Methods for Power Factor Improvement
The following devices and equipments are used for power factor Improvement.
1. Static Capacitor
2. Synchronous Condenser
3. Phase Advancer
1. Static Capacitor
Capacitors are connected in parallel with the equipments operating at low power factor for the purpose of power factor improvement. These static capacitors provide leading current which partially or totally neutralize the lagging inductive component of load current, thus power factor of the load circuit is improved. These capacitors are installed in vicinity of large inductive load e.g. Induction motors and transformers etc, and improve the load circuit power factor to improve the system or devises efficiency.
Suppose, here is a single phase inductive load which is taking lagging current (I_{1}) and the load power factor is cosφ_{1} as shown in fig-(i).
In fig-(iii), a capacitor (C) has been connected in parallel with load. Now a current (I_{C}) is flowing through capacitor which lead 90° from the supply voltage. The load current is (I_{1}). The vectors combination of (I_{1}) and (I_{C}) is (I_{2}) which is lagging from voltage at φ_{2} as shown in fig-(iv).
It’s clear from fig-(iv) that angle of φ_{2 }< φ_{1 }i.e. angle of φ_{2 }is less than from angle of φ_{1}. Therefore cosφ_{2} is less than from cosφ_{1} (cosφ_{2}> cosφ_{1}). Hence the load power factor is improved by capacitor.
Also note that after the power factor improvement, the circuit current would be less than from the low power factor circuit current. Also, before and after the power factor improvement, the active component of current would be same in that circuit because capacitor eliminates only the reactive component of current. Also, the active power (in watts) would be same after and before power factor improvement.
Calculation of capacitor value
Let VAR_{1 }and VAR_{2 }are reactive power corresponding to p.f.1 and p.f.2 respectively. Since,
VAR_{1} = W tanφ_{1} = W tan (cos^{-1} p.f._{1})
VAR_{2} = W tanφ_{2} = W tan (cos^{-1} p.f._{2})
So, VAR = VAR_{1} - VAR_{2}
Hence, the required reactive power (Q_{C}) to improve p.f. from cosφ_{1} to cosφ_{2} is
Q_{C} = VAR_{1} - VAR_{2}
Now, Since, Q_{C} = V I_{C} & I_{C} = V / X_{C} = VωC i.e., Q_{C} = V^{2 }/ X_{C}
or, Q_{C } = V^{2}ωC
or, C (in farad) = Q_{C} / ωV^{2} = Q_{C} / 2πf V^{2}
For three phase delta connection, V_{P} = V_{L}
Farad rating of each capacitor, C_{D }= Q_{C} / ω (V_{L})^{2}
For three phase star connection, V_{P} = V_{L} / √3
Farad rating of each capacitor, C_{Y }= 3Q_{C} / ω (V_{L})^{2}
Note that:
1. Q_{C} = VAR per phase
2. For 3 ph system, given power is 3 ph power, so first of all find out single phase power.
3. For 3 ph system, given voltage is line voltage (V_{L}).
Advantages:
1. Capacitor bank offers several advantages over other methods of power factor improvement.
2. Losses are low in static capacitors
3. There is no moving part, therefore need low maintenance
4. It can work in normal air conditions (i.e. ordinary atmospheric conditions)
5. Do not require a foundation for installation
6. They are lightweight so it is can be easy to installed
Disadvantages:
1. The age of static capacitor bank is less (8 – 10 years)
2. With changing load, we have to ON or OFF the capacitor bank, which causes switching surges on the system
3. If the rated voltage increases, then it causes damage it
4. Once the capacitors spoiled, then repairing is costly