Question #178306

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


1
Expert's answer
2021-04-12T01:55:03-0400

Consider the diagram below

R1=RA=LkA=0.012×0.12×1=0.040C/WR _1=R_A=\frac{L}{kA}=\frac{0.01}{2 \times 0.12 \times 1}=0.04 ^0C/W

R2=R4=RC=LkA=0.0520×0.04×1=0.060C/WR _2=R_4=R_C=\frac{L}{kA}=\frac{0.05}{20 \times 0.04 \times 1}=0.06 ^0C/W

R3=RB=LkA=0.058×0.04×1=0.160C/WR _3=R_B=\frac{L}{kA}=\frac{0.05}{8 \times 0.04 \times 1}=0.16 ^0C/W

R5=RD=LkA=0.0115×0.06×1=0.110C/WR _5=R_D=\frac{L}{kA}=\frac{0.01}{15 \times 0.06 \times 1}=0.11 ^0C/W

R6=RE=LkA=0.0135×0.06×1=0.050C/WR _6=R_E=\frac{L}{kA}=\frac{0.01}{35 \times 0.06 \times 1}=0.05 ^0C/W

R7=RF=LkA=0.062×0.12×1=0.250C/WR _7=R_F=\frac{L}{kA}=\frac{0.06}{2 \times 0.12 \times 1}=0.25 ^0C/W

R8=0.000120.12=0.0010C/WR _8=\frac{0.00012}{ 0.12}=0.001 ^0C/W

1Rmid,1=1R2+1R3+1R4=10.06+10.16+10.06\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}

Rmid,1=0.0250C/WR_{mid,1}=0.025^0C/W

1Rmid,2=1R5+1R6=10.11+10.05\frac{1}{R_{mid,2}}=\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{0.11}+\frac{1}{0.05}

Rmid,2=0.0340C/WR_{mid,2}=0.034^0C/W

RTotal=0.04+0.025+0.034+0.25+0.001=0.350C/WR_{Total}=0.04+0.025+0.034+0.25+0.001=0.35^0C/W

Q=3001000.35=571.43WQ=\frac{300-100}{0.35}=571.43 W

QTotal=571.435×80.12×1=190476.67WQ_{Total}=571.43\frac{5 \times8}{0.12 \times1}=190476.67 W


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