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  1. #1
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    Sep 2004
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    If a 3.5 ton heat pump has a COP of 3.5 at 47 degrees and the cost of a KWH is .o8 cents. Would you divide the .08 by the COP of 3.5 to get .022 per actual KWH used?

    There are 3400 btu's in a KWH, if I need 34,000 btu's that is then 10 KWH. Would I divide that 10 by the COP and then multiply by my energy charge?

    So 10 KWh X .o8 equals 80 cents per hour or

    10 KWH divided by 3.5 (COP) equals 2.85 KWH X .08 equals 22 cents per hour?

    Am I on the right track here or not??

  2. #2
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    Pertty much.

    It won't work ouot ezact but, you'll be in the ball park.
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  3. #3
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    Originally posted by beenthere

    Better rephrase what I said.

    You way of doing the math might get you close.

    You need to know the btu output of your heat pump at 47.

    Your duct work and registers may be restricting air flow that you aren't getting that 3.5 cop.



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  4. #4
    Join Date
    Feb 2005
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    Ya'll correct me if I'm wrong

    The COP is a term that has to do with the fact that a heat pump will produce less capacity at lower temps. I believe the COP is just a reference to electric heat with a COP of 1.
    But, I don't think it can be related as you are trying to do. The COP is more of an output term rather than an input term. You will still be paying the same amount on your utility bill, but you won't be getting as much out.

    North, this is probably more of a question myself, than an answer.
    At 47 degrees, yur heat pump will be using ( pretty much ) the same electricity, but you will not be getting the same BTU's out, and the unit will run longer, costing you more money

    [Edited by bornriding on 03-18-2005 at 06:20 PM]

  5. #5
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    The COP, is an output rating.
    Copared to the 100% eff. of electric resistance heat.
    A 3.5 cop at 47 means that compared to resistance heat, he will be getting 3.5 times more heat per kw consumed, then resistance heat.

    Or 10KWH of resistance heat =34130btu
    And 2.857KWH of heat pump =34128.94btu at 3.5 cop provided that is his heat pumps btu, and cop rating.

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  6. #6
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    Feb 2004
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    N..of49.

    Yep, you are on track. There are some other variables to consider, primarily defrost cycle.

    On my own house I have the electric resistance heat cutoff with a big relay that only closes if the outside temp is below 25F (almost never in Seattle area).

    So, during a defrost cycle, there is cold air coming out of the registers, but DW and I live with that. What really puts a "hurt" on your simple COP calculation is that 5 minutes every 1/2 hour that 10kW of resistance heat kicks in. If you live witht the short cold air blast, that expense is also reduced by the COP ratio. If you are in a dry area, you may never go into defrost anyway, but the HP sure does here in PNW.

    Moral of story, you are $$ ahead to turn the breaker for the resistance heat section of your air handler off, turn it on only when its so cold outside you cannot heat otherwise. On current HP on own house (with the outside thermo control as above) in Seattle area, have never had resistance heat kick in except for a test.

  7. #7

    energy

    Originally posted by northof49
    If a 3.5 ton heat pump has a COP of 3.5 at 47 degrees and the cost of a KWH is .o8 cents. Would you divide the .08 by the COP of 3.5 to get .022 per actual KWH used?

    There are 3400 btu's in a KWH, if I need 34,000 btu's that is then 10 KWH. Would I divide that 10 by the COP and then multiply by my energy charge?

    So 10 KWh X .o8 equals 80 cents per hour or

    10 KWH divided by 3.5 (COP) equals 2.85 KWH X .08 equals 22 cents per hour?

    Am I on the right track here or not??
    COP is short for coefficient of performance. So this means that the HP is working at 350% efficient compared to electric strips when it is 47deg. So for $.08 you get 34,130 btus with strips. When the heat pump is working in 47 degree air it will give you 34,130 x 3.5 = 119,455 btus for the $.08. The higher the air temp. the higher the COP, the lower the air temp the lower the COP. Most air sourse HP have a 1 COP at about -10 degrees.

  8. #8
    Join Date
    Sep 2004
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    67
    geoexchangeman, Thank you, that is exactly what I was looking for.

    Thanks everyone else as well. I realize that the COP drops as the outside temp. drops. This heat pump (Amana) has a COP of 2.4 at 17 degrees. I also realize that there are some variables in there that will affect the actual COP such as defrost cycle etc.

    But the crux of the matter is that compared to a 95 % efficient gas furnace that the heat pump will be in the neigborhood of 200 to 300 % efficient even if natural gas is a little cheaper initially than electricity the heat pump will be cheaper to operate year in year out by a fairly wide marging I would guess.

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