# Electrical usage

• posted on March 29, 2010, 12:11 pm

What is the formula to determine the amount of electricity used by different size motors or appliances.
Thanks Mike
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• posted on March 29, 2010, 12:17 pm

I'm not aware of a formula. You need to know stuff about the motor as well as the load that is on it. Unless you mean volts x amps = watts?
You're better off getting a clamp on amp meter and noting the usage while the appliance is running.
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• posted on March 29, 2010, 12:27 pm

There is no formula for figuring out appliances like a refrigerator, a Kill-A-Watt meter works well and is accurate, [although i just bought a defect a HD] my old KAW works well. New apliances like friges can be 75% more efficent than units made maybe 20 years ago, with defrost cycles and old compressors pulling more at end of life you need something to measure consumption over a day at least to get an accurate overall picture. For a motor use a clamp on amp meter.
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• posted on March 29, 2010, 1:05 pm
ransley wrote:

...
Or read the nameplate???
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• posted on March 29, 2010, 2:52 pm
For a general idea. But, individual things like dry berrings change the actual use.
--
Christopher A. Young
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• posted on March 29, 2010, 4:11 pm
On 3/29/2010 6:52 AM Stormin Mormon spake thus:

*Berrings*???
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- a Usenet "apology"
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• posted on March 29, 2010, 4:13 pm
Yes, typically a braas sleeve that supports the irin or steal armiteur shaft. You should goggle it.
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Christopher A. Young
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• posted on March 29, 2010, 4:39 pm
dpb wrote the following:

You mean the Energy Star tag? ::-)
--

Bill
In Hamptonburgh, NY
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• posted on March 29, 2010, 1:17 pm
c1gmlm wrote:

For single-phase motors,
HP = V I Eff Pf/746
where V = input voltage (V), I = current drawn (A), Eff = efficiency (fraction), and Pf = power factor (fraction)
For theoretical purposes you can use nameplate values and nominal voltage and get estimate of current draw assuming the nameplate has necessary data for Eff/Pf; (if they aren't given it's often because they're low... :) )
If you're after normal operation consumption you'll have to measure the inputs as nameplate ratings are at max and you'll undoubtedly not know the other factors precisely...
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• posted on March 29, 2010, 1:24 pm
c1gmlm wrote:

Depends on if you want to know the power consumption, amperage or kilowatt hours.
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LSMFT

I'm trying to think but nothing happens.........
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• posted on March 29, 2010, 1:29 pm
wrote:

My formula is- (plug in Kill-o-watt meter), read display.
In the case of refrigerators, freezers, washing machines, or anything that doesn't use a constant amount of electricity, I leave it on for a few days and note high usage & overall usage.
Is this just an intellectual exercise, or do you have a specific problem?
Jim
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<%-name%>
• posted on March 29, 2010, 2:51 pm
Amps x volts = watts
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Christopher A. Young
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<%-name%>
• posted on March 29, 2010, 4:33 pm
On 3/29/2010 6:51 AM Stormin Mormon spake thus:

Well, sorta.
That formula (P = V x A) works for DC, but needs more terms to be accurate for AC (power factor, etc.) But it gets you into the ballpark.
--
You were wrong, and I'm man enough to admit it.

- a Usenet "apology"
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• posted on March 29, 2010, 4:13 pm
Red hot with ketchup, and a diet Coke for me. Side of fries. While I'm in the ball park.
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Christopher A. Young
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<%-name%>
• posted on March 29, 2010, 10:14 pm
On 3/29/2010 8:13 AM Stormin Mormon spake thus:

OK, wise guy: since you brought it up (or ran with it in this thread), try this on for size:
Try to sing the following well-known song that's been phase-changed just a little bit (a music teacher friend used to use this to teach his grade-school kids simple music theory):
Me out to the ball game, take Me out to the game, buy Me some peanuts and crackerjacks, I Don't care if I never get back, 'cause it's Root, root, root for the home team if They don't win it's a shame 'cause It's one, two, three strikes you're out of The old ball game ...
Not easy, eh?
(Some of the words may not be exactly right, but the meter is basically correct)
--
You were wrong, and I'm man enough to admit it.

- a Usenet "apology"
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• posted on March 29, 2010, 2:58 pm

Mike; the amount of electricity used by anything electrical is measured in watts.
For example a 100 watt electric light bulb switched on for one hour consumes 100 watt hours which is equal to one tenth of a kilowatt hour. If there are ten 100 watt bulbs on for the hour they will use 10 x 100 = 1000 watt hours = One kilowatt hour.
When it comes to electric motors, fridges etc. the rating (under normal load conditions) will be either in Horsepower or the volts and amps will be on the nameplate. A table saw for example may have a half- horsepower motor? One horsepower (HP) is very close to 750 watts, so again if the half-horse motor runs for one hour it will use 0.375 kilowatt hours of electricity.
A coffee maker may be rated at at say 3 amps; amps times the voltage = watts. So if the coffeemaker bubbles away plugged into a regular 115 volt North American wall outlet, continuously for one hour (strong coffee eh!) it will use 115 x 3 = 345 watt hours or 0.345 kilowatt hours of electrcity.
We pay about ten cents per kilowatt hour for electrcity here; your rate/mileage may vary!
So the 100 watt bulb would cost me one cent per hour. Ten 100 watt bulbs would cost me ten cents per hour. The half-horse motor will cost about 3.75 cents per hour; of course if I'm cutting oak instead of pine the harder wood will probably use more power! The coffee maker about 3.45 cents per hour.
Not everything runs by the hour of course, so if I only run my table saw for six minutes (one tenth of an hour) ............. Or if it takes twenty minutes (one third of an hour) to make coffee .................... Well work it out.
Any help?
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• posted on March 29, 2010, 5:02 pm

Kwh is exactly what it says. Kw (watts/1000) X hours run.Watts is only Amps X Volts in the case of resistive loads (heating or tunsten/ incandescent lamps.
If you have a inductive load, Kw = volts X amps X the cosine of the phase angle between them. The cosine of the angle between them is known as the power factor, it may be marked on the motor, it's usually around 0.7
With resistive loads, the power factor is 1 (ie Volts and Amps are in phase) and so is ignored. So, if you ran a 500w (0.5Kw) motor for 2 hours it would use 1 Kwh. If you ran a 750w (0.75Kw) motor for 1 hour it would use 0.75Kwh. Assuming they were fully loaded up that is, the Kw given is the maximum power they can draw.
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• posted on March 29, 2010, 6:17 pm
In typed:

Depends: Startup current/watts, or running current/watts? What do you YOU consider to be an AMOUNT of electricity? Watts? Watt-hours? Etc.?
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• posted on March 30, 2010, 2:46 am
wrote:

Electricity is purchased by measuring kilowatt-hours. For example, a 1000 watt appliance (on high) for one hour will have used 1 kilowatt-hour, or a 500 watt appliance on for 2 hours will also use 1 kilowatt-hour, etc. If you have several items to check, get a killowatt meter.