Motor PF & Resultant Capacitive Equivalents 1

[Fraser21]

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?

 

[desertfox]

Hi Fraser21

This site might help:-

http://www.allaboutcircuits.com/vol_2/chpt_11/3.html

desertfox

 

[rbulsara]

Yes, so far. To me remembering Q=S* sin(x) works better.

Just remmber that motor and its circuit losses are not that signinficant to begin with.

Rafiq Bulsara

http://www.srengineersct.com

 

[wareagle]

Quote"This guy claims his "little black box" will reduce power losses in motors 10-200 H.P. by 6-12%."

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.

 

[wareagle]

Quote"This guy claims his "little black box" will reduce power losses in motors 10-200 H.P. by 6-12%."

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.

 

[Fraser21]

Ok the next step is where I get hung up. I need to solve for my resistance / reactance values.

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.

 

[rbulsara]

Electrical losses are only I^2*R. You need to find or measure resistance of cables and motor windings. Use per wire and per phase values, along with per phase voltage (VLL/1.732).

Rafiq Bulsara

http://www.srengineersct.com

 

[wareagle]

You don't solve for the resistance. If you are using the NEC, look in table 9 in the back and read the AC resistance from the table. The table is R per 1000 ft or meters. If you are using multiple cables per phase, take the Resistance and divide by the number of cables.

 

[electricpete]

Motor resistance should be irrelevant - motor current depends only on terminal voltage and load... nothing to do with any capacitor connected external to the motor.

So (unless you change terminal voltage), you do nothing to reduce any losses in the motor. Only supply system losses, which are small.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[waross]

Reactive load = 89 KVARs, buy 75 KVARs of capacitors.That should bring your power factopr up to 99% plus.
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

 

[rbulsara]

electricpete is very correct...current downstream of the capacitor connection does not change.


Rafiq Bulsara

http://www.srengineersct.com

 

[Fraser21]

Guys attached are the survey results from a couple of motors in the building. From what I understand the intention is to install this little black box on a couple motors to see if the system works and from that there are almost 20 buildings full of motors to proceed with. This guy who surveyed never took any impedance or resistance values and i'm assuming he's guessing at his "optimized" pf after his invention is installed, unless you guys can figure how he came up with these #'s not knowing. Again he claims that it will save 6-12% on losses for motors 10-200H.P. What do you guys think?

 

[Fraser21]

Waross, if i'm correct here. If we add 75kVARs to that motor to bring the PF up to what would be about .998PF. That drop in 37A would result in the following kVA due to a smaller reactive vector:

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?

 

[ccjersey]

I think you're back to figuring kVA and calling it kW. The power factor correction will not change the kW drawn by the motor and billed on the utility bill

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.

 

[DRWeig]

Fraser21,

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 ?/kft). Total resistance is 3 * 200 * 0.05 / 1000 or 0.03?. 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

 

[Fraser21]

Ahhh, your right I was looking at it all wrong, those savings are not even worth looking at. Did you happen to see the sheet I attached above? I'm trying to figure where this guy is coming from and if this "black box" is worth installing throughout these buildings.

thanks ccjersey & DWR much appreciated, i knew i was doing something wrong there.

 

[rbulsara]

Fraser:

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

 

[PQDoctor]

There are two ways to reduce the current consumed by motors:

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://www.energycentral.com/enduse...le-Costs-using-Sinusoidal-Motor-Controllers/.

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:

http://www.PowerQualityDoctor.com

 

[waross]

An anecdote that may save you some time.>
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

 

[DRWeig]

Thanks Bill!

I've had to do that recently all too often. The energy saving gizmo marketers are everywhere.

Good on y'all,

Goober Dave