View single post by Keinokuorma
 Posted: Mon Mar 3rd, 2008 12:28 am
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Keinokuorma



Joined: Mon Jun 26th, 2006
Location: Finland
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Reg wrote:
...if your fridge draws 1.6a @ 120v = 192w 900btu/192w = EER 4.7

1 kilowatt hour of electricity = 3413 British thermal units (Btus) = EER 3.143...

Hmm. And converting the transferred heat to Wh units... that would give us

900BTU/h = 286W
286W / 192W = 1,49 (this would be Wh[heat]/Wh[power] ratio)

Note that the 192W was calculated assuming that it would be the total power of the compressor... it isn't... more likely to be in the class of COS(fi)=0,6 and therefore P[tot] would be about 115W, and the efficiency ratio would be...

286W / 115W = 2,4... sounds like a plausible efficiency ratio...

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Well, there is a lot of confusion among people, about what is power, what is energy, how do they correlate and convert...

WARNING: THE FORTHCOMING TEXT IS MEANT TO BE ILLUSTRATIVE. IT MAY BE INACCURATE OR CONTAIN ERRORS. DO NOT ASSUME THIS TO BE THE ABSOLUTE TRUTH. I HAVE TRIED MY BEST TO USE AS MUCH LAYMAN TERMINOLOGY AS POSSIBLE. MY INTENTION IS NOT TO TRY AND APPEAR SMART, WHICH I AM NOT. CAVEAT READER.

- BTU, J, kJ, Cal, kCal, and kWh are ENERGY units: They are the product of POWER and WORKING TIME, and correlate to the amount of energy consumed or work done by the device in a given time frame. For example, a 1 kW heater that is on for one hour consumes 1 kWh energy and will produce 3143 BTU heat. Those are equal amounts of energy, and as well equal to 3600 kJ or 857 kCal (1 kCal is 1 centigrade in 1 kg of water).

- W, kW, and HP are POWER units, correlating to the power demand or work capability of the device, but unless total working time is known, they don't correlate to the energy consumed or work done by the device. For example, a 1 kW heater will draw 1 kW power and exhaust 3143 BTU heat per hour, but will not consume 1kWh or exhaust 3143 BTU unless left on for a full hour.

- MEAN POWER (W, kW, BTU/h etc) of a device can be calculated by measuring the input or output energy (Ws, Wh, kWh, BTU) and dividing the result by the working time (h, s, etc). For example, a 1kW heater with thermostat, assumed to be more than sufficient to heat your room to the desired temperature, will not be on all of the time, and will not draw 1kW or exhaust 3143 BTU/h mean power.

- MOMENTARY POWER (W, kW etc) can be, and mostly is, higher than mean power. Assume again the 1 kW heater with thermostat, being sufficient for heating your room. It will draw 1 kW power when on, but none when off. If it cycles off for 30% of the time, the ratio between mean and momentary power is 70% (or 0,7) and therefore the mean power would be 0,7 kW. In this example, 2200 BTU/h would be sufficient to keep the temperature while 3143 BTU/h are available if needed.

Obviously, an efficiency ratio is the ratio between output/input power, OR output/input energy when the working time is known. For direct electric heating, you can consider the ratio to be close to but not greater than 1,0. For a heat pump or refrigeration system, their energy transfer capacity is higher than their power demand, and their efficiency ratio (in heating at least) will be greater than 1,0.

Last edited on Mon Mar 3rd, 2008 09:43 am by Keinokuorma



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