Poor Power Factor Effects 2

[Matador]

I am a mechanical engineer with an electrical problem. The power calculation using amperage, voltage, motor power factor and efficiency when combined with pressure readings on the pump performance curve gives a flow whiich is 20% higher (5000 GPM) than other measurements indicate.

The overall power factor for the plant site is corrected to .99 using a couple of large 18,000 Hp motors designed for the purpose but the power factor may not be .99 where the 700 Hp motor is connected. In fact it could be as low as 0.80

Can my horsepower calculation be the apparent power rather than the real power which actually used by the pump? I measure 96 amps, use 4160 Volts and interpolate for motor pf & eff from the test report.

I'm using the formula hp=1.73*E*I*pf*eff/746. The motor test report indicated that at full load the values are E=4000V, I=100.7 amps, Eff=94.8, Pf=78.9, Syn speed=900 rpm, Slip @ fL=1.43.

How would a supply power factor of 0.80 affect the Hp calculation. It isn't accounted for anywhere?

Any help would be appreciated.

 

[electricpete]

If I understand you correctly your measurements of mechanical power (corrected for motor and pump efficiency) indicate 20% more than your measurments of electrical power. Is that what you're saying? Some numbers might be interesting.

To go directly to your question, any power factor correction should be irrelevant to your power calculation UNLESS the power factor correction capacitors are connected at a point downstream of where you measure your motor current. ie plant power factor correction capacitors have not effect. If Power factor correction capacitors (or surge capacitors) were switched connected near motor terminals such that they could be switched with the motor, and you measured current at the breaker, then those capacitors would effect your calculation. To summarize, assuming you have only power factor correction capacitors, then your formula for shaft horsepower should be shp=1.73*E*I*pf*eff/746 as you stated.

When faced with an apparent contradiction, you have to start looking at what you have assumed vs what you have measured, and also consider the accuracy of what you have measured.

For the electrical power calc, best case is to measure the current (try all three phases and use the average). Measure the voltage at motor terminals or if measured at some remote point upstream, then correct for the voltage drop along the cable (forgetting to do this can give you a few percent error).

It sounds like you have assumed a value of power factor from the test data... but power factor can be measured directly with a power analyser, or with any two-channel oscilloscope (need to determine phase angle between voltage and current). If you're hooking up at secondaries of PT's and CT's, then consider the phase-angle shift introduced here.

Motor efficiency... you're stuck there. No way to measure that. But as a double check on the motor power you can measure the speed of motor with tach. Then you can estimate output power from the slip as compared to nameplate... with correction for voltage^2 and estimated rotor temp---- see recent threat on estimating Power from slip... temperatre correction.

The fluid parameters are likely to have more errors, even if you measure them. You need to know densitiy of your fluid (varies with temp). You need to know flow... presumably you're measuring dp across some flow element (annubar, orifice, venturi etc) to estimate flow... if it has clogged than your dp has gone up and it looks like you're drawing more mechanical power than you really are. Have your gages been calibrated? A good independent check on the flow is to use ultrasonic flow measurement. Likewise you might consider whether your measurement of DP is accurate... is it directly across the pump? Is there an elevation difference between the two pressure taps? Have elevation differences from tap to gage been properly accounted for?

The large slip makes me want to doublecheck... this is not a wound-rotor motor is it?

For more ideas do a search of this forum on efficiency... it comes up every month or so

 

[rhatcher]

The horsepower output of the motor will be based on the formula you are using. The power factor at the terminals of the motor will be determined by the motor design and load. This is independent of the power factor of the main bus of your plant.

One way to look at this is with a one line diagram. The main bus will be connected to the utility tranformer and will supply feeders to the loads in the plant (such as the 700hp pump motor) and to the pf correcting motors. The power factor of the main bus measured at the utility transformer may be 0.99. But, each feeder from that bus will have a current draw, and power factor, based on the load connected to the feeder. According to Kirkoff's Current Law, the sum of the feeder currents, each with different magnitude and power factor, will be the bus current at magnitude I_bus with a power factor of 0.99. Clearly for this to take place the feeders to the pf correcting motors will have a leading power factor to offset the lagging power factors of all of the inductive feeders.

This being the case, the equation you are using is valid with the motor power factor as the variable. However, when you look at the motor's nameplate data, you must understand that it is based on full load. The power factor and efficiency of an induction motor vary with load. As load decreases, power factor and efficiency decrease.

The manufacturer should be able to provide curves (or tables) showing how current, speed, power factor, and efficiency vary with load for your motor. You should try to get these as the improved accuracy of your calculation may clear up the discrepancy.

 

[electricpete]

I hope I'm not beating a dead horse here, but it strikes me there are many more opportunities for error on the fluid/mechanical side of the calculation.

Density of course changes with temperature and if temperature is varying among different points in your loop, that needs to be considered.

Density at the flow element must be known to determine either mass flow rate or volumetric flow rate from flow element dp.

Density at the pump must be known to determine fluid horspower delivered by pump.

If density at flow element differs from density at the pump, then you have to do your power calculation based on mass flow rate, since volumetric flow rate will differ at those two locations.

In determining a dp across the pump based on two pressure measurements.... I mentioned that the height of fluid between those pressure taps and the pressure meter plays a role which must be accounted for. Also the temperature of the fluid in the sensing lines determines density of that fluid and impacts the indicated pressure.

 

[Matador]

More info to clarify:
(1)- We manufacture pulp for papermaking. There are no capacitor banks for power factor correction but in the plant design we knew that without any correction the overall Pf would be ~0.80 Therefore when we purchased the main refiner motors (~18,000 Hp) they were designed to give us the correction required to offset the inductive load from the other motors (~150) located elsewhere in the plant. When you look at the single line diagram there is one main feeder at 13.6 KV supplying the large 18,000 Hp motors and individual 4160 V & 600V feeders to the rest of the plant.
(2) - The manufacturer curves for centricleaners and plant mass balances agree but the capacity from the pump performance curve using Hp and discharge pressure indicates an additional 20% or 5000 GPM. The cenricleaner curves are used everywhere for pump sizing and a 20% error would mean major mistakes in pump sizing throughout the industry, which is not the case. In addition, if our mass balance calculations for the entire plant were out by 20% for the past 9 years, it would be embarassing. This leaves the pump flow calculation as the main source of error. The pressure readings were obtained from pressure transmitters with manual gauges used to confirm the readings. That leaves the power reading as the suspect.
(3) - Next week we will rent a more sofisticated device to measure the amps, voltage, & power factor. Should the pf obtained from the meter reading be the resultant of the plant plus the motor or simply the motor pf from the manufacturers test report at that load?
(4) - I put a strip chart recorder and recorded 96 amps for 12 hours. Just a straight line close to the full load value.

At this point in time we are concentrating on the power calculation. The ramifications of a low pf supply is not fully understood by me.

 

[electricpete]

Matador - I get the hint, I'll try to focus more on electrical side ;-)

Your large synchrnous motors, for the purposes of this discussion can be treated the same as capacitors. Namely, if the current flowing to those sync motors is NOT flowing past the point where you measure your inducation motor current (which I understand to be the case), then they are irrelevant to the calcualtion of power in your motor.

The only thing your motor knows is that you give it a voltage. Then the motor draws a current which will be displaced by a certain power factor angle from the voltage you gave it. That power factor angle depends only on the motor (and of course the mech load on the motor and the voltage magnitude), but it does NOT depend on any other electrical load or power factor correction device in the plant.

When you talk about a power factor, it is the angle between voltage and current and a certain point in your power distribution system. The real power flowing at that point in the distribution system with be P=|V||I|*cos(theta) where theta is angle betwee that V and that I.

We can apply the equation either at the plant level or at the motor level, but we will be using a different differnt V a different I and a different cos(theta) and calculating a different power P. There is really no reason to bring the plant power factor into any discussion of the motor power factor.

The syncronous motors affect the plant power factor, In effect, they will increase cos(theta) to 1, so you can deliver the same P with lower |I|. The same equation is also true at the motor terminals where we are using a differnt V a different I and a different cos(theta). Once you give the motor a |V|, it will determine it's own |I| and cos(theta) REGARDLESS of what is going on at the plant level. Sorry for repeating myself... maybe repetition will make up for any lack of clarity ;-)

Can you tell us whether the power calculated electrically is more or less than the power calculated mechanically? ie does it seem like we're getting free energy or does it seem like we're losing energy somewere?

If it seems like we're losing energy, then the following types of items might be investigated:
Rotor bar problems
Voltage at motor terminals is lower than voltage at poitn of measurement due to volt dro in the motor cables.
Motor power factor is low due to core problems.
Pump has wearing ring problems or unmetered bypass flow (whoops sorry ... back to electrical).
harmonics in the system cause |I| or |V| to be larger for the same power.
Motor is running hot... more I^2R losses (but not that 20% of output more).
Electrical measurement errors - see below.

If it looks like we're gaining energy, the list is much shorter. I can't particularly see efficiency or power factor improving that dramatically, even at reduced load. Off-hand all that comes to mind is measurement errors:
Calibration of ammeter
calibration of voltmeter
Unbalanced voltages or currents and you read the highest one.
Perhaps you have assumed 4kv and your actual voltage is higher?
Perhaps you are measuring on the secondary of a voltage or current instrument transformer which has a ratio error... or you have mis-read the ratio?

OK, before I go any further, which case are we looking at?

 

[electricpete]

Whoops. The equation I used for power left out the 1.73 factor used for 3-phase calcs. Your equation was of course correct.

 

[BJC]

Shouldn't the formula hp=1.73*E*I*pf*eff/746
be hp= 1.72*E*I*pf/eff*746 ??
It's Friday and I'm tired so I may be off.

 

[electricpete]

BJC - I'm not sure exactly which part of the formula you're questioning.

1.73 is closer to the sqrt(3) than 1.72. Sqrt 3 is the actual number.

Motor efficiency should go in the numerator (remembering that the P we are calculating is shaft horsepower output delivered by the motor).

I'm pretty sure that the 732 goes in the denominator.

 

[BJC]

The only change I made was to divide by the efficiency not multiply

 

[Matador]

Since everything is based on the power reading and we are not getting enough flow through the centri cleaners to consume that much energy, I would have to say the enrgy reading is too high. I am responding from home so I don't have my reference books to verfiy the formula but here goes. The Bhp required at the pump coupling = (Flow (GPM)*Head(ft)*Specific Gravity)/(3960* Pump eff) To consume that much power the flow to each centricleaner would have to be 140 GPM instead of the 117 GPM specified for the operating pressure. We have 144 centricleaners connected to the pump. The above power formula must equal the motor output if you neglect the coupling losses which is a small percentage when we talk about 700 Hp being consumed.

We have two identical pumps but only one is operated at any given time, the other is a spare. One pump has a 700 Hp motor & the other has a 900 Hp motor. The reason for the 900 Hp had to do with expansion. Since both pumps have the same size impeller they find a point on the curve to operate and only take as much power as required. That is the case with the measurements. The Hp consumed is about the same after you apply the different eff & PF from the manufacturers test report for the given current flows.

One other thing, they measured one current and calculated a percent motor load using the full load amperage number, which checked out on the DCS. After that they just took % motor load readings for any given flow and calculated the HP. I believe DCS assumes the % power taken from the CTs follow a linear line when in reality it could be a curve. This will be checked next week.

Thanks for the feedback & I'll update things next week after another round of tests are complete.

 

[electricpete]

You talked about checking linearity of motor crrent vs power. As rhatcher mentioned the nameplate values of p.f. and efficiency apply at full load. One thing I read from your numbers is that the motor is near full load (96A measured 101A full load), so I'd expect these numbers to hold pretty well. The other thing you'll expect to see is that at low load, current is much higher than predicted from linear fraction of nameplate values. This is due to lower power factor and lower efficiency, particularly below approx 50% (varies by motor). I tried to find web-based figure of typical efficiency and power factor vs load but couldn't (anyone else have one to post?).


In your case where motor input power corrected for motor efficiency and power factor is still greater than pump output power corrected for efficiency, almost everything discussed above holds as possible source of error. It'll be tough to narrow it down from this end, maybe you have a better feel where to start. For instance, have we been measuring mechanical power all along and it suddenly went down. Or have we been measuring electrical power and it suddenly went up.

Broadly, the 4 types of factors can be divided into:
#1 - Errors in measuring mech power out (discussed above).
#2 - Errors in measuring elect power out (discussed above)
#3 - Things that make mechanical efficiency go down (worn wearing rings, debris in motor etc)
#4 - Things that make motor efficiency or p.f. go down (hi temperature, broken rotor bars, high voltage which causes low power factor).

 

[Matador]

Thanks for the info. I have the manufacturers test report for the motor which I have been using.

For example at 0, 25, 50, 75, 100, & 125 % full load.
Current= 46.5, 51.6, 64.0, 80.9, 100.7, 122.8
Efficiency= 0, 90.1, 93.9, 94.8, 94.8, 94.5
Power Factor= 4.2, 40.5, 62.7, 73.7, 78.9, 81.1
Slip= 0, 0.33, 0.68, 1.04, 1.43, 1.86

The Locked rotor torque= 129.9
Locked rotor current = 656.5 amps
Breakdown torque = 264.3%
Pull up torque = 103.9%

The no load test had values of:
Volts=4000
Amperes=46.5
Watts=13500

A couple of years ago they put a full size impeller in the pump to maximize the flow. According to the pump performance curves it should not have overloaded but it did and the impeller had to be removed. It appears that the power we measure is higher than expected. Either we read the wrong value and use that value for control purposes or the excess power is being dissipated as friction or wear. The power, if measured correctly, is being consumed, which is the reason I started thinking about the apparent, real, & imaginary components of power. I was hoping to explain the difference by saying I measured the apparent power when in actual fact I should be subtracting the imaginary component to get the real power.

Very interesting challenge which will get solved. Thanks for the input.

 

[peterb]

I suggest that the DCS be taught the correct calculation - install a power transducer to measure the kilowatts drawn by the motor and do the math from there -
(Pout (HP) = Pin(kW)*eff/0.746), rather than extrapolate from a single phase current reading. Should be worth the investment in a 2-element watt transducer to get the right numbers.

 

[rhatcher]

electricpete's post of 7/30/01 was pretty good (as usual!).

The 96A reading is for all practical purposes equal to the motor full load, so the values of power factor, efficiency, and slip should match up between the motor in service and the manufacturer's data. Power factor and efficiency are difficult to measure, but slip isn't. I would recommend measuring the shaft speed of the motor and comparing it to the full load slip in the data. By the way, if I am reading the slip data right you are saying a slip of 1.43% which is not very much at all. Either way, if you can determine that the amps and slip match the full load data, then you can make a safer assumption of the motor output being full load.

Measure the shaft speed and see what you get. If it is running at a slip of 1.43%, or at 887 rpm, then you may follow with the assumption that bhp at the pump input is correct and the losses are downstream of the motor (in the pump etc.). If the motor is running at a lower value of slip (at a higher speed) than indicated by the current, then you may follow with the assumption that the bhp input is less than indicated by the current and the problem is in the motor or upstream of the motor.

 

[BJC]

Have you checked the temperature of the water into and out of the pump?

 

British horses are smaller than us Yanks have, so you use the 732 (I thought it was 736, but I usually use american horses, so I'll stand corrected) in their horsepower calculations, vice our 746. I also use 1.732 for root 3, but that's a personal preference, I guess. The correct values are:

power = I * E * 1.732 * pf * eff

This is in Watts. Divide by 746 for horsepower. Efficiency cannot exceed 1.0, nor can power factor.

I think the discussion is only being complicated by the plant pfc synchronous motor references. Best thing is to get a cheap power meter to measure the power factor and apparent kw close to the motor terminals. Please.... I realize "apparent kw" is poorly worded.

Anyway, since the FLA appears to be in the ballpark, I think a harmonic analysis (FFT) needs to be done to determine how much trash is on his line. Chances are this motor is not causing it, but I wouldn't be surprised to find some motor drives or precipitators or similar harmonic source on the same line... What about it matador? If this is the case, you're reading a lot of amperage noise that isn't contributing to anything but the errors in your calculations.

I've been in a few paper mills and pulp plants, and I'd guess E-P's 7-27 comment on harmonics is being overlooked in this case. Break out an oscilloscope for a quick-and-dirty check. If you don't see sinewaves off some HED's/current probes, this is the right track to follow. If you do see sinewaves, then I never suggested this, and I'll blame it on rhatcher when he's not looking.

Don, who may not always be right but is always opinionated

 

[BJC]

Matador
I have had a chance to print this out and read it more carefully. A couple of questions.
I did not see the differential pressure across the pump. Did you state what it is? Sorry if I missed it. The efficiency of you system is Pump efficiency x Motor efficiency. Pump efficiencies can run as low as 30% for things like trash pumps. The loss in motor efficiency shows up as heat in the motor. The loss in pump efficiency shows up as heat in the water.
5000 GPM is 41,700 lbs of water moving through the pump per minute. One horsepower is 42.4 BTU/ min. I calculate that one lost Horsepower would raise 41,700 lbs of water about 0.001 degrees F. Someone please check me on that. You could be loosing power as heat in the pump and the temperature rise in the water may be very slight ( to small to measure with instruments of normal range).
I do know that pumps can be very good water heaters. The hydro test on BWR pressure vessels was done by running the recir pumps- It didn't take long to get a very large vessel up to 1000 degrees and 800 PSI.

 

[Matador]

Yes the pressure differentail across the pump was checked and we plotted calculated HP vs pump TDH. SeveraL tests were done at different flows. When the points were joined a line could be drawn which was parallel to other lines for different impeller sizes. The pressure transmitter readings were checked with manual gauge readings.

Measurement errors (power or preesure) would move the curve left or right. In this case we are hoping to measure less power which will give us less flow for a given pressure.

We rented an instrument and it was damaged when it arrived at the site. A replacement unit has been shipped.

 

[Matador]

Our E/I people finally got the meter hooked up & the data recorded doesn't clear much up.

The meter will stay on all night. One set of reading are:

Phase Voltage Current Measured KVA Watts Measured (W) KVAR Measured PF Measured

A 3759 101.3 381100 194100 328000 0.51
B 4821 97.83 471600 180100 435000 0.38
C 4414 101.9 450100 215600 395000 0.48
Totals 1300000 589900 1159000 0.45


The motor PF should be .78 for the load & we are getting an average of .45 Comments please.