As we talked last Saturday NZ / Friday nite USA time I said that I will make a thread re a worked example, so I get my old tech. notes out and have a look and refresh my own memory a bit too. there I don't deal much with PF controls at work.

Example:
A 5 kW motor has a PF of 0.6 with a 240 Volts 50 Hz supply.

Find capacitance required to:
a) increase the PF to unity.
b) increase the PF to 0.9.

Impedance Triangle

......../I
......./.I
....../..I
VA./...I VAr
..../....I
.../.....I
../......I
./.......I
/........I
----------
..Watts..

When the PF is improved, the power of the inductive load remains constant.

PF = true power / apparent power = W / VA.

a) P = 5000 Watts, PF = 0.6, U = 240 Volts.

therefore VA = W / PF = 5000 / 0.6 = 8333.3

and apply pythagoras on the triangle will yield

sq.rt(8333.3²-5000²) = 6666.6 VAr.

to get the PF to unity a capacitance of 6666.6 VAr is required.

VAr = U * Icap. 6666.6 = 240 * Icap. therefore I cap is 27.7 Amps.

Xc = U / I = 240 / 27.7 = 8.64 ohms.
where Xc is reactance of the Capacitor.

C = 1 / (2*pi*f*Xc) = 1 / 2*3.1415*50*8.64 = 368 µF.


[This message has been edited by RODALCO (edited 02-28-2006).]

b) To improve to 0.9.

W / PF = VA so 5000 / 0.9 = 5555.5

sq.rt (5555.5² - 5000²) = 2420 VAr

Required VAr is 6666.6 - 2420 = 4246.2

4246.2 = 240 * Icap. therefore Icap = 17.69 Amps.

Xc = U / Icap = 240 / 17.69 = 13.56 ohms.

C = 1 / 2*3.1415*50*13.56 = 234.7 µF.

Currents to motor at different PF values:

P = U*I*PF

PF 0.6 then I = 34.72 Amps
PF 0.9 then I = 23.15 Amps
PF 1.0 then I = 20.83 Amps

As you can see is that the current drops considerably between PF of 0.6 and 0.9
Also the heating in cables will be reduced as are losses in motorwindings with an improved PF.

To improve from 0.9 to 1.0 the capacitor bank needs to be about1½ times bigger and probably about 1½ times more expensive.

In general the aim is for around 0.95 depending upon utility tariffs.

Unity power factor may cause resonance in the circuit which can lead to higher voltages, as a large amount of energy is maintained in oscillation between L and C.

(Big smile)

Rodalco, that is awesome. You just refreshed and jogged my brain this morning..

Thanks for taking the time to do that, it helps out more than you know....

Dnk..

VAr = U * Icap. 6666.6 = 240 * Icap. therefore I cap is 27.7 Amps

I got a spreadsheet figured out based upon your calculations above, and so far it all checks out, mind us this is a single phase application right now.
But can you explain the above formula?

Is this (reactive current) equals (the applied voltage, multiplied by the current in the capacitor)?

I don't fully understand this one?

Dnk..



[This message has been edited by Dnkldorf (edited 02-28-2006).]

We need to know the current which flows to the cap. to work out its reactance, and with Xc we can calculate it's value in µF.

There are probably spreadsheets out for PF calculations anyway, or wholesalers who can provide assistance re the calcs.

In 3 situations the total value can also be calculated this way as long the phase voltages are used.

the total µF value found can be divided by 3 to spread the Caps over the 3 phases of the system.

In case of an installation 'unknown' the best is to get a datalogger installed for a week and analyse the PF value from the graph. Then work out what value to aim for as per utility tariffs and either do the calcs or get someone to do it to improve the PF and reduce the losses.

Cheers Ray (RODALCO)

I think I see what you are doing now, you first are figuring what it takes to go to unity, and then you back off?

Is this right?

Dnk..

[This message has been edited by Dnkldorf (edited 02-28-2006).]

Yep, there are other ways of doing these calculations, but I liked the step by step approach more.

You can also do it via the Tan (delta) approach, although I have to do some reading re this and you require a scientific calculator to work it out.