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.
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
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
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
RE: Low power factor, recip pump
RE: Low power factor, recip pump
See this useful link on pf correction and some no-no's.
http:
RE: Low power factor, recip pump
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
Keith Cress
kcress - http://www.flaminsystems.com
RE: Low power factor, recip pump
RE: Low power factor, recip pump
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
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
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
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
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
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
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RE: Low power factor, recip pump
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
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RE: Low power factor, recip pump
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.
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?
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
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
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
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RE: Low power factor, recip pump
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 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
<-------- 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
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RE: Low power factor, recip pump
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RE: Low power factor, recip pump
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
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RE: Low power factor, recip pump
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RE: Low power factor, recip pump
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
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RE: Low power factor, recip pump
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
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
===============================
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 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
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
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RE: Low power factor, recip pump
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
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Low power factor, recip pump
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
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
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.
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
RE: Low power factor, recip pump
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
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
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"Why not the best?"
Jimmy Carter
RE: Low power factor, recip pump
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RE: Low power factor, recip pump
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
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Low power factor, recip pump
Agreed. Please point to any statement of mine that you think implies otherwise.
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?
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?
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.
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.
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.).
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).
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
#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
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RE: Low power factor, recip pump
You owe it to yourself - and us.
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Low power factor, recip pump
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
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
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.
#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
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
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
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
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
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.
ht
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
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
"The reactive current has a little distortion".
Why would the reactive current be distorted ?
RE: Low power factor, recip pump
http:
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
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
RE: Low power factor, recip pump
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
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
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.
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
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
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
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
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
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
RE: Low power factor, recip pump
Design "C"s preferred.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Low power factor, recip pump
=====================
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.
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
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RE: Low power factor, recip pump
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
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RE: Low power factor, recip pump
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RE: Low power factor, recip pump
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
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Low power factor, recip pump
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RE: Low power factor, recip pump
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RE: Low power factor, recip pump
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
RE: Low power factor, recip pump
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
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Low power factor, recip pump
Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Low power factor, recip pump
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.
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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
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Eng-tips forums: The best place on the web for engineering discussions.
RE: Low power factor, recip pump
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.
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
RE: Low power factor, recip pump
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
RE: Low power factor, recip pump
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RE: Low power factor, recip pump
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
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"Why not the best?"
Jimmy Carter
RE: Low power factor, recip pump
"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
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
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
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Eng-tips forums: The best place on the web for engineering discussions.
RE: Low power factor, recip pump
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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