How to Calculate Heat Loss in a Hot Water Pipe?

[DerekWeir29]

Homework Statement

A thick wall cylinder having a thermal conductivity of 60W/mK has a 75mm inner diameter and is 2.5mm thick. The pipe carries water at 90 degrees celsius inside and the outside is surrounded by air at 20 degrees celsius.

The heat transfer coefficients inside and outside the hot water pipe are 500W m2 K and 35 W m2 K respectively. Natural convection can be ignored.

Calculate the heat loss in W per unit length of the pipe

Homework Equations

q= K Delta U T1-T2/r1-r2 = T1-T2/r2-r/Kal

Alm = 2Pie Lr2- 2Pie Lr1/Ln(2(2{PieLr2/2pieLr1 [/B]

= A2-A1/ln(A2/A1)

The Attempt at a Solution

q = 90-20/0.03875-0.037
60 * (2Pie x 1 x 0.03875- 2Pie x 1 x 0.03875)/
ln (2Pie x 1 x0.3875/2Pie x 1 x0.3875/2Pie x 0.3875

 

[Chestermiller]

Your equations are unreadable. Please use the symbols provided with the ∑ sign on the tool bar, or use Latex. There is a LaTex tutorial available on our sight.

Regarding your calculations, you seem to have omitted the heat transfer resistance on the air side and on the fluid side, which are in series with the pipe wall. Also, for such a thin wall, there is no need to use the log-mean.