Question

A thin metal pipe of 1 meter length and 1 cm radius carries steam at ${100}^{\circ }C$. This is covered by two layers of lagging. The thermal conductivity of outer layer, which is 2 cm thick is $3.6\times {10}^{-4}cal/c{m}^{\circ }C$ sec while that of inner layer, which is 1 cm thick is $1.2\times {10}^{-4}cal/m^{\circ }C$ sec. If the outer surface of the laggings is at ${30}^{\circ }C$, find (a) temperature of the cylindrical interface of two lagging materials (b) the mass of steam condensed per second. Given $lo{g}_e=0.6931$.

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

The situation is shown in figure. The radial rate of flow of heat through cylindrical tube is given by,
$Q=\frac{K2\pi l({\theta }_1-{\theta }_2)}{lo{g}_e\left(\frac{r_2}{r_1}\right)}$
Let $\theta$ be the temperature of the interface of two lagging materials,
The rate of flow of heat through inner layer is given by,
$Q_1=\frac{K_{1}2\pi l({\theta }_1-\theta )}{lo{g}_e\left(\frac{r_2}{r_1}\right)}$
$\frac{(1.2\times {10}^{-4})\times (2\times 3.14\times 100)\times (100-\theta )}{lo{g}_{e}2}$
The rate of flow of heat through outer layer
$Q_2=\frac{K_{2}2\pi l(\theta -{\theta }_2)}{lo{g}_e\left(\frac{r_2}{r_1}\right)}$
$\frac{(3.6\times {10}^{-4})\times (2\times 3.14\times 100)\times (\theta -30)}{lo{g}_e\left(\frac{4}{2}\right)}Insteadystate,Q_1=Q_2,\therefore \frac{(1.2\times {10}^{-4})\times (2\times 3.14\times 100)\times (\theta -30)}{lo{g}_{e}2}\frac{(3.6\times {10}^{-4})\times (2\times 3.14\times 100)\times (\theta -30)}{lo{g}_{e}2}\therefore \theta ={47.5}^{\circ }C$
Let the mass of steam condensed per second be m gm. The heat taken away from the steam per second = $m\times 540cal$. This is equal to $Q_1$ or $Q_2$. Thus
$\frac{(1.2\times {10}^{-4})\times (2\times 3.14\times 100)\times (100-47.5)}{lo{g}_{e}2}=m\times 540or,m=0.0106gm$