Why do Soleco units not have much better COPs than equivalent units for example the Kensa C120 T1H to C300T3H?
Soleco are the only heat pump designed specifically for use with borehole water rather than a closed brine or water/glycol loop. Given the underground water represents a huge constantly refreshed heat source this should achieve a better COP than the closed loop alternative.
Heat Pumps have been available for many years often sold with wildly exaggerated COPs. To tackle this we welcomed the introduction of EN14511 which dictates the way in which the formal COP should be calculated. For open loop systems it should be calculated at a source water on 10C and a source water off of 7C, and for closed loop the figures are 0C source water on and -3 source water off. The key measure being that the difference (TD) in the in and out temperature is the same at 3C.
The formula for heat taken from water is
Kw = 4.18 x TD (°C) x Flow (L/Sec)
Soleco publish on these pages their official MCS certificate giving the COP at these conditions. The Kensa data sheet states the COP is calculated using the methodology in EN14511 but using design conditions namely brine in at 0C and water out at -4C, giving an increased TD of 4C. This increases the relative heat generated from the water by 33% massively improving the COP calculation. It is interesting to note why they do not publish their MCS certificate in full with the formally COP? Is it because it is much lower than Soleco?
The accrediting agency when asked would only comment that all the Kensa units achieved the minimum standard namely a COP of at least 3.5.
So I am unable to answer the question due to a lack of clarity around the statistics offered by Kensa.
Soleco units, we believe are very efficient and better than many equivalent units on the market, in our informal testing our data suggests that an open loop heat pump has at least a 10% better COP than the exact equivalent closed loop system.
Can you please explain the maximum amount of heat that you can generate from 20m3 of source water? M - Surrey
A residential household is allowed to draw unto 20m3 per day from a borehole without and abstraction license
The amount of usable heat that can be taken from 20m3 of source water by the GSHP per day is a factor of --
1. The GSHP's COP
2. The site set up and GSHP's TD (the difference between the source water in temperature and source water out temperature)
Please remember that the COP will be effected by the source water TD. (The more the TD is increased, the smaller the COP)
For the purpose of this explanation, we have made the following assumptions
1. COP = 4.12 (MCS Certificate figure for Soleco 26KW)
2. Source water inlet temp = 11°C
3. Source water return Temp = 6°C
4. Source water TD = 5°C
The basis for calculating COP is Total Heat Rejected ÷ Total engery in (electrical power to drive GSHP).
∴ For every 4.12Kw of usable heat Rejected by the GSHP, there will be 1Kw of electric power input. So the remaining 3.12Kw of heat is taken from the source water.
The calculation for heat transfer from the source water will be:-
Kw = 4.18 x TD (°C) x Flow (L/Sec)
so typically a heat pump will run for up to 6 hours per day, assuming 18,000 litres used (within the 20,000 limit) is 3,000 per hour of .83l/s
(17.3Kw = 4.18 x 5 x .83) = heat that can be taken from the source in 1 day
∴ Using a COP of 4.12 to move 17.3Kw for six hours of source heat , there will be an electrical input of 5.5kw less losses/efficiency factors of 0.5kw will give a total heat rejection of 22.8Kw. For the six hours the heat pump is running the heat generated is 136.8
So you will see that by changing the TD, the calculation will change, and so will the COP.
As we have said, the GSHP must be sized to accommodate initial heating of the property and calculated heat loss. This is the design consultants responsibility! As an example the Energy Performance Certificate for my colleagues house which is 270m2 property (five bedroom family home) level D. This supports an annual heating requirement of 25,246KWH on average 69KWHours per day. So the 26KW heat pump could generate 2 times the average space heating requirement for the home.
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