Page 1 Chapter 4 Transient Heat Conduction 4-55 Transient Heat Conduction in Multidimensional Systems 4-69C The product solution enables us to determine the dimensionless temperature of two- or three- dimensional heat transfer problems as the product of dimensionless temperatures of one-dimensional heat transfer problems. The dimensionless temperature for a two-dimensional problem is determined by determining the dimensionless temperatures in both directions, and taking their product. 4-70C The dimensionless temperature for a three-dimensional heat transfer is determined by determining the dimensionless temperatures of one-dimensional geometries whose intersection is the three dimensional geometry, and taking their product. 4-71C This short cylinder is physically formed by the intersection of a long cylinder and a plane wall. The dimensionless temperatures at the center of plane wall and at the center of the cylinder are determined first. Their product yields the dimensionless temperature at the center of the short cylinder. 4-72C The heat transfer in this short cylinder is one-dimensional since there is no heat transfer in the axial direction. The temperature will vary in the radial direction only. Page 2 Chapter 4 Transient Heat Conduction 4-55 Transient Heat Conduction in Multidimensional Systems 4-69C The product solution enables us to determine the dimensionless temperature of two- or three- dimensional heat transfer problems as the product of dimensionless temperatures of one-dimensional heat transfer problems. The dimensionless temperature for a two-dimensional problem is determined by determining the dimensionless temperatures in both directions, and taking their product. 4-70C The dimensionless temperature for a three-dimensional heat transfer is determined by determining the dimensionless temperatures of one-dimensional geometries whose intersection is the three dimensional geometry, and taking their product. 4-71C This short cylinder is physically formed by the intersection of a long cylinder and a plane wall. The dimensionless temperatures at the center of plane wall and at the center of the cylinder are determined first. Their product yields the dimensionless temperature at the center of the short cylinder. 4-72C The heat transfer in this short cylinder is one-dimensional since there is no heat transfer in the axial direction. The temperature will vary in the radial direction only. Chapter 4 Transient Heat Conduction 4-56 4-73 A short cylinder is allowed to cool in atmospheric air. The temperatures at the centers of the cylinder and the top surface as well as the total heat transfer from the cylinder for 15 min of cooling are to be determined. Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. 2 The thermal properties of the cylinder are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of brass are given to be ? ? 8530 kg/ m 3 , C kJ/kg 389 . 0 ? ? ? p C , C W/m 110 ? ? ? k , and ? ? ? ? 3 39 10 5 . m / s 2 . Analysis This short cylinder can physically be formed by the intersection of a long cylinder of radius D/2 = 4 cm and a plane wall of thickness 2L = 15 cm. We measure x from the midplane. (a) The Biot number is calculated for the plane wall to be 02727 . 0 ) C W/m. 110 ( ) m 075 . 0 )( C . W/m 40 ( 2 ? ? ? ? ? k hL Bi The constants ? 1 1 and A corresponding to this Biot number are, from Table 4-1, 0050 . 1 and 164 . 0 1 1 ? ? A ? The Fourier number is ? ? ? ? ? ? ? ? ? t L 2 5 339 10 0 075 5424 0 2 ( . ( . . . m / s)(15 min 60 s / min) m) 2 2 Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the dimensionless temperature at the center of the plane wall is determined from 869 . 0 ) 0050 . 1 ( ) 424 . 5 ( ) 164 . 0 ( 1 0 , 2 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? e e A T T T T i wall o We repeat the same calculations for the long cylinder, 01455 . 0 ) C W/m. 110 ( ) m 04 . 0 )( C . W/m 40 ( 2 0 ? ? ? ? ? k hr Bi ? 1 1 01704 10038 ? ? . . and A ? ? ? ? ? ? ? ? ? t r o 2 5 339 10 0 04 19 069 0 2 ( . ( . . . m / s)(15 60 s) m) 2 2 ? ? ? o cyl i T T T T A e e , ( . ) ( . ) ( . ) . ? ? ? ? ? ? ? ? ? ? 0 1 0 1704 19 069 1 2 2 10038 0577 Then the center temperature of the short cylinder becomes C 85.1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ) , 0 , 0 ( 501 . 0 20 150 20 ) , 0 , 0 ( 501 . 0 577 . 0 869 . 0 ) , 0 , 0 ( , , t T t T T T T t T cyl o wall o cylinder short i D0 = 8 cm L = 15 cm z Brass cylinder Ti = 150 ?C r Air T ? = 20 ?C Page 3 Chapter 4 Transient Heat Conduction 4-55 Transient Heat Conduction in Multidimensional Systems 4-69C The product solution enables us to determine the dimensionless temperature of two- or three- dimensional heat transfer problems as the product of dimensionless temperatures of one-dimensional heat transfer problems. The dimensionless temperature for a two-dimensional problem is determined by determining the dimensionless temperatures in both directions, and taking their product. 4-70C The dimensionless temperature for a three-dimensional heat transfer is determined by determining the dimensionless temperatures of one-dimensional geometries whose intersection is the three dimensional geometry, and taking their product. 4-71C This short cylinder is physically formed by the intersection of a long cylinder and a plane wall. The dimensionless temperatures at the center of plane wall and at the center of the cylinder are determined first. Their product yields the dimensionless temperature at the center of the short cylinder. 4-72C The heat transfer in this short cylinder is one-dimensional since there is no heat transfer in the axial direction. The temperature will vary in the radial direction only. Chapter 4 Transient Heat Conduction 4-56 4-73 A short cylinder is allowed to cool in atmospheric air. The temperatures at the centers of the cylinder and the top surface as well as the total heat transfer from the cylinder for 15 min of cooling are to be determined. Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. 2 The thermal properties of the cylinder are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of brass are given to be ? ? 8530 kg/ m 3 , C kJ/kg 389 . 0 ? ? ? p C , C W/m 110 ? ? ? k , and ? ? ? ? 3 39 10 5 . m / s 2 . Analysis This short cylinder can physically be formed by the intersection of a long cylinder of radius D/2 = 4 cm and a plane wall of thickness 2L = 15 cm. We measure x from the midplane. (a) The Biot number is calculated for the plane wall to be 02727 . 0 ) C W/m. 110 ( ) m 075 . 0 )( C . W/m 40 ( 2 ? ? ? ? ? k hL Bi The constants ? 1 1 and A corresponding to this Biot number are, from Table 4-1, 0050 . 1 and 164 . 0 1 1 ? ? A ? The Fourier number is ? ? ? ? ? ? ? ? ? t L 2 5 339 10 0 075 5424 0 2 ( . ( . . . m / s)(15 min 60 s / min) m) 2 2 Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the dimensionless temperature at the center of the plane wall is determined from 869 . 0 ) 0050 . 1 ( ) 424 . 5 ( ) 164 . 0 ( 1 0 , 2 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? e e A T T T T i wall o We repeat the same calculations for the long cylinder, 01455 . 0 ) C W/m. 110 ( ) m 04 . 0 )( C . W/m 40 ( 2 0 ? ? ? ? ? k hr Bi ? 1 1 01704 10038 ? ? . . and A ? ? ? ? ? ? ? ? ? t r o 2 5 339 10 0 04 19 069 0 2 ( . ( . . . m / s)(15 60 s) m) 2 2 ? ? ? o cyl i T T T T A e e , ( . ) ( . ) ( . ) . ? ? ? ? ? ? ? ? ? ? 0 1 0 1704 19 069 1 2 2 10038 0577 Then the center temperature of the short cylinder becomes C 85.1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ) , 0 , 0 ( 501 . 0 20 150 20 ) , 0 , 0 ( 501 . 0 577 . 0 869 . 0 ) , 0 , 0 ( , , t T t T T T T t T cyl o wall o cylinder short i D0 = 8 cm L = 15 cm z Brass cylinder Ti = 150 ?C r Air T ? = 20 ?C Chapter 4 Transient Heat Conduction 4-57 (b) The center of the top surface of the cylinder is still at the center of the long cylinder (r = 0), but at the outer surface of the plane wall (x = L). Therefore, we first need to determine the dimensionless temperature at the surface of the wall. 857 . 0 ) 164 . 0 cos( ) 0050 . 1 ( ) / cos( ) , ( ) , ( ) 424 . 5 ( ) 164 . 0 ( 1 1 2 2 1 ? ? ? ? ? ? ? ? ? ? e L L e A T T T t x T t L i wall ? ? ? ? Then the center temperature of the top surface of the cylinder becomes C 84.2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ) , 0 , ( 494 . 0 20 150 20 ) , 0 , ( 494 . 0 577 . 0 857 . 0 ) , ( ) , 0 , ( , t L T t L T t L T T T t L T cyl o wall cylinder short i ? ? (c) We first need to determine the maximum heat can be transferred from the cylinder ? ? kJ 325 C ) 20 150 )( C kJ/kg. 389 . 0 )( kg 43 . 6 ( ) ( kg 43 . 6 m) 15 . 0 ( m) 04 . 0 ( ) kg/m 8530 ( max 2. 3 2 ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? T T mC Q L r V m i p o Then we determine the dimensionless heat transfer ratios for both geometries as 135 . 0 164 . 0 ) 164 . 0 sin( ) 869 . 0 ( 1 ) sin( 1 1 1 , max ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? wall o wall Q Q 427 . 0 1704 . 0 0846 . 0 ) 577 . 0 ( 2 1 ) ( 2 1 1 1 1 , max ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? J Q Q cyl o cyl The heat transfer ratio for the short cylinder is 504 . 0 ) 135 . 0 1 )( 427 . 0 ( 135 . 0 1 max max max max ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? wall plane cylinder long wall plane cylinder short Q Q Q Q Q Q Q Q Then the total heat transfer from the short cylinder during the first 15 minutes of cooling becomes kJ 164 ? ? ? kJ) 325 )( 504 . 0 ( 503 . 0 max Q Q Page 4 Chapter 4 Transient Heat Conduction 4-55 Transient Heat Conduction in Multidimensional Systems 4-69C The product solution enables us to determine the dimensionless temperature of two- or three- dimensional heat transfer problems as the product of dimensionless temperatures of one-dimensional heat transfer problems. The dimensionless temperature for a two-dimensional problem is determined by determining the dimensionless temperatures in both directions, and taking their product. 4-70C The dimensionless temperature for a three-dimensional heat transfer is determined by determining the dimensionless temperatures of one-dimensional geometries whose intersection is the three dimensional geometry, and taking their product. 4-71C This short cylinder is physically formed by the intersection of a long cylinder and a plane wall. The dimensionless temperatures at the center of plane wall and at the center of the cylinder are determined first. Their product yields the dimensionless temperature at the center of the short cylinder. 4-72C The heat transfer in this short cylinder is one-dimensional since there is no heat transfer in the axial direction. The temperature will vary in the radial direction only. Chapter 4 Transient Heat Conduction 4-56 4-73 A short cylinder is allowed to cool in atmospheric air. The temperatures at the centers of the cylinder and the top surface as well as the total heat transfer from the cylinder for 15 min of cooling are to be determined. Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. 2 The thermal properties of the cylinder are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of brass are given to be ? ? 8530 kg/ m 3 , C kJ/kg 389 . 0 ? ? ? p C , C W/m 110 ? ? ? k , and ? ? ? ? 3 39 10 5 . m / s 2 . Analysis This short cylinder can physically be formed by the intersection of a long cylinder of radius D/2 = 4 cm and a plane wall of thickness 2L = 15 cm. We measure x from the midplane. (a) The Biot number is calculated for the plane wall to be 02727 . 0 ) C W/m. 110 ( ) m 075 . 0 )( C . W/m 40 ( 2 ? ? ? ? ? k hL Bi The constants ? 1 1 and A corresponding to this Biot number are, from Table 4-1, 0050 . 1 and 164 . 0 1 1 ? ? A ? The Fourier number is ? ? ? ? ? ? ? ? ? t L 2 5 339 10 0 075 5424 0 2 ( . ( . . . m / s)(15 min 60 s / min) m) 2 2 Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the dimensionless temperature at the center of the plane wall is determined from 869 . 0 ) 0050 . 1 ( ) 424 . 5 ( ) 164 . 0 ( 1 0 , 2 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? e e A T T T T i wall o We repeat the same calculations for the long cylinder, 01455 . 0 ) C W/m. 110 ( ) m 04 . 0 )( C . W/m 40 ( 2 0 ? ? ? ? ? k hr Bi ? 1 1 01704 10038 ? ? . . and A ? ? ? ? ? ? ? ? ? t r o 2 5 339 10 0 04 19 069 0 2 ( . ( . . . m / s)(15 60 s) m) 2 2 ? ? ? o cyl i T T T T A e e , ( . ) ( . ) ( . ) . ? ? ? ? ? ? ? ? ? ? 0 1 0 1704 19 069 1 2 2 10038 0577 Then the center temperature of the short cylinder becomes C 85.1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ) , 0 , 0 ( 501 . 0 20 150 20 ) , 0 , 0 ( 501 . 0 577 . 0 869 . 0 ) , 0 , 0 ( , , t T t T T T T t T cyl o wall o cylinder short i D0 = 8 cm L = 15 cm z Brass cylinder Ti = 150 ?C r Air T ? = 20 ?C Chapter 4 Transient Heat Conduction 4-57 (b) The center of the top surface of the cylinder is still at the center of the long cylinder (r = 0), but at the outer surface of the plane wall (x = L). Therefore, we first need to determine the dimensionless temperature at the surface of the wall. 857 . 0 ) 164 . 0 cos( ) 0050 . 1 ( ) / cos( ) , ( ) , ( ) 424 . 5 ( ) 164 . 0 ( 1 1 2 2 1 ? ? ? ? ? ? ? ? ? ? e L L e A T T T t x T t L i wall ? ? ? ? Then the center temperature of the top surface of the cylinder becomes C 84.2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ) , 0 , ( 494 . 0 20 150 20 ) , 0 , ( 494 . 0 577 . 0 857 . 0 ) , ( ) , 0 , ( , t L T t L T t L T T T t L T cyl o wall cylinder short i ? ? (c) We first need to determine the maximum heat can be transferred from the cylinder ? ? kJ 325 C ) 20 150 )( C kJ/kg. 389 . 0 )( kg 43 . 6 ( ) ( kg 43 . 6 m) 15 . 0 ( m) 04 . 0 ( ) kg/m 8530 ( max 2. 3 2 ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? T T mC Q L r V m i p o Then we determine the dimensionless heat transfer ratios for both geometries as 135 . 0 164 . 0 ) 164 . 0 sin( ) 869 . 0 ( 1 ) sin( 1 1 1 , max ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? wall o wall Q Q 427 . 0 1704 . 0 0846 . 0 ) 577 . 0 ( 2 1 ) ( 2 1 1 1 1 , max ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? J Q Q cyl o cyl The heat transfer ratio for the short cylinder is 504 . 0 ) 135 . 0 1 )( 427 . 0 ( 135 . 0 1 max max max max ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? wall plane cylinder long wall plane cylinder short Q Q Q Q Q Q Q Q Then the total heat transfer from the short cylinder during the first 15 minutes of cooling becomes kJ 164 ? ? ? kJ) 325 )( 504 . 0 ( 503 . 0 max Q Q Chapter 4 Transient Heat Conduction 4-58 4-74 "!PROBLEM 4-74" "GIVEN" D=0.08 "[m]" r_o=D/2 height=0.15 "[m]" L=height/2 T_i=150 "[C]" T_infinity=20 "[C]" h=40 "[W/m^2-C]" "time=15 [min], parameter to be varied" "PROPERTIES" k=110 "[W/m-C]" rho=8530 "[kg/m^3]" C_p=0.389 "[kJ/kg-C]" alpha=3.39E-5 "[m^2/s]" "ANALYSIS" "(a)" "This short cylinder can physically be formed by the intersection of a long cylinder of radius r_o and a plane wall of thickness 2L" "For plane wall" Bi_w=(h*L)/k "From Table 4-1 corresponding to this Bi number, we read" lambda_1_w=0.2282 "w stands for wall" A_1_w=1.0060 tau_w=(alpha*time*Convert(min, s))/L^2 theta_o_w=A_1_w*exp(-lambda_1_w^2*tau_w) "theta_o_w=(T_o_w-T_infinity)/(T_i-T_infinity)" "For long cylinder" Bi_c=(h*r_o)/k "c stands for cylinder" "From Table 4-1 corresponding to this Bi number, we read" lambda_1_c=0.1704 A_1_c=1.0038 tau_c=(alpha*time*Convert(min, s))/r_o^2 theta_o_c=A_1_c*exp(-lambda_1_c^2*tau_c) "theta_o_c=(T_o_c-T_infinity)/(T_i-T_infinity)" (T_o_o-T_infinity)/(T_i-T_infinity)=theta_o_w*theta_o_c "center temperature of short cylinder" "(b)" theta_L_w=A_1_w*exp(-lambda_1_w^2*tau_w)*Cos(lambda_1_w*L/L) "theta_L_w=(T_L_w- T_infinity)/(T_i-T_infinity)" (T_L_o-T_infinity)/(T_i-T_infinity)=theta_L_w*theta_o_c "center temperature of the top surface" "(c)" V=pi*r_o^2*(2*L) m=rho*V Q_max=m*C_p*(T_i-T_infinity) Q_w=1-theta_o_w*Sin(lambda_1_w)/lambda_1_w "Q_w=(Q/Q_max)_w" Q_c=1-2*theta_o_c*J_1/lambda_1_c "Q_c=(Q/Q_max)_c" J_1=0.0846 "From Table 4-2, at lambda_1_c" Q/Q_max=Q_w+Q_c*(1-Q_w) "total heat transfer" Page 5 Chapter 4 Transient Heat Conduction 4-55 Transient Heat Conduction in Multidimensional Systems 4-69C The product solution enables us to determine the dimensionless temperature of two- or three- dimensional heat transfer problems as the product of dimensionless temperatures of one-dimensional heat transfer problems. The dimensionless temperature for a two-dimensional problem is determined by determining the dimensionless temperatures in both directions, and taking their product. 4-70C The dimensionless temperature for a three-dimensional heat transfer is determined by determining the dimensionless temperatures of one-dimensional geometries whose intersection is the three dimensional geometry, and taking their product. 4-71C This short cylinder is physically formed by the intersection of a long cylinder and a plane wall. The dimensionless temperatures at the center of plane wall and at the center of the cylinder are determined first. Their product yields the dimensionless temperature at the center of the short cylinder. 4-72C The heat transfer in this short cylinder is one-dimensional since there is no heat transfer in the axial direction. The temperature will vary in the radial direction only. Chapter 4 Transient Heat Conduction 4-56 4-73 A short cylinder is allowed to cool in atmospheric air. The temperatures at the centers of the cylinder and the top surface as well as the total heat transfer from the cylinder for 15 min of cooling are to be determined. Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. 2 The thermal properties of the cylinder are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of brass are given to be ? ? 8530 kg/ m 3 , C kJ/kg 389 . 0 ? ? ? p C , C W/m 110 ? ? ? k , and ? ? ? ? 3 39 10 5 . m / s 2 . Analysis This short cylinder can physically be formed by the intersection of a long cylinder of radius D/2 = 4 cm and a plane wall of thickness 2L = 15 cm. We measure x from the midplane. (a) The Biot number is calculated for the plane wall to be 02727 . 0 ) C W/m. 110 ( ) m 075 . 0 )( C . W/m 40 ( 2 ? ? ? ? ? k hL Bi The constants ? 1 1 and A corresponding to this Biot number are, from Table 4-1, 0050 . 1 and 164 . 0 1 1 ? ? A ? The Fourier number is ? ? ? ? ? ? ? ? ? t L 2 5 339 10 0 075 5424 0 2 ( . ( . . . m / s)(15 min 60 s / min) m) 2 2 Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the dimensionless temperature at the center of the plane wall is determined from 869 . 0 ) 0050 . 1 ( ) 424 . 5 ( ) 164 . 0 ( 1 0 , 2 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? e e A T T T T i wall o We repeat the same calculations for the long cylinder, 01455 . 0 ) C W/m. 110 ( ) m 04 . 0 )( C . W/m 40 ( 2 0 ? ? ? ? ? k hr Bi ? 1 1 01704 10038 ? ? . . and A ? ? ? ? ? ? ? ? ? t r o 2 5 339 10 0 04 19 069 0 2 ( . ( . . . m / s)(15 60 s) m) 2 2 ? ? ? o cyl i T T T T A e e , ( . ) ( . ) ( . ) . ? ? ? ? ? ? ? ? ? ? 0 1 0 1704 19 069 1 2 2 10038 0577 Then the center temperature of the short cylinder becomes C 85.1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ) , 0 , 0 ( 501 . 0 20 150 20 ) , 0 , 0 ( 501 . 0 577 . 0 869 . 0 ) , 0 , 0 ( , , t T t T T T T t T cyl o wall o cylinder short i D0 = 8 cm L = 15 cm z Brass cylinder Ti = 150 ?C r Air T ? = 20 ?C Chapter 4 Transient Heat Conduction 4-57 (b) The center of the top surface of the cylinder is still at the center of the long cylinder (r = 0), but at the outer surface of the plane wall (x = L). Therefore, we first need to determine the dimensionless temperature at the surface of the wall. 857 . 0 ) 164 . 0 cos( ) 0050 . 1 ( ) / cos( ) , ( ) , ( ) 424 . 5 ( ) 164 . 0 ( 1 1 2 2 1 ? ? ? ? ? ? ? ? ? ? e L L e A T T T t x T t L i wall ? ? ? ? Then the center temperature of the top surface of the cylinder becomes C 84.2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ) , 0 , ( 494 . 0 20 150 20 ) , 0 , ( 494 . 0 577 . 0 857 . 0 ) , ( ) , 0 , ( , t L T t L T t L T T T t L T cyl o wall cylinder short i ? ? (c) We first need to determine the maximum heat can be transferred from the cylinder ? ? kJ 325 C ) 20 150 )( C kJ/kg. 389 . 0 )( kg 43 . 6 ( ) ( kg 43 . 6 m) 15 . 0 ( m) 04 . 0 ( ) kg/m 8530 ( max 2. 3 2 ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? T T mC Q L r V m i p o Then we determine the dimensionless heat transfer ratios for both geometries as 135 . 0 164 . 0 ) 164 . 0 sin( ) 869 . 0 ( 1 ) sin( 1 1 1 , max ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? wall o wall Q Q 427 . 0 1704 . 0 0846 . 0 ) 577 . 0 ( 2 1 ) ( 2 1 1 1 1 , max ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? J Q Q cyl o cyl The heat transfer ratio for the short cylinder is 504 . 0 ) 135 . 0 1 )( 427 . 0 ( 135 . 0 1 max max max max ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? wall plane cylinder long wall plane cylinder short Q Q Q Q Q Q Q Q Then the total heat transfer from the short cylinder during the first 15 minutes of cooling becomes kJ 164 ? ? ? kJ) 325 )( 504 . 0 ( 503 . 0 max Q Q Chapter 4 Transient Heat Conduction 4-58 4-74 "!PROBLEM 4-74" "GIVEN" D=0.08 "[m]" r_o=D/2 height=0.15 "[m]" L=height/2 T_i=150 "[C]" T_infinity=20 "[C]" h=40 "[W/m^2-C]" "time=15 [min], parameter to be varied" "PROPERTIES" k=110 "[W/m-C]" rho=8530 "[kg/m^3]" C_p=0.389 "[kJ/kg-C]" alpha=3.39E-5 "[m^2/s]" "ANALYSIS" "(a)" "This short cylinder can physically be formed by the intersection of a long cylinder of radius r_o and a plane wall of thickness 2L" "For plane wall" Bi_w=(h*L)/k "From Table 4-1 corresponding to this Bi number, we read" lambda_1_w=0.2282 "w stands for wall" A_1_w=1.0060 tau_w=(alpha*time*Convert(min, s))/L^2 theta_o_w=A_1_w*exp(-lambda_1_w^2*tau_w) "theta_o_w=(T_o_w-T_infinity)/(T_i-T_infinity)" "For long cylinder" Bi_c=(h*r_o)/k "c stands for cylinder" "From Table 4-1 corresponding to this Bi number, we read" lambda_1_c=0.1704 A_1_c=1.0038 tau_c=(alpha*time*Convert(min, s))/r_o^2 theta_o_c=A_1_c*exp(-lambda_1_c^2*tau_c) "theta_o_c=(T_o_c-T_infinity)/(T_i-T_infinity)" (T_o_o-T_infinity)/(T_i-T_infinity)=theta_o_w*theta_o_c "center temperature of short cylinder" "(b)" theta_L_w=A_1_w*exp(-lambda_1_w^2*tau_w)*Cos(lambda_1_w*L/L) "theta_L_w=(T_L_w- T_infinity)/(T_i-T_infinity)" (T_L_o-T_infinity)/(T_i-T_infinity)=theta_L_w*theta_o_c "center temperature of the top surface" "(c)" V=pi*r_o^2*(2*L) m=rho*V Q_max=m*C_p*(T_i-T_infinity) Q_w=1-theta_o_w*Sin(lambda_1_w)/lambda_1_w "Q_w=(Q/Q_max)_w" Q_c=1-2*theta_o_c*J_1/lambda_1_c "Q_c=(Q/Q_max)_c" J_1=0.0846 "From Table 4-2, at lambda_1_c" Q/Q_max=Q_w+Q_c*(1-Q_w) "total heat transfer" Chapter 4 Transient Heat Conduction 4-59 time [min] T [C] T [C] Q [kJ] 5 119.3 116.8 80.58 10 95.18 93.23 140.1 15 76.89 75.42 185.1 20 63.05 61.94 219.2 25 52.58 51.74 245 30 44.66 44.02 264.5 35 38.66 38.18 279.3 40 34.12 33.75 290.5 45 30.69 30.41 298.9 50 28.09 27.88 305.3 55 26.12 25.96 310.2 60 24.63 24.51 313.8 0 10 20 30 40 50 60 20 40 60 80 100 120 50 100 150 200 250 300 350 time [min] T o,o [C] Q [kJ] temperature heatRead More

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