Polaris Pump Electricity Usage Increase: From 1.32 kWh to 5.92 kWh

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I think that the main issue is that you are not accurately identifying what is using power during the time in question.

The main pump uses a lot of power (2,250 to 2,400 watts) and if it is on due to Freeze Protection, then that adds up fast.

If the main pump ran 24/7, the cost would be 57.6 kwh per day ($8.64) or 1,728 kwh per month ($259.2).
 
LoneStarPool said:
I think the breaker is 30 AMP (3rd one from the bottom on the left) - I assume it's 30 per pump?
did not use a meter to measure the power usage. I was going off electricity usage on my electric utilities company's website. Each screenshot represents 1 day of use, with each bar being a 15 minute interval. Now that I look at my calculations, I guess it's possible I'm not supposed to be multiplying the values by 4 (I did it to translate 15 minute measurements into 1 hour ones, but there's a good chance it's wrong). If that's the case, the spike would still be x5, but it would be from 0.33 kWh to 1.48 kWh.

Previous owner set the system to turn on at 8 AM and shut off at 4 PM. The vacuum pump used to run for a few hours inside that period, which can be seen from that graph. I lowered the times to run only from 10AM to 4PM and vacuum only for 1 hour inside that to cut down on electric bill.

This is usage from December (before the spike):
  1. Baseline (no pump running): ~0.04 * 4 => 0.16 kWh
  2. Main pump: (0.57 - 0.04) kWh * 4 => 0.53 kWh * 4 = 2.12 kWh
  3. Vacuum pump: (0.90 - 0.57) kWh * 4 => 0.33 kWh * 4 = 1.32 kWh



This is usage from February (after usage spike):
  1. Baseline: ~0.07 kWh
  2. Main pump: (0.64 - 0.07) kWh * 4 => 0.57 kWh * 4 = 2.28 kWh
  3. Vacuum pump: (2.12 - 0.64) kWh * 4 => 1.48 kWh * 4 = 5.92 kWh
Click to expand...
Please note that if your Polaris pump, which is rated for an amp draw of 6.1A, +/- 1,400 watts/hr., were to be drawing 4 times that much, 24.6A at 240V, to get to the 5,920 watts/hr. It would have completely burned up the first day, probably the first few minutes. The internal wiring is not large enough to carry that much current. It has to be an error somewhere.
 
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So, I finally got the meter and took some current measurements (see attached pics). The results were as follows:
  1. Main pump: 9.88 Amps → 2.17 kWh
  2. Vacuum pump: 6.88 Amps → 1.5 kWh
Recap of My Original Calculations Based on the Electric Company’s Usage Chart:
  • Baseline: ~0.07 kWh
  • Main pump: (0.64 - 0.07) kWh × 4 → 0.57 kWh × 4 = 2.28 kWh
  • Vacuum pump: (2.12 - 0.64) kWh × 4 → 1.48 kWh × 4 = 5.92 kWh
While I was pretty close on the main pump (2.17 kWh actual vs. 2.28 kWh calculated), I was way off on the vacuum pump (1.5 kWh actual vs. 5.92 kWh calculated). I'm chalking this up to the run time being too short (less than an hour) and the possibility of another device running—someone mentioned a space heater, which could explain the discrepancy.

That said, I’m confident in the calculations from before I moved in:
  • Baseline (no pump running): ~0.04 × 4 → 0.16 kWh
  • Main pump: (0.57 - 0.04) kWh × 4 → 0.53 kWh × 4 = 2.12 kWh
  • Vacuum pump: (0.90 - 0.57) kWh × 4 → 0.33 kWh × 4 = 1.32 kWh
Someone in the thread mentioned that if the pump lines were clogged, the pump would actually draw less current. This seemed counterintuitive to me since I imagined a vacuum cleaner struggling when clogged, but I guess that shows how much I know about pumps. This could explain why the vacuum pump usage increased from 1.32 kWh before I moved in to 1.5 kWh afterward. When I first moved in, the pool hadn't been maintained for almost two months, so it was full of leaves, and the Polaris bag was also packed with them. I assume this reduced flow enough to lower the current draw by about 10%.

To verify my findings, I plan to run the vacuum for 4 hours while ensuring nothing else is running in the house. This should help confirm whether the electric company’s usage chart aligns with my actual calculations - which it really should.

I guess my only question at this point is if these measured usage values seem correct? The vacuum pump seems to draw a lot of power compared to the main pump, is that expected for this particular model?
  • 2.17 kWh for a 2HP Century B130 pump
  • 1.5 kWh for a PB4-60 pump
1739903283779.jpeg1739903312187.jpeg
 
Yes, your measured amps are correct and should agree with the motor data plates on the pumps.
 
LoneStarPool said:
Main pump: 9.88 Amps
Click to expand...
JamesW said:
The main pump uses a lot of power (2,250 to 2,400 watts) and if it is on due to Freeze Protection, then that adds up fast.
Click to expand...
9.88 x 235 = 2,322 watts, which is about what is expected depending on the pump model.

Note: The Century B130 is the motor model number and not the pump model number.

Can you show the actual pump?

1739904842612.png
LoneStarPool said:
Vacuum pump: 6.88 Amps
Click to expand...
That is higher than it should be.

The amperage should never be higher than the S.F Amps (6.4) on the motor label.

Assuming 235 volts at the motor, the power is about 1,617 watts.

That is about 1.4 times higher than it should be.

This can happen with the robot unplugged and low resistance/High Flow.

The other reason might be overheating and power being converted to heat.

The Booster pump is using about 492 watts more than expected.

The Maximum Power at 6.4 amps and 230 volts is 1,472 watts.

The motor will begin to overheat and thermally overload at somewhere between about 6.5 and 7.5 amps depending on the ambient conditions like air temperature and sunlight exposure.

JamesW said:
The motor uses about 1,125 watts of power.

At 230 volts, the amps should be about 4.9 amps.

To get kwh, multiply by number of hours running/1000.

1 hour On uses 1.125 kwh.

2 hours On = 2.25 kwh.
Click to expand...


1739905313521.png
1739904897421.png


Service factor amps (S.F.A.) is the amount of current a motor will use when running at its full service factor.

It's a rating that shows how much extra work a motor can handle without overloading or failing for short periods of time.
 
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I posted all the relevant pics in the original post. As far as I can tell the booster pump is sucking water from the main return. Is that how it's supposed to be set up? The booster return seems to just be a dedicated pipe that goes to the pool:
1739909680278.jpeg1739909735591.jpeg
 
LoneStarPool said:
I'm pretty sure the meter was just flipping from 2 to 3 :). I can't see anything identifying the pump model. Here are the pics from all sides:
View attachment 628729
View attachment 628730
View attachment 628731
Click to expand...
You have a Hayward SuperPump 2.
A good replacement motor to save energy would be the Nidec Neptune NPTT225.
Your booster pump is well within its amperage range. The full-load rating on the motor can be +/-10% and be fine.
Your motor is probably rated at 6.4A. If you are concerned about that overage, install a restrictor washer in the wall fitting. A pump moving less water draws less amperage, but you may not like the performance of the cleaner.
At 6.88A you should be fine, that's within the motor's designed performance.
 
The black disc is a pressure relief.

If that is allowing a lot of water out, then the amps can increase due to extra water flow.

You can replace the pressure relief and other parts.

Check the amperage with the red or blue restrictor disc in place.

1739913151140.png
 
"Full Load Amps" (FLA) represents the amount of current a motor draws when operating at its rated horsepower, while "Service Factor Amps" (SFA) indicates the current the motor will draw when operating at its full service factor, which allows for a short-term overload beyond the rated power, meaning SFA is typically higher than FLA when a motor is under increased load conditions.

________________________________________________________________________

When a motor is running above its full load amps and in the service factor range, it may function, but its operational life will be shorter. It will also generally run at lower efficiency and power factor.

In other words, only go into the “service factor” range when necessary, not as a matter of normal operation.

hvacrschool.com

Motor Service Factor - HVAC School

“Service factor” is an interesting motor rating that you will see on many motor data tags. It simply means how much additional “work” a motor can do or how much “load” it may be placed under for short periods of time without failure or overload. For example, the FLA or full load amps of the […]
hvacrschool.com hvacrschool.com

________________________________________________________________________
 
One thing that you are overlooking is that the current drawn multiplied by the voltage applied doesn't necessarily equal the power consumed. This is always true if you were dealing with direct current, but pumps are AC line powered and to accurately measure the real power consumed (what you get charged for) you need to know the power factor of the load, in this case the pump.

Single speed motors are inherently inductive, and that means the current waveform is out of phase from the applied voltage. The amount that the two waveforms are out of phase is called the power factor. When the power factor is one, both waveforms are in phase and then the current times the voltage equals the power consumed. A typical single speed induction motor has a power factor of 85 to 90 percent when the motor is running at full load.

So you really should be multiplying your power consumption numbers by .85 to .90 to account for the power factor of the motor.

If a motor is not fully loaded the power factor will even be lower than this. It is likely that your booster pump for vacuum is lightly loaded so it's power factor is running at a lower number. This would mean that the real power is considerably less than number you are coming up with.

It's the unknown power factor number that accounts for the differences in the real power drawn, your calculations will always be higher than real number because your calculations are using a power factor of one.
 
A motor should not normally run at its "service factor amps" as this represents the current draw when the motor is operating under a temporary overload condition, and continuously running at this level can significantly shorten the motor's lifespan; it's designed to handle occasional overloads for short periods, not sustained operation at that level.

To put it most simply, operating your electric motor at the service factor rated load continuously will reduce the motor's speed and efficiency and ultimately reduce the motor’s lifespan.

You can safely run your motor at Service Factor 'intermittently', according to the National Electrical Manufacturers Association (NEMA). This leaves some room for interpretation.

https://www.emotorsdirect.ca/knowledge-center/article/what-is-service-factor-and-how-is-it-used
 
Check for Proper RPM.

Before operating the cleaner, check for proper wheel revolutions per minute (RPM).

For maximum efficiency, the cleaner should operate between 28 and 32 RPM.

1. Mark the single-wheel side tire.

2. Turn pump on, hold cleaner below water level and count wheel revolutions for one minute.

If count is less than 28 RPM:

• Check the filter screen in the in-line filter for debris that restricts water flow.

• Clean the skimmer, filter and pump basket, clearing debris that restricts water flow.

• Check the hoses, connections and swivels for leaks that cause loss of water pressure.

• Remove the blue restrictor disk from the UWF.

• If an adjustable valve is installed on the booster pump or cleaner line, open it completely so water can flow freely to the cleaner.

If count is more than 32 RPM:

• Replace the blue restrictor disk in the UWF with the red restrictor. If the flow is still too high, unscrew the pressure relief valve until the proper RPM is reached.

Note: The pressure relief valve should only be adjusted when a restrictor is installed.

• If an adjustable valve is installed after the booster pump, adjust it to reduce the water flow to the cleaner.
 

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