We have just replaced our existing LPG heating system with a Ground Source Heat Pump. The heat pump comes with a design CoP of 4.41. I would like to see if there is any way to do some calculations to see how the heat pump is performing.
The unit is a CTC GSi 12 manufactured in Sweden. I contacted the manufacturer to see if this info is included in the system. But they said no, too complicated to calculate with too many variables. The system does give me some data, key of this is cold source temperature extracted and returned, the hot temperatures sent to the radiators and the return temperature and also the energy consumed in electric (on its own meter).
Now thermodynamics is probably one of the reasons I didn’t become a mechanical engineer, but if I’m correct I can calculate a CoP for the carnot cycle based on hot and cold temperature, at least for an idealised system. But having done this the values are way too high.
Would there be any method I could use to approximate the CoP?
COP = useful heat provided / energy used.
You’ll need the flow rate of the heating fluid.
The company wasn’t wrong. It does depend on a lot of variables.
But I think you can at least get a comparative value based on the design COP. Temperatures of your hot source and your cold source alone can get you a theoretical COP, if you assume perfect efficiency. If you take:
- the 4.41 design COP as “ideal”
- assume your thermostat’s temperature setting is uniform and constant
- and the temperature readings you said are available from the unit
I think you can at least see if it is living up to the 4.41 COP or not.
Its with perfect efficiency.
So, if you know your temps, you can calculate a theoretical COP that ignores a lot of the troublesome variables of the real world. I would assume your number will be well above the 4.41 right now, but that may change over time. You can then use that same calculation and compare it to the original to see the COP change over time as the unit ages.
Mint, the manual gives a nominal flow rate if 0.28 l/s.
SS, That is the fudge I did. Temperature at the source is 279K, Temperature output to heating is 308K. This gives me 10.6.
It will be complicated by the factor that the unit provides hot water at 323K as well, but it will be a much lower volume.
Another way to use as a first-order check. You will need to know a lot about the home’s construction. Skip the equipment specs and go straight to MIntJulep’s definition of COP:
Choose a fixed time period (several hours), preferably on a cloudy night (minimize radiant heat transfer) when the the outdoor temperature is more or less constant and little or no wind (minimize air leaks and convective cooling).
Determine the house’s heat loss rate by making careful calcs of insulating properties of all surfaces.
Don’t open any doors or windows or run any bathroom vents during the test. No showers or baths. No washing machine or clothes dryer use.
Determine incidental heat generation inside the house from things in operation, like appliances, lighting, and humans occupants.
Run the heat pump to keep the home’s interior at a steady state temperature.
Do the math to calculate total heat loss over the fixed time period (useful heat provided in, say BTU).
Read the instrumentation to get the unit’s electrical consumption (energy used in, say KWH).
Make the conversion and divide for the answer.
The heating system we removed was powered by LPG. I know what my annual heat demand was based on my LPG usage, approximately 17000 kWh per annum . Assuming the heat demand for my house is unchanged (because the house is unchanged) The I could use that as a measure of the heat loss for my house.
The LPG provides 7.08 kWh per litre, and costs me 44.1p per litre, or 6.23p per kWh. The electric to run the GSHP costs me 16.2p per kWh. So simplistically, if my CoP is 2.6 or better then cost will be the same.
I suppose that once I have a months worth of electric readings I can see what the comparison is.
Keep it simple.
You’ll need the actual water flow rate, not a “nominal” datasheet number.
If there is no flow meter perhaps you could use the pump power and performance curve if that is available.
Measure over shortish periods of near-steady state operation.
Do not include time when the thing is not making heat because 0/0.
Unfortunately, there is no flow meter. And, while I have managed to find the pump curves, the pump is a variable speed. The heat pump only tells you if the heating pump is running on or off, I don’t think it give me any speed or % data. I’ll have a dig.
As a civil engineer, farting about in the mechanical engineers pool, I had a stab at determining a heat output…