Heat lost in polypro 4" pipe .25" wall temperture diff 40 degrees F?

Question

Material in pipe is air, rate of flow is 400 cu ft/min

Answer 1

Q/t = h*A*(Tsubstance-Tenvironment)

where

Q/t is the heat rate per foot

h is the material thermal conductivity k over thickness t (k/t)

A is the area of the pipe (ft2 for 1 ft length is what I'm using)

The difference in T is given to you.

You have to look up the k for polypro, it'll be in whatever textbook you have; I used .76

Plugging everything into place and watching your units carefully, you get

Q/t = (.76/.25)*(2*pi*radius of pipe in feet*1 foot)*(40 degrees)

This should get you an answer in Btu/hr*ft.