1 edit | reply to djr777
Re: Pool questions....pump, power usage, how to find a leak? said by djr777:The pump is a Hayward Super II Pump Series. I was running this 4-6 hours a day and my pool looked great. It is either 1 or 1 1/2 HP » www.haywardnet.com/inground/prod···r_ii.cfmDo you know how much this might run monthly at 4-6 hours a day? A 1hp pump uses 746 watts.
746 watts x 6 hours per day = 4.476 KWh per day.
If we multiply that by 30 you get 134.28 KWh per month.
According to the EIA the average cost of electricity in California is 14 cents per kWh.
If we assume she's on the high side and is paying 20 cents per kWh it would only cost $27 per month if the pump runs for 6 hours per day.
If the timer isn't working and it's actually running 24 hours per day then it would cost around $107 per month for the 1 hp pump.
Add 50% to those figures if it's a 1.5hp pump. |
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 whizkid3Premium,MVM
join:2002-02-21
Queens, NY
kudos:8 | said by x_:A 1hp pump uses 746 watts. That is not correct. What is true, is that 1hp equals 746 watts (both are measurements of power). A motor that puts out 1 hp of mechanical energy, puts out 746 watts (of mechanical energy). All motors have innefficiencies, so a motor that puts out 1 hp, draws more than 746 watts. The efficiency of any motor depends on its load, as well.
This pump, depending on its load profile, could be drawing anywhere from 825 watts to 1500 watts or more (90% to 50% efficient).
Their electric bill is $500 a month, because they are wasting energy and everyone's electric bills have gone up. Their bill is high because they are using too much air-conditioning. Its almost winter - tell them to turn it off. They are damaging your pool. Either way, time to find a new tenant. |
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 djr777Premium
join:2005-01-25
Pacific Grove, CA | reply to x_
I talked to Hayward pump tech support and Ike gave me the max amount that the pump could draw. It could be less but I would need a clamp to check it exactly. So here are the max amounts.
9.3 amps x 240v = 2232 watts divided by 1000 is 2.32 kilowatts.
2.32 kilowatts x .14 cents = .31 cents an hour.
If she runs it for 6hrs a day it would cost 1.86 per day.
1.86 x 30 days = 55.80 a day.
Not a lot to pay if you enjoy a good splash now and then. 
--
...there will be an answer. let it be |
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robbinPremium,MVM
join:2000-09-21
Leander, TX
kudos:1 | said by djr777:I talked to Hayward pump tech support and Ike gave me the max amount that the pump could draw. It could be less but I would need a clamp to check it exactly. So here are the max amounts. 9.3 amps x 240v = 2232 watts divided by 1000 is 2.32 kilowatts. 2.32 kilowatts x .14 cents = .31 cents an hour. If she runs it for 6hrs a day it would cost 1.86 per day. 1.86 x 30 days = 55.80 a month. Not a lot to pay if you enjoy a good splash now and then. Fixed it for you.  |
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whizkid3Premium,MVM
join:2002-02-21
Queens, NY
kudos:8 | reply to djr777
9.3 amps x 240v x power factor= ???? watts
The power factor of the pump is likely 0.8 or less. This means that it is likely using less energy, and not costing nearly that much, although its still not a bargain.
55.80 x 0.8 pf = $44.64 a month |
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4 edits | Edit: Removed post |
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whizkid3Premium,MVM
join:2002-02-21
Queens, NY
kudos:8
4 edits | said by x_:Lower power factors results in more power required not less. Power factor has nothing to do with efficiency. Power factor (excluding non-linear components) is lower where more reactive power is required. A motor has coils and and power must be stored in its magnetic field. This power flows on the wires, shows up in a reading of amps, but it does no work and is not used. It simply flows from the grid into the magnetic field and back every cycle.
quote:
If the pump's electrical usage is 9.3A x 240v that's 2,232W
The formula 9.3A x 240v is 2,232 Volt-Amps (VA) is missing its power factor term. A volt-amp is a measure of Apparent Power, and inlcudes reactive (stored) power and real power, which is the power used by the pump motor. Real power can only be calculated by knowing the power factor, (pf). So,
9.3 amps x 240v x power factor= X watts
Lets say the power factor is 0.8; a fair estimate. The pump motor then uses only: 9.3 amps x 240v x 0.8 = 1785.6 watts
There is innefficiency, but its not in the reactive power necessary to run the motor, but rather in losses such as friction, heat losses, magnetization losses, and more. The efficency, 'N', is nessasary to calculate the actual power delivered to the mechanical load - the pump shaft.
If the efficiency were 75%, for example, then: 1785.6 watts x 75% = 1339.2 Watts,
or 1339.2 / 746 hp = 1.795 horsepower delivered to the pump shaft.
Note that the reactive power does indeed flow in the wires, the distribution system, the breaker, the motor, etc. All have to be sized to handle it. It shows up in any reading or specification of amps; like the one djr777 got from the phone call to Hayward pump tech support. But the reactive power does not do any work; does not supply power to either the pump shaft; and in a residence you do not pay money to the electic utility for this power. |
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1 edit | You're correct.
I got this backwards when I was reading it:
quote:
For example, to get 1 kW of real power, if the power factor is unity, 1 kVA of apparent power needs to be transferred (1 kW ÷ 1 = 1 kVA). At low values of power factor, more apparent power needs to be transferred to get the same real power. To get 1 kW of real power at 0.2 power factor, 5 kVA of apparent power needs to be transferred (1 kW ÷ 0.2 = 5 kVA). This apparent power must be produced and transmitted to the load in the conventional fashion, and is subject to the usual distributed losses in the production and transmission processes.
»en.wikipedia.org/wiki/Power_factor
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