Answer to Question #170647 in Thermal Power Engineering for Md Asif Iqbal

Question

Water at an average temperature of 10°C flows inside a

horizontal smooth tube of 5 mm diameter with a velocity of

10 cm/sec. The surface of the tube is maintained at 70°C and

tube is 1 m long. Calculate the rate of heat transfer to the

water.

Answer 1

$m=\rho AU_m=\rho (\frac{\pi}{4}d^2)U_m=996(\frac{\pi}{4} \times 0.005^2) \times 0.1= 0.00195564143$

At $10^oC , \mu=0.001308 kg/m.s$

$Re=\frac{\rho U_m d}{\mu}=\frac{996 \times0.1 \times 0.005}{0.001308}=380.7339$

This is a laminar flow

At $10^oC, Pr=9.46$

$Gz^{-1}=(\frac{1}{380.7339 \times9.46})(\frac{1}{0.005})=0.0555$

At $Gz^{-1}=0.0555$, $Nu=3.66$ and $k=0.580 W/m.k$

$h=\frac{Nu \times k}{d}=\frac{3.66 \times 0.580}{0.005}=424.56 m^2 C$

$q=hA(T_w-T_c)=424.56 \times2 \pi (0.005)(70-10)=800.28 W$