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Author Topic: how would I work out the COP of an installed system in practice?  (Read 3464 times)
mpooley
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« on: February 11, 2011, 03:33:52 PM »

how would I work out the COP of an installed system in practice?

one can know how much electricity used but how do you know how much heat was generated?

Mike
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HalcyonRichard
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« Reply #1 on: February 11, 2011, 04:42:04 PM »

Hi Mike,
           You need three measurements. Input temperature, output temperature and flow. You can then work out the energy required to raise the temperature of the amount of water heated per unit time. You can then compare this to the electrical input. i.e. cop = energy out/energy in (energy out includes some or all of the electrical input energy as heat). I think an electric fire has a cop of 1.

Regards Richard
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mpooley
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« Reply #2 on: February 11, 2011, 04:58:34 PM »

Thanks Richard

As I dont have a heat pump at the moment, are these measurements on a display on the pump?
if not how would you obtain them?
the input temp say on an ASHP would that be the air temp?
the flow temp would be fairly easy to obtain (displayed on the one i saw) but Flow?

I assumed that heatpumps worked this out and displayed it for you but I saw my first one today and it didn't!

Mike
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It's not easy having a good time. Even smiling makes my face ache.

“Science is the belief in the ignorance of experts.” Richard Feynman
HalcyonRichard
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« Reply #3 on: February 11, 2011, 05:38:04 PM »

Hi Mike,
           You would only be interested in the electrical energy input. And the heat energy output. If the heat pump has not got these built in you need some instrumentation. Accurate flow meters can cost thousands. Accurate temperature probes such as PT100's can cost hundreds. If you are after a rough measurement then if you know the house heat input requirements you could measure the electrical input and do a guesstimate of the output. You would probably need to do this over a long period say weeks. The accuracy depends on how well you estimate the house's requirements/temperature rise.

Regards Richard
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Laws are for the guidance of wise men and the obeyance of fools - Richard Burton upon Trent
desperate
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« Reply #4 on: February 11, 2011, 05:47:09 PM »

Could you temporarily plumb up a tank of water to heat up, a known volume, a known temp rise, and a known time...............bingo

Desp
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KLD
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« Reply #5 on: February 11, 2011, 10:06:34 PM »

Thanks Richard

As I dont have a heat pump at the moment, are these measurements on a display on the pump?
if not how would you obtain them?
the input temp say on an ASHP would that be the air temp?
the flow temp would be fairly easy to obtain (displayed on the one i saw) but Flow?

I assumed that heatpumps worked this out and displayed it for you but I saw my first one today and it didn't!

Mike

Mike,
 The "input" and "output" temperatures Richard mentioned are both on the water circuit through the HP.
Richards "cop = energy out/energy" could be expanded a little to read: cop is the ratio of the total amount of energy the heat pump sends to the house (or wherever), divided by the electrical energy you have to put in.
While the electrical input is easy to measure, doing so for the total thermal energy output is not so easy. 

One way to measure the heat output is to measure how much water flows through the heatpump's output circuit, and by how many degrees Celsius the temperature is raised. You'd need an accurate measurement of the temperatures in the pipe leading to the HP, a second thermometer in the pipe leading away from the HP. The flow rate can be measured either by a flow meter, but as Richard said, good equipment is not cheap. If you have a circulation pump that's always at the same setting (i.e. no speed modulation), then it might be sufficient to measure the flow-rate just once. If you have access to the pipework, measure a stretch of say 10m along the circuit. At one point you heat the pipe rapidly (blowtorch?), while feeling the temperature at the 10m further away point, and measure the time that it takes for the heat pulse to arrive. The rest is just maths (pi * pipe diameter^2 * length = volume, and volume / time interval = flow-rate). E voila!

Klaus
rushing back to Egypt to build another pyramid  tumble
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mpooley
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« Reply #6 on: February 11, 2011, 10:42:27 PM »

Ok Thanks everyone that is very interesting.  Smiley

reading lots  of forum post's where people talk glibly about COP I 'd assumed it was somthing easy to find out.

I would probably take the route of trying to estimate COP after a whole year.
I am pretty certain of my House thermal requirements so using the recorded degree days in my area I could make a good stab at it.

Thanks


Mike
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It's not easy having a good time. Even smiling makes my face ache.

“Science is the belief in the ignorance of experts.” Richard Feynman
EccentricAnomaly
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« Reply #7 on: February 11, 2011, 11:10:59 PM »

The rest is just maths (pi * pipe diameter^2 * length = volume, and volume / time interval = flow-rate). E voila!

radius
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KLD
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« Reply #8 on: February 11, 2011, 11:13:20 PM »

Blush!
WHere is that edit button when you need it? (pi*D^2)/4 or (pi*R^2) .

Sorry
Klaus
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dhaslam
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« Reply #9 on: February 12, 2011, 10:37:18 AM »

It isn't easy to get a reliable answer without having some kind of tank  that can be heated.   Mine is around  860 litres  so  it takes one kWh to raise the temperature by one degree.   On this test on a warm day the  temperature went up buy twenty  degrees  in four hours with  5 kWh  input which would be a COP of 4.   Unfortunately  the  day was a bit too warm and  some heat was dumped from the DHW cylinder  which  heated the tank a little but also  made the heat pump reduce its  output temperature so the result wasn't very conclusive.      This exercise  also needs to be done at  lower ambient temperatures.      At this time of the year the tank gains   between two and three degrees  per hour  which is between  1.6 and 2.4 COP. 


* COPCalc.jpg (78.34 KB, 900x585 - viewed 406 times.)
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DHW 250 litre cylinder  60 X 47mm tubes
Heating  180,000 litre straw insulated seasonal store, 90X58mm tubes + 7 sqm flat collectors, 1 kW VAWT, 3 kW heatpump plus Walltherm gasifying stove
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