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

The heat transfer from the not insulated pipe is given by:

$\dot{Q} {rad}=\varepsilon \sigma A(T {skin}^{4}-T_{sur}^{4})$

The convective heat transfer coefficient for a cylinder can be obtained from:

$\dot{Q}=\frac{423-293}{\frac{1}{2\pi \times 0.1 \times 5}ln(\frac{0.06}{0.04})}=19.1W$ The heat transfer from the not insulated pipe

However we are interested to solve problem from the begining

$h=\frac{Nu_{D}k}{D}=\frac{10 \times 0.025}{0.004}=62.5W/m^{2}K$

$\dot{Q}_{rad}=1 \times 5.67 \times 10^{-8} \times 1.5 \times (305^{4}-293^{4})=41.9W$ The heat transfer from the not insulated pipe

Solution:

Assuming $k=50W/mK$ for the wire material,

The rate of heat transfer is:

Solution:

$\dot{Q}=h A(T_{s}-T_{\infty})$

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

The heat transfer due to conduction through inhaled air is given by:

(c) Conduction: