Answer:
rate of heat transfer = 9085708.80 W
Explanation:
Given:
Inside diameter, D = 5.1 cm
= 5.1 x [tex]10^{-2}[/tex] m
Average velocity, V = 7 m/s
Mean temperature, T = (66+38) /2
= 52°C
Therefore kinematic viscosity at 52°C is ν = 0.104 X [tex]10^{-6}[/tex] [tex]m^{2}[/tex] / s
Prandtl no., Pr = 0.021
We know Renold No. is
Re = [tex]\frac{V\times D}{\nu }[/tex]
Re = [tex]\frac{7\times 5.1\times 10^{-2}}{0.104\times 10^{-6}}[/tex]
= 3.432 X [tex]10^{6}[/tex]
Therefore the flow is turbulent.
Since the flow is turbulent and the ratio of L/D is greater than 60 we can use Dittua-Boelter equation.
Nu = 0.023 [tex]Re^{0.8}[/tex].[tex]Pr^{0.3}[/tex]
= 0.023 x [tex](3.432 \times10^{6})^{0.8}[/tex] x [tex](0.021)^{0.3}[/tex]
= 1221.52
Since Nu = [tex]\frac{h.D}{k}[/tex]
h = [tex]\frac{k\times Nu}{D}[/tex]
= [tex]\frac{9.4\times 1221.52}{5.1\times 10^{-2}}[/tex]
= 225143.3
Therefore rate of heat transfer, q = h.A(T-[tex]T_{\infty }[/tex]
q= 225143.3 x 2πrh ( 66-38)
= 225143.3 X 2π X [tex]\frac{5.1\times10^{-2}}{2}\times 9\times 28[/tex]
= 9085708.80 W