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liquid cooled vs ari cooled 1

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JimboJones21

Electrical
Mar 21, 2005
55
Hi all, I'm an ee looking to understand some thermo so that I can do some calculations. A colleague at work said that a heat sink could remove more heat than a water cooled system with a radiator, and he said the heat sink would be smaller than the radiator? (both have fans) I've never thought about it becasue I always assumed water cooled was more efficient than air cooled? Is there a way to calculate which setup would remove more heat and rates at which it would remove the heat? I want to know more about this subject so I can know if he's knows what's going on. Thanks in advance for pointing me in the right direction.
 
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For my sake, so I can better understand the query, let me state that long before there were mainframes on campus and EE's took thermodynamics the term "heat sink" was well known and used in thermo as a generic word to describe a colder state within a system that provoked a heat driving force - something you "dumped" your Btus into. Now, with the new "computer" engineers we are hearing the term "heat sink" in many different and distinct ways.

Are you naming an air-cooled electronic device a "heat sink"? Or is it the air that is the heat sink? This is very confusing because it tends to sway the basic laws of heat transfer if you don't describe how the heat is being transferred. Heat exchange is heat exchange. It doesn't get any fancier or avant garde because it is applied to an electronic board or printed circuit. It works in the same basic manner as it does in a heat exchanger. Whether a water-cooled system transfers more heat than a direct, air-cooled system depends on the sizes of each of the individual units. Its very difficult to compare one with the other unless you state what you holding fixed: the heat transfer area? the amount of air circulated? I presume the heat load is the same on both systems and so are the working temperatures - or are they?

If I were to rely on the most efficient heat transfer system, I would employ the water cooled radiator - mainly because I have unlimited availability of radiator fin surface, while if I rely on putting fins on the electronic gizmo itself and fan-cooling it, I'm limited by its external surface area. Plus I can create more turbulence with water circulation and circulate more water if I have to. However, the "footprint" of the water-radiator system is bigger and it's bulkier. But it certainly can be more efficient than direct, air-cooled units.
 
Are there any links you can point me too? Basically I'm looking at leds on a block of al with fins "heat sink" or a smaller block with ports to run water through to a heat exchanger "radiator". I am limited to one box size, small. So I was wondering if the radiator setup would take the same space and move the heat away faster?
 
The answer depends on what you're looking for.

For a given cold plate area available for heat removal, a liquid system will remove more heat from that cold plate for a given temperature delta.

From a system perspective a liquid system, by definition, is less efficient, since the liquid system incorporates additional losses in the piping friction and losses in the circulation pump; a liquid system must have more heat to exchange to the air than a direct cooling system, thus making it more inefficient. The heat removal from the liquid system is essentially the same as a heat sink, so given the same radiator parameters, the liquid system is worse off.

TTFN
 
For the air cooled system you have:

conduction from the device into the the "heat sink" ----> conduction through the heat sink ----> convection into the air + radiation to the surroundings.

For the water cooled system:

Conduction from the device into the water block ----> conduction through the water block ---------> convection from the water block into the water ----> convection from the water to the water side surface of the "radiator" ----> conduction through the radiator fin material -----> convection into the air + radiation to the surroundings.

Each heat exchange mechanism is in effect a layer of insulation. Of course the value of the insulation is a function of a bunch of factors at each interface.

So, the answer clearly is "it depends".
 
If an water cooled system is more inefficient than an air cooled system why water cool a car engine?
 
That's NOT what anyone said.

Please RE-READ the postings.

TTFN
 
What I have is a thermocouple attached to a heat sink and I want to be able to measure the heat sink temperature and predict the juction temperature for the LED. I have the thermal resistance for the LED and a plot of the heat sink thermal resistance, but the grease I'm using only has the conductance. I want to use this formual, adding the thermal resistnaces to find the junction temp, Tj=Ta+(Rjc+Rcg+Rgs)P.
I was looking at a site that has an equation for thermal resistance, R=thickness/k*area which gives units degK/W, I have the resistnace of the heat sink and led package but it's in degC/W. Can I simply add the degC/W value and degK/W because these units are related to changes in temperature?
Also could someone help me understand the Heat Sink Thermal Resistance plot better


It's thermal resistance goes down with increasing the amount of heat you put into it? If I was to assume that the LED was 0% efficient and I put 10 Watts of power into it can I use this chart to predict what the heat sink temp will be? Then measure the sink temp and figure out how efficient the LED is? Thanks for the help.
 
You're not making any sense whatsover.

You already have the thermal resistance of the LED, then you talk about effectively measuring the same thermal resistance. And if you already have the thermal resistance of the LED, then "predicting" the junction temperature is simply a matter of plugging in the power dissipation into the equation.

TTFN
 
Jimbo,

Yes, in your example C/W and K/W are identical.

Yes, increasing the power can cause sink resistance to decrease. In natural convection, air movement is caused by a temperature difference between the sink and the air. At higher power, the sink temperature is higher and the air movement increases. Of course, adding power still causes the sink temperature to rise, but just not linearly.

I don't have much confidence that you can reasonably estimate LED efficiency by measuring the sink temperature. Some power will conduct through the LED leads, and the values you have for interface and sink resistances are only approximations.

ko (
 
Thanks ko, I needed to know if the calculation was right first. What I want to know how far I can drive the LED without damaging it. The specs for the LED say that the junction temp must be kept below 125C, and I have in the thermocouple on the sink. Is there a way to estimate from the sink temp what the junction temp will be? Can I just look at the sink temp and read the thermal resistance off the sink plot and use than in the junction temp eqn I'm using? If so is there an eqn that relates sinks thermal resistance to sink temp so I can plot junction temperature against sink temperature?

I used the junction temp eqn (Tj) above in excel, and the resistance eqn to find the grease Tr, then put 0-30 watts into the eqn to see where my junction temp went above 125C. I then looked at the worst case for the sink thermal resistance put it along side the Tj results with the same wattage so that I could see what sink temp would correspond to the Tj. This is obviously inaccurate but should be safe since the sink become less resistive as temp increases and there are other losses in the LED package, the major one being light, but the problem I'm having with the experiment is that the temperature I'm predicting for the sink with the amount of power I'm putting into the LED is 10deg higher? I figured since I was using the worst case resistance for the sink and that the LED should be around 15% efficient the temperature should be lower? Are there simple ways to model sink resistance vs temp i.e. something that could be done in excel? Ideally if I could put a thermocouple on the LED substrate then I could measure the temperature, but the LED package is very small, fraction of a millimeter. Ultimately I want to be able to fully drive the LED with the above heat sink. Thanks again.

 
Then, you should be designing for the worst-case. Pick the highest thermal resistance, do the calc and move on.

Trying to shave the numbers will only get you in trouble later on, when the convection is restricted for other reasons or your thermal grease poops out.

If you need to play this sort of game with the numbers, then you've undersized the heat sink.

TTFN
 
Jimbo,

Your formula "Tj=Ta+(Rjc+Rcg+Rgs)P" appears to be missing the sink-to-air loss.
It should be: Tj = Ta + (Rjc + Rcs + Rsa)* Pjs

Tj = LED die, C
Rjc = die-to-case loss spec'd by LED manufacturer, C/W
Rcs = loss at case-to-sink interface (grease), C/W
Rsa = loss from sink to air, C/W
Pjs = power going from die to sink, W.
Unless you have better data, conservatively assume Pjs = 90% of total power

If you have actual measurements of the sink temperature:
Tj = Ts + (Rjc + Rcs) * Pjs

Ts = sink temp measured just above die, C

ko (
 
Thanks ko, I understand fully now. A friend loaned me his thermo text, Reynolds and Perkins. I will use this as a reference now. Do you have any good links for thermodynamics info?
 
If you want to be conservative, you should assume 100% of the power is dissipated. Most LEDs are only a few percent efficient.

see thread391-109133 for a downloadable HT textbook

TTFN
 
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