Hi guys...... A client of mine is interested in installing pfc to a joinery factory but Im a bit unsure of exactly how it all works. I have found out that his mdi reading is 265kva. I also know that the factory is drawing 185kw peak. The energy supplier is saying that the power factor is currently at 0.7. I can see the logic in the 185 kw @0.7 gives a correct kva of 265. If we improve the power factor to 0.9, the kva will drop to 206kva and thus the supply costs will decrease. (tell me if im wrong)

Now the tricky part for me. The supplier of the pfc unit says that we will also decrease the amount of current being used and thus will not have to install bigger mains for new machinery. I dont understand how the current actually decreases when the total kw will stay the same and the line voltage will stay the same.

Thanks

MICK

# Questions About Power Factor Correction

Started by mikspark, May 02 2007 12:10 PM

4 replies to this topic

### #1

Posted 02 May 2007 - 12:10 PM

### #2

Posted 02 May 2007 - 06:02 PM

Read this

http://www.lmphotoni...om/pwrfact1.htm

then come back with whatever you didn't understand.

In general though, you should look at correcting power factor as close to the inductive loads as you can, i.e. near the motors as opposed to bulk correction. Bulk correction leads to overcorrection if motor loads are switched on and off through the work day. A PFC control system can mitigate that, but adds in a risk of creating spikes in the line every time a cap bank switches in and out as the inductive loading changes. If you correct your PF at each motor load with caps that are only switched on with the motors, you avoid all those issues.

http://www.lmphotoni...om/pwrfact1.htm

then come back with whatever you didn't understand.

In general though, you should look at correcting power factor as close to the inductive loads as you can, i.e. near the motors as opposed to bulk correction. Bulk correction leads to overcorrection if motor loads are switched on and off through the work day. A PFC control system can mitigate that, but adds in a risk of creating spikes in the line every time a cap bank switches in and out as the inductive loading changes. If you correct your PF at each motor load with caps that are only switched on with the motors, you avoid all those issues.

"He's not dead, he's just pinin' for the fjords!"

### #3

Posted 09 May 2007 - 07:26 AM

well i'm still not 100%. if your drawing 185 kw of power. i dont understand how this can just go downwith pfc. the motors are still rated at 185kw.

Anyway, more questions. In working out your kva, I've found 2 calculations. Which one is correct. Also would the line voltage be 240volt or 415volt.

KVA= I x V/ PF x 1000

or

KVA = I x V x 1.732/1000

in my case I want to find out the total amps being drawn

I= 265 x1000/ 415 x 1.732

= 370 amps

the other calc.

I= kva x 1000 x 0.7/415

= 265 x 1000 x 0.7 /415

= 446 amps

I think this last one is wrong because as the pf goes up so does the amps.

AS for local caps on motors for pfc, our supplier reccommends using a unit at the Main switch board. And the energy suppliers in Aus seem to reccommend it as well.

Anyway, more questions. In working out your kva, I've found 2 calculations. Which one is correct. Also would the line voltage be 240volt or 415volt.

KVA= I x V/ PF x 1000

or

KVA = I x V x 1.732/1000

in my case I want to find out the total amps being drawn

I= 265 x1000/ 415 x 1.732

= 370 amps

the other calc.

I= kva x 1000 x 0.7/415

= 265 x 1000 x 0.7 /415

= 446 amps

I think this last one is wrong because as the pf goes up so does the amps.

AS for local caps on motors for pfc, our supplier reccommends using a unit at the Main switch board. And the energy suppliers in Aus seem to reccommend it as well.

### #4

Posted 09 May 2007 - 07:53 AM

Hello mikspark

If you measure the current drawn from the supply, you are actually measuring two currents. One current is the work current that goes into KW, and the other is the reactive current. The two currents are 90 degrees out of phase. The work current is in phase with the supply voltage.

The total current that you measure is the vector sum of the two currents and is equal to the square root of the sum or the squares of the two currents.

You can alter the reactive current without changing the work current. If you increase the reactive current and keep the work current constant, the resultant measured current will increase. If you reduce the reactive current and keep the work current constant, the resultant current will reduce. If the reactive current is zero, then the resultant current will be equal to the work current.

The power factor is the cosine of the angle between the work current and the resultant current.

Increasing the power factor, reduces the reactive current.

Adding power factor correction to improve the power factor will reduce the reactive current and therefore the resultant current.

The correct formula for KVA = I x V x rt3 / 1000 where I equals the line current, V equals the phase to phase line voltage (i.e. 415 V)

At 265KVA, the current will be 265000 / 415 / 1.732 = 369A

At a power factor of 1.0 and 185KW, the current would be 185000 / 415 / 1.732 / 1.0 = 257A

In other words, if you corrected the load from 0.7 to 1.0, the current would drop from 369A to 257A

Bulk correction is usually set up to keep the power factor better than 0.95.

The current at 185KW and 0.95 pF would be 185000 / 415 / 1.732 / 0.95 = 271A

Best regards,

If you measure the current drawn from the supply, you are actually measuring two currents. One current is the work current that goes into KW, and the other is the reactive current. The two currents are 90 degrees out of phase. The work current is in phase with the supply voltage.

The total current that you measure is the vector sum of the two currents and is equal to the square root of the sum or the squares of the two currents.

You can alter the reactive current without changing the work current. If you increase the reactive current and keep the work current constant, the resultant measured current will increase. If you reduce the reactive current and keep the work current constant, the resultant current will reduce. If the reactive current is zero, then the resultant current will be equal to the work current.

The power factor is the cosine of the angle between the work current and the resultant current.

Increasing the power factor, reduces the reactive current.

Adding power factor correction to improve the power factor will reduce the reactive current and therefore the resultant current.

The correct formula for KVA = I x V x rt3 / 1000 where I equals the line current, V equals the phase to phase line voltage (i.e. 415 V)

At 265KVA, the current will be 265000 / 415 / 1.732 = 369A

At a power factor of 1.0 and 185KW, the current would be 185000 / 415 / 1.732 / 1.0 = 257A

In other words, if you corrected the load from 0.7 to 1.0, the current would drop from 369A to 257A

Bulk correction is usually set up to keep the power factor better than 0.95.

The current at 185KW and 0.95 pF would be 185000 / 415 / 1.732 / 0.95 = 271A

Best regards,

Mark Empson | administrator

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

### #5

Posted 09 May 2007 - 10:48 AM

Thanks Marke

You wrote that explanation perfectly for someone of my intelligence to understand. If I have any more issues when I do the installation, I know who to ask. Thanks again.

Mikspark.

You wrote that explanation perfectly for someone of my intelligence to understand. If I have any more issues when I do the installation, I know who to ask. Thanks again.

Mikspark.

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