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).]
