Question

A steel casting cools to 90% of the original temperature difference in 30 min in still air. The same casting is cooled to 90% of the original temperature difference in a moving air stream. The convective heat transfer coefficient in moving air is 5 times that of still air. The time (in minutes) to cool the casting in moving air is__________.

• A. 10 min
• B. 16 min
• C. 6 min
• D. 5 min

Answer 1

The correct option is C 6 min

$\frac{T_i-T_{\infty }}{T-T_{\infty }}=e^{(\frac{hA}{\rho V{C}_p}\times \tau )}\text{For both the cases, the ratio of temperature}\text{difference in the LHS of the above equation is constant.}\therefore e^{\left(\frac{hA}{\rho V{C}_p}\tau \right)}=\text{constant}\left(\frac{hA}{\rho V{C}_p}\tau \right)=\text{constant}A,V,\rho ,C_p=\text{constant (because same casting}\text{Only variable is h}{h}_1{\tau }_1=h_2{\tau }_2{h}_2=5h_1\left(given\right)h_1\times 30=5h_1\times {\tau }_2\Rightarrow {\tau }_2=6min$