Re: Power Factor Correction Equipment: advantages and disadvantages

[Spir Georges GHALI]

@Al-Firasah said:
Before installs of capacitor
kVAri = kW*tan(thetai) ……… (1)

After installs of p.f.correction
kVArf = kW*tan(thetaf) …….. (2)

The value of capacitor, kVAr = kVAri – kVArf

equation (1) – (2)
kVAri – KVArf = kW*tan(thetai) – kW*tan(thetaf)
kVAr = kW*(tan(thetai) – tan(thetaf)) …… (3)

i = initial
f = final
theta = angle of power factor
kVAr = reactive power = 117.6
kW = real power = 100

assume that,
cos(thetai) = 0.85, therefore tan(thetai) = 0.6197
cos(thetaf) = 1, therefore tan(thetaf) = 0.0000

where
thetai = initial angle
thetaf = final angle
kVAri = initial reactive power
kVArf – final reactive power
kW = real power of electrical load

The reduction in current after installs of capacitor, Ir is:

Ir = [kVAi – kW*cos(thetai)]/(sqrt(3)*V) ……. (4)

where V in kV = nominal voltage supply from supply authority = 0.415kV

The reduction of power in cable = Ir^2*resistance of cable ………… (5)
assume that resistance of cable 0.005 ohm

From equation (3): kVAr = 100*(0.6197 – 0)
= 61.97

From equation (4): Ir = [117.6 – 100*0.85]/(1.732*0.415)
= 32.6/0.7188
= 45.4 Amperes

From equation (5): The reduction in power = 45.5^2*0.005
= 10.4 Watts

Dear Al-Firasah

refer to your equations mentioned above, i have the following remarks :

1- For any application, we don’t, never ever, correct the power factor to ” Cos φ = 1 ” that means ” tg φ = 0 “. 

2- In the equation No. 4, you multiply ” kW x Cos φ ” that means ” P x Cos φ “, what is the kind of this result ?

3- For any inductive load we have the following values : P, U, Cos φ, so the nominal current will be : ” I = P/1.732 x U x Cos φ “, where : P ( Watt ), U ( Volt ), and I ( Amp ), and this Current’s value doesn’t change, because that current is one of the load’s specifications, but if we installed the Capacitors bi-side the load, in this case, the current runs in the cable supplying the load & capacitor will be changed.