Natural convection question

[smukthiHT]

Hi All,

I hope all are doing good!


I had our lab do a test for heat transfer for a simple scenario. It is heat transfer from a top(outside) surface of a horizontal plate. The temperature of the ambient is around 23 C and temperature measured at 2 thermo couple points on the top surface of part is 165 C for a heat source of 15 watts on the inside of the horizontal flat plate. The other side of the horizontal plate has a thick insulation to make sure heat flow only towards the top surface of the part.

I have used analytical equation from standard literature for a horizonatl plate with HT from the top surface to find out the heat transfer due to convection for this configuration for the above parameters and I got a h_conv=13.1 W/m2 and heat transfed_convection=8.5 Watts by using Newton’s law of cooling.
The heat transfer because of radiation is 0.75W/m2 as the emissivity of polished aluminum is 0.1.
The sum of heat transfers because of convection and radiation(8.5+0.75) does not even come close to the input in the test=15 Watts.
I am trying to unsderatnd the what could hav lead to this devaition?
Is it because, I am computing the heat transfer_covection using an average film coefficient of convection value but not localized one?
Can you please comment and share your thoughts?

Thanks
Sridhar

 

[IRstuff]

How big is the plate?

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[ZDR1985]

Why you add 8.5 Watts to 0.75 W/m2?

 

[IRstuff]

sure, 13.1+0.75 = 13.85 would make more sense, assuming the units were correct, and is tolerably close to the measurement.

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[smukthiHT]

@IRStuff,
The plate is 8cmX6cm


@ZDR1985
Sorry I have type the incorrect units above for 0.75. The correct units are watts.
0.75 is heat trasfer due to radiation. Its uints are Watts.0.75 watts. So I have added it to 8.5 watts.

Thank you both for your responses.

Thanks
Sridhar

 

[IRstuff]

But your input is 15 W, not 15 W/m^2

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[smukthiHT]

Yes.

So that's the question I was asking. What could have lead to difference between between test and analytical equation.

 

[btrueblood]

Well, let's see...
1. you assumed there is no heat flux on bottom or sides of plate (where you found perfect insulation is what I want to know), and
2. you assumed the convection h coefficient to be some value, presumably from an equation that matched the scale of your plate and for similar temperature gradient, and you checked that there were no cross-flow conditions occurring to spoil the natural convection boundary layer development, and
3. you assumed the plate emissivity to be 0.1...

I've listed them in the order that I think the major errors are occurring, but do you see where some of the problem occurs?

More, you say the ambient is "about" 23 C, and the input power is 15 W...did you measure either parameter, and if not why not?

 

[IRstuff]

The big issue is that you didn't say what heat transfer coefficient you used, but it looks you used only 12.5W/m^2-K, but the heat transfer coefficient is a function of the temperature, i.e., the higher the temperature, the higher the heat transfer coefficient. A value of 21 W/m^2-K might be a better choice:

http://www.brighthubengineering.com...transfer-coefficient-estimation-calculations/

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[prex]

Your average heat transfer coeff (conv.+rad.) is 22 W/m²K.
I cannot check this figure in the literature right now, but it appears to me quite reasonable.
In calm air and for a low ΔT, I normally take 10 W/m²K; this includes convection and radiation, roughly contributing 50% each.
Here you have a fairly high ΔT, this increases convection, possibly creating a chimney effect around and over the plate, and increases also radiation.
Your estimate of the radiative exchange is too low: the emissivity is possibly as low as 0.1, but the overall result appears underestimated to me. Also I wonder whether alu fully maintains its brightness at 165 degC temperature.

prex
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[smukthiHT]

Thanks for the response btrueblood and prex.

@btrueblood
I have done the diagnosis on the same lines.
I am comfortable with assumptions 1 and 3, but not sure about 2.

1. The heat source is a set of resistors that are attached on the back of the plate and the resistors are glued to the plate and a thick foam of 5 cm is placed on the other side to make sure heat trasnsfer happens through the top surface of the part. There is nothing perfect, It think it a reasonable assumption to say that heat is trasfred only through the top surface of thepart.

2. The computed the film co-efficinet of convection h_con from the Raleigh number for the configuration and the realion between Nusselt number and Raleigh number for a horzionatl plate form literature. You can find this any stanadrd heat transfer testbook.
You can find this calculation from the below link too, by using inputs W=0.06m, L=0.08m, Tp=165°C, Ta=23°C and e(emissivity)=0.1

http://www.thermal-wizard.com/tmwiz/convect/natural/hup-isot/hup-isot.htm

This matches with my hand calculation.
My reason for devaition
The test is done in a draft free room. But 6 samples are tested side by side with a gap of around 20 cms between each other. This could lead to cross-flow conditions occurring to spoil the natural convection boundary layer development, which might render these equations not to be applicable. Would this lead to the devaition I see in the resulst from testing and analytical euqation? Could you please share your insights on that?

3. The plate is polished Alluminum. We have test data that emissvity of of polished Allumiium is between 0.05 -0.15 and from literature too.
I have used the equation q_rad=sigms(Stefan Boltzmann constant)*e_part(emissivity of the paty)*(T_part^4-T_ambient) and got q_rad= 0.75 watts, for T_part=(135+273)K , T_ambient= (23+273)K , e_part=0.1.
I am comfartable with this assuption too.

@@PREX
How did you come up with a heat transfer co-effcint of 22W/m2. Could you please share?

I have expertise in structural analyis, but pretty new to Thermal CFD analysis. IU really appreciate all you help.

Thanks
Sridhar

 

[IRstuff]

That's high school math -- 15W/area/deltaT

I'm not sure that particular calculator is correct. Other calculators have Rayleigh and Grashof numbers higher by an order of magnitude

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[btrueblood]

Regarding assumption #1: how certain are you that there is no convection participation from the area of the insulation surrounding the plate?

The assumption is fairly simple to check, you just need to measure the temperature of the insulation at its various boundaries.

When you've carefully measured all the variables but one (h is going to be tough to measure directly), you can start talking about correlation. SciencE!

 

[smukthiHT]

@Irstuff,

I understand that h=22W/m2 will correspond to 15 Watts.
but when h is computed from the relationship for a horizontal plate using Raleigh number and Nusselt number for the given configuration we get a h of 13.1 W/m2.
So started this thread with what could lead to the deviation in h.
I am euqation used by the calculator are pretty standard from literature for a Horizontal plate and I beleive they are correct.

@btrueblood: Thank you for your inputs.

I have total hemispherical emissivity measured for this plosihed suraface(AL 3003) at 300k to be 0.1. I know hemispherical emissivity changes with temperature. I am working at @450K. Would there be an significant chnage emissivity between both the temperatures. Can any one please comment?

Thanks
Sridhar

 

[btrueblood]

Yes, I would expect the emissivity to change a few points for that relative change in temperature. There is a pretty good discussion of this in section II of this report:

http://www.aztechnology.com/pdfs/nasa-tp-2005-212792-lauder.pdf

Finally, to evaluate cross-flow effects from the multiple cells, could you turn off the power on some cells, and see what the effect is? Also, how difficult would it be to put a Schlieren camera system across the plane?

 

[IRstuff]

I've never seen an analysis on what the impact of scale is to the convection correlations, so I'd imagine that the correlations potentially break down for a horizontal plate, in particular, because there are some inherent assumptions in the equations that deal the characteristic dimension that's used in the correlations.

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[IRstuff]

This article seems to suggest that physically smaller plates might indeed have a higher convection coefficient than what might be otherwise calculated

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[smukthiHT]

Sorry for the late response.Thank you all for the responses. It was helpful.

can anyone refer me to some white papers or journal papers on an ideal test set up for predicting natural convection behavior?

Thanks!

 

[IRstuff]

I think your lab already did that. The only question is whether their test was accurate to your conditions, e.g., did they adequately protect against outside air currents.

There are lots of resources on the web:

http://c.ymcdn.com/sites/

http://www.saimeche.org.za/resource...ger-2002_01__600_dpi_-_2002__18_3___49-54.pdf

http://help.solidworks.com/2012/English/SolidWorks/cworks/Convection_Heat_Coefficient.htm

http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=566779

http://www.astm.org/DIGITAL_LIBRARY/MNL/PAGES/MNL12069M.htm

http://www.tufts.edu/as/tampl/en43/lecture_notes/ch5.html

Most good textbooks cover the theoretical subject quite well.

http://web.mit.edu/lienhard/

http://www/ahtt.html

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[smukthiHT]

Thanks for the information.

What is an ideal configuation to conduct a natural convection test. Any prefered orientation? I know that I have to minimize the wall effects due to the surroundings on the part. But the part has to be suspended on a fixture for the test to be done. How do the minimize the effect of the fixture on the part? Any coments pls..

Thanks
Sridhar