"Solid State" Thermoelectric AC

caps04 said:

I was looking up Thermoelectric cooling (without refrigerant) and wondering if anyone knows if it can be applied to car ACs? Thermoelectric cooling - Wikipedia, the free encyclopedia
Thermoelectric Technical Reference - Advantages of Thermoelectric Cooling

Advantages: Good for the environment, no moving parts, less complexity, super quite and more reliable.
Disadvantages: Not as efficient?

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Got it pretty much right - but the difference in efficiency is rather large, and I believe that Peltier stacks are also heavier (though likely more compact) for a given cooling power, and I suspect they are more expensive.

AFAIK there's no reason one couldn't design an automotive HVAC system using Peltiers if one were so inclined.

There aren't many locations where they seem to make functional and economic sense right now, though that may change as the technology improves.
Walter

The most common way to express efficiency of cooling is the "Coefficient of Performance". If COP were 100%, then every unit of energy expended (like a watt) would result in the same amount of heat removal. A true 100% cannot be achieved in the physical world in which we live... still, the closer the better.

• Compressor/refrigerant based systems have demonstrated COPs as high as 60%. A more typical real world value of an economically viable implementation is 45%.
• Peltier (solid state) have demonstrated 5%. A more typical real world deployed number is 2%.

So, yeah, there is a reason. I don't know the exact capacity of a Tesla A/C; many sedan sized vehicles are in the 24000 to 36000 BTU per hour area, max capacity. Of course, they don't run all the time; the peak is used to quickly cool a hot car and after that, they cycle, and they cycle a lot. Let's assume (pulling this out of thin air) they run 20% of the time. Assuming the same for a Tesla:

• 7 to 10kW * .2 (for cycling) / .45 (for efficiency) means a compressor based A/C will take about 3 to 4 kW from the pack spread over one hour, which makes it very easy to say 3 to 4 kWh. That's about 4 to 5 % of the total pack (i.e. reduction in range, per hour).
• If Tesla invested in a very efficient compressor based system, like 55% efficient, that could be as low as 2 or 3% of the pack.
• 7 to 10kW * .2 (for cycling) / .02 (for efficiency) means a solid state based A/C will take about 70 to 100 kW from the pack spread over one hour. That's the entire pack.

Those numbers are based on a lot of assumptions and averages. Reality in a Tesla may be different... but the relative efficiency will be somewhere in that ballpark, namely "a few percent" vs. "pretty much everything".

The Tesla uses a variable speed scroll compressor rather than a piston type so it doesn't really cycle, it just goes slower or faster. This uses far less energy than the cycling of a pulley driven system. While it can use 3-4 kW, it almost never does so (at least in my car).

Short video explanation of a scroll compressor:
How a Copeland Scrollâ„¢ Compressor Works - YouTube
~Larry

Danal said:

I don't know the exact capacity of a Tesla A/C; many sedan sized vehicles are in the 24000 to 36000 BTU per hour area, max capacity...

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I think you would need a heck of a lot of Peltier devices (space & $$$) to try to match that...

How many BTUs does a Peltier cooling chip generate? | Answerbag

Q:How many BTUs does a Peltier cooling chip generate?
A: That depends on the specific chip in question. For the sake of argument, the most powerful peltier cooling chip i've seen, generates about 374 btu's in one hour. To put this into perspective, a 2ton central air unit in a small home, will put out 400 btu's in one minute. It has its uses and applications, but clearly they are limited to cooling electronic circuitry and the like.

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jerry33 said:

The Tesla uses a variable speed scroll compressor rather than a piston type so it doesn't really cycle, it just goes slower or faster. This uses far less energy than the cycling of a pulley driven system. While it can use 3-4 kW, it almost never does so (at least in my car).

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Cool! (Pun Intended).

So the gap is likely to be even larger than my assumption-laden estimates above.

So I think you're thinking of the related (but different) concept of thermal efficiency as it relates to heat engines, primarily because practical heat pumps (i.e., air conditioners) usually have COP well over 100% [1]. Thermal efficiency, of course, can never exceed 100%. COP measures the amount of heat removed from a cold reservoir per amount of work used to run the pump. The sum of those two quantities is pumped into the hot reservoir.

This differs from measurement of thermal efficiency, which is measuring work performed per amount of heat entering from the hot reservoir. The difference of these quantities is exhausted as waste into the cold reservoir.

A heat engine run backwards is the same as a heat pump. However, people conventionally measure the "waste heat" (cold) side of the machine when talking about heat pumps and the "combustion" (hot) side of the machine when talking about heat engines. People also flip the fraction (work on top for engine vs. bottom for pump) depending on the context. Hence the confusion. In both cases, the heat flux on the hot side of the engine/pump is the sum of the work in/out and heat flux on the cold side.

[1] Trane, the first hit for "SEER air conditioner" on Google, advertises units with a SEER rating of 21, which corresponds to a COP of around 6 (600%). I suspect that's at the high end of "economically practical heat pumps for stationary applications."

Danal said:

The most common way to express efficiency of cooling is the "Coefficient of Performance". If COP were 100%, then every unit of energy expended (like a watt) would result in the same amount of heat removal. A true 100% cannot be achieved in the physical world in which we live... still, the closer the better.

• Compressor/refrigerant based systems have demonstrated COPs as high as 60%. A more typical real world value of an economically viable implementation is 45%.
• Peltier (solid state) have demonstrated 5%. A more typical real world deployed number is 2%.

So, yeah, there is a reason. I don't know the exact capacity of a Tesla A/C; many sedan sized vehicles are in the 24000 to 36000 BTU per hour area, max capacity. Of course, they don't run all the time; the peak is used to quickly cool a hot car and after that, they cycle, and they cycle a lot. Let's assume (pulling this out of thin air) they run 20% of the time. Assuming the same for a Tesla:

• 7 to 10kW * .2 (for cycling) / .45 (for efficiency) means a compressor based A/C will take about 3 to 4 kW from the pack spread over one hour, which makes it very easy to say 3 to 4 kWh. That's about 4 to 5 % of the total pack (i.e. reduction in range, per hour).
• If Tesla invested in a very efficient compressor based system, like 55% efficient, that could be as low as 2 or 3% of the pack.
• 7 to 10kW * .2 (for cycling) / .02 (for efficiency) means a solid state based A/C will take about 70 to 100 kW from the pack spread over one hour. That's the entire pack.

Those numbers are based on a lot of assumptions and averages. Reality in a Tesla may be different... but the relative efficiency will be somewhere in that ballpark, namely "a few percent" vs. "pretty much everything".

Click to expand...

Good point. Bad terminology on my part... yet the math still holds. Thermal efficiency of compressor based systems vs. Peltier on the same scale is so far apart that it does represent the difference between a few percent of the traction pack for cooling vs. all (or almost all) of it.