Electrical costs for my pool

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Originally Posted by JohnT

KWH is approximately (Amps X Volts divided by 1000) X Hours. At max current, your motor would use 12AX120V=1440W Dividing by 1000 gives 1.44KW X 24Hours= 34.56KWH per day, so at $0.083 per KWH, you are using $2.86 worth of electricity a day, or roughly $86 per month. Double that if that is a 220 motor.

The plate current is max, and you probably aren't using that. The only way to know how much current you are using is to measure it.

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For the same HP motor operating at 230v, it will use half the amps as 115v so the total power is the same.

JohnT, you would not have to double the cost for a 220 volt motor, as a 220 volt would use about half the amps. Therefore the cost would be approximately the same regardless of 120 or 220 volt motor.

Cheers

Quote:

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Originally Posted by hancop

JohnT, you would not have to double the cost for a 220 volt motor, as a 220 volt would use about half the amps. Therefore the cost would be approximately the same regardless of 120 or 220 volt motor.

Cheers

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I was referring to the question that only gave amps and didn't mention whether it was 120 or 220. You are correct, but at a given current, 220 would essentially be double the power.

Quote:

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Originally Posted by JohnT

KWH is approximately (Amps X Volts divided by 1000) X Hours. At max current, your motor would use 12AX120V=1440W Dividing by 1000 gives 1.44KW X 24Hours= 34.56KWH per day, so at $0.083 per KWH, you are using $2.86 worth of electricity a day, or roughly $86 per month. Double that if that is a 220 motor.

The plate current is max, and you probably aren't using that. The only way to know how much current you are using is to measure it.

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Thanks.

I was always under the impression that I was spending around $35 per month. This makes 12 hour run time look much more attractive.

If you decide to use your house electric meter, here is a formula you can use:

kwh = 3.6 * Kh * R / T

Kh is the meter constant and usually printed on the meter. Most are 7.2.
R is the number revolutions you are timing. I use 10 when the pump is on and 1 when it is off.
T is the time in seconds for R revolutions.

If you measure with the pump off and then on and subtract the two, you get a pretty accurate number for just the pump.

my electric meter gives the reading for KWhrs but I have so much stuff going on generally it wouldn't be very practical to try and shut everything down and monitor the pool pump or AC. Currently we are using about 100 KWhrs a day, my last electric bill was $811 a personal record :)

Based on the estimated cost arrived at above, my pump accounts for 34% of my electric bill.

Timer here we come :)

Wow. Glad I posted a popular topic. I wasn't aware of the Power Factor. And it sounds like I need to do some more research on calculating motor costs.

I've found several articles on the web to read, but alas, my work is calling me... In addition to the PF number, there also seems to be a .00173 number that is also being thrown in. This was used to calculate kVA. Siting from http://www.cop.com/SEUhtmDOCS/SEU17.htm, the formulas are:
kVA = 0.00173 * V * amps
kw = kVA * PF * .01 (PF being % between 1-100)
running for one hour would be kwh and then multiply by number of hours times cost per kwh.

Now, looking at the calculations above and the linked site, they are using the stated amps on the nameplate, where I am using the measured amps on my amp meter. It looks like the only place they're using the measured amps is to help calculate the PF.

So what is the 0.00173 number? If I throw in the .00173 number and a 90% power factor, I'm actually increasing my kwh because the .00173 increases more than the .9 decreases it.

Robert

Just FYI at 60 GPM on my Intelliflo I run 1445 watts per hr. and it takes approx 11 hours to turn the pool over 1 time per day. If I drop it down to 40 GPM then I use 440 watts. The problem is the Trio Pure needs 55 GPM to work well.
No Bragging just sharing some options, the big issue is the $1250.00 price tag for the pump.

Does the intelliflow give you the watts or are you measuring it somehow?

Thanks,
Robert