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Low power factor, recip pump
8

Low power factor, recip pump

Low power factor, recip pump

(OP)
I have a recip pump and the motor seems to have an exceedingly low power factor of roughly 0.24.  Here's the details:
Motor: 100 hp TEFC, 3 ph, 460 V, 1750 RPM
Pump: Recip, 356 RPM
Belt drive

Motor is way oversized, power necessary is only about 40 hp.  The recip has a small flywheel but not huge.  We recently doubled the flywheel inertia by adding some metal plates but peak current readings didn't change (around 125 amps as read from hand held meter).

To solve this problem we installed a device to capture current and voltage data.  I'm a mechanical engineer, so I'm not familiar enough with the electrical aspect of all this.  

Attached is a graph of current and voltage as taken by this device we installed.  The time scale is 0.25 seconds which is sufficient to see more than 1 full rotation of the piston.  

1) Why would current lag the voltage like this?  Is there simply insufficient flywheel energy provided?  Or is there also a problem with the motor being oversized?  

2) When the voltage and current are in opposite sides of the 0 line, does that mean the current going through the windings is 'slowing down' or retarding the motor?  What happens for example, when current goes negative but the voltage is positive?

I'm really looking to understand why the power factor is so low and what can be done about it.  Thanks for your help.

 

RE: Low power factor, recip pump

Induction motor pf increases with load. At low loads, the magnetic component of the current prevails over the power component of the current and hence low pf.

To improve your pf (to about 0.95), connect a 30 KVAR, 460 V capacitor in parallel with the motor, perferably right at the motor terminals, if possible.

RE: Low power factor, recip pump

The key is probably that the motor is oversized.  The reactive power needed by the motor stays fairly constant over its load range.  So at low loads, the power factor can be quite poor.  

I have taken measurements at sawmills with power factors of 0.3 to 0.4 because the motors are unloaded much of the time.

If you have a way to increase the load on the motor briefly, you may see the pf improve pretty dramatically.   

"Theory is when you know all and nothing works. Practice is when all works and nobody knows why. In this case we have put together theory and practice: nothing works... and nobody knows why! (Albert Einstein)

RE: Low power factor, recip pump

(OP)
Thanks edison.  I've heard the capacitor trick can work.  So you think a PF of .95 is doable?  What if the capacitors are ~40 feet away?  They would need to be at least that far unfortunately.

 

RE: Low power factor, recip pump

(OP)
Also, can anyone explain what's really happening when voltage and current are on opposite sides of the 0 axis?  Does this retard the motor?

RE: Low power factor, recip pump

Yes, a pf of 0.95 doable, even with capacitor connected away from the motor. The idea of having capactor at motor head is to reduce the current through the supply cables too.

See this useful link on pf correction and some no-no's.

http://www.elec-toolbox.com/usefulinfo/pfcorrection.htm

RE: Low power factor, recip pump

(OP)
To answer my own question:

Quote:

Also, can anyone explain what's really happening when voltage and current are on opposite sides of the 0 axis?  Does this retard the motor?

I heard today that when the voltage and current are on opposite sides of the 0 line (ie: voltage is positive when current is negative OR voltage negative while current is positive) the motor is actually acting like a generator and putting power back into the grid.  That power comes from the inertia of the windings and driven equipment.  

Does this seem reasonable?

RE: Low power factor, recip pump

(OP)
Thanks Keith.

RE: Low power factor, recip pump

2
Well, it's correct if you have matched up your currents and voltages correctly, but you have not.  You plotted phase to phase voltage and phase current.  The product is not instantaneous power.  It would be instantaneous power for a single phase of the motor if you could measure voltage to neutral point of the motor and multiply it by the same phase current.  There are ways to compute power from phase voltages and line currents but not simple multiplication.

There is something quite strange going on with those current readings. They appear to be oscillating in both phase and magnitude.   I guess you are attributing this to torsional oscillation and that's why you increased the flywheel size?  Perhaps it needs to increase more.  Or maybe belt is slipping erratically?  Does it squeal? Which side of the belt is the flywheel on?  (I guess you get more effect out of large flywheel putting it on the motor side of the belt, but smoother torque on the belt if you put it on the compressor side of the belt).  By the way there is no vfd or special power supply on this is there?

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RE: Low power factor, recip pump

Quote:

I heard today that when the voltage and current are on opposite sides of the 0 line (ie: voltage is positive when current is negative OR voltage negative while current is positive) the motor is actually acting like a generator and putting power back into the grid.  That power comes from the inertia of the windings and driven equipment.  
IF your motor were running at steady state driving a smooth load and had a power factor of less than 1 (let's say 0.8pf),  THEN, if we examine the power put in by a single phase of the motor, we find it is putting power in for portion of the cycles and taking power out for portion of the cycle (smaller portion).  This is due to interchange of energy with the field of the motor, not the inertia (again in steady state with smooth load... not your situation... your situation is more complex).  Contiuing with the steady state smooth running motor, when we add all three phase powers together, the average is a constant even though each is adding energy for portion of cycle and remvoing energy for portion of cycle.

Your situation more complex and interesting.  I haven't measured recip currents before but looks awfully strange to me.

 

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RE: Low power factor, recip pump

(OP)
Hi electricpete, thanks for the lengthy responce.

Quote:

Well, it's correct if you have matched up your currents and voltages correctly, but you have not.  You plotted phase to phase voltage and phase current.  The product is not instantaneous power.  It would be instantaneous power for a single phase of the motor if you could measure voltage to neutral point of the motor and multiply it by the same phase current.  There are ways to compute power from phase voltages and line currents but not simple multiplication.
Sorry, I don't understand this.  Please check my assumptions:  I'm assuming the data I was provided is of a single phase of a 3 phase motor.  I'm assuming the voltage shows the typical, 60 hz wave form, and the current trace indicates the magnitude over time, but what has me confused is the positive and negative.  I'm assuming the voltage is simply changing direction, so I'm also assuming the current direction is indicated by the positive or negative value.  If current is in the same direction as voltage, I'm assuming there is power going into the motor which is taken out by the pump/flywheel.  When current and voltage is in opposite directions, I'm assuming power is being 'taken out' of the system.  Isn't there always some amount of power that needs to be added or removed when voltage and current are not zero?  If there is a non-zero voltage and current, then isn't there power that needs to be accounted for?  Maybe I'm not understanding what phase to phase voltage is.  I keep hearing funny things like the power isn't real, but if there is voltage and current, then I fail to see how there isn't power being either generated or disipated.  The power has to go somewhere, right?  And if it's not leaving the system as useful work, the only other option is for it to be disipated as heat, and I'm not seeing significant heating.

Regarding the belt and flywheel - the belt is a toothed style, it doesn't slip at all.  The flywheel is on the pump crankshaft which rotates at 356 RPM.  Yes, flywheel energy is a function of rotational speed squared, so it would be much more effective on the motor (about 25 times!) but the electric motor already has a large amount of rotational inertia.  The flywheel certainly could be larger and if it were extreamly large, the motor wouldn't see any fluctuation from the pump.  Unfortunately, increasing flywheel size isn't very practical.  And yes, I would assume a major issue here is that the recip pump requires all of it's energy during a very short period of time which means there's a rapidly fluctuating torque on the motor.

There is no VFD or special power supply.  I also wonder if it would help to put a VFD, synchronous motor or other motor on this with a different number of poles?

Regarding the current trace, this is actually a double acting piston, but the second stage does the vast majority of the work.  The second stage peaks (ie: discharges) at 0.036 and 0.200 seconds and the first stage peaks at 0.118 seconds, which is the much smaller hump you see near the middle of the graph.

Bottom line - looks like I have a few options:
1. Add capacitors
2. Reduce motor size
3. Change type of motor?  Would this help at all?
4. Increase flywheel inertia (not practical).

Thanks again for all the help.

RE: Low power factor, recip pump

Sorry - I was wrong about phase to phase voltage - I think you are monitoring phase to neutral.  Never mind winky smile

My best guess is you have severe torsional oscillation.  I guess the problem I would be worried about is equipment reliability (assuming this is an important piece of equipment).  I wouldn't necessarily worry about the poor power factor (are you being charged for reactive power... or just concerned about power factor as a symptom?).

 

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RE: Low power factor, recip pump

(OP)

Quote:

I guess the problem I would be worried about is equipment reliability (assuming this is an important piece of equipment).
Pump (mechanical side) reliability is unaffected by this power issue.  Not sure if motor reliability is affected though.  Anyone?

Quote:

I wouldn't necessarily worry about the poor power factor (are you being charged for reactive power... or just concerned about power factor as a symptom?).
As the pump designer (mechanical side), I'm concerned that we've matched the wrong motor to the pump and at least some of our customers, if not most, will be concerned about power usage.  

RE: Low power factor, recip pump

FLA I'm guessing is around 115A (rms)

Let's examine the effective magnitude of your current:  You said peak current around 100A and from the graph that is true peak.  If it was a sinusoidal waveform with 100A peak, it would be 70A rms  (your rms would be even lower).   Since the peaks vary down to about 60% of that value, we might guess rms current is 42A.  

Let's examine the effective displacement power factor:  Based on looking at phase angle, power factor ranges from somewhere around 0.9 I'm guessing (near the peak around 0.03 sec), down to 0 at the point where the current lags voltage by 90 degrees (near the peak around 0.07 sec), and a little bit beyond 0 (shall we call it negative power factor.... generator action) in the areas where current lags voltage by more than 90 degrees (near the peak around 0.011 sec)

It's tough to estimate from such a bizarre waveform.    I couldn't argue 0.24 looking at the waveform.  But if you're really drawing 40hp as you said, I'd think power factor would have to be at least 0.4 to draw that much power without exceeding nameplate amps 115A, and perhaps 0.8 to draw that much power without exceeding your an effective current which appears somewhere in the neighborhoold of 50% of nameplate.  So if the 40hp number is correct then I would question the 0.24 power factor. Where did you come up with 40hp?  Does you analyser compute the actual power drawn?

Regardless of the above, easy to see the motor is oversized for steady state thermal rating... just a question of how much.   If you opt for a smaller motor, you still have to look at:
1 – smaller motor may well have more oscillation.  Are you sure the mech system will be reliable?    Out of curiosity, why are you not concerned about effect of the oscillations on the belt?  Do you have some experience with successful running of this machines for lets say a year?
2 – Can the smaller motor accelerate the load including flywheel
3 – keep a margin to breakdown torque.

Now you had a very specific question about the significance of current and voltage in opposite directions.  I muddled the answer a little last time with my confusion about phase-to-phase vs phase to neutral.   Let me try again.  I think it is again instructive to look again at a motor feeding a NON-FLUCTUATING load in steady state.   Examine the voltage to neutral for a single phase and current for a single phase as you have done.   ANY motor feeding a non-fluctuating load in steady state with power factor less than 1 will have periods when current and votlage are on the same side of zero and periods where they are on opposite sides of zero.    Assuming the motor is acting in motor mode, , they spend more time on the same side of zero then the opposite side of zero and so the AVERAGE power for that phase is positive.   So does a motor feeding non-pulsating load in steady state see an oscillating motor torque based on this reversal?  No!... .when we add all three phases together the power delivered is constant over time (no fluctuating).  So for the steady state motor feeding non-fluctuating load, there is no back and forth interchange with the kinetic energy.... speed is constant.   Each phase interchanges power with the power system and the magnetic field.   So the steady state motor feeding non-fluctuating load does not transition between motor and generator action even though there are periods in each cycle when a given phase is returning power to the power system from the field energy.

Your motor on the other hand is transitioning between motor and generator action based on analysis of the phase relationship between voltage and current.  When the current phase lags the voltage by more than 90 degrees, the average power direction changes to generator action.
 

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RE: Low power factor, recip pump

One of the reliability questions would be fatigue of the shafts. If you have a machine which is changing between motor and generator action every 6 power cycles, there are reversing stress on the shafts which can lead to fatigue.

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RE: Low power factor, recip pump

I did some playing around with simulations.   This is based on dynamic model in Krauss' "Analysis of Electric Machinery".   It's a new toy for me - take it for what it's worth.  I would be glad to post the matlab code if you are interested.

My model of your motor (all parameters referred to stator):
Rs = 0.08 % ohms
Xls = 0.4 % stator leakage reactance ohms
Xms = 20 % magnetizing reactance ohms
Xlr = 0.4 % rotor leakage reactance
Rr = 0.08 % rotor resistance
J = 1  % kg *m^2 - represents MOTOR J only
Poles = 2

I assumed the motor acts like a rigid mass.  But it would not be reasonable to make the same assumption about the entire system since the belt has a spring constant which may be low.  Since the dyanamics of the system are unknown I did not model the system beyond the torque seen at the motor shaft which includes effects of system dynamics.  I asumed the torque at the motor shaft of the form:
Tload = FLT * Pav * [1 + Pmult cos(theta_slow)]
where
theta_slow is 1/10 the rotor angle of the motor.  (theta_slow varies at one tenth of motor speed).
FLT = Full Load Torque of motor  = 110 N-m

Note again the varying term represents not only direct variation in load torque but any amplification produced by system dynamics (Like resonant amplification).

I swept through several combinations of Pav and Pmult.  The closest match I could find to your current waveform and phase is shown on slide 2.  It is Pav = 0.3 and Pmult = 5 which would correspond to a torque profile at motor shaft of:
Tload = 110 * 0.3 * [1 + 5 cos(theta_slow)]
Tload = 33 + * 0.3 * 165*(theta_slow)   [N-m]

Considering the full load torque is 110 N-m, a 330 N-m swing in shaft torque 10 times per second seems pretty large to me.  Note the electrical torque shown on slide 3 is does not swing as much – part of the shaft torque oscillation goes into accelerating the motor inertia.  The swing in motor speed shown on slide 4 is roughly 58.8hz to 60.5 hz.

If your machine is really acting like that, I would be concerned about fatigue as mentioned.  Perhaps you can monitor with strobe to see if you really have such large oscillations.  Or else use some kind of optical encoder.

The remainder of the slides from 5 on are other combinations of Pav and Pmult simulated:  Start with a low value of Pavg and sweep through Pmult from low to high.  Then start with next higher Pavg and again sweep Pmult from low to high.  

It is interesting to pick one series and watch the evolution as we increase Pmult (increases the oscillating term as a multiple of the constant term).  At low Pmult (for example slide 16), the only variation present is w_slow (a period of around 0.16 seconds)... but as you increase Pmult (slide 17), there appears halfway in the middle of  the two large current peaks another smaller peak, similar to what you have around 0.11 sec in your waveform.  Comparing the location of the smaller peak in slide 2 to the torque in slide 3, we see the smaller peak corresponds to the negative peak in torque (highest generator action).   If Pmult is small the torque stays positive and there is no small peak between.  As Pmult grows we get have significant negative torque peaks between the positive peaks and we get that small current peak between.  It is for me another confirmation that your torque swings at least enough to provide generator action.
 

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RE: Low power factor, recip pump

Correction:

Quote:

Tload = 33 + * 0.3 * 165*(theta_slow)   [N-m]

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RE: Low power factor, recip pump

(OP)
Hi e-pete.  Thanks very much for all your time and effort.   

Quote:

I assumed the motor acts like a rigid mass.  But it would not be reasonable to make the same assumption about the entire system since the belt has a spring constant which may be low.  Since the dyanamics of the system are unknown I did not model the system beyond the torque seen at the motor shaft which includes effects of system dynamics.  I asumed the torque at the motor shaft of the form:
Tload = FLT * Pav * [1 + Pmult cos(theta_slow)]
where
theta_slow is 1/10 the rotor angle of the motor.  (theta_slow varies at one tenth of motor speed).
FLT = Full Load Torque of motor  = 110 N-m
Most of it is shooting overhead, but here's a few thoughts. Yes, the belt is certainly going to act as a very rigid spring, the inertia of the flywheel, and the force produced by the piston will radically affect any analysis of the motor as you've noted.  In addition, the ineria of the motor is large compared to the flywheel and pump, and must also be taken into consideration to determine rotational acceleration.  It's the rotational acceleration of the pump flywheel that will determine the stress on the crankshaft of the pump.  So without all these factors incorporated into a model, one wouldn't be able to determine the actual torque fluctuations on the cranshaft.  Which leads in to the observation about fatigue.  

Regarding fatigue, the pump is designed for infinite fatigue life and has readily exceeded the number of cycles needed to ensure that no material is stressed above its infinite fatigue life.  Fatigue life can be plotted on a Goodman diagram.  What's most interesting about this analysis is that once 100 million cycles have been exceeded, then (in short) the parts will not break from fatigue.  This number of cycles is seen fairly quickly in most compression equipment, so metal fatigue is carefully analyzed for and well understood.  I'm not concerned about fatigue of any metal parts.

Quote:

  I couldn't argue 0.24 looking at the waveform.  But if you're really drawing 40hp as you said, I'd think power factor would have to be at least 0.4 to draw that much power without exceeding nameplate amps 115A, and perhaps 0.8 to draw that much power without exceeding your an effective current which appears somewhere in the neighborhoold of 50% of nameplate.  So if the 40hp number is correct then I would question the 0.24 power factor. Where did you come up with 40hp?  Does you analyser compute the actual power drawn?
This is a very good observation.  It occurred to me at the time but I really didn't consider it carefully enough, so I appreciate you bringing this up.  I had to think about it quite a bit actually, but here's what I came up with.  At the time the data was taken, the controls engineer had mentioned the PF = 0.24 as this was a readout on his computer as we took the data.  That value, and the readings posted earlier showing peak amps being less than 100 A, were at a point when discharge pressure was significantly lower than maximum (maybe 70% of peak pressure) so power required was roughly 70% of the 40 hp I'd estimated as being the peak.  At peak outlet pressure which corresponds to peak power required for the compressor, the current draw (peak current) was about 128 amps, which is quite a bit higher than FLA.  FLA as I recall was around 115 A (as you already noted).  Also, the 40 hp number is a 'ballpark' value which could easily be off by up to 20%.  

I think what all this says, is that PF must have been lower when the power required by the pump was lower, and as the power demanded by the pump increased, the PF also increased.  Would you agree?  

Quote:

So the steady state motor feeding non-fluctuating load does not transition between motor and generator action even though there are periods in each cycle when a given phase is returning power to the power system from the field energy.

Your motor on the other hand is transitioning between motor and generator action based on analysis of the phase relationship between voltage and current.  When the current phase lags the voltage by more than 90 degrees, the average power direction changes to generator action.
Thanks for confirming this.  This is the part that really confused me for the longest time.  The concept that a motor can also act as a 'generator' was something I was totally unaware of.






 

RE: Low power factor, recip pump

iainuts

The motor will work as induction generator when its speed goes beyond the synch speed, in your case beyond 1800 RPM.

As for the pf, yes, it increases with the load. To ensure a good pf over the entire load range, you need the pf correcting capacitor connected in parallel, to offset the no-load current of the motor. The no-load cuurent is the villain here.

RE: Low power factor, recip pump

This is not a typical steady state case and I don't think  power factor correction capacitors will cause much improvement.  Power factor = real power / apparent power.  During the portion of the cycle when the machine is acting like a generator, we are decreasing the average real power (numerator of power factor) and still contributing to the apparent power (denominator).    During the generating portion the current is almost in phase with the voltage... if you move the current further to the left by adding caps you may have a leading power factor during that portion of the cycle.

Quote:

In addition, the ineria of the motor is large compared to the flywheel and pump, and must also be taken into consideration to determine rotational acceleration.
I'm not saying the model is perfect, but I did take that into account.  I assumed 1 kg-m^2 inertia for a 3600rpm 100hp motor based on a formula in NEMA MG-1 (used for dynamic braking calculations).   There are two torques acting on that inertia:  the electrical torque (varies -40 N-to 100 N-m) and the shaft torque (varies from -130 N-m to 190 N-m).    The shaft torque was assumed as the first two terms of a fourier series periodic at w_slow (the constant component and the fundamental component) and varied until the patern matched.   

Another thing to consider if you decrease the motor size is where does the torsional resonance end up.  Fresonant = sqrt(Jeffective/K) / (2*Pi) where K is torsional spring constnat - mostly related to the belt and Jeffective = (Jmotor*Jload)/(Jmotor+Jload).   Jload is of course sum of flywheel and pump corrected for speed ratio ^2.    Easy to ask questions - harder to answer but it is something to consider.  

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RE: Low power factor, recip pump

You may get slight improvement with power factor caps, but nowhere near what you'd get correcting a steady state machine for reasons discussed above (downward influence on power factor due to generator action)

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RE: Low power factor, recip pump

I'm not sure a synchronous motor would be a good idea. A synchronous motor very closely locks the rotor to the stator rotating field. Any attempt to change the rotation speed of the rotor will result is a very rapid peak in the torque the motor produces. You will likely find much higher torque peaks with a synchronous motor, to the point of being damaging.

An induction motor has more of a "spring" in the coupling between the rotor and the stator and will allow the rotor speed to vary or oscillate more before the torque motor really changes. Basically, the oscillation of the rotor speed produces a softer torque response.
 

RE: Low power factor, recip pump

I think that going to smaller motor will have limited effectiveness in improving this power factor just as adding caps.  There is low average power factor inherent in the change of power direction when you work out the math (although it may not be intuitive).

I will use my simulated waveform to compute power factor and check the effect on power factor of adding caps.

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RE: Low power factor, recip pump

Let's say you have the following system:

                                   <-------- Box ---------------->
Ground ==-Vs+ =======R ==== +Vbox-===+====Ground
                                                |                     |
                                                L===||=====|



We have a source voltage Vs=10vacrms external to a box.  
The box contains a resistor R=1 ohm and a voltage source Vbox=15vacrms connected in opposite polarity to the external voltage source.

The box voltage source has a shorting switch which bypasses the internal voltage source for one second out of every two second period.


What is the power factor of the power flowing from Vs to the box?  

Analyse the 1 second period when the switch is closed (Vbox bypassed)
I1 = Vs/R = 10A.
P1 = Vs * I = 100 watts.
S1 = |P| = 100VA.

Analyse the 1 second period when the switch is open (Vbox in the ckt)
I2 = (Vs-Vbox)/R = (10-15)/1 = -5A.
P2 = Vs * I = 10 * -5 = -50 watts.
S2 = |P| = 50 VA.

Now look at the two second period encompassing both of the above periods.
P = (P1*1sec + P2*1sec)/2sec = (P1 + P2)/2
S = (S1*1sec + S2*1sec)/2sec = (S1 + S2)/2
PF = P/S = (P1*1 + P2) / (S1 + S2) = (100 – 50) / (100 + 50) = 50/150

PF = 0.3333 even though we had no reactive components!

The low power factor was a result of the power reversal.  We have the same pheneomenon going on in the motor with reversing power.

We may attribute a portion of the low power factor to the fact that the motor is oversized, but there is also a large portion of the low power factor attributable to reversing power.
 

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RE: Low power factor, recip pump

Whoops.  I guess I had a math error on the 2nd half second period.  Should've been -25 watts and -25va.  

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RE: Low power factor, recip pump

Correction in bold:

Quote:

What is the power factor of the power flowing from Vs to the box?  

Analyse the 1 second period when the switch is closed (Vbox bypassed)
I1 = Vs/R = 10A.
P1 = Vs * I = 100 watts.
S1 = |P| = 100VA.

Analyse the 1 second period when the switch is open (Vbox in the ckt)
I2 = (Vs-Vbox)/R = (10-15)/1 = -5A.
P2 = Vs * I = 5 * -5 = -25 watts.
S2 = |P| = |25| VA.

Now look at the two second period encompassing both of the above periods.
P = (P1*1sec + P2*1sec)/2sec = (P1 + P2)/2
S = (S1*1sec + S2*1sec)/2sec = (S1 + S2)/2
PF = P/S = (P1*1 + P2) / (S1 + S2) = (100 – 25) / (100 + 25) = 75/125


PF = 0.6 even though we had no reactive components!

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RE: Low power factor, recip pump

pete

I admit I havn't followed your math. Are you implying that this motor is somehow working as a generator ?

RE: Low power factor, recip pump

Yes, the motor acts as a generator during the portion of the cycle where the current lags voltage by more than 90 degrees (see attachment to the original post by iainuts).

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RE: Low power factor, recip pump

portions of the "cycle" was a poor choice of words.  Make that portions of the "waveform"...

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RE: Low power factor, recip pump

Attached is an overview of power factor related to this type waveform.

I started with the current and voltage from my previous simulation.  Then I multipled by 125% to scale the current approximately the same as iainuts' waveform.

Then I per-unitized current and voltage based on the FLA and line-to-neutral voltage expressed in peak basis.  Results are plotted as Ipu(t) [dark blue] and Vpu(t) [magenta]

Then I calculated Ppu = I*V/[Pnp/3]  averaged over a rolling 1-cyle (16.7 msec) period   [yellow]
Likewise I calculated Spu = Irms * Vrms/[Pnp/3]  over a rolling 1-cycle period. [cyan]

Then I calculated Pavg = average of Ppu over the ~ 10 cycle period from 4.52sec to 4.69 sec (roughly one revolution of the pump)  [green].   {Note you can see the green Pavg line is halfway up the "sinusoidally" varying yellow Ppu]

Then I plotted Savg =average of Spu over the same 10 cycle period

Finally I computed PF = Pavg/Savg = 0.21/0.36 = 0.58.

The results are not exactly dramatic enough to illustrate my previous points, but I think they match the original waveform fairly closely.  Which brings me again to suspect the originally reported power factor 0.25 was low (should be 0.58 by my estimation)... do you know over what period the power factor was computed?  I'll bet it was computed over a single cycle.  I think you can see that the results of a power factor calculation over a one cycle period could be all over the place.

The power is about 21% of nameplate by this waveform – very low.  We expect low power factor at low load as edison and others have pointed out.   I think for a 2-pole motor at 20% load we might typically expect PF ~ 0.7 and the difference between 0.7 and 0.58 is attributable to the reversing nature of the load.

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RE: Low power factor, recip pump

Quote (electricpete):

I think you can see that the results of a power factor calculation over a one cycle period could be all over the place.
I forgot to mention: PF over a one-cycle period is plotted in brown.  You could get any value in between if you calculated over a one-cycle period.  I'm not sure the instrument will be smart enough to know what the proper period is.  It is only by inspecting the waveform periodicity that we can recognize that 10 cycles is the correct period to compute power factor.  I don't think most meters are that smart.

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RE: Low power factor, recip pump

I should mention that Ppu and Pavg plotted are motor input powers, even though I chose to normalize them by an output power (motor nameplate).   Since Pavg = 0.21 on the input, the corresponding output would be 5 or 10% lower due to losses (0.19 - 0.20).

I still need to add in the effects of power factor correction capacitor when I get a chance.
 

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RE: Low power factor, recip pump

(OP)

Quote:

Finally I computed PF = Pavg/Savg = 0.21/0.36 = 0.58.

The results are not exactly dramatic enough to illustrate my previous points, but I think they match the original waveform fairly closely.  Which brings me again to suspect the originally reported power factor 0.25 was low (should be 0.58 by my estimation)... do you know over what period the power factor was computed?  I'll bet it was computed over a single cycle.  I think you can see that the results of a power factor calculation over a one cycle period could be all over the place.

...

I forgot to mention: PF over a one-cycle period is plotted in brown.  You could get any value in between if you calculated over a one-cycle period.  I'm not sure the instrument will be smart enough to know what the proper period is.  It is only by inspecting the waveform periodicity that we can recognize that 10 cycles is the correct period to compute power factor.  I don't think most meters are that smart.
Interesting... I'll have to do some digging into this.  I'll see if the meter supplier can lend any insight.  

Can you tell from the graph how much power the motor actually put out (ie: how much power the pump was using at the time)?  

Thanks again for all your effort on this.   

RE: Low power factor, recip pump

Quote:

Can you tell from the graph how much power the motor actually put out (ie: how much power the pump was using at the time)?  
My recreation of your numbers is certainly not exact. FWIW my model  predicted 21hp on the motor input which  would be in the ballpark 17 - 20hp on the motor output depending on efficiency at low load.  If you have your data in digital format I can do the same calc for your data (or I can upload my spreadsheet for you to do it),.

===============================

Attached are my results for adding pf correction caps to this motor (assuming voltage is unchanged, the only effect is to add a capacitive/leading current to the motor current).

KVAR = 0, PF = 0.58
KVAR = 10, PF = 0.65
KVAR = 20, PF 0.60
KVAR = 30, PF = 0.49
KVAR = 40, PF = 0.40

The first slide is same as previous presentation.  In each of the later slides you see an increasing leading capacitive current (pink) which is added to motor current (blue) to give total current (red).

As expected, there is no change to average power (Pavg = 0.21, light purple).
However the apparent power S initially decreases and then increases as the overall current becomes capacitive (in the motor region) and inductive (in the generator region).    It transitions to capacitive in motor region earlier than we expect.  A related observation is that the power factor in motor region of slide 1 (no caps) is already quite near 1.0.  Moreso than the data posted by the original poster.   I'm not sure exactly why that is.     To be sure we can't make predictions about the transient power factor using the steady state equivalent circuit.  

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RE: Low power factor, recip pump

The motor does not act as a generator due to power factor. The motor will only act as a generator if the reciprocating load over speeds the motor during part of the mechanical cycle.
The motor will produce VARs during part of the cycle but that is a result of applying an alternating current to an inductor and has nothing to do with rotation.  

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Low power factor, recip pump

Quote:

The motor does not act as a generator due to power factor.
I did not use the phrase "due to".    What I did say is that the gradual change in current lag angle from ~ 0 to ~ 180 (and in power factor from ~+1 to ~-1 with positive power factor representing motor action and negative representing generator action in this case) corresponds to the reversal in real power flow.    

In my analysis I have not referred to the concept of vars (except as a means of describing  pf capacitors).  I have instead used real power and apparent power which can be determined directly from the waveforms.  Additionally I have used the relationship:  PF = Real Power / Apparent Power.  I have tried to stay away from vars and the  "power triangle" as these are steady state concepts and I think will lead us astray during transient analysis (*).  For example my post 13 Jul 09 19:46  illustrates a situation where power factor is 0.6 in a circuit including only resistances and sinusoidal voltage sources and a time-varying switch.   Try to explain that using the power triangle!  You cannot.  But we can easily explain (compute) the power factor for that circuit from real power and apparent power.

*  I have not been 100% rigorous in staying away from the power triangle and the sinusoidal steady state mentality.  I have referred to the "phase angle" between voltage and current and inferred a relationship between phase angle and power factor.  But phase angle is a steady state concept!  And p.f. = cos(phi) is inherently a steady-state concept.  In my defense, I believe  it helps with the intuition to talk about the phase angles of these waveforms which are ROUGHLY  sinusoidal over a time interval of a cycle (if you squint!), but it is not as rigorous as computing the real and apparent power as was done in the spreadsheet.

If you think there is a specific statement which is wrong, please quote it directly.

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RE: Low power factor, recip pump

To be more precise I should have used the term "sinusoidal steady state" instead of "steady state" throughout my post directly above.   Also to clarify, cos(Phi) is the sinusoidal steady state concept, whereas power factor is not limited to the sss context.

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RE: Low power factor, recip pump

Bill - I did state several times that the reversal of direction of real power flow corresponds to current lag angle changing from below 90 degrees to above 90 degrees.  Is that what you disagree with?

For sinusoidal steady state it can easily be shown in the either the time domain or phasor domain to be 100% correct.   For this particular waveform which ROUGHLY resembles a sinusoid over one period, the conclusions about power flow direction based on phase angle are roughly correct.  The exact time of power reversal can be computed and is shown graphically but not as intuitive as the conclusion we draw by eyeballing the phase angle.
 

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RE: Low power factor, recip pump

Are you trying to say that a low power factor is normal on a lightly loaded motor? Sorry, I got lost somewhere.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Low power factor, recip pump

Low motor power factor is certainly normal on a lightly loaded motor in sinusoidal steady state.  I think everyone will agree on this and it is obvious from our normal steady state equivalent circuit.   This type of low power factor can be improved by adding power factor capacitors.

Low motor power factor will also occur if the motor has an oscillating load such that power reverses direction.  It is not so obvious why that is the case if you try to visualize it using impedances of the equivalent circuit.     But again I think the principle is well illustrated in my post 13 Jul 09 19:46 (resistor, time-varying switch etc).     This type of low power factor can not be improved by adding power factor capacitors.

Both causes of low power factor are at work in the motor posted at the beginning of this thread.    
 

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RE: Low power factor, recip pump

Check the difference between an oscillating load and an overhauling load. As the piston compresses the gas, the flywheel is slowing down. When the piston reaches top dead center most of the gas is discharged. There is not compresed gas energy left in the cylinder to accelerate the motor over speed. The flywheel is being accelerated so as to assist with the next compression cycle. No regeneration. The load is varying during each revolution but not as much as may be thought. As the load varies the power factor varies, but capacitors may still correct to unity PF despite the varying load. Before you dissolve in calculations, draw a couple of power triangles and remember that the VARs remain relatively constant regardless of load. Those VARs may be supplied by capacitors.
Come on, many utilities charge PF penalties based on VAR consumption per month. You can certainly average VARs over a revolution, and when the PF is corrected to unity, the average of zero VARs doesn't take too much math.
Please recheck your rebuttal twice before posting.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Low power factor, recip pump

(OP)
Hi Bill,

Quote:

The motor does not act as a generator due to power factor. The motor will only act as a generator if the reciprocating load over speeds the motor during part of the mechanical cycle.
Thanks for speaking up.  How do you interpret the graph in the OP which shows voltage and current on opposite sides of the 0 line?  From this thread, and from my limited understanding of what's going on inside an electric motor, electric power is not going into the pump, it is actually going 'back onto the grid' and yet the customer is being charged for that power.  

Quote:

As the piston compresses the gas, the flywheel is slowing down. When the piston reaches top dead center most of the gas is discharged. There is not compresed gas energy left in the cylinder to accelerate the motor over speed. The flywheel is being accelerated so as to assist with the next compression cycle. No regeneration.
I'd agree with all this.  But although the pump (compressor) itself is not accelerating the flywheel, I suspect that as the motor is slowed during compression, more current is necessary to overcome that resistance which is seen where the current peaks (ex: time = 0.03).  The reciprocating nature of compression results in a very sudden load, and this sudden load causes a momentary dip in motor RPM, and it's this momentary dip in RPM that causes the current to increase very rapidly.  So, similar to an inrush, there's a step change in current needed for the compression.

Now if you look a bit further out along this timeline, between 0.08 and 0.13, the peak current and peak voltage are on opposite sides of the 0 line.  I'm assuming, when the motor had this surge around 0.03, it oversped the pump and now has to slow back down, and that's when power is actually coming out and it's acting like a generator. Thoughts?

 

RE: Low power factor, recip pump

(OP)

Quote:

If you have your data in digital format I can do the same calc for your data (or I can upload my spreadsheet for you to do it),.
Thanks for the offer.  I've requested the data in Excel format, which I can post.  I'll keep this thread updated as we work through the issue for future reference.

RE: Low power factor, recip pump

I'm not sure how instructive it is to say the motor is generating for part of the cycle. It is true that motors can generate, and that energy is stored by the inertia. But you don't need a motor to cause a negative instantaneous power for part of the cycle; any RC or RL circuit will have the same effect.

Does low pf retard the motor: The motor is retarded by the load, and the pf is simply is simply matching the load.

What can be done about low pf: Low pf is not a problem until you put a cost on it. You can raise the pf at your metering point by adding capacitors downstream, but the pf at the motor terminals will be unchanged. If your utility charges a penalty for low pf, it may make economic sense to add capacitors.  

 

RE: Low power factor, recip pump

The load is a combination of the pumping load and the flywheel load. As the compression is slowing the motor, both the motor and the flywheel are contributing energy to the load. When the piston passes top dead center, the load is accelerating the motor, flywheel and pump back to maximum speed.
There are a couple of second order effects that I will ignore to keep it simple.
The current is a combination of in phase current and reactive current. The power is the product of the voltage and the in phase current.
The reactive current is at 90 degrees to the voltage. As a result, there is no power consumed or generated by the reactive current.
The actual or apparent current is the vector sum of the real and the apparent current.
Now, keeping it simple, let's look at the power triangle:
This is a right triangle.
The altitude may represent  either the reactive current or the VARs. Let's work with VARS.
The hypotenuse represents the apparent power or VA (Volt Amps). This is what you see when you measure the voltage and the current and take the product.
The base represents the Watts or real power.
The VARs are constant for a running motor.
The Watts depend on the load on the motor plus the losses.
Let's take a hypothetical motor and say that the KVARs are 40. The altitude of our triangle will be a scale of 40.
There is little load on the motor and the kW are 10.
From Pythagoris, the hypotenuse will be 41.23 KVA
10/41.23 = 0.24  Power factor is 24%
Now increase the load until the kW are 100. Kvars are still 40.
The hypotenuse or apparent power is 108 KVA
100/108 = 0.93 or 93%
Imagine a wooden board about 8 inches wide and 24 inches long. Imagine a horizontal line near the closer edge and a vertical line at 90 degrees near one end. Drive in a nail at the intersection of the lines and another nail about 6 inches above it on the vertical line. Now loop an elastic around both nails and pull the corner of the loop along the horizontal line. The elastic will form a triangle. pulling the elastic back and forth will demonstrate the power triangle of a motor with a changing load. The KVARs remain constant, but the real power, the apparent power, the angle between the real power and the apparent power and the ratio between them will be changing. I find this hard to visualize from waveforms from a recording device. Once I grasped an understanding and became comfortable with power triangles I found waveforms much easier to analyze.
Now inductive KVARs (motors) may be cancelled by capacitive KVARs (capacitors). The cancellation or correction may be full or partial.
Changes in the system voltage will affect the balance between Inductive reactance and capacitive reactance.
I'm tired and it's hot here. I'll pause and give you a chance to throw in any questions.
 

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Low power factor, recip pump

Hi Bill - Needless to say we have a disagreement.  I see your transient analysis rests solidly on the foundation of applying the concept of vars to the power triangle.  Please apply this analysis to the scenaro that I posed 13 Jul 09 19:46.  I will look forward to seeing how you compute power factor = 0.6 when the vars are 0.

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RE: Low power factor, recip pump

Didn't see the part about customers paying for power going back to the grid. The sub-cycle variations of energy use are fully accounted for in any watt-hour meter built to ANSI accuracy. These meters use instantaneous power and time.

What customers may pay for, depending on the utility and rate class, is the pf penalty I mentioned above. This is to recover the investment required to serve low pf loads while keeping acceptable voltage levels to all loads.

RE: Low power factor, recip pump

Pete I don't try to mislabel circulating currents resulting from the interconnection of two separate sources as power factor, so I don't have to reconcile it. And, what has that to do with a motor?

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Low power factor, recip pump

Bill – I have to admit that the tone of your posts to me starting 14 Jul 09 14:39 caught me by surprise.     It is time to start clearing this stuff up.

Quote:

The motor does not act as a generator due to power factor.
Agreed. Please point to any statement of mine that you think implies otherwise.

Quote (waross):

The motor will only act as a generator if the reciprocating load over speeds the motor during part of the mechanical cycle.
Was it not clear to you that I was talking about a reciprocating load?  That was after all  the subject of the thread.  Did you see in my simulation that the speed went above synchronous?   So... what statement of mine are you possibly disgareeing with?

Quote (waross):

Check the difference between an oscillating load and an overhauling load. As the piston compresses the gas, the flywheel is slowing down. When the piston reaches top dead center most of the gas is discharged. There is not compresed gas energy left in the cylinder to accelerate the motor over speed. The flywheel is being accelerated so as to assist with the next compression cycle
Similar as above – I have shown the assumptions of my model -  they include a load torque which oscillates into negative territory.  The assumptions are  transparent and the resulting waveform matches the posted waveform.   So what it is that I am supposed to "check" and why do you seem to think that your description of reciprocating loads conflicts with anything I said?

Quote (waross):

No regeneration.
I'm not sure the context intended here but in the waveform posted in the OP has several consecutive cycles per revolution over which the average power flows from pump to motor.  i.e. the motor is acting like a generator during that period.  

Quote (waross):

the VARs remain relatively constant regardless of load
Vars are a sinusoidal steady state concept which is why you may notice I did not use them at all in my analysis..   You can get in trouble trying to apply the concept of vars simplistically during a transient which deviates significantly from SSS.    This is illustrated in the circuit scenario that I posed above (resistor, switch etc).     I don't think you can find any definitions of var which do not rely on sinusiodal steady state concepts.   In contrast  there exist precise definitions of average power, apparent power, and power factor which do not in the least rely on being anything close to sinusoidal steady state.

Quote (waross):

Come on, many utilities charge PF penalties based on VAR consumption per month.  You can certainly average VARs over a revolution, and when the PF is corrected to unity, the average of zero VARs doesn't take too much math.
Here again, I have not used vars in my analysis or made any claim about any change in vars, so you're not contradicting anything I said.  How the waveform will be perceived varies by type of monitoring equipment and I suspect the lower power factor due to oscillating torque load would not show up on a a rotating var meter, just as low power factor due to current harmonics wouldn't show up in a rotating wattmeters. ... but does that mean the sub-unity distortion power factor from current harmonics doesn't exist ?  (rhetorical question – the answer is no).   The point is again vars are not the whole picture in presence of this non-sinusoidal waveform.  With a modern digital  instrument monitoring the motor input current and voltage, the increased apparent power and associated lower power factor would be apparent (no pun intended).   The effect of monitoring equipment, rate structures, combined loads etc might be interesting topics to explore, but does not conflict with anything that I said (I have in my previous posts said absolutely nothing about the rate structure or monitoring equipment.).  

Quote:

Imagine a wooden board about 8 inches wide and 24 inches long. Imagine a horizontal line near the closer edge and a vertical line at 90 degrees near one end. Drive in a nail at the intersection of the lines and another nail about 6 inches above it on the vertical line. Now loop an elastic around both nails and pull the corner of the loop along the horizontal line. The elastic will form a triangle. pulling the elastic back and forth will demonstrate the power triangle of a motor with a changing load. The KVARs remain constant, but the real power, the apparent power, the angle between the real power and the apparent power and the ratio between them will be changing
As I said vars don't apply to this analysis.   And we need to be cautious about vectors also (a phasor representation of SSS quantities).    So I would not necessarily agree this is a productive model to spent too much time on.   Nevertheless I think there are some modifications to the model to make it illustrate something resembling average power factor.  You have to pull the rubber band both the the left and right of the vertical line to reflect the alternative periods of motor/generator action, and when you're done you have to come up with a single number (power factor) representing the integral of signed (algebraic) real power during the interval divided by the integral of the unsigned (always positive) apparent power during the interval.   The pulling in opposite directions along the horizontal axis tends to decrease the nuemrator of that fraction (integral of real power) due to the opposing direction, but it does not decrease the denominator of that fraction (integral of apparent power).      The net result is that pulling in alternately in opposite directions reduces the ratio (power factor).

Quote (waross):

Pete I don't try to mislabel circulating currents resulting from the interconnection of two separate sources as power factor, so I don't have to reconcile it. And, what has that to do with a motor?
What mislabeling? What circulating currents?   Let's break it down:
* For the period when switch is closed, we have only one source to worry about.
* For the period when switch is open, we have two voltage sources in series, so we add their voltages  algebraically.
* With the voltage sources simplified during each period as above we can solve the currents I = V/R.   From currents and known voltages we compute the real and apparent powers as shown in my prior post.  It's not a trick.    

I picked a constant external source and an intermittent internal source because it has some resemblance to our motor.    But if two sources causes you heartache for some strange reason, we can certainly substitute a slightly different circuit that has only one power supply (external) whose voltage varies between +10vac for 1 second and –5vac for a second ... repeat...

First second:  Vext = 10vac
P1 = 10^2 / 1 = 100 watts;   S1 = 100VA

Second second: Vext = -5 vac
P2  = -(5)^2 / 1 = -25 watts;    S2 = 25VA

Overall PF = (P1 + P2)/(S1+S2) = (100-25)/(100+25) = 0.6

Quote (waross):

What does it have to do with a motor:
#1 – Motors are required to obey the same laws of physics as any other device.   If you  are using a circuit analysis approach that cannot satisfactorily explain transient behavior in a very very simple circuit, then I would think you would be more than a little bit worried about trying to apply the same circuit analysis technique toward analysing the transient behavior of a much more complicated device (induction motor experiencing torque oscillations that cause severe amplitude and phase modulation of the current).
#2 – There are definite similariities.  Reversing power results in lower power factor in both cases.   I have removed reactive elements from the simple circuit to make it even more obvious that the low power factor in that circuit has absolutely nothing to do with reactive elements or reactive power.
#3 – It was intended to lead you to think very carefully about  the limitations of the vars / power triangle approach for determining power factor in a non-sinusoidal-steady- state circuit.  Those limitations are real and they are very obvious in the simple circuit example.  

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RE: Low power factor, recip pump

Quote:

I picked a constant external source and an intermittent internal source because it has some resemblance to our motor.    But if two sources causes you heartache for some strange reason, we can certainly substitute a slightly different circuit that has only one power supply (external) whose voltage varies between +10vac for 1 second and –5vac for a second ... repeat...First second:  Vext = 10vacP1 = 10^2 / 1 = 100 watts;   S1 = 100VASecond second: Vext = -5 vacP2  = -(5)^2 / 1 = -25 watts;    S2 = 25VAOverall PF = (P1 + P2)/(S1+S2) = (100-25)/(100+25) = 0.
No, that's not a good analysis.  Power does not switch polarity.  It was late when I typed that.  Let's stick with the scenario posted 13 Jul 09 19:46

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RE: Low power factor, recip pump

Relax, Pete!

You owe it to yourself - and us.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: Low power factor, recip pump

Thanks for the concern Gunnar.  As I had posted 14 Jul 09 14:25:

Quote (electricpete):

Low motor power factor is certainly normal on a lightly loaded motor in sinusoidal steady state.  I think everyone will agree on this and it is obvious from our normal steady state equivalent circuit.   This type of low power factor can be improved by adding power factor capacitors.

Low motor power factor will also occur if the motor has an oscillating load such that power reverses direction.  It is not so obvious why that is the case if you try to visualize it using impedances of the equivalent circuit.     But again I think the principle is well illustrated in my post 13 Jul 09 19:46 (resistor, time-varying switch etc).     This type of low power factor can not be improved by adding power factor capacitors.

Both causes of low power factor are at work in the motor posted at the beginning of this thread.
That sums up my position which is well supported by my previous comments and attachments.    
 

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RE: Low power factor, recip pump

Quote (waross):

As the load varies the power factor varies, but capacitors may still correct to unity PF despite the varying load.
I have shown the effect of adding capacitors in my post 14 Jul 09 9:16,  and they do not correct to unity.

Quote (waross):

The actual or apparent current is the vector sum of the real and the apparent current.
I assume you meant to say apparent is the vector sum of real and reactive. (obvious typo - not a concern).  But that is a sinusoidal steady state concept which does not apply to this transient analysis.  It is illogical to use a sinusoidal steady state analysis of a transient phenomenon to override conclusions which are based on transient analysis of a transient phenomenon.  Especially when the sinusoidal steady state analysis has been shown to fail for other transients.....

Which leads me back to my simple circuit scenario posted.   At the risk of being repetitive,  since I revised one aspect of the discussion (retracted the new scenario with only one source), I want to make it crystal clear how waross' objections to my simple circuit scenario posted 13 Jul 09 19:46 are addressed

Quote:

Pete I don't try to mislabel circulating currents resulting from the interconnection of two separate sources as power factor, so I don't have to reconcile it
There is no mislabeling or circulating currents:  
* For the period when switch is closed, we have only one source to worry about.
* For the period when switch is open, we have two voltage sources in series, so we add their voltages  algebraically.
* With the voltage sources simplified during each period as above we can solve the currents I = V/R.   From currents and known voltages we compute the real and apparent powers as shown in my prior post.  It shows a power factor far less than 1 even though no reactive elements are in the circuit.

Quote:

And, what has that to do with a motor?
#1 – Motors are required to obey the same laws of physics as any other device.   If you  are using a circuit analysis approach that cannot satisfactorily explain transient behavior in a very simple circuit, then I would think you would be more than a little bit worried about trying to apply the same circuit analysis technique toward analysing the transient behavior of a much more complicated device (induction motor experiencing torque oscillations that cause severe amplitude and phase modulation of the current).
#2 – There are definite similarities.  Reversing power results in lower power factor in both cases.   I have removed reactive elements from the simple circuit to make it even more obvious that the low power factor in that circuit has absolutely nothing to do with reactive elements or reactive power.
#3 – It was intended to lead you to think very carefully about  the limitations of the vars / power triangle approach for determining power factor in a non-sinusoidal-steady- state circuit.  Those limitations are real and they are very obvious in the simple circuit scenario posed 13 Jul 09 19:46.
 

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RE: Low power factor, recip pump

(OP)
Hi electricpete, Thanks again for the input.  I still need to pick your brain some more, so don't get too wound up.  Your advice is highly valued!  :)

What I'm still struggling with is what's going on when "the motor is acting like a generator during [the] period" when current and voltage are of opposite sign.  I believe that's the point you are making here, so please correct that if I'm wrong. You mentioned that in your simulation, motor speed went above synchronous speed, and I'm assuming it's at this time when the voltage and current flip sign.

- Can you explain how that happens? Are you suggesting that the compressor is actually the source of this overspeed?
It's possible that re-expansion of gas in the compressor's void volume (the small volume in the head that can't be completely displaced during compression) actually contributes to some power returning to the motor, in which case I could actually analyze that and see how much but I have to believe it's very small.

- Once above synchronous speed, I understood you and others here to say that current is actually being 'generated' at this point in time.  Is this true or false?

- The current being measured, is that "apparant current"?  And is apparant current a combination of in phase current and reactive current with power being the product of the voltage and the in phase current?  Ultimately, I'd like to understand how customers are being charged for the power that's used by this motor.  This is still confusing me.

Sorry for all the dumb questions,
Dave.

RE: Low power factor, recip pump

iainuts

Once the motor goes beyond synchronous speed of 1800 RPM, it acts an induction generator and returns power to the grid. But in your case, I don't think that is happening even with the flywheel.  

RE: Low power factor, recip pump

Last time I checked, a motor had a fairly constant inductive VAr component and that component could be cancelled out using a capacitor.

Electricpete, I think you're somewhat falsely calling the cyclic reversal of power in your 13Jul09 19:51 post a varying power factor. The power factor is always unity but the direction of power flow keeps changing. As stevenal posted the meter will read the instantaneous wattage and should just end up calculating power as power used - power returned over the long term. I have read that some power meters will not ignore power returned to the grid but that is not my area of expertise.

I've still run into engineers who insist that a switching capacitor bank is required to correct the power factor of a motor with a changing load. The last one was willing to pay about $90k more than necessary for the capacitors because they didn't understand, and at that point I lost all desire to help them.

I'm with Bill, even if the motor is generating, adding capacitors should correct the motor so the power factor has a smaller fluctuation around unity.

Using real motor data for a 50hp motor I get the following.

CODE

motor only;

load   power factor
50%     -0.77
75%     -0.83
100%    -0.86

after adding 15kVAR

load   power factor
50%     -0.996
75%     -0.986
100%    -0.977

I then will postulate that the motor will basically mirror the motoring conditions when it's generating. In other words, the operation is mirrored around the synchronous speed at least for the range of speed we are discussing. So, for the above motor with a 15kVAR capacitor, this would mean the power factor would fluctuate between about -0.996 to 0.996 if the motor went from 50% load to 50% regenerating.

I think using a long term calculation using the power reversals as part of the power factor calculation is flawed from a field point of view as the hydro meter won't care.
 

RE: Low power factor, recip pump

Quote:

- Can you explain how that happens? Are you suggesting that the compressor is actually the source of this overspeed?
It's possible that re-expansion of gas in the compressor's void volume (the small volume in the head that can't be completely displaced during compression) actually contributes to some power returning to the motor, in which case I could actually analyze that and see how much but I have to believe it's very small.
Theoretically, yes, but not likely to happen in practice.
You are concerned with a low power factor on an oversized motor.
Some possible issues:
Starting; There may be instances when a reciprocating load needs more torque to start then available from a standard motor. More likely with a liquid pump than with a compressor with a functioning unloader.
The majority of pumps and compressors DO NOT have oversized motors.
However you may need a design "C" or a design "D" motor to handle the starting torque.
See the Cowern papers for accurate and well explained information of motors.
http://baldor.com/support/literature_load.asp?LitNumber=PR2525
Check the section on RMS loading. While it is not meant to apply to reciprocating loads, the concept may be helpful.
Power usage by an oversized motor:
You pay for the power used plus the losses. It is common for motors to exhibit maximum efficiency at less than 100% load, so the oversized motor MAY actually be more efficient than a smaller motor.
Power factor:
Keep it simple. The real current drawn by a motor is basically sinusoidal. The reactive current has a little distortion but not enough to worry about. (nonsinusoidal distortion becomes a serious issue with rectifier loads where the current only flows during the peaks of the voltage wave.)
I suspect that your instruments are misleading you as to the actual power factor.
You mentioned reading the peak current. I am unsure whether you mean the peak of the varying load current or the peak of the current waveform. You may be happier to step back in technology and measure the current with an analogue ammeter.
A motor draws VARs. The VARs may come from the utility supply or from a capacitor bank. The VARs (or KVARs) are a product or the Voltage and the Reactive Amps. The motor will always draw VARs. The issue is the source of the VARs. If the motor is allowed to draw VARs from the utility supply, the utility [b[MAY[/b] impose a penalty. There are differing methods of calculating the penalty depending where you are in the world. Some calculations are very harsh, some are more forgiving. Even the most forgiving calculations usually make it cost effective to install PF correction, but that is a matter for a separate thread.
It is usually possible to install correction equipment that will reduce to zero the VARs metered by the utility. Some of us call that 100% correction.
Some utilities only charge penalties when the monthly average PF drops below 90%. Some of us will install 9% correction in such cases.
Correcting power factor is about avoiding penalties. The method and amount of correction depends to a large extent on the penalty rate structure. For 90% correction, PF correction as an art Vs PF correction as a science may be an issue. PF as an art is much more economical than PF as a science but is not always possible.
But Mr. Cowern said it all much better than I can. Check the Cowern papers. You may enjoy them.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Low power factor, recip pump

The original question asked for an explanation of power factor measured on a digital recording device (not computed from a wattmeter and a varmeter).   That is what all of my comments have addressed.  I have used the concepts of  average power, apparent power power factor.   I have not made any claims about vars.

There is another question that has emerged what will be seen by various types of  utility metering and specifically a device like a rotating var meter (rotating watt meter reconnected to measure vars).   It is a very practical and interesting question.  I am sure there are many people that can join that discussion.   I will not be one of them.  I have my hands full defending out-of-context criticisms of my original comments and so I will focus on the original context.

That's it until tonight.  I have work to do.
 

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RE: Low power factor, recip pump

Bill

"The reactive current has a little distortion".

Why would the reactive current be distorted ?

RE: Low power factor, recip pump

An apology Pete. I went back and took a close look at the original graph. For two cycles per pump revolution there is a very small regeneration. They proof is in a few cycles in the center of the low current part of the graph.
http://files.engineering.com/getfile.aspx?folder=8b5a4a9e-01c8-4e49-9a0e-59
Time stamps 0.08, 0.10, 0.12 and 0.14.
If there was just a lagging phase angle on a graph where I was not privy to the setup of the equipment I would dismiss it as measurement errors, but here, while the phase angle stays lagging, the current increases slightly. As the load on the motor decreases the current decreases until the minimum current is reactive current plus negligible losses. The slight increase in current is indicative of regeneration.
Electrically this is so small as to be negligible. Use a heavier flywheel and the effect will go away.
By the way, you don't have a leaky discharge valve do you.
A smaller motor may also mitigate the regeneration.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Low power factor, recip pump

This is getting way off track, folks. iainuts (Mechanical) is simply asking about about the sub-cycle negative going instantaneous power that occurs in any low pf load. Try doing your analysis on an RL circuit. No regeneration or distortion is needed, even if you happen to see these effects in the badly scaled waveforms he provided.

He is also seeing a problem where no problem may in fact exist.

Any chance we can steer this guy straight? I tried above, but I'm afraid my posts were lost in all the weeds.

RE: Low power factor, recip pump

(OP)

Quote:

waross: An apology Pete. I went back and took a close look at the original graph. For two cycles per pump revolution there is a very small regeneration. They proof is in a few cycles in the center of the low current part of the graph.
http://files.engineering.com/getfile.aspx?folder=8b5a4a9e-01c8-4e49-9a0e-59
Time stamps 0.08, 0.10, 0.12 and 0.14.
If there was just a lagging phase angle on a graph where I was not privy to the setup of the equipment I would dismiss it as measurement errors, but here, while the phase angle stays lagging, the current increases slightly. As the load on the motor decreases the current decreases until the minimum current is reactive current plus negligible losses. The slight increase in current is indicative of regeneration.
Electrically this is so small as to be negligible. Use a heavier flywheel and the effect will go away.
By the way, you don't have a leaky discharge valve do you.
A smaller motor may also mitigate the regeneration.
Interesting...  Leakage isn't likely.  On shut down, there is a very small (a few cubic inches) section of pipe downstream of the compressor which stays pressurized.  It wouldn't stay pressurized if the discharge valve leaked as it would blow down through some high pressure shaft seals.  Also, tests done in the shop at ambient indicate this valve is bubble tight.  I'm wondering however if at least part of this issue is re-expansion of gas inside the cylinder since there is always a void volume of trapped gas that doesn't exit the chamber.  That affect should be exceedingly small.



 

RE: Low power factor, recip pump

(OP)

Quote:

stevenal: This is getting way off track, folks. iainuts (Mechanical) is simply asking about about the sub-cycle negative going instantaneous power that occurs in any low pf load. Try doing your analysis on an RL circuit. No regeneration or distortion is needed, even if you happen to see these effects in the badly scaled waveforms he provided.

He is also seeing a problem where no problem may in fact exist.

Any chance we can steer this guy straight? I tried above, but I'm afraid my posts were lost in all the weeds.
Thanks for the input, your comments aren't getting lost.  I guess what I'm hearing from everyone is that there are some straightforward things that could be done to improve this.
1. Smaller motor: Dropping to 60 or 75 hp may be doable, but there's a side issue with this.  An upset condition would result in the motor stalling and tripping the breakers.  The 100 hp motor would probably keep going.  That has to be evaluated.
2. Add capacitors: Sounds like this helps reduce the current up to the capacitors but not through the motor.  Good solution to reduce current draw at the customer's pay meter perhaps but doesn't help motor current.  Is that right?
3. Add flywheel inertia:  We actually tried doubling the inertia but hand held amp meters saw absolutly no difference whatsoever!  Note - this was prior to putting in the meter that gave us the print out attached, so I'm not sure if the added flywheel plates helped or not but it doesn't seem as if they did anything.  There's not much room left on the crankshaft, so I'm considering adding a balanced wheel to the motor, especially if we go smaller.  Thoughts?
4. Different types of motors (different RPM, poles, ?) probably won't get us anywhere.

Anything else?

Thanks for everyone's input.

RE: Low power factor, recip pump

Electricpete - you are correct in that the origional post asked why the power factor measured so low and your data shows the effect which could be messing up the meter.

But, you ended up "This type of low power factor can not be improved by adding power factor capacitors." I'll agree that the capacitor may not fix the power factor the meter is reading but the capacitor will still correct the motor power factor even when the load is cyclic.

The origional poster appears to be concerned about the motor power factor and correcting it. It also appears he attempted to measure the motor to determine what size of capacitor is required but the meter readings were confusing.


Stevenal - We could. If iainuts is concerned about the motor having a low power factor then add a capacitor the same way as he would do with any other motor. Use motor efficiency and motor pf to calculate the required capacitor size. 30kVAR, 480V as posted by edison123 is a good starting capacitor if more data is not known.

 

RE: Low power factor, recip pump

Read the Cowern papers. If you consider changing motors, consider a high torque type motor. A design "C" will probably suffice. Design "D" motors are for extreme situations like alligator presses and large sheet steel shears.
I am surprised that the trapped gas can supply enough energy to accelerate the motor into regeneration. As for power usage, it cancels out. The power used will mainly be a function of the amount of gas compressed, and of course, temperature and pressures. Losses may be a little different with different flywheel inertia but when you work out the difference in dollars over a year of slightly different losses when the load profile is changed by changing flywheel inertia, you won't waste time like that a second time.
(A portion of the losses are related to the square of the current. If more flywheel inertia lowers the peak current, the peak losses will drop but lower losses will be spread over more time for a small net reduction in losses.)
The power returned during the regen has been put into the system as the flywheel was accelerating and as the piston was compressing. For power factor I am concerned with the net power, and the power factor over several revolutions, not on a cycle by cycle basis. If you want to use cycle per cycle calculations, you must calculate all cycles and then combine the results.
One thing that has not been mentioned is the value of your graph to determine PF correction.
Take the lowest current value, when the current is 90 degrees displaced from the voltage. Multiply by the voltage and by root three. The answer will be in VAR. Select the closest size capacitor bank below this level. NOTE this will not work for most installations, but with a 90 degree displacement and evidence of regeneration it should serve to determine reactive current in this instance.
I scaled your minimum current as 40 amps.
I get 33240 VAR, use a 20 KVAR or 30 KVAR capacitor bank.

Quote (Pete):

"The reactive current has a little distortion".

Why would the reactive current be distorted ?
I remember the good old days when you remembered that the hysteresis  of the magnetic core rendered the magnetizing curve nonlinear. A little magnetic nonlinearity results in a little current nonlinearity equals a little distortion of the reactive current, usually ignored.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Low power factor, recip pump

OK, now I have no idea what you want.

At the end of the post you questioned the power factor. To correct the power factor, just add a capacitor. Now, it seems you want to take further actions.

"Good solution to reduce current draw at the customer's pay meter perhaps but doesn't help motor current.  Is that right?"

Not really, the motor current due to poor power factor does not really matter.

Also, there was a post here not too long ago that showed that a 2 times oversized motor was actually more energy efficient. I believe it was comparing a 5hp and a 10hp doing the work of a 5hp motor.
 

RE: Low power factor, recip pump

I disagree with the flywheel. Waross suggested it for some regeneration he observed. Won't have much effect on overall pf. Capacitors, changing motor, changing load can all change the pf. Yes, capacitors will not affect motor current.

Still don't know what the problem is. Low pf by itself is not a problem. It only becomes a problem if it causes excessive voltage drop, overloads equipment, or causes unacceptably high utility bills via a pf penalty.

RE: Low power factor, recip pump

(OP)
Had a long meeting on this in the afternoon. Now I understand why PF isn't an impact on a typical power meter.  Also got a much better understanding of why volts and amps are on opposite sides of the 0 line and what reactive current, apparant current, etc... is all about.  Very enlightening!  lol

Still, the problem with PF is twofold.  If we can't reduce the motor size, we have to buy motors that are much larger than necessary.  That's capital cost on all new units.  Second is that customers now have to install a larger electric supply in order to put this machine in.  That's a cost to our customer.  Third, there's a soft issue - it looks like the machine is inefficient and needs a motor that's much larger than actually necessary.  Because these are green projects, it doesn't look good to have inefficient machinery, even though it isn't really inefficient.

The recommendation by the electrical engineer was not to use a smaller motor because the peak torque necessary for this machine may cause a smaller motor to stall or trip breakers.  He also recommended not using capacitors because the current trace is too 'odd' - he thinks there's some unusual interaction between the motor and compressor that may just get worse if capacitors are installed.  He's concerned that current could actually be adversely affected, and any simulation would be far too difficult.  (I think he'd be impressed by e-pete's analysis.)

The final recommendation was to add a flywheel to the motor pully since motor RPM is 5 times compressor RPM and therefore 25 times more effective.  I have no problem with the resolution.  There really isn't any additional stress on any of the machine's components doing this, and we'll be getting additional data to see how this solution works so that we may in the future be able to reduce motor size.  

Thanks again for all the input.  
Dave.

 

RE: Low power factor, recip pump

I agree with stevenal. And does this 100 hp reciprocating pump probably qualified with that problem criteria? Maybe, or maybe not.
The OP sees it a problem! How much was the penalty slapped on his plant? I got billed around $2000 more with a 0.35 lagging pf once (around 100 welder units) and that was a real problem.

RE: Low power factor, recip pump

Look at design "c" and design "B" motors. I doubt that your engineer has much industrial experience. The varying current is the real component of the current. The capacitors supply the reactive component of the current which is relatively constant. No problem.
stevenal; I did not make myself clear. My bad. An increased flywheel may reduce or eliminate the regeneration. It will not have an effect on PF as you noted.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Low power factor, recip pump

(OP)
Thanks Bill.  I'll do that.

RE: Low power factor, recip pump

Sorry for the typo. Should have read design "C" and "D".
Design "C"s preferred.
 

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Low power factor, recip pump

waross – my apologies – I think I had misunderstood some of your comments.  Particularly about the role of the flywheel -  that was a good point.

=====================

Quote (Lionel):

1Electricpete, I think you're somewhat falsely calling the cyclic reversal of power in your 13Jul09 19:51 post a varying power factor. 2The power factor is always unity but the direction of power flow keeps changing. 3As stevenal posted the meter will read the instantaneous wattage and should just end up calculating power as power used - power returned over the long term.
Regarding statement 2 – there is no logic to chop up the interval into pieces and ignore how those pieces interact with each other.  It would be misleading in that it would not at all reflect the true relationship between average/real power and apparent power.    Imagine if you were a power provider whose cost to transmit the power depended on apparent power (*),  would you still be content to call this PF=1.0?   (* yes I know transmission costs are more complciated than that).
Regarding statement 3 – it relates to real power.  I have made no comments about metering or billing of real power.  

Quote (Lionel):

Electricpete - you are correct in that the origional post asked why the power factor measured so low and your data shows the effect which could be messing up the meter.

But, you ended up "This type of low power factor can not be improved by adding power factor capacitors."   I'll agree that the capacitor may not fix the power factor the meter is reading but the capacitor will still correct the motor power factor even when the load is cyclic.
There are two important aspects to discuss:  1 – Terminolgy;  2 – Billing-meter considerations

1 – Terminology.  Let's create the terms "true power factor" and "indicated power factor".     The meanings are typical of "true" and "indicated"...  the true one is the correct one, the indicated one may or may not reflect the true  power factor due to limitations of sensing or computation method.     So which is which?  The only way I can make sense of your comment is if you consider my calculated numbers as an "indicated power factor" and what might be read by someone computing power factor from a varmeter and wattmeter as the "true power factor".    But that is utterly backwards from what I would call it.  We are talking about a non-sinusoidal waveform.  The true power factor is average power over apparent power which is precisely what I have calculated and discussed (I am pretty sure I could find an IEEE document to back that up if such is needed).    In this context of a (non-sinusoidal steady state ) any other power factor we have discussed would be indicated (for example indicated power factor computed by output of watt and var meter is a SSS-formula-based approximation of power factor, but not the true power factor).   I am not hung up on the terminology but I do think it is important enough to clarify this particular term (power factor) in this discussion.

2 – Billing-meter considerations.  I don't think you were hung up on the terminology either, and I think your real point was a practical one related to what would actually be sensed by certain instruments used for billing (as I call it – the indicated power factor for those instruments).    It is a very good point as raised by others previously and I am pretty sure that rotary devices would give an indicated power factor corresponding to the SSS-formla based version of power factor i.e. watts / sqrt(watt^2 + var^2).     Like you, I am not a big metering guy.   So I have to ask - is this the only type of device used for billing?  Aren't things headed digital and wouldn't  a sophisticated micro-processor based smart meter sample the current and voltage and compute the true power factor sqrt(S^2-P^2)?  
===============
Now that I have been touting the importance of true power factor and the effect of lowered true power factor due to reversing load, I will have to acknowledge one factor which will certainly diminish it significance in certain circumstances.   And that is cancellation by diversity.  i.e. what happens when you put it together with a very large number of other plant loads that may also be reversing, perhaps oscillating, perhaps starting, stopping, or otherwise changing over time.   These large number of time variations are not synchronized to each other and statistically will tend to cancel each other out (especiallly as the number of contributors grow).    In contrast the other flavor of low power factor due to reactive power demand does not cancel with multiple equipment (excluding leading loads)  because the reactive power among all the loads is synchronized together via the line frequency.    

====================


OSCILLATION

I still don't have my arms around what is causing the oscillation.  I have some wandering thoughts.

First since you are a mechanical guy, I would like to suggest a simple 2DOF mechanical mass spring model which I think you can realte to:

Ground ===Spring1 ==== Mass1 ===Spring2 ===Mass2

Now translate that to our motor system (torsional mass spring system):
Stator === Field === MotorRotor === Belt=== Pump/Flywheel

By analogy to the mechanical 2DOF mass spring system, there are two modes.  

In-phase mode – this is the lower frequency mode.  The two masses  (MotorRotor and Pump/Flywheel) move together.
l
Out-of-phase mode – this is the higher frequency mode.  The two masses (MotorRotor and Pump/Flywheel) move opposite each other.

If I believe my simulation, the motor is dramatically speeding and slowing every 10 cycles.  That would seem to limit us to the in-phase mode.  After alll – there is only so much stretch in the belt... not enough to accmodate all that motor speed change if the flywheel wasn't changing the same direction.

BUT, if I continue to believe my simulation, the oscillating torque associated with the field (I called it electrical torque in the simulation) is smaller than the torque associated with the Belt (I called it motor shaft torque envisioning the torque on the output shaft extention).    But  that relative relationship between torques is not consistent with the in-phase mode.  In the in-phase mode, I think spring 1 (the field/electrical torque) should see more force/torque than the belt (spring 2) .

Two contradicting conclusions from the same simulation makes me begin to question my simulation.  But at the same time I thought that the agreement of the current waveform was a pretty good sign that the simulation was on-target.    Not sure what to make of that.

I am going to go back and review my Krauss textbook.  I think there is some very relevant eigenvalue analysis in there.

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RE: Low power factor, recip pump

By the way do you have a strobe light or any other means to check for speed variation?

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RE: Low power factor, recip pump

Do you know the peak torque required by the load?

Do you know the average power draw of the motor? You could measure kWh for a certain time (say an hour) and then divide by the time to get the wattage (divide kWh by 1hr = kW).

Did someone size this motor to match the load or is it mostly guesswork?

I don't agree with the worry that a capacitor is a problem.

I also don't agree with adding more flywheel weight. Why do you want to fix this cyclic loading again? It won't correct the power factor. Also, adding a large mass to the motor will incerse some stresses. It will be harder to start and will increase the peak forces the drive belt is subjected to.
 

RE: Low power factor, recip pump

Quote:

In the in-phase mode, I think spring 1 (the field/electrical torque) should see more force/torque than the belt (spring 2) .
I think this is where I went wrong.  If it was unforced free resonant oscillation it would be true.  But if was forced oscillation with the oscillating force is applied to the pump/flywheel, it might not be true.

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RE: Low power factor, recip pump

It is valid to ask why the oscillation needs to be fixed.  It is also a valid question for the OP to ask of us what does this current waveform mean and is it normal.  I for one have never seen a recip waveform. Has anyone out there seen a recip current waveform before?   What did it look like?

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RE: Low power factor, recip pump

Pete,

There is one very typical recording in the OP from 9 Jul 09 11:55. I printed it out and saw nothing special with it. It is a very typical recording showing that the motor is working against varying torque that is, also, typical for a reciprocal pump.

It is the OP's limited experience in industrial matters coupled with his inability to understand the difference between electric cycles and the periodicity of the load (the pump) that has caused most of the confusion.

The local electrical guy has also added to the confusion with his mentioning of "recommended not using capacitors because the current trace is too 'odd' - he thinks there's some unusual interaction between the motor and compressor that may just get worse if capacitors are installed" has made the whole thing stray away from engineering into superstition. The current shape is not odd at all and a normal grid does not let the capacitors interact with the motor - the capacitors just deliver the reactive current. That's all.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: Low power factor, recip pump

It is not just a varying torque, but a reversing torque.  You have seen that on motors driving other recips?

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RE: Low power factor, recip pump

Is this a gas compressor or a liquid pump?

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RE: Low power factor, recip pump

Here is another idea to throw out regarding the oscillation.

Krauss' Analysis of Electric Machinery talks about 5 eigenvalues in his motor model.  Two of them are a damped complex pole pair associated with "rotor electrical transients".  An example of excitation of this eigenvalue apparently occurs when starting the motor.   If you look at my attachment 8 Jul 09 23:07 in the thread thread237-248895: quiz - can a DOL-start unloaded induction motor "overshoot" sync speed you see an example of rotor speed oscillations around 7-8 hz upon starting a large induction motors.

If this were a mechanical resonant system, I picture this is similar to impacting the system to excite the resonant frequency.

But we can also excite a resonance by applying sinusoidal excitation near the resonant frequency.  If the sinusoidal excitation frequency is very close to the resonance and the damping is small, then a small exitation causes a large response.

Maybe your rotor eignevalue is near 6hz and that is causing resonant amplification of a realitvely small 6hz torque pulsation (excitation).  

Under this scenario the force that drives the motor above sync speed is a transient motor electrical torque (not predicted by the steady state torque speed curve, but it can exist as shown in the start ranseint).  The power reversal comes when the transient forward torque subsides and rotor inertia (above sync speed) feeds energy back to the system.

Just a thought to explain where the power for generator action might come from.   

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RE: Low power factor, recip pump

This is all much ado about a simple matter, IMO. Fix the capacitor anywhere in the system and solve the problem. I do not believe the motor is generating regardless what the "meter recordings" say.

RE: Low power factor, recip pump

Yes, I see that when I print an enlarged picture and measure the phase angles. Assuming that the traces represent phase-neutral voltage and phase current (plausible, since the max/min voltages represent peak of a phase-neutral 480 V grid), then the phase angle is going from around 30 at 10 ms to 120 at 100 ms. And everything above 90 degrees is generating.

I have no idea what the mechanical situation is, but seeing this kind of oscillation usually means either flexible (rubber, usually) coupling or that the pump is mounted on rubber pads. Still nothing to worry about.

This view is supported by the fact that the voltage peaks show no variation over the load cycle. So, there is no interaction voltage-wise.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: Low power factor, recip pump

My previous posting was in respones to Pete's observation that there is actually a power reversal.
 

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: Low power factor, recip pump

Good comments.

Edison  - I can understand your comment "much ado about nothing".  For all I know this is normal for a compressor (I haven't looked at any compressor waveforms before).  I am just throwing out thoughts and I have a completely open mind about the cause of the oscillation.  The only thing I know for sure is that I don't understand it.    It is an interesting discussion topic for me since it overlaps the area of vibration/dynamics that I am interested  in, as well as a chance to use dynamic computer simulation of motors using Krauss model that is a new toy for me which yields results that I never would have dreamed of (like the DOL start speed overshoot).   At any rate I  personally would like to continue to discuss and try to understand it because I think I learn from the process.  I am certainly not suggesting it needs to be a research project or drastic action for the original poster.   I would think a company that manufactures compressors has access to "normal" compressor current waveforms.  I may have an opportunity to get some myself but not in the immediate future.

I guess you are suggesting some kind of measurement error.  But it would be very coincidental for both the phase lag and the magnitude to tell the same story.  And they do tell the same story – as power factor passes through zero (based on lag angle), the current magnitude is at a minium.   On either side of this zero-power-factor point, the magnitude is higher due to the added real load current.   

=============

Gunnar – thanks for your response.    Flexible rubber coupling means a torsionally compliant connection between motor and driver.  This machine has something similar to that in the form of the belt (which is more compliant than a rigid coupling and probably more torsionally compliant than most flexible couplings).   This could certainly affect or  enhance (*) torsional oscillations in the "out of phase" mode that I described above (* more specifically, it would lower the out-of-phase torsional resonant frequency).     That is completely inconsistent with my simulation model (which shows changing speeds that are inconsistent with out of phase motion), but that data point doesn't count for much (maybe my simulation is out to lunch) and I think out-of-phase torsional resonance is something that should be included high in the list possibilities.

The connection to flexible feet is interesting. I guess it could be torsional movement of the stator frame facilitated by that flimsy base?   Another completely new idea for me.  Never heard of that or thought of that but it seems plausible.


Iainuts.  

If I get a chance I am going to do some more simulation.  I would like to "tweak" the model parameters to see the effect of moving the rotor eigenvalues close to the excitation frequency (6hz) and far from the excitation frequency.  It will be interesting to see if I can get a large response out of a small excitation when the eigenvalue is close tot the excitation frequency.

I would like some more info about the motor if you have any of the following available:
Nameplate data:  FLA,  KVA code, Efficiency, PowerFactor Nameplate speed (i.e. 3575rpm).
In the unlikely event you have stator winding resistance recorded (along with temperature at time of the measurement) that would be helpful.  Also in the unlikely event you have recorded no-load current that would be helpful.     Again I don't need all that info, but just what you can easily get.  One of those pieces of info would help if that's all you've got.
 

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RE: Low power factor, recip pump

Also if you can describe the rough dimensions/material of the flywheel before/after the change, that would be interesting.  I know it is small compared to the motor inertia when adjusted for speed, but if it's available I'd be interested.

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RE: Low power factor, recip pump

Hello pete

My comment was addressed to the OP about his original problem. I appreciate your efforts to do the math part, which none in this site can ever hope to do at the level you're doing. (of course, jbartos comes to mind but then you know that story. bigcheeks

And yes, I still have doubts about the OP's data. I would like a cross-check with another meter/recorder or even repeatability with the same equipment.

RE: Low power factor, recip pump

(OP)

Quote:

e-pete: By the way do you have a strobe light or any other means to check for speed variation?
Yes, very good suggestion.  I'm looking into some way of monitoring instantaneous RPM.

RE: Low power factor, recip pump

(OP)

Quote:

e-pete: Is this a gas compressor or a liquid pump?
Ok, I'll fess up.  It's a liquid hydrogen pump that can do 100% liquid, 100% cryogenic gas or any 2-phase mixture.  At this particular site, we're using it for gas which requires much less power per my calculations.  With 100% liquid, it needs just over 100 hp, thus the selection of motor gives us the opportunity to test on full liquid at the same site, though flow rate goes up by a factor of 10.

Flywheel on pump is now at roughly 420 lb ft2.
Motor is 27.5 lb ft2

Motor cut sheet attached.

RE: Low power factor, recip pump

(OP)
Current draw at no load ~ 31 amps.

RE: Low power factor, recip pump

Thanks that is more data than I had hoped for.  Should be very helpful to get a good estimate of the motor parameters.

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RE: Low power factor, recip pump

Pete;
The load profile is fairly simple. Ignoring heat losses, the force on the piston increases directly as the pressure increases. You will have to apply a sine function modified by the length of the con rod to determine the torque on the shaft. When the cylinder/piston pressure equals the discharge pressure, the pressure into a large reservoir stays the same. When the piston reaches top dead center the discharge flow ceases. As the piston progresses past top dead center, the gas trapped in the dead space and any resiliance in the mounting and any stretch in the drive belt all give the motor a little kick, and with the high pressures common with hydrogen compression, I now see where the "Kick" from trapped gas is greater than normal. I suspect that a multi-phase pump may have a greater dead space than a shop air compressor and a larger dead space will develop more "Kick".
A comment in support of Gunnar. When I have not set up or seen the instrumentation myself, I am slow to accept critical phase angle measurements. The recording trace as posted looks quite good, but I like a second look when possible. I am always concerned that errors such as a small DC offset may affect accuracy when scaling critical phase angles from a graph. And it is critical to the question of regeneration as to whether the angle of the current trace is greater than 90 degrees. I also looked at the current peaks. When I saw a slight increase in current where there should be a minimum point (without regeneration) with no corresponding reduction in phase angle, I accept that as my "double check" that the motor is regenerating and the graph is error free.
 

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Low power factor, recip pump

I have quoted the key questions from how I interpreted the origional post.

"1) Why would current lag the voltage like this?"

It's an induction motor. That's what happens. The current lags the voltage.

"I'm really looking to understand why the power factor is so low and what can be done about it."

The inductive reactive component of every 3-phase squirrel cage induction motor causes a lagging power factor. The power factor lags more when the motor load is less. A properly sized 3-phase capacitor is used to cancel most of the motors reactive component improving the power factor as seen at the utility.

You do understand that the poor power factor only matters when you get billed for a poor power factor?

You do also understand that the only way to get a motor to have a better power factor is to increase the load? The flywheel will not increase the load.
 

RE: Low power factor, recip pump

My first attempt to develop a model to match your motor data sheet gave the following parameters:
R_NL    202.4852054
R_1    0.043934
X_1    0.335347
R_2    0.025981
X_2    0.338430
X_M    8.433329

R_NL is a resistance to be connected in parallel with the input of the stator to simulate portion of the no-load losses.  The other parameters are standard induction motor equivalent circuit parameters.

Here is a comparison of the performance of these parameters against the "targets" from the data sheet:
Parameter    //    Model    //    Units    //    Target    //    FractionalError
FullLoadAmps    //    112.5    //    Amps    //    113.0    //    -0.0045
FullLoadEff    //    0.9548    //    none    //    0.954    //    0.0008
FullLoadPF    //    0.8674    //    none    //    0.87    //    -0.0029
FullLoadPower    //    74234    //    watts    //    74600    //    -0.0049
FullLoadTorque    //    398    //    N*m    //    400.0    //    -0.0043
HLEfficiency    //    0.957    //    none    //    1.0    //    -0.0009
HLPowerFactor    //    0.818    //    none    //    0.815    //    0.0042
NoLoadCurrent    //    30.3    //    Amps    //    28    //    0.0829
BD_Tq    //    738.5    //    N-m    //    969.4    //    -0.2382
LRC    //    399.9    //    Amps    //    675    //    -0.4075
LRT    //    61.1    //    N-m    //    603.3    //    -0.8987
(this summary is more legible in rows 57-69 of the main tab)

The fit to the performance data was developed using the attached spreadsheet.  You can get slightly different results by adjusting the weights of the various performance measures.  This model fits very well at the full load and half-load points.  Reasonably well at the no-load current point (slightly above the target 28A on the data sheet but below the 35A you said was meased).   Not so good at break-down torque point and miserable at the locked rotor point.  However at starting deep bar effect will cause X2 to decrease and R2 to increase which causes LRC to increase and locked rotor torque per current to increase.  The intent is to model conditions running (perhaps near 25% load), so the lack of match at locked rotor and breakdown torque conditions is not a concern.

I used the above parameter values along with your inertia values (assume motor inertia and speed-corrected flywheel inertia act as a single inertia) to calculate the eigenvalues of the linearized system at 25% load using the approach described by Lipo and Krause here:
http://www.ece.wisc.edu/~lipo/1970s%20pubs/T07.pdf

The result are 5 eigenvalues (two pairs and a real)
  -25.23 +/- 376.08i
  -7.28 +/- 40.19i
  -14.74  

The first pair of complex eigenvalues are the stator eigenvalues very close to 60hz.  The second pair of complex eigenvalues is the rotor eigenvalues in this case is calculated as 6.4hz.  That happens to fall very close to your load torque oscillation frequency.   If the model accurately reflects your machine (and that's a mighty big if),  it would certainly be expected that small load torque oscillations will give larger oscillations in speed, electrical torque and current.   It is also noted that these eigenvalues are not particularly lightly damped.  Additionaly there machine may provide mechanical damping for which credit was not taken.

I did just a little informal sensitivity analysis (playing around with the parameters) and the eigenvalues seemed surprisingly stable to me to variations in loading and  the fitted parameters.  For example at full load (instead of 25%)  the rotor eigenvalues were 6.1hz (instead of 6.4hz).   One thing I did note was the rotor eigenvalue frequency varies roughly as 1/sqrt(J).  This is consistent with the results which would be seen for a single degree of freedom mass spring system.   

The next step for me would be to do a simulation with these parameters and perhaps repeat without the flywheel – resonant frequency would be around 25% higher and your oscillation should be much lower (for a given load torque oscillation) according to this model.

The matlab file which was used to compute these eigenvalues is included in a tab in the worksheet.  The Matlab code along with the article should be  everything you need to try to validate or disprove it for yourself if you have the time, energy, and interest.   I think this code will also run on the free matlab-like programs:  Sci-lab and Octave.
 

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RE: Low power factor, recip pump

Quote:

  The Matlab code along with the article should be  everything you need to try to validate or disprove it for yourself if you have the time, energy, and interest.
Sorry -  didn't mean to suggest this is enough to validate or disprove the applicability of this model to your machine.  I meant to say this is enough for you to check my work.

Applicability to your machine depends on a whole lot more of course  (is Krause's model sound, do the rotor and flywheel move as a single rigid mass or is the flexibility of the belt enough to make it act like a 2dof system, etc.  An interesting experiement if you have the ability to record speed would be to record the oscillations about no-load  speed immediately following a no-load start with flywheel connected but without the compressor.  Those oscillations should be at a frequency corresponding to the rotor eigenvalue.  These oscillations following start were the main topic of discussion in the thread thread237-248895: quiz - can a DOL-start unloaded induction motor "overshoot" sync speed
 

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RE: Low power factor, recip pump

Quote:

An interesting experiement if you have the ability to record speed would be to record the oscillations about no-load  speed immediately following a no-load start with flywheel connected but without the compressor.  Those oscillations should be at a frequency corresponding to the rotor eigenvalue.
Recording speed does require some specialized equipment.  An easier alternative would be to record the three currents, then perform a dq transform and the oscillation at rotor eigenvalue frequency should show up in the dq ccurrents (does not show up as well in the a abc currents).  If voltages are also recorded, a little more detailed analysis is possible (such as aligning the dq reference frame to the voltages, in which case the oscillation should show up more strongly in the q current only based on the simulations I have seen).  If you  upload those waveforms (abc curretns from start without compressor) I would be more than happy to do that analysis.

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RE: Low power factor, recip pump

Two more comments and then I'll  leave this alone for awhile:
1 - Wherever I said "dq transformation", I should have said "dq transformation" to the synch ref frame.

2 - The summary of my recent comments is this:  I suspect you might have a torsional resonance of your electromechanical system near your torsional excitation frequency (6hz) which causes resonant amplification of your torsional excitation from the compressor.  One way to test this hypothesis is to remove the excitation (compressor),  pertub the system (DOL  start), and measure the oscillation frequency to see if it is in fact near 6hz.

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