Your home AC prep for somer

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It seems that every once awhile some uniform prospective buyers are giving wrong information on efficiency of Air Condition systems. Let me inject piece of my mind on efficiency of any Refrigeration unit, Air-Condition or Low temperature equipment. I will estimate and go on lame to say that 80% of efficiency comes from Condenser the other 20% is in compression ratio at which compressor operates. Because R-410 operating at much higher pressures, compressors it self pumping ratio is reduced, there for compressor efficiency is increased, not only that but compression ratio playing large part in life expectancy of compressor itself. For this to be achieved condenser must have adequate proper ratings to get ready of heat generated by compressors pump. From my 40 years in field I have found just about on all older system condensers "are" under size, why well it is lot chipper to manufacture. By switching to new systems, gains are almost immediately noticeable. Remember the efficiency does not come from Refrigerant; efficiency is increased because compressor is running at much lower compression ratio. Efficiency come from system at hole and not from Refrigerant alone or Compressor. We are going in winter we are not need of AC but every one should keep in mind that spring will be here soon enough, make sure that your condensers are clean from debris and not blocked for free air circulation.
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There is no such thing as "efficiency " for ANY heat pump. There is a Coefficient Of Performance, a different thing. http://en.wikipedia.org/wiki/Coefficient_of_performance
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Wrong again. Actually there is such a thing. In the USA all such systems have a Seasonal Energy Efficiency Rating which is a key spec listed on every system. It's a guide to how much cooling cooling you get for the amount of electric power consumed. A higher SEER system is more efficient.
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I helped change an outdoor unit, maybe ten years ago. The old one coming out was a heat pump unit with a piston compressor. We put in AC outdoor unit with new rotary scroll compressor. The customer remarked how quiet it was. The old unit, so noisy you could hardly be outdoors when it was running. I arrived after the old unit died, so I didn't get to take amp draw reading. I'm sure the new unit was more efficient.
Christopher A. Young Learn more about Jesus www.lds.org .
In the USA all such systems have a Seasonal Energy Efficiency Rating which is a key spec listed on every system. It's a guide to how much cooling cooling you get for the amount of electric power consumed. A higher SEER system is more efficient.
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On Nov 30, 1:21 pm, "Stormin Mormon"

No, not more efficient. The extra energy does not arise from converting electricity to heat. The "extra" output heat energy is not created from another sort of energy, it is transferred.
So all the electricity put into the system is converted to heat regardless. (100% efficient in all cases.) The extra heat is transferred through the system.
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On Nov 30, 1:21 pm, "Stormin Mormon"

No it wasn't. In all cases, all the electricity was converted to heat. All electricity>heat conversions are 100% efficient.
The coefficient of performance may have been better.
You clearly don't understand the technology.
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Google SEER moron and tell us what the second E stands for. The efficiency being measured is how much COLD you get for a given amount of electricity put into the system. COP is one such measure. In fact the root, "efficient" appears in "coefficient" of performance too.
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wrote:

I assume it's for the benefit of ignorant USAians who were never educated. http://en.wikipedia.org/wiki/Coefficient_of_performance.
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The unit I replaced, was using electricity to move refrigerant, used for air conditioning. I didn't get any numbers, but I'm sure the new unit was much more efficient.
You, Harry, are describing filament heaters.
Who didn't understand the technology, Harry?
Christopher A. Young Learn more about Jesus www.lds.org .

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On Dec 1, 5:56 pm, "Stormin Mormon"

It makes no difference whether it's an absorption or compressor system. The principle remains.
And absorption systems are pretty useless.
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Harry is talking about filament heaters. I'm talking about air conditioning. Perhaps Trader and I can agree that Harry doesn't understand the technology?
Christopher A. Young Learn more about Jesus www.lds.org .

Google SEER moron and tell us what the second E stands for. The efficiency being measured is how much COLD you get for a given amount of electricity put into the system. COP is one such measure. In fact the root, "efficient" appears in "coefficient" of performance too.
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On Dec 1, 12:57 pm, "Stormin Mormon"

Harry has proven time and time again that he doesn't know his backside from a hole in the ground. The best part on this one is he's running around saying there is no such thing as efficiency of an AC unit. Then he cites "coefficient of performance". What root does "coefficient" come from? answer: efficient COP is one measure, SEER is another.
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On Fri, 30 Nov 2012 23:09:28 -0800 (PST), harry

Seasonal Energy Efficiency Ratio - and is THE accepted measurement for air conditioning. The subject is "your home AC prop for somer (mis-spelled)" - so EFFICIENCY (SEER) is correct. And SEER IS also the accepted rating for COOLING efficiency of heat pumps.
The OPsaid he replaced the heat pump with an " AC outdoor unit with a new rotary scroll compressor. " - not a heat pump so any of the accepted ratings would be correct.
Other ratings for HVAC equipment are: EER (energy efficiency Ratio) and HSPF (Heating Season PerformanceFactor) and , yes, COP (Coefficient ofPerformance).
****The Coefficient of Performance - COP - is the ratio of heat output to the amount of energy input of a heat pump. COP can be expressed as
COP = hh / hw (1)
where
COP = Coefficient of Performance
hh = heat produced (Btu/h)
hw = equivalent electric energy input (Btu/h) = 3413 Pw
where
Pw = electrical input energy (W)
If a heat pump delivers 3 units of heat for every unit of energy input - the COP is 3.
•1 kW = 1000 W = 3413 Btu/h
**Example - COP Heat Pump *Cooling Cycle A heat pump delivering 60000 Btu/h with a total input of 9 kW:
COP = 60000 (Btu/h) / (3413 9 (kW))
= 1.95
*Heating Cycle A heat pump delivering 50000 Btu/h with a total input of 7 kW:
COP = 50000 (Btu/h) / (3413 7 (kW))
= 2.1
****The Energy Efficiency Ratio - EER - measures the cooling efficiency of a heat pump.
EER can be expressed as
EER = hc / Pw (2)
where
EER = Energy Efficiency Rating
hc = cooling heat (Btu/h)
Pw = electrical power (W)
***Example - EER An air conditioner or heat pump in cooling modus draws 1000 W to produce 10000 Btu/h cooling. The EER can be calculated as
EER = 10000 (Btu/h) / 1000 (W)
= 10
****The Heating Season Performance Factor - HSPF - is a measure of the overall heating efficiency of a heat pump during a season.
HSPF = hs / 1000 Pws (3)
where
hs = heat produced during the season (Btu)
Pws = electrical power consumed during the season (kWh)
The HSPF can be regarded as an "average" COP for an entire heating season. It is common to compare BTUs of heat output to watts of electrical energy input. HSPF of 6.8 can be compared with an average COP of 2 and a HSPF in the range of 5-7 is acceptable.
***Example - Heat Pump Heating Season Performance Factor For a heat pump delivering 120,000,000 Btu during the season when consuming 15,000 kWh the HSPF can be calculated as
HSPF = 120000000 (Btu) / (1000 15000 (kWh))
= 8
****Seasonal Energy Efficiency Ratio is a measure of the seasonal cooling efficiency of a heat pump or a consumer central air conditioning system. Basically EER factored over the entire cooling season.
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On Dec 1, 1:49 pm, snipped-for-privacy@snyder.on.ca wrote:

Ain't all that just a long way of saying that harry is a big dummy?
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He may well have a lot of talents, in other field. But, on this list, he hasn't done well.
Christopher A. Young Learn more about Jesus www.lds.org .
Ain't all that just a long way of saying that harry is a big dummy?
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If memory serves, the OP, Grumpy, said to keep leaves away from the outdoor unit.
As for me, the outdoor unit I replaced was pretty ancient. The new one was much quieter, and I believe far less current draw.
Christopher A. Young Learn more about Jesus www.lds.org .

Seasonal Energy Efficiency Ratio - and is THE accepted measurement for air conditioning. The subject is "your home AC prop for somer (mis-spelled)" - so EFFICIENCY (SEER) is correct. And SEER IS also the accepted rating for COOLING efficiency of heat pumps.
The OPsaid he replaced the heat pump with an " AC outdoor unit with a new rotary scroll compressor. " - not a heat pump so any of the accepted ratings would be correct.
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On Dec 1, 10:11 pm, "Stormin Mormon"

The OP is another American Idiot. There is no difference in principle between any sort of compressor, scroll/reciprocating/rotary. They all do the same thing. Some are cheaper/more compact/last longer than others. But all the electricity they use ends up as heat and is therefore 100% efficient. (As it is in an absorption system too) No other heat is created/can be created, it is merely moved around. So no efficiency is involved.
Read the link in the previous post and try to comprehend.
The COP is limited by the temperature difference indoor/outdoor and the refrigerant gas, assuming correct installation..
And there is no difference between heating and cooling roles. Any apparent difference is caused by the size of the heat exchangers installed. (Evaporator/condensor)
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But anyone with a brain isn't interested in the heat that's wasted. We're interested in how much cooling we get compared to how much electricity is used. With the old system Stormin was talking about, we might get 40,000BTUs/hr using $1 of electricity. Replace it with a newer, higher efficiency system, and we get 40,000BTU/hr using only $.70 The new system is more efficient at producing cooling. Capiche?
(As it is in an absorption system too)

Coefficient of Performance is one measure of efficiency. In fact the word "coefficient" is derived from the root "efficient. Have you Googled SEER, as requested? What is the third word? EFFICIENCY
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On Sat, 1 Dec 2012 23:05:46 -0800 (PST), harry

And you, sir, are thick-skilled and arguementative.
when one is attempting to do anything other than heat, conversion of electricity to heat lowers the efficiency of the "machine" or "system" - and when running an external compressor as a heat pump, any heat produced by the compressor and motor which is located outside the insulated envelope you are attemptintg to heat is a TOTAL LOSS, and therefore reduces the efficiency of the system. If you believe otherwise, YOU are the "non-american" idiot. Sorry, but I have to call a spade a spade.

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I know your English is poor but read the link I posted and try to understand.
If you put a gallon of petrol in your car you might get out 7% of useful energy. Efficiency =7%
If you put a gallon of oil in your heating system, you might get 75% of available energy as heat out of it. Efficiency =75%
If you put in one KWh of electricity into a heat pump, you might get out four KWh. So all the heat is not coming from the electricity, clearly you can't get out more than you put in.
So this is not a matter of efficiency. Is this hard to understand?
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