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Power Factor
Correction
ABDUR REHMAN
13004102003
Objectives
•Define power factor.
•Calculate the capacitor value required to correct AC
series parallel networks to the desired apparent power.
•Compare currents, voltages, and power in AC series
parallel networks before and after power factor
correction.
Importance of Power Factor
•Consider the following example: A generator is
rated at 600 V and supplies one of two possible
loads.
•
Load 1: P = 120 kW, P.F = 1
Load 2: P = 120 kW, P.F = 0.6
•Generator current required to supplythe load
Importance of Power Factor
• For the load with Fp = 0.6, the generator
had to supply 133 more amperes in order to
do the same work .
• Larger current means larger equipment
(wires, transformers, generators) which cost
more.
• Larger current also means larger
transmission losses (think I2
R).
POWER-FACTOR CORRECTION
FIG. shows the impact of power-factor
correction on the power triangle.
Power factor correction capacitors for A, B, and C
phases at the Crofton , MD substation
Rating: 230 kV, 360 MVAR
size comparison Capacitor banks
Power Factor Correction
•Almost all loads are inductive.
•To cancel the reactive component of power, we
must add reactance of the opposite type. This is
called power factor correction.
Capacitor bank in
shipboard power panel
for FP correction
Power Factor Correction
•Almost all loads (commercial, industrial and
residential) look inductive (due to motors,
fluorescent lamp).
•Hence, almost all power factor correction consists of
adding capacitance.
Power Factor Correction Solution Steps
1. Calculate the reactive power (Q) of the load
2. Insert a component in parallel of the load that will cancel out that
reactive power
e.g. If the load has QLD=512 VAR, insert a capacitor with QC=-512
VAR.
3. Calculate the reactance (X) that will give this value of Q
Q=V2
/X
1. Calculate the component value (F ) required to provide that
reactance.
Power Factor Correction
•Transmission lines and generators must be sized to
handle the larger current requirements of an
unbalanced load.
•Industrial customers are frequently fined by the
utility if their power factor deviates from the
prescribed value established by the utility.
Thank you

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Power factor correction

  • 2. Objectives •Define power factor. •Calculate the capacitor value required to correct AC series parallel networks to the desired apparent power. •Compare currents, voltages, and power in AC series parallel networks before and after power factor correction.
  • 3. Importance of Power Factor •Consider the following example: A generator is rated at 600 V and supplies one of two possible loads. • Load 1: P = 120 kW, P.F = 1 Load 2: P = 120 kW, P.F = 0.6 •Generator current required to supplythe load
  • 4. Importance of Power Factor • For the load with Fp = 0.6, the generator had to supply 133 more amperes in order to do the same work . • Larger current means larger equipment (wires, transformers, generators) which cost more. • Larger current also means larger transmission losses (think I2 R).
  • 5. POWER-FACTOR CORRECTION FIG. shows the impact of power-factor correction on the power triangle.
  • 6. Power factor correction capacitors for A, B, and C phases at the Crofton , MD substation Rating: 230 kV, 360 MVAR size comparison Capacitor banks
  • 7. Power Factor Correction •Almost all loads are inductive. •To cancel the reactive component of power, we must add reactance of the opposite type. This is called power factor correction. Capacitor bank in shipboard power panel for FP correction
  • 8. Power Factor Correction •Almost all loads (commercial, industrial and residential) look inductive (due to motors, fluorescent lamp). •Hence, almost all power factor correction consists of adding capacitance.
  • 9. Power Factor Correction Solution Steps 1. Calculate the reactive power (Q) of the load 2. Insert a component in parallel of the load that will cancel out that reactive power e.g. If the load has QLD=512 VAR, insert a capacitor with QC=-512 VAR. 3. Calculate the reactance (X) that will give this value of Q Q=V2 /X 1. Calculate the component value (F ) required to provide that reactance.
  • 10. Power Factor Correction •Transmission lines and generators must be sized to handle the larger current requirements of an unbalanced load. •Industrial customers are frequently fined by the utility if their power factor deviates from the prescribed value established by the utility.