I want that, but how many watts would I have to save to cover 320 bucks?!Get this. You know that you want to know the truth. It's made by a Flir company, so you know it's good quality.
380976-K: Single Phase/Three Phase 1000A AC Power Clamp Meter Kit | Extech Instruments
380976-K: Measures Current, Resistance, Temperature (Type K) and Power. Experience the Extech Advantagewww.extech.com
Pentair VS Watts Mystery Not Equal to Volts x Amps Why?
- Thread starter EnoughToBeDangerous
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Sometimes you just gotta know.
Hopefully, you will be able to sleep tonight not knowing the real truth.
At 3:00 am you will be staring into the darkness wondering what the real numbers are.
Just saying.
Its going to be very hard to know the PF on the drive input side as it relates to the motor. There are some "significant" losses using a single phase input to a 3 phase output. The drive uses a PWM output to the motor. The input side will be relatively clean. Some of the notching will be on the primary. Depending on the filtering a True RMS meter may work on the input of the drive.
Most if not all utility companies charge home owners on DGS (Demand General Service). This means we ONLY pay for kWh or real work. Shaft HP is Shaft HP, regardless of the pump motor size. The only thing that will change is the motors efficiency. The lower the efficiency the more kWh we pay for the same output.
The reactive power or VARS is the magnetizing current required to excite the stator core and rotor. The higher the HP the more magnetizing current required. Volt Amperes is equal to line voltage times line amps for single phase. These values are really "meaningless" for the average consumer unless you pay demand penalties.
The pump motor can't operate at unity PF unless the drive has an active front end which can do some sophisticated things with the DC bus (don't see that being a standard option on a $1500 pump motor & inverter).
If you don't have a true watt meter you CAN look at the pump curve to determine your shaft HP, if you know flow rate and headloss. From there you'll take that number and multiply it by 0.746. That'll be output kW. Most likely the motor is in the mid 90s for efficiency. So you can then take your output kW and divide it by 0.925. This will get you input kW to the motor. Since it is a VFD with a single phase input the power section isn't going to be 98% efficient like most. You could estimate it to be around 95%. So take that input kW and divide it by 0.95. Now you have the input kW to the drive which will be the kW you pay for the pump motors operation.
Shaft HP x 0.746kW/HP x 1/motor eff% × 1/drive efficiency = consumer kW consumption.
The affinity laws are in our favor for pumpage at lower speeds. Since most of us don't pay for VA the motor HP isn't as important as it's efficiency at the load points we use it at. All VS pump motors use permanent magnet rotors which makes then more efficient than a standard squirrel cage motor since they don't have rotor bars to excite.
If I can help anymore just let me know. Electric motors are part of my career.
Most if not all utility companies charge home owners on DGS (Demand General Service). This means we ONLY pay for kWh or real work. Shaft HP is Shaft HP, regardless of the pump motor size. The only thing that will change is the motors efficiency. The lower the efficiency the more kWh we pay for the same output.
The reactive power or VARS is the magnetizing current required to excite the stator core and rotor. The higher the HP the more magnetizing current required. Volt Amperes is equal to line voltage times line amps for single phase. These values are really "meaningless" for the average consumer unless you pay demand penalties.
The pump motor can't operate at unity PF unless the drive has an active front end which can do some sophisticated things with the DC bus (don't see that being a standard option on a $1500 pump motor & inverter).
If you don't have a true watt meter you CAN look at the pump curve to determine your shaft HP, if you know flow rate and headloss. From there you'll take that number and multiply it by 0.746. That'll be output kW. Most likely the motor is in the mid 90s for efficiency. So you can then take your output kW and divide it by 0.925. This will get you input kW to the motor. Since it is a VFD with a single phase input the power section isn't going to be 98% efficient like most. You could estimate it to be around 95%. So take that input kW and divide it by 0.95. Now you have the input kW to the drive which will be the kW you pay for the pump motors operation.
Shaft HP x 0.746kW/HP x 1/motor eff% × 1/drive efficiency = consumer kW consumption.
The affinity laws are in our favor for pumpage at lower speeds. Since most of us don't pay for VA the motor HP isn't as important as it's efficiency at the load points we use it at. All VS pump motors use permanent magnet rotors which makes then more efficient than a standard squirrel cage motor since they don't have rotor bars to excite.
If I can help anymore just let me know. Electric motors are part of my career.
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Knowing the actual number isn’t really going to make much difference because you are going to run the pump based on whatever flow is necessary to provide the function you want.
Running at the slowest speed possible while achieving the function is going to use the least amount of power and energy.
The pump power reading should be accurate enough.
If you want more accuracy, you need to get the right test equipment.
Running at the slowest speed possible while achieving the function is going to use the least amount of power and energy.
The pump power reading should be accurate enough.
If you want more accuracy, you need to get the right test equipment.
Instead of buying an expensive meter turn off all your breakers to your house except the pool pump then use your utility meter to measure the power.
Thanks, I like that idea...a lot. But is the meter sensitive enough for me to get a reading in a few minutes? I guess I'll find out.
Depends on the meter. Our's is a smart meter and the display cycles through various parameters. One parameter is the instantaneous power draw shown down to a 10W resolution.
If you have too rely on the power "odometer" it might not be accurate enough for a short term measurement. Our's only shows down to the single kWh precision.
Ok so I have the true RMS meter from amazon. Amazon.com: RockSeed True-RMS Clamp Meter, Digital Multimeter with Temperature, AC/DC Voltage & AC Current, Resistance, Continuity, Diode Detect; Auto-ranging Amp Volt Ohm Meter Voltage Tester with NCV (CM1): Home Improvement
It does not report the power factor.
It is not sold as "rockseed" anymore but amazon is still selling now under other brands. It's basically a generic unit. The amazon reviewers seem happy with it for whatever that's worth. Ok let's take some readings. The pump is wired through an intermatic timer, a vestige from an older setup from a previous owner who upgraded to the variable speed pump and left the timer in place. The timer is not used from timing the pump, but it gives easy access to attach the clamp meter, which plugged into a 240v outlet. So that's how I'm measuring the amps in case it matters. I checked both legs of the circuit and got very similar readings.
The pump is running at 1900 RPM and reports watts of about 412. The clamp on ammeter is reporting 1.89 amps. The Emporia Vue is reporting consumption of 1.71 amps and 406 watts. This actually seems pretty good. So if I understand correctly we can calculate the implied power factor as PF=Watts/Volts x Amps = 412/(240V * 1.89A) -= .908. Which makes sense, although I assume it would change at different RPMs, I may do some testing of this later.
Now heres the weird thing. The emporia vue is tying out with the pump reading, almost perfectly today. But the vue tracks the history of the power draw. So when I look at the history, although the pump is has been running at 1900, and reporting about 410 watts, there are periods of time where it increases the draw to over 600 watts without any apparent change in what is happening at the pump. However I think I see the cause of this. It looks like the SWG is wired into the intermatic as well, although I'm not sure if it's on the same circuit, I'll have to take a closer look. Could that explain extra 200 watts of use or so?
Thanks!
ETBD
It does not report the power factor.
It is not sold as "rockseed" anymore but amazon is still selling now under other brands. It's basically a generic unit. The amazon reviewers seem happy with it for whatever that's worth. Ok let's take some readings. The pump is wired through an intermatic timer, a vestige from an older setup from a previous owner who upgraded to the variable speed pump and left the timer in place. The timer is not used from timing the pump, but it gives easy access to attach the clamp meter, which plugged into a 240v outlet. So that's how I'm measuring the amps in case it matters. I checked both legs of the circuit and got very similar readings.
The pump is running at 1900 RPM and reports watts of about 412. The clamp on ammeter is reporting 1.89 amps. The Emporia Vue is reporting consumption of 1.71 amps and 406 watts. This actually seems pretty good. So if I understand correctly we can calculate the implied power factor as PF=Watts/Volts x Amps = 412/(240V * 1.89A) -= .908. Which makes sense, although I assume it would change at different RPMs, I may do some testing of this later.
Now heres the weird thing. The emporia vue is tying out with the pump reading, almost perfectly today. But the vue tracks the history of the power draw. So when I look at the history, although the pump is has been running at 1900, and reporting about 410 watts, there are periods of time where it increases the draw to over 600 watts without any apparent change in what is happening at the pump. However I think I see the cause of this. It looks like the SWG is wired into the intermatic as well, although I'm not sure if it's on the same circuit, I'll have to take a closer look. Could that explain extra 200 watts of use or so?
Thanks!
ETBD
If it's on the same circuit, the 200 watts is about what a SWG pulls.
You should measure the exact voltage instead of using a standard 240 volts.
Ok, I think my mystery may be solved. The SWG(autopilot digital) pulls an extra .73 amps when I set it to 100% vs 0%. It pulls about .19A on zero vs .92 amps at 100%. So 0.73*240V = 175 watts. Should have no issue with phases on the SWG, I'd expect that to be a resistive load, so I think the emporia is actually tying out pretty well! I'm going to have to examine the meter to see if I can test my results against that. It's a digitial meter, but I don't think it's a smart meter.
I don't think that's right because we know there's a power factor. Emporia told me the voltage they are seeing is 243.9. It could be different based on the load balance on the system then vs. now, but I'm not sure I want to open the panel right now just to check that, but next time I do I'll retest everything and report the results. But let's redo the PF calculation with 243.9
So we have PF = 412/(243.9V * 1.89A) = .894. So the implied power factor is comes out slightly worse than with the 240V assumption, but in a range that seems plausible given the manufacturers claims, and the difference in readings from the clamp on meter(apparent power) and the pump itself. Would you expect the power factor to be higher or lower at a higher RPM?
Here are the power factor readings from the wattmeter in Steve's SuperFlo VS video.
Rpm......PF
1000....0.56
1500....0.79
2000....0.91
2500....0.95
3000....0.97
3450....0.98
Based on this, it looks like the power factor improves with speed.
That's at the panel. You have to account for voltage drop from the panel to the pump. The voltage at the pump will always be less than at the panel.
The NEC recommends that the maximum combined voltage drop for both the feeder and branch circuit shouldn't exceed 5%, and the maximum on the feeder or branch circuit shouldn't exceed 3%.
A 3% voltage drop would mean that the voltage at the pump would be about 236.58 volts.
3% is 7.32 volts x 1.89 amps is 13.83 watts lost in the wiring as heat.
Assuming a wire size of 12 AWG at 100 feet and 1.89 amps, the voltage drop would be about 0.7 volts.
At 14 amps, the voltage drop would be about 5.5 volts
At 14 amps, the voltage drop would be about 5.5 volts
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The panel is only about 30 feet from the pump, so the voltage drop is probably less than 3%, but you're right the pump PF is probably a better than I estimated. But at the end of the day, I still have to pay for the power lost to resistance in the wires. So if it's 13 watts, that's .013 kwh/hr = .312 kwh/day = 114 kwh/year, about 22 dollars per year at my rates, half that since I close the pool for winter. 8/2 nm-b wire costs about 1.50 a foot, so 45 dollars, probbly wouldn't save enough to justify upgrading, but it might make sense to oversize the wires a bit when doing the wiring initially since the pump runs so much.
Ok well thanks everyone who commented, this is helpful, and I feel more confident that I'm getting useable readings from my vue. I also learned that the SWG is hooked up to the pool pump circuit, which I did not know. I'm excited about being able to see how much I'm spending to operate the pool and especially to heat the pool with the aquacal variable speed heat pump. My first observation, is that it hasn't stepped down the current draw, but it's still not very warm here so hopefully I'll get some interesting data on that when the weather gets warmer. Thanks again, this forum is an amazing resource, and I really appreciate all the helpful responses.
I was using the maximum amperage for the wire to calculate the 3% drop.
Since the amperage will be well below 20 amps, the voltage drop will be much less.
Overall, between the VUE and the pump readout, you should have a relatively good estimate of the actual power draw.
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