Motor PF & Resultant Capacitive Equivalents
Motor PF & Resultant Capacitive Equivalents
(OP)
Ok guys, little rusty over here..project management too long and not enough engineering in my life I guess.
We recently surveyed at Pump 150 H.P. 480V 3P. Initial Current was 175A @ .79PF We have a guy who claims he makes a "little black box" that will change the intial PF to .96 reducing the amperage to 144A. I'm assuming this little black box simply has a capacitor in it to counter the inductive load by magnetizing the coils in the motor. Anywho, i'm just working what little info I have to try and figure the reactive power. And what capacitance will bring the reactive power close to 0 leaving almost a totally resistive load. Just for fun. This guy claims his "little black box" will reduce power losses in motors 10-200 H.P. by 6-12%. I'm going to install a Dranetz PX5 before and after this box is installed to see if this will work.
I figure the from what we monitored (175A @ 480 3P)
Your Real or True Power (P) would be (175*480*1.73*.79)/(1000) = 114.8kW
Angle = cos^-1 (.79) = 37.8 Deg.
From this your Apparent power (S) would be the following: 114.8kW / (cos 37.8) = 145kVA
And your Reactive Power (Q) would be 114.8kW(tan 37.8) = 89kVARs
First off, am I right so far?
We recently surveyed at Pump 150 H.P. 480V 3P. Initial Current was 175A @ .79PF We have a guy who claims he makes a "little black box" that will change the intial PF to .96 reducing the amperage to 144A. I'm assuming this little black box simply has a capacitor in it to counter the inductive load by magnetizing the coils in the motor. Anywho, i'm just working what little info I have to try and figure the reactive power. And what capacitance will bring the reactive power close to 0 leaving almost a totally resistive load. Just for fun. This guy claims his "little black box" will reduce power losses in motors 10-200 H.P. by 6-12%. I'm going to install a Dranetz PX5 before and after this box is installed to see if this will work.
I figure the from what we monitored (175A @ 480 3P)
Your Real or True Power (P) would be (175*480*1.73*.79)/(1000) = 114.8kW
Angle = cos^-1 (.79) = 37.8 Deg.
From this your Apparent power (S) would be the following: 114.8kW / (cos 37.8) = 145kVA
And your Reactive Power (Q) would be 114.8kW(tan 37.8) = 89kVARs
First off, am I right so far?






RE: Motor PF & Resultant Capacitive Equivalents
This site might help:-
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desertfox
RE: Motor PF & Resultant Capacitive Equivalents
Just remmber that motor and its circuit losses are not that signinficant to begin with.
Rafiq Bulsara
http://www.srengineersct.com
RE: Motor PF & Resultant Capacitive Equivalents
Some time this can be of significant value. You are reducing the I²R losses by reducing the motor current. This can make a significant reduction depending on the motor circuit length and the number of hrs that the motors run. You can calculate the KWH's saved from from the reduction of the amps with with the new power factor.
RE: Motor PF & Resultant Capacitive Equivalents
Forgot this. You can do the same thing by contacting the MFG of the motor and getting the suggested capacitor size for the motor. The Capacitor is mounted on the motor and is energized when the motor starts.
RE: Motor PF & Resultant Capacitive Equivalents
Q = E^2/X which leads to X = E^2/Q but is this the same for a 3-Phase system. I always mess up the equations here, with where to include 1.73 for 3-phases etc. etc. I know this sounds dumb...
On another note this motor is part of a 2-Motor system that alternates. So there's actually 2 150H.P. motors that run at 12-Hour intervals 24 hrs a day. I'll take a length on the feeders, but I think the run is pretty significant as well.
RE: Motor PF & Resultant Capacitive Equivalents
Rafiq Bulsara
http://www.srengineersct.com
RE: Motor PF & Resultant Capacitive Equivalents
RE: Motor PF & Resultant Capacitive Equivalents
So (unless you change terminal voltage), you do nothing to reduce any losses in the motor. Only supply system losses, which are small.
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RE: Motor PF & Resultant Capacitive Equivalents
Correcting the power factor to 100% will reduce the current by about 37 amps.
This is the figure to use to find the I2R savings in the feeder.
By sure to get the correct table in the code. You want AC resistance, not impedance.
Two possible solutions:
1> Connect a set of capacitors to each motor.
2> Buy one set of capacitors and connect them to a contactor at the head end of the feeders. Run control wiring from each motor starter to the capacitor contactor and have the contactor put the capacitors on line when either motor is running.
Option 1 will save some I2R losses in the motor feeders but possibly not as much as you think.
Option 2 may be cheaper to purchase and install.
BUT!!!
Are you paying power factor penalties? If not why bother? The payback from I2R saving may be a very long time.
If you are paying power factor penalties, you shouln't be worrying about one or two motors, you should be doing a total plant survey, or having it done by an expert.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Motor PF & Resultant Capacitive Equivalents
Rafiq Bulsara
http://www.srengineersct.com
RE: Motor PF & Resultant Capacitive Equivalents
RE: Motor PF & Resultant Capacitive Equivalents
Before 175A (145kVA/(480*1.73)
After 138A (115kVA/(480*1.73)
Gives you the 37A Delta
At this motors operating time:
145kVA x 12Hrs x 30 days/month = 52,200kWH
115kVA x 12Hrs x 30 days/month = 41,400kWH
a savings of 10,800kWH @ Approx. ($0.12/kWH) Cost is a savings of $1,296.00 per month or $15,552.00/year.
Do this for a whole building worth of motors, and there would be some significant savings if my math is correct?
RE: Motor PF & Resultant Capacitive Equivalents
You need to start with the utility billing for each of these buildings. Like Waross mentione above, if you aren't paying for VAR or paying a penalty for poor power factor, then you are left with only reducing the losses in feeders which winds up as heat. This would significantly extend the payback period for the correction equipment.
RE: Motor PF & Resultant Capacitive Equivalents
No, not correct. You only save the I-squared-R losses in the circuit to the motor. The motor power (and current) remain the same.
To prove to yourself, measure current between the motor and the capacitor. You'll see that it's very nearly the same as before.
Let's say your motor circuit is 200 feet and it's 4/0 wire (0.05 Ω/kft). Total resistance is 3 * 200 * 0.05 / 1000 or 0.03Ω. If you're saving 37A, you save 37^2 * 0.03 or 0.041 kW. At 12 hours and 30 days/month, that's 177 kWh saved in a year's time, or about $21 per year.
Somebody will check my math, I'm sure. Anyway, if you're not eliminating a power factor penalty, you're not saving big bucks.
Good on ya,
Goober Dave
RE: Motor PF & Resultant Capacitive Equivalents
thanks ccjersey & DWR much appreciated, i knew i was doing something wrong there.
RE: Motor PF & Resultant Capacitive Equivalents
NO. Read what DRWeig said.
kWH is Kw x hrs not kVA x hrs.
When power factor changes, only line kVA changes not motor kW. Your kW remains essentially the same. If your meter is at the motor you will not see any change in kW. If your meter is away from the motor, you will see some reduction in I^28R losses in the cables. This is usually insignificant.
Rafiq Bulsara
http://www.srengineersct.com
RE: Motor PF & Resultant Capacitive Equivalents
1. Power Factor Correction - adding capacitors to improve PF and reduce current. Note that there are technical limitation to add capacitors to motors driven by Inverters or Soft Starters.
2. Voltage regulators - reducing the voltage to partially loaded motors to improve their efficiency and inherently improve their power factor. You can read this article - http://ww
Note that the measurement information that you provided shows that the motor is fully loaded and this means you cannot save using this technology.
In addition, there are two different technologies to control the voltage:
1. Electronic - similar to soft starter - cut part of the waveform to reduce the RMS voltage.
2. Electro-mechanical - smart auto transformer to reduce the voltage.
The advantage of the first method is that it is continuous (no limit on voltage levels) while the advantage of the second is that it does not generate harmonics (the main problem of the first method). The bottom line is that the second method is better as the first one cause you harmonics which reduces your profits. I would suggest use your Dranetz to measure also harmonics before and after. Also make sure to measure upstream the box to include its losses in your calculations.
My Power Quality and Energy Efficiency blog:
www.PowerQualityDoctor.com
RE: Motor PF & Resultant Capacitive Equivalents
Back when I was young, I was tasked with correcting the power factor for a large heavy fabricating plant.
With lots of theory and little experience I measured the PF of most of the motors in the plant. Then I estimated the cost of correcting every motor, versus bulk correction at the panels. It took a lot of time but the time spent gave me a good feel for power factor and was time well spent.
The point that makes this relevant was that the only instrument that I had to measure the power factor was a clamp-on ammeter.
I took a 5 KVAR capacitor bank and connected it to each motor in turn. I recorded the feeder current, the capacitor current and the motor current for each motor.
Back in the office I used the data to construct a scalene triangle for each motor and from that I was able to scale the KW, KVA, KVAR and power factor for each motor.
After two days of measurements I am confident to report that the addition of capacitors does not change the first order current of a motor. There may be some obscure second order effects that are small compared to the $21 per year already calculated.
That is how to measure power factor without expensive instruments.
Then I found an easier, more accurate, faster way.
Now I call the customers accounting department and request copies of the original power bills for the last year at least and the last two years if possible. I use the bulk monthly figures for KWHr and KVARHr and determine the number of KVAHr of capacity to correct to just above the penalty cut-in point. 90% PF in the old days.
ARE YOU PAYING A PF PENALTY? THE PAYBACK TO AVOID PF PENALTIES WAS USUALLY MUCH LESS THAN A YEAR.
I you are paying a penalty, we can suggest steps to analyze and correct your power factor.
DRWeig; Thanks for doing the math. My attitude is that the saving is going to be so little that I won't waste my time calculating it. You have confirmed my sloth with your figure of $21 per year. grin.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Motor PF & Resultant Capacitive Equivalents
I've had to do that recently all too often. The energy saving gizmo marketers are everywhere.
Good on y'all,
Goober Dave
RE: Motor PF & Resultant Capacitive Equivalents
http
Rafiq Bulsara
http://www.srengineersct.com
RE: Motor PF & Resultant Capacitive Equivalents
I'm gettting the feeling that this black box is a type of sinosoidal motor controller as PQ Doctor said above. Its too small for a capacitor set-up.
And again, the motor is fully loaded deeming it useless in this application?
Guys I appreciate the help on this stuff, i'm still pretty green on the engineering side of things...I know enough to get me in trouble haha.
RE: Motor PF & Resultant Capacitive Equivalents
Rafiq Bulsara
http://www.srengineersct.com
RE: Motor PF & Resultant Capacitive Equivalents
This guy is doing the same thing you were doing in your calculations. He takes the Amps and volts before the caps and gets KWHs. Then he uses the new Amps x volts to again calculate KWHs. Adding the caps does not reduce the KWHs used by the motor. It only reduces the KVAH used by the motor. As previously stated your only gain will be a reduction of the I²R losses. However, if you are being billed a penalty for PF, adding capacitors can be a big help in reducing you monthly bill. Again you don't need this guy to get this done. By adding the correct size capacitors to each motor you will reduce the KVAR demand and I²R losses. Do it yourself. You know more that this salesman.
RE: Motor PF & Resultant Capacitive Equivalents
RE: Motor PF & Resultant Capacitive Equivalents
Installation was completed on a sewer plant and several of the local power monitors indicated load reduction as the mystery boxes were connected, however; the power company primary metering located about 1000' away indicated no change. Eventually the very expensive little black boxes were removed.
I was not able to verify why the local meters showed a reduction. The units might have been manufactured in Isreal, and from photographs looked like they had capacitors in them, but it was never clearly specified. I thought there might have been some circuit imposing a harmonic to frustrate the local CT's or metering, but once again this is speculation. If the vendor cannot clearly communicate what is going to happen in terms understandable to other engineers it might be a hoax.
RE: Motor PF & Resultant Capacitive Equivalents
Muthu
www.edison.co.in
RE: Motor PF & Resultant Capacitive Equivalents
We always suggest bulk correction for clients with high kVA demand charges or PF penalties, mainly due to reduced maintenance and installation costs.
RE: Motor PF & Resultant Capacitive Equivalents
If you don't give us some feedback we will have to ban you from the forum for an unspecified time period yet to be detemined.
RE: Motor PF & Resultant Capacitive Equivalents
You can not lower the kWh used by the motor with a capacitor. You would need a more efficient motor to lower the kWh used by the motor.
The capacitors will have about 0.3W of losses per kVAR. So, 75kVAR will have about 22.5W of losses. This would eat up just over half of that 41W of savings calculated above, meaning the savings in that example would really be about $10 a year.