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

The heat transfer from the wire can also be calculated by:

(c) Conduction:

$r_{o}+t=0.04+0.02=0.06m$

$\dot{Q}=\frac{V^{2}}{R}=\frac{I^{2}R}{R}=I^{2}R$

$\dot{Q} {rad}=\varepsilon \sigma A(T {skin}^{4}-T_{sur}^{4})$ The heat transfer from the wire can also

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

$\dot{Q}=h \pi D L(T_{s}-T_{\infty})$

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