Question

Steam at a temperature of 280 DegreeC flows in a steel pipe (k=70 W/m K) having an inner and outer diameter of 5 cm and 5.6 cm. The pipe is covered with glass wool insulation (k=0.05 W/m K) to a thickness of 4 cm. On a windy day heat is lost from the lagging surface with a convective heat transfer coefficient of 50 W/m² K. The surroundings are at a temperature of 5 DegreeC and the internal heat transfer coefficient for heat transfer between the steam and the steam pipe is 600 W/m² K.

What are the rate of heat loss from pipe per unit length and the outer surface temperature of the lagging?

Answer 1

Thermal circuit diagram

Ri = resistance inside the pipe

R1 = resistance between inside and outside the pipe

R2 = resistance between outside and insulation of the pipe

R0 = resistance between insulation and surroundings of the pipe

Heat transfer rate

heat transfer rate per unit length

= 0.0106 m-K/W

Radius of insulation = outer radius of pipe + insulation thickness

= (5.6/2) + (4) = 6.8 cm

Diameter of insulation = 13.6cm = 0.136 m

1/(h_o*3.14*D₀) = 1/(50*3.14*0.136) = 0.0468 m-K/W

= (1/2*3.14*70)* ln (5.6/5)

= 0.0002578 m-K/W

= (1/2*3.14*0.05)* ln (13.6/5.6)

= 2.8258 m-K/W

rate of heat loss from pipe per unit length

Q/L = (280-5)/(0.0106+0.0468+0.0002578+2.8258)

= 95.37 W/m

T_ins =( Q/L)*Rins

= 95.37*2.8258

= 269.5 C

T_Lag = 269.5 - 5 = 264.5 °C