[DEnd]

Originally Posted by ecomodded

This answers my previous posts COP question

I see why now , remember reading that they measure COP in heat pumps by the inlet and outlet temps / indoor heat gain vs outdoor air temps.

So this units COP *should* be a accurate representation of its efficiency at around 7*c / 44*f or there about like with the outdoor splits COP measurement

That makes sense or the Btu numbers are next to useless / a advertising stunt.

It might not be 100% accurate but should be in the realm of its heat output COP @ 44*f outdoor air temps

I'm not sure of that but it sounds reasonable

Edit

I read the last statement wrong " The system COP will follow outdoor air temperature minus exhaust air temperature "

That throws a loop into my theory I thought I would measure outdoor air as the inlet and Btu heat as the outlet temperature

its sounding like a reverse COP math
I should be able to figure out Btu gain at 44* (or there about) with this system like with a mini split.


New math plan

I'll keep it simple and follow what was wrote above in the highlighted area

Lets say its 7*c outside and the vent blows out 2*c , divided is 3.5 COP

I'll try it again with more numbers

12*c divided by outlet temp of 2*c = COP of 6

Might be onto something I will have to take real numbers when its running and see what those are

That math my well be totally wrong Im far from a HVAC guy

Edit I need to convert it to Kelvin first

12*c to K = 285
7 to K= 280
2 to K= 275

COP @ 12*c outside and 2* from cold outlet = 12 - 2 = 10 COP

7 - 2= 5 COP

I'll try that with real numbers when I have them and do more research in the meanwhile

You're missing energy flows. The unit will operate at its COP regardless of outdoor temp. however in reality the added BTUs are only there if the temperature difference of outdoor air and the discharge air tempreatures is about 1/2 the temperature difference of the outdoor air vs the desired indoor temp (assuming there isn't a phase change of something in the air needed in there somewhere). Lets say outdoor temps are 72ºF and you desire a 76ºF indoor temp and the unit discharges at 32ºF The unit will operate at a 2.7COP but the actual COP of the house (with just this unit) will be ~2.48, because 10% of that energy will be needed to heat the infiltrated outdoor air to the desired indoor temps. Yes the BTUs output by the unit will remain constant regardless of the outdoor air temps but a part of those BTUs are unusable because they are "used" to increase the temperature of the infiltrated outdoor air. Take that energy flow into account and you will get much closer to what the COP of the system actually is.