# Thermal conductivity of airplane

• joemama69
In summary, the rate of heat flow through the plane is kA*dQ/dt, where k is the heat flow rate and A is the cross-sectional area of the inside or outside of the plane.

## Homework Statement

The passenger section of an airplane can be thought to have a shape of a cylindrical tube. For a small airplane, the tube is 35 m long and has an inner radius of 2.5 m. The exterior of the tubular wall is lined up with a 6- cm thick layer of insulating material of thermal conductivity 4 X 10-5 cal/s- cm-K. If the inside temperature is to be maintained at 25 °C, and the outside temperature is -35 °C, determine the rate that heat must be delivered to maintain this temperature difference.

## The Attempt at a Solution

dQ/dt = kA(T(inside) - T(outside))/l

I a having trouble identifying y variables

k = 4E^-5, A = this is my confusion, is it the cross-sectional area of the inside or outside of the plane, l = .006 m

You'll have to use calculus. Consider an infinitely very thin layer of material. It will have only one area, 2*pi*r*dr. Consider the heat flow through this thin layer, then use integration to get the rate of heat flow through the whole thing.

kinda confused but r u saying that i have to find the area of the entire outside cylinder

am i suppose to integrate 2pi r dr fro the inside radius to the outside radius

joemama69 said:
kinda confused but r u saying that i have to find the area of the entire outside cylinder

No.

am i suppose to integrate 2pi r dr fro the inside radius to the outside radius

Oops, I meant 2pi*r*L, which would be the surface area of a thin slice of the cylinder at radius r. H=kA*delta-T/delta-x, which you can rewrite as H=k*2pi*r*L*dT/dx. Do you know how to get the answer from here?

what do i use for my radius, the inside or inside plus the insulation.

also why did you put the delta-t on the other side, the answer is ging to be in jewls/seconds right.