Solution Manual Heat And Mass Transfer Cengel 5th Edition Chapter 3 [patched] [ TRUSTED · WORKFLOW ]

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

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

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

The heat transfer due to radiation is given by: $\dot{Q} {cond}=\dot{m} {air}c_{p

The heat transfer from the not insulated pipe is given by: $\dot{Q}=10 \times \pi \times 0