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The most common method for improving power factor is to add
capacitors banks to the system. Capacitors are attractive
because they're economical and easy to maintain. Not only
that, they have no moving parts, unlike some other devices
used for the same purpose.

supplies the reactive power needed by the load. If you size
and select the capacitor bank to compensate to a unity
power factor, it can supply all the reactive power needed
by the load, and no reactive power is demanded from the
utility. If you design the capacitor bank to improve the
power factor to a quantity less than 1.0, the reactive
power supplied by the bank will be its rated kVARs (or
MVARs), while the rest of the reactive power needed by the
load will be supplied by the utility.

 Is This Answer Correct ? 132 Yes 11 No

guys this question does't mean that to specify your
general how to calculate your required capacity of

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200Kvar

 Is This Answer Correct ? 35 Yes 8 No

17 no's 25 kvar

 Is This Answer Correct ? 26 Yes 9 No

The addition of capacitance depend upon your power factor.suppose u may get more than 0.97 or unity Pf,u no need to add capacitor.becoz it may act like load

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50 kvra eacth amount

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Power factor is only present in the ac power supply systems. In dc there is no usage of power factor terms. Ac supply is having frequency and potential is alternating with respective to time as such the load on the ac power source is of mix like capacitance, resistance and inductance

Capacitance is offered by ug cables and heater and filament lamps offer the resistive loads and hid and fluorescent lamps offers inductance value

Inductive reactance (xl) and capacitive reactance (xc) is counter each other and resultant effect will reflect in ac system in addition to normal resistance. This cumulative resistive effect is called as impedence.

Now power factor is equel to pf = r/z.

Now if we draw a power circle the you know the differnce of kw and kva is made by kvar that is reactive power if we reduce this reactive power then the kva demand and kw i mean apparent power and actual power will became more or less equal. Now power factor that is quality power.

So we can improve the power systems by adding counter reactive power by adding capacitors in kvar ratings.

Apfc panels are best suitable for proper monitoring as static correction will not perform well in massive load centres as failure and degradation of capacitor is hard to find out and rectify.

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I Need to know the how much capacitor bank that i have to put for these Load KW or HP or KVA
For Example: 15HP we can use 2kVAR of capacitor
like that?...

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6300kvar & 2100kvar 11kv

 Is This Answer Correct ? 6 Yes 3 No

The correct method for PFC is to calculate how much correct you need.
If you know the PF and Pd of a motor being used on average, and want to improve to 90%p.f. then compute then angle of PF before and after from theta=cos^-1(PF) then take difference in tan (theta) * Pd

Good to Know:
Important formulas which is used for Power factor improvement calculation as well as used in the above calculation

Power in Watts
kW = kVA x Cosθ
kW = HP x 0.746 or (HP x 0.746) / Efficiency … (HP = Motor Power)
kW = √ ( kVA2– kVAR2)
kW = P = VI Cosθ … (Single Phase)
kW = P =√3x V x I Cosθ … (Three Phase)

Apparent Power in VA
kVA= √(kW2+ kVAR2)
kVA = kW/ Cosθ

Reactive Power in VA
kVAR= √(kVA2– kW2)
kVAR = C x (2 π f V2)

Power factor (from 0.1 to 1)
Power Factor = Cosθ = P / V I … (Single Phase)
Power Factor = Cosθ = P / (√3x V x I) … (Three Phase)
Power Factor = Cosθ = kW / kVA … (Both Single Phase & Three Phase)
Power Factor = Cosθ = R/Z … (Resistance / Impedance)

XC = 1/ (2 π f C) … (XC = Capacitive reactance)
IC = V/ XC … (I = V / R)

C = kVAR / (2 π f V2) in microfarad

Required Capacity of Capacitor in kVAR
kVAR = C x (2 π f V2)

 Is This Answer Correct ? 6 Yes 3 No

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