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

Question #170647

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.


1
Expert's answer
2021-03-11T06:56:54-0500

m=ρAUm=ρ(π4d2)Um=996(π4×0.0052)×0.1=0.00195564143m=\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 10oC,μ=0.001308kg/m.s10^oC , \mu=0.001308 kg/m.s

Re=ρUmdμ=996×0.1×0.0050.001308=380.7339Re=\frac{\rho U_m d}{\mu}=\frac{996 \times0.1 \times 0.005}{0.001308}=380.7339

This is a laminar flow

At 10oC,Pr=9.4610^oC, Pr=9.46

Gz1=(1380.7339×9.46)(10.005)=0.0555Gz^{-1}=(\frac{1}{380.7339 \times9.46})(\frac{1}{0.005})=0.0555

At Gz1=0.0555Gz^{-1}=0.0555, Nu=3.66Nu=3.66 and k=0.580W/m.kk=0.580 W/m.k

h=Nu×kd=3.66×0.5800.005=424.56m2Ch=\frac{Nu \times k}{d}=\frac{3.66 \times 0.580}{0.005}=424.56 m^2 C

q=hA(TwTc)=424.56×2π(0.005)(7010)=800.28Wq=hA(T_w-T_c)=424.56 \times2 \pi (0.005)(70-10)=800.28 W



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