Hi guys,
Probably simple but I am in mental block territory here. I have a smallish house with two in wall AC 12,000BTU units and they do a
pretty good job as I can zone them and not have to run both at once.
I am trying to get an idea of cost so I timed the wheel on the meter and it took 350 seconds for 10 rotations with no AC running. I then turned on one AC and it took 150 seconds for 10 turns.
Dividing my electrical bill by the total number of kWh for last month gives me \$0.092 per kWh.
So how do I calculate the cost of that AC load?
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On Mon, 13 Jul 2009 16:24:12 -0700, Dave, I can't do that wrote:

Feeling the heat?. Wouldn't mind that problem here in NZ at the moment!. Revolutions per kWh are marked on the faceplate of your meter. The reciprocal of this is kWh/revolution. Ten times that is what you measured. Of your total load as measured, the portion drawn by the AC unit when running is 150/(150+350)= 0.3. How much the unit costs to run depends on duty cycle or, if it has a variable speed compressor, on compressor speed.
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Hi Guys,
Thanks for that information but this Westinghouse type D4 S Form 25 meter does not have the kWh/Rev on it. The wheel is graduated from 0 to 100 though. It is probably an old meter as it has the 5 little round dials for the readings. It says 240V 3W and to one side it says Kh 7.2. To the right of the wheel it says 2A. To the left it has 200 CL and at the top right it shows 13 8/9
It has cooled down now so tomorrow I will measure the time for x- revolutions with AC off and again on as Malcolm suggests and will post the results.
Because the heat load was so high when I measured this today, the compressor did not cycle and was full on during the measuring. The inside temp was 86 when I started measuring and it did not cycle off until it was down to 80 about 40 minutes later. By then I had taken all the timings.
Roger, sorry it is a bit brisk down there in NZ. I skied at Mt Cook many years ago and had a great time. You Kiwis know how to party.
Dave
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Is there a plate or sticker on the AC units listing the electrical specifications? If yes, that will give you the power requirements. KW = V*A/1000. That number running for 1 hour is your KWH usage. Art
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On Mon, 13 Jul 2009 19:19:24 -0700 (PDT), David41616

Hi again, according to
<http://www.cdfa.ca.gov/dms/programs/ctep/Checklists/WatthourChecklist.pdf
Kh is "watt-hour constant (disk constant). The watt-hour constant of a meter is the registration of one revolution of its disk expressed in watt-hours. The constant is usually identified by the symbol Kh"
so
1/Kh is the same as revs/Wh.
1/Kh = 1/7.2 = 0.139 revs/Wh
which equals 139 revs/kWh
Calculating as in my previous post but using the correct figure for your meter
energy used in an hour equals 136.8 / 139 = 0.98 kWh
You're paying \$0.092/kWh so it costs 0.092 x 0.98 = \$0.0902 per hour
You don't need to count again for 350 seconds, because I previously converted that from your 150 seconds count. -- Regards Malcolm Remove sharp objects to get a valid e-mail address
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David41616 wrote:

Kh is the watt-hours per turn.

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Paul Hovnanian mailto: snipped-for-privacy@Hovnanian.com
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On Mon, 13 Jul 2009 16:24:12 -0700 (PDT), "Dave, I can't do that"

You need to know how much energy each revolution of the disc represents.
The name plate of the meter will have that stated on it. eg my meter here says "187.5 revs per kWh".
It's easier if you measure the time for a given number of revolutions. Using your figures above and converting, in the second case if you had counted the number of turns in 350 seconds you would have got 23.3 turns. The load due to the AC is the difference 23.3 - 10 = 13.3 turns
350 seconds is 350/3600 of an hour so if you had counted for an hour you should have got 136.8 turns.
Pretending that figure came from my meter (use your meters figure here) energy used in an hour equals 136.8 / 187.5 = 0.73 kWh
You're paying \$0.092/kWh so it costs 0.092 x 0.73 = \$0.0671 per hour
Of course if the AC cycles on & off the cost will be lower.
-- Regards Malcolm Remove sharp objects to get a valid e-mail address
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Malcolm Moore wrote:

Sometimes refered to as Rr. The meter might have a Kh constant, which is watt-hours per turn. So Rr of 187.5 is Kh 5.333.
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Paul Hovnanian mailto: snipped-for-privacy@Hovnanian.com
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I'd say that you need to look at the bigger picture, looking at all the details leaves too many uncertainties, especially duty cycle. You really want to know what does it costs to run your AC for month or season. So, plug it into a Kill-a-Watt & measure the power (err, "energy" (kwh)) it uses directly. Anything else is going to be a WAG.
Bob
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Bob beat me too it.
Your problem is that the actual current draw when running is an interesting number but your cost is a function of indoor vs outdoor temps over a 24 hour period.
If you need a rough estimate, a fairly new (less than 5 years old) 12,000 BTU unit will pull around 1400 or 1500 watts. That would draw about 6 amps on 240 volts, double that on 120 volts. For more on that http://en.wikipedia.org/wiki/Energy_efficiency_ratio
Get a Kill-a-Watt power meter. Run it for measured 24 hour periods. Multiply the results by your \$/kwh from your electrical bill. (Amazon.com product link shortened)
Dave, I can't do that wrote:

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Thanks Guys, especially Malcolm.
I realize the cycle times (forced by temps) will influence the average cost but I want a worst case scenario and that is with the compressor on all the time.
I am trying to decide if I should continue with the AC units or install a swamp cooler. The swamp I need is 5000cfm and about \$500 to buy. That 500-bucks buys a lot of power for the AC as I usually only have one of the 2 units running at any time. I am trying to work out a rough break-even guide. I am not including the cost of running the swamp as at this point I am only interested in ball park figures.
I can't find the compliance plate so I am guessing I would need to remove the unit from the wall box so will pass on that one. I also cannot find any model number without pulling it out of the wall but I am pretty sure it is 12,000BTU. It is running off a standard 110v outlet with 20A breaker, it is a Haier unit and 2 years old.
I do not want to buy a kill-a-watt just for this one shot use as normally I don't care how much power something is using. If I want it on I want it on. :)
Another option is to split one of the wires out of the dryer-style flat cable and use a clip on meter to measure current. However for what I want I think Malcolm's calculations will suffice. Having said that though, I am surprised that it is only costing \$0.09 an hour to run. That would make buying the swamp redundant. By around 6:00pm it is cooling down enough to turn off the AC so that unit is costing me about \$0.50 a day.
At that rate a break even would be 40 days x \$0.50/day = \$20.00 per year and \$500 / \$20 = 25 years so AC it is.
Thanks to everyone who helped. As usual a great group.
Dave
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1 BTU equals 0.0002928 kilowatt-hour
1 BTU/minute equals 0.01757 kilowatt.
12,000 BTU = 3.5136 KwH (14.64 amps @ 240 volts).
So worst case about 30 cents/hr BUT a heat pump (which is what this is) is more effective than straight consumption in, say, an electric radiator, so less than that. Still, a whole house fan is cheaper to run and often as effective at night.
BTW, it wasn't the best idea for the Kiwis to sell everything off to the wide boys in private enterprise it seems. All the dams and distribution built and paid for with taxpayer dollars and now who gets the profits?
And 240 volts? It used to be 230 in the land of the long white cloud! The Ockers ran 240.
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Guys, just as a follow up at 120.6V (measured) the clamp on is showing 10.9A with everything under full load.
120.6V x 10.9A = 1314W 1314W / 1000 = 1.314 kW 1.314kW x \$0.092/kWh = \$1.21/hour
Is that correct?
Thanks
Dave
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You slipped a decimal point. \$0.12/hr Art
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wrote:

Also,
120.6V x 10.9A = 1314VA NOT 1314W
The watthour meter measures real watts, which is what is being charged fo by the power company. The clampmeter is measuring total current. Because the AC is a motor it has a lagging power factor, which means the VA is greater than the watts.
-- Regards Malcolm Remove sharp objects to get a valid e-mail address
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Thanks for the corrections Malcolm and Artemus,
Pretty obvious where my skills and talents ain't. :)
That skipped decimal did have me scratching my head as the swamp was suddenly looking a lot better BUT I could not see how the meter and Malcolm's calculations could be that far off.
Here I sit in air conditioned comfort typing this reply. :D
Many thanks, again.
Dave
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Swamp coolers are AWFUL!!! They are cheap to buy, constant maintenance, are not very effective at many times of the year.
I would have calculated about \$.13 per hour without any measurements so your other calcs are right in line with that.
David41616 wrote:

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Thanks Roy,but where I live max humidity is around 28% and swamps work quite well here. I only need electric cooling for about 40 days a year. All of July and a few days either side but at \$0.12/hour I will stay with the AC units.
Dave
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I have a friend in Albuquerque who was doing the same calcs on his swamp cooler vs a new central air. He finally opted for the central air, then kept asking me why I wasn't MORE vocal about doing it sooner.
I don't care what the relative humidity is, what's important is the dew point. You need to have a dew point low enough to be able to make the temperature drop vs dew point rise equation to work out with a reasonable final temp and reasonable final dew point. Albuquerque is fine most of the hot months but the prevailing winds shift around for a couple of months, make the swamp cooler equation not work well.
David41616 wrote:

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Hi,     A few years back I was looking to replace the rusted out swamp cooler and the 20 year old AC. After a lot of thinking, I decided against having a Swamp cooler and went with a very high efficiency AC/ furnace unit.
Strikes against the swap cooler were: They use a fair amount of electrical power because they run constantly. They use a fair amount of water. They always need attention.
The better new AC units are pretty good. It use to cost over \$150-200 to run the AC in July when the swamp cooler does not work well, now it costs less then \$40. This is in Tucson in a moderately well insulated house. We keep the house cooler during the day now, because my wife is no longer working. The reduced power bills were almost shocking, but the AC unit was expensive.
Thanks Roger Haar Tucson
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