Page 1 Chapter 2 Heat Conduction Equation 2-58 Variable Thermal Conductivity 2-94C 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. 2-95C The thermal conductivity of a medium, in general, varies with temperature. 2-96C 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. 2-97C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies with temperature. 2-98C 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. 2-99 A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer through the plate is to be determined. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies quadratically. 3 There is no heat generation. Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ? 0 2 1 ? . 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 T T 1 2 and can be determined from ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 1 2 1 2 2 0 1 2 3 1 3 2 1 2 0 1 2 3 0 1 2 2 0 1 2 ave 3 1 3 3 ) 1 ( ) ( 2 1 2 1 2 1 T T T T k T T T T T T k T T T T k T T dT T k T T dT T k k T T T T T T ? ? ? ? This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal conductivity k ave 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 ? ? L T T A T T T T k L T T A k Q 2 1 2 1 2 1 2 2 0 2 1 ave 3 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 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. T 2 x k(T) L T 1 Page 2 Chapter 2 Heat Conduction Equation 2-58 Variable Thermal Conductivity 2-94C 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. 2-95C The thermal conductivity of a medium, in general, varies with temperature. 2-96C 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. 2-97C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies with temperature. 2-98C 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. 2-99 A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer through the plate is to be determined. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies quadratically. 3 There is no heat generation. Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ? 0 2 1 ? . 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 T T 1 2 and can be determined from ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 1 2 1 2 2 0 1 2 3 1 3 2 1 2 0 1 2 3 0 1 2 2 0 1 2 ave 3 1 3 3 ) 1 ( ) ( 2 1 2 1 2 1 T T T T k T T T T T T k T T T T k T T dT T k T T dT T k k T T T T T T ? ? ? ? This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal conductivity k ave 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 ? ? L T T A T T T T k L T T A k Q 2 1 2 1 2 1 2 2 0 2 1 ave 3 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 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. T 2 x k(T) L T 1 Chapter 2 Heat Conduction Equation 2-59 2-100 A cylindrical 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 T ( ) ( ) ? ? 0 1 ? . Solution (a) The rate of heat transfer through the shell is expressed as ? ln( / ) Q k L T T r r cylinder ave ? ? 2 1 2 2 1 ? where L is the length of the cylinder, r 1 is the inner radius, and r 2 is the outer radius, and ? ? ? ? ? ? ? ? ? ? 2 1 ) ( 1 2 0 ave ave T T k T k k ? is the average thermal conductivity. (b) To determine the temperature distribution in the shell, we begin with the Fourierâ€™s law of heat conduction expressed as ? ( ) Q k T A dT dr ? ? where the rate of conduction heat transfer ? Q is constant and the heat conduction area A = 2 ?rL is variable. Separating the variables in the above equation and integrating from r = r 1 where T r T ( ) 1 1 ? to any r where T r T ( ) ? , we get ? ? ? ? T T r r dT T k L r dr Q 1 1 ) ( 2 ? ? Substituting k T k T ( ) ( ) ? ? 0 1 ? and performing the integrations gives ? ln [( ) ( ) / ] Q r r Lk T T T T 1 0 1 2 1 2 2 2 ? ? ? ? ? ? ? Substituting the ? Q expression from part (a) and rearranging give T T k k r r r r T T T T 2 0 1 2 1 1 2 1 2 1 2 2 2 0 ? ? ? ? ? ? ? ? ? ave ln( / ) ln( / ) ( ) 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 k k r r r r T T T T ( ) ln( / ) ln( / ) ( ) ? ? ? ? ? ? ? 1 1 2 2 2 0 1 2 1 1 2 1 2 1 ? ? ? ? ave 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 T T 1 2 and . r 2 T 2 r r 1 T 1 k(T) Page 3 Chapter 2 Heat Conduction Equation 2-58 Variable Thermal Conductivity 2-94C 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. 2-95C The thermal conductivity of a medium, in general, varies with temperature. 2-96C 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. 2-97C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies with temperature. 2-98C 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. 2-99 A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer through the plate is to be determined. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies quadratically. 3 There is no heat generation. Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ? 0 2 1 ? . 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 T T 1 2 and can be determined from ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 1 2 1 2 2 0 1 2 3 1 3 2 1 2 0 1 2 3 0 1 2 2 0 1 2 ave 3 1 3 3 ) 1 ( ) ( 2 1 2 1 2 1 T T T T k T T T T T T k T T T T k T T dT T k T T dT T k k T T T T T T ? ? ? ? This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal conductivity k ave 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 ? ? L T T A T T T T k L T T A k Q 2 1 2 1 2 1 2 2 0 2 1 ave 3 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 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. T 2 x k(T) L T 1 Chapter 2 Heat Conduction Equation 2-59 2-100 A cylindrical 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 T ( ) ( ) ? ? 0 1 ? . Solution (a) The rate of heat transfer through the shell is expressed as ? ln( / ) Q k L T T r r cylinder ave ? ? 2 1 2 2 1 ? where L is the length of the cylinder, r 1 is the inner radius, and r 2 is the outer radius, and ? ? ? ? ? ? ? ? ? ? 2 1 ) ( 1 2 0 ave ave T T k T k k ? is the average thermal conductivity. (b) To determine the temperature distribution in the shell, we begin with the Fourierâ€™s law of heat conduction expressed as ? ( ) Q k T A dT dr ? ? where the rate of conduction heat transfer ? Q is constant and the heat conduction area A = 2 ?rL is variable. Separating the variables in the above equation and integrating from r = r 1 where T r T ( ) 1 1 ? to any r where T r T ( ) ? , we get ? ? ? ? T T r r dT T k L r dr Q 1 1 ) ( 2 ? ? Substituting k T k T ( ) ( ) ? ? 0 1 ? and performing the integrations gives ? ln [( ) ( ) / ] Q r r Lk T T T T 1 0 1 2 1 2 2 2 ? ? ? ? ? ? ? Substituting the ? Q expression from part (a) and rearranging give T T k k r r r r T T T T 2 0 1 2 1 1 2 1 2 1 2 2 2 0 ? ? ? ? ? ? ? ? ? ave ln( / ) ln( / ) ( ) 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 k k r r r r T T T T ( ) ln( / ) ln( / ) ( ) ? ? ? ? ? ? ? 1 1 2 2 2 0 1 2 1 1 2 1 2 1 ? ? ? ? ave 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 T T 1 2 and . r 2 T 2 r r 1 T 1 k(T) Chapter 2 Heat Conduction Equation 2-60 2-101 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 T ( ) ( ) ? ? 0 1 ? . Solution (a) The rate of heat transfer through the shell is expressed as ? Q k r r T T r r sphere ave ? ? ? 4 1 2 1 2 2 1 ? where r 1 is the inner radius, r 2 is the outer radius, and ? ? ? ? ? ? ? ? ? ? 2 1 ) ( 1 2 0 ave ave T T k T k k ? is the average thermal conductivity. (b) To determine the temperature distribution in the shell, we begin with the Fourierâ€™s law of heat conduction expressed as ? ( ) Q k T A dT dr ? ? where the rate of conduction heat transfer ? Q is constant and the heat conduction area A = 4 ?r 2 is variable. Separating the variables in the above equation and integrating from r = r 1 where T r T ( ) 1 1 ? to any r where T r T ( ) ? , we get ? ? ? ? T T r r dT T k r dr Q 1 1 ) ( 4 2 ? ? Substituting k T k T ( ) ( ) ? ? 0 1 ? and performing the integrations gives ] 2 / ) ( ) [( 4 1 1 2 1 2 1 0 1 T T T T k r r Q ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Substituting the ? Q expression from part (a) and rearranging give T T k k r r r r r r T T T T 2 0 2 1 2 1 1 2 1 2 1 2 2 2 0 ? ? ? ? ? ? ? ? ? ? ? ave ( ) ( ) ( ) 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 k k r r r r r r T T T T ( ) ( ) ( ) ( ) ? ? ? ? ? ? ? ? ? 1 1 2 2 2 0 2 1 2 1 1 2 1 2 1 ? ? ? ? ave 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 T T 1 2 and . r 1 r 2 T 1 k(T) r T 2 Page 4 Chapter 2 Heat Conduction Equation 2-58 Variable Thermal Conductivity 2-94C 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. 2-95C The thermal conductivity of a medium, in general, varies with temperature. 2-96C 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. 2-97C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies with temperature. 2-98C 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. 2-99 A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer through the plate is to be determined. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies quadratically. 3 There is no heat generation. Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ? 0 2 1 ? . 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 T T 1 2 and can be determined from ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 1 2 1 2 2 0 1 2 3 1 3 2 1 2 0 1 2 3 0 1 2 2 0 1 2 ave 3 1 3 3 ) 1 ( ) ( 2 1 2 1 2 1 T T T T k T T T T T T k T T T T k T T dT T k T T dT T k k T T T T T T ? ? ? ? This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal conductivity k ave 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 ? ? L T T A T T T T k L T T A k Q 2 1 2 1 2 1 2 2 0 2 1 ave 3 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 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. T 2 x k(T) L T 1 Chapter 2 Heat Conduction Equation 2-59 2-100 A cylindrical 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 T ( ) ( ) ? ? 0 1 ? . Solution (a) The rate of heat transfer through the shell is expressed as ? ln( / ) Q k L T T r r cylinder ave ? ? 2 1 2 2 1 ? where L is the length of the cylinder, r 1 is the inner radius, and r 2 is the outer radius, and ? ? ? ? ? ? ? ? ? ? 2 1 ) ( 1 2 0 ave ave T T k T k k ? is the average thermal conductivity. (b) To determine the temperature distribution in the shell, we begin with the Fourierâ€™s law of heat conduction expressed as ? ( ) Q k T A dT dr ? ? where the rate of conduction heat transfer ? Q is constant and the heat conduction area A = 2 ?rL is variable. Separating the variables in the above equation and integrating from r = r 1 where T r T ( ) 1 1 ? to any r where T r T ( ) ? , we get ? ? ? ? T T r r dT T k L r dr Q 1 1 ) ( 2 ? ? Substituting k T k T ( ) ( ) ? ? 0 1 ? and performing the integrations gives ? ln [( ) ( ) / ] Q r r Lk T T T T 1 0 1 2 1 2 2 2 ? ? ? ? ? ? ? Substituting the ? Q expression from part (a) and rearranging give T T k k r r r r T T T T 2 0 1 2 1 1 2 1 2 1 2 2 2 0 ? ? ? ? ? ? ? ? ? ave ln( / ) ln( / ) ( ) 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 k k r r r r T T T T ( ) ln( / ) ln( / ) ( ) ? ? ? ? ? ? ? 1 1 2 2 2 0 1 2 1 1 2 1 2 1 ? ? ? ? ave 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 T T 1 2 and . r 2 T 2 r r 1 T 1 k(T) Chapter 2 Heat Conduction Equation 2-60 2-101 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 T ( ) ( ) ? ? 0 1 ? . Solution (a) The rate of heat transfer through the shell is expressed as ? Q k r r T T r r sphere ave ? ? ? 4 1 2 1 2 2 1 ? where r 1 is the inner radius, r 2 is the outer radius, and ? ? ? ? ? ? ? ? ? ? 2 1 ) ( 1 2 0 ave ave T T k T k k ? is the average thermal conductivity. (b) To determine the temperature distribution in the shell, we begin with the Fourierâ€™s law of heat conduction expressed as ? ( ) Q k T A dT dr ? ? where the rate of conduction heat transfer ? Q is constant and the heat conduction area A = 4 ?r 2 is variable. Separating the variables in the above equation and integrating from r = r 1 where T r T ( ) 1 1 ? to any r where T r T ( ) ? , we get ? ? ? ? T T r r dT T k r dr Q 1 1 ) ( 4 2 ? ? Substituting k T k T ( ) ( ) ? ? 0 1 ? and performing the integrations gives ] 2 / ) ( ) [( 4 1 1 2 1 2 1 0 1 T T T T k r r Q ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Substituting the ? Q expression from part (a) and rearranging give T T k k r r r r r r T T T T 2 0 2 1 2 1 1 2 1 2 1 2 2 2 0 ? ? ? ? ? ? ? ? ? ? ? ave ( ) ( ) ( ) 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 k k r r r r r r T T T T ( ) ( ) ( ) ( ) ? ? ? ? ? ? ? ? ? 1 1 2 2 2 0 2 1 2 1 1 2 1 2 1 ? ? ? ? ave 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 T T 1 2 and . r 1 r 2 T 1 k(T) r T 2 Chapter 2 Heat Conduction Equation 2-61 2-102 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 T ( ) ( ) ? ? 0 1 ? . Analysis The average thermal conductivity of the medium in this case is simply the conductivity value at the average temperature since the thermal conductivity varies linearly with temperature, and is determined to be K W/m 24 . 34 2 K 350) + (500 ) K 10 (8.7 + 1 K) W/m 25 ( 2 1 ) ( 1 - 4 - 1 2 0 ave ave ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? T T k T k k Then the rate of heat conduction through the plate becomes W 30,820 ? ? ? ? ? ? ? m 15 . 0 0)K 35 (500 m) 0.6 m K)(1.5 W/m 24 . 34 ( 2 1 ave L T T A k Q ? 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. T 2 k(T) T 1 L Page 5 Chapter 2 Heat Conduction Equation 2-58 Variable Thermal Conductivity 2-94C 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. 2-95C The thermal conductivity of a medium, in general, varies with temperature. 2-96C 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. 2-97C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies with temperature. 2-98C 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. 2-99 A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer through the plate is to be determined. Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies quadratically. 3 There is no heat generation. Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ? 0 2 1 ? . 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 T T 1 2 and can be determined from ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 1 2 1 2 2 0 1 2 3 1 3 2 1 2 0 1 2 3 0 1 2 2 0 1 2 ave 3 1 3 3 ) 1 ( ) ( 2 1 2 1 2 1 T T T T k T T T T T T k T T T T k T T dT T k T T dT T k k T T T T T T ? ? ? ? This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal conductivity k ave 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 ? ? L T T A T T T T k L T T A k Q 2 1 2 1 2 1 2 2 0 2 1 ave 3 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 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. T 2 x k(T) L T 1 Chapter 2 Heat Conduction Equation 2-59 2-100 A cylindrical 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 T ( ) ( ) ? ? 0 1 ? . Solution (a) The rate of heat transfer through the shell is expressed as ? ln( / ) Q k L T T r r cylinder ave ? ? 2 1 2 2 1 ? where L is the length of the cylinder, r 1 is the inner radius, and r 2 is the outer radius, and ? ? ? ? ? ? ? ? ? ? 2 1 ) ( 1 2 0 ave ave T T k T k k ? is the average thermal conductivity. (b) To determine the temperature distribution in the shell, we begin with the Fourierâ€™s law of heat conduction expressed as ? ( ) Q k T A dT dr ? ? where the rate of conduction heat transfer ? Q is constant and the heat conduction area A = 2 ?rL is variable. Separating the variables in the above equation and integrating from r = r 1 where T r T ( ) 1 1 ? to any r where T r T ( ) ? , we get ? ? ? ? T T r r dT T k L r dr Q 1 1 ) ( 2 ? ? Substituting k T k T ( ) ( ) ? ? 0 1 ? and performing the integrations gives ? ln [( ) ( ) / ] Q r r Lk T T T T 1 0 1 2 1 2 2 2 ? ? ? ? ? ? ? Substituting the ? Q expression from part (a) and rearranging give T T k k r r r r T T T T 2 0 1 2 1 1 2 1 2 1 2 2 2 0 ? ? ? ? ? ? ? ? ? ave ln( / ) ln( / ) ( ) 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 k k r r r r T T T T ( ) ln( / ) ln( / ) ( ) ? ? ? ? ? ? ? 1 1 2 2 2 0 1 2 1 1 2 1 2 1 ? ? ? ? ave 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 T T 1 2 and . r 2 T 2 r r 1 T 1 k(T) Chapter 2 Heat Conduction Equation 2-60 2-101 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 T ( ) ( ) ? ? 0 1 ? . Solution (a) The rate of heat transfer through the shell is expressed as ? Q k r r T T r r sphere ave ? ? ? 4 1 2 1 2 2 1 ? where r 1 is the inner radius, r 2 is the outer radius, and ? ? ? ? ? ? ? ? ? ? 2 1 ) ( 1 2 0 ave ave T T k T k k ? is the average thermal conductivity. (b) To determine the temperature distribution in the shell, we begin with the Fourierâ€™s law of heat conduction expressed as ? ( ) Q k T A dT dr ? ? where the rate of conduction heat transfer ? Q is constant and the heat conduction area A = 4 ?r 2 is variable. Separating the variables in the above equation and integrating from r = r 1 where T r T ( ) 1 1 ? to any r where T r T ( ) ? , we get ? ? ? ? T T r r dT T k r dr Q 1 1 ) ( 4 2 ? ? Substituting k T k T ( ) ( ) ? ? 0 1 ? and performing the integrations gives ] 2 / ) ( ) [( 4 1 1 2 1 2 1 0 1 T T T T k r r Q ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Substituting the ? Q expression from part (a) and rearranging give T T k k r r r r r r T T T T 2 0 2 1 2 1 1 2 1 2 1 2 2 2 0 ? ? ? ? ? ? ? ? ? ? ? ave ( ) ( ) ( ) 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 k k r r r r r r T T T T ( ) ( ) ( ) ( ) ? ? ? ? ? ? ? ? ? 1 1 2 2 2 0 2 1 2 1 1 2 1 2 1 ? ? ? ? ave 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 T T 1 2 and . r 1 r 2 T 1 k(T) r T 2 Chapter 2 Heat Conduction Equation 2-61 2-102 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 T ( ) ( ) ? ? 0 1 ? . Analysis The average thermal conductivity of the medium in this case is simply the conductivity value at the average temperature since the thermal conductivity varies linearly with temperature, and is determined to be K W/m 24 . 34 2 K 350) + (500 ) K 10 (8.7 + 1 K) W/m 25 ( 2 1 ) ( 1 - 4 - 1 2 0 ave ave ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? T T k T k k Then the rate of heat conduction through the plate becomes W 30,820 ? ? ? ? ? ? ? m 15 . 0 0)K 35 (500 m) 0.6 m K)(1.5 W/m 24 . 34 ( 2 1 ave L T T A k Q ? 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. T 2 k(T) T 1 L Chapter 2 Heat Conduction Equation 2-62 2-103 "GIVEN" A=1.5*0.6 "[m^2]" L=0.15 "[m]" "T_1=500 [K], parameter to be varied" T_2=350 "[K]" k_0=25 "[W/m-K]" beta=8.7E-4 "[1/K]" "ANALYSIS" k=k_0*(1+beta*T) T=1/2*(T_1+T_2) Q_dot=k*A*(T_1-T_2)/L T1 [W] Q [W] 400 9947 425 15043 450 20220 475 25479 500 30819 525 36241 550 41745 575 47330 600 52997 625 58745 650 64575 675 70486 700 76479 400 450 500 550 600 650 700 0 10000 20000 30000 40000 50000 60000 70000 80000 T 1 [K] Q [W]Read More

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