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Conduction in Semiinfinite Solid
- Thread starter swguru
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McAdams-Heat Transmission has some worked out solutions for your case, that seem to come from an unpublished reference.
He doesn't give the temperature distributions, only the thermal resistance between the buried body and the surface R=(t1-t2)/q, t1 being the isothermal temperature of the body, t2 the temperature of the soil at the surface. The latter can't of course be uniform, so, though this is unclear in the book, I suppose that t2 is to be considered as an average temperature by which the subsequent flow of heat to the air may be calculated.
For a sphere of diameter D with center at distance z below surface:
R=(1-D/(4z))/(2[π]Dk)
prex
: Online tools for structural design
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He doesn't give the temperature distributions, only the thermal resistance between the buried body and the surface R=(t1-t2)/q, t1 being the isothermal temperature of the body, t2 the temperature of the soil at the surface. The latter can't of course be uniform, so, though this is unclear in the book, I suppose that t2 is to be considered as an average temperature by which the subsequent flow of heat to the air may be calculated.
For a sphere of diameter D with center at distance z below surface:
R=(1-D/(4z))/(2[π]Dk)
prex
: Online tools for structural design
: Magnetic brakes for fun rides
: Air bearing pads
- Thread starter
- #3
Thanks. I used a similar equation for a cylinder that is buried. R=(ln(4*z/D))/2*pi*D*k)
To verify calculations, I conducted a simple experiment where a resistor was placed soil. The resistor dissipated 4W. A thermocouple was attached to the resistor. After equilibrium, the resistor temperature was 89C. Based on this the therm resistance was 16.7C. Thus the calculated k for the soild was 1.19watt/m-K.
However, after further verification using an FEA model, I found that this k value is not correct. The K value that yielded model results that agreed with the experiment was 0.39watt/m-K. I further verified model results by placing another TC approx 1in form the resistor and the measured value agreed with the temperature probed in the model.
So, the equation initially used to predict k was incorrect. Is there another form of the euation that would yield better results (or another way to calculate k)
swguru
To verify calculations, I conducted a simple experiment where a resistor was placed soil. The resistor dissipated 4W. A thermocouple was attached to the resistor. After equilibrium, the resistor temperature was 89C. Based on this the therm resistance was 16.7C. Thus the calculated k for the soild was 1.19watt/m-K.
However, after further verification using an FEA model, I found that this k value is not correct. The K value that yielded model results that agreed with the experiment was 0.39watt/m-K. I further verified model results by placing another TC approx 1in form the resistor and the measured value agreed with the temperature probed in the model.
So, the equation initially used to predict k was incorrect. Is there another form of the euation that would yield better results (or another way to calculate k)
swguru
swguru
The equation you used is different than the one prex suggested. Because you didn't list your z and D values, I couldn't calculate which one you based your final k value on.
Prex's formula gives much smaller values, provided, of course, the diameter is less than one.
However, I think you hit on an important point. Most all of heat transfer is based on experimental results with the "formulas" being those that best fit the obtained data. I don't know how hard it would be for you to run additional tests. Then you could try variations on the R= [ƒ](4z, D)/2[π]Dk formula until you found one that best fit your data.
Patricia Lougheed
Please see FAQ731-376 for tips on how to make the best use of the Eng-Tips Forums.
The equation you used is different than the one prex suggested. Because you didn't list your z and D values, I couldn't calculate which one you based your final k value on.
Prex's formula gives much smaller values, provided, of course, the diameter is less than one.
However, I think you hit on an important point. Most all of heat transfer is based on experimental results with the "formulas" being those that best fit the obtained data. I don't know how hard it would be for you to run additional tests. Then you could try variations on the R= [ƒ](4z, D)/2[π]Dk formula until you found one that best fit your data.
Patricia Lougheed
Please see FAQ731-376 for tips on how to make the best use of the Eng-Tips Forums.
Your results may simply be a function of the fact that soil is, by it's very nature, variable. Moreover, just the simple fact that you disturbed the soil to bury the resistor.
Additionally, if you left the setup alone, the results 6 months from now would be different than what you got, because the soil would have compacted and settled more.
TTFN
FAQ731-376
Additionally, if you left the setup alone, the results 6 months from now would be different than what you got, because the soil would have compacted and settled more.
TTFN
FAQ731-376
swguru,
McAdams gives the following equation for a horizontal buried cylinder of length L, diameter D and axis at distance z below surface:
R=ln(4z/D)/(2[π]Lk)
with z>>D and z<<L
or more exactly:
R=cosh-1(2z/D)/(2[π]Lk)
with z<<L
What is unclear to me is the meaning of temperature t2 at surface of ground. If it is assumed variable, then it would be some average value, but then the exchange to air cannot be calculated as the area over which to calculate is not provided. It is possibly taken as constant (soil surface isothermal, as isothermal is the surface of the body), but McAdams doesn't clear up this point.
prex
: Online tools for structural design
: Magnetic brakes for fun rides
: Air bearing pads
McAdams gives the following equation for a horizontal buried cylinder of length L, diameter D and axis at distance z below surface:
R=ln(4z/D)/(2[π]Lk)
with z>>D and z<<L
or more exactly:
R=cosh-1(2z/D)/(2[π]Lk)
with z<<L
What is unclear to me is the meaning of temperature t2 at surface of ground. If it is assumed variable, then it would be some average value, but then the exchange to air cannot be calculated as the area over which to calculate is not provided. It is possibly taken as constant (soil surface isothermal, as isothermal is the surface of the body), but McAdams doesn't clear up this point.
prex
: Online tools for structural design
: Magnetic brakes for fun rides
: Air bearing pads
- Thread starter
- #7
Correct. I typed in D instead of L in the equation. In my experiment z=3in, D=0.5in and L=1in. Im my case z>D, but not <L, so maybe this is why the result from this calculation does not agree with FEA model and experiment results?
I assumed room temperature for t2 since my soil experiment is in a lab.
Agree with IRstuff about k variability of soil due to compacting over time. Thsi is one of my big concerns for my actual application. Also moisture content would play a significant role.
swguru
I assumed room temperature for t2 since my soil experiment is in a lab.
Agree with IRstuff about k variability of soil due to compacting over time. Thsi is one of my big concerns for my actual application. Also moisture content would play a significant role.
swguru
sailoday28
Mechanical
If t2 is the surrounding ambient and somehow you can account for thermal radiation, AND since the thermal flux at the surface is known, then, the REAL t2 -surface temperature can be calculated.
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