Some guidance about heating calculations (homemade heat-pump)

[ThMihov]

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

 

[jrmichler]

You are doing it right. There are some things that you can do to improve the accuracy of your heat calculations:

Measure temperature vs time, plot the data, draw a smooth curve through the data, then calculate rate of temperature rise from the slope of the curve at the normal operating temperature.

Add the heat capacity of the tubing and tank to the heat capacity of the water.

The heat pump motor almost certainly has a power factor less than one, so the correct power consumption is less than the 1.3 kW that you measured. You need a wattmeter to measure power. Kill A Watt is a low cost wattmeter, but it's only for 120 volts. Google kill a watt 240 volt, maybe you can find something affordable.

The water circulating pump is putting heat into the water. That heat is equal to the power drawn by the pump motor, minus the inefficiency of the motor itself.

You can cross check the heat flow from water to air by measuring the air flow and temperature rise. HVAC air flow meters can be expensive, something like this will do the job nicely: http://www.forestry-suppliers.com/p...82&title=Kestrel® 2000 Pocket Wind Meter Plus. Or google HVAC air flow meter.

The people at http://ecorenovator.org/forum/ would love to hear about your project.

 

[ThMihov]

jrmichler,thanks for the replay :).Plotting temp vs time is a good idea,which i missed :).As for the heat capacity of the tank-it weights about 30kg,cp of the iron is 0.45kJ/Kg.K so it requires about 40kJ to heat ,which is negligible compared with the water :).As for the piping it's even negligible than the tank - they are PEX-AL-PEX, 20mm diameter with total length of about 4m,because the AC is at the outer side of the wall and the indoor unit is at the inner side of the same wall :)
For the power consumption-you are right,since the motor is inductive load,it's power factor should be less than 1,although it's DC inveter and have some kind of PFC board inside.From the "scientific" side it's more accurate to read the power with wattmeter as you mentioned,but from the user side,the important thing is the RMS value of the current,for which the user pays at the end of the month :)
My pump (Grundfos UP 25-50 130) is 80W ,and since strangely i can't find any data about it,i compare it with similar models and it's efficiency is awfully low at my flow rate - less than 10%,so ignoring heat losses from it,ideally about 70W comes from it.Good point,thanks :)
I forgot to mention that i have checked the heat flow from the indoor unit using temperature difference of the inlet and return air as you said-inlet temperature was 14C,outlet temp was 25C and the air flow from the indoor unit at maximum fan speed according to the sticker on it is 420m3/h so if we assume 1m3 of air weights 1kg,the heat transfer is 420*11=4620kJ or 1.28kW which is close to the measured with the water,so there is really some truth in my actions :D
Thanks a lot again :)

 

[Tom.G]

Would this be the pump?
https://product-selection.grundfos.com/product-detail.product-detail.html?productnumber=96281424&qcid=323224578