Capacitor Bank Rating in Farad

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To find out Farad or Microfarad, you shall also define following:

1. System:1 phase or 3 phase

2. Service Voltage

3. System Frequency.

For example for 3 phase system, at 415V line voltage and 50Hz system frequency, to produce 1kVAR of Capacitive Reactance we require 3 Numbers of 11Micro farad (0.000011Farad) capacitors in delta connection. Voltage rating of each capacitors shall be at least 415V.

Calculations are:

Xc (capacitive reactance) = 1/2ΠfC (Π = 22/7, f = 50 Hz, C =0.000011 F)

= 298.29 Ohms

Capacitive Power = 1.732*415*415/298.29

= 1000 VAR

= 1 kVAR

Similarly one can calculate capacitance at any voltage and frequency.

For example at new 3 phase line voltage = V volts And Frequency = F Hz, Capacitance for 1kVAR will be = 0.000011*(415/V)²*(50/F)

Regards,

Ramesh

You cannot find a direct relation between kVAr and Farad. The higher the voltage the lower value capacitor will result the same kVAr value. The grid works as an inductive load. Its impedance X_L. You have to install the same value capacitive impedance X_C.

X_LGrid = X_CBank

In this case the capacitor bank will fully compensate the power factor. It's a sort of resonance.

Guys, stop answering this guy he says he is a graduated engineer in electrical working as an electrical design engineer in the construction field. He has been told enough times to perform simple internet or CR4 site searches first.

This is a discussion forum not a free education site for the plain lazy, especially when the questions are of a transmission or distribution level (high voltage = very dangerous).

Muthuvel, have you considered taking a proper power generation and distribution course rather than asking us?

If I can only discuss things that I know, and not can ask about things I don't know, then I won't stay here any longer.

Things I know, I don't have to discuss, because I know them already.

I come here to learn and to help other people.

I don't have a problem to share my information with people who

• - have no time to take a course about the their unsolved question
• - have no money to take that course
• - have only a small problem and don't want to take the complete course
• - have examined their problem and ask a second opinion from other members.

And I don't think people are helped just by telling them to google their problem.

I'm sure they already did some research on google before they come here.

If this is directed at my comment (which I think it is), firstly thank you for being the first to ask a very valid question I hoped someone would ask.

Secondly, have a look at the numerous previous posts by Muthuvel to get an idea of where I am coming from as an actual power and transmission Engineer and long time CR4 user. This is directed at a number of people on CR4 (not just Muthuvel).

At some point questions stop being general interest and start becoming potentially dangerous 'homework-like' spam, especially when most if not all can be answered with a simple internet search he obviously did not bother to perform (as we have mentioned to him and others time and time again). He says he is a graduate Engineer, he should know better.

Learning is all well and good and encouraged by most of us here, but trying to teach yourself transmission and distribution theory by asking us simple questions instead of performing basic internet searches yourself or taking a proper course is just plain lazy, potentially dangerous and NOT (in my opinion) what CR4 is intended for.

I don't think we would be having this conversation if this was a medical discussion site and he was a graduate Doctor, but is high voltage specification and design any less dangerous?

I hope most of us act like a bonus pater familias, and know to judge the dangers they are in.

As an Engineer, ethics are ever present when we answer (or fail to answer) questions posted on CR4. Some don't understand this sometimes, and often we don't clearly explain our reasons for not answering the question properly or at all.

Finally, a word of thank to you, jack of all trades, and all the others who share their knowledge in this forum.

We're here to help and learn but sometimes rather than answering the question completely we judge links, short answers or 'tough love' responses to be the most appropriate.

Finally I got the Answer

• Capacitor current(Amps)=Icap= (Reactive power/Nominal voltage)=Q/E.
• Reactance across the capacitor Xc=E/Icap=(1/2∏f C)
• f is the frequency So C=(1/2∏f Xc)