If your interest is in knowing the performance of the building, then yes, have the heat metering where the primary pipework enters the house.
If your interest is in the entire heating system then it surely must be most appropriate to include all the primary pipework, including that which is outside the envelope of the house.
What doesn’t make sense to me is to have the heat sensors inside the house, and the electricity meter “outside” the house. In that scenario, if you accept that some heat is lost in external pipework then when calculating CoPs you should subtract the electricity used to deliver the heat that you are not including.
Surely, the various CoPs relate to total heat delivered by the heatpump against total electricity used to deliver the heat.
I’m fortunate here in that I was able to site my heatpump so that the primaries come directly out of the back of the heatpump through to an (unheated) attached garage and the heat meter is in the garage about 1 metre away from the pump. I’m working on double-insuating the garage pipework to minimise heat lost into that space.
My data reports how much heat goes into my house and how much electricity it took to do it.
That is what most of us with heat meters do.
As heat meters are the most accurate thing we have, for comparison purposes it would be best if we all measure the same thing.
Subtracting some electricity that was used to move water outside? I don’t think so, all the electricity was used to produce the heat that actually entered the house.
Anyway, the good news is that systems that aren’t MID monitored are kept somewhere else.
As for Chris’s system, I only mentioned it as as a claimed 1.0c to 1.5c heat loss on the flow with nothing on the return would have made a massive hole in his COP, something that may have needed attention.
It turns out that his temperature sensors just aren’t very accurate.
The task is to determine how much heat is lost by 1 meter of pipe laid in the floor slab at a given outdoor air temperature.
To do this, you need to measure the temperature drop over a section of pipe of a known length.
Next, you need to somehow determine the water flow rate in this pipe.
Knowing this, you can calculate the heat loss of the pipe using the formula:
Q = Cpw • ρw • L • Δt / 3600, kW
Where:
ρw – water density, kg/m³ (Approximately 991 kg/m³)
Cpw – water specific heat capacity, kJ/(kg • K) (Approximately 4.19 kJ/(kg • K))
L – Fluid flow, m³/h
Δt – Temperature difference in a section of pipe, °C
Accordingly, this calculation needs to be done for both the supply and return pipelines.
For example, you might obtain heat loss of 0.3 kW, while the total power of the heat pump at that moment is 5 kW. This would mean that heat losses amount to 0.3 / 5 = 0.06 = 6%.
The challenge lies in determining the fluid flow rate. If this is a main pipe section and you have a heat meter, you might be able to see the fluid flow rate there.
Additionally, if you happen to have a differential manometer, you can measure the pressure drop on any valve, then consult pressure drop tables to calculate the fluid flow rate. (However, I doubt you have a differential manometer for this purpose.)