# Heat Loss per Linear Ft

For an assignment, I was given:

Calculate the heat loss per linear ft from $2$ $in$ nominal pipe. ($2.375$ $in$ outside diameter covered with $1$ $in$ of an insulating material having an average thermal conductivity of $0.0375$ $Btu/hrft^oF$. Assume that the inner and outer surface temperatures of the insulation are $380^oF$ and $80^oF$ respectively.

The correct answer is $116$ $Btu/lb^oF$

I used the formula from conduction through pipes:

$${ Q = \frac{\triangle t}{R_{total}} = \frac{\triangle t}{ \frac{ln \frac{r_2}{r_1} }{2 \times \pi \times k \times L} } = \frac{\triangle t}{ \frac{ln \frac{d_2}{d_1} }{2 \times \pi \times k \times L} } }$$

Where:

$${ \triangle t = 380-80= 300^oF }$$ $${ d_2 = 3.375 }$$ $${ d_1 = 2 }$$ $${ k = 0.0375 }$$

And I get close, but not correct. I calculated it to be: $135.090464$ $Btu$

What am I doing wrong?

I get 115.647

What you are doing wrong (in this problem and your other recent one) is going too fast and not visualising how the data given translates into the real world situation. You are applying formulae correctly and seem to have a good understanding of what is involved to turn given parameters into correctly formulated expressions. Now all you have to do is slow down a bit and think about what you are doing. Do that and you'll do well ! :-)

The pipe is described as

• 2in nominal pipe. (2.375 in outside diameter covered with 1in of an insulating material).

One interpretation of that would be a pipe of 2.375 " finished OD and 1" thick insulation under the outer surface. That gives a 0.375" internal pipe ID. Doesn't sound likely.

A second interpretation is a 2.375" ID internal pipe with 1" of external insulation over it for an external OD of ???

A third interpretation is the one you used.

I used the 2nd one and get the correct answer.
Look at the 2nd version above.
What is the OD?
plug in the results and see what you get.