egen (ro2 − r 2 ) + egen ro 4k 2h which is the desired solution for the temperature distribution in the wire as a function of r. Then the temperature at the center line (r = 0) is determined by substituting the known quantities to be egen 2 egen ro T (0) = T + ro + 4k 2h (16.4  10 6 W/m 3 )( 0.006 m) 2 (16.4  10 6 W/m 3 )( 0.006 m) = 100 C + + = 125C 4  (15 .2 W/m  K) 2  (3200 W/m 2  K) Thus the centerline temperature will be 25°C above the temperature of the surface of the wire. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-60 2-103 A long resistance heater wire is subjected to convection at its outer surface. The surface temperature of the wire is to be determined using the applicable relations directly and by solving the applicable differential equation. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the wire is uniform. Properties The thermal conductivity is given to be k = 15.1 W/m°C. Analysis (a) The heat generation per unit volume of the wire is E gen E gen 2000 W egen = = = = 1.061 10 8 W/m 3 V wire ro 2 L  (0.001 m) 2 (6 m) The surface temperature of the wire is then (Eq. 2-68) egen ro (1.061 10 8 W/m 3 )( 0.001 m) Ts = T + = 20 C + = 323C 2h 2(175 W/m 2  C) T h (b) The mathematical formulation of this problem can be expressed as 1 d  dT  egen =0 r + r dr  dr  k and − k T h k egen 0 ro r dT (ro ) = h[T (ro ) − T ] (convection at the outer surface) dr dT (0) = 0 (thermal symmetry about the centerline) dr Multiplying both sides of the differential equation by r and integrating gives egen egen r 2 d  dT  dT (a) r →r =− + C1 r =− dr k 2 dr  dr  k Applying the boundary condition at the center line, egen dT (0) B.C. at r = 0: 0 =−  0 + C1 → C1 = 0 dr 2k Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating, egen 2 egen dT → (b) T (r ) = − r + C2 =− r 4k dr 2k Applying the boundary condition at r = ro , B. C. at r = ro : −k egen ro 2k egen ro egen 2  egen 2  = h − ro + C 2 − T  → C 2 = T + + ro 2h 4k  4k  Substituting this C2 relation into Eq. (b) and rearranging give T (r ) = T + egen (ro2 − r 2 ) + egen ro 4k 2h which is the temperature distribution in the wire as a function of r. Then the temperature of the wire at the surface (r = ro ) is determined by substituting the known quantities to be egen 2 egen r0 egen ro (1.061 10 8 W/m 3 )(0.001 m) T (r0 ) = T + (ro − ro2 ) + = T + = 20 C + = 323C 4k 2h 2h 2(175 W/m 2  C) Note that both approaches give the same result. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-61 2-104 A cylindrical fuel rod is cooled by water flowing through its encased concentric tube while generating a uniform heat. The variation of temperature in the fuel rod and the center and surface temperatures are to be determined for steady onedimensional heat transfer. Assumptions 1 Heat transfer is steady and one-dimensional with thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 The rod surface at r = ro is subjected convection. 4 Heat generation in the rod is uniform. Properties The thermal conductivity is given to be 30 W/mK. Analysis For one-dimensional heat transfer in the radial r direction, the differential equation for heat conduction in cylindrical coordinate with heat generation can be expressed as egen d  dT  1 d  dT  egen = 0 or r r =− r + dr  dr  k r dr  dr  k Integrating the differential equation twice with respect to r yields egen 2 egen dT dT C r =− r + C1 or =− r+ 1 dr 2k dr 2k r egen 2 T (r ) = − r + C1 ln r + C2 4k where C1 and C2 are arbitrary constants. Applying the boundary conditions give dT (0) → r = 0: C1 = 0 =0 dr r = ro : −k dT (ro ) = h[T (ro ) − T ] dr → k  egen 2  ro = h − r0 + C2 − T  2k  4k  egen Solving for C2 gives egen egen 2 C2 = ro + ro + T 2h 4k Substituting C1 and C2 into the general solution, the variation of temperature is determined to be egen 2 egen 2 egen egen 2 egen 2 egen → T (r ) = − r + C2 = − r + ro + ro + T T (r ) = (ro − r 2 ) + ro + T 4k 4k 2h 4k 4k 2h The temperature at r = 0 (the centerline of the rod) is T (0) = egen 4k ro2 + egen 2h ro + T = 100  10 6 W/m3 100  10 6 W/m3 (0.01 m) 2 + (0.01 m) + 75C 4(30 W/m  K) 2(2500 W/m2  K) T (0) = 358C The temperature at r = ro = 0.01 m (the surface of the rod) is egen 100  10 6 W/m 3 T (ro ) = ro + T = (0.01 m) + 75 C = 275C 2h 2(2500 W/m 2  K) Fuel rod surface not cooled adequately. Discussion The temperature of the fuel rod surface is 75°C higher than the temperature necessary to prevent the cooling water from reaching the CHF. To keep the temperature of the fuel rod surface below 200°C, the convection heat transfer coefficient of the cooling water should be kept above 4000 W/m 2∙K. This can be done either by increasing the mass flow rate of the cooling water or by decreasing the inlet temperature of the cooling water. The topic of critical heat flux is covered in Chapter 10 (Boiling and Condensation). Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-62 2-105 A long electrical resistance wire that is generating heat uniformly is covered with polyethylene insulation. Formulate the temperature profiles for the wire and the polyethylene insulation. Determine the temperature at the interface of the wire and the insulation, and the temperature at the center of the wire. Conclude whether the polyethylene insulation for the wire meets the ASTM D1351 standard. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivities are constant. 3 Heat generation in the wire is uniform. 4 There is no contact resistance at the interface of the wire and the insulation, r = r1. 5 At the center of the wire, r = 0, is a symmetry boundary. 6 The outer surface of the insulation, r = r2, is subjected to convection and radiation. PropertiesThe thermal conductivities of the wire and the polyethylene insulation are given to be kwire = 15 W/m∙K and kins = 0.4 W/m∙K, respectively. AnalysisFor one-dimensional heat transfer in the radial r direction with uniform heat generation, the differential equation for heat conduction in cylindrical coordinate for the wire can be expressed as ?̇gen 1? ??wire (? )+ =0 ? ?? ?? ?wire or ?̇gen ? ??wire (? )=− ? ?? ?? ?wire Integrating the differential equation twice with respect to r yields ? ?̇gen 2 ??wire =− ? + ?1 ?? 2?wire ?wire (?) = − ?̇gen 2 ? + ?1 ln ? + ?2 4?wire whereC1 and C2 are arbitrary constants. Applying the boundary conditions give ? = 0: ??wire (0) =0 ?? → ? = ?1 : ?wire (?1 ) = ?? = − ?̇gen 2 ? + ?2 4?wire 1 ?1 = 0 → ?2 = ?? + ?̇gen 2 ? 4?wire 1 Substituting C1 and C2 into the general solution, the temperature profile in the wire is determined to be ?wire (?) = ?? + ?̇ gen (?? − ?? )for ??wire ? 0 ≤ ? ≤ ?1 The insulation layer does not involve any heat generation, the heat conduction equation in the insulation layer is ? ??ins (? )=0 ?? ?? Integrating the differential equation twice with respect to r yields ? ??ins ??ins ?3 = ?3 or = ?? ?? ? ?ins (?) = ?3 ln ? + ?4 whereC3 and C4 are arbitrary constants. Applying the boundary conditions yields ? = ?1 : − ?wire ??wire (?1 ) ??ins (?1 ) = −?ins ?? ?? ? = ?2 : − ?ins ??ins (?2 ) ?3 = −?ins = ℎcombined [?ins (?2 ) − ?surr ] ?? ?2 → ?̇gen ?3 ?1 = −?ins 2 ?1 Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-63 Note that ℎcombined = ℎconv + ℎrad . The arbitrary constants C3 and C4 can be expressed as ?3 = − ?̇gen ?12 2 ?ins ?̇gen ?12 ?ins 1 ( + ln ?2 ) 2?ins ℎcombined ?2 Substituting C3 and C4 into the general solution, the temperature profile in the insulation layer is determined to be ?4 = ?surr + ?ins (?) = ?̇ gen ??? ?ins ? ?? ( + ?? ) + ?surr for?1 ≤ ? ≤ ?2 ??ins ?combined ?? ? At the interface of the wire and the insulation, r = r1, we have ?? = ?ins (?1 ) = ?? = ?̇gen ?12 ?ins 1 ?2 ( + ln ) + ?surr 2?ins ℎcombined ?2 ?1 W ) (0.002 m)2 0.4 W/mK 1 0.007 m m3 ( × + ln ) + 20C = ??. ?℃ > 75℃ W 2(0.4 W/mK) 0.007 m 0.002 m 7 2 m K (1.2 × 106 whereand r2 = r1 + wall thickness = 0.002 m + 0.005 m = 0.007 m. The temperature at the center of the wire, r = 0, is W (1.2 × 106 3 ) (0.002 m)2 ?̇gen 2 m ?wire (0) = ?? + ? = 76.5C + = ??. ?℃ 4?wire 1 4(15 W/mK ) Discussion With the temperature at the interface of the wire and the insulation being 1.5°C higher than the specification of the ASTM D1351 standard for polyethylene insulation, the ASTM standard is not met. We can consider using a different insulation material with a higher temperature rating. From the ASTM database, the crosslinked polyethylene insulation (ASTM D2655) is rated up to 90°C for normal operation. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-64 2-106 Heat is generated uniformly in a spherical radioactive material with specified surface temperature. The mathematical formulation, the variation of temperature in the sphere, and the center temperature are to be determined for steady onedimensional heat transfer. Assumptions 1 Heat transfer is steady since there is no indication of any changes with time. 2 Heat transfer is onedimensional since there is thermal symmetry about the mid point. 3 Thermal conductivity is constant. 4 Heat generation is uniform. Properties The thermal conductivity is given to be k = 15 W/m°C. Analysis (a) Noting that heat transfer is steady and one-dimensional in the radial Ts=80°C r direction, the mathematical formulation of this problem can be expressed as k egen 1 d  2 dT  egen  r + = 0 with e = constant   gen dr  k r r 2 dr  0 ro and T (ro ) = Ts = 80C (specified surface temperature) dT (0) = 0 (thermal symmetry about the mid point) dr (b) Multiplying both sides of the differential equation by r2 and rearranging gives egen 2 d  2 dT  r r =− dr  dr  k Integrating with respect to r gives egen r 3 dT (a) r2 =− + C1 dr k 3 Applying the boundary condition at the mid point, egen dT (0) B.C. at r = 0: 0 =−  0 + C1 → C1 = 0 dr 3k Dividing both sides of Eq. (a) by r2 to bring it to a readily integrable form and integrating, egen dT =− r dr 3k and T (r ) = − egen 6k r 2 + C2 (b) Applying the other boundary condition at r = r0 , B. C. at r = ro : Ts = − egen 6k ro2 + C 2 → C 2 = Ts + egen 6k ro2 Substituting this C 2 relation into Eq. (b) and rearranging give T (r ) = Ts + egen (ro2 − r 2 ) 6k which is the desired solution for the temperature distribution in the wire as a function of r. (c) The temperature at the center of the sphere (r = 0) is determined by substituting the known quantities to be T (0) = Ts + egen 6k (ro2 − 0 2 ) = Ts + egen ro2 6k = 80 C + (4  10 7 W/m 3 )( 0.04 m) 2 = 791C 6  (15 W/ m  C) Thus the temperature at center will be about 711°C above the temperature of the outer surface of the sphere. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-65 2-107 Prob. 2-106 is reconsidered. The temperature as a function of the radius is to be plotted. Also, the center temperature of the sphere as a function of the thermal conductivity is to be plotted. Analysis The problem is solved using EES, and the solution is given below. “GIVEN” r_0=0.04 [m] g_dot=4E7 [W/m^3] T_s=80 [C] k=15 [W/m-C] “ANALYSIS” T=T_s+g_dot/(6*k)*(r_0^2-r^2) “Temperature distribution as a function of r” T_0=T_s+g_dot/(6*k)*r_0^2 “Temperature at the center (r=0)” k [W/m.C] 10 30.53 51.05 71.58 92.11 112.6 133.2 153.7 174.2 194.7 215.3 235.8 256.3 276.8 297.4 317.9 338.4 358.9 379.5 400 T0 [C] 1147 429.4 288.9 229 195.8 174.7 160.1 149.4 141.2 134.8 129.6 125.2 121.6 118.5 115.9 113.6 111.5 109.7 108.1 106.7 800 700 600 500 T [C] T [C] 791.1 789.1 783.2 773.4 759.6 741.9 720.2 694.6 665 631.6 594.1 552.8 507.5 458.2 405 347.9 286.8 221.8 152.9 80 400 300 200 100 0 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 r [m] 1200 1000 800 T0 [C] r [m] 0 0.002105 0.004211 0.006316 0.008421 0.01053 0.01263 0.01474 0.01684 0.01895 0.02105 0.02316 0.02526 0.02737 0.02947 0.03158 0.03368 0.03579 0.03789 0.04 600 400 200 0 0 50 100 150 200 250 300 350 400 k [W/m-C] Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-66 Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-67 2-108 A spherical communication satellite orbiting in space absorbs solar radiation while losing heat to deep space by thermal radiation. The heat generation rate and the surface temperature of the satellite are to be determined. Assumptions 1 Heat transfer is steady and one-dimensional. 2 Heat generation is uniform. 3 Thermal properties are constant. Properties The properties of the satellite are given to be  = 0.75,  = 0.10, and k = 5 W/m ∙ K. Analysis For steady one-dimensional heat conduction in sphere, the differential equation is 1 d  2 dT  egen =0 r + k r 2 dr  dr  and T (0) = T0 = 273 K dT (0) =0 dr (midpoint temperature of the satellite) (thermal symmetry about the midpoint) Multiply both sides of the differential equation by r 2 and rearranging gives egen 2 d  2 dT  r r =− dr  dr  k Integrating with respect to r gives egen r 3 dT (a) =− + C1 dr k 3 Applying the boundary condition at the midpoint (thermal symmetry about the midpoint), egen dT (0) 0 =−  0 + C1 → r = 0: C1 = 0 dr k r2 Dividing both sides of Eq. (a) by r 2 and integrating, egen dT =− r dr 3k egen 2 T (r ) = − r + C2 and (b) 6k Applying the boundary condition at the midpoint (midpoint temperature of the satellite), egen T0 = −  0 + C2 → r = 0: C 2 = T0 6k Substituting C 2 into Eq. (b), the variation of temperature is determined to be T (r ) = − egen 6k r 2 + T0 At the satellite surface ( r = ro ), the temperature is Ts = − egen ro2 + T0 (c) 6k Also, the rate of heat transfer at the surface of the satellite can be expressed as 4  4 egen   ro3  = As  (Ts4 − Tspace ) − As s q solar 3  where Tspace = 0 The surface temperature of the satellite can be explicitly expressed as  1 4  3 Ts =    ro egen + As s q solar  A  3   s  1/ 4 1/ 4  egen ro / 3 +  s q solar   =      (d) Substituting Eq. (c) into Eq. (d) 1/ 4  egen ro / 3 +  s q solar         =− egen 6k ro2 + T0  egen (1.25 m) / 3 + (0.10 )(1000 W/m 2 )     (0.75 )(5.67  10 −8 W/m 2  K 4 )  1/ 4 =− egen (1.25 m) 2 6(5 W/m  K ) + 273 K Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-68 Copy the following line and paste on a blank EES screen to solve the above equation: ((e_gen*1.25/3+0.10*1000)/(0.75*5.67e-8))^(1/4)=-e_gen*1.25^2/(6*5)+273 Solving by EES software, the heat generation rate is egen = 233 W/m 3 Using Eq. (c), the surface temperature of the satellite is determined to be Ts = − (233 W/m3 ) (1.25 m) 2 + 273 K = 261 K 6(5 W/m  K) Discussion The surface temperature of the satellite in space is well below freezing point of water. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-69 Variable Thermal Conductivity, k(T) 2-109C The thermal conductivity of a medium, in general, varies with temperature. 2-110C Yes, when the thermal conductivity of a medium varies linearly with temperature, the average thermal conductivity is always equivalent to the conductivity value at the average temperature. 2-111C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies with temperature. 2-112C During steady one-dimensional heat conduction in a plane wall in which the thermal conductivity varies linearly, the error involved in heat transfer calculation by assuming constant thermal conductivity at the average temperature is (a) none. 2-113C During steady one-dimensional heat conduction in a plane wall, long cylinder, and sphere with constant thermal conductivity and no heat generation, the temperature in only the plane wall will vary linearly. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-70 2-114 A silicon wafer with variable thermal conductivity is subjected to uniform heat flux at the lower surface. The maximum allowable heat flux such that the temperature difference across the wafer thickness does not exceed 2 °C is to be determined. Assumptions 1 Heat conduction is steady and one-dimensional. 2 There is no heat generation. 3 Thermal conductivity varies with temperature. Properties The thermal conductivity is given to be k(T) = (a + bT + cT2) W/m ∙ K. Analysis For steady heat transfer, the Fourier’s law of heat conduction can be expressed as dT dT = −(a + bT + cT 2 ) dx dx Separating variable and integrating from x = 0 where T (0) = T1 to x = L where T ( L) = T2 , we obtain q = −k (T ) L T2 0 T1  qdx = − (a + bT + cT )dT 2 Performing the integration gives b c   qL = −a(T2 − T1 ) + (T22 − T12 ) + (T23 − T13 ) 2 3   The maximum allowable heat flux such that the temperature difference across the wafer thickness does not exceeding 2°C (where T1 = 602 K and T2 = 600 K ) is 1.29 0.00111   2 2 (600 3 − 602 3 ) W/m 437 (600 − 602 ) − 2 (600 − 602 ) + 3  q = −  (925  10 − 6 m) = 1.35 105 W/m 2 Discussion For heat flux less than 135 kW/m2, the temperature difference across the silicon wafer thickness will be maintained below 2°C. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-71 2-115 A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer through the plate is to be determined. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies linearly. 3 There is no heat generation. Properties The thermal conductivity is given to be k (T ) = k 0 (1 + T ) . Analysis The average thermal conductivity of the medium in this case is simply the conductivity value at the average temperature since the thermal conductivity varies linearly with temperature, and is determined to be k(T) T1 T2 T +T   k ave = k (Tavg ) = k 0 1 +  2 1  2   (500 + 350) K   L = (25 W/m  K) 1 + (8.7 10 -4 K -1 )  2   = 34 .24 W/m  K Then the rate of heat conduction through the plate becomes T − T2 (500 − 35 0)K Q = k avg A 1 = (34 .24 W/m  K)(1.5 m  0.6 m) = 30,820 W = 30.8 kW L 0.15 m Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of Eq, 2-76, and performed the indicated integration. 2-116 On one side, a steel plate is subjected to a uniform heat flux and maintained at a constant temperature. On the other side, the temperature is maintained at a lower temperature. The plate thickness is to be determined. Assumptions 1 Heat transfer is steady and one-dimensional. 2 There is no heat generation. 3 Thermal conductivity varies with temperature. Properties The thermal conductivity is given to be k(T) = k0 (1 + βT). Analysis For steady heat transfer, the Fourier’s law of heat conduction can be expressed as T −T q = kavg 1 2 L Solving for the plate thickness from the above equation T − T2 L = k avg 1 q The average thermal conductivity of the steel plate is T +T  (600 + 800 ) K    kavg = k0 1 +  2 1  = (9.14 W/m  K)1 + (0.0023 K -1 )  = 23.86 W/m  K 2  2    Substituting into the equation for the plate thickness, L = (23 .86 W/m  K ) (800 − 600 ) K 50000 W/m2 = 0 .095 m Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of Eq. 2-76, and performed the indicated integration. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-72 2-117 A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer through the plate is to be determined. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies quadratically. 3 There is no heat generation. Properties The thermal conductivity is given to be k (T ) = k 0 (1 + T 2 ) . Analysis When the variation of thermal conductivity with temperature k(T) is known, the average value of the thermal conductivity in the temperature range between T1 and T2 can be determined from T2 kavg = T2 ò k(T )dT ò k (1+ b T )dT T1 T1 = 0 T2 – T1 T2 – T1 é ù b 3 k0 ê(T2 – T1)+ T2 – T13 ú êë ú 3 û ( = 2 = k(T) T2 T1 T æ b ö2 k0 ççT + T 3÷ ÷ çè 3 ÷ øT 1 T2 – T1 L ) x T2 – T1 é b 2 ù = k0 ê1+ T + T1T2 + T12 ú êë 3 2 ú û ( ) This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal conductivity kavg equals the rate of heat transfer through the same medium with variable conductivity k(T). Then the rate of heat conduction through the plate can be determined to be ( ) T −T    T −T Q = k avg A 1 2 = k 0 1 + T22 + T1T2 + T12  A 1 2 L L  3  Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of Eq. 2-76, and performed the indicated integration. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-73 2-118 The thermal conductivity of stainless steel has been characterized experimentally to vary with temperature. The average thermal conductivity over a given temperature range and the k(T) = k0 (1 + βT) expression are to be determined. Assumptions 1 Thermal conductivity varies with temperature. Properties The thermal conductivity is given to be k(T) = 9.14 + 0.021T for 273 < T < 1500 K. Analysis The average thermal conductivity can be determined using kavg T2 1200 T1 300  k (T ) dT =  (9.14 + 0.021T ) dT = (9.14T + 0.0105 T ) = T2 − T1 2 1200 − 300 1200 − 300 1200 300 = 24.9 W/m  K To express k(T) = 9.14 + 0.021T as k(T) = k0 (1 + βT), we have k (T ) = k0 + k0  T and comparing with k(T) = 9.14 + 0.021T, we have k0 = 9.14 W/m  K and k0  = 0.021 W/m K 2 which gives = 0.021 W/m  K 2 0.021 W/m  K 2 = = 0.0023 K-1 k0 9.14 W/m  K Thus, k (T ) = k0 (1 + T ) where k0 = 9.14 W/m  K and  = 0.0023 K -1 Discussion The average thermal conductivity can also be determined using the average temperature:  1200 + 300  kavg = k (Tavg ) = 9.14 + 0.021  = 24.9 W/m  K 2   Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-74 2-119 A pipe outer surface is subjected to a uniform heat flux and has a known temperature. The metal pipe has a variable thermal conductivity. The inner surface temperature of the pipe is to be determined. Assumptions 1 Heat transfer is steady and one-dimensional. 2 There is no heat generation. 3 Thermal conductivity varies with temperature. Properties The thermal conductivity is given to be k(T) = k0 (1 + βT). Analysis For steady heat transfer, the heat conduction through a cylindrical layer can be expressed as 2 L kavg T2 − T1 kavg T2 − T1 Q q = = = 2 r2 L 2 r2 L ln( r2 / r1 ) r2 ln( r2 / r1 ) The inner and outer radii of the pipe are r1 = 0.1 / 2 m = 0.05 m and r2 = (0.05 + 0.01) m = 0.06 m The average thermal conductivity is T + T1  (773 ) K + T1    k avg = k 0 1 +  2  = (7.5 W/m  K ) 1 + (0.0012 K -1 )  2 2     = [7.5 + 0.0045 (773 + T1 )] W/m  K Thus, 5000 W/m2 =  [7.5 + 0.0045 (773 + T1 )] W/m  K  773 − T1 K  0.06 m ln( 0 . 06 / 0 . 05 )   Solving for the inner pipe temperature T1, T1 = 769 .21 K = 496.2C Discussion There is about 4°C drop in temperature across the pipe wall. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-75 2-120 A pipe is used for transporting boiling water with a known inner surface temperature in surroundings of cooler ambient temperature and known convection heat transfer coefficient. The pipe wall has a variable thermal conductivity. The outer surface temperature of the pipe is to be determined to ensure that it is below 50°C. Assumptions 1 Heat transfer is steady and one-dimensional. 2 There is no heat generation. 3 Thermal conductivity varies with temperature. 4 Inner pipe surface temperature is constant at 100°C. Properties The thermal conductivity is given to be k(T) = k0 (1 + βT). Analysis The inner and outer radii of the pipe are r1 = 0.030 / 2 m = 0.015 m r2 = (0.015 + 0.003 ) m = 0.018 m The rate of heat transfer at the pipe’s outer surface can be expressed as Q = Q cylinder 2 kavg L conv T1 − T2 = h(2 r2 L)(T2 − T ) ln(r2 / r1 ) kavg T1 − T2 = h(T2 − T ) r2 ln( r2 / r1 ) (1) where h = 70 W/m2 K, T1 = 373 K, and T∞ = 283 K The average thermal conductivity is T +T  T + (373 K )    kavg = k0 1 +  2 1  = (1.23 W/m  K ) 1 + (0.002 K -1 ) 2  2  2    kavg = [1.23 + 0.00123( T2 + 373)] W/m  K (2) Solving Eqs. (1) & (2) for the outer surface temperature yields T2 = 364.3 K = 91.3C Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. "GIVEN" h=70 [W/(m^2*K)] "convection heat transfer coefficient" r_1=0.030/2 [m] "inner radius" r_2=r_1+0.003 [m] "outer radius" T_1=100+273 [K] "inner surface temperature" T_inf=10+273 [K] "ambient temperature" k_0=1.23 [W/(m*K)] beta=0.002 [K^-1] "SOLVING FOR OUTER SURFACE TEMPERATURE" k_avg=k_0*(1+beta*(T_2+T_1)/2) Q_dot_cylinder=2*pi*k_avg*(T_1-T_2)/ln(r_2/r_1) "heat rate through the cylindrical layer" Q_dot_conv=h*2*pi*r_2*(T_2-T_inf) "heat rate by convection" Q_dot_cylinder=Q_dot_conv The outer surface temperature of the pipe is more than 40°C above the safe temperature of 50°C to prevent thermal burn on skin tissues. Discussion It is necessary to wrap the pipe with insulation to prevent thermal burn. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-76 2-121 A pipe is used for transporting hot fluid with a known inner surface temperature. The pipe wall has a variable thermal conductivity. The pipe’s outer surface is subjected to radiation and convection heat transfer. The outer surface temperature of the pipe is to be determined. Assumptions 1 Heat transfer is steady and one-dimensional. 2 There is no heat generation. 3 Thermal conductivity varies with temperature. Properties The thermal conductivity is given to be k(T) = k0 (1 + βT), α = ε = 0.9 at the outer pipe surface. Analysis The inner and outer radii of the pipe are r1 = 0.15 / 2 m = 0.075 m r2 = (0.075 + 0.005 ) m = 0.08 m The rate of heat transfer at the pipe’s outer surface can be expressed as Q = Q + Q − Q cyl 2 kavg L conv rad abs T1 − T2 4 = h(2 r2 L)(T2 − T ) +  (2 r2 L)(T24 − Tsurr ) −  (2 r2 L)qsolar ln(r2 / r1 ) kavg T1 − T2 4 = h(T2 − T ) +  (T24 − Tsurr ) − qsolar r2 ln( r2 / r1 ) (1) where h = 60 W/m2 K, q∙ solar = 100 W/m2, T1 = 423 K, and T∞ = Tsurr = 273 K The average thermal conductivity is T +T  T + (423 K)    kavg = k0 1 +  2 1  = ( 8.5 W/m  K )1 + (0.001 K -1 ) 2  2  2    kavg = [8.5 + 0.00425( T2 + 423)] W/m  K (2) Solving Eqs. (1) & (2) for the outer surface temperature yields T2 = 418.8 K = 145.8C Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. "GIVEN" h=60 [W/(m^2*K)] "outer surface h" r_1=0.15/2 [m] "inner radius" r_2=r_1+0.005 [m] "outer radius" T_1=423 [K] "inner surface T" T_inf=273 [K] "ambient T" T_surr=273 [K] "surrounding surface T" alpha=0.9 "outer surface absorptivity" epsilon=0.9 "outer surface emissivity" q_dot_solar=100 [W/m^2] "incident solar radiation" k_0=8.5 [W/(m*K)] beta=0.001 [K^-1] "SOLVING FOR OUTER SURFACE TEMPERATURE" k_avg=k_0*(1+beta*(T_2+T_1)/2) q_dot_cyl=k_avg/r_2*(T_1-T_2)/ln(r_2/r_1) "heat flux through the cylindrical layer" q_dot_conv=h*(T_2-T_inf) "heat flux by convection" q_dot_rad=epsilon*sigma#*(T_2^4-T_surr^4) "heat flux by radiation emission" q_dot_abs=alpha*q_dot_solar "heat flux by radiation absorption" q_dot_cyl-q_dot_conv-q_dot_rad+q_dot_abs=0 Discussion Increasing h or decreasing kavg would decrease the pipe’s outer surface temperature. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-77 2-122 A spherical container has its inner surface subjected to a uniform heat flux and its outer surface is at a known temperature. The container wall has a variable thermal conductivity. The temperature drop across the container wall thickness is to be determined. Assumptions 1 Heat transfer is steady and one-dimensional. 2 There is no heat generation. 3 Thermal conductivity varies with temperature. Properties The thermal conductivity is given to be k(T) = k0 (1 + βT). Analysis For steady heat transfer, the heat conduction through a spherical layer can be expressed as 4 kavg r1 r2 T1 − T2 Q r T −T q = = = kavg 2 1 2 2 2 4 r1 4 r1 r2 − r1 r1 r2 − r1 The inner and outer radii of the container are r1 = 1 m r2 = 1 m + 0.005 m = 1.005 m The average thermal conductivity is T +T  ( 293 ) K + T1    kavg = k0 1 +  2 1  = ( 1.33 W/m  K )1 + ( 0.0023 K -1 )  2 2     = [1.33 + 0.00153(29 3 + T1 )] W/m  K Thus,  1.005 m  T1 − 293 K   7000 W/m2 = [1.33 + 0.00153(29 3 + T1 )] W/m  K   1m  0.005 m  Solving for the inner pipe temperature T1, T1 = 308 .5 K The temperature drop across the container wall is, T1 − T2 = 308 .5 K − 293 K = 15.5C Discussion The temperature drop across the container wall would decrease if a material with a higher kavg value is used. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-78 2-123 A spherical shell with variable conductivity is subjected to specified temperatures on both sides. The variation of temperature and the rate of heat transfer through the shell are to be determined. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies linearly. 3 There is no heat generation. Properties The thermal conductivity is given to be k (T ) = k 0 (1 + T ) . T2 Analysis (a) The rate of heat transfer through the shell is expressed as k(T) T − T2 Q sphere = 4k avg r1 r2 1 r2 − r1 r1 T1 r2 where r1 is the inner radius, r2 is the outer radius, and r T +T   kavg = k (Tavg ) = k0 1 +  2 1  2   is the average thermal conductivity. (b) To determine the temperature distribution in the shell, we begin with the Fourier’s law of heat conduction expressed as dT Q = −k (T ) A dr where the rate of conduction heat transfer Q is constant and the heat conduction area A = 4r2 is variable. Separating the variables in the above equation and integrating from r = r1 where T (r1 ) = T1 to any r where T (r ) = T , we get Q r dr T 2 T1  r = −4  k (T )dT r1 Substituting k (T ) = k 0 (1 + T ) and performing the integrations gives  1 1 Q  −  = −4k 0 [(T − T1 ) +  (T 2 − T12 ) / 2]  r1 r  Substituting the Q expression from part (a) and rearranging give T2 + 2  T+ 2k avg r2 (r − r1 ) k 0 r (r2 − r1 ) (T1 − T2 ) − T12 − 2  T1 = 0 which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature distribution T(r) in the cylindrical shell is determined to be T (r ) = − 1   1  2 − 2k avg r2 (r − r1 ) k 0 r (r2 − r1 ) (T1 − T2 ) + T12 + 2  T1 Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the temperature at any point within the medium must remain between T1 and T2 . Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-79 2-124 A spherical vessel, filled with chemicals undergoing an exothermic reaction, has a known inner surface temperature. The wall of the vessel has a variable thermal conductivity. Convection heat transfer occurs on the outer surface of the vessel. The minimum wall thickness of the vessel is to be determined so that the outer surface temperature is 50°C or lower. Assumptions 1 Heat transfer is steady and one-dimensional. 2 There is no heat generation. 3 Thermal conductivity varies with temperature. Properties The thermal conductivity is given to be k(T) = k0 (1 + βT). Analysis The inner and outer radii of the vessel are r1 = 5 / 2 m = 2.5 m and r2 = (r1 + t ) where t = wall thickness The rate of heat transfer at the vessel’s outer surface can be expressed as Q = Q sph conv 4 kavg r1 r2 T1 − T2 = h(4 r22 )(T2 − T ) r2 − r1 r T −T kavg 1 1 2 = h(T2 − T ) r2 r2 − r1 (1) where h = 80 W/m2 K, T1 = 393 K, T2 = 323 K, and T∞ = 288 K The average thermal conductivity is T +T  (323 K) + (393 K)    kavg = k0 1 +  2 1  = ( 1.01 W/m  K )1 + (0.0018 K -1 )  = 1.6611 W/m  K 2 2     Solving Eq. (1) for r2 yields r2 = 2.541 m Thus, the minimum wall thickness of the vessel should be t = r2 − r1 = 0.041 m = 41 mm Discussion To prevent the outer surface temperature of the vessel from causing thermal burn, the wall thickness should be at least 41 mm. As the wall thickness increases, it would decrease the outer surface temperature. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-80 2-125 A spherical tank, filled with ice slurry, has a known inner surface temperature. The tank wall has a variable thermal conductivity. The tank’s outer surface is subjected to radiation and convection heat transfer. The outer surface temperature of the tank is to be determined. Assumptions 1 Heat transfer is steady and one-dimensional. 2 There is no heat generation. 3 Thermal conductivity varies with temperature. Properties The thermal conductivity is given to be k(T) = k0 (1 + βT), α = ε = 0.75 at the outer tank surface. Analysis The inner and outer radii of the tank are r1 = 9 / 2 m = 4.5 m and r2 = (4.5 + 0.02) m = 4.52 m The rate of heat transfer at the tank’s outer surface can be expressed as Q = Q + Q + Q sph conv 4 k avg r1 r2 k avg rad abs T1 − T2 4 = h(4 r22 )(T − T2 ) +  (4 r22 )(Tsurr − T24 ) +  (4 r22 )q solar r2 − r1 r1 T1 − T2 4 = h(T − T2 ) +  (Tsurr − T24 ) + q solar r2 r2 − r1 (1) where h = 70 W/m2 K, q∙ solar = 150 W/m2, T1 = 273 K, and T∞ = Tsurr = 308 K The average thermal conductivity is T +T  T + (273 .15 K)    kavg = k0 1 +  2 1  = (0.33 W/m  K)1 + (0.0025 K -1 ) 2  2 2     kavg = [0.33 + 0.0004125( T2 + 273)] W/m  K (2) Solving Eqs. (1) & (2) for the outer surface temperature yields T2 = 299.5 K = 26.5C Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. "GIVEN" h=70 [W/(m^2*K)] "outer surface h" r_1=9/2 [m] "inner radius" r_2=r_1+0.020 [m] "outer radius" T_1=273 [K] "inner surface T" T_inf=308 [K] "ambient T" T_surr=308 [K] "surrounding surface T" alpha=0.75 "outer surface absorptivity" epsilon=0.75 "outer surface emissivity" q_dot_solar=150 [W/m^2] "incident solar radiation" k_0=0.33 [W/(m*K)] beta=0.0025 [K^-1] "SOLVING FOR OUTER SURFACE TEMPERATURE" k_avg=k_0*(1+beta*(T_2+T_1)/2) q_dot_sph=k_avg*r_1/r_2*(T_1-T_2)/(r_2-r_1) "heat flux through the spherical layer" q_dot_conv=h*(T_inf-T_2) "heat flux by convection" q_dot_rad=epsilon*sigma#*(T_surr^4-T_2^4) "heat flux by radiation emission" q_dot_abs=alpha*q_dot_solar "heat flux by radiation absorption" q_dot_sph+q_dot_conv+q_dot_rad+q_dot_abs=0 Discussion Increasing the tank wall thickness would increase the tanks’ outer surface temperature. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-81 Special Topic: Review of Differential equations 2-126C We utilize appropriate simplifying assumptions when deriving differential equations to obtain an equation that we can deal with and solve. 2-127C A variable is a quantity which may assume various values during a study. A variable whose value can be changed arbitrarily is called an independent variable (or argument). A variable whose value depends on the value of other variables and thus cannot be varied independently is called a dependent variable (or a function). 2-128C A differential equation may involve more than one dependent or independent variable. For example, the equation  2 T ( x, t ) e gen 1 T ( x, t ) + = has one dependent (T) and 2 independent variables (x and t). the equation k  t x 2  2 T ( x, t ) x 2 + W ( x, t ) 1 T ( x, t ) 1 W ( x, t ) = + has 2 dependent (T and W) and 2 independent variables (x and t). x  t  t 2-129C Geometrically, the derivative of a function y(x) at a point represents the slope of the tangent line to the graph of the function at that point. The derivative of a function that depends on two or more independent variables with respect to one variable while holding the other variables constant is called the partial derivative. Ordinary and partial derivatives are equivalent for functions that depend on a single independent variable. 2-130C The order of a derivative represents the number of times a function is differentiated, whereas the degree of a derivative represents how many times a derivative is multiplied by itself. For example, y  is the third order derivative of y, whereas ( y ) 3 is the third degree of the first derivative of y. 2-131C For a function f ( x, y) , the partial derivative f / x will be equal to the ordinary derivative df / dx when f does not depend on y or this dependence is negligible. 2-132C For a function f (x) , the derivative df / dx does not have to be a function of x. The derivative will be a constant when the f is a linear function of x. 2-133C Integration is the inverse of derivation. Derivation increases the order of a derivative by one, integration reduces it by one. 2-134C A differential equation involves derivatives, an algebraic equation does not. 2-135C A differential equation that involves only ordinary derivatives is called an ordinary differential equation, and a differential equation that involves partial derivatives is called a partial differential equation. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-82 2-136C The order of a differential equation is the order of the highest order derivative in the equation. 2-137C A differential equation is said to be linear if the dependent variable and all of its derivatives are of the first degree, and their coefficients depend on the independent variable only. In other words, a differential equation is linear if it can be written in a form which does not involve (1) any powers of the dependent variable or its derivatives such as y 3 or ( y ) 2 , (2) any products of the dependent variable or its derivatives such as yy  or y y  , and (3) any other nonlinear functions of the dependent variable such as sin y or e y . Otherwise, it is nonlinear. 2-138C A linear homogeneous differential equation of order n is expressed in the most general form as y (n) + f1 ( x) y (n−1) +  + f n−1 ( x) y  + f n ( x) y = 0 Each term in a linear homogeneous equation contains the dependent variable or one of its derivatives after the equation is cleared of any common factors. The equation y  − 4x 2 y = 0 is linear and homogeneous since each term is linear in y, and contains the dependent variable or one of its derivatives. 2-139C A differential equation is said to have constant coefficients if the coefficients of all the terms which involve the dependent variable or its derivatives are constants. If, after cleared of any common factors, any of the terms with the dependent variable or its derivatives involve the independent variable as a coefficient, that equation is said to have variable coefficients The equation y  − 4x 2 y = 0 has variable coefficients whereas the equation y  − 4 y = 0 has constant coefficients. 2-140C A linear differential equation that involves a single term with the derivatives can be solved by direct integration. 2-141C The general solution of a 3rd order linear and homogeneous differential equation will involve 3 arbitrary constants. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-83 Review Problems 2-142 A plane wall is subjected to uniform heat flux on the left surface, while the right surface is subjected to convection and radiation heat transfer. The boundary conditions and the differential equation of this heat conduction problem are to be obtained. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation in the wall. 4 The left surface at x = 0 is subjected to uniform heat flux while the right surface at x = L is subjected to convection and radiation. 5 The surrounding temperature is T∞ = Tsurr. Analysis Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, the differential equation for heat conduction can be expressed as d 2T =0 dx 2 The boundary conditions for the left and right surfaces are dT (0) x = 0: −k = q0 dx dT ( L) 4 x = L: −k = h[T ( L) − T ] +  [T ( L) 4 − Tsurr ] dx where T∞ = Tsurr Discussion Due to the radiation heat transfer equation, all temperatures are expressed in absolute temperatures, i.e. K or °R. 2-143 A long rectangular bar is initially at a uniform temperature of Ti. The surfaces of the bar at x = 0 and y = 0 are insulated while heat is lost from the other two surfaces by convection. The mathematical formulation of this heat conduction problem is to be expressed for transient two-dimensional heat transfer with no heat generation. Assumptions 1 Heat transfer is transient and two-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. Analysis The differential equation and the boundary conditions for this heat conduction problem can be expressed as  2T  2T x y 2 + 2 = 1 T  t T ( x,0, t ) =0 x T (0, y , t ) =0 y T ( a, y, t ) = h[T ( a, y, t ) − T ] y T ( x, b, t ) −k = h[T ( x, b, t ) − T ] x h, T b h, T −k a Insulated T ( x, y,0) = Ti Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-84 2-144E A large plane wall is subjected to a specified temperature on the left (inner) surface and solar radiation and heat loss by radiation to space on the right (outer) surface. The temperature of the right surface of the wall and the rate of heat transfer are to be determined when steady operating conditions are reached. Assumptions 1 Steady operating conditions are reached. 2 Heat transfer is one-dimensional since the wall is large relative to its thickness, and the thermal conditions on both sides of the wall are uniform. 3 Thermal T2 properties are constant. 4 There is no heat generation in the wall. Properties The properties of the plate are given to be k = 1.2 Btu/hft°F and  = 0.80, and  s = 0.60 . 520 R qsolar Analysis In steady operation, heat conduction through the wall must be equal to net heat transfer from the outer surface. Therefore, taking the outer surface temperature of the plate to be T2 (absolute, in R), T1 − T2 =  AsT24 −  s As qsolar L Canceling the area A and substituting the known quantities, kAs (1.2 Btu/h  ft  F) L x (520 R) − T2 = 0.8(0.1714  10 −8 Btu/h  ft 2  R 4 )T24 − 0.60 (300 Btu/h  ft 2 ) 0.8 ft Solving for T2 gives the outer surface temperature to be T2 = 553.9 R Then the rate of heat transfer through the wall becomes q = k T1 − T2 (520 − 553 .9) R = (1.2 Btu/h  ft  F) = −50.9 Btu/h  ft 2 (per unit area) L 0.8 ft Discussion The negative sign indicates that the direction of heat transfer is from the outside to the inside. Therefore, the structure is gaining heat. 2-145 A spherical vessel is subjected to uniform heat flux on the inner surface, while the outer surface is subjected to convection and radiation heat transfer. The boundary conditions and the differential equation of this heat conduction problem are to be obtained. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation in the wall. 4 The inner surface at r = r1 is subjected to uniform heat flux while the outer surface at r = r2 is subjected to convection and radiation. 5 The surrounding temperature is T∞ = Tsurr. Analysis For one-dimensional heat transfer in the radial r direction, the differential equation for heat conduction in spherical coordinate can be expressed as d  2 dT  r =0 dr  dr  The boundary conditions for the inner and outer surfaces are dT (r1 ) r = r1 : − k = q1 dr dT (r2 ) 4 r = r2 : − k = h[T (r2 ) − T ] +  [T (r2 ) 4 − Tsurr ] dr where T∞ = Tsurr Discussion Due to the radiation heat transfer equation, all temperatures are expressed in absolute temperatures, i.e. K or °R. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-85 2-146 Heat is generated at a constant rate in a short cylinder. Heat is lost from the cylindrical surface at r = ro by convection to the surrounding medium at temperature T with a heat transfer coefficient of h. The bottom surface of the cylinder at r = 0 is insulated, the top surface at z = H is subjected to uniform heat flux q h , and the cylindrical surface at r = ro is subjected to convection. The mathematical formulation of this problem is to be expressed for steady two-dimensional heat transfer. Assumptions 1 Heat transfer is given to be steady and two-dimensional. 2 Thermal conductivity is constant. 3 Heat is generated uniformly. Analysis The differential equation and the boundary conditions for this heat conduction problem can be expressed as 1   T   2T egen + =0 r + r r  r  z 2 k T (r ,0) =0 z T (r , H ) k = q H z T (0, z ) =0 r T (ro , z ) −k = h[T (ro , z ) − T ] r qH egen h T z ro 2-147 A small hot metal object is allowed to cool in an environment by convection. The differential equation that describes the variation of temperature of the ball with time is to be derived. Assumptions 1 The temperature of the metal object changes uniformly with time during cooling so that T = T(t). 2 The density, specific heat, and thermal conductivity of the body are constant. 3 There is no heat generation. Analysis Consider a body of arbitrary shape of mass m, volume V, surface area A, density , and specific heat cp initially at a uniform temperature Ti. At time t = 0, the body is placed into a medium at temperature T , and heat transfer takes place between the body and its environment with a heat transfer coefficient h. During a differential time interval dt, the temperature of the body rises by a A differential amount dT. Noting that the temperature changes with time only, an energy balance of the solid for the time interval dt can be expressed as h  Heat transfer from the body  The decrease in the energy  m, c, Ti T   =   T=T(t) during dt of the body during dt     or hAs (T − T )dt = mc p ( −dT ) Noting that m = V and dT = d (T − T ) since T = constant, the equation above can be rearranged as hAs d (T − T ) =− dt T − T Vc p which is the desired differential equation. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-86 2-148 A large plane wall is subjected to convection on the inner and outer surfaces. The mathematical formulation, the variation of temperature, and the temperatures at the inner and outer surfaces to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. Properties The thermal conductivity is given to be k = 0.77 W/m°C. Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the inner surface, the mathematical formulation of this problem can be expressed as d 2T =0 dx2 and h1 [T1 − T (0)] = −k dT (0) dx dT ( L) −k = h2 [T ( L) − T 2 ] dx (b) Integrating the differential equation twice with respect to x yields k h2 T2 h1 T1 L dT = C1 dx T ( x) = C1x + C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give x = 0: h1[T1 − (C1  0 + C 2 )] = −kC1 x = L: −kC1 = h2 [(C1 L + C 2 ) − T2 ] Substituting the given values, these equations can be written as 8(22 − C2 ) = −0.77C1 −0.77C1 = (12)(0.2C1 + C 2 − 8) Solving these equations simultaneously give C1 = −38.84 C2 = 18.26 Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be T ( x) = 18.26 − 38.84x (c) The temperatures at the inner and outer surfaces are T (0) = 18.26 − 38.84  0 = 18.3C T ( L) = 18.26 − 38.84  0.2 = 10.5C Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-87 2-149 The base plate of an iron is subjected to specified heat flux on the left surface and convection and radiation on the right surface. The mathematical formulation, and an expression for the outer surface temperature and its value are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. 4 Heat loss through the upper part of the iron is negligible. Properties The thermal conductivity and emissivity are given to be k = 18 W/m°C and  = 0.7. Analysis (a) Noting that the upper part of the iron is well insulated and thus the entire heat generated in the resistance wires is transferred to the base plate, the heat flux through the inner surface is determined to be Tsurr Q 1000 W 2 q 0 = 0 = = 66 , 667 W/m q Abase 150 10 −4 m 2  Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, the mathematical formulation of this problem can be expressed as h T d 2T =0 2 dx and −k dT (0) = q 0 = 66 ,667 W/m 2 dx L x dT ( L) 4 4 = h[T ( L) − T ] +  [T ( L)4 − Tsurr ] = h[T2 − T ] +  [(T2 + 273 )4 − Tsurr ] dx (b) Integrating the differential equation twice with respect to x yields −k dT = C1 dx T ( x) = C1x + C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give q x = 0: − kC1 = q 0 → C1 = − 0 k x = L: 4 − kC1 = h[T2 − T ] +  [(T2 + 273)4 − Tsurr ] Eliminating the constant C1 from the two relations above gives the following expression for the outer surface temperature T2, 4 h(T2 − T ) +  [(T2 + 273)4 − Tsurr ] = q0 (c) Substituting the known quantities into the implicit relation above gives (30 W/m2  C)(T2 − 26) + 0.7(5.67 10−8 W/m2  K 4 )[(T2 + 273) 4 − 2954 ] = 66,667 W/m2 Using an equation solver (or a trial and error approach), the outer surface temperature is determined from the relation above to be T2 = 759C Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-88 2-150 A 30 m2 concrete slab with embedded heating cable melts snow at a rate of 0.1 kg/s. Formulate the temperature profile in the concrete slab in terms of the snow melt rate. The power density for the embedded heater is to be determined whether it is in compliance with the NFPA 70 code. Assumptions1 Heat transfer is steady. 2 One dimensional heat conduction through the concrete slab. 3 The bottom surface at x = 0 is subjected to uniform heat flux from the heating cable. 4 The upper surface at x = L is at a constant temperature of 0°C from the snow melt. 5 There is no heat generation in the concrete slab. 6 Thermal properties are constant. Properties The latent heat of fusion for water is 333.7 kJ/kg (Table A-2). AnalysisTaking the direction normal to the surface of the concrete slab to be the x direction with x = 0 at the bottom surface (the surface that is in contact with the heater surface), the differential equation for heat conduction can be expressed as ?2? =0 ?? 2 Integrating the differential equation twice with respect to x yields ?? = ?1 ?? ?(?) = ?1 ? + ?2 whereC1 and C2 are arbitrary constants. Applying the boundary conditions yields ??(0) ?̇ 0 ? = 0: −? = ?̇ 0 = −??1 → ?1 = − ?? ? ?̇ 0 ? = ?: ?(?) = ?? = ?1 ? + ?2 → ?2 = ?? − ?1 ? = ?? + ? ? Substituting C1 and C2 into the general solution yields ?̇ 0 ?(?) = (? − ?) + ?? ? The heat rate required for melting snow can be determined from the latent heat of fusion for water, ?̇ = ?̇ice ℎ?? For a surface area of 30 m2, the heat flux is determined using ?̇ice ℎ?? ?̇ ?̇ 0 = = ?? ?? Therefore, the temperature profile in the concrete slab in terms of the snow melt rate is ?̇ice ??? (? − ?) + ?? ?(?) = ?? ? The power density (heat flux) for the heater to melt snow at 0.1 kg/s is ?̇ice ℎ?? (0.1 kg/s)(333700 J/kg) ?̇ 0 = = = ???? W/m2 75℃ 2(0.4 W/mK) 2 0.005 m 0.0025 m (5 × 105 whereand r2 = r1 + wall thickness = 0.0025 m + 0.0025 m = 0.005 m. The temperature at the center of the wire, r = 0, is W (5 × 105 3 ) (0.0025 m)2 ?̇gen 2 m ?wire (0) = ?? + ? = 178.96 + = ???. ??℃ 4?wire 1 4(15 W/mK) Discussion With the temperature at the interface of the wire and the insulation being about 104°C higher than the specification of the ASTM D1351 standard for polyethylene insulation, the ASTM standard is not met. If the convection heat transfer coefficient at the outer surface of the insulation is increase to 6 W/m 2∙K or higher, the temperature at the interface of the wire and the insulation would be lower than 75°C. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-98 2-158 In a quenching process, steel ball bearings at a given instant have a rate of temperature decrease of 50 K/s. The rate of heat loss is to be determined. Assumptions 1 Heat conduction is one-dimensional. 2 There is no heat generation. 3 Thermal properties are constant. Properties The properties of the steel ball bearings are given to be c = 500 J/kg ∙ K, k = 60 W/m ∙ K, and  = 7900 kg/m3. Analysis The thermal diffusivity on the steel ball bearing is k 60 W/m  K = = = 15 .19  10 −6 m 2 /s c (7900 kg/m 3 )(500 J/kg  K) The given rate of temperature decrease can be expressed as dT (r ) = −50 K/s dt For one-dimensional transient heat conduction in a sphere with no heat generation, the differential equation is 1   2 T  1 T r = r 2 r  r   t Substituting the thermal diffusivity and the rate of temperature decrease, the differential equation can be written as 1 d  2 dT  − 50 K/s r = r 2 dr  dr  15.19  10 −6 m 2 /s 2 Multiply both sides of the differential equation by r and rearranging gives d  2 dT  − 50 K/s r2 r = − 6 2 dr  dr  15.19  10 m /s Integrating with respect to r gives r2  r3  dT − 50 K/s   + C1 = dr 15 .19 10 −6 m 2 /s  3  (a) Applying the boundary condition at the midpoint (thermal symmetry about the midpoint), r = 0: 0 dT (0) − 50 K/s 0 =   + C1 dr 15.19  10 −6 m 2 /s  3  → C1 = 0 Dividing both sides of Eq. (a) by r 2 gives dT − 50 K/s r =   dr 15.19  10 −6 m 2 /s  3  The rate of heat loss through the steel ball bearing surface can be determined from Fourier’s law to be dT Q loss = −kA dr = −k (4 ro2 ) dT (ro )  ro  50 K/s = k (4 ro2 )   − 6 2 dr 15 .19 10 m /s  3  = (60 W/m  K )( 4 )( 0.125 m) 2 50 K/s −6  0.125 m    3 m /s   15 .19 10 = 1.62 kW Discussion The rate of heat loss through the steel ball bearing surface determined here is for the given instant when the rate of temperature decrease is 50 K/s. 2 Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-99 2-159 A spherical reactor of 5-cm diameter operating at steady condition has its heat generation suddenly set to 9 MW/m 3. The time rate of temperature change in the reactor is to be determined. Assumptions 1 Heat conduction is one-dimensional. 2 Heat generation is uniform. 3 Thermal properties are constant. Properties The properties of the reactor are given to be c = 200 J/kg∙°C, k = 40 W/m∙°C, and  = 9000 kg/m3. Analysis The thermal diffusivity of the reactor is k 40 W/m  C = = = 22 .22  10 −6 m 2 /s c (9000 kg/m 3 )( 200 J/kg  C) For one-dimensional transient heat conduction in a sphere with heat generation, the differential equation is 1   2 T  egen 1 T = r + 2 r k  t r  r  or  1   2 T  egen  T =  2 r +  t k   r r  r  At the instant when the heat generation of reactor is suddenly set to 90 MW/m 3 (t = 0), the temperature variation can be expressed by the given T(r) = a – br2, hence   egen  1   2  1  2 T  egen  =  2 r (a − br 2 ) + r (−2br ) +  =  2   t k  k    r r  r  r r egen  egen  1   =   2 (−6br 2 ) +  =   − 6b + k  k  r  The time rate of temperature change in the reactor when the heat generation suddenly set to 9 MW/m 3 is determined to be egen    T 9  10 6 W/m3   = (22 .22  10 −6 m 2 /s) − 6(5  10 5 C/m 2 ) + =   − 6b +  t k  40 W/m  C    = −61.7 C/s Discussion Since the time rate of temperature change is a negative value, this indicates that the heat generation of reactor is suddenly decreased to 9 MW/m3. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-100 2-160 A cylindrical shell with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer through the shell is to be determined. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies quadratically. 3 There is no heat generation. Properties The thermal conductivity is given to be k (T ) = k 0 (1 + T 2 ) . Analysis When the variation of thermal conductivity with temperature k(T) is known, the average value of the thermal conductivity in the temperature range between T1 and T2 is determined from T2 k avg  k (T )dT = T1 T2 − T1 T2  k (1 + T )dT = T1 k(T) 2 T1 0 T2 − T1 T2 T =    2 k0 T + T 3  3   T1 T2 − T1 r1 r2 r ( )    k 0 (T2 − T1 ) + T23 − T13  3   = T2 − T1 ( )    = k 0 1 + T22 + T1T2 + T12  3   This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal conductivity k avg equals the rate of heat transfer through the same medium with variable conductivity k(T). Then the rate of heat conduction through the cylindrical shell can be determined from Eq. 2-77 to be T − T2 Q cylinder = 2k avg L 1 ln( r2 / r1 ) ( )    T − T2 = 2k 0 1 + T22 + T1T2 + T12  L 1 3   ln( r2 / r1 ) Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of Eq. 2-77, and performed the indicated integration. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-101 2-161 A pipe is used for transporting boiling water with a known inner surface temperature in a surrounding of cooler ambient temperature and known convection heat transfer coefficient. The pipe wall has a variable thermal conductivity. The outer surface temperature of the pipe is to be determined. Assumptions 1 Heat transfer is steady and one-dimensional. 2 There is no heat generation. 3 Thermal conductivity varies with temperature. 4 Inner pipe surface temperature is constant at 100°C. Properties The thermal conductivity is given to be k(T) = k0 (1 + βT). Analysis The inner and outer radii of the pipe are r1 = 0.025 / 2 m = 0.0125 m and r2 = (0.0125 + 0.003 ) m = 0.0155 m The rate of heat transfer at the pipe’s outer surface can be expressed as Q = Q cylinder 2 kavg L conv T1 − T2 = h(2 r2 L)(T2 − T ) ln(r2 / r1 ) kavg T1 − T2 = h(T2 − T ) r2 ln( r2 / r1 ) (1) where h = 50 W/m2 K, T1 = 373 K, and T∞ = 293 K The average thermal conductivity is T +T  T + (373 K)    kavg = k0 1 +  2 1  = ( 1.5 W/m  K )1 + (0.003 K -1 ) 2  2  2    kavg = [1.5 + 0.00225( T2 + 373)] W/m  K (2) Solving Eqs. (1) & (2) for the outer surface temperature yields T2 = 369 K = 96C Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. “GIVEN” h=50 [W/(m^2*K)] “convection heat transfer coefficient” r_1=0.025/2 [m] “inner radius” r_2=r_1+0.003 [m] “outer radius” T_1=373 [K] “inner surface temperature” T_inf=293 [K] “ambient temperature” k_0=1.5 [W/(m*K)] beta=0.003 [K^-1] “SOLVING FOR OUTER SURFACE TEMPERATURE” k_avg=k_0*(1+beta*(T_2+T_1)/2) Q_dot_cylinder=2*pi*k_avg*(T_1-T_2)/ln(r_2/r_1) “heat rate through the cylindrical layer” Q_dot_conv=h*2*pi*r_2*(T_2-T_inf) “heat rate by convection” Q_dot_cylinder=Q_dot_conv Discussion Increasing h or decreasing kavg would decrease the pipe’s outer surface temperature. Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-102 2-162 A metal spherical tank, filled with chemicals undergoing an exothermic reaction, has a known inner surface temperature. The tank wall has a variable thermal conductivity. Convection heat transfer occurs on the outer tank surface. The heat flux on the inner surface of the tank is to be determined. Assumptions 1 Heat transfer is steady and one-dimensional. 2 There is no heat generation. 3 Thermal conductivity varies with temperature. Properties The thermal conductivity is given to be k(T) = k0 (1 + βT). Analysis The inner and outer radii of the tank are r1 = 5 / 2 m = 2.5 m and r2 = (2.5 + 0.01) m = 2.51 m The rate of heat transfer at the tank’s outer surface can be expressed as Q = Q sph conv T −T 4 kavg r1 r2 1 2 = h(4 r22 )(T2 − T ) r2 − r1 r T −T kavg 1 1 2 = h(T2 − T ) r2 r2 − r1 (1) where h = 80 W/m2 K, T1 = 393 K, and T∞ = 288 K The average thermal conductivity is T +T  T + (393 K)    kavg = k0 1 +  2 1  = ( 9.1 W/m  K )1 + (0.0018 K -1 ) 2  2  2    kavg = [9.1 + 0.00819( T2 + 393)] W/m  K (2) Solving Eqs. (1) & (2) for T2 and kavg yields T2 = 387 .8 K and kavg = 15.5 W/m  K Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. “GIVEN” h=80 [W/(m^2*K)] “outer surface h” r_1=5/2 [m] “inner radius” r_2=r_1+0.010 [m] “outer radius” T_1=120+273 [K] “inner surface T” T_inf=15+273 [K] “ambient T” k_0=9.1 [W/(m*K)] beta=0.0018 [K^-1] “SOLVING FOR OUTER SURFACE TEMPERATURE AND k_avg” k_avg=k_0*(1+beta*(T_2+T_1)/2) q_dot_sph=k_avg*r_1/r_2*(T_1-T_2)/(r_2-r_1) “heat flux through the spherical layer” q_dot_conv=h*(T_inf-T_2) “heat flux by convection” q_dot_sph+q_dot_conv=0 Thus, the heat flux on the inner surface of the tank is Q 4 kavg r1 r2 T1 − T2  2.51  (393 − 387.8) K r T −T  q1 = sph2 = = kavg 2 1 2 = (15.5 W/m  K)  2 4 r1 4 r1 r2 − r1 r1 r2 − r1 0.01 m  2.5  q1 = 8092.2 W/m2 Discussion The inner-to-outer surface heat flux ratio can be related to r1 and r2: q1 / q2 = (r2 / r1 ) 2 . Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-103 Fundamentals of Engineering (FE) Exam Problems 2-163 The heat conduction equation in a medium is given in its simplest form as 1 d  dT   rk  + e gen = 0 Select the wrong r dr  dr  statement below. (a) the medium is of cylindrical shape. (b) the thermal conductivity of the medium is constant. (c) heat transfer through the medium is steady. (d) there is heat generation within the medium. (e) heat conduction through the medium is one-dimensional. Answer (b) thermal conductivity of the medium is constant 2-164 Consider a medium in which the heat conduction equation is given in its simplest form as 1   2 T  1 T r = r   t r 2 r  (a) Is heat transfer steady or transient? (b) Is heat transfer one-, two-, or three-dimensional? (c) Is there heat generation in the medium? (d) Is the thermal conductivity of the medium constant or variable? (e) Is the medium a plane wall, a cylinder, or a sphere? (f) Is this differential equation for heat conduction linear or nonlinear? Answers: (a) transient, (b) one-dimensional, (c) no, (d) constant, (e) sphere, (f) linear 2-165 Consider a large plane wall of thickness L, thermal conductivity k, and surface area A. The left surface of the wall is exposed to the ambient air at T with a heat transfer coefficient of h while the right surface is insulated. The variation of temperature in the wall for steady one-dimensional heat conduction with no heat generation is (a) T ( x) = h( L − x ) T k (b) T ( x) = k T h( x + 0.5L)  xh  (c) T ( x) = 1 − T k   (d) T ( x) = ( L − x)T (e) T ( x) = T Answer (e) T ( x) = T Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-104 2-166 A solar heat flux q s is incident on a sidewalk whose thermal conductivity is k, solar absorptivity is s and convective heat transfer coefficient is h. Taking the positive x direction to be towards the sky and disregarding radiation exchange with the surroundings surfaces, the correct boundary condition for this sidewalk surface is dT =  s q s dx (d) h(T − T ) =  s q s (a) − k Answer (c) − k dT = h(T − T ) dx (e) None of them (b) − k (c) − k dT = h(T − T ) −  s q s dx dT = h(T − T ) −  s q s dx 2-167 A plane wall of thickness L is subjected to convection at both surfaces with ambient temperature T1 and heat transfer coefficient h1 at inner surface, and corresponding T2 and h2 values at the outer surface. Taking the positive direction of x to be from the inner surface to the outer surface, the correct expression for the convection boundary condition is dT (0) = h1 T (0) − T1 ) dx dT (0) = h1 T1 − T 2 ) (c) − k dx (a) k Answer (a) k dT ( L) = h2 T ( L) − T 2 ) dx dT ( L) = h2 T1 − T 2 ) (d) − k dx (b) k (e) None of them dT (0) = h1 T (0) − T1 ) dx 2-168 Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the variation of temperature in the direction of heat transfer be linear is (a) plane wall (b) cylindrical shell (c) spherical shell (d) all of them (e) none of them Answer (a) plane wall 2-169 The conduction equation boundary condition for an adiabatic surface with direction n being normal to the surface is (a) T = 0 (b) dT/dn = 0 (c) d2T/dn2 = 0 (d) d3T/dn3 = 0 (e) -kdT/dn = 1 Answer (b) dT/dn = 0 Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-105 2-170 The variation of temperature in a plane wall is determined to be T(x)=65x+25 where x is in m and T is in °C. If the temperature at one surface is 38ºC, the thickness of the wall is (a) 2 m (b) 0.4 m (c) 0.2 m (d) 0.1 m (e) 0.05 m Answer (c) 0.2 m Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 38=65*L+25 2-171 The variation of temperature in a plane wall is determined to be T(x)=110-48x where x is in m and T is in °C. If the thickness of the wall is 0.75 m, the temperature difference between the inner and outer surfaces of the wall is (a) 110ºC (b) 74ºC (c) 55ºC (d) 36ºC (e) 18ºC Answer (d) 36ºC Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. T1=110 [C] L=0.75 T2=110-48*L DELTAT=T1-T2 2-172 The temperatures at the inner and outer surfaces of a 15-cm-thick plane wall are measured to be 40ºC and 28ºC, respectively. The expression for steady, one-dimensional variation of temperature in the wall is (a) T ( x) = 28 x + 40 (b) T ( x) = −40 x + 28 (d) T ( x) = −80 x + 40 (e) T ( x) = 40 x − 80 (c) T ( x) = 40 x + 28 Answer (d) T ( x) = −80 x + 40 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. T1=40 [C] T2=28 [C] L=0.15 [m] “T(x)=C1x+C2” C2=T1 T2=C1*L+T1 Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-106 2-173 The thermal conductivity of a solid depends upon the solid’s temperature as k = aT + b where a and b are constants. The temperature in a planar layer of this solid as it conducts heat is given by (a) aT + b = x + C2 (b) aT + b = C1x2 + C2 (c) aT2 + bT = C1x + C2 2 2 (d) aT + bT = C1x + C2 (e) None of them Answer (c) aT2 + bT = C1x + C2 2-174 Hot water flows through a PVC (k = 0.092 W/mK) pipe whose inner diameter is 2 cm and outer diameter is 2.5 cm. The temperature of the interior surface of this pipe is 35oC and the temperature of the exterior surface is 20 oC. The rate of heat transfer per unit of pipe length is (a) 22.8 W/m (b) 38.9 W/m (c) 48.7 W/m (d) 63.6 W/m (e) 72.6 W/m Answer (b) 38.9 W/m Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. do=2.5 [cm] di=2.0 [cm] k=0.092 [W/m-C] T2=35 [C] T1=20 [C] Q=2*pi*k*(T2-T1)/LN(do/di) 2-175 Heat is generated in a long 0.3-cm-diameter cylindrical electric heater at a rate of 150 W/cm 3. The heat flux at the surface of the heater in steady operation is (a) 42.7 W/cm2 (b) 159 W/cm2 (c) 150 W/cm2 (d) 10.6 W/cm2 (e) 11.3 W/cm2 Answer (e) 11.3 W/cm2 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. “Consider a 1-cm long heater:” L=1 [cm] e=150 [W/cm^3] D=0.3 [cm] V=pi*(D^2/4)*L A=pi*D*L “[cm^2]” Egen=e*V “[W]” Qflux=Egen/A “[W/cm^2]” “Some Wrong Solutions with Common Mistakes:” W1=Egen “Ignoring area effect and using the total” W2=e/A “Threating g as total generation rate” W3=e “ignoring volume and area effects” Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-107 2-176 Heat is generated uniformly in a 4-cm-diameter, 16-cm-long solid bar (k = 2.4 W/mºC). The temperatures at the center and at the surface of the bar are measured to be 210ºC and 45ºC, respectively. The rate of heat generation within the bar is (a) 240 W (b) 796 W b) 1013 W (c) 79,620 W (d) 3.96106 W Answer (b) 796 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. D=0.04 [m] L=0.16 [m] k=2.4 [W/m-C] T0=210 [C] T_s=45 [C] T0-T_s=(e*(D/2)^2)/(4*k) V=pi*D^2/4*L E_dot_gen=e*V “Some Wrong Solutions with Common Mistakes” W1_V=pi*D*L “Using surface area equation for volume” W1_E_dot_gen=e*W14_1 T0=(W2_e*(D/2)^2)/(4*k) “Using center temperature instead of temperature difference” W2_Q_dot_gen=W2_e*V W3_Q_dot_gen=e “Using heat generation per unit volume instead of total heat generation as the result” 2-177 Heat is generated in a 8-cm-diameter spherical radioactive material whose thermal conductivity is 25 W/m.C uniformly at a rate of 15 W/cm3. If the surface temperature of the material is measured to be 120C, the center temperature of the material during steady operation is (a) 160C (b) 280C (c) 212C (d) 360C (e) 600C Answer (b) 280C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. D=0.08 Ts=120 k=25 e_gen=15E+6 T=Ts+g*(D/2)^2/(6*k) “Some Wrong Solutions with Common Mistakes:” W1_T= e_gen*(D/2)^2/(6*k) “Not using Ts” W2_T= Ts+e_gen*(D/2)^2/(4*k) “Using the relation for cylinder” W3_T= Ts+e_gen*(D/2)^2/(2*k) “Using the relation for slab” Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2-108 3 2-178 Heat is generated in a 3-cm-diameter spherical radioactive material uniformly at a rate of 15 W/cm . Heat is dissipated to the surrounding medium at 25C with a heat transfer coefficient of 120 W/m2C. The surface temperature of the material in steady operation is (a) 56C (b) 84C (c) 494C (d) 650C (e) 108C Answer (d) 650C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. h=120 [W/m^2-C] e=15 [W/cm^3] Tinf=25 [C] D=3 [cm] V=pi*D^3/6 “[cm^3]” A=pi*D^2/10000 “[m^2]” Egen=e*V “[W]” Qgen=h*A*(Ts-Tinf) 2-179 …. 2-181 Design and Essay Problems  Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.