Question

Consider steady-state heat conduction across the thickness in a plane composite wall (as shown in the figure) exposed to convection conditions on both sides.
Given :$h_i=20W/m^{2}K;h_{\circ }=50W/m^{2}K;T_{\infty ,i}={20}^{\circ }C,T_{\infty ,\circ }=-2^{\circ }C{k}_1=20W/mK;k_2=50W/mK;L_1=0.30mand{L}_2=0.15m$
Assuming negligible contact resistance between the wall surfaces, the interface temperature, $T\left(in{}^{\circ }C\right),$ of the two walls will be

• A.
  $-0.50$
• B.
  $2.75$
• C.
  $3.75$
• D.
  $4.50$

Answer 1

The correct option is C

$3.75$
Given data:
$h_i=20W/m^{2}K;h_{\circ }=50W/m^{2}K;T_{\infty ,\circ }={20}^{\circ }C{k}_1=20W/mK;k_2=50W/mK;L_1=0.30m;L_2=0.15m$


Heat flow:

$Q=\frac{T_{\infty ,i}-T_{\infty ,\circ }}{R_{t_i}+R_{t_1}+R_{t_2}+R_{t_{\circ }}}$

$Q=\frac{T_{\infty ,i}-T_{\infty ,\circ }}{\frac{1}{h_{i}A}+\frac{L_1}{k_{1}A}+\frac{L_2}{k_{2}A}+\frac{1}{h_{\circ }A}}$

$\frac{Q}{A}=\frac{T_{\infty ,i}-T_{\infty ,\circ }}{\frac{1}{h_i}+\frac{L_1}{k_1}+\frac{L_2}{k_2}+\frac{1}{h_{\circ }}}$

$=\frac{20-(-2)}{\frac{1}{20}+\frac{0.30}{20}+\frac{0.15}{50}+\frac{1}{50}}$
$=\frac{20+2}{\frac{5+1.50+0.3+2}{100}}=\frac{22\times 100}{8.8}=250W$

also $\frac{Q}{A}=\frac{T_{\infty ,i}-T}{\frac{1}{h_i}+\frac{L_1}{k_1}}$

$\therefore 250=\frac{20-T}{\frac{1}{20}+\frac{0.3}{20}}=\frac{20-T}{\frac{0.3}{20}}$
$250\times \frac{0.3}{20}=20-T$
$16.25=20-T$
or $T=20-16.25$
$={3.75}^{\circ }C$