Calculation of thermal output???

[Dogsbody]

I've just had an air to water heat pump installed at my house, and I wanted to get a rough idea of what it's performance is. Can anyone help with a reasonably simple way to calculate the real world COP as a cross-check against the manufacturers BS?

I appreciate that COP is a crude measure, that it's only a snapshot, and that the average across the whole season is more useful etc etc, but indulge me for the sake of my curiosity.

The heat pump control tells me the inlet and outlet water temperatures, the Wilo circulation pump shows the water flow rate, and I have the electricity consumption metered separately. What would the formula be, and do I need any other measurements? Anyone?

 

[doitmyself]

The heat pump control tells me the inlet and outlet water temperatures, the Wilo circulation pump shows the water flow rate, and I have the electricity consumption metered separately. What would the formula be, and do I need any other measurements? Anyone?

The heat output is calculated using the formula: kW = flow x ΔT x 4.18

Flow is measured in litres/sec
ΔT is the difference between inlet and outlet temperature.

e.g

flow =0.5 l/sec
ΔT = 10°C

kW = 0.5 x 10 x 4.18 = 20.9kW

If the system produces this for one hour then 20.9 kWh will have been produced.

As well as knowing the electricity consumption you also need the time over which it was consumed. Only the time when the system is running is relevant.

 

[Dogsbody]

doitmyself: Thanks for taking the time to help. So, the 4.18 is the specific heat of water I think? More importantly, can you expand a bit on your last point? How does the time get plugged into the equation? Come to that, exactly what is the equation? And am I right in thinking that COP is a snapshot taken at a point in time?

 

[Nostrum]

Which Wilo pump is it? Are you sure it's displaying flow and not power consumption or metres head?

Also, which heat pumps have you had installed?

 

[Dogsbody]

It's the Stratos Pico circulation pump, and yes, it definitely shows flow rate. I'd like to keep to myself the heat pump make for the time being as I don't want the debate to get sidetracked by pros and cons of different brands. I don't think it's relevant to my original question is it?

 

[JonG1]

Might be easier to get a heat meter installed along with an electricity meter to calculate seasonal COP that would be more useful for you.

 

[Dogsbody]

Might be easier to get a heat meter installed along with an electricity meter to calculate seasonal COP that would be more useful for you.

Cheers. Yeah, I may do that eventually, but at the moment I just want a quick fix as a starting point.

 

[doitmyself]

So, the 4.18 is the specific heat of water I think? More importantly, can you expand a bit on your last point? How does the time get plugged into the equation? Come to that, exactly what is the equation?

You are right about 4.18.

As for time, remember that one watt is actually one Joule per second. So one kWh is therefore 3.6 Megajoules (1000 x 60 x 60).

If the system in my previous example runs for one minute it will produce 20.9 x 60 = 1254 Joules of energy. Let's say the electricity consumption over the minute is 0.1 kWh = 0.1 X 60 x 60 = 360 Joules. Then the COP is 1254/360 = 3.48.

I have no idea about the COP part of your question, but a snapshot seems the most likely method. It would be similar to the way a gas boiler is rated by measuring the gas consumption over two minutes. Measuring over a long time would be difficult as the temperatures would not be constant.

 

[doitmyself]

If the system in my previous example runs for one minute it will produce 20.9 x 60 = 1254 Joules of energy. Let's say the electricity consumption over the minute is 0.1 kWh = 0.1 X 60 x 60 = 360 Joules. Then the COP is 1254/360 = 3.48.

Sorry, but that's not correct. I forgot the kilo prefix. The heat produced is 1254 kJ and electricity used is 360 kJ. The COP doesn't change.

 

[Dogsbody]

......As for time, remember that one watt is actually one Joule per second. So one kWh is therefore 3.6 Megajoules (1000 x 60 x 60).

If the system in my previous example runs for one minute it will produce 20.9 x 60 = 1254 Joules of energy. Let's say the electricity consumption over the minute is 0.1 kWh = 0.1 X 60 x 60 = 360 Joules. Then the COP is 1254/360 = 3.48.

OK, I'm struggling a bit with the maths.... My figures are as follows:

Flow rate: 66.66 litres per minute
DeltaT: 14 C
Energy produced = 66.66 x 14 x 4.18 = 3900 Joules
Electricity consumed: .067 kW per minute x 60 x 60 = 241 Joules
COP = 3900/241 = 16.2

A COP of 16.2 seems a little unlikely. Can you see what I am doing wrong? I'm sure it's really basic. I think all the 60s are confusing me. (Didn't see your second post, but as you say, it doesn't change the COP)