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

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

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

The heat transfer due to radiation is given by: $\dot{Q}=10 \times \pi \times 0

$Nu_{D}=hD/k$

Solution:

The heat transfer from the wire can also be calculated by: $\dot{Q}=10 \times \pi \times 0

$\dot{Q} {cond}=\dot{m} {air}c_{p,air}(T_{air}-T_{skin})$ $\dot{Q}=10 \times \pi \times 0