Heat loss = (outsideT - insideT) * coefficient - constant

Flow = (heat loss / radiator output) ^ (1/1.3) x 50 + insideT + dT / 2 Min flow is the same but with heat loss bounded by minimum heat pump output

The lowest temperature the heat pump can run at is 37 C, which matches the stable heating periods I’ve observed. If I’m willing to accept a duty cycle of 66% when it’s not freezing weather, then this drops to 33 C.

Here’s my spreadsheet if you’d like to do the same - make a copy and input your own numbers.
Adding more radiator output moves the line downwards. Reducing heat loss flattens the curve.

Hi Tim,
I did something similar in a slightly different way, armed with little more than Q=U.A.dT in a recent test on my most significant room (the lounge).
Happily, the external temp was constant for a whole day (12degC), so the ASHP exit sat at the WC temp (42degC) all day too, and the room reached a constant temp (i.e. steady state with heat lost = heat gained).
Based on the MCS-based U values (as used by my installer in his heat loss calcs) for extl/intl walls, ceiling, floor and windows, and their respective areas and dTs, I calculated my steady-state heat loss Q.
Then I used this Q in the same equation for my radiators (known A) to calculate the U for them. This was 48W/m2/degC, which compares with Vendor’s 61W/m2/degC (I can believe my figure as I have furniture items near the rads which limit convection).
Armed with these U values for rads and losses, it’s easy (spreadsheet) to look at different outside temps and different rad temps (which I assume to be very close to ASHP exit temp), so I will be able to tune the WC accordingly as we go into my first winter.