Question 7.8: Determination of the Heat Transfer Coefficient The heat tran......
Determination of the Heat Transfer Coefficient
The heat transfer coefficients for flow of air at 28°𝐶 over a 12.5𝑚𝑚 diameter sphere are measured by observing the temperature – time history of a copper ball of the same dimension. The temperature of the copper ball ( c_{p} = 0.4𝑘𝑗/𝑘𝑔 𝐾 and 𝜌 = 8850𝑘𝑔/m³) was measured by two thermo – couples, one located in the center and the other near the surface. Both the thermocouples registered the same temperature at a given instant. In one test the initial temperature of the ball was 65°𝐶, and in 1.15 minutes the temperature decreased by 11°𝐶 . Calculate the heat transfer coefficient for this case.
Given: \;T_{\infty}=28^{\circ}C;r({\mathrm{sphere}})={\frac{12.5}{2}}=6.25m m=0.00625m;\;\;c_{p}=0.4k j/k g\;K;
{{\rho=8850k g/m^{3};\;T_{o}=65^{\circ}C;\;T(t)=65-11=54^{\circ}C;\;\tau=1.15m i n=69\;s;}}
B i={\frac{h L}{k}}; characteristic length L=\frac{V}{A_{s}}=\frac{r}{3}=\frac{0.00625}{3}=2.083\times10^{-3}
Since, heat transfer coefficient has to be calculated, so assume that the internal resistance is negligible and 𝐵𝑖 is much less than 0.1 .
B i=\frac{hr}{3k}=\frac{0.00625h}{3k}=\frac{2.083\times10^{-3}h}{k}
F o=\frac{k}{\rho c_{P}L^{2}}\cdot\tau=\frac{k}{8850\times0.4\times10^{3}\times(2.083\times10^{-3})^{2}}\times270
=0.0651\;k\times69=4.5\;k
B i\times F o=\frac{0.00625h}{3k}\times4.5k=9.375\times10^{-3}h
\frac{\theta}{\theta_{o}}=\frac{T(t)-T_{\infty}}{T_{o}-T_{\infty}}=e^{-Bi\times 𝐹𝑜}
\frac{\theta}{\theta^{}_{o}}=\frac{54-28}{65-28}=e^{-9.375\times10^{-3}h}
0.7027=e^{-0.009375h}
\ln0.7027=-0.009375h\ \ln e
∴ h=\frac{{ \ln}\,0.7027}{-0.009375\times1}=37.63w/{m}^{2}K