Question

A spherical vessel 2m diameter is lagged to a depth of 300mm. The thermal conductivity of the lagging is 0.1W/mK. The temperature of the inside and outside of the lagging is 180°C and 40°C respectively. Calculate the heat loss.

Answer 1

Q = - 762.3598 W

Step-by-step explanation:

given data:

thermal conductivity k = 0.1 W/mK

Diameter of vessel = 2 m

hence radius R1 is 1 m

R2 = R1 + Lag distance = 1 +.3 = 1.3 m

outside Temperature T2 = 40 Degree C

Inside Temperature T1 = 180 Degree C

heat loss or gain is given as

`Q = 4* \pi k*R1*R2[(T2-T1)/(R2-R1)]`

`Q = 4* \pi*0.1*1*1.3*(-140)/(0.3)`

Q = - 762.3598 W

here negative indicate heat loss