Question 7.6: Determining the Temperature of a Solid Copper Sphere at a Gi......
Determining the Temperature of a Solid Copper
Sphere at a Given Time after the Immersion in a Well – Stirred Fluid
A solid copper sphere of 10𝑐𝑚 diameter (𝜌 = 8954𝑘𝑔/m³, c_{p} = 383𝐽/𝑘𝑔, 𝐾 = 386𝑤/𝑚𝐾) initially at a uniform temperature of 𝑇𝑜 = 250°𝐶, is suddenly immersed in a well – stirred fluid which is maintained at a uniform temperature T_{\infty} = 50°𝐶. The heat transfer coefficient between the sphere and the fluid is ℎ = 200𝑤/𝑚²𝐾. Determine the temperature of the copper block at 𝜏 = 5 𝑚𝑖𝑛. after the immersion.
Given: d=10c m = 0.1m; \ p=8954kg/m^{3}; \ c{}_{p}=383J/kgk; \ k=386w/mK,
T_{o}=250^{\circ}C;\;T_{\infty}=50^{\circ}C;\;h=200w/m^{2}K;\;\;\tau=5m i n=5\times60=300\,s
Temperature of the copper block, 𝑇(𝑡) = ?
The characteristic length of the sphere is,
L=\frac{V}{A^{}_{s}}=\frac{\frac{4}{3} \pi r^{3}}{4\pi r^{2}}=\frac{r}{3}=\frac{d}{6}=\frac{0.1}{6}=0.01667 \ m
B i=\frac{h L}{k}=\frac{209\times0.01667}{386}=8.64\times10^{-3}
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^{}_{0}}=\frac{T(t) – T^{}_{∞}}{T^{}_{o}-T^{}_{∞}}=e^{-Bi\times 𝐹𝑜}
F_{O}=\frac{k}{\rho c_{P}L^{2}}\cdot\tau=\frac{386}{8954\times383\times0.01667^{2}\times3000}=121.513
B i\times F o=8.64\times10^{-3}\times121.513=1.05
∴ \frac{\theta}{\theta_{o}}=\frac{-T(t)-50}{250-50}=e^{-1.05}
T(t)-50=200\;e^{-1.05}
∴ T(t)=50+200\ e^{-1.05}=50+70=120^{o}c