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

∴ 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)

(17.3

∴ 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 22.8

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.

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*(17.3

**Kw**= 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 22.8

*Kw. 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.