Topic: Heat Transfer
A spherical shaped vessel of 1.4 m outer diameter is 90 mm thick. Find the rate of heat leakage, if the temperature difference between the inner and outer surfaces is $220^{\circ}C$. Thermal conductivity of the material of the sphere is 0.083 W/m K.
- A. 0.2 kW
- B. 0.5 kW
- C. 1.088 kW
- D. 1.6 kW
Answer: C. 1.088 kW
Explanation: Given, Outer diameter $(d_{2})=1.4m$, $r_{2}=0.7m$. $r_{1}=(0.7-0.09)m=0.61~m$. $dT=220^{\circ}C$, $k=0.083~W/mK$. Rate of heat leakage $Q=\frac{dT}{R_{th}}$. For a spherical shell, $R_{th}=\frac{r_{2}-r_{1}}{4\pi k r_{1}r_{2}} = \frac{0.09}{4\times3.14\times0.083\times0.7\times0.61} = 0.20218$. $Q=\frac{220}{0.20218} = 1.088~kW$.