Question 7.9: Calculation of the Initial Rate of Cooling of a Steel Ball A......
Calculation of the Initial Rate of Cooling of a Steel Ball
A steel ball 50mm in diameter and at 900°𝐶 is placed in still atmosphere of 30°𝐶. Calculate the initial rate of cooling of the ball in °𝐶/𝑚𝑖𝑛, if the duration of cooling is 1 minute.
Take: 𝜌 = 7800𝑘𝑔/m³; c_{p} = 2𝑘𝑗/𝑘𝑔°𝐶 (for steel); ℎ = 30𝑤/ m^{2^{\circ} } 𝐶.
Neglect internal thermal resistance.
Given: r={\frac{50}{2}}=25m m=0.025m;\;T_{o}=900^{\circ}C;\;T_{\infty}=30^{\circ}C;\;P=7800k g/m^{3}
c_{p}=\,2k j/k g^{\circ}C;\,\,\,\,h=\,30w/m^{2^{\circ}}C;\,\,\,\tau=\,1m i n=\,60\,\,s;
The temperature variation in the ball (with respect to time), neglecting internal thermal resistance, is given by:
\frac{\theta}{\theta_{o}}=\frac{T(t)-T_{\infty}}{T_{o}-T_{\infty}}= e^{−𝐵𝑖×𝐹𝑜}
B i={\frac{h L}{k}}, L \ of~a~ball=\frac{r}{3}=\frac{0.025}{3}
∴ B i=\frac{h r}{3k}=\frac{30\times0.025}{3k}=\frac{0.25}{k}
F_{O}=\frac{k}{\rho c_{P}L^{2}}\cdot\tau=\frac{k}{7800\times2\times10^{3}\times(0.025/3)^{2}}\cdot\tau
F_{O}=9.23\times10^{-4}\times60=0.0554k
B i\times F o=\frac{0.25}{k}\times0.0554k=0.01385
\frac{\theta}{\theta_{o}}=\frac{T(t)-30}{900-\,30}=\,e^{-0.01385}
T(t)=870\;e^{-0.01385}+30=888^{\circ}C
∴ rate of cooling {}\,=\,\frac{900-888}{1\,\displaystyle m i n}=\,12^{\circ}C/m i n