Question 7.14: Determining the Total Time Required for a Cylindrical Ingot ......
Determining the Total Time Required for a Cylindrical Ingot to be heated to a Given Temperature
A hot cylinder ingot of 50mm diameter and 200mm long is taken out from the furnace at 800°𝐶 and dipped in water till its temperature fall to 500°𝐶 . Then, it is directly exposed to air till its temperature falls to 100°𝐶. Find the total time required for the ingot to reach the temperature from 800°𝐶 to 100°𝐶. Take the following:
𝑘(thermal conductivity of ingot) = 60𝑤/𝑚°𝐶;
𝑐(specific heat of ingot) = 200𝐽/𝑚°𝐶;
𝜌(density of ingot material) = 800𝑘𝑔/𝑚³;
h_w (heat transfer coefficient in water) = 200𝑤/ m^{2^{\circ} } 𝐶;
h_a (heat transfer coefficient in air) = 20𝑤/ m^{2^{\circ} } 𝐶;
Temperature of air or water = 30°𝐶
Given: r=\frac{50}{2}=25m m=0.025m;{ L}=200m m\mathrm{~or~}0.2m
The characteristic length of a cylinder,
L_{c}=\frac{r}{2}=\frac{0.025}{2}=0.0125m
B i=\frac{h\,L}{k}=\frac{200\times0.0125}{60}=0.04167
As Bi is less than 0.1, the internal thermal resistance can be neglected, and lumped capacitance theory can be used. The total time (𝜏) can be calculated by calculating \tau^{}_{1} (time required in water) and \tau^{}_{2} (time required in air) and adding them such that
\tau=\,\tau_{1}+\tau_{2}(a) The temperature variation with respect to time when cooled in water is given by: (see Fig. (7.2) below)
\frac{T(t)-T_{w}}{T_{o}-T_{w}}=e^{-B i\times F o}
F_{O}=\frac{k}{\rho c_{P}L^{2}}\cdot\tau_{1}=\frac{60}{800\times200\times(0.0125)^{2}}\cdot\tau_{1}=2.4\,\tau_{1}
B i\times F o\ \:=\:0.04167\times2.4\:\tau_{1}=0.1\:\tau_{1}
∴ \frac{\theta}{\theta_{o}}=\frac{T(t)-T_{w}}{T_{o}-T_{w}}=e^{−𝐵𝑖×𝐹𝑜}
={\frac{500-30}{800-30}}=e^{-0.1\ \tau_{1}}
0.61\,=\,e^{-\,0.1\,\tau_{1}}
-0.1\;\tau_{1}\ ln\ e=\ln \ 0.61
\tau_{1}=\frac{{ln \ 0.61}}{-0.1}=-4.94\,s
(b) The temperature variation with respect to time when cooled in air is given by: (see Fig. (7.3) below)
∴ \frac{\theta}{\theta_{o}}=\frac{T(t)-T_{a}}{T_{o}-T_{a}}=e^{-B i\times F o}
B i=\frac{h\,L}{k}=\frac{20\times0.0125}{60}=0.004167
F_{o}=\frac{k}{\rho c^{}_{P}L^{2}}\cdot\tau_{2}=2.4\,\ \tau_{2}
B i\times F o=0.004167\times2.4\;\tau_{2} =0.01\;\tau_{2}
\frac{\theta}{\theta^{}_{o}}=\frac{100-30}{500-30}=\frac{70}{470}=e^{-0.01 \ \tau^{}_{2}}
0.149=e^{-0.01\,\tau_{2}}
-0.01\;\tau_{2} \ \ln \ e=\ln \ 0.149
∴ \tau_{2}=\frac{\ln \ 0.149}{-0.01\times1}=190.42\,s
Total time (𝜏) is given by:
\tau=\tau_{1}+\tau_{2}=4.94+\,190.42=\,195.36\,s=\,3.256\,m i n.
