face26
2022-12-15 11:27:37
- #1
That actually sounds quite reasonable. I just can't figure out how 7000 kWh of heat should become 5700 kWh of electricity. That would give the heat pump an annual performance factor of 1.22. Do you have the electricity consumption for the individual months this year?
No idea how valid the estimate is. But I'm just guessing now that the heating rod is actively involved there. The question is how exactly the setup looks. Integrated heating rod, possibly a heating rod built into the storage tank?
238 hours of heating rod might be about 2000 kWh. If the heat pump had done that, it would have been maybe 666 kWh (with an annual performance factor of 3).
But the description of the 2nd heat generator also sounds strange to me (besides the flange heater). For me, the 2nd heat generator is the heating rod integrated into the heat pump. It ran for 7 hours in my case, mostly by accident on the first day of installation.
So I would first find out what is installed here and how. Then clarify which heat generator does what. And then see if that’s necessary. Your pump has enough power to manage without the heating rod. The question would then be whether simply deactivating the heating rod would be enough. But for a new system I would simply approach my heating technician about that.