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ThMihov
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Hi all
In general I'm experimenting with an old split air conditioner which i would turn into some kind of homemade heat-pump.The basic idea is to replace the indoor unit with some kind of heat exchanger so the AC heats up water instead of air ,which in turn circulates over radiators in house-some kind of central air conditioning.
Since my AC runs on R410A and water to refrigerant heat exchangers are very expensive (new one costs more than the AC unit!),I chose to wind up a copper tube coil inside a vessel to heat the water in it.That way as positive side effect the vessel works as buffer.So a friend of mine gave me a old 60 liter tank to experiment with.I cut it ,wound in it a coil of 7or 8 meters of 8mm copper tube,and welded back.
Since the low temperature of the water which is used for heating requires bigger radiators,i chose to solder some plumbing fittings to the indoor unit (which is not used anyway) and use it as radiator because of the forced convection which should compensate for the bigger size (maybe?).
Today i went for some experiments and now i have some questions about power and so on.
First-when i started the AC,the initial water temp inside the water tank was about 12C (I'm measuring it at the inlet of the indoor unit inside the house).I waited some time before i took the measurements,because in addition to the water,the tank and all the metal plumbing parts needs to be heated also.So i waited until the water was 32C and started to measure the time in which the temperature rises with 4C to 36C,and current consumption of the AC,trying to estimate roughly the power which the AC outputs to the tank.The time was 6min and 20sec.The tank is insulated with 5cm mineral wool and the pipings to the house with 9mm rubber piping insulation.
So according to my knowledge heat transfer is a product of specific heat capacity of the liquid times mass of liquid times temp difference or in other words Q=cp*m*dT.Roughly estimated cp of water is 4.178kJ/kg.K, 1 liter of water is almost 1kg so Q=4.178*60*4=1002kJ,which was gained for 380sec or 1002/380=2.63kW
The average current consumption of the AC at that time was 5.7A or at 230V mains that gives about 1.3kW
So the COP should be 2.63/1.3=2 ?
At the indoor unit a have attached a cheap flow sensor,but although the calibration which i made i still don't believe it because the readings changes with it's position and who knows what else,so i decided to check the heat output of the indoor unit with the same method as above.So when the water at the inlet was 36C i turned off the AC with the indoor unit fan running and waited to see how long it takes to cool the water with 3C.It took 9minutes and 30seconds so again according to the upper equation Q=4.178*60*3=752kJ for 570 seconds or 1.31kW
My main question is that way of roughly estimating powers right or I'm doing something wrong?
Thanks in advance
In general I'm experimenting with an old split air conditioner which i would turn into some kind of homemade heat-pump.The basic idea is to replace the indoor unit with some kind of heat exchanger so the AC heats up water instead of air ,which in turn circulates over radiators in house-some kind of central air conditioning.
Since my AC runs on R410A and water to refrigerant heat exchangers are very expensive (new one costs more than the AC unit!),I chose to wind up a copper tube coil inside a vessel to heat the water in it.That way as positive side effect the vessel works as buffer.So a friend of mine gave me a old 60 liter tank to experiment with.I cut it ,wound in it a coil of 7or 8 meters of 8mm copper tube,and welded back.
Since the low temperature of the water which is used for heating requires bigger radiators,i chose to solder some plumbing fittings to the indoor unit (which is not used anyway) and use it as radiator because of the forced convection which should compensate for the bigger size (maybe?).
Today i went for some experiments and now i have some questions about power and so on.
First-when i started the AC,the initial water temp inside the water tank was about 12C (I'm measuring it at the inlet of the indoor unit inside the house).I waited some time before i took the measurements,because in addition to the water,the tank and all the metal plumbing parts needs to be heated also.So i waited until the water was 32C and started to measure the time in which the temperature rises with 4C to 36C,and current consumption of the AC,trying to estimate roughly the power which the AC outputs to the tank.The time was 6min and 20sec.The tank is insulated with 5cm mineral wool and the pipings to the house with 9mm rubber piping insulation.
So according to my knowledge heat transfer is a product of specific heat capacity of the liquid times mass of liquid times temp difference or in other words Q=cp*m*dT.Roughly estimated cp of water is 4.178kJ/kg.K, 1 liter of water is almost 1kg so Q=4.178*60*4=1002kJ,which was gained for 380sec or 1002/380=2.63kW
The average current consumption of the AC at that time was 5.7A or at 230V mains that gives about 1.3kW
So the COP should be 2.63/1.3=2 ?
At the indoor unit a have attached a cheap flow sensor,but although the calibration which i made i still don't believe it because the readings changes with it's position and who knows what else,so i decided to check the heat output of the indoor unit with the same method as above.So when the water at the inlet was 36C i turned off the AC with the indoor unit fan running and waited to see how long it takes to cool the water with 3C.It took 9minutes and 30seconds so again according to the upper equation Q=4.178*60*3=752kJ for 570 seconds or 1.31kW
My main question is that way of roughly estimating powers right or I'm doing something wrong?
Thanks in advance
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