Heat transfer through a cylinder

[marko11]

Hey Guys,

i've an interesting question for you! I'm currently designing a test bench with a focus on heat transfer of bearings. Therefore I've studied this topic and want to calculate following example:

A horizontal cylinder out of steel is heat insulated on the fronts (so that no heat transfer occurs). The heat flow into the inner diameter is given with 241 Watts. The outer diameter gives heat to the environment by Radiation (polished steel)and natural Convection (Environmetal fluid is dry air). Environment temperature is 20 degree. Now I want to Calculate the temperature on the surface of inner and outer diameter. Well actually i'e done the Calculation with MAthcad iterative. But the temperatures I calculated differ strongly from the expected values. On the picture I draw the geometry an the heat flow.

http://files.engineering.com/getfil...2e2-4f3a-befa-b2b0cfb4f1a5&file=heat_flow.PNG

On the other ones are my Calculations. On the first one I modify the surface temperature of the outer diameter. On the secondone surface temperature of the outer diameter ist calculated and the two values are subtracted, which results in an Deviation. This is repeated iterative until the Deviation is less than 1K.

http://files.engineering.com/getfil...-422a-a8b2-18d504462326&file=calculation1.PNG

http://files.engineering.com/getfil...-4842-8a4e-838779e32a21&file=calculation2.PNG

I expected surface temperatures below 100° but calculated 1077K and 791°C. Has anyone an idea where the mistake could be? I repeatedly checked al the input values. They also differ only very slightly with Temperature.

Thanks in advance
Marko

 

[IRstuff]

Kudos for using Mathcad! BUT, the actual worksheet would have been more useful that a picture of the worksheet.

Nevertheless, your answers seem to be in the general ballpark, given that you are trying to dissipate 241W from less than 1/100th of a square meter. The area you've used is roughly comparable to the surface area of a light bulb, which we know from experience is not touchable by hand for any power level above 40W. If the surface temperature of the bulb is 100C, that's a delta of 80K. Multiply that by 6 to get 240W, you'd get a delta of at least 480K

> Your construct of blending the conductive and convective is incorrect. They are not in series in the sense that you are using them, since you know the 241W must leave the outer surface of the cylinder. Therefore, only the convective and radiative coefficients are useful for calculating the outer surface temperature. The conductive coefficient is only of interest if you need to calculate the inner surface.

> your last calculations don't make sense, but they're irrelevant since the initial setup is incorrect. your 241W, divided by 41W/m^2-K, divided by the area should result in about 780K delta, so a surface temperature of about 800C. Since this lower than the temperature that you used for blackbody calculation, you'd have to iterate on that, of construct a solve block to do that automatically.

TTFN
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[marko11]

Hey IRStuff,

Thank you very much for your fast and helpful response. You're right that I would be more sensefull to share the worksheet so I translated it in english for a better understanding and attached it on the response.

Well your second advice is that my construct of blending conductive and convective is incorrect. But I calculate the temperature on the inner surface it's 1077K. The outer surface is only calculated from the inner surface temperature. In the mathcad sheet I now added the calculation fronm the environmental temperature by conductive and radiation coefficient. It's the same result. Is this truely wrong?

Please help me in addition to understand your last advice: " Since this lower than the temperature that you used for blackbody calculation, you'd have to iterate on that, of construct a solve block to do that automatically. " Sry I don't get it?

 

[marko11]

Greetings,

Marko

 

[IRstuff]

Oh, OK, I wasn't absolute positive, since you had the conductive and convective coefficients merged together at one spot in the sheet.

Mathcad Prime is only on my home computer, so that will have to wait.

You can work backwards from the convection and radiation, since conservation of energy applies. So, you would have roughly 800C outside surface temperature, and the inner surface would be roughly 817 C, since the thermal conductivity results in about 14 W/K. With 241 W input, the temperature rise across the cylinder wall would be about 17C



TTFN
I can do absolutely anything. I'm an expert!
faq731-376 forum1529