Question 7.5: Determination of the Time Required for the Plate to Reach a ......
Determination of the Time Required for the Plate to Reach a Given Temperature
An aluminum alloy plate of 400𝑚𝑚 × 400𝑚𝑚 × 4𝑚𝑚 size at 200°C is suddenly quenched into liquid oxygen at −183°C. Starting from fundamentals or deriving the necessary expression, determine the time required for the plate to reach a temperature of −70°C. Assume h=20000kJ/m^{2}\cdot h r\cdot^{\circ}C c_{P}=0.8kJ/k g^{\circ}C\;, and 𝜌 = 3000𝑘𝑔/𝑚³, 𝑘 for aluminum at low temperature may be taken as 214w/m°C 𝑜𝑟 770.4 kJ/mh°C
Surface area of the plate,
A_{s}=2\times0.4\times0.4=0.32m^{2}
Volume of the plate,
V=0.4\times0.4\times0.004=0.00064m^{3}
Characteristic length,
L=\frac{t}{2}=\frac{0.004}{2}=0.002\;m
Or L=\frac{V}{A_{s}}=\frac{0.00064}{0.32}=0.002\;m
𝑘 for aluminum, at low temperature may be taken as 214w/m°C or 770.4𝑘𝐽/𝑚ℎ° 𝑐
∴ B i=\frac{h L}{k}=\frac{2000\times0.002}{770.4}=0.0519
Since, 𝐵𝑖 ≪ 0.1, hence lumped capacitance method may be applied for the solution of the problem.
The temperature distribution is given by:
\frac{\theta}{\theta_{o}}=\frac{T(t)-T_{\infty}}{T_{o}-T_{\infty}}= e^{-B i\times F o}
F_{O}=\frac{k}{\rho c_{P}L^{2}}\cdot\tau=\frac{214}{3000\times0.8\times10^{3}\times0.002^{2}}\cdot\tau=22.3\,\tau
B i\times F o\,=\,0.0519\times22.3\,=\,1.15737\,\tau
∴ \frac{\theta}{\theta_{o}}=\frac{-70-(-183)}{200-(-183)}==\,e^{-1.15737\,\tau}
\frac{113}{383}=e^{-1.15737 \ \tau}
0.295=e^{-1.15737\,\tau}
\ln0.295=-1.15737\,\tau\,\ln e
\tau=\frac{{ln}\ 0.295}{-1.15737\times1}=\,1.055\,s