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03-06-22, 04:12 AM | #1 |
Lurking Renovator
Join Date: Mar 2018
Location: the Netherlands
Posts: 10
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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 |
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