Power Factor Correction
A low lagging power factor can be corrected, or brought nearer to unity,
by adding capacitance parallel to the load (similarly, a low leading power
factor can be corrected, or brought nearer to unity, by adding inductance
parallel to the load, but this an unusual situation).
Consider the following phasor diagrams:
Ic
V Supply
IL
Θ1
IR
V Supply
Θ2
IR
IL
=Ic
1.
Uncorrected Circuit
2.
Corrected Circuit
It can be seen that the addition of capacitance to the inductive circuit
decreases the phase angle and therefore increases the power factor.
Ideally, power factor correction capacitors are connected at the terminals
of the inductive load. This reduces the current up to the load and hence,
the size of cabling and switchgear within the installation.
Power factor capacitors are rated by their reactive power. That is, they
are rated in VArs (Volt-Amperes Reactive) or kVArs.
Single phase power factor correction capacitors are simply connected
across the phase and neutral conductors.
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© Motorpol
Smaller Phase Power Factor Capacitors
Power factor correction capacitors are fitted to inductive loads in order to
improve the low lagging power factor.
It is desirable to achieve a power factor of about 0·95. If the power factor
is corrected to a value any higher than this, several problems are
encountered:
The amount of capacitance required to bring the power factor to above
this value is substantial. The extra capacitance not only takes a large
space to install, but is also more expensive than the savings gains
achieved by installing it.
As the power factor approaches unity, the electrical system approaches
resonance. This causes large currents to flow and can damage
switching equipment and present a shock hazard to personnel.
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3 Phase Power Factor Capacitors
Three phase capacitors are normally internally connected in a delta
configuration as this is the most economical use of capacitance. While
the capacitors must be rated to a higher voltage when delta connected,
the capacitance needs only be ⅓ the amount necessary for a lower
voltage star connection. However, individual capacitors may be
connected in star across the load terminals.
Connecting the capacitor directly across the terminals of the load also
allows the capacitor to discharge into the load when the supply is
removed, thus removing the potential shock hazard associated with
charged capacitors.
If the inductive load is a motor, then the overload settings in the motor
starter need to be adjusted to compensate for the reduced current due to
the improved power factor.
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Power factor capacitors should never be fitted to the output of an
electronic motor controller (VSD or soft starter). If these devices are
used then they usually present a near-unity power factor to the supply
but if any capacitors are required then they must be fitted on the supply
side of the starter or motor controller.
starter
Phase
M
1∼
Neutral
a. Capacitors Fitted To a Single Phase Motor
starter
M
3∼
b. Delta –Connected Capacitors on a Three Phase Motor
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starter
M
3∼
c. Star-Connected Capacitors on a Three Phase Motor
Larger installations often ignore the economies of internal infrastructure
obtained by fitting individual capacitors to each device and instead install
automatically controlled power factor correction equipment at the
switchboard of an installation.
This system of control involves having several banks of capacitors which
can be switched in and out of circuit as required. The switching is
controlled by an automatic power factor monitoring system. As loads
around the installation are switched on or off, the banks of capacitors
can also be switched on or off to compensate for the change in power
factor. Such a system can be more economical in the cost of capacitors.
It also more effectively compensates for the variable nature of motor
power factor. Motors have an extremely low power factor (typically 0·15 0·2) when lightly loaded and have a reasonably high power factor when
running at full load (typically 0·70 - 0·85). If an installation has a lot of
motors running at variable loads, an automatic power faction correction
system can be more effective than fixed capacitance at each motor
which is usually sized for the full load of the motor.
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In this situation it is important to install an automatic facility to discharge
the disconnected capacitors so that there is reduced risk of shock or fire
hazard from charged capacitors within the switchboard enclosure. This
is usually achieved by switching discharge resistors across the capacitor
terminals when they are disconnected from the supply.
Charged capacitors must never be short-circuited to discharge them.
Charged capacitors can store an enormous amount of energy and the
high current that results from this action can permanently damage the
capacitor and also cause a fire and safety hazard. A solenoid-type live
line tester such as a Duspol brand tester can be used to discharge a
capacitor as the charge will dissipate slowly through the solenoid coil of
the tester without causing damage to the capacitor.
Calculating Capacitor Requirements
Consider the power triangle for a motor.
P
θ
Q
S
To calculate the capacitive power required to correct the power factor,
we only need to consider P and Q, and angle θ.
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The relationship between these is that:
Q
P
∴ Q = P × Tanθ
Tanθ =
To improve the power factor, we need to reduce the length of Q by
adding capacitive power. This will reduce the size of angle θ, the phase
angle, and also reduce the length of S, the apparent power, thus
affecting a decrease in the current drawn from the supply.
P
Θ2
Q2
Θ1
S2
Q1
S1
Qc
To reduce the phase angle from θ1 to θ2, the reactive power must be
reduced from Q1 to Q2. This can be achieved by adding capacitive power
to the value of Qc. The task is therefore to calculate the value of Qc:
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Q1 = P × Tanθ1 and Q2 = P × Tanθ 2
∴ QC = Q1 − Q2 = (P × Tanθ1 ) − (P × Tanθ 2 )
∴ QC = P × (Tanθ1 − Tanθ 2 )
So we can use this formula to calculate the capacitive power required to
reduce the phase angle from θ1 to θ2.
Example:
Calculate the capacitive power required to increase the power factor of a
215 kW load from 0·65 lagging to 0·95 lagging.
Solution:
1. Calculate the value of the two phase angles:
λ1 = Cosθ1 = 0 ⋅ 65
λ2 = Cosθ 2 = 0 ⋅ 95
∴θ1 = Cos −1 0 ⋅ 65
∴ θ 2 = Cos −1 0 ⋅ 95
∴θ1 = 49 ⋅ 46°
∴ θ 2 = 18 ⋅ 20°
2. Calculate the size of the required capacitor:
QC = P(Tanθ1 − Tanθ 2 )
∴ QC = 215(Tan49 ⋅ 46° − Tan18 ⋅ 2°)
∴ QC = 215(1 ⋅169 − 0 ⋅ 329)
∴ QC = 215 × 0 ⋅ 84
∴ QC = 180 ⋅ 6kVAr
If you are confident in the use of your calculator you can achieve the
above result in one operation by using the formula:
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QC = P(Tan (Cos −1λ1 ) − Tan (Cos −1λ 2 ))
So the calculation is performed:
QC = P(Tan(Cos −1λ1 ) − Tan(Cos −1λ2 ))
∴ QC = 215(Tan(Cos −1 0 ⋅ 65) − Tan (Cos −1 0 ⋅ 95))
∴ QC = 180 ⋅ 695kVAr
This method is more accurate as it eliminates the “rounding” of results
throughout the calculation.
Except for small single phase applications such as light fittings, power
factor capacitors are sold commercially by their power rating. They are
available in a range of sizes from 0·25kVAr to 330 kVAr.
Larger Capacitors
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Calculating the value of capacitance in microfarads is generally not a
requirement but can easily be achieved. The following formula is used:
C=
Where
QC
2.π . f .V 2
C is the capacitance in Farads
QC is the reactive power of the capacitor in VArs
f is the supply frequency in Hertz
V is the voltage on the capacitor in Volts
To calculate the capacitance from the previous problem if the supply is
400 V, 50 Hz:
If the capacitors are connected in Delta:
C=
QC
2.π . f .V 2
∴C =
180 ⋅ 695
2 × π × 50 × 400 2
∴ C = 3 ⋅ 595µF
If the capacitors are connected in Star:
C=
QC
2.π . f .V 2
∴C =
180 ⋅ 695
2 × π × 50 × 230 2
∴ C = 10 ⋅ 873µF
Example Problems:
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(a)
A 315 kW motor is supplied from a 400V, 50 Hz, three-phase
supply. It has an efficiency of 90% and a full-load power
factor of 0·78.
Draw a power triangle for the motor.
Calculate the amount of capacitance required to increase the
power factor to 0·95, and the reduction in line current that
this will achieve.
Solution:
Step1
As motor power is always output power, we must first find
the input power.
Pin =
Pout
η
315,000
0⋅9
∴ Pin = 350,000W
∴ Pin =
We must also calculate the input current.
P = 3 × VL × I L × λ
∴IL =
P
3 × VL × λ
350,000
3 × 400 × 0 ⋅ 78
∴ I L = 647 ⋅ 67 Amps
∴IL =
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Step 2
Construct a power triangle.
P=350kW
‫ג‬1=0·78
‫ג‬2=0·95
Q2
S2
Q1
S1
Step 3
Qc
Calculate QC.
QC = P(Tan(Cos −1λ1 ) − Tan(Cos −1λ2 ))
∴ QC = 350(Tan (Cos −1 0 ⋅ 78) − Tan(Cos −1 0 ⋅ 95))
∴ QC = 165 ⋅ 759kVAr
Step 4
Calculate new input current.
P = 3 × VL × I L × λ
∴IL =
P
3 × VL × λ
350,000
3 × 400 × 0 ⋅ 95
∴ I L = 531⋅ 77 Amps
∴IL =
Step 4
Calculate the reduction in input current.
I = I1 − I 2
∴ I = 647 ⋅ 67 − 531⋅ 77
∴ I = 115 ⋅ 9 Amps
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(b)
A 400Volt 4-wire installation consists of the following balanced 3
phase loads:
• 120 kW of Resistive heating.
• 210 kVA of inductive lighting and transformers with an overall
0.85 lagging power factor.
• 280 kW of motors with an average efficiency of 87% and an
overall 0.78 lagging power factor.
Calculate:
1 The total (true) power used by the installation.
2 The total kVA.
3 The line current.
4 The overall power factor.
5 The amount of capacitance required to correct the power factor
to 0.95 lagging.
6 The new line current with the capacitors in circuit.
Solution.
Step 1:
Motors are rated in OUTPUT POWER so it is necessary to
work out the INPUT POWER.
INPUT POWER WILL BE GREATER THAN OUTPUT POWER
Powerout
efficiency
280 kW
=
0.87
= 321.84kW
Powerin =
Step 2:
Calculate true power (kW) of inductive load.
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True Power(kW) will be LESS THAN Apparent Power(kVA)
kW = kVA × pf
= 210 × 0.85
= 178.5kW
Step 3:
Draw a diagram of the power triangles.
Total kW
Heating
120kW
Overall
Phase Angle
Lighting
178.5kW
31.79
210kVA
Motors
321.84kW
38.74
Total kVA
Step 4:
Total kVAr
Calculate the total kW.
kWtotal = kW1 + kW2 + kW3
= 120 + 178.5 + 321.84
= 620.34kW
This is answer 1.
Step 5:
Calculate the reactive power (kVAr) for the reactive circuits.
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Lighting circuit:
θ = cos −1 0.85
= 31.79°
kVAr = kVA × sin θ
= 210 × sin 31.79°
= 210 × 0.53
= 111.3kVAr
2.
Motor circuit:
θ = cos −1 0.78
= 38.74°
kW
kVA =
cos θ
321.84
=
0.78
= 412.62kVA
kVAr = kVA 2 − kW 2
= 412.62 2 − 321.84
= 258.21kVAr
Step 6:
To calculate the total kVA you must first calculate the total
kVAr.
kVArtotal = kVAr1 + kVAr2 + kVAr3
= 0 + 111.3 + 258.21
= 369.51kVAr
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Now calculate the total kVA
kVA = kW 2 + kVAr 2
= 620.34 2 + 369.512
= 722.05kVA
This is answer 2.
Step 7:
Calculate the line current
kVA = 3 × V L × I L
kVA
IL =
=
3 × Vl
722.05 × 10 3
3 × 400
= 1042.19 A
This is answer 3.
Step 8:
Calculate the overall power factor
kW
kVA
620.34
=
722.05
= 0.86
pf =
This is answer 4.
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Step 9:
Calculate the kVAr to be added to raise the pf to 0.95
kVAr = kW( tan θ1 − tan θ 2 )
= 620.34( tan ( cos −1 0.86 ) − tan ( cos −1 0.95 ))
= 164.19kVAr
This is answer 5
Step 10:
Calculate the new line current
kW = 3 × V L × I L × cos θ
IL =
=
kW
3 × Vl × cos θ
620.34
3 × 400 × 0.95
= 942.51A
This is answer 6.
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Student Exercise 4
(a)
A 255 kW motor is supplied from a 400V, 50 Hz, three-phase
supply. It has an efficiency of 92% and a full-load power
factor of 0·72.
Draw a power triangle for the motor.
Calculate the amount of capacitance required to increase the
power factor to 0·93, and the reduction in line current that
this will achieve.
(b)
A 400Volt 4-wire installation consists of the following balanced 3
phase loads:
• 120 kW of Resistive heating.
• 210 kVA of inductive lighting and transformers with an overall
0.85 lagging power factor.
• 280 kW of motors with an average efficiency of 87% and an
overall 0.78 lagging power factor.
Draw a power triangle showing all loads, and the complete
installation.
Calculate:
1 The total (true) power used by the installation.
2 The total kVA.
3 The line current.
4 The overall power factor.
5 The amount of capacitance required to correct the
power factor to 0.95 lagging.
6 The new line current with the capacitors in circuit.
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