Question

An oven is made of a composite wall of three materials A, B and C. Under steady-state operating conditions, the outer surface temperature $T_{so}$ is 20°C, the inner surface temperature $T_{si}$ is 600°C, and the oven air temperature is $T_{\infty }$ = 800°C.
Further data are given as follows:
Thermal conductivities, $k_A=20W/m-Kand{k}_C$ = 50 W/m-K
Thickness, $L_A=0.3m,L_B=0.15mand{L}_C=0.15m$

Inner wall heat transfer coefficients, $h=25W/m^2-K$

The thermal conductivity, $k_B(W/m-K)$ of the material B, is calculated as

• A. 0.66
• B. 0.03
• C. 35
• D. 1.53

Answer 1

The correct option is D 1.53

$Heattransfer\left(Q\right)=h\times A(T_{\infty }-T_{si})$

$q=h(T_{\infty }-T_{si})$

$q=25(200-600)=25\times 200=5000W/m^2$

Thermal circuit:

$Q=\frac{T_{Si}-T_{So}}{\frac{L_A}{k_A\times A}+\frac{L_B}{k_B\times B}+\frac{L_C}{k_C\times C}}$

$q=\frac{600-20}{\frac{0.30}{20}+\frac{0.15}{k_B}+\frac{0.15}{50}}$

$5000=\frac{580}{\frac{0.3}{20}+\frac{0.15}{k_B}+\frac{0.15}{50}}$

$\Rightarrow k_B=1.53W/m-K$