Power Factor Correction Equipment: advantages and disadvantages

[Steven Mill]

Normally, the power factor of the whole load on a large generating station is in the region of 0•8 to 0•9. However, sometimes it is lower and in such cases it is generally desirable to take special steps to improve the power factor. This can be achieved by the following equipment:

Static capacitor

The power factor can be improved by connecting capacitors in parallel with the equipment operating at lagging power factor. The capacitor (generally known as static capacitor) draws a leading current and partly or completely neutralizes the lagging reactive component of load current. This raises the power factor of the load. For three-phase loads, the capacitors can be connected in delta or star.

Advantages

1. They have low losses
2. They require little maintenance as there are no rotating parts
3. They can be easily installed as they are light and require no foundation
4. They can work under ordinary atmospheric conditions

Disadvantages

1. They have short service life ranging from 8 to 10 years
2. They are easily damaged if the voltage exceeds the rated value
3. Once the capacitors are damaged, their repair is uneconomical

Synchronous condenser

A synchronous motor takes a leading current when over-excited and, therefore, behaves as a capacitor. An over-excited synchronous motor running on no load is known as synchronous condenser. When such a machine is connected in parallel with the supply, it takes a leading current which partly neutralizes the lagging reactive component of the load. Thus the power factor is improved.

Advantages

1. By varying the field excitation, the magnitude of current drawn by the motor can be changed
   by any amount. This helps in achieving step less †control of power factor
2. The motor windings have high thermal stability to short circuit currents
3. The faults can be removed easily

Disadvantages

1. There are considerable losses in the motor
2. The maintenance cost is high
3. It produces noise
4. Except in sizes above 500 kVA, the cost is greater than that of static capacitors of the same
   rating
5. As a synchronous motor has no self-starting torque, therefore, an auxiliary equipment has to
   be provided for this purpose

Phase advancers

Phase advancers are used to improve the power factor of induction motors. The low power factor of an induction motor is due to the fact that its stator winding draws exciting current which lags be-hind the supply voltage by 90 degrees. If the exciting ampere turns can be provided from some other a.c. source, then the stator winding will be relieved of exciting current and the power factor of the motor can be improved. This job is accomplished by the phase advancer which is simply an a.c. exciter. The phase advancer is mounted on the same shaft as the main motor and is connected in the rotor circuit of the motor. It provides exciting ampere turns to the rotor circuit at slip frequency. By providing more ampere turns than required, the induction motor can be made to operate on leading power factor like an over-excited synchronous motor.

Advantages

1. As the exciting ampere turns are sup-plied at slip frequency, therefore, lagging kVAR drawn by the motor are considerably reduced
2. The phase advancer can be conveniently used where the use of synchronous motors is inadmissible

However, the major disadvantage of phase advancers is that they are not economical for motors below 200 H.P.

What do you think about it? Do you have others advantages or disadvantages in mind? If yes, just add a comment so that we can discuss about it.

[Al-Firasah]

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 = 100

assume that,
cos(thetai) = 0.85, therefore tan(thetai) = 0.6197
cos(thetaf) = 1, therefore tan(thetaf) = 0.0000

where
thetai = initial angle
thetaf = final angle
kVAri = initial reactive power
kVArf – final reactive power
kW = real power of electrical load

The 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 ohm

From equation (3): kVAr = 100*(0.6197 – 0)
= 61.97

From equation (4): Ir = [117.6 – 100*0.85]/(1.732*0.415)
= 32.6/0.7188
= 45.4 Amperes

From equation (5): The reduction in power = 45.5^2*0.005
= 10.4 Watts

[Anonymous]

Please give a chart for selecting capacitors for PFC of individual motors of different hp,speed,design code,starting kVa/kW etc
Also another chart for selecting capacitors for shunt capacitor type starting of induction motors.

[Anonymous]

everything has been explained by a fellow member, no comment on that.

[Spir Georges GHALI]

@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 = 100

assume that,
cos(thetai) = 0.85, therefore tan(thetai) = 0.6197
cos(thetaf) = 1, therefore tan(thetaf) = 0.0000

where
thetai = initial angle
thetaf = final angle
kVAri = initial reactive power
kVArf – final reactive power
kW = real power of electrical load

The 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 ohm

From equation (3): kVAr = 100*(0.6197 – 0)
= 61.97

From equation (4): Ir = [117.6 – 100*0.85]/(1.732*0.415)
= 32.6/0.7188
= 45.4 Amperes

From 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.

[Anonymous]

By using equation: kVAr = kW*(tan(acos(thetai))-tan(acos(thetaf)))

Assume kW = 1.00
cosΦf 0.85 0.86 0.87 0.88 0.89 0.9 0.91
cosΦi from 0.02 49.37 49.40 49.42 49.45 49.48 49.51 49.53
0.03 32.70 32.72 32.75 32.78 32.81 32.83 32.86
0.04 24.36 24.39 24.41 24.44 24.47 24.50 24.52
0.05 19.36 19.38 19.41 19.44 19.46 19.49 19.52
0.06 16.02 16.04 16.07 16.10 16.12 16.15 16.18
0.07 13.63 13.66 13.68 13.71 13.74 13.77 13.80
0.08 11.84 11.87 11.89 11.92 11.95 11.98 12.00
0.09 10.45 10.47 10.50 10.53 10.55 10.58 10.61
0.1 9.33 9.36 9.38 9.41 9.44 9.47 9.49
0.11 8.42 8.44 8.47 8.50 8.52 8.55 8.58
0.12 7.65 7.68 7.71 7.73 7.76 7.79 7.82
0.13 7.01 7.03 7.06 7.09 7.11 7.14 7.17
0.14 6.45 6.48 6.51 6.53 6.56 6.59 6.62
0.15 5.97 6.00 6.02 6.05 6.08 6.11 6.14
0.16 5.55 5.58 5.60 5.63 5.66 5.69 5.71
0.17 5.18 5.20 5.23 5.26 5.28 5.31 5.34
0.18 4.85 4.87 4.90 4.93 4.95 4.98 5.01
0.19 4.55 4.57 4.60 4.63 4.65 4.68 4.71
0.2 4.28 4.31 4.33 4.36 4.39 4.41 4.44
0.21 4.04 4.06 4.09 4.12 4.14 4.17 4.20
0.22 3.81 3.84 3.87 3.89 3.92 3.95 3.98
0.23 3.61 3.64 3.66 3.69 3.72 3.75 3.78
0.24 3.43 3.45 3.48 3.51 3.53 3.56 3.59
0.25 3.25 3.28 3.31 3.33 3.36 3.39 3.42
0.26 3.09 3.12 3.15 3.17 3.20 3.23 3.26
0.27 2.95 2.97 3.00 3.03 3.05 3.08 3.11
0.28 2.81 2.84 2.86 2.89 2.92 2.94 2.97
0.29 2.68 2.71 2.73 2.76 2.79 2.82 2.84
0.3 2.56 2.59 2.61 2.64 2.67 2.70 2.72
0.31 2.45 2.47 2.50 2.53 2.55 2.58 2.61
0.32 2.34 2.37 2.39 2.42 2.45 2.48 2.51
0.33 2.24 2.27 2.29 2.32 2.35 2.38 2.40
0.34 2.15 2.17 2.20 2.23 2.25 2.28 2.31
0.35 2.06 2.08 2.11 2.14 2.16 2.19 2.22
0.36 1.97 2.00 2.02 2.05 2.08 2.11 2.14
0.37 1.89 1.92 1.94 1.97 2.00 2.03 2.06
0.38 1.81 1.84 1.87 1.89 1.92 1.95 1.98
0.39 1.74 1.77 1.79 1.82 1.85 1.88 1.91
0.4 1.67 1.70 1.72 1.75 1.78 1.81 1.84
0.41 1.60 1.63 1.66 1.68 1.71 1.74 1.77
0.42 1.54 1.57 1.59 1.62 1.65 1.68 1.71
0.43 1.48 1.51 1.53 1.56 1.59 1.62 1.64
0.44 1.42 1.45 1.47 1.50 1.53 1.56 1.59
0.45 1.36 1.39 1.42 1.44 1.47 1.50 1.53
0.46 1.31 1.34 1.36 1.39 1.42 1.45 1.47
0.47 1.26 1.28 1.31 1.34 1.37 1.39 1.42
0.48 1.21 1.23 1.26 1.29 1.32 1.34 1.37
0.49 1.16 1.19 1.21 1.24 1.27 1.29 1.32
0.5 1.11 1.14 1.17 1.19 1.22 1.25 1.28
0.51 1.07 1.09 1.12 1.15 1.17 1.20 1.23
0.52 1.02 1.05 1.08 1.10 1.13 1.16 1.19
0.53 0.98 1.01 1.03 1.06 1.09 1.12 1.14
0.54 0.94 0.97 0.99 1.02 1.05 1.07 1.10
0.55 0.90 0.93 0.95 0.98 1.01 1.03 1.06
0.56 0.86 0.89 0.91 0.94 0.97 1.00 1.02
0.57 0.82 0.85 0.87 0.90 0.93 0.96 0.99
0.58 0.78 0.81 0.84 0.86 0.89 0.92 0.95
0.59 0.75 0.78 0.80 0.83 0.86 0.88 0.91
0.6 0.71 0.74 0.77 0.79 0.82 0.85 0.88
0.61 0.68 0.71 0.73 0.76 0.79 0.81 0.84
0.62 0.65 0.67 0.70 0.73 0.75 0.78 0.81
0.63 0.61 0.64 0.67 0.69 0.72 0.75 0.78
0.64 0.58 0.61 0.63 0.66 0.69 0.72 0.74
0.65 0.55 0.58 0.60 0.63 0.66 0.68 0.71
0.66 0.52 0.54 0.57 0.60 0.63 0.65 0.68
0.67 0.49 0.51 0.54 0.57 0.60 0.62 0.65
0.68 0.46 0.48 0.51 0.54 0.57 0.59 0.62
0.69 0.43 0.46 0.48 0.51 0.54 0.56 0.59
0.7 0.40 0.43 0.45 0.48 0.51 0.54 0.56
0.71 0.37 0.40 0.43 0.45 0.48 0.51 0.54
0.72 0.34 0.37 0.40 0.42 0.45 0.48 0.51
0.73 0.32 0.34 0.37 0.40 0.42 0.45 0.48
0.74 0.29 0.32 0.34 0.37 0.40 0.42 0.45
0.75 0.26 0.29 0.32 0.34 0.37 0.40 0.43
0.76 0.24 0.26 0.29 0.32 0.34 0.37 0.40
0.77 0.21 0.24 0.26 0.29 0.32 0.34 0.37
0.78 0.18 0.21 0.24 0.26 0.29 0.32 0.35
0.79 0.16 0.18 0.21 0.24 0.26 0.29 0.32
0.8 0.13 0.16 0.18 0.21 0.24 0.27 0.29
0.81 0.10 0.13 0.16 0.18 0.21 0.24 0.27
0.82 0.08 0.10 0.13 0.16 0.19 0.21 0.24
0.83 0.05 0.08 0.11 0.13 0.16 0.19 0.22
0.84 0.03 0.05 0.08 0.11 0.13 0.16 0.19

[Spir Georges GHALI]

Dear Al-Firasah ;

 

My remarks in reply No. 6 were about some points in your reply No. 3, and your reply No. 7 is totally different, please reply to my remarks mentioned in No. 6

Regards.   

[Anonymous]

Have you been introduced to the 3DFS Power Controller?  This technology completely restores electricity to a perfect sine wave in true real time 24/7.

[Bilal Aslam]

These steps are very useful , but but we can also better the power factor by replacing the induction motor (if being used ) with same rating of synchronous motor . it would draws same load with less power.

[Anonymous]

What will be the impact of torque/speed curve on the load characteristics compared to induction motor?..

[Anonymous]

gud mrng ,i  am doing project on “measurement and automatic control of power factor by microprocessor”.in this project  loads are varied.according to value of phase angle between between voltage and current,capacitorrs are to be switched.can u please tell me the microprocessor program for this project.this program should calculate the phase difference and should swich suitable number of caapacitors and display power factor accordingly.please send 

me email.

[Anonymous]

Power factor correction and harmonic mitigation go hand in hand and should be tackeled in totality

[Spir Georges GHALI]

@guest said:
Power factor correction and harmonic mitigation go hand in hand and should be tackeled in totality

Dear ;

 

You’re 100% right, as I say always, before installing and even calculating the Capacitors’s Power, we should know all about the existed Harmonics Currents in the Network, then decide if we can install the Capacitors without any Network’s treatment and in this case what Type of Capacitors should be used, or we should do the Harmonics’ treatment before than calculate and install the Capacitors.   

[Anonymous]

Pretty! This was an extremely wonderful article. Thank you for supplying these details.

[Anonymous]

How do you select harmonic filters ,by datalogger recording harmonics over a period or some other means?.
What are the advantages/disadvantages of switching capacitors by contactors & thyristers?.

[Author]

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