$h=\frac{Nu_{D}k}{D}=\frac{2152.5 \times 0.597}{2}=643.3W/m^{2}K$
Solution:
$\dot{Q} {cond}=\dot{m} {air}c_{p,air}(T_{air}-T_{skin})$ $h=\frac{Nu_{D}k}{D}=\frac{2152
$\dot{Q}=\frac{V^{2}}{R}=\frac{I^{2}R}{R}=I^{2}R$
$\dot{Q} {conv}=h A(T {skin}-T_{\infty})$ $h=\frac{Nu_{D}k}{D}=\frac{2152
$\dot{Q}=10 \times \pi \times 0.08 \times 5 \times (150-20)=3719W$
The heat transfer due to conduction through inhaled air is given by: $h=\frac{Nu_{D}k}{D}=\frac{2152
$T_{c}=800+\frac{2000}{4\pi \times 50 \times 0.5}=806.37K$
