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03-06-22, 05:12 AM | #1 |
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Where is this idea flawed / how to guestimate efficiency?
Hi,
I just had a weird idea. I am thinking of different concepts of heatpumps, and I realised I have one in the shed. It's just not considered a heat pump. It's an air compressor. It compresses the air it sucks in to 6 bars, and that compressed air is very hot. If I would lead that through a heat-exchanger of some sort (metal pipe through boiler tank) it could heat water. Of course with a 5,9 bar blow-off valve at the end of the pipe. I just haven't found how I can calculate the temperature delta per bar per litre of air, and how much kW it takes to produce that. I think the flaw in my system is right there, more kW electricity used than kW generated to warm the water. Can someone tell me how to calculate / guestimate those numbers? If it is efficient (coefficiency of more than 1), it is a very simple system to make. And it would give a big delta T which I need, as I want to take in outside air (minus 20 C at the lowest) and still give about 60 C water max water temperature. As my compressor sucking in ambient air (15 degrees) already heats the outfeed pipe to 140 C at 6 bars, this seems feasible and easy. Anyway, please feel free teach me, Best regards, Hugo |
03-06-22, 03:53 PM | #2 |
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I think the main thing you are missing to a heat pump, and its efficiency, is that you are only compressing/decompressing, but there is no phase change. It is the evaporation/condensing which allows taking or giving up heat from/to the environment.
I will now excuse myself to free the stage for more knowledgeable users...
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03-06-22, 06:35 PM | #3 |
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Agreed! I am not an expert, sorry, but I understand just enough to be dangerous.
All normal heatpumps use a closed loop system with a compressor, condensor, pressure drop, evaporator. The working fluid is operated in the region where it is changing from gas to liquid and back -thats where the serious energy carrying capacity is. I’ve added a pic showing water temperature versus enthalpy (heat, energy transfer, it’s what you need lots of). You can see it takes way more heat to actually boil water to vapour than it does to merely heat it up all the way from freezing to boiling. I picked water, as we’re used to this phenomenon with it. I think you could use air, but it phase changes at much colder temperatures, so the efficiency will be awful for normal use. It will give hot air, but hardly any heat before it cools down too much. Somebody could perhaps use a nitrogen pressure-enthalpy diagram to help guesstimate efficiency. Pick a better gas! CO2, propane, ammonia, they work at normal temps and pressures. Last edited by Daox; 03-08-22 at 09:48 AM.. |
03-10-22, 08:18 PM | #4 |
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The electricity to an air compressor motor ends up as heat distributed as follows:
From the motor (about 5% to 10% of the electricity in). From the compressor itself. From the hot compressed air. The total heat from all of those is equal to the amount of electricity in. One kilowatt-hour of electricity equals 3412 BTU's, so an air compressor drawing 1.00 kW will make 3412 BTU/hr total heat. If the compressor is located in the heated space, then the end result is exactly the same as an electric heater that draws the same amount of electric power. The only way to get more BTU's out than electricity in is to use the electricity to drive a heat pump. Then the BTU's out is equal to the electricity in plus the amount of heat pumped from the cold side to the hot side. |
The Following User Says Thank You to JRMichler For This Useful Post: | jeff5may (03-17-22) |
03-19-22, 06:49 PM | #5 |
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The puzzle piece that's missing here is the phase change. This is how a heat pump "cheats the system". Read the "heat pumps for dummies" post, it's super easy to understand and comprehend.
The key term here is the specific heat of condensation/vaporization. When compression happens, a certain amount of heat is used to move a certain mass of gas. This heat is released from the condenser at a constant mass conversion rate, and at constant temperature, determined by the condenser pressure. The gas must release this heat in order to become a liquid. So it does, and does, and does, and does... Ok now, the refrigerant is in a high pressure, condensed state. This means it's packed tightly into a shrunken state. It needs energy to be turned back into a gas. Let's just do the easy thing here to illustrate and agree with your scenario. The carbon dioxide fire extinguisher experiment! When a co2 extinguisher is directed at a beer cooler, it freezes the beer! This is because the heat taken to vaporize the co2 is around 200 joules per gram of co2. Its normal "specific heat" is a single digit, so the escaping co2 falls to it's natural state (cold dry ice) before it can absorb enough heat to exist as a gas at low pressure. This heat is absorbed from the beer, which freezes pretty fast. Might not freeze solid, but that depends on the beer's phase change heat capacity. So you see, it would take many hundreds of times as much mass of a gas following the ideal gas law to draw out as much heat as a single vessel of liquid CO2 fire extinguisher. Same principle with phase change refrigeration. The compression doesn't magically manufacture heat flow, it's the phase change of the refrigerant that absorbs and releases the massive energy. |
03-30-22, 04:02 PM | #6 |
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Phase change isn't a necessity it's a change in pressure. The air in the tank and airline is hot but when discharged through an orifice the pressure drops and the air will absorb heat as the pressure has dropped.
Refrigeration works the same way but for it to be efficient you also need the phase change as the most heat is absorbed or consumed depending on the situation. Now to use an air compressor for heating or cooling yes compressed air is used to do that in industrial applications but it's extremely inefficient and noisy. You can get a cold gun which uses shop air to create cold air and hot air as well. It's voodoo magic in a tube. https://www.exair.com/products/cold-...t-systems.html In short air compressors are the biggest energy hogs in the shop. They are noisy produce allot of heat and use allot of electricity. And I need a much larger one for my shop as my customized 60 gallon doesn't keep up with my needs anymore. Couldn't imagine what it would take solar panel and inverter wise to run an air compressor... |
03-31-22, 06:41 AM | #7 | |
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See vortex tube, https://en.wikipedia.org/wiki/Vortex_tube
Commercial vortex tubes are designed for industrial applications to produce a temperature drop of up to 71 °C (160 °F). With no moving parts, no electricity, and no refrigerant, a vortex tube can produce refrigeration up to 1,800 W (6,000 BTU/h) using 100 standard cubic feet per minute (2.832 m3/min) of filtered compressed air at 100 psi (6.9 bar). A control valve in the hot air exhaust adjusts temperatures, flows and refrigeration over a wide range.[23][24] Quote:
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04-03-22, 10:10 AM | #8 |
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And 100CFM is allot of air. The typical 60 gallon compressor puts out around 10-12cfm @90psi while pulling 17-20 amps.
I love air tools and pneumatic stuff but it's incredibly inefficient. I can't think of anything that is less efficient. |
04-05-22, 09:25 PM | #9 | |
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Quote:
That's 18.65 kW of power to get 1.8 kW (6000 BTUH) of cooling, so the EER is 0.3. Window air conditioners have EER about 10, so use about 1/30 as much power for the same cooling. |
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04-06-22, 02:42 AM | #10 | |
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