Heat Transfer Mechanisms — Question 5

Question 5

Convective cooling reveals a heat-transfer coefficient

A small body remains nearly uniform in temperature and has heat capacity $C=600\,\mathrm{J/K}$ and exposed area $.0500\,\mathrm{m^2}$. It cools by convection alone in air at 290 K, obeying $C\,dT/dt=-hA(T-290)$ with constant $h$. Starting at 350 K, it reaches 320 K after 120 s.

1. Infer $h$ and the thermal time constant.

2. Find the time from the start at which the body reaches 300 K and the heat lost by that time.

Question 5 - Solution

Convective cooling reveals a heat-transfer coefficient

(a) Method and calculation

$$T-290=(350-290)e^{-t/\tau},\quad\tau=\frac C{hA}.$$$$\frac{320-290}{350-290}=\frac12=e^{-120/\tau}\ \Rightarrow\ \tau=120/\ln2=173.123\,\mathrm s.$$$$h=\frac{600\ln2}{.0500(120)}=69.3147\,\mathrm{W/(m^2K)}.$$

(b) Interpretation and further calculation

$$\frac{300-290}{60}=\frac16=e^{-t/\tau}\ \Rightarrow\ t=\frac{120\ln6}{\ln2}=310.196\,\mathrm s.$$$$Q_{\mathrm{out}}=C(350-300)=30000\,\mathrm J.$$The uniform-temperature assumption concerns fast internal equilibration; it is separate from the external convection law.

Physical check

The body approaches the ambient temperature asymptotically; the exponential model does not reach it in finite time.