Average power factor values for the most commonly-used equipment and appliances

Equipment and appliancescos φtan φ
  • Common induction motor
loaded at
0%
25%
50%
75%
100%
0.17
0.55
0.73
0.80
0.85
5.80
1.52
0.94
0.75
0.62
  • Incandescent lamps
  • Fluorescent lamps (uncompensated)
  • Fluorescent lamps (compensated)
  • Discharge lamps
  
1.0
0.5
0.93
0.4 to 0.6
0
1.73
0.39
2.29 to 1.33
  • Ovens using resistance elements
  • Induction heating ovens (compensated)
  • Dielectric type heating ovens
  
1.0
0.85
0.85
0
0.62
0.62
  • Resistance-type soldering machines
  • Fixed 1-phase arc-welding set
  • Arc-welding motor-generating set
  • Arc-welding transformer-rectifier set
  
0.8 to 0.9
0.5
0.7 to 0.9
0.7 to 0.8
0.75 to 0.48
1.73
1.02 to 0.48
1.02 to 0.75
  • Arc furnace
  0.80.75

Fig. L6: Values of cos φ and tan φ for commonly-used equipment

The nature of reactive power

All inductive (i.e. electromagnetic) machines and devices that operate on AC systems convert electrical energy from the power system generators into mechanical work and heat. This energy is measured by kWh meters, and is referred to as “active” energy.
In order to perform this conversion, magnetic fields have to be established in the machines. The magnetic field is created by the circulation of current in coils, which are mainly inductive. The current in these coils is therefore lagging by 90° relative to the voltage, and represent the reactive current absorbed by the machine.
It should be noted that while reactive current does not draw power from the system, it does cause power losses in transmission and distribution systems by heating the conductors.
In practical power systems, load currents are invariably inductive, and impedances of transmission and distribution systems predominantly inductive as well. The combination of inductive current passing through an inductive reactance produces the worst possible conditions of voltage drop (i.e. in direct phase opposition to the system voltage).
For these two reasons (transmission power losses and voltage drop), the Network Operators work for reducing the amount of reactive (inductive) current as much as possible.

FigL01.jpg

Fig. L4: An electric motor requires active power P and reactive power Q from the power system

Reactive power of capacitors
The current flowing through capacitors is leading the voltage by 90°. The corresponding current vector is then in opposition to the current vector of inductive loads. This why capacitors are commonly used in the electrical systems, in order to compensate the reactive power absorbed by inductive loads such as motors.
Inductive-reactive power is conventionally positive (absorbed by an inductive load), while capacitive-reactive power is negative (supplied by a capacitive load).
As reactive-inductive loads and line reactance are responsible for voltage drops, reactive-capacitive currents have the reverse effect on voltage levels and produce voltage-rises in power systems.

Equipment and appliances requiring reactive energy
All AC equipment and appliances that include electromagnetic devices, or depend on magnetically coupled windings, require some degree of reactive current to create magnetic flux.
The most common items in this class are transformers, reactors, motors and discharge lamps with magnetic ballasts (see Fig. L5).
The proportion of reactive power (kvar) with respect to active power (kW) when a piece of equipment is fully loaded varies according to the item concerned being:
  • 65-75% for asynchronous motors (corresponding to a Power Factor 0.8 – 0.85)
  • 5-10% for transformers (corresponding to a Power Factor close to 0.995)

FigL02.jpg

Fig. L5: Power consuming items that also require reactive energy

Definition of reactive power

For most electrical loads like motors, the current I is lagging behind the voltage V by an angle φ.
If currents and voltages are perfectly sinusoidal signals, a vector diagram can be used for representation.
In this vector diagram, the current vector can be split into two components: one in phase with the voltage vector (component Ia), one in quadrature (lagging by 90 degrees) with the voltage vector (component Ir). See Fig. L1.
Ia is called the active component of the current.
Ir is called the reactive component of the current.

Fig L01.jpg

Fig. L1 : Current vector diagram

The previous diagram drawn up for currents also applies to powers, by multiplying each current by the common voltage V. See Fig. L2.
We thus define:
  • Apparent power: S = V x I (kVA) 
  • Active power: P = V x Ia (kW)
  • Reactive power: Q = V x Ir (kvar)
Fig L02.jpg
Fig. L2 : Power vector diagram

In this diagram, we can see that:
  • Power Factor: P/S = cos φ

This formula is applicable for sinusoidal voltage and current. This is why the Power Factor is then designated as "Displacement Power Factor".
  • Q/S = sinφ
  • Q/P = tanφ
A simple formula is obtained, linking apparent, active and reactive power:
  • S² = P² + Q²

A power factor close to unity means that the apparent power S is minimal. This means that the electrical equipment rating is minimal for the transmission of a given active power P to the load. The reactive power is then small compared with the active power.
A low value of power factor indicates the opposite condition.

Useful formulae (for balanced and near-balanced loads on 4-wire systems):
  • Active power P (in kW)
  -  Single phase (1 phase and neutral): P = V.I.cos φ
  -  Single phase (phase to phase): P = U.I.cos φ
  -  Three phase (3 wires or 3 wires + neutral): P = √3.U.I.cos φ
  • Reactive power Q (in kvar)
  -  Single phase (1 phase and neutral): P = V.I.sin φ
  -  Single phase (phase to phase): Q = U.I.sin φ
  -  Three phase (3 wires or 3 wires + neutral): P = √3.U.I.sin φ
  • Apparent power S (in kVA)
  -  Single phase (1 phase and neutral): S = V.I
  -  Single phase (phase to phase): S = U.I
  -  Three phase (3 wires or 3 wires + neutral): P = √3.U.I
where:
V= Voltage between phase and neutral
U = Voltage between phases
I = Line current
φ = Phase angle between vectors V and I.

An example of power calculations (see Fig. L3)

Type of circuitApparent power S (kVA)Active power P (kW)Reactive power Q (kvar)
Single-phase (phase and neutral)  S = VIP = VI cos φQ = VI sin φ
Single-phase (phase to phase)  S = UIP = UI cos φQ = UI sin φ
Example        5 kW of load10 kVA5 kW8.7 kvar
cos φ = 0.5
Three phase 3-wires or 3-wires + neutralS = \definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}\sqrt 3  UIP = \definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}\sqrt 3  UI cos φQ = \definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}\sqrt 3  UI sin φ
ExampleMotor Pn = 51 kW65 kVA56 kW33 kvar
cos φ= 0.86
ρ= 0.91 (motor efficiency)
Fig. L3 : Example in the calculation of active and reactive power

The calculations for the three-phase example above are as follows:
Pn = delivered shaft power = 51 kW
P = active power consumed
P=\frac {Pn}{\rho}=\frac{51}{0.91}=56\, kW
S = apparent power
S=\frac{P}{cos \phi}=\frac {56}{0.86}= 65\, kVA
So that, on referring to diagram Figure L3 or using a pocket calculator, the value of tan φ corresponding to a cos φ of 0.86 is found to be 0.59
Q = P tan φ = 56 x 0.59 = 33 kvar (see Figure L15).
Alternatively:
Q=\sqrt{S^2 - P^2}=\sqrt{65^2 - 56^2}=33\, kvar

FigL05.jpg
Fig. L2b : Calculation power diagram

Definition of Power Factor

The Power Factor is an indicator of the quality of design and management of an electrical installation. It relies on two very basic notions: active and apparent power.
The active power P (kW) is the real power transmitted to loads such as motors, lamps, heaters, and computers. The electrical active power is transformed into mechanical power, heat or light.
In a circuit where the applied r.m.s. voltage is Vrms and the circulating r.m.s. current is Irms, the apparent power S (kVA) is the product: Vrms x Irms.
The apparent power is the basis for electrical equipment rating.
The Power Factor λ is the ratio of the active power P (kW) to the apparent power S (kVA):

 \lambda = \frac {P(kW)}{S(kVA)}
The load may be a single power-consuming item, or a number of items (for example an entire installation).
The value of power factor will range from 0 to 1

Parallel operation of transformers

  • 1Total power (kVA)
  • 2Conditions necessary for parallel operation
  • 3Common winding arrangements

The need for operation of two or more transformers in parallel often arises due to:
§  Load growth, which exceeds the capactiy of an existing transformer
§  Lack of space (height) for one large transformer
§  A measure of security (the probability of two transformers failing at the same time is very small)
§  The adoption of a standard size of transformer throughout an installation
Total power (kVA)
The total power (kVA) available when two or more transformers of the same kVA rating are connected in parallel, is equal to the sum of the individual ratings, providing that the percentage impedances are all equal and the voltage ratios are identical.
Transformers of unequal kVA ratings will share a load practically (but not exactly) in proportion to their ratings, providing that the voltage ratios are identical and the percentage impedances (at their own kVA rating) are identical, or very nearly so. In these cases, a total of more than 90% of the sum of the two ratings is normally available.
It is recommended that transformers, the kVA ratings of which differ by more than 2:1, should not be operated permanently in parallel.
Conditions necessary for parallel operation
All paralleled units must be supplied from the same network.
The inevitable circulating currents exchanged between the secondary circuits of paralleled transformers will be negligibly small providing that:
§  Secondary cabling from the transformers to the point of paralleling have approximately equal lengths and characteristics
§  The transformer manufacturer is fully informed of the duty intended for the transformers, so that:
  - The winding configurations (star, delta, zigzag star) of the several transformers have the same phase change between primary and
    secondary voltages
  -
 The short-circuit impedances are equal, or differ by less than 10%
 
 - Voltage differences between corresponding phases must not exceed 0.4%
 
 - All possible information on the conditions of use, expected load cycles, etc. should be given to the manufacturer with a view to
    optimizing load and no-load losses 
 
Common winding arrangements
As described in 4.4 "Electrical characteristics-winding configurations" the relationships between primary, secondary, and tertiary windings depend on: 
§  Type of windings (delta, star, zigzag)
§  Connection of the phase windings
Depending on which ends of the windings form the star point (for example), a star winding will produce voltages which are 180° displaced with respect to those produced if the opposite ends had been joined to form the star point. Similar 180° changes occur in the two possible ways of connecting phase-to-phase coils to form delta windings, while four different combinations of zigzag connections are possible.
§  The phase displacement of the secondary phase voltages with respect to the corresponding primary phase voltages.
As previously noted, this displacement (if not zero) will always be a multiple of 30° and will depend on the two factors mentioned above, viz type of windings and connection (i.e. polarity) of the phase windings.
By far the most common type of distribution transformer winding configuration is the Dyn 11 connection (see Fig. B21). 



Fig. B21: Phase change through a Dyn 11 transformer



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