Thermal conductivity of airplane

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SUMMARY

The discussion focuses on calculating the rate of heat transfer required to maintain a temperature difference in an airplane's passenger section, modeled as a cylindrical tube. The inner radius is 2.5 m, the tube length is 35 m, and the insulating material has a thermal conductivity of 4 x 10-5 cal/s-cm-K with a thickness of 6 cm. The heat transfer equation used is dQ/dt = kA(Tinside - Toutside)/l, where the area A is derived from the surface area of the cylindrical layers. The integration of the area from the inner to the outer radius is essential for accurate calculations.

PREREQUISITES
  • Understanding of heat transfer principles, specifically Fourier's law.
  • Familiarity with cylindrical geometry and surface area calculations.
  • Basic calculus skills for integration.
  • Knowledge of thermal conductivity and its units.
NEXT STEPS
  • Study Fourier's law of heat conduction in detail.
  • Learn how to calculate the surface area of cylindrical objects.
  • Practice integration techniques for calculating heat transfer through varying materials.
  • Explore thermal conductivity values and their implications in engineering materials.
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Students and professionals in mechanical engineering, aerospace engineering, and thermal management who are involved in heat transfer analysis and insulation design for aircraft structures.

joemama69
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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.


Homework Equations





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
 
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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.
 

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