Therefore, the heat flow rate under these conditions is 19600 Btu/hr.

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SUMMARY

The discussion centers on calculating the required insulation thickness for a 4-inch outer diameter pipe transporting liquid metals, with an outer surface temperature of 1400°F and a maximum allowable insulation surface temperature of 300°F. The thermal conductivity of the insulation is defined as k = 0.08(1-0.0003T) [Btu/hr*ft*F]. The heat flow rate under these conditions is determined to be 19600 Btu/hr. Participants explore the integration of heat transfer equations and the implications of insulation thickness on temperature control.

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  • Understanding of heat transfer principles, specifically conduction.
  • Familiarity with thermal conductivity equations and their applications.
  • Basic knowledge of calculus for integration in heat transfer equations.
  • Experience with thermal insulation materials and their properties.
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  • Study the derivation of the heat transfer equation q/A = k dT/dx in detail.
  • Learn about the impact of varying thermal conductivity on insulation performance.
  • Explore numerical methods for solving heat transfer problems, including spreadsheet applications.
  • Investigate the effects of different insulation materials on thermal efficiency.
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Mechanical engineers, thermal system designers, and students studying heat transfer who are involved in insulation design and thermal management of piping systems.

orangeklingon
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Homework Statement


A 4-in.-OD pipe is used to transport liquid metals and will have an outside surface temperature of 1400 F under operating conditions. Insulation 6-in. thick and having a thermal conductivity expressed as k = 0.08(1-0.0003T) where k is in [Btu/hr*ft*F] and T is in [F], is applied to the outside surface of the pipe.
(a)What thickness of insulation would be required for the outside insulation temperature to be no higher than 300 F.
(b) What heat-flow rate will occur under these conditions?


Homework Equations


q/A = k dT/dx
A = 2*(pi)*r*L, Assume L is 1 ft.

The Attempt at a Solution


q/A dr = k dT
Integrate on both sides
q/2*(pi)( ln(r-pipe)-ln(r-insulation)) = 0.08(T-0.00015T^2)(evaluate from 1400 to 300)

Is this a Trick question? Do I assume that the insulation thickness is 6-in? Or is there a trick that I am missing?

I would like to have help on this by Wednesday November 25th by 4PM PST. Thanks
 
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I may be able to help a bit..I had a question simmilar to this, but I've handed it in. I made and used a spreadsheet to solve it, not calculus.

With all of your given information, you can solve for the temperature of the fluid inside the pipe. (Is the thermal conductivity of the pipe neglected?)

Knowing the fluid temperature and the surface temperature, with the k value for each composite could you not solve for the thickness required to limit the temp. to 300?
 

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