Topic: Heat Transfer
A mild steel tank of wall thickness 12 mm contains water at $100^{0}C.$ The atmospheric temperature is $20^{0}C$. The thermal conductivity of mild steel is $50~W/mK,$ and the heat transfer co-efficients for inside and outside the tank are 2850 and $10~W/m^{2}K,$ respectively. Calculate the rate of heat loss per $m^{2}$ of tank surface and the temperature of the outside surface of the tank.
- A. $300.5~W/m^{2}$ $45.5^{0}C$
- B. $495.2~W/m^{2}$ $67.6^{0}C$
- C. $602.6~W/m^{2},$ $80.6^{0}C$
- D. $795.2~W/m^{2}$ $99.52^{0}C$
Answer: D. $795.2~W/m^{2}$ $99.52^{0}C$
Explanation: Thermal Resistance $R_{th}=\frac{1}{h_{i}}+\frac{t}{k}+\frac{1}{h_{o}} = \frac{1}{2850} + \frac{0.012}{50} + \frac{1}{10} = 0.10059~K m^{2}/W$. Heat loss per unit area $q = \frac{\Delta T}{R_{th}} = \frac{100-20}{0.10059} = 795.2~W/m^{2}$. To find outside surface temperature T, $\frac{T-20}{1/10} = 795.2 \Rightarrow 10(T-20) = 795.2 \Rightarrow T-20 = 79.52 \Rightarrow T = 99.52^{\circ}C$.