MX-3.com

In the case you put above, I'd say there's a possibility that the dead battery (on the other car) was pulling so much current off your battery that it got it too low for the fans to work properly which, combined with the engine's heavy load caused it to overheat. If on top of that you had, for example, an A/C system with not enough gas, that could've caused the A/C fan to not work at all, making the problem even worse.
The hot booster wires don't actually have much to do with the alty. The 70 to 90 Amps a normal alt delivers are nothing for a normal booster cable. But if the other car's battery was dead, your battery would've been delivering hundreds of amps on top of the alt's current.

Inodoro Pereyra wrote:Daninski: Nightfire is right. The car doesn't have an additional radiator. That's the A/C condenser, and it actually hinders (when the A/C is in use) engine cooling (being that it heats up the air right in front of the radiator), so if you get rid of it, your issue would probably be the engine overcooling, not overheating.

Nightfire: I hate to be the bearer of bad news, but, as far as I know, Peltier chips have an efficiency of 5 to 10%, and that's 5-10% of an ideal refrigerator's CARNOT NUMBER, not its electrical consumption.

True, but a standard A/C (compressor) cycle is only 15-30% is it not? I'm not looking for efficiency.. just cold air.

If 2kW isn't enough (should transfer ~500W @ 30C delta according to the graphs) I guess I'm hosed. Still a fun project!

nightfire wrote: True, but a standard A/C (compressor) cycle is only 15-30% is it not? I'm not looking for efficiency.. just cold air.

If 2kW isn't enough (should transfer ~500W @ 30C delta according to the graphs) I guess I'm hosed. Still a fun project!

The compressor cycle has nothing to do with the efficiency of the system. The compressor on an A/C system will stay on the minimum time needed to keep the gas under enough pressure that it stays liquid before the expansion valve.
The Carnot number is a theoretical (therefore unreachable) number that tells you the energy efficiency a given device would have under perfect circumstances.
For example, let's suppose we have a heater with a Carnot number of 50%, and we want to make a heater for a car with a 1000 W heating power. That means we will need to "feed" 2000 W to the heater. Now, assuming we have an alternator with 100% efficiency, and knowing 1 HP equals roughly 745 W, we know we'll have to apply about 2.685 HP to the alty to power the heater.
Now, going back to the A/C.
Let's suppose for a second that the Carnot number for a perfect A/C is 50%, and that the efficiency of a real A/C is 60% the Carnot number.
Let's also suppose the CN for a Peltier system is 10% (remember that's the CN compared to a perfect A/C, not to a 100% efficiency), and that you have a 100% efficient alty.
A normal A/C system draws about 15 HP from the engine at full power. That means a Peltier A/C would draw, for the same output, 6 times as much, or...90 HP!!!
And that's assuming a 100% efficient alternator, which, of course, is far from real.

I understand where you come from. I also like to experiment.
But I think you're setting up for a ton of work, and for a bitter disappointment at the end of it.
I'd highly recommend you get the parts you need to get your A/C running.

Well something doesn't add up here, because the graphs they supplied with the device I ordered shows that it can transfer 120W of heat with a 30C delta, when fed 30A @ 16V (~500W). Now it could be BS, sure.. but if it's accurate, 2kW of input should "generate" 2480W of heat (water cooled) and remove 480W of heat from the car. 480W of heat removed from the car should drop its temperature at least a few degrees, no?

The big variables (assuming the above is factual and not marketing hype) are:

- Heat generated in the car (from electronics, exhaust, firewall, etc) + heat absorbed from sunlight
- Heat absorbed from ambient air transfer (ie. through cracks)
- Delta between cold and hot side (can I keep the radiator <20C over ambient?)

Sounds like marketing BS to me, and coming from an Ebay seller...
Do you have a link to the module's spec sheets? I'd love to have a close look at them.

Datasheet is here: http://new-electronics.gr/peltier/CP1-1 ... ATIONS.pdf

Mine's the CP1-12730.

nightfire wrote:Datasheet is here: http://new-electronics.gr/peltier/CP1-1 ... ATIONS.pdf

Mine's the CP1-12730.

Well, for the last 2+ hours I've been studying the datasheet and brushing up on my (not so good to start with) Peltier knowledge, and I have to say I'm more skeptical than ever.

For starters, the datasheet itself is way too incomplete. No circuit diagram, no working principle, nothing. Just a max parameter table, and 2 diagrams per module.
Then, the diagrams are also incomplete, and the one on the right is downright funny.
A negative ∆T? What does that mean, that the cold side gets hot, and the hot side gets cold?
And then you have the scale...Let's suppose the "0" is between -13 and 7, right? So then you have 7, 27 and..."0"? The relation between ∆T and power consumption is supposed to be fairly linear, so how can they have straight current lines when the cited ∆T parameters are nothing if not chaotic?
And what's the ∆T unit? °C? °F? °K?

Anyways, let's work (or try to work) with what we have. The diagram on the right gives you for a 25 A consumption on 12.5 V (approx), a ∆T of 27. Now, assuming a 10% error (very conveniently... ), we can go for the 30 ∆T in the diagram on the left, and see that for a 25 A current we get a Qc of 120 W, Which is ALL you can get from one TEC on 12.5 V. To get 480 W Qc you'd need 4 TEC's in parallel.
But here's where it gets good: ∆T is the temperature difference between cold and hot side of the TEC, not its cooling capacity. What I mean is, if you are gonna water cool the TEC (which is by far the best, more stable way), the water will, inevitably, get hot to an extent. You will have to take that extra heat, as well as the thermal resistance of the heatsinks into the equation.

So, to make it short:

I don't trust that datasheet. The information is incomplete, and badly presented.

You're about to embark in a built that's a lot more complicated than it seems. I'd recommend you to build a prototype at home, MEASURE EVERYTHING, including power consumption, real ∆T when in use, and how much time it takes to cool down a small room in how many degrees (the consensus seems to be that it takes more cooling power to cool down a car than a small room). Cool down the TEC with HOT water from the faucet, to mimic what you'll have in a closed system in the car.
Then, if the manufacturer's claims are accurate (I doubt it), you can make an informed decision as to what you want to do.

Besides, look at their address: "2 shady"? What's that, a coincidence or a warning?

lol.. ya I know.

It's totally sketch. But I only ordered one at $20, 4 high density heatsinks for a combined $27, and a small water pump ($40). I'm gonna try with one first, inside, measuring heat transfer using 2 1L water tanks. See how fast it raises/cools the temperatures at 12.8V. Then I'll run one big coolant tank at 50C, and see how much cold air it makes on the other side. If it's anything substantial, my hope is that 4 of them can blow cool air in my face, at least.

If not, maybe a little mini desktop A/C. :p

That sounds good.
Worst case scenario, you have a nice toy, but, who knows? Maybe everything goes well, and you become the forum's top solid state A/C provider...
Keep us posted.

But of course.

Well, I ordered the remaining parts I need to finish my electric A/C. Everything except the alternator, that is. However, if it works on a test bench, I found someone who can make me a 230A alternator for $350.

Here's what I got:

4 62mm peltier devices (16V / 30A peak):


4 heat sinks for the air cooling side (60mm, ultra high density):


A big heatsink for the water cooling side:


A 500L/h water pump (from a PC cooling rig):


A motorcycle radiator (13" x 6" x 1.5"):


And two high speed 6" fans, some assorted nuts and bolts, gaskets & gasket maker, hose, fittings, etc.

I plan to drill channels in the water cooling heatsink and install 3/8" fittings for water flow, then seal it with plates on the top and sides of the fins. Surface area shouldn't be a problem since water draws heat very well.

Then I'll create an airflow chamber using a 2x2 air exchange heatsink layout, with air flowing top to bottom. I'll put a little trap at the bottom with a drain hose for humidity removal. I plan to use the MX-3's existing blower motor, though I haven't looked under the dash to see if I could make the unit fit. Final size will be about 14cm x 14cm x 12cm. Plus room for the trap, air hoses, etc. But those can be selected based on how much room there is in the area.

So I'll sandwich the peltier devices between the two heatsink sets, bolt it under some pressure, hook up the water hoses, mount the radiator, run the drain hose, wire it to a 150A relay, and that should be it!

Only question is... will it significantly cool the car?

[edit] Paging wytbishop! Does my math look right for the water flow?

Assuming a draw of 1200W for heat removal of 300W:

1500W * 1h = 5.4MJ
4.186kJ would raise 1kG of water 1'C
5.4MJ would raise 500kG of water ~2.6'C

So there should be a 2.6'C rise at 500L/h, right? Assuming the radiator could dissipate >1500W, 500L/h should be sufficient right?

That is, the radiator should always be ~2.6'C higher than the warm side of the A/C unit?

Your math looks fine but I think I don't understand the problem correctly because I don't see you dissipating 1500W of power through the cooling system.

I would approach the problem differently.

If you consider Q as the rate at which the hot side of the Peltier device dissipates heat, and I’m assuming these are 75W Peltiers you have only 300W to remove.

Q = m*Cp*T

where Q and m are rates and T is the desired temperature change of the cooling water...say 5ºC...

The required mass flow rate of water is:

m = Q / [Cp * T]

= [300J/s] / [4180 J/kgºC * 5ºC]

= 0.014kg/s

= 0.86kg/min

= 51.7kg/Hour or 52L/hr

I’m not a thermodynamicist but I’m sure that the 1200W you’re adding to the heat flow equation does not belong there.

You use 1200W to create a 300W heat flow. That's the efficiency of the Peltier (I also think you're giving it too much credit. It's eficiency will be less than 25%). I don’t think those values are to be added as you have done.

If you use a 500L/hr pump you will only be able to raise the temperature of the water by 0.5ºC or so. Not enough to cool the Peltier. I think you have too much pump.

But this is certainly not my strength. I do mechanics.

Well the peltier transfers some amount of heat (they claim 300W, but yeah, I doubt it too). However in order to do this, it must also convert electricity to waste heat. Each of the units will draw ~25A @ 14V, and that energy has to go somewhere.

It's not going to the cold side, so there's only one other place.

So the hot side is heated by 1200W of electricity, plus 300W of heat (transferred from the car), meaning I have to dissipate a total of 1500W.

But we arrive at the same number (when multiplied), so that fills me with confidence. A 2.5'C @ 500L/hr rise should be fine; assuming the radiator is able to dissipate 1500W at 10C over ambient (a big if) then we're looking at ~12.5C across the plates, plus the desired drop. At 35C ambient, a 15C air temperature would be achieved with a 32.5C delta between across the plates, and peltiers are "reasonably" efficient at that delta.

I think I'll have to add thermal sensors on both the hot and cold sides to make sure the cold side never drops below 1C (to avoid frost) and the hot side never rises above 60C.

One nice thing is that even if the air chamber surface area is insufficient, the cold side will drop further and the delta will naturally increase, which will cause a greater heat transfer from the air to occur. That will heat the cold side more, and eventually it'll settle in at whatever the maximum transfer temperature is, or 1'C.

The excitement is too much!

No see I disagree.

That 1200W is not all going to the hot side. It is creating the potential difference between the cold and the hot side. That is the work done by that energy. The cold side gets cold because of that expendature of energy. Ultimately it's the flow of energy from cold to hot that is the Peltier effect. The 300W which it releases as heat is a reflection of the poor efficiency of the unit as a cooling/heating device. If it was totally efficient it would release all 1200W as heat.

I could be way wrong. I haven't done any real Thermo since I learned it. But that's how I see it.

You can easily measure the heat flux coming off the Peltier. Here's what I would do...Set it up on the bench and run it just as a heater with a known mass of water, say several liters in a bucket. I would use a metal container and glue the Peltiers to the heat sinks and place the container right on top of the heat sinks. Try to insulate the water from the surroundings like in a styrofoam cooler or something so you don't lose a lot of heat to the atmosphere. You don't even need the pump, just stir it every so often and measure the temperature every 10 minutes or so. Use all 4 Peltiers and run it for a while and you should see a steady consistent rise in temperature.

Q = m*Cp*T

Once you know the actual change in temperature of a given mass of water you will know exactly how much heat energy is coming off the Peltiers. Then you can calculate how much flow you need to cool them.

wytbishop wrote:No see I disagree.

That 1200W is not all going to the hot side. It is creating the potential difference between the cold and the hot side. That is the work done by that energy. The cold side gets cold because of that expendature of energy. Ultimately it's the flow of energy from cold to hot that is the Peltier effect. The 300W which it releases as heat is a reflection of the poor efficiency of the unit as a cooling/heating device. If it was totally efficient it would release all 1200W as heat.

But, by definition it has to release all the electricity it draws as either heat, or some other form of energy. Energy is never created nor destroyed.

So you're removing heat from one area, which means you're removing energy. That has to go somewhere else (ie. the hot side).

But you've also drawn power from the battery. Where else could that energy go? If it's being used for work, it still has to ultimately end up somewhere. A motor lifting an elevator converts electricity to heat and potential energy (lifting the elevator against gravity; you could drop it to get that energy back). If the motor draws 100J to lift the elevator, adding a potential energy of 50J (ie. E=mv^2/2, v=Fgh), then by definition 50J must have been released as heat.

In the case of the peltier device, it uses 100J to move 20J of heat from one place to another. That's a total of 120J that is needs to radiate: 100J are missing from the battery, and 20J are missing from the cabin of the car. Those 120J have to go somewhere, so why not the hot side of the plate?

You can easily measure the heat flux coming off the Peltier. Here's what I would do...Set it up on the bench and run it just as a heater with a known mass of water, say several liters in a bucket. I would use a metal container and glue the Peltiers to the heat sinks and place the container right on top of the heat sinks. Try to insulate the water from the surroundings like in a styrofoam cooler or something so you don't lose a lot of heat to the atmosphere. You don't even need the pump, just stir it every so often and measure the temperature every 10 minutes or so. Use all 4 Peltiers and run it for a while and you should see a steady consistent rise in temperature.

Q = m*Cp*T

Once you know the actual change in temperature of a given mass of water you will know exactly how much heat energy is coming off the Peltiers. Then you can calculate how much flow you need to cool them.

Yup that's what I'll do the minute it comes in. Question is ... where do I find a 12V / 100A power supply?