Question 7.4: Calculation of the Time Required for the Plate to Reach a Gi......
Calculation of the Time Required for the Plate to Reach a Given Temperature
A 50𝑐𝑚 × 50𝑐𝑚 copper slab 6.25mm thick has a uniform temperature of 300°C. Its temperature is suddenly lowered to 36°C. Calculate the time required for the plate to reach the temperature of 108°C.
Take: 𝜌 = 9000𝑘𝑔/𝑚³ ; c_{P}=0.38kJ/k g^{\circ}C; ; 𝑘 = 370w/m^{\circ}C and h=90w/m^{2^{\circ}}C
Surface area of plate (two sides),
A_{s}=\,2\times0.5\times0.5\,=\,0.5m^{2}
Volume of plate,
V=0.5\times0.5\times0.00625=0.0015625m^{3}
Characteristic length,
L=\frac{V}{A_{s}}=\frac{0.0015625}{0.5}=0.003125\;m
B i=\frac{h L}{k}=\frac{90\times0.003125}{370}=7.6\times10^{-4}
Since, 𝐵𝑖 ≪ 0.1 , hence lumped capacitance method (Newtonian heating or cooling) 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{370}{900\times0.38\times10^{3}\times0.003125^{2}}\cdot\tau=11.0784\,\tau
\frac{\theta}{\theta_{o}}=\frac{108-36}{300-36}=\,e^{-7.6\times10^{-4}\times11.0784\,\tau}=\,\,e^{-8.42\times10^{-3}\,\tau}
0.27273\,=\,e^{-8.42\times10^{-3}\;\tau}
\ln0.27273=-8.42\times10^{-3}\;\tau\;\ln e
\tau=\frac{ln \ 0.27273}{-8.42\times10^{-3} \ ln \ e}=\,154.31\,s