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

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

$\dot{Q}=\frac{423-293}{\frac{1}{2\pi \times 0.1 \times 5}ln(\frac{0.06}{0.04})}=19.1W$

Assuming $Nu_{D}=10$ for a cylinder in crossflow,

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

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