Questions About Power Factor Correction
Posted 02 May 2007 - 12:10 PM
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.
Posted 02 May 2007 - 06:02 PM
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.
Posted 09 May 2007 - 07:26 AM
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
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.
Posted 09 May 2007 - 07:53 AM
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
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Posted 09 May 2007 - 10:48 AM
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.
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