Aspherical 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.

Q = - 762.3598 W

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 = - 762.3598 W

here negative indicate heat loss

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explanation:

answer:

an example of solubility is salt or sugar dissolving in water.

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