Question

36. Steam at 373K is passed through a tube of radius 10cm and length 2m. The thickness of tube is 5mm and thermal conductivity of the material is 390 wm 'k', calculate the heat loss per second. The outside temperature is 0°C​

Answer 1

The heat loss per second=10,044.821 KJ/s.

1. Steam temperature T1 (say) is 373 K.

Outside temperature T2 (say) is 0C=273K.  [T centigrade+273= T kelvin]

Thermal conductivity k=390 W-m/K.

Length L=2m.

inner radius r2= 10 cm=0.1 m

Thckness t=5mm=.005 m    [milli corresponds to 10^(-3)]

outer radius r1=r2+ t=0.105m    [from figure]

2. Formula for heat loss by conduction in cylinders is given by

Heat loss Q=2*(pi)*k*L*(temperature difference)/(ln(r1/r2))

Q=2*3.14*390*2*(373-273)/(ln(0.105/0.1))

Q=10,044,820.76 J/s=10,044.821 KJ/s

Answer 2

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