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

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )