A horizontal cylinder, 3.0 cm in diameter and 0.8 m length, is suspended in water at 20^{o}C. Calculate the rate of heat transfer if the cylinder surface is at 55^{o}C. Given Nu = 0.53 (Gr x Pr)^{1/4}

The properties of water at average temperature are as follows:

Density, r = 990 kg/m^{3}

Viscosity, m = 2.47 kg/hr.m

Thermal conductivity, k = 0.534 kcal/hr.m.^{o}C

C_{p} = 1 kcal/kg.^{o}C.

Calculations:

Gr = gbD^{3}(T_{w} - T_{µ} )r^{2} / m^{2}

b = 1/T_{f}

T_{f} = 0.5 (T_{w} + Tµ ) = 0.5 x (55 + 20) = 37.5^{o}C = 310.5 K

b = 1/310.5 K^{-1}

m = 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 0.03^{3} x (55 - 20) x 990^{2} / (6.861 x 10^{-4})^{2}

= 62.176 x 10^{6}

Pr = C_{p}m/k

k = 0.534 kcal/hr.m.^{o}C = 0.534 x 4184 / 3600 W/m.^{o}C = 0.6206 W/m.^{o}C

C_{p} = 1 kcal/kg.^{o}C = 4184 J/kg.^{o}C

Pr = 4184 x 6.861 x 10^{-4} / 0.6206 = 4.6254

Nu = 0.53 x (62.176 x 10^{6 }x 4.6254)^{1/4} = 69.02

Nu = hD/k = 69.02

h = 69.02 x 0.6206 / 0.03 = 1427.8 W/m^{2}.^{o}C

Rate of heat transfer Q = hA(T_{w} - Tµ ) = h p D L (T_{w} - Tµ )

= 1427.8 x 3.142 x 3 x 10^{-2} x 0.8 x (55 - 20) = 3767.8 W