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

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

$\dot{Q}_{rad}=1 \times 5.67 \times 10^{-8} \times 1.5 \times (305^{4}-293^{4})=41.9W$ $r_{o}+t=0

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

$T_{c}=800+\frac{2000}{4\pi \times 50 \times 0.5}=806.37K$ $r_{o}+t=0

$\dot{Q}_{conv}=150-41.9-0=108.1W$