A horizontal cylinder, 3.0 cm in diameter and 0.8 m length, is suspended in water at 20oC. Calculate the rate of heat transfer if the cylinder surface is at 55oC. Given Nu = 0.53 (Gr x Pr)1/4
The properties of water at average temperature are as follows:
Density, ρ = 990 kg/m3
Viscosity, μ = 2.47 kg/hr.m
Thermal conductivity, k = 0.534 kcal/hr.m.oC
Cp = 1 kcal/kg.oC.
Calculations:
Gr = gβD3(Tw - T∞)ρ2/μ2
β = 1/Tf
Tf = 0.5 (Tw + T∞) = 0.5 x (55 + 20) = 37.5oC = 310.5 K
β = 1/310.5 K-1
μ = 2.47 kg/hr.m = (2.47 / 3600) kg/m.sec = 6.861 x 10-4 kg/m.sec
Substituting for the variables,
Gr = 9.812 x (1/310.5) x 0.033 x (55 - 20) x 9902 / (6.861 x 10-4)2
= 62.176 x 106
Pr = Cpμ/k
k = 0.534 kcal/hr.m.oC = 0.534 x 4184 / 3600 W/m.oC = 0.6206 W/m.oC
Cp = 1 kcal/kg.oC = 4184 J/kg.oC
Pr = 4184 x 6.861 x 10-4 / 0.6206 = 4.6254
Nu = 0.53 x (62.176 x 106 x 4.6254)1/4 = 69.02
Nu = hD/k = 69.02
h = 69.02 x 0.6206 / 0.03 = 1427.8 W/m2.oC
Rate of heat transfer
Q = hA(Tw - T∞) = h π D L (Tw - T∞)= 1427.8 x 3.142 x 3 x 10-2 x 0.8 x (55 - 20) = 3767.8 W
Last Modified on: 04-Feb-2022
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