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Thermal Resistance with differing Q

Thermal Resistance with differing Q

Thermal Resistance with differing Q

(OP)
Hi,

Can any one help. I am trying to work out the thermal resistance of a heatsink. I have all the equations for doing so, however I want to plot a graph of the thermal resistance against different values of heat transfered Q.

However, non of the equations I have for working out Thermal resistance (these include working out Raleigh number, Nusselt number and a value of h) seem to include Q with in them.

Any help for a poor novice will be mostly appriciated.

many thanks

RE: Thermal Resistance with differing Q

You seem to be confused.  None of the equations should contain Q.  The only effect Q has is indirect, since it would be manifested as a temperature delta, and some of the dimensionless numbers are temperature dependent.  

TTFN

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RE: Thermal Resistance with differing Q

(OP)
Hi IRstuff,

I think you are correct, I have that many equations floating about that Im heavily confused and need a pointer.

I have all the equations (based on the work of simmons and also using text book: Heat Transfer by J.P Holman) to calculate thermal resistance.

However, I have seen graphs of heatsink performance that plot the thermal resistance against the watts being input into the base of the heatsink. From asking around, I understand these graphs are created using the simple equation of newtons law of cooling.

Q = h.A(Tw - Ta)

what I do not understand is that if you know Q (from the value you want to enter on your graph), you can work out Tw, which should increase with more Q. Tw can then be used in the Resistance calculation to find the thermal resistance at that value of Q.

However, h also require you know Tw. You cant work out Tw using Newton law of cool cause you need to know h which also need Tw.

I hope this makes sense and you can help

many thanks

RE: Thermal Resistance with differing Q

There are cases where Q will affect h or R.

In the case of fouling , as in heating milk , etc, a high Q will increase R due to fouling.See papers by Somerscales

In the case of boiling of pure water, a high Q may increase h and decrease R, but once one exceeds critical heat flux, then R increases dramatically.See papers by KWU authors  Kohler, Hein, etc.

RE: Thermal Resistance with differing Q

IR is right.  R for natural convection (if this is the case) is R=1/(hA) and h = 0.29 (deltaT/L)^0.25.  If Q = (1/R) deltaT, then Q = h A deltaT, sub in h and you finally get Q = (0.29 A deltaT^1.25)/L^0.25.  So Q is a function of either delta T or L.

Tobalcane
"If you avoid failure, you also avoid success."

RE: Thermal Resistance with differing Q

It's an iterative process.  Tw will not drastically change with Q.  You start with a guess value for T, crank for htc, get a new T, crank htc again, etc.  If you have Excel, Mathcad, Matlab, or TKSolver, you can get any of them to iteration and solve numerically.

TTFN

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RE: Thermal Resistance with differing Q

To add to twoballcane's note about the calculation for h, that h is for laminar flow, across a vertical plate.

RE: Thermal Resistance with differing Q

(OP)
Hi Guys,

Thanks for the input, Im still not sure if I follow how the best method of doing this is.

If you follow the below link, it shows a set of graphs for heatsink performance. The first one, top left, shows the thermal resistance against Q. The graphs update in real time with user changed length.

http://www.thermaflo.com/bin/cr_exdatasht.pl?Pnum=e1074&LengthUnits=in&ExLength=3.00

Having spoken with a few people that understand some of the back ground of these graphs, it is understood that the program uses the parellel plate theory as a base and modifies it based on some emperical results (Newtons Law of Cooling off Efunda.
Obviously I do not know these modifications, but want to use something standard that will get me close.

Now I have followed a number of examples for calculating that thermal resistance of a heatsink, mainly being the links below. However, I do not under stand how you do this for each different value of Q as per the graph.

http://www.electronics-cooling.com/articles/2002/2002_february_calccorner.php

http://www.electronics-cooling.com/articles/2000/2000_sep_a4.php

http://www.electronics-cooling.com/articles/2003/2003_february_calccorner.php

I hope this helps explain my confusion a little more, and Im really sorry for the noob questions.

Many thanks


RE: Thermal Resistance with differing Q

In simple terms, the higher the power, the higher the temperature delta.  This makes the air more buoyant, so it rises faster, thereby requiring cooler air to move in quicker, thereby moving more heat out of the system.

But, note that the thermal resistance barely doubles for an 8x increase in power dissipation.  Note that the temperature delta is less than linear with power, precisely because of the slightly increased htc.

This says that you can start with a midrange value for htc, determine the temperature delta, and use the log-mean, or whatever, temperature to recalculate the htc, and run the calculation again.  Since the slope is so low, the solution should converge pretty rapidly.

TTFN

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RE: Thermal Resistance with differing Q

(OP)
hmmmm,

I see what you mean yes, but is there not a way of calculating Tw with out the need for R or h.

For example, if you take in to consideration the simplifed equations for a flat plate.

h = C(Tw - Ta / L)^n
where C and n are constants

q = h.A.(Tw - Ta)

then through algebraic manipulation we can cancel out many of the variables to leave.

Tw = (q / C.A)^(1/(n+1)) . L^(n/(n+1)) . Ta
therefore we can find Tw from knowing our heat dissipation....which is usually the case.

However this is based on a single flat plate.

Now in order to consider an array of flat plates, we have to look at elenbass correlation that starts to take in to consideration that the plates may be spaced so close that boundary layers on adjacent plates will merge. Therefore modifed eqs for Nusselt and Rayliegh are,

Nu = h.b / k = Ra/24 (1 - e^(-35/Ra))^0.75
and
Ra = rho^2 . g.Beta.Cp.b^4.(Tw-Ta) / u.k.L

However, I do not know how you can manipulate this to leave only Tw based on just q, A, L and Ta. Or if this is even possible.
Unfortunately, this is way past my skill of trying algerbraic manipulation.

Am I look down the wrong path and this is not even possible, or is there a way of working out Tw when you know the q.

Many thanks

RE: Thermal Resistance with differing Q

I don't think there's a closed form solution, if that's what you're asking.  If there were, it would be published or in textbooks.  

That said, the answer to, "is there a way of working out Tw when you know the q," is solve numerically.  As I mentioned earlier, there are a number of math problems for which this is a simple solver problem.

TTFN

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RE: Thermal Resistance with differing Q

Hello Navier Stokes , im new at this forum im  thermal engineer and i am working on the heat sink and cold plate since 7 month , so i think i'm  specialised on that.
I haven't read ur question completly but u can aswer me about the characterisation of the heat sink ( i have done my final year specialization on thermal electronic)

 

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