Question 7.12: Estimation of the Time Required to Cool a Decorative Plastic......
Estimation of the Time Required to Cool a Decorative Plastic Film on Copper Sphere to a Given Temperature using Lumped Capacitance Theory
The decorative plastic film on copper sphere 10mm in diameter is cured in an oven at 75°𝐶. After removal from oven, the sphere is exposed to an air stream at 10m/s and 23°𝐶. Estimate the time taken to cool the sphere to 35°𝐶 using lumped capacitance theory.
Use the following correlation:
For determination of correlation coefficient ℎ, use the following properties of air and copper:
For copper: 𝜌 = 8933𝑘𝑔/𝑚³; 𝑘 = 400𝑤/𝑚 𝐾; c_{p} = 380𝐽/𝑘𝑔°𝐶
For air at 23°𝐶: \mu_{a}=\mathrm{18.6}\times\mathrm{10^{-6}}N s/{m^{2}},\nu=\mathrm{15.36}\times\mathrm{10^{-6}}m^{2}\mathrm{/s}
𝑘 = 0.0258𝑤/𝑚 𝐾, 𝑝𝑟 = 0.709, and \mu_{s}=\mathrm{19.7}\times\mathrm{10^{-6}}Ns/{m^{2}}, at 35°𝐶
Time taken to cool the sphere, 𝜏 = ?
R e={\frac{v\;d}{\nu}}={\frac{10\times0.01}{15.36\times10^{-6}}}=6510
{ N}u=2+\left[0.4(6510)^{0.5}+0.06(6510)^{2/3}\right]\times(0.709)^{0.4}\times\left[\frac{18.16\times10^{-6}}{19.78\times10^{-6}}\right]^{0.25}
=2+[32.27+20.92]\times{0.87\times 0.979}=47.3
or N{{\mathit{u}}}=\frac{h\;d}{k}=47.3
∴ h=\frac{k}{d}N u=\frac{0.0258}{0.01}\times47.3=\,122{w/m}^{2^{o}}C
The time taken to cool from 75°𝐶 to 35°𝐶 may be found from the following relation:
\frac{\theta}{\theta_{o}}=\frac{T(t)-T_{\infty}}{T_{o}-T_{\infty}}=e^{−𝐵𝑖×𝐹𝑜}
B i={\frac{h\,L}{K}}
The characteristic length of a sphere, L\,=\,\frac{r}{3}=\frac{0.005}{3}m
B i=\frac{h \ L}{\mathit{k}}=\frac{122\times0.005}{3\times400}=5.083\times10^{-4}
Since, 𝐵𝑖 ≪ 0.1 , so we can use the lumped capacitance theory to solve this problem.
F o=\frac{k}{\rho c_{P}L^{2}}\cdot\tau=\frac{40}{893\times380\times\left(\frac{0.005}{3}\right)^{2}}\cdot\tau=42.421\,\,\tau
B i\times F o=5.083\times10^{-4}\times42.421\,\tau=0.0216\,\tau
\frac{\theta}{\theta_{o}}=\frac{35-23}{75-23}=e^{-0.0216\;\tau}
0.2308=e^{-0.0216\,\tau}
\ln0.2308=-0.0216\tau\ \mathrm{ln}\,e
\tau=\frac{\mathrm{ln}\,0.2308}{-0.0216}=67.9 ≃ \,68\,s