Question 12.4: The tube in Example 12.3 is maintained at 350 K and air is p......

Taking the physical properties of air at 310 K and assuming that fully developed flow exists in the pipe, then:

R e=0.0250\times3.5\times{\frac{(29/22.4)(273/310)}{0.018\times10^{-3}}} = 5535

P r={\frac{1.003\times1000\times0.018\times10^{-3}}{0.024}} = 0.75

The heat transfer coefficients and final temperatures are then calculated as in Example 12.3 to give:

In this case the result obtained using the Reynolds analogy agrees much more closely with the other three methods.