Answer to Question #178306 in Mechanical Engineering for luke

Question

Consider a 5m high 8m long and 0.22m thick wall whose representative cross section is given in the figure. The thermal conductivities of various materials used in W/m-C are kA = kF = 2, kB = 8, kC = 20, kD = 15, and kE = 35. The left and right surfaces of the wall are maintained at uniform temperature s of 300C and 100C respectively. Assuming heat transfer through the wall to be one-dimensional, Determine the rate of heat transfer through the wall in W

Answer 1

Consider the diagram below

$R _1=R_A=\frac{L}{kA}=\frac{0.01}{2 \times 0.12 \times 1}=0.04 ^0C/W$

$R _2=R_4=R_C=\frac{L}{kA}=\frac{0.05}{20 \times 0.04 \times 1}=0.06 ^0C/W$

$R _3=R_B=\frac{L}{kA}=\frac{0.05}{8 \times 0.04 \times 1}=0.16 ^0C/W$

$R _5=R_D=\frac{L}{kA}=\frac{0.01}{15 \times 0.06 \times 1}=0.11 ^0C/W$

$R _6=R_E=\frac{L}{kA}=\frac{0.01}{35 \times 0.06 \times 1}=0.05 ^0C/W$

$R _7=R_F=\frac{L}{kA}=\frac{0.06}{2 \times 0.12 \times 1}=0.25 ^0C/W$

$R _8=\frac{0.00012}{ 0.12}=0.001 ^0C/W$

$\frac{1}{R_{mid,1}}=\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{0.06}+\frac{1}{0.16}+\frac{1}{0.06}$

$R_{mid,1}=0.025^0C/W$

$\frac{1}{R_{mid,2}}=\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{0.11}+\frac{1}{0.05}$

$R_{mid,2}=0.034^0C/W$

$R_{Total}=0.04+0.025+0.034+0.25+0.001=0.35^0C/W$

$Q=\frac{300-100}{0.35}=571.43 W$

$Q_{Total}=571.43\frac{5 \times8}{0.12 \times1}=190476.67 W$