Question

A 30 cm thick furnace wall (k = 1.5 W/mK) has inner surface temperature of 1250${}^o$C. If the heat transfer coefficient on the outer surface is prescribed by the relation$h_0=8+0.09\Delta t$, where, $\Delta$t is the temperature difference between the outer wall surface and surrounding. What is the value of outer surface heat transfer coefficient at steady state condition?
Take surrounding temperature as 30${}^o$C.

• A. 26 W/m${}^2$K
• B. 57 W/m${}^2$K
• C. 110 W/m${}^2$K
• D. 37 W/m${}^2$K

Answer 1

The correct option is A 26 W/m${}^2$K


Considering unit area of the furnace wall

$Q=\frac{t_1-t_0}{\frac{{\delta }_1}{k}+\frac{1}{h_0}}=\frac{t_2-t_0}{\frac{1}{h_o}}$

$\Rightarrow [t_2-t_0]=\frac{1}{h_o}\left[\frac{t_1-t_0}{\frac{{\delta }_1}{k}+\frac{1}{h_0}}\right]=C[\frac{1250-30}{\frac{0.3}{1.5}}+C]$, where $C=\frac{1}{h_0}$

Substituing $[t_2-t_0]$ in the given expression of $h_0$ i.e $h_0=8+0.09\Delta t$

$\Rightarrow h_0=8+0.09C\times \left[\frac{1220}{0.2+C}\right]$

$\Rightarrow \frac{1}{C}=8+0.09C\times \frac{1220}{0.2+C}$

Upon simplification,
$\Rightarrow 117.8C^2+0.6C-0.=0$

On solving
$\Rightarrow C=0.03874$

Now,
$h_0=\frac{1}{C}=\frac{1}{0.03874}=25.82W/m^{2}K\simeq 26W/m^{2}K$