$\dot{Q}=\frac{423-293}{\frac{1}{2\pi \times 0.1 \times 5}ln(\frac{0.06}{0.04})}=19.1W$
The heat transfer from the wire can also be calculated by: $\dot{Q}=\frac{423-293}{\frac{1}{2\pi \times 0
lets first try to focus on
$Re_{D}=\frac{\rho V D}{\mu}=\frac{999.1 \times 3.5 \times 2}{1.138 \times 10^{-3}}=6.14 \times 10^{6}$ $\dot{Q}=\frac{423-293}{\frac{1}{2\pi \times 0
The convective heat transfer coefficient for a cylinder can be obtained from: $\dot{Q}=\frac{423-293}{\frac{1}{2\pi \times 0
(b) Not insulated:
$\dot{Q}=10 \times \pi \times 0.08 \times 5 \times (150-20)=3719W$