sizing power factor
- Thread starter jaykool
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Usually the mfg will give you the proper size. Are you trying to correct the facilities power factor?
mivey
Senior Member
Would be good but:
With some motor frame size data, rpm, etc you could use a look-up table, like from ANSI/NEMA.
This is an existing application, the caps are shot and we are replacing. We only have 12.5KVAR caps...therefore install 3 for a total of 37.5KVAR. I know this is not a "big" deal. Just wondering if there was a quick way to ballpark KVAR needed without looking up in tables.
There are certainly "easy" ways like a spreadsheet on this site... but as an earlier poster replied, for a machine this size you need some real data if there is no nameplate. Undercorrecting doesn't get the job done in terms of money saved if you are trying to reduce a low PF charges from POCO and overcorrecting can cause serious technical consequences and damage equipment.
I guess if you were in an emergency, if you made some reasonable assumptions like a motor FL efficiency of 95% (which I looked up for a new 4 pole 200HP motor, your's may be different depending on a number of factors) and a existing power factor of 0.87, then 35KVAR of caps corrects you back to 0.95 which is your goal.
I can't stress enough the need to
1. Get some real data on the situation instead of guessing or assuming
2. It takes maybe 3 minutes to draw a power triangle and perform the above calculation.... until you are adept at this, I would stay away from "easy" methods such as spreadsheets or tables
I guess if you were in an emergency, if you made some reasonable assumptions like a motor FL efficiency of 95% (which I looked up for a new 4 pole 200HP motor, your's may be different depending on a number of factors) and a existing power factor of 0.87, then 35KVAR of caps corrects you back to 0.95 which is your goal.
I can't stress enough the need to
1. Get some real data on the situation instead of guessing or assuming
2. It takes maybe 3 minutes to draw a power triangle and perform the above calculation.... until you are adept at this, I would stay away from "easy" methods such as spreadsheets or tables
I don't think so.
As an example, on one of our current projects there are two 225kW (300 hp) motors driving pumps. We are building the starters for those motors.
The main contractor is supplying the motors and pumps. Some way into the project, the proposed motor was changed. The first had a power factor of 0.84. On the second (also 225kW), it was 0.76.
Our PFC capacitors went from 70kVAr to 120kVAr because of that change. So not really in the same ballpark.
One further thought on this.....
If there is no nameplate data, how can you be sure it is a 200 hp motor?
drbond24
Senior Member
- Location
- Barboursville, West Virginia
I'm with Besoeker on his question; if there is no nameplate data how do you know it is 480 V, 3 phase, 200 hp?
Anyway, in order to figure what power factor you are correcting to, you must know what power factor you are starting at. If you don't have it on the nameplate you'll have to measure it. You could assume a starting power factor, but then this is just getting silly. Your answer wouldn't mean anything. We need to know where you are starting to tell you where you are going on this one.
ohmhead
Senior Member
- Location
- ORLANDO FLA
Well use your motor name plate E X I X 1.73/1000=KVA
Now measure your watts with wattmeter on that motor running this is your true power KW .
KW/KVA = power factor
glene77is
Senior Member
- Location
- Memphis, TN
Jay,
Are you saying that you plan to connect the 3 caps in series to obtain a 37.5KVAR rating?
That would make the effective Capacitance
to be 1/3 of what you think.
And that would be 1/3 of the required PF correction.
Given that you know from the nameplate that it's 200hp or 150kW output*, you don't need to measure the power. In any case, the motor might not be running at full rated power at the time of the measurements.
*plus about 5% for losses to get input power.
I would assume that each 12.5 kVAr unit is three-phase.
If they are, each unit is most likely three capacitors in a delta configuration.
Not easy to connect three such units in series.
ohmhead
Senior Member
- Location
- ORLANDO FLA
Well i agree he doesnt have to with no load on that motor it would be a inductance with little resistance but when a loaded it would give him a wattage to go by .
Increased current while fully loaded and running makes watts the old right triangle formula to get VARS needed .
If he had a wattmeter on that motor it would give him a way to check his power factor .
Let me understand this more what would you do in this case to test a motor for power factor correction .
Iam here to learn also dont take this the wrong way iam interested in this .
Iam just a electrician and may need help here also ?
Just wondering about how one knows about what losses are know inside that motor without it running ?
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I'm not quite sure what you are asking so I'll do my best.
The nameplate data gives rated current and voltage. From that you can calculate input kVA.
The nameplate also gives rated output power.
Add 5% or thereabouts for losses abd you have input kW.
Input kW/input kVA gives you input pf.
ohmhead
Senior Member
- Location
- ORLANDO FLA
Well Besoeker i guess iam a old time electrician ii always thought load needed to be a factor with a motor .
I know amperage draw of a motor is not a good test for power factor.
I always thought you really needed to run at full load to check all losses ?
We would measure fully loaded motors and test & record A&B VOLTS B&C VOLTS C&A VOLTS and also IA IB IC given PF- A PF- B PF- C
We then add v1+v2+v3/3= volts
IA+IB+IC+/3= AMPS
PFA+PFB+PFC/3=PF Then use PF=EXIXPFX1.73/1000=KVA
Then used hp x746 kw
KW/KVA =PF by measuring i did not know you could do it any other way i know today they have test equipment to do all this but just wondering how to do it on paper with out a motor running is different ?
Iam interested in your thoughts i have read your posts and know you deal with drives and design work with motors so help me understand this better .
input /output ?
I know amperage draw of a motor is not a good test for power factor.
I always thought you really needed to run at full load to check all losses ?
We would measure fully loaded motors and test & record A&B VOLTS B&C VOLTS C&A VOLTS and also IA IB IC given PF- A PF- B PF- C
We then add v1+v2+v3/3= volts
IA+IB+IC+/3= AMPS
PFA+PFB+PFC/3=PF Then use PF=EXIXPFX1.73/1000=KVA
Then used hp x746 kw
KW/KVA =PF by measuring i did not know you could do it any other way i know today they have test equipment to do all this but just wondering how to do it on paper with out a motor running is different ?
Iam interested in your thoughts i have read your posts and know you deal with drives and design work with motors so help me understand this better .
input /output ?
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On most, or possibly all, of the larger motors I have dealt with (500kW/750hp up) there would be a works test at full load and performance noted. Among other things, that would give losses and power factor usually at 2/4, 3/4, and 4/4 of full load rating.
For smaller machines, there is usually published manufacturer's data if it is a standard machine. Some will have been given a full test but it would be uneconomic to do every single one.
This gives you kVA which you could have determined directly from nameplate voltage and current.
Sqrt(3)*VL*IL/1000
Then there is the question of how you know the motor is fully loaded. It is quite difficult to measure output power after the motor is installed. Output is torque multiplied by rotational speed. You can fairly easily measure rotational speed to within about 1 rpm. Torque is a whole lot more difficult. So you'd probably take full load to be when the motor is drawing full load current - which you got from the nameplate. Along with nameplate voltage.
Again, where did you get the hp rating?
I'm sure you see what I'm getting at.
I appreciate that, thank you.
But you make me more than I am.
petersonra
Senior Member
- Location
- Northern illinois
- Occupation
- engineer
One of the problems with using anything off the nameplate is that it states the PF at full load. At lighter loads the PF drops, sometimes by a whole lot.
Its not impossible to have 80% of FLC at 20% of mechanical full load. You can imagine what the PF is in such a case.
Best bet is to size off the nameplate and hope for the best. Its a bit of a crap shoot, and to get the whole plant in line at 95% is going to probably require some automatic capacitor banks somewhere.
Its not impossible to have 80% of FLC at 20% of mechanical full load. You can imagine what the PF is in such a case.
Best bet is to size off the nameplate and hope for the best. Its a bit of a crap shoot, and to get the whole plant in line at 95% is going to probably require some automatic capacitor banks somewhere.
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