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

The rate of heat transfer is:

$\dot{Q}=62.5 \times \pi \times 0.004 \times 2 \times (80-20)=100.53W$

$h=\frac{Nu_{D}k}{D}=\frac{2152.5 \times 0.597}{2}=643.3W/m^{2}K$ The rate of heat transfer is: $\dot{Q}=62

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

$\dot{Q}_{cond}=0.0006 \times 1005 \times (20-32)=-1.806W$ The rate of heat transfer is: $\dot{Q}=62

lets first try to focus on

$h=\frac{Nu_{D}k}{D}=\frac{10 \times 0.025}{0.004}=62.5W/m^{2}K$ The rate of heat transfer is: $\dot{Q}=62

The convective heat transfer coefficient is: