24v 172w pelt @ 12v what would I get?

[qwerty57]

would I loose 50% so 87w??

[Since87]

With a 172Watt TEC at half Vmax you would be at:

80% of dTmax. (i.e. dT would be 55.2 C with no heatload)

70% of Qmax. (i.e. if the TEC was pumping 120.4 Watts of heat, the hotside and coldside of the TEC would be at the same temperature.)

If cooling a processor, your actual operating point would be between these extremes.

[progamer]

You will lose 50% heat pump capacity
but you get more efficiency

[Since87]

Originally posted by progamer
You will lose 50% heat pump capacity

And why do you say this?

It's contrary to the technical literature as well as Kryotherm simulation results.

qwerty57,

I saw you posted the same question over at Overclockers. Don't pay too much attention to the bozo's over there. Doc seriously understands TEC's, and has had substantial experience using them professionally, but he doesn't say much. wymjym has a fair amount of practical experience. The rest are just making noise.

[redleader]

You will lose 50% heat pump capacity
but you get more efficiency

If you dropped power by 50% and increased effciency, wouldn't you end up at >50% of intial capcity?

[muffy]

i have a 24volt pet and on a 12 volt rail it draws 62w's of power

[]JR[]

Imax vs. Vmax is very linear, so using y=mx + c with c = 0, since @ 0V it draws 0A, therefore knowing that tecs are 60% efficent you can work out the Q or pumped heat simply.

my website has a little script that estimates Q from imax vmax and operating voltage.
Looks like @ 12v you end up with ~40watts of cooling power, dont forget that its a quadratic loss since i-v are linearly related. Itll draw ~ 5.3amps.

Edit: i just noticed someone used the kryotherm simulator in the other thread, pretty close to my estimates

]JR[

[Since87]

Originally posted by ]JR[
Imax vs. Vmax is very linear, so using y=mx + c with c = 0, since @ 0V it draws 0A, therefore knowing that tecs are 60% efficent you can work out the Q or pumped heat simply.

The efficiency of TEC's varies substantially depending on operating conditions. Treating them as if they had a fixed efficiency of 60% will not lead to accurate calculations of cooling performance.

Also, the current draw for a TEC is dependent on the hot and coldside temperatures as well as the voltage. TEC's do not behave much like a typical resistor.

[winewood]

And the answer you may have been looking for.. It sucks. The temperature goes up slowly on an amd 1800+ and it doesn't stop. I thought it was broke, but I was just naive. I then ordered a 24V meanwell. PM me and I can tell you how to get one for $50 the list price. (normally $118)
Ill be gone for a week though. So don't get upset if I don't respond untill next monday.
Now if you were using it for an older gpu it may do fine.

[MadDogMe]

Imax vs. Vmax is very linear,

If it was linear the TEC would be most efficient at 100%voltage would'nt it?. It's most efficient at 75% of v~max though not 100% is'nt it?...

Or did I take a wrong turn?...

[]JR[]

I vs V is very linear, but, S, or whatever you want to call it, the efficency changes, although you are correctish around 70% the efficency is the greatest.

From my arguments sake however the efficency never rises above 70% that ive seen? (could be wrong) so the 100% linear assumptions i made are fairly accurate, and infact more accurate than 99.8% of end users can measure too.

]JR[

[bigben2k]

I think that what ]JR[ is trying to say here is that V and I are linear, actual efficiency is not.

But that's wrong: I and V do not have a straight linear relationship (V=RI), as it is with many other electrical devices, like heaters and lightbulbs. This is easily measured, with the right instrument.

We've seen it over and over. ]JR[, I suggest that you take some measurements.

[Since87]

One big problem with discussing TEC's and efficiency is deciding what "efficiency" refers to.

One commonly used measure of "efficiency" is called the Coefficient Of Performance. (COP) It is the heat pumped (Q) divided by the power supplied to the TEC. (I * V)

Here is an example of the Kryotherm simulator showing a system with a COP of 5.56:



In this case the 172 Watt TEC is powered with 3V, and is cooling a 25 Watt heatload. (Note that Tob, the temperature of the object being cooled, is only 0.2 °C below ambient.)

]JR[,
It appears to me, that your calculator is off by a factor of 10 in this case.

Here's an example of a more practical setup:



In this case, the TEC is powered with 18V, (73% of Vmax) and the TEC is cooling a heatload of 70 Watts. All the other input parameters are the same as before. The calculated results show that the heatsource is cooled to 13 °C and the TEC draws 8.87 Amps for a power consumption of 159.7 Watts and a COP of 0.482.

The following simulation is identical to the last except in this case the heatsource is only 35 Watts.



Notice that the current draw (I) is less than in the previous image, even though the voltage is the same in both cases. In this case the object is cooled to 0.2 °C, and the COP is 0.358.

Finally a simulation with the voltage to the TEC at Vmax:



Notice that the inputs are identical to those of the second image, except for the applied voltage. Also notice, that even though the voltage is higher, (24.6V) the temperature of the heatsource is actually higher than when the TEC was powered with 18V. (14.1 °C vs 13.0 °C)

I don't know if this will make anything clearer to anyone, but hopefully it gave those interested in thinking about this stuff something to thing about.

[]JR[]

I-V are linear (enough) on a drift 172, i know they are. And if your being pedantic the maximum heat it could ever pump would be IxV is you assumed 100% efficency...

And @ 3v the program is calculating a heat pumped of > 100% to the input power, which is impossible, it has to be.




I know the lowest temp attainable isnt @ vmax/imax but you need to play with each setup to find whats best for your system as there a too many parameters to say which voltage is best.

]JR[

[Since87]

Originally posted by ]JR[
I-V are linear (enough) on a drift 172, i know they are. And if your being pedantic the maximum heat it could ever pump would be IxV is you assumed 100% efficency...

And @ 3v the program is calculating a heat pumped of > 100% to the input power, which is impossible, it has to be.

Please define what you mean by "efficiency".

Why do you think the power consumed by a TEC must be >= the heat pumped by a TEC? Do you think that a fan that is pumping away 100 Watts of heat, in the form of heated air, must draw at least 100 Watts?

Why do you think your ability to calculate TEC performance is better than Kryotherm's?

[]JR[]

I dont, but why dont we all have tec's in our fridges and freezers @ home if they can do more work (i.e the heat pumped through the device) than the power input to do the work? Since they are better than phase change wrt moving parts/maintanance?

Also thought experiment:
I take the coldsides of a large number of tecs and place them, for arguments sake into the sea, then i power them up, they then remove heat from the sea , the heat i get from the sea has more energy than i need to run the tec*. Correct and with me so far?

So I now boil water with my heat removed from the sea (and the hotside of the tec), generating electricity, which is more than enough to power the tec* and have electricity left over to power my computer.

*Since in your regieme its possible for a tec to pump more energy than the input energy

Hopefully that illustrates why you cannot get more work out of a system than the input work/energy/power (all one and the same) And also it illustrates why we have phase change rather than tecs in our fridges since phase change is ~75% efficent whereas tecs are less.

]JR[

[Since87]

Originally posted by ]JR[
I dont, but why dont we all have tec's in our fridges and freezers @ home if they can do more work (i.e the heat pumped through the device) than the power input to do the work? Since they are better than phase change wrt moving parts/maintanance?

The reason we don't use TEC's in our homes is because they are less efficient than phase change for the tasks you mention. I didn't say that the first simulation I posted showed a system that was of any real use to anybody.

What do you mean by "work"? The physics definition of work involves moving a mass. There is no net movement of mass involved in the operation of a TEC.

Originally posted by ]JR[

Also thought experiment:
I take the coldsides of a large number of tecs and place them, for arguments sake into the sea, then i power them up, they then remove heat from the sea , the heat i get from the sea has more energy than i need to run the tec*. Correct and with me so far?

Yes, the energy contained in the heat removed from the sea, is in the form of material at a slightly higher temperature than the temperature of the sea.

Originally posted by ]JR[

So I now boil water with my heat removed from the sea (and the hotside of the tec), generating electricity, which is more than enough to power the tec* and have electricity left over to power my computer.

WHOA!!!

Boil water? Look at that simulation again. Th (the hotside temperature) is 23C. Even if you pull a vacuum I'm not sure you could get a phase change turbine system to run at 23C using water. But suppose you find another liquid to boil - so what? Can you design a system that can extract more than 1% of the available energy? If you've got a means for everyone living near a body of water to generate significant amounts of energy whenever there is a differential between the air and water temperature, you need to be getting in touch with a patent attorney.

Here's a link. It's not a particularly good link, but it was the first I found with a statement such as...

Quote:
"High efficiency air conditioning systems may require as little as 0.33 BTUs of energy to remove 1 BTU of heat."

Air conditioners dissipate heat by taking in outside air and putting out hotter air. By your reasoning, the energy contained in the temperature differential between masses of air, can be extracted and used to power the airconditioner and more. Why don't airconditioner manufacturers do this?

Originally posted by ]JR[

Hopefully that illustrates why you cannot get more work out of a system than the input work/energy/power (all one and the same) And also it illustrates why we have phase change rather than tecs in our fridges since phase change is ~75% efficent whereas tecs are less.
]JR[

There is not more energy coming out of the system than is going into it. The energy coming out of the TEC is exactly equal to the energy going into the TEC. Some of that energy is coming in as electricity, and some of it is coming in as heat from the heatsource. As far as "work" is concerned, I don't know what "work" you are talking about.

"Energy" is not the same thing as "Power". Power is a rate of energy dissipation/usage/whatever.

The reason phase change is used more commonly than TEC's is that phase change is usually more practical and efficient for jobs people want to do. Nobody is likely to want to go to the expense of using a TEC as a means of 'efficiently' cooling a 25 Watt heatsource to 0.2 °C below ambient. The fact that no one wants to do it, doesn't mean it can't be done though.

[muffy]

jR is right

it can not be more than 100% efficient because this is braking physics laws.
In reality it wont be anywhere near 100%

A 172 watt TEC will not and can not move 172 watts of heat from one side to another.

We know this for sure because the TEC’s hot side gets hotter that the cold side gets cold

It’s basic physics really

[muffy]

the imformation that i can find says that they are between 40 and 70 % efficient meaning for X watts 40 to 70 % of that power is transformend in to heat transfer.

heat pumped / input power = efficient

JR is not saying it is more than 100% efficient . he was using this example of the sea as away of explaining how silling the idea was

[Since87]

Originally posted by muffy
jR is right

No he wasn't, but he seems like a fairly bright guy, and he may have already realized that he was wrong since reading my last post.

I see I'm going to have to take things reallllllllllll slooooooooow for you though.

Originally posted by muffy
it can not be more than 100% efficient because this is braking physics laws.
In reality it wont be anywhere near 100%

Here's a link. And a quote from it:

Air conditioners and heat pumps are heat engines like the refrigerator. They make good use of the high quality and flexibility of electric energy in that they can use one unit of electric energy to transfer more than one unit of energy from a cold area to a hot area. For example, an electric resistance heater using one kilowatt-hour of electric energy can transfer only 1 kWh of energy to heat your house at 100% efficiency. But 1 kWh of energy used in an electric heat pump could "pump" 3 kWh of energy from the cooler outside environment into your house for heating. The ratio of the energy transferred to the electric energy used in the process is called its coefficient of performance (CP). A typical CP for a commercial heat pump is between 3 and 4 units transferred per unit of electric energy supplied.

There is no law of physics which applies specially to TEC's.

Heatpumps can move more energy than they consume.

A TEC is a heatpump.

Originally posted by muffy
A 172 watt TEC will not and can not move 172 watts of heat from one side to another.

A Drift-0.8 TEC is called a 172 Watt TEC because its Qmax spec is 172 Watts.

A TEC's Qmax spec indicates the maximum heat it CAN pump with a specific hotside temperature, (typically 300 °K) under ideal conditions. (typically 300 °K)

A Drift-0.8 can actually pump more than 172 Watts at hotside temperatures higher than 300 °K.

Originally posted by muffy
It’s basic physics really

Uh huh.

[muffy]

How efficient is a thermoelectric cooler?
Technically, the word efficiency relates to the amount of energy one gets out of a machine versus how much energy one puts into it. In heat pumping applications, this term is rarely used because the energy-in is very different from the service provided. We supply electrical energy to a TEC, but we get heat pumping. For TECs, it is standard to use "coefficient of performance" not efficiency. The coefficient of performance (COP) is the amount of heat pumping divided by the amount of supplied electrical power. In other words, COP tells you how many units of heat pumping you will get for each unit of electrical power you supply. It is possible, in special situations, to pump more watts of heat than the watts of electrical power input. COP depends on the application, heat pumped, and temperature differential required. Typically, the coefficient of performance, heat pumped then divided by input power, is between 0.4 and 0.7 for single stage applications. However, higher COPs can be achieved with optimized custom TECs.

http://www.marlow.com/TechnicalInfo/..._faqs.htm#th07


well these guys make them and thats what they have to say on the subject .

they do say it is possible to move more heat then watts of power however they do state the norm is no where near that

so your both right and both wrong i guess

[Since87]

Originally posted by muffy

so your both right and both wrong i guess

Where was I wrong?

[muffy]

oh and most of what i said wasn't true about it moving more heat than power used . IN theory , sorry about that. oh well you learn new things

in our application it's going to be far less that 100 % efficient which i believe was JRs point.

oh well have fun guys