Answer:
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
[tex]Q = 4* \pi k*R1*R2[\frac{T2-T1}{R2-R1}][/tex]
[tex]Q = 4* \pi*0.1*1*1.3*\frac{-140}{0.3}[/tex]
Q = - 762.3598 W
here negative indicate heat loss
