Solution Manual Heat And Mass Transfer Cengel 5th Edition Chapter 3 -

$h=\frac{\dot{Q} {conv}}{A(T {skin}-T_{\infty})}=\frac{108.1}{1.5 \times (32-20)}=3.01W/m^{2}K$

$\dot{Q}=\frac{V^{2}}{R}=\frac{I^{2}R}{R}=I^{2}R$ $h=\frac{\dot{Q} {conv}}{A(T {skin}-T_{\infty})}=\frac{108

Assuming $h=10W/m^{2}K$,

$I=\sqrt{\frac{\dot{Q}}{R}}$

For a cylinder in crossflow, $C=0.26, m=0.6, n=0.35$ $I=\sqrt{\frac{\dot{Q}}{R}}$ For a cylinder in crossflow

The convective heat transfer coefficient is: $h=\frac{\dot{Q} {conv}}{A(T {skin}-T_{\infty})}=\frac{108

$\dot{Q}=10 \times \pi \times 0.08 \times 5 \times (150-20)=3719W$