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@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 = 100assume that,

cos(thetai) = 0.85, therefore tan(thetai) = 0.6197

cos(thetaf) = 1, therefore tan(thetaf) = 0.0000where

thetai = initial angle

thetaf = final angle

kVAri = initial reactive power

kVArf – final reactive power

kW = real power of electrical loadThe 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 ohmFrom equation (3): kVAr = 100*(0.6197 – 0)

= 61.97From equation (4): Ir = [117.6 – 100*0.85]/(1.732*0.415)

= 32.6/0.7188

= 45.4 AmperesFrom 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.