COP Calculation

[Potatoman]

Hi, is there a calculation you can make to determine the daily COP of a system. Thanks

[JohnMo]

Measured heat output or calculated for flow/return temp and actual flow rate.

 

All electrical energy use - timer, controllers, all pump, immersions etc. Divide heat out by electric in.

[SteamyTea]

Usually the system has a built in display that can show it. They are sometimes a bit misleading though.

 

You can calculate it yourself, but you really need some monitoring equipment.  This would usually be data loggers on the flow and return pipes to log the temperature, a flow sensor to measure the mass of fluid pumped around and an electrical logger to measure the energy input.

 

Then the arithmetic starts.

 

Calculate the output.

 

Specific heat capacity [SHC] of fluid, 4 is close enough for this [kJ.kg^-1.K^-1]

Flow rate [kg.h^-1]

Temperature Delta between flow and return in kelvin [ΔT]

This will give you the kilojoules over the time period measured [kJ].  You can convert kilojoules to kilowatt.hour by multiplying by 0.0002778.

It is usual to place a subscript to distinguish the type of energy delivered, in this case a _t for thermal is used.

 

 kJ.kg^-1.K^-1 x kg.h^-1 x ΔT x 0.0002778 = kWh_t

 

Electrical logging is usually calculated in mean power [W] or total energy [Wh].  Care must be taken here as they are not the same thing.  Energy is power multiplied by time.  Energy is usually purchased by the kilowatt.hour so divide by 1000.  Again a subscript _e is used to show that it is electrical energy.

 

(W x h) / 1000 = kWh_e

 

The coefficient of performance [CoP] is the ratio of the output to the input energies.

 

CoP = kWh_t / kWh_e

 

 

 

 

 

 

 

[Bramco]

47 minutes ago, SteamyTea said:

you really need some monitoring equipment

 

The open energy monitor guys do a kit -> https://shop.openenergymonitor.com/level-3-heat-pump-monitoring-bundle-emonhp/

And they have a site showing the data from people who have installed the kit and made their data public -> https://heatpumpmonitor.org/   

[FarmerN]

3 hours ago, SteamyTea said:

Usually the system has a built in display that can show it. They are sometimes a bit misleading though.

 

 

 

 

 

 

 

 

 

 

Feel my display must be misleading , showing a COP of 6 for December so far,  Too good to be true ??

My GSHP is a NIBE S1155 12 KW , with 600M ground loop, heating a 215sqM new build+ DHW

December            179 kWh in         1073 kWh out.                 

Brine in at 8.8 Deg C , main living space 21Deg C ,  bedrooms 19 Deg C

DHW 48 Deg C

[PhilT]

3 minutes ago, FarmerN said:

Feel my display must be misleading , showing a COP of 6 for December so far,  Too good to be true ??

My GSHP is a NIBE S1155 12 KW , with 600M ground loop, heating a 215sqM new build+ DHW

December            179 kWh in         1073 kWh out.                 

Brine in at 8.8 Deg C , main living space 21Deg C ,  bedrooms 19 Deg C

DHW 48 Deg C

If you can get the data, John Cantor has a nice easy calc template, so you can at least sense check the output power.

https://heatpumps.co.uk/heat-pump-resources/numbers-and-calculations/

 

 

[Beau]

28 minutes ago, FarmerN said:

Feel my display must be misleading , showing a COP of 6 for December so far,  Too good to be true ??

My GSHP is a NIBE S1155 12 KW , with 600M ground loop, heating a 215sqM new build+ DHW

December            179 kWh in         1073 kWh out.                 

Brine in at 8.8 Deg C , main living space 21Deg C ,  bedrooms 19 Deg C

DHW 48 Deg C

Thats a very good brine temp. If your heating flow temps are low I would have thought a COP of 6 is within reach. 

[Potatoman]

Can anybody tell me if this calculation holds water.

Heat loss calculation of my house is 5.8Kw at -2 for 21 degrees inside.

So to raise the temperature of my house I require 252w/degree      5800 divided by temp difference 23

So if you take average outside temp - average inside temp 0-21 =21      21 x 252 = 5292w

5.3kW x 24 = 127 kWhr/day divided by the actual kWHr used, in this case 41 kWHr  would  this give a COP of 3.097 ?

 

[JohnMo]

In theory it may be ok, but heat loss is best guess, and the -2 is unlikely to stay constant all day, so could be correct, but may not be. At that temp outside you could have a bunch of defrosting happening also to confuse things.