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Heat Transfer at different ambients vs thermal efficiency

Heat Transfer at different ambients vs thermal efficiency

Heat Transfer at different ambients vs thermal efficiency

(OP)
I am a EE and had 1 heat transfer class 29 years ago so bare with me.
If I test a transformer in a 25C ambient and a 50C ambient. In both cases to equilibrium. Will the temperature rise be different? My memory is telling me that the higher temperature will yield a lower rise.
Thanks

Neil

RE: Heat Transfer at different ambients vs thermal efficiency

Nope your temp rise will be the same from ambient.   

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

RE: Heat Transfer at different ambients vs thermal efficiency

That assumes that the heat transfer coefficients are constant over temperature, which is not strictly true.  

The thermal conductivity of air is roughly proportional to the square root of absolute temperature, so the amount heat removed at a higher ambient temperature will be higher, but it's not even a percent for the temperatures cited.

TTFN

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RE: Heat Transfer at different ambients vs thermal efficiency

(OP)
Thanks,
I have a customer who is demanding we test our product in a 50C ambient. They don't believe our test in a 25C ambient even though we had more than 40 deg of head room in the rise. They feel parts could still faile a full temperature.
Any thoughts on how to convince this person that so long as the temp rise over ambient is at least 25 deg below the rise limit, we are not likely to have thermal failures?

Neil

RE: Heat Transfer at different ambients vs thermal efficiency

This is a tough question because your tranformer has probably fins.  Also as the oil temperature goes up from the increase in ambient temperature, radiation cooling may become a factor depending on the transormer skin temprature.

RE: Heat Transfer at different ambients vs thermal efficiency

(OP)
Actually it isn't a transformer. I just used that example for the sake of simplicity. It's a more complex problem. We have an industrial control panel with inductor, resistor, caps, fan, control xfmr, terminal blocks, wire, etc. But the only component throwing off significant heat energy is the inductor and resistor.

Neil

RE: Heat Transfer at different ambients vs thermal efficiency

Without knowing your heat transfer circuit, I kinda gave you a very simplistic answer.  But, to answer your last post, I guess I'll answer it with another simplistic answer.  Heat transfer's equation Q=(1/R)deltaT where deltaT is your temp rise which is Tfinal-Tamb.  If we apply some algebra we get Tfinal=QR+Tamb.  Now for conduction if QR is kept constant than deltaT will always be the same no matter the ambient temperature.  However, for convection, well the heat transfer coefficient may change due to the Reynolds number, but I think these are small variants and should not scientifically change the temperature rise.  A few ways to change the rise is to change the Q, R, or the speed of the flow of air going over the unit, however, even the flow or air has it's limitations.

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

RE: Heat Transfer at different ambients vs thermal efficiency

That seems to be a totally different question than what was implied in the OP.  If the system is intended to operate at 50ºC, then a 25ºC test is NOT a sufficient condition.  

You may have other, hidden, thermal problems unrelated to the heat dissipation, per se.  Issues with thermally induced noise, race conditions, etc., would certainly be more evident at 50ºC.

TTFN

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RE: Heat Transfer at different ambients vs thermal efficiency

Well as I always say, analysis is only 80% of your answer; the other 20% comes from test.  Build one and test, if it pass, you can quickly quite down the peanut gallery.  If not, you have your analysis to correlate with to find a fix, if somebody did the analysis.  I hope you're not shooting from the hip and just putting something together and hopping for the best.   

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

RE: Heat Transfer at different ambients vs thermal efficiency

(OP)
We're trying to avoid building a relatively expensive test setup. based on my previous experiences I have tested transformer that failed rise by 2 degrees in room temp ambients, about 23C. Then built an internal insulated room with separate uniform heat source to create 40C ambient. The xfrmr then passed by 1 degree rise. Thus my memory about higher thermal efficiency at higher total temp.

As far as other problems go. Noise would not be an issue in this application anyway. Don't know what "race conditions" refers to. In the end it just seems we have such a large amount of head room between the max rises (for 40C) and tested rises that to add another 10C should not cause serious consequence. If we were within a few deg of the max rise under room ambient conditions I might be more concerned.

Neil

RE: Heat Transfer at different ambients vs thermal efficiency

Well sooner or later you have to build one to pass qual.  Once you get a test item, just do a prequal and see what happens.  You want the failure to happen in house instead in front of the customer.  Another advice would be to model the unit in Flotherm or TAS bring the amb to 50C and see if there are any temp concentrations and make sure everything in the area has a higher temp requirement.  With this you can show some pretty pics to the customer and maybe get thru this round of questions.  Other than that, your customer has a good point which has to be satisfied either by test or analysis not by speculation.

 

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

RE: Heat Transfer at different ambients vs thermal efficiency

Your problem can best solved by first drawing a sketch of the apparatus with all compoments particularly that generate heat.  Then a solution can be devised applying famous first law od thermo,being:
 rate of heat generated - rate of heat loss = increase in internal energy of all components in the cabinet.
 You already have an idea of the heat generatyed being an EE.
Heat loss will be time dependent when starting cold and basically will have this general appearance:
    qloss=UA(Tb(t)- Ta)
Increase in internal energy of components+air will have this general appearance:
     u(t)= m*Cp*(Tb(t)-Tb(t-1))/ dt
There is a lot to consider in these equations: shape of components and cabinet, coefficient of heat transfer inside and outside of cabinet, perhaps infiltration, mass of components and cabinets, and a selection for dt.
After identifying all relevant items then you want to itinerate with a spread sheet to determine a steady value for Tb(t).

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