Ordered a Intelliflow VF to replace Jandy Stealth 1.5hp

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G

-Greg-

0
Mar 25, 2008
49
Palm Springs, CA.
Due my $477 electric bill last month, I completed a comprehensive breakdown of all of my appliances/equipment etc. In short, here are my pool equipment findings:

Jandy Stealth 1 1/2 hp pump
7.7 amps, 885 watts, 2.12 Kwh

Polaris booster pump:
5.6 amps, 644 watts, 1.54 Kwh

We pay .25 per kwh from edison, for the Jandy pump this translates into $95.40 for 30 days @ 6hrs per day. The booster pump costs $45.60 for 30 days @ 4 hrs. per day. We ordered the Intelliflow from our pool builder, $1350.00 installed, plus we will apply for a $200 rebate from Edison.

I plan on listing the Jandy pump, the Polaris booster pump, and my Polaris 280 on our local craiglist. We are going to wait on ordering the the intelliomm 4 since we have dual Jandy RS one touch controllers, our pool builder indicated that we could add it later with out any problems. Our pool builder is also going to let me try out a handful of suction side cleaners to determine which one will work best on our pool, we are going to try out the pentair great white and the pentair kreepy krauly to start with, and then he told me we could also try the Jandy rayvac and the Polaris ATV. Hopefully this plan will help out our bottom line!


Greg
 
-Greg- said:
Jandy Stealth 1 1/2 hp pump
7.7 amps, 885 watts, 2.12 Kwh

Polaris booster pump:
5.6 amps, 644 watts, 1.54 Kwh


We pay .25 per kwh from edison, for the Jandy pump this translates into $95.40 for 30 days @ 6hrs per day. The booster pump costs $45.60 for 30 days @ 4 hrs. per day. We ordered the Intelliflow from our pool builder, $1350.00 installed, plus we will apply for a $200 rebate from Edison.
Click to expand...


I don't follow your numbers:

885W for 6 hours is 5.3KWH. Multiply by 0.25 and you get $1.33 per day, or $39.90 per 30 days.
644W for 4 hours is 2.6KWH. Multiply by 0.25 and you get $0.64 per day, or $19.20 per 30 days.
 
Also, a Jandy 1 1/2 HP pump would not be 885 watts. Probably closer to 2 kw.

I think his second number is the actual energy use 2.12 kw and 1.54 kw. I am not sure what those numbers labeled as 885 & 664 watts really are.
 
Jandy:
885 watts x 24 hours = 21,240 watt hours per day x .0001 = 2.12 Kilo watt hours or Kwh

2.12 Kwh x $.25 = $.53 per Kwh

$.53 x 6 hours = $3.18 per day

$3.18 x 30 days = $95.40 per 30 days


Polaris:
644 watts x 24 hours = 15,456 watt hours per day x .0001 = 1.54 Kilo watt hours or Kwh

1.54 Kwh x $.25 = $.38 per Kwh

$.38 x 4 hrs = $1.52 per day

$1.52 x 30 days = $45.60 per month

According to my pool builder this was the formula I was supposed to use. Is this not correct?

Thanks!

Greg
 
Are you sure it isn't 230 volts? A Jandy 1 1/2 HP usually runs on 230v not 115v. And the wattage would be more like 2x885 watts for a 1 1/2 HP.

And to find daily totals it should be.

1.770 kw x 6 hours / day = 11 kwh/day or 330 kwh/month
 
Mark, you are correct about the 230 volts, however the amps are not doubled when measuring only one leg. If the motor was wired with only 115 volts, the amps would approximately double. I measured each leg, each one pulled 7.7 amps, and each leg measured 115 volts. When applying the volts and amps into the equation, it is my understanding that since the motor is wired using 230 volts, I would only use 7.7 amps. At least this is my understanding.

Thanks again for the help!

Greg
 
What did you use to measure the current? A standard hand held amp meter doesn't measure quite the same thing as the electric meter measures. Often the electrical usage is (EDIT)lower(/EDIT) than the hand held amp meters indicate.

Also your math is a little off:
7.7 amps * 230 volts is 1771 watts.
1771 watts x 6 hours/day = 10,626 watt hours per day = 10.6 Kilo watt hours per day or Kwh/day
10.6 Kwh/day * $0.25/Kwh = $2.65/day or about $81/month

Your voltage might be different, if so replace 230 by your actual voltage.
 
Ohms Law states that Power = Voltage x Amps, but you have to use total voltage in the circuit, not the voltage on each leg. Jason's calculation is correct: 7.7 amps * 230 volts is 1771 watts.

While there is only 7.7 amps on each leg, the power company is sending you 2 separate 115 volt circuits, each on an opposite phase of the incoming power. The load (the motor in this example) is using 7.7 amps at 230 volts, for a total of 1771 watts, or 1.771 Kwh for each hour of operation.
 

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Jandy
7.7 amps x 230 volts = 1771 watts
1.77 Kwh per hour used.
1.77 x $.25 = $.44 per hour
$.44 x 6 hrs per day = $2.64 per day
$2.64 per day x 30 days = $79.20 per month

Polars:

5.6 amps x 230 volts = 1288 watts
1.28 Kwh per hour used
1.28 x $.25 = $.32 per hour
$.32 x 4 hrs per day = $1.28 per day
$1.28 per day x 30 days = $38.40

$79.20 + $38.40 = $117.60 per month to operate my pool.




Seems much better, thanks Harry for the help!

Greg
 
A clamp on meter like the one you are using is ideal for resistive loads, like a standard incandescent light bulb, but doesn't tell the whole story for inductive loads like a motor. An inductive load creates a phase difference between the current and voltage waves, which changes the watts in interesting ways. This is taken account of by knowing the power factor of the motor. The power factor of a pool pump motor will be sort of close to one, say 0.8, but never exactly one, so the total you get charged for is actually (EDIT)less(/EDIT) than amps*volts.

The best solution is to use your power company electric meter to measure the power used by your pump. It takes account of the power factor, which an amp meter does not.

If you want to think about each leg of the power independently you would take the current of that leg times the voltage of that leg and then total all the legs. Almost all pool pumps are single phase motors where the current will be the same on both legs and all you need is a single current reading and the total voltage across both legs. For example measuring each leg would give you 7.7*115 + 7.7*115, which is the same as 7.7*230. For a three phase motor things can get more complex and you sometimes have to measure the current and voltage on each leg and sum the products.
 
Thanks Jason for the explanation, that refreshes my memory somewhat! I think I will check the meter as you said to see how close I am, I will shut all of the breakers off except for the pool equipment. Count the revolutions against a stop watch, and see if my Kwh match up.

Again thanks for your help!

Greg
 
I thought power factor ONLY comes into play for industrial customers that are billed based of KVAR not KWH. Since a homeowner is not billed for KVARs power factor should be of no concern.

Further, as for the cleaner scrap the polaris and get a blue diamond or blue pearl. They use much less energy and clean much better.
 
The power company bills consumers for watts. At power factors other than 1, volts*amps is not equal to watts. Some industrial customers are billed in kVAR, which is based on volts*amps, not watts. When you are being billed in kVAR you can measure volts and amps and be done. When you are being billed in Kwh you can measure watts and be done. If you want to measure watts with kVAR billing or measure volts and amps with Kwh billing then you need to worry about power factors.
 
Just a quick note: for power factors less than one (ie motors), the watts will be LESS than the VA (which is volts times amps). For example a motor drawing 7.7 amps at 230 volts draws 1771 VA. At a PF 0.8 that will be 1417 watts. Most induction motors at or near full load will usually have a PF of 0.8
 
Right you are, the actual electrical usage should be less that what is measured by an amp meter because motors have power factors lower than one. Ah well, I had everything else right.

Another thing to note. Capacitor run motors, which have historically been rare in pool motors but are getting more common because of electrical efficiency concerns, have higher power factors than ordinary motors.
 
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