 bobbagelsJust Another Scorpion MechwarriorPremium
join:2000-11-15
Matawan, NJ
kudos:2
 ·Verizon Online DSL
 ·Verizon FiOS
| AC /KW? Question for the electricians. Hello hello,
Just a quick question here,
How to figure out kilawats for HVAC
The unit is 30A and 220V
15,000 BTU
I know there is a formula for this but cant remember it.
THX in advance.
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Last night I saw a naked cowgirl, she was floatin across the ceilin |
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 SmokChsrWho let the magic smoke out?Premium
join:2006-03-17
Saint Augustine, FL
 ·Clearwire Wireless
| There are two answers.. Is this for sizing a generator, or just wanting to know approximately how much power it uses?
For a generator you have to account for starting current along with any additional load.
For general load you are looking at around 4.5 to 5 KW for the HVAC depending on it's efficiency.
You'd probably need something like a 10+KW generator to run it.
Formula KW= (BTU/3412) * (1/efficiency) |
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bobbagelsJust Another Scorpion MechwarriorPremium
join:2000-11-15
Matawan, NJ
kudos:2 | Its for just wanting to know approx how many KW it will use.
No generator involved.
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Last night I saw a naked cowgirl, she was floatin across the ceilin |
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TheMGPremium
join:2007-09-04
Canada
kudos:1 | reply to bobbagels
You will need to find information on the unit's efficiency.
Power consumption can not be calculated from the amperage rating, nor can it be calculated from the BTU rating if you don't know the efficiency. |
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 whizkid3Premium,MVM
join:2002-02-21
Queens, NY
kudos:8
4 edits | reply to bobbagels
said by bobbagels:Hello hello,
Just a quick question here,
How to figure out kilawats for HVAC
The unit is 30A and 220V
15,000 BTU The BTU is actually BTU-per-hour. In other words, it provides cooling of 15,000 BTU per hour. This can be converted into watts; but it is the mechanical power - not the electrical power needed to run the unit.
30 amps x 220 volts / (1000 / 1 kilo) = 6.6 kilo-volt-amps (kVA).
For your purposes, 6.6 kVA is approximately equal to 6.6 kW.
There are a few issues.
1. You didn't tell us whether this was 3-phase or single-phase power. I assumed single-phase with the above calculation.
2. 220 volts is not a nominal voltage used in North America. Its more likely 240V single-phase, which is a nominal distribution voltage. If so, 240V should be used in the calculation as per the NEC. The unit is most likely rated for use at 230V. This is the 'nameplate' utilization voltage.
3. You didn't tell use what you are using the result for. If you are trying to determine the amount of power consumed; the answer really doesn't work. The unit will use less than 6.6 kW at full load; because we haven't applied a power factor to convert from kVA to kW. (And we wouldn't know what that is - it can be estimated or tested. The manufacturer will have a curve showing power factor at different loads; but these are almost impossible to get, especially for small units like this.) kVA is excellent for designing electrical systems; because it includes the reactive power that the wiring must be sized for. But the reactive power is not consumed. If you want to know the correct formula (single-phase):
Amps x Volts x power factor / 1000 = kW
said by TheMG:Power consumption can not be calculated from the amperage rating, nor can it be calculated from the BTU rating if you don't know the efficiency.
Exactly. I couldn't have said it better myself.
1 watt = 3.412 BTU/h
15,000 BTU/h / (3412) = 4.40 kW
Note the big difference between what was calculated above. That is the cooling power; not electrical power consumed (at full load). If you can get the efficiency (in percentage); the right formula will be (with 'E' meaning efficiency):
15,000 BTU/h / (3412 x efficiency) = Power consumed (at full load)
Because the air-conditioner does not run at full load continuously - it cycles - this formula will not really tell you anything worthwhile (except for the designers perhaps). There is a much better and easier formula to use:
(BTU/h) / (SEER) = Average power consumed in watts.
This is as good as its going to get for the consumer. Problem is; it doesn't take into account a few critical factors such as:
1. what temperature you like to maintain indoors
2. what are the typical temperatures outdoors over a cooling season
To figure this stuff out, you either need a mechanical engineer or you can the regional 'bin' data, and study the Relationship of SEER to EER and COP section at this link:
»en.wikipedia.org/wiki/Seasonal_e···cy_ratio
Then you can actually figure out what you're usage will be. |
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 tp0dyabbazooiePremium
join:2001-02-13
Carnegie, PA
kudos:2 | Sorry to be a smart ass, but the formula is lookin on the side of the outdoor cooling unit and seeing how many watts its rated at. typically a 30a device does not exceed 24 amps.
-j
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if it aint broke, tweak it!!
currently on FiOS (kick aZZ!) |
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SmokChsrWho let the magic smoke out?Premium
join:2006-03-17
Saint Augustine, FL Reviews:
·Clearwire Wireless
| reply to whizkid3
said by whizkid3:The BTU is actually BTU-per-hour. In other words, it provides cooling of 15,000 BTU per hour. This can be converted into watts; but it is the mechanical power - not the electrical power needed to run the unit.
Humm.. I got to somewhere around 5KW and it only too a few lines.
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whizkid3Premium,MVM
join:2002-02-21
Queens, NY
kudos:8 | reply to tp0d
said by tp0d:but the formula is lookin on the side of the outdoor cooling unit and seeing how many watts its rated at. Not really a formula, but very true.
said by tp0d:typically a 30a device does not exceed 24 amps. 24 amps of continuous load, yes. But air-conditioners such as these are not continuous load devices and will draw more than 24 amps at times; and draw a lot more than 30 when the motor kicks in. So the plug size or even 'MCA' will not tell you much if one is trying to figure out how much energy it consumes. Same thing for the kW rating on the nameplate. It will tell you the full-load real power requirements; but not how much energy it consumes on an average basis, which is the only useful thing in determining how much it costs to run. |
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 eightball0098-Ball Strikes AgainPremium
join:2003-11-03
Lexington Park, MD
 ·Metrocast Commun..
| reply to bobbagels
900 Watts
Power = Current * Voltage
Find out what you are paying per kWh for power and that will be about how much it will cost you to run your HVAC for an hour.
That 900 W is probably just for the indoor unit, what about the outdoor unit?
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For those who fight for freedom, life has a special flavor the protected will never know |
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 SparkChaserR.I.P. DocPremium
join:2000-06-06
Downingtown, PA
kudos:3 | Where'd 900 come from? |
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SmokChsrWho let the magic smoke out?Premium
join:2006-03-17
Saint Augustine, FL Reviews:
·Clearwire Wireless
| reply to eightball009
Ummm.. Not sure where you came up with 900W. The 220V & 30A the OP gave would come out to 6,600W, at least on my calculator.
To go a bit further, it's most likely a 240V 30A circuit. If it's on a 30A circuit, that would say that the unit should draw less than 30 Amps. (WhizKid3 will know the % of overrate the circuit needs to be respective to the load)
Also P=I*E is correct for DC circuits and AC circuits with a Power Factor (PF) of 1.00. Otherwise to determine the power used by an AC circuit it's P=I*E*PF. An AC unit with a compressor and fan motor (both highly inductive loads) is not going to have a PF of 1.
While it could be a separate unit, at 15,000 BTU (1.25 tons) it's most likely a window or portable unit. I don't even know if they make central units that small, except for refrigeration applications. |
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eightball0098-Ball Strikes AgainPremium
join:2003-11-03
Lexington Park, MD Reviews:
·Metrocast Commun..
| reply to SparkChaser
I don't know where I got that number. I didn't recheck it before I posted.
SmokeChaser, how does the power factor fit in? That is the efficiency of the system. Wouldn't the RMS voltage be the right voltage to use?
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For those who fight for freedom, life has a special flavor the protected will never know |
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SparkChaserR.I.P. DocPremium
join:2000-06-06
Downingtown, PA
kudos:3 | PF and RMS are 2 totally different things. PF is the COS of the phase shift between V and I |
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SmokChsrWho let the magic smoke out?Premium
join:2006-03-17
Saint Augustine, FL Reviews:
·Clearwire Wireless
| reply to eightball009
said by eightball009:SmokeChaser, how does the power factor fit in? That is the efficiency of the system. Wouldn't the RMS voltage be the right voltage to use?
No, Power factor and efficiency are totally different things. If the load is capacitive or inductive it will have a power factor of less than 1. Only a circuit that appears purely resistive will have a PF of 1. The PF does not have anything to do with the efficiency at all. This is why in AC circuits we refer to "apparent power", that is the amount of power it looks like the circuit is using going by the P=IE formula. To determine the "True Power" being consumed you also have to determine the PF and put that into the equation. The power factor is determined by the amount of phase shift between the voltage & current. To explain in detail would take quite a bit of typing.
Yes, RMS voltage is correct. I was just saying, as a correction to the OP, if it's in the USA, it's actually 240V not 220V that should be the nameplate rating. I was not using the peak voltage (339V) just merely correcting the stated RMS value. |
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eightball0098-Ball Strikes AgainPremium
join:2003-11-03
Lexington Park, MD | Sorry Bobbagles for the hijack.
Smoke,
I know what you are talking about now and am a little familiar with that calculation.
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For those who fight for freedom, life has a special flavor the protected will never know |
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1 edit | reply to bobbagels
As previously said, you need to know the efficiency.
Assuming a typical 14[SEER]
P=15,000/14=1071W (1.07kW)
This varies a bit with the temperature in/out, but you have a ballpark number.
Power factor doesn't matter in this case. |
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| reply to bobbagels
You could turn off everything in your house except the air-conditioner, then (with the air-conditioner running, of course) time how long it takes the wheel in your electric meter to make one revolution. Then Google for a spreadsheet that makes the conversion between dial speed and kW.
I timed 10 revolutions to make it easier. My air-conditioning uses 6 kW. |
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harald
join:2010-10-22
Columbus, OH
kudos:1 | reply to bobbagels
It also depends upon the phase of the moon and what astrological sign your mother was born under.
I can see that y'all are having trouble making this complicated enough for the techies. |
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 Jack_in_VAPremium
join:2007-11-26
Mathews, VA
kudos:1 | said by harald:It also depends upon the phase of the moon and what astrological sign your mother was born under.
I can see that y'all are having trouble making this complicated enough for the techies.
I was thinking the very same thing.  |
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ncbillPremium
join:2007-01-23
Winston Salem, NC | reply to bobbagels
Why worry about power factor?
Residential users don't pay based on PF. |
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