Solution Manual For Heat And Mass Transfer: Fundamentals And Applications, 5th Edition

2-1 Solutions Manual for Heat and Mass Transfer: Fundamentals & Applications 5th Edition Yunus A. Cengel & Afshin J. Ghajar McGraw-Hill, 2015 Chapter 2 HEAT CONDUCTION EQUATION PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) 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: 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 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 onedimensional. 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 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.) PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 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 0.1C  (0.345 W/m C)  17.3 W/m 2 L 0.002 m 2-16 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. Analysis We consider a unit surface area of 1 m2. The total rate of heat generation in this section of the plate is E gen  egenV plate  egen ( A  L)  (5 10 6 W/m3 )(1 m 2 )(0.03 m)  1.510 5 W e L Noting that this heat will be dissipated from both sides of the plate, the heat flux on either surface of the plate becomes q  E gen Aplate  1.5 10 5 W 2 1 m 2  75,000 W/m2  75 kW/m 2 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-5 2-17 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 = 2108 W/m3 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 D = 5 cm L=1m E gen  egenV rod  egen (D 2 / 4) L  (2  108 W/m3 )[ (0.05 m) 2 / 4](1 m)  3.93  105 W = 393 kW 2-18 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 e bx ( Adx)  Ae0 b L  0 Ae 0 (1  e bL ) b 2-19E 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. Analysis An 800 W iron will convert electrical energy into heat in the wire at a rate of 800 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 V wire  E gen (D / 4) L 2  q = 800 W D = 0.08 in L = 15 in  3.412 Btu/h  7 3    6.256  10 Btu/h  ft 1 W [ (0.08 / 12 ft) / 4](15 / 12 ft)   800 W 2 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 q  E gen Awire  E gen DL  800 W  3.412 Btu/h  5 2    1.043  10 Btu/h  ft  (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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-6 Heat Conduction Equation 2-20C 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. 2-21C 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. 2-22 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 x  x  Q x Q   T      kA  x 0 x x x  x  lim 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-7 2-23 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 r  r  Q r Q   T      kA  r 0 r r r  r  lim 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-8 2-24 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 r  r  Q r Q   T      kA  r 0 r r r  r  lim 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. 2-25 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-9  T 2 2-26 For a medium in which the heat conduction equation is given by x 2   T 2 y 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. 2-27 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. 2-28 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. 2-29 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. 2-30 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-10 1   2 T  1  T 1 T  r  2 2 r 2 r  r  r sin   2  t 2 2-31 For a medium in which the heat conduction equation is given by (a) Heat transfer is transient, (b) it is two-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is constant. 2-32 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        or E element Q x  Q y  Q x  x  Q y  y  t 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 Ay  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     kyz    k   k 2 x 0 yz x yz x yz x  x  x  x  x lim   1 Q y  y  Q y 1 Q y 1   T    T   2T   kxz     k   k   y 0 xz y xz y xz y  y  y  y  y 2 lim Here the property   k / c is the thermal diffusivity of the material. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-11 2-33 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 ) rr r+r Substituting, 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  2   c r r  r  r     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      k (2rz )   kr  r 0 2rz r 2rz r 2rz r  r  r r  r  lim lim 1 z 0 2rr Q z  z  Q z 1 Qz 1   T    T      k (2rr )    k  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 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-12 2-34 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   theelement   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 z  z  Q z Q   T      kA  z 0 z z z  z  lim 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-13 Boundary and Initial Conditions; Formulation of Heat Conduction Problems 2-35C 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. 2-36C 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. 2-37C 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 . 2-38C 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 2-39C Yes, the temperature profile in a medium must be perpendicular to an insulated surface since the slope T / x  0 at that surface. 2-40C 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-14 2-41 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 q s  E gen Q s 0.90  (900 W)    31,831 W/m2 As D 2 / 4  (0.18 m) 2 / 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/m2 dx T ( L)  TL  108C 2-42 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 q s  E gen Q s 0.85  (1250 W)    33,820 W/m2 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 dx 2 dT (0)  q s  33,280 W/m2 dx dT ( L) k  h[T ( L)  T ] dx k PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-15 2-43 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 surface at x = 0 is subjected to convection only. Tsky Analysis Expressing all the temperatures in Kelvin, the differential equation and the boundary conditions for this heat conduction problem can be expressed as T1 h1 T2 h2 d 2T 0 dx 2 k dT (0)  h1[T1  T (0)] dx k dT ( L) 4  h1 [T ( L)  T 2 ]   2 T ( L) 4  Tsky dx L  x  2-44 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 gen  egenV wire  egen (ro2 L) E gen egen (ro2 L) egen ro Q q s  s    A A (2ro ) L 2 D egen L 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 2-45 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 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-16 2-46E A 2-kW resistance heater wire is used for space heating. 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 Heat is generated uniformly in the wire. 2 kW Analysis The heat flux at the surface of the wire is q s  E gen Q s 2000 W    353.7 W/in2 As 2ro L 2 (0.06 in)(15 in) D = 0.12 in L = 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/in2 dr 2-47 Water flows through a pipe whose outer surface is wrapped with a thin electric heater that consumes 400 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 s  Q s Q s 400 W    979.4 W/m2 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 d  dT  r 0 dr  dr  Q = 400 W h T r1 r2 dT (r1 )  h[T (ri )  T ]  85[T (ri )  90] dr dT (r2 ) k  q s  734.6 W/m2 dr k PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-17 2-48 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 45 W/m2 towards the center: k r1 dT (r1 )  45 W/m2 dr (c) Convection to a medium at T with a heat transfer coefficient of h: k dT (r1 )  h[T (r1 )  T ] dr 2-49 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  k r1 Tsurr r2  2-50 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 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-18 2-51 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 2-52 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 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-19 Solution of Steady One-Dimensional Heat Conduction Problems 2-53C 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. 2-54C 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. 2-55C 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. 2-56C 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-20 2-57 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 dx 2 and k q=700 W/m2 T1=80°C L=0.3 m dT (0) k  q 0  700 W/m2 dx T (0)  T1  80C x (b) Integrating the differential equation twice with respect to x yields dT dx  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/m2 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-21 2-58 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 = 60 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 0  Q 0 800 W   50,000 W/m2 Abase 160 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 Q =800 W A=160 cm2 d 2T 0 dx 2 and k k T2 =112°C L=0.6 cm dT (0)  q 0  50,000 W/m2 dx T ( L)  T2  112C x (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 q 0 k x = 0:  kC1  q 0  C1   x = L: T ( L)  C1 L  C 2  T2  C 2  T2  C1 L  C 2  T2  q 0 L k 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/m2 )( 0.006  x)m   112C 60 W/m  C  833.3(0.006  x)  112 T ( x)   (c) The temperature at x = 0 (the inner surface of the plate) is T (0)  833.3(0.006  0)  112  117C Note that the inner surface temperature is higher than the exposed surface temperature, as expected. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-22 2-59 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 = 1.8 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 T1=90°C A=30 m2 T (0)  T1  90C k dT ( L)  h[T ( L)  T ] dx L=0.4 m T =25°C h=24 W/m2.°C (b) Integrating the differential equation twice with respect to x yields dT  C1 dx x T ( x)  C1x  C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give x = 0: T (0)  C1  0  C2  C2  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/m2  C)(90  25)C (1.8 W/m  C)  (24 W/m2  C)( 0.4 m)  90  90.3x x  90C (c) The rate of heat conduction through the wall is h(T1  T ) dT Q wall  kA  kAC1  kA dx k  hL (24 W/m2  C)(90  25)C  (1.8 W/m  C)(30 m 2 ) (1.8 W/m  C)  (24 W/m2  C)( 0.4 m)  7389 W Note that under steady conditions the rate of heat conduction through a plain wall is constant. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-23 2-60 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 dx 2 The boundary conditions for this problem are: h1[T1  T (0)]  k k dT (0) dx 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 L dT  C1 dx T ( x)  C1 x  C2 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  C2 )  0.77C1 0.77C1  (12)(0.2C1  C2  8) Solving these equations simultaneously give C1  45.45 C2  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 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-24 2-61 An engine housing (plane wall) is subjected to a uniform heat flux on the inner surface, while the outer surface is subjected to convection heat transfer. The variation of temperature in the engine housing and the temperatures of the inner and outer surfaces 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 engine housing (plane wall). 4 The inner surface at x = 0 is subjected to uniform heat flux while the outer surface at x = L is subjected to convection. Properties Thermal conductivity 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 inner 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)  C1 x  C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give q0 k x  0: k dT (0)  q 0  kC1 dx x  L: k dT ( L)  h[T ( L)  T ]  h(C1 L  C 2  T ) dx  C1    kC1  h(C1 L  C2  T ) Solving for C2 gives q  k k   C2  C1   L   T  0   L   T h k h     Substituting C1 and C2 into the general solution, the variation of temperature is determined to be k  T ( x)  C1 x  C1   L   T h   T ( x)  q 0  k    L  x   T k h  The temperature at x = 0 (the inner surface) is T (0)  q 0  k 6000 W/m2 13.5 W/m  K    0.010 m  35C  339C   L   T   2 k h 13.5 W/m  K  20 W/m  K   The temperature at x = L = 0.01 m (the outer surface) is T ( L)  q 0 6000 W/m2  T   35C  335C h 20 W/m2  K Discussion The outer surface temperature of the engine is 135°C higher than the safe temperature of 200°C. The outer surface of the engine should be covered with protective insulation to prevent fire hazard in the event of oil leakage. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-25 2-62 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 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)  C1 x  C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give q dT (0)  q0  kC1  C1   0 dx k x  0: k x  L: T ( L)  TL  C1L  C2  C2  C1L  TL  q0 L  TL k Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T ( x)   q0 q x  0 L  TL k k  T ( x)  q0 ( L  x)  TL k 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  5128 W/m2 The temperature at x = 0 (the left surface of the wall) is T (0)  q 0 5128 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-26 2-63 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)  C1 x  C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give dT (0)  q0  kC1 dx x  0: k x  0: T (0)  T0  C2  C1   q0 k Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T ( x)   q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)(500 W/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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-27 2-64 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 q 0  k dT dx Knowing that the heat flux and thermal conductivity are constant, integrating the differential equation once with respect to x yields T ( x)   q 0 x  C1 k Applying the boundary condition gives x  L: T ( L )  TL   q 0 L  C1 k  C1  q 0 L  TL k Substituting C1 into the general solution, the variation of temperature in the furnace front is determined to be T ( x)  q 0 ( L  x )  TL k The inside surface temperature of the furnace front is T (0)  T0  q 0 5000 W/m2 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-28 2-65E 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 Btu/hft°F and  = 0.7. Tsky Analysis (a) Taking the direction normal to the surface of the plate to be the x direction with x = 0 at the bottom surface, and the mathematical formulation of this problem can be expressed as k T h L d 2T 0 dx 2 and 80°F  x dT ( L) 4 4  h[T ( L)  T ]   [T ( L) 4  Tsky ]  h[T2  T ]   [(T2  460) 4  Tsky ] dx T ( L)  T2  80F (b) Integrating the differential equation twice with respect to x yields dT dx  C1 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 ] k ( L  x) (12 Btu/h  ft 2  F)(80  90)F + 0.7(0.1714  10-8 Btu/h  ft 2  R 4 )[( 540 R )4  (480 R)4 ] (4 / 12  x) ft 7.2 Btu/h  ft  F  80  11.3(1 / 3  x)  80F  (c) The temperature at x = 0 (the bottom surface of the plate) is T (0)  80  11.3  (1 / 3  0)  76.2F PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-29 2-66 The top and bottom surfaces of a solid cylindrical rod are maintained at constant temperatures of 20C and 95C while the side surface is perfectly insulated. The rate of heat transfer through the rod is to be determined for the cases of copper, steel, and granite rod. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. Properties The thermal conductivities are given to be k = 380 W/m°C for copper, k = 18 W/m°C for steel, and k = 1.2 W/m°C for granite. Analysis Noting that the heat transfer area (the area normal to the direction of heat transfer) is constant, the rate of heat transfer along the rod is determined from T T Q  kA 1 2 L Insulated D = 0.05 m T1=25°C T2=95°C where L = 0.15 m and the heat transfer area A is A  D 2 / 4   (0.05 m) 2 / 4  1.964 10 3 m 2 L=0.15 m Then the heat transfer rate for each case is determined as follows: (a) Copper: T T (95  20)C Q  kA 1 2  (380 W/m C)(1.964 10 3 m 2 )  373.1 W L 0.15 m (b) Steel: T T (95  20)C Q  kA 1 2  (18 W/m C)(1.964 10 3 m 2 )  17.7 W L 0.15 m (c) Granite: T T (95  20)C Q  kA 1 2  (1.2 W/m C)(1.964 10 3 m 2 )  1.2 W L 0.15 m Discussion: The steady rate of heat conduction can differ by orders of magnitude, depending on the thermal conductivity of the material. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-30 2-67 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 Insulated d  dT  r 0 dr  dr  and dT (r1 ) k  h[T f  T (r1 )] dr Water Tf r2 r1 dT (r2 ) 0 dr L (b) Integrating the differential equation once with respect to r gives r 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-31 2-68E 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 =175F Steam 300F h=12.5 dT (r1 )  h[T  T (r1 )] dr T (r2 )  T2  175F (b) Integrating the differential equation once with respect to r gives r L = 30 ft 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 C1  h[T  (C1 ln r1  C 2 )] r1 r = r1: k r = r2: T (r2 )  C1 ln r2  C 2  T2 Solving for C1 and C2 simultaneously gives C1  T2  T r k ln 2  r1 hr1 and C 2  T2  C1 ln r2  T2  T2  T ln r2 r k ln 2  r1 hr1 Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T (r )  C1 ln r  T2  C1 ln r2  C1 (ln r  ln r2 )  T2   T2  T r ln  T2 r2 k r2 ln  r1 hr1 (175  300)F r r ln  175F  34.36 ln  175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 T  T dT Q  kA  k (2rL) 1  2Lk 2 r k dr r ln 2  r1 hr1  2 (30 ft)( 7.2 Btu/h  ft  F) (175  300)F  46,630 Btu/h 2.4 7.2 Btu/h  ft  F ln  2 (12.5 Btu/h  ft 2  F)( 2 / 12 ft ) PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-32 2-69 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  c p (Tin  Tout )  (0.3 kg/s)( 2190 J/kg  C)(7) C  4599 W Q loss  m 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 r  r1 : T (r1 )   1 Q loss ln r1  C 2 2 kL  C1   1 Q loss 2 kL  C2  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 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-33 1 Q loss 1 Q loss T (r )   ln r  ln r1  T (r1 ) 2 kL 2 kL 1 Q loss  ln(r / r1 )  T (r1 ) 2 kL The outer pipe surface temperature is 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 T (r2 )   From Newton’s law of cooling, the rate of heat loss at the outer pipe surface by convection is Q loss  h(2 r2 L)T (r2 )  T  Rearranging and the convection heat transfer coefficient is determined to be h Q loss 4599 W   25.1 W/m2  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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-34 2-70 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) k dT (r2 )  h2 [T (r2 )  T,2 ] dr (convection at the outer pipeline surface) 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 : 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 ) PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-35 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  (b) The inner surface temperature of the pipeline is T (r1 )   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/m2  C)  0.25  (0.258 m)(150 W/m2  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 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 Lh2  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 m   0.258    ln  (0.25 m)( 250 W/m2  C)  0.25  (0.258 m)(150 W/m2  C)  5947 W/m 2 Discussion Knowledge of the inner pipeline surface temperature can be used to control wax deposition blockages in the pipeline. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-36 2-71 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  r1 : k dT (r1 )  h[T (r1 )  T ] dr   C1  q s k r2 k C1  hC1 ln r1  C2  T  r1 Solving for C2 gives  r k 1 C2  q s 2   ln r1   T k  h r1  Substituting C1 and C2 into the general solution, the variation of temperature is determined to be  r r k 1 T (r )  C1 ln r  C2  q s 2 ln r  q s 2   ln r1   T k k  h r1   r k 1 r  T (r )  q s 2   ln 1   T k  h r1 r 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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 0.015  T (r2 )  q s 2   ln 1   T  (1000 W/m2 )  ln  10C   k  h r1 r2  15 W/m K  50 W/m2  K  0.015 m 0.018  T (r2 )  33.8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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-37 2-72 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 T (r )   or dT C1  dr r 2 C1  C2 r where C1 and C2 are arbitrary constants. Applying the boundary conditions give 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-38 2-73 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 T (r )   or dT C1  dr r 2 C1  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 r2  r2  Solving for C2 gives r2  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 r2  k 1 1  C2  q1 1  q1 1     T 2 r k r k  h r2 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 T (r2 )  q1  r1     T h  r2  Discussion Increasing the convection heat transfer coefficient h would decrease the outer surface temperature T(r2). PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-39 2-74 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 r1 T (r1 )  T1  0C k T1 k r2 T h dT (r2 )  h[T (r2 )  T ] dr (b) Integrating the differential equation once with respect to r gives dT  C1 dr Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, r2 dT C1  dr r 2 C1  C2 r where C1 and C2 are arbitrary constants. Applying the boundary conditions give T (r )   r = r1: T (r1 )   r = r2: k C1  C 2  T1 r1  C  C1  h  1  C2  T  2 r2  r2  Solving for C1 and C2 simultaneously gives C1  r2 (T1  T ) r k 1 2  r1 hr2 and C 2  T1  C1  T1  r1 T1  T r2 r k r1 1 2  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      T1 r2 k  r1 r  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/m2  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) PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-40 2-75 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 r2 dT  C1 dr T (r )   or dT C1  dr r 2 C1  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 r2  r2  Solving for C2 gives r2  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 r2  k 1 1  C2  q1 1  q1 1     T 2 r k r k  h r2 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/m2  0.5 m     T   23C  72.6C   h  r2  1000 W/m2  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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-41 2-76 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 800 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  800 W q s  s  s 2   340.8 W/m2 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 k T1 k T (r1 )  T1  120C Heater dT (r2 )  q s dr 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 )   C1  q s  C1  2 r = r2: k r = r1: T (r1 )  T1   r2 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  (340.8 W/m2 )( 0.41 m) 2 1  1   120C      120  38.19 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 )  120  38.19 2.5    120  38.19 2.5    122.3C r2  0.41    Noting that the maximum rate of heat supply to the water is 0.9  800 W = 720 W, water can be heated from 20 to 100C at a rate of Q 0.720 kJ/s  c p T  m  Q  m   0.002151 kg/s = 7.74 kg/h c p T (4.185 kJ/kg  C)(100  20)C PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-42 2-77 Prob. 2-76 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_1=0.40 [m] r_2=0.41 [m] k=1.5 [W/m-C] T_1=120 [C] Q_dot=800 [W] f_loss=0.10 “ANALYSIS” q_dot_s=((1-f_loss)*Q_dot)/A A=4*pi*r_2^2 T=T_1+(1/r_1-1/r)*(q_dot_s*r_2^2)/k “Variation of temperature” T [C] 120 120.3 120.5 120.8 121 121.3 121.6 121.8 122.1 122.3 122.5 122 121.5 T [C] r [m] 0.4 0.4011 0.4022 0.4033 0.4044 0.4056 0.4067 0.4078 0.4089 0.41 121 120.5 120 0.4 0.402 0.404 0.406 0.408 0.41 r [m] PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-43 Heat Generation in a Solid 2-78C 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. 2-79C 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. 2-80C 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. 2-81C 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. 2-82C No, it is not possible since the highest temperature in the plate will occur at its center, and heat cannot flow “uphill.” 2-83 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 Ts  T  To  Ts  egen L h egen L2 2k  25C  (2 10 5 W/m3 )(0.05 m)  252.3C  44 W/m2  C k egen Insulated L=5 cm T =25°C h=44 W/m2.°C  252.3C (2 10 5 W/m3 )( 0.05 m) 2  254.6C 2(111 W/m C) PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-44 2-84 Prob. 2-83 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) h [W/m2.C] 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 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 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 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-45 550 500 Tmin [C] 450 400 350 300 250 200 150 100 20 30 40 50 60 70 80 90 100 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 h [W/m2-C] PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-46 2-85 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 Ts  T  To  Ts  egen L h egen L2 2k  30C  (5 10 5 W/m3 )(0.015 m)  155C  60 W/m2  C k egen T =30°C h=60 W/m2°C T =30°C h=60 W/m2.°C 2L=3 cm  155C (5 10 5 W/m3 )( 0.015 m) 2  158.7C 2(15.1 W/m C) PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-47 2-86 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 dx 2  egen egen dT  x  C1 dx k  k L x Integrating one more time, T ( x)   egen x 2 2k  C1 x  C 2 (1) Applying the boundary conditions: egen dT (0) 0 (0)  C1  0  C1  0 dx k B.C. at x = 0:   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 2k  egen L h  egen L2 2k  T  egen 2k ( L2  x 2 )  egen L h  T 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 ( L)  e gen L2 2k e gen L h  e gen L h  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). PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-48 2-87E 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 where d 2T egen ( x) dx k  2 0 egen  ax 2 d 2T egen ( x) dx k  2  a 2 x k The boundary conditions for this problem are: T (0)  T0 (specified surface temperature at x = 0) dT ( L) 0 dx (insulated surface at x = L) (b) Rearranging the differential equation and integrating, d 2T a dT 1a 3   x2   x  C1 2 k dx 3k dx Integrating one more time, T ( x)   1 a 4 x  C1 x  C 2 12 k (1) Applying the boundary conditions: B.C. at x = 0: T (0)  T0  C2 B.C. at x = L: dT ( L) 1a 3 aL3  L  C1  0  C1  dx 3k 3k Substituting the C1 and C2 relations into Eq. (1) and rearranging gives T ( x)   1 a 4 aL3 x  x  T0 12 k 3k (2) (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 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) T ( L)  Tmax    760 F PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-49 2-88 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 0 k egen where egen  e0 e 0.5 x / L and e 0 = 8106 W/m3 and dT (0)  0 (insulated surface at x = 0) dx 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: B.C. at x = 0: 2e L 2e L dT (0) 2e0 L 0.50 / L  e  C1  0  0  C1  C1   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 T (0)  T2  e0 L2 [4(e 0.5  e 0 )  (2  0 / L)] k  30C  (8  10 6 W/m3 )( 0.05 m) 2 [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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-50 2-89 Prob. 2-88 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” “x is the parameter to be varied” x [m] 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 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 e [W/m3] 7.000×106 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] PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-51 2-90 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. 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. 220°C egen Analysis The center temperature of the rod is determined from To  Ts  egen ro2 4k  220C  (4 10 7 W/m3 )( 0.005 m) 2  228C 4(29.5 W/m.C) Uranium rod 2-91E 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 generation in the heater is uniform. 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 V wire  E gen ro2 L  (3  3412.14 Btu/h)  (0.04 / 12 ft) (1 ft) 2  2.933 10 8 Btu/h  ft 3 r Ts ro 0 Heater Then the temperature difference between the centerline and the surface becomes Tmax  egen ro2 4k  (2.933 10 8 Btu/h  ft 3 )( 0.04 / 12 ft) 2  140.5F 4(5.8 Btu/h  ft  F) PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-52 2-92 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. Properties The thermal conductivity is given to be k = 20 W/m°C. 230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 egen  E gen V wire  E gen ro2 L r  2000 W  (0.002 m) 2 (0.9 m)  1.768 10 8 W/m3 D The center temperature of the wire is then determined from Eq. 2-71 to be To  Ts  egen ro2 4k  230C  (1.768  108 W/m3 )( 0.002 m) 2  238.8C 4(20 W/m.C) 2-93 Heat is generated in a long solid cylinder with a specified surface temperature. The variation of temperature in the cylinder is given by T (r )  2 egen ro2   r   1      Ts k   ro     (a) Heat conduction is steady since there is no time t variable involved. 80C (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 k  e gen ro2  egen D q s  k dT (ro )  k  dr  k    2r    r 2   o  r  r0 r  e gen ro2  2r    o   2e gen ro  2(35 W/cm3 )( 4 cm) = 280 W/cm 2  k   k  ro2  PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-53 2-94 Prob. 2-93 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” r [m] 0 0.004444 0.008889 0.01333 0.01778 0.02222 0.02667 0.03111 0.03556 0.04 T [C] 2320 2292 2209 2071 1878 1629 1324 964.9 550.1 80 2500 2000 T [C] 1500 1000 500 0 0 0.01 0.02 0.03 0.04 r [m] PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-54 2-95 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: Ts ,rod  T  egen r1 2h1 (b) 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/m3 )( 0.005 m)     40 C 0.010 m  2(2000 W/m2  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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-55 2-96 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 egen  Vwire  I 2 Re  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  1010 W / m 3 )0.0005m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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-56 2-97 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 the radial r direction, the mathematical formulation of this problem can be expressed as r T h Water ro 1 d  dT  egen 0 r  r dr  dr  k and k 0 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 r egen r 2 dT   C1 dr k 2 (a) 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)   0  C1 dr 2k 0 B.C. at r = 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 r 2  C2 4k (b) 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 4k (ro2  r 2 )  egen ro 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  100C + (16.4  10 6 W/m3 )( 0.006 m) 2 (16.4  10 6 W/m3 )( 0.006 m)   125C 4  (15.2 W/m  K) 2  (3200 W/m2  K) Thus the centerline temperature will be 25°C above the temperature of the surface of the wire. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-57 2-98 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 V wire  E gen 3000 W  ro L 2  (0.001 m) 2 (6 m)  1.592  10 8 W/m3 T h The surface temperature of the wire is then (Eq. 2-68) Ts  T  egen ro 2h  20C  (1.592  10 8 W/m3 )(0.001 m) 2(175 W/m  C) 2 T h k egen 0  475C ro r (b) The mathematical formulation of this problem can be expressed as 1 d  dT  egen 0 r  r dr  dr  k and k 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 d  dT  r r  dr  dr  k egen r 2 dT   C1 dr k 2 r (a) Applying the boundary condition at the center line, B.C. at r = 0: 0 egen dT (0)   0  C1 dr 2k  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  T (r )   egen 4k r 2  C2 (b) 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 4k (ro2  r 2 )  egen ro 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 T (r0 )  T  egen 4k (ro2  ro2 )  egen r0 2h  T  egen ro 2h  20C  (1.592  108 W/m3 )(0.001 m) 2(175 W/m2  C)  475C Note that both approaches give the same result. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-58 2-99 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 = 6 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 r 1 d  dT  egen 0 r  r dr  dr  k and 180°C ro T (ro )  Ts  180C (specified surface temperature) egen 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 r egen r 2 dT   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 ]  207C 4  (6 W/ m  C) Thus the temperature at that location will be about 20°C above the temperature of the outer surface of the wire. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-59 2-100 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 1 d  dT  egen 0 r  r dr  dr  k egen d  dT  r r  dr  dr  k or Integrating the differential equation twice with respect to r yields r egen 2 dT  r  C1 dr 2k T (r )   egen 4k egen dT C  r 1 dr 2k r or r 2  C1 ln r  C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give r  0: dT (0) 0 dr r  ro : k  C1  0 dT (ro )  h[T (ro )  T ] dr  k  egen 2  ro  h  r0  C2  T  2k  4k  egen Solving for C2 gives C2  egen 2h ro  egen 4k ro2  T Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T (r )   egen r 2  C2   4k egen 4k r2  egen 2h ro  egen 4k ro2  T  T (r )  egen 4k (ro2  r 2 )  egen 2h ro  T The temperature at r = 0 (the centerline of the rod) is T (0)  egen 4k ro2  egen 2h ro  T  100  106 W/m3 100  106 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 T (ro )  egen 2h ro  T  100  10 6 W/m3 2(2500 W/m2  K) (0.01 m)  75C  275C 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/m2∙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). PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-60 2-101 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 r direction, the mathematical formulation of this problem can be expressed as 1 d  2 dT  egen 0 r  dr  k r 2 dr  and Ts=110°C k with egen  constant egen T (ro )  Ts  110C (specified surface temperature) 0 ro r 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 r2 egen r 3 dT   C1 dr k 3 (a) Applying the boundary condition at the mid point, B.C. at r = 0: egen dT (0)   0  C1 dr 3k 0  C1  0 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 6k (ro2  r 2 ) 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  110C + (5  10 7 W/m3 )( 0.04 m) 2  999C 6  (15 W/ m  C) Thus the temperature at center will be 999°C above the temperature of the outer surface of the sphere. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-61 2-102 Prob. 2-101 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=5E7 [W/m^3] T_s=110 [C] k=15 [W/m-C] r=0 [m] “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)” T [C] 998.9 996.4 989 976.7 959.5 937.3 910.2 878.2 841.3 799.4 752.7 701 644.3 582.8 516.3 444.9 368.5 287.3 201.1 110 1000 900 800 700 T [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 0.04 600 500 400 300 200 100 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 r [m] PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-62 T0 [C] 1443 546.8 371.2 296.3 254.8 228.4 210.1 196.8 186.5 178.5 171.9 166.5 162 158.2 154.8 151.9 149.4 147.1 145.1 143.3 1600 1400 1200 1000 T0 [C] k [W/m.C] 10 30.53 51.05 71.58 92.11 112.6 133.2 153.7 174.2 194.7 215.3 235.8 256.3 276.8 297.4 317.9 338.4 358.9 379.5 400 800 600 400 200 0 0 50 100 150 200 250 300 350 400 k [W/m-C] PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-63 2-103 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 r2 egen r 3 dT   C1 dr k 3 (a) Applying the boundary condition at the midpoint (thermal symmetry about the midpoint), r  0: 0 egen dT (0)   0  C1 dr k  C1  0 Dividing both sides of Eq. (a) by r 2 and integrating, egen dT  r dr 3k and T (r )   egen 6k r 2  C2 (b) Applying the boundary condition at the midpoint (midpoint temperature of the satellite), r  0: T0   egen 6k  0  C2  C2  T0 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 6k ro2  T0 (c) 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    As   3 1/ 4 1/ 4  egen ro / 3   s q solar         (d) PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-64 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/m2 )     (0.75)(5.67  10 8 W/m2  K 4 )  1/ 4  egen (1.25 m) 2 6(5 W/m K)  273 K 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/m3 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-65 Variable Thermal Conductivity, k(T) 2-104C The thermal conductivity of a medium, in general, varies with temperature. 2-105C 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. 2-106C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies with temperature. 2-107C 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. 2-108C 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-66 2-109 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 q  k (T ) 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  L qdx   0 T2  (a  bT  cT )dT 2 T1 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 is 1.29 0.00111   2 2 (5983  600 3 ) W/m 437(598  600)  2 (598  600 )  3  q    6 (925  10 m)  1.35 10 5 W/m2 Discussion For heat flux less than 135 kW/m2, the temperature difference across the silicon wafer thickness will be maintained below 2 °C. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-67 2-110 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 T  T1   k ave  k (Tavg )  k 0 1   2  2   (500 + 350) K    (18 W/m  K)1 + (8.7  10 -4 K -1 )  2    24.66 W/m  K k(T) T1 T2 L Then the rate of heat conduction through the plate becomes T  T2 (500  350)K Q  k avg A 1  (24.66 W/m K)(1.5 m  0.6 m)  22,190 W  22.2 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. 2-111 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 q  kavg T1  T2 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-68 2-112 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. k(T) 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 T T2 k avg T2  k (T )dT   k (1  T )dT   T1 T2  T1 T1  2 0 T2  T1     2 k 0 T  T 3  3   T1 T2  T1 T2 T1      k 0 T2  T1   T23  T13  3    T2  T1 L     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 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 3 L   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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. x 2-69 2-113 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  0.0105T )  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   PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-70 2-114 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 q  2 L kavg T2  T1 kavg T2  T1 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-71 2-115 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 cylinder  Q conv 2 kavg L 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-72 2-116 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 cyl  Q conv  Q rad  Q abs 2 kavg L 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 ) where h = 60 W/m2 K, (1) 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  (0.001K -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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-73 2-117 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 q  4 kavg r1 r2 T1  T2 Q r T T   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(293  T1 )] W/m  K Thus,  1.005 m  T1  293 K   7000 W/m2  [1.33  0.00153(293  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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-74 2-118 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 T Q sphere  4k avg r1 r2 1 2 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 r r1  4 2 T  k (T )dT T1 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 . PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-75 2-119 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 where and r2  (r1  t ) t = wall thickness The rate of heat transfer at the vessel’s outer surface can be expressed as Q sph  Q 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   ( 1.01W/m  K )1  (0.0018 K -1 )   1.6611W/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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-76 2-120 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 sph  Q conv  Q rad  Q abs 4 k avg r1 r2 k avg 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-77 Special Topic: Review of Differential equations 2-121C We utilize appropriate simplifying assumptions when deriving differential equations to obtain an equation that we can deal with and solve. 2-122C 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). 2-123C A differential equation may involve more than one dependent or independent variable. For example, the equation  2T ( 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  2T ( 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 2-124C 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. 2-125C 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. 2-126C 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. 2-127C 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. 2-128C Integration is the inverse of derivation. Derivation increases the order of a derivative by one, integration reduces it by one. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-78 2-129C A differential equation involves derivatives, an algebraic equation does not. 2-130C 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. 2-131C The order of a differential equation is the order of the highest order derivative in the equation. 2-132C 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. 2-133C 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   4 x 2 y  0 is linear and homogeneous since each term is linear in y, and contains the dependent variable or one of its derivatives. 2-134C 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   4 x 2 y  0 has variable coefficients whereas the equation y   4 y  0 has constant coefficients. 2-135C A linear differential equation that involves a single term with the derivatives can be solved by direct integration. 2-136C The general solution of a 3rd order linear and homogeneous differential equation will involve 3 arbitrary constants. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-79 Review Problems 2-137 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 x  0: k dT (0)  q0 dx x  L: k dT ( L) 4  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. 2-138 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 x 2   2T y 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 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-80 2-139E 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 onedimensional since the wall is large relative to its thickness, and the thermal conditions on both sides of the wall are uniform. 3 Thermal 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 . 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), kAs T2 520 R T1  T2  AsT24   s As qsolar L qsolar L x Canceling the area A and substituting the known quantities, (1.2 Btu/h  ft  F) (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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-81 2-140 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 r  r1 : k dT (r1 )  q1 dr r  r2 : k dT (r2 ) 4  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. 2-141 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   2 T 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 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-82 2-142 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. A During a differential time interval dt, the temperature of the body rises by a differential amount dT. Noting that the temperature changes with time only, an energy h balance of the solid for the time interval dt can be expressed as m, c, Ti T  Heat trans fer from the body   The decrease in the energy  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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-83 2-143 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 q 0  Q 0 1200 W   80,0000 W/m2 Abase 150  10 4 m 2 Tsurr q  h T 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 and k 0 L x dT (0)  q 0  80,000 W/m2 dx 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 0 k x = 0:  kC1  q 0  C1   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  295 4 ]  80,000 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 = 819C PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-84 2-144 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 dx 2 and k dT (0) h1 [T1  T (0)]  k dx k h2 T2 h1 T1 dT ( L)  h2 [T ( L)  T 2 ] dx L (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 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 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-85 2-145 A steam pipe is subjected to convection on both the inner and outer surfaces. The mathematical formulation of the problem and expressions for the variation of temperature in the pipe and on the outer surface temperature are to be obtained 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 steady and one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as d  dT  r 0 dr  dr  k and k dT (r1)  hi [Ti  T (r1)] dr Ti hi dT (r2 )  ho[T (r2 )  To ] dr r1 r2 r To ho (b) Integrating the differential equation once with respect to r gives r 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 = r1: k C1  hi [Ti  (C1 ln r1  C2 )] r1 r = r2: k C1  ho [(C1 ln r2  C2 )  To ] r2 Solving for C1 and C2 simultaneously gives C1  T0  Ti r2 k k ln   r1 hi r1 ho r2 and   k  T0  Ti k    Ti   ln r1   C2  Ti  C1 ln r1    r k k  hi r1  hi r1   ln 2   r1 hi r1 ho r2 Substituting C1 and C 2 into the general solution and simplifying, we get the variation of temperature to be r k  k r1 hi r1 T (r )  C1 ln r  Ti  C1 (ln r1  )  Ti  r2 k k hi r1 ln   r1 hi r1 ho r2 (T0  Ti ) ln (c) The outer surface temperature is determined by simply replacing r in the relation above by r2. We get r k (T0  Ti ) ln 2  r1 hi r1 T (r2 )  Ti  r k k ln 2   r1 hi r1 ho r2 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-86 2-146 A 10-m tall exhaust stack discharging exhaust gases at a rate of 1.2 kg/s is subjected to solar radiation and convection at the outer surface. The variation of temperature in the exhaust stack and the inner surface temperature of the exhaust stack 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 pipe. Properties The constant pressure specific heat of exhaust gases is given to be 1600 J/kg ∙ °C and the pipe thermal conductivity is 40 W/m ∙ K. Both the emissivity and solar absorptivity of the exhaust stack outer surface are 0.9. Analysis The outer and inner radii of the pipe are r2  1 m / 2  0.5 m r1  0.5 m  0.1 m  0.4 m The outer surface area of the exhaust stack is As,2  2 r2 L  2 (0.5 m)(10 m)  31.42 m 2 The rate of heat loss from the exhaust gases in the exhaust stack can be determined from  c p (Tin  Tout )  (1.2 kg/s)(1600 J/kg  C)(30) C  57600 W Q loss  m The heat loss on the outer surface of the exhaust stack by radiation and convection can be expressed as Q loss 4  h [T (r2 )  T ]   [T (r2 ) 4  Tsurr ]   s q solar As, 2 57600 W 31.42 m 2  (8 W/m2  K )[T (r2 )  (27  273)] K  (0.9)(5.67  10 8 W/m2  K 4 )[T (r2 ) 4  (27  273) 4 ] K 4  (0.9)(150 W/m2 ) Copy the following line and paste on a blank EES screen to solve the above equation: 57600/31.42=8*(T_r2-(27+273))+0.9*5.67e-8*(T_r2^4-(27+273)^4)-0.9*150 Solving by EES software, the outside surface temperature of the furnace front is T (r2 )  412.7 K (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 Q dT (r1 ) Q loss   loss dr As ,1 2 r1 L T (r2 )  412.7 K (heat flux at the inner exhaust stack surface) (outer exhaust stack surface temperature) PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-87 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 r  r2 : T (r2 )   1 Q loss ln r2  C 2 2 kL   C1   C2  1 Q loss 2 kL 1 Q loss ln r2  T (r2 ) 2 kL Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be 1 Q loss 1 Q loss ln r  ln r2  T (r2 ) 2 kL 2 kL 1 Q loss  ln(r / r2 )  T (r2 ) 2 kL T (r )   (b) The inner surface temperature of the exhaust stack is 1 Q loss ln(r1 / r2 )  T (r2 ) 2 kL 1 57600 W  0.4   ln   412.7 K 2 (40 W/m  K )(10 m)  0.5   417.7 K  418 K T (r1 )   Discussion There is a temperature drop of 5 °C from the inner to the outer surface of the exhaust stack. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-88 2-147E 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 = 8 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  T =160F The boundary conditions for this problem are: k Steam 250F h=12.5 dT (r1 )  h[T  T (r1 )] dr T ( r2 )  T2  160F L = 35 ft (b) Integrating the differential equation once with respect to r gives r 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  C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give C1  h[T  (C1 ln r1  C 2 )] r1 r = r1: k r = r2: T (r2 )  C1 ln r2  C2  T2 Solving for C1 and C2 simultaneously gives C1  T2  T r k ln 2  r1 hr1 and C2  T2  C1 ln r2  T2  T2  T ln r2 r k ln 2  r1 hr1 Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T (r )  C1 ln r  T2  C1 ln r2  C1 (ln r  ln r2 )  T2   T2  T r ln  T2 r2 k r2 ln  r1 hr1 (160  250)F r r ln  160F  26.61 ln  160F 2.4 8 Btu/h  ft  F 2.4 in 2.4 in ln  2 (15 Btu/h  ft 2  F)(2/12 ft) (c) The rate of heat conduction through the pipe is C T  T dT Q  kA  k (2rL) 1  2Lk 2 r k dr r ln 2  r1 hr1  2 (35 ft)( 8 Btu/h  ft  F) ln (160  250)F 8 Btu/h  ft  F 2.4  2 (15 Btu/h  ft 2  F)(2/12 ft)  46,813 Btu/h PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-89 2-148 A compressed air pipe is subjected to uniform heat flux on the outer surface and convection on the inner surface. The mathematical formulation, the variation of temperature in the pipe, and the surface temperatures 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 = 14 W/mK. Analysis (a) Noting that the 85% of the 300 W generated by the strip heater is transferred to the pipe, the heat flux through the outer surface is determined to be qs  Q s Q s 0.85  300 W    169.1 W/m2 A2 2r2 L 2 (0.04 m)(6 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  dT  r 0 dr  dr  r k k dT (r1 )  h[T  T (r1 )] dr Heater r2 The boundary conditions for this problem are: Air, -10°C dT (r2 )  q s dr r1 L=6 m (b) Integrating the differential equation once with respect to r gives r 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  C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give C1 q r  qs  C1  s 2 r2 k r = r2: k r = r1: k   C1 k  k  q s r2 C1 = T   ln r1    h[T  (C1 ln r1  C 2 )]  C 2  T   ln r1  r1 hr1  hr1  k   Substituting C1 and C2 into the general solution, the variation of temperature is determined to be    r k  k  k  q s r2 C1  T   ln r  ln r1  C1  T   ln   T (r )  C1 ln r  T   ln r1  hr1  hr1     r1 hr1  k  r  (169.1 W/m2 )(0.04 m)  r  14 W/m  K   10C   ln   10  0.483 ln  12.61 2  r  14 W/m  K  r1   1 (30 W/m  K)(0.037 m)  (c) The inner and outer surface temperatures are determined by direct substitution to be  r  Inner surface (r = r1): T (r1 )  10  0.483 ln 1  12.61  10  0.4830  12.61  3.91C  r1   r   0.04   12.61  3.87C Outer surface (r = r2): T (r2 )  10  0.483 ln 2  12.61  10  0.483 ln r 0 . 037    1  Discussion Note that the pipe is essentially isothermal at a temperature of about -3.9C. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-90 2-149 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 c  60 W/m K (7900 kg/m )(500 J/kg  K) 3  15.19  10 6 m 2 /s 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  2 dr r  dr  15.19  10 6 m 2 /s Multiply both sides of the differential equation by r 2 and rearranging gives d  2 dT   50 K/s r2 r  dr  dr  15.19  10 6 m 2 /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 6 2 dr 15.19  10 m /s  3   C1  0 Dividing both sides of Eq. (a) by r 2 gives dT  50 K/s r     6 2 dr 15.19  10 m /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  3  dr 15.19  10 m /s    (60 W/m K )( 4 )( 0.125 m) 2 50 K/s 15.19  10 6  0.125 m    3 m /s   2  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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-91 2-150 A hollow pipe is subjected to specified temperatures at the inner and outer surfaces. There is also heat generation in the pipe. The variation of temperature in the pipe and the center surface temperature of 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 centerline. 2 Thermal conductivity is constant. Properties The thermal conductivity is given to be k = 14 W/m°C. Analysis The rate of heat generation is determined from W W 25,000 W egen     26,750 W/m3 2 2 2 V  ( D2  D1 ) L / 4  (0.4 m)  (0.3 m) 2 (17 m) / 4   Noting that heat transfer is 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 egen and T (r1 )  T1  60C T1 T (r2 )  T2  80C T2 Rearranging the differential equation d  dT  egen r r1 r2 0 r  dr  dr  k r and then integrating once with respect to r, 2 dT  egen r r   C1 dr 2k Rearranging the differential equation again dT egen r C1   dr 2k r and finally integrating again with respect to r, we obtain T (r )   egen r 2  C1 ln r  C 2 4k where C1 and C2 are arbitrary constants. Applying the boundary conditions give T (r1 )  r = r1:  egen r1 2 T (r2 )  4k  C1 ln r1  C 2  egen r2 2  C1 ln r2  C 2 4k Substituting the given values, these equations can be written as r = r2: 60   (26,750)(0.15) 2  C1 ln(0.15)  C 2 4(14) 80   (26,750)( 0.20) 2  C1 ln(0.20)  C 2 4(14) Solving for C1 and C 2 simultaneously gives C1  98.58 C 2  257.8 Substituting C1 and C 2 into the general solution, the variation of temperature is determined to be T (r )   26,750r 2  98.58 ln r  257.8  257.8  477.7r 2  98.58 ln r 4(14) The temperature at the center surface of the pipe is determined by setting radius r to be 17.5 cm, which is the average of the inner radius and outer radius. T (r )  257.8  477.7(0.175) 2  98.58 ln(0.175)  71.3C PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-92 2-151 A spherical ball in which heat is generated uniformly is exposed to iced-water. The temperatures at the center and at the surface of the ball are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional., and there is thermal symmetry about the center point. 3 Thermal conductivity is constant. 4 Heat generation is uniform. Properties The thermal conductivity is given to be k = 45 W/m°C. Analysis The temperatures at the center and at the surface of the ball are determined directly from Ts  T  T0  Ts  egen ro 3h egen ro2 6k  0C  (4.2  10 W/m )(0.12 m)  140C  6 3 3(1200 W/m2 .C) D  140C h T egen (4.2  10 6 W/m3 )( 0.12 m) 2  364C 6(45 W/m.C) 2-152 A spherical reactor of 5-cm diameter operating at steady condition has its heat generation suddenly set to 9 MW/m3. 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/m3 )( 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  k  t r 2 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/m3 (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/m3 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-93 2-153 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(T)  k (1  T )dT  T1 2 T1 0 T2 T2  T1 T     2 k 0 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-94 2-154 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 cylinder  Q conv 2 kavg L 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. PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-95 2-155 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 sph  Q 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, and T∞ = 288 K The average thermal conductivity is T T  T  (393 K)    kavg  k0 1   2 1   ( 9.1W/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1  Q sph 4 r 2 1  4 kavg r1 r2 T1  T2 4 r12 r2  r1  2.51  (393  387.8) K r T T   kavg 2 1 2  (15.5 W/m  K) 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 . PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-96 Fundamentals of Engineering (FE) Exam Problems 2-156 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 2-157 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 2-158 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 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-97 2-159 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 2-160 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) (c)  k  h1 T1  T 2 ) dx (a) k Answer (a) k dT ( L)  h2 T ( L)  T 2 ) dx dT ( L) (d)  k  h2 T1  T 2 ) dx (b) k (e) None of them dT (0)  h1 T (0)  T1 ) dx 2-161 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 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-98 2-162 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 2-163 The variation of temperature in a plane wall is determined to be T(x)=52x+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) 0.10 m (b) 0.20 m (c) 0.25 m (d) 0.40 m (e) 0.50 m Answer (c) 0.25 m Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 38=52*L+25 2-164 The variation of temperature in a plane wall is determined to be T(x)=110 – 60x 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) 30ºC (b) 45ºC (c) 60ºC (d) 75ºC (e) 84ºC Answer (b) 45ºC 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=110-60*L DELTAT=T1-T2 PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-99 2-165 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)  28x  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 2-166 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 2-167 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 50oC and the temperature of the exterior surface is 20oC. The rate of heat transfer per unit of pipe length is (a) 77.7 W/m (b) 89.5 W/m (c) 98.0 W/m (d) 112 W/m (e) 168 W/m Answer (a) 77.7 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=50 [C] T1=20 [C] Q=2*pi*k*(T2-T1)/LN(do/di) PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-100 2-168 Heat is generated in a long 0.3-cm-diameter cylindrical electric heater at a rate of 180 W/cm3. The heat flux at the surface of the heater in steady operation is (a) 12.7 W/cm2 (b) 13.5 W/cm2 (c) 64.7 W/cm2 (d) 180 W/cm2 (e) 191 W/cm2 Answer (b) 13.5 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=180 [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” 2-169 Heat is generated uniformly in a 4-cm-diameter, 12-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) 597 W (b) 760 W b) 826 W (c) 928 W (d) 1020 W Answer (a) 597 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=0.12 [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*W1_V 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" PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-101 2-170 Heat is generated in a 10-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) 205C (c) 280C (d) 370C (e) 495C Answer (d) 370C D=0.10 Ts=120 k=25 e_gen=15E+6 T=Ts+e_gen*(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" 2-171 Heat is generated in a 3-cm-diameter spherical radioactive material uniformly at a rate of 15 W/cm3. 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) 2-172 …. 2-174 Design and Essay Problems  PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.