Heat Transfer
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Re: Heat Transfer
Re: Heat Transfer -- rookie eng.	Post Reply	Top of thread	Forum

Posted by: mechaic ® 04/04/2006, 11:05:17 Author Profile eMail author Edit

I think you can find the formula concerning the conduction across cylindrical shell in the heat transfer text book. (formula deployed across cyclindrical coordinate) If your problem is only heat source would a core conductor and heat path a few cylindrical layers, you can apply this heat resistance formala, R=ln(r2/r1)/(2pikL), where r2=outer radius, r1=inner radius, and k=thermal conductivity, L= thickness of a layer to your problem, establing the appropriate THERMAL CIRCUIT. Of course, you need to check feasibility in the real situation considering extra effects, such as 'surface resistance' and convection as well. But, I'm not sure whether your problem is just heat resistance formula or not. Take care~

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Re: Heat Transfer
Re: Re: Heat Transfer -- mechaic	Post Reply	Top of thread	Forum

Posted by: zekeman ® 04/04/2006, 11:36:49 Author Profile eMail author Edit

Draw the crossection giving the dimensions and thermal conductivity of each layer. I would ignore the copper braid since only the insulating materials and air gaps contribute to the thermal resistance. Also state whether the 20 deg is the temperature of ambient air. Are you sure of the 100 deg centgrade at the wire? I would think that if you would not know this unless this is an academic problem.

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Re: Heat Transfer
Re: Re: Heat Transfer -- zekeman	Post Reply	Top of thread	Forum

Posted by: rookie eng. ® 04/04/2006, 13:52:38 Author Profile eMail author Edit

The room temperature is ambient at 20 degrees C. It is estimated that the center conductor will be 100 degrees C. I'm stuck finding the temperature on the outside of the jacket, I've found out the resistance in the materials, but am not confident with my answer. Im not too sure on the free convection process. Where I will add in 1/ha to get help get the heat transfer rate. I'm having technical difficulties with the picture. The center conductor has a diameter of .00184m and the thermal conductivity of 386 w/m K. Second is the layer of PTFE. The diameter is .00518m The thermal conductivity is .25 w/m K The jacket is FEP with a diameter of .0062m The thermal conductivity is .19 w/m K. The length of the cable is 12.7 m Thanks

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Re: Heat Transfer
Re: Re: Heat Transfer -- rookie eng.	Post Reply	Top of thread	Forum

Posted by: devitg ® 04/04/2006, 14:07:50 Author Profile eMail author Edit

Could you put a thermocuple at the jacket???? It seem extrange that a conductor will work at 100°C, but it can be, what's the application?

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Re: Heat Transfer
Re: Re: Heat Transfer -- devitg	Post Reply	Top of thread	Forum

Posted by: rookie eng. ® 04/04/2006, 14:24:22 Author Profile eMail author Edit

We are going to test it eventually, I'm a co-op student and it was a little challenge to see if we could figure it out with the math...too bad I haven't taken heat transfer yet. This type of cable is used for semiconductor apps.

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Re: Heat Transfer
Re: Re: Heat Transfer -- rookie eng.	Post Reply	Top of thread	Forum

Posted by: rookie eng. ® 04/05/2006, 09:12:42 Author Profile eMail author Edit

Thanks a lot, I've seen too many different formulas for the resistance but wasn't sure which one to apply. When the resistance is found I can apply it to the formula: Q=Change in Temp /Resistance then figure my temperature at the jacket

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Re: Heat Transfer
Re: Re: Heat Transfer -- rookie eng.	Post Reply	Top of thread	Forum

Posted by: zekeman ® 04/04/2006, 17:57:05 Author Profile eMail author Edit

The resistance of annular cylindrical layers per unit length of cylinder (since the clinder is very long) is given by 1/Ki*Ln[r(i+1)/r(i)] and the surface resistance is1/h*ro where the h units I use are ro= outside radius for ambient air at those temperatures h is the order of 1BTU/hr/ft^2/deg F= 6 W/m^2degC For example the resistance of PTFE would be 1/.25*ln(.00518/.00184)=4.14 and outer jacket 1/.19*ln(00620/00518)=0.94 and 1/(h*ro)=1/6(.0062)=27 As you know, you treat this like a series electrical circuit where 100-20 represents the temperature drop is the analogue of voltage and the resistances of each of the 1/Ki*Ln(ri+1/r1) Note:The conversion of watts/degC/m^2 would be=3.14BTU/hr/(39/12)^2ft^2/1.8 deg F=0.165 i.e. you multiply your units of K by 0.165 to get the BTU-hr-ft units of K

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