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Calculating minisplit COP?

Eric_Old | Posted in General Questions on

I live in southwest NH (zone 5a) and recently had a Mitsubishi MXZ-36NAHZ hyper heat pump with four ductless indoor units installed at/in my house. 

I read countless questions and answers here regarding individual units versus multi units efficiency. I had multiple quotes, perspectives, reading, calculations, etc. They all came in with different results. In the end, I had to analyze all the information and come up with my own conclusion. 

Now that I have had them running through the fall, winter, and now into the coming spring, I want to actually calculate their COP efficiency. I have let them heat my home exclusively. My home has 2400 sq feet living (two floors) and 900 sq ft walk out finished basement, built in 2002. They have worked amazingly well (comfort wise). One unit in basement. two on main level, one upstairs. I have an open entryway that acts as a chimney for the upstairs. The house is very open. 

I would love to figure out their COP and help other people on this site with their questions about the multi versus single head. I have not heard enough actual case studies of the multi head units in larger homes. I think that this would be useful for people as these units become more popular/essential in our fight against climate change. 

How would I get some efficiency data using my energy consumption in kwh?

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Replies

  1. MattJF | | #1

    The full on approach can be seen here: https://www.nrel.gov/docs/fy11osti/49881.pdf

    Do you have power logging on the minisplit sytem?

    Another approach would use a load estimate for your house vs the power consumption. The load of the house can extracted from a good manual J as (BTU/h)/(deg) combined with heating degree day data https://www.degreedays.net/. You can also measure the house load if you have another heat source such as a gas furnace or electric. The power consumption for a gas furnace or electric heat will be proportional to the outside temp, so you can back out the (BTU/h)/(deg) of your house from that data. Then use that value and indoor/outdoor temp data to estimate the COP. There are a lot of caveats with these approaches. Wind and solar impacts the house load significantly. The impact can be minimized by only looking at data in the very early morning, say 2am-just before daybreak.

    There is more to get into with this, but that is a rough overview of what is involved.

  2. Eric_Old | | #2

    How about this thinking?

    I have oil consumption (boiler furnace) data from years past . From the oil, I can estimate the BTUs of energy used to heat the home for a season. Example:

    800 gallons of oil x 139,000 = 11,200,000 BTUs * (0.8 - efficiency of boiler) = 88,960,000 BTUs of energy used to heat home for season

    88,960,000 BTUs = ~26,000 kwh

    I know my kwh from the heat pump to be = ~ 8500 kwh for the heating season

    26,000/8,5000 = 3.05 COP average seasonal COP rating?

    Now this would be for the 2019-2020 season which was very mild.

    But does this work? On an electric resistance heater the COP is 1 right?

    I am only looking for a rough estimate. These things have already proved their efficiency/money savings, just want a little rough data to throw around when people start to discuss practicals of these systems.

    1. Expert Member
      Dana Dorsett | | #4

      >"I have oil consumption (boiler furnace) data from years past ."

      Don't use anything but WINTERTIME fuel use in the calculations, and correlate that use to heating degree-day data for that fuel use. If you were on a regular fill-up service that stamps a "K-factor" (= HDD65F per gallon), a few wintertime fill-up K-factors will do.

      Shoulder season oil use has errors too large to be very accurate. The very low duty of the boiler under low-load period yields a much lower than nameplate average efficiency, with a much larger fraction of the fuel use being standby loss, increasing the apparent heat load in a fuel use calc, and springtime solar gains can skew it strongly in the other direction in some cases.

      More on fuel-use based load calculations here:

      https://www.greenbuildingadvisor.com/article/out-with-the-old-in-with-the-new

      With a BTU/HDD constant derived from wintertime oil use numbers you should be able to look at electricity use between meter reading dates, make some adjustments subtracting out the presumptive "other uses", which over a winter would likely yield an average COP estimate accurate good to 1.5 significant digits , but probably not 2. If the 4C36NAHZ is sub-metered it could be even more precise.

      The COP of air source heat pumps goes up during the shoulder seasons, down during colder weather. A particularly cold January might only yield something the low-2s, but a mild March could easily hit 3.5-4, though it would be tough to properly adjust out the solar gain error which can be pretty big in March.

      1. RickyJ_GDE | | #6

        Hi Dana, this is an interesting consideration. I'm curious what is the main thing driving higher standby boiler losses during low duty? I think my boiler operates under low duty pretty much all the time since its 100k BTU/hr driving like a 25k BTU/hr load. It heats its 15 gallons of water until high temp cutoff and then that water circulates through the underfloor heating after mixing valves until the water temp drops low enough to start the boiler again. So the distribution circulator runs at a pretty high duty relative to the boiler. I would think that the standby loss you refer to means heating up the boiler, then a long period with no heating call during which all the thermal mass is released into an unheated space?

  3. _jt | | #3

    Us an IR meter to measure outlet temperature and then read the CFM of the blower. Then the HVAC formula BTU=1.08*CFM*Delta T.

    Divide by measured power use to get COP. I now use an Emporia energy monitor - they are around $50. Software is bad - but you get the number you need.

    Seems to match the owner's guide COP pretty well at default fan speeds. If you turn the fan up on the indoor side you'll typically get COP bump for a 5-10 extra watts.

  4. walta100 | | #5

    COP is calculated using numbers collected in a laboratory under controlled conditions. The manufacturers of HVAC equipment do everything they can to be sure there equipment is going to produce the best number it possibly can given the test conditions. It seems unlikely you could ever get a COP above the manufacturers published number. In the field getting an accurate estimate of the number of BTUs a system is moving seems impossible.

    Maybe the real question is there a set of data one could log to monitor your systems performance over time so one would know you have a leak or something else that is affecting the performance lowering the COP.

    My Rheem heat pump will display a lot of information. I tend to pay attention to the superheat number on my display.

    Walta

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