05-04-12, 07:09 PM | #71 |
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heatpump has been done for a few weeks working great my loopfeild temp returns at 82-84 *F which is kind of high and leaves the heatpump at 90-92*F but i have been keeping my house at 72*F most of the time
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05-04-12, 07:23 PM | #72 | |
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Quote:
It would probably be a good idea to do a weekly log of your temps that you just reported. If you see a significant rise over time (several years) you might want to look into upgrading the house insulation, reflective window coverings, roof color, natural or artificial shade, etc. -AC
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05-05-12, 10:31 PM | #73 |
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my heat pump draws 2367.8 watts with the circulator pump running. as measured with my clamp on meter
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05-07-12, 07:19 AM | #74 |
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Any idea what kind of COP you're getting?
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05-07-12, 06:09 PM | #75 |
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how do i calculate COP
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05-07-12, 09:09 PM | #76 |
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You have to calculate energy you get out of the system versus the energy you put in. So, you'll need to figure out how much power your pumps and things use. Then, figure out how many BTUs of output you're getting.
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05-07-12, 10:13 PM | #77 | |
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Quote:
2.3678 kWh is equal to 8076.539 BTUh. (like heat from a resistive load) So, if 16,000 BTUh (4.69 kWh) is being pumped.. 4.69 / 2.3678 = 1.98 COP?? Humm, 16,000 / 8076.539 = 1.98 !! The problem is, on a good day you might be getting a lot more output than 16,000 BTU. And on a bad day, less than 16,000. But the real problem is being able to measure your BTUh output! That's one project that I've decided to skip, until I'm in my late 70s (Maybe 80s) and I'm bored beyond all hope of recovery..
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05-07-12, 11:40 PM | #78 |
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Measuring the COP of an air conditioner is not as straight forward as measuring the COP of a water source HP... in fact, it can be kind of tricky.
The formula: COP = (energy out) divided by (energy in) ... still holds. The 'energy in' part is pretty easy, it's the power going into the compressor + power going into your water pump(s) + power going into your fan(s). The 'power out' part is where the going gets tricky... You will need to know the volume of air coming into the A/C, the temperature of the air and the humidity of the air. You will also need to know the volume of the air coming out (it will be different), the temperature and the humidity. From this you can determine the amount of energy required to cool and dehumidify the air. This is not so trivial. You might get a fair approximation if you know someone who has a similar house, with a similar pattern of power consumption, and who has an A/C that they know the COP and power consumption of. Then you could divide their power consumption by your power consumption and multiply that number (hopefully bigger than 1) by the other person's COP, to approximate your COP. Do the this in late July, early August because then A/C will be the biggest part of the total electric load. * * * When I was calculating the COP of my first heat pump, I had two 55 gallon barrels of water, so all I had to do was measure the change of temperature of the water (it was easy to calculate the weight)... a BTU is the amount of energy to raise one pound of water one degree F. I used a Kill-a-Watt to measure power in. All very straight forward. -AC
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05-12-12, 03:46 PM | #79 |
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im not ever going to be bored enough to calculate cop lol i was hoping it would draw less power but ohh well
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05-12-12, 04:40 PM | #80 |
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How long does it run at 2367.8 watts when it's cooling on a normal day??
If it's on 24/7, then it's going to be costly (to say the least). But, if it only runs for a few hours a day..?. Not a big deal..
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