Question

Temperature distribution at certain instant of time across a wall of thickness 1 m is given by $T=800-300x-50x^2$. The wall has a constant heat generation of $1000{\text{W/m}}^3$. Density of material is $1600{\text{kg/m}}^3$, thermal conductivity is 40 W/mK and specific heat of material is 3 kJ/kgK. What will be the rate of change of internal energy of wall for per unit cross-sectional area ?

• A. 5 kJ/s, increasing
• B. Can't be determined from given data
• C. 3 kJ/s, decreasing
• D. 3 kJ/s, increasing

Answer 1

The correct option is C 3 kJ/s, decreasing

$\frac{{\partial }^{2}T}{\partial x^2}+\frac{q}{k}=\frac{1}{\alpha }\left(\frac{\partial T}{\partial \tau }\right)$

$\alpha =\left(\frac{k}{\rho c}\right)=\frac{40}{1600\times 3000}$

$=8.33\times {10}^{-6}{\text{m}}^2/\text{s}$

$\left(\frac{\partial T}{\partial \tau }\right)=\alpha \{\frac{{\partial }^{2}T}{\partial x^2}+\frac{q}{k}\}=\alpha \left\{\right(-100)+\frac{1000}{40}\}$

$\left(\frac{\partial T}{\partial \tau }\right)=-75\alpha$

Rate of charge of internal energy
$=mc\frac{\partial T}{\partial \tau }$

$=1600\times 1\times 1\times 3000\times (-75\alpha )$

$=-1600\times 3000\times 75\times 8.33\times {10}^{-6}$

$=-2998.8\text{J/s}$ per ${\text{m}}^2$ cross-section

$\simeq 3\text{kJ/s}$ (decreasing)