Theoretical principles to improve power factor: Difference between revisions
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An inductive load having a low power factor requires the generators and transmission/distribution systems to pass reactive current (lagging the system voltage by 90 degrees) with associated power losses and exaggerated voltage drops. If a bank of shunt capacitors is added to the load, its (capacitive) reactive current will take the same path through the power system as that of the load reactive current. Since, as pointed out in | An inductive load having a low power factor requires the generators and transmission/distribution systems to pass reactive current (lagging the system voltage by 90 degrees) with associated power losses and exaggerated voltage drops, as noted in sub-clause 1.3. If a bank of shunt capacitors is added to the load, its (capacitive) reactive current will take the same path through the power system as that of the load reactive current. Since, as pointed out in sub-clause 1.3, this capacitive current Ic (which leads the system voltage by 90 degrees) is in direct phase opposition to the load reactive current (IL). The two components flowing through the same path will cancel each other, such that if the capacitor bank is sufficiently large and Ic = IL, there will be no reactive current flow in the system upstream of the capacitors.<br>This is indicated in '''Figure L8 (a)''' and '''(b)''' which show the flow of the reactive components of current only.<br>In this figure:<br>R represents the active-power elements of the load<br>L represents the (inductive) reactive-power elements of the load<br>C represents the (capacitive) reactive-power elements of the power-factor correction equipment (i.e. capacitors). | ||
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It will be seen from diagram '''(b)''' of '''Figure L9''', that the capacitor bank C appears to be supplying all the reactive current of the load. For this reason, capacitors are sometimes referred to as “generators of | It will be seen from diagram '''(b)''' of '''Figure L9''', that the capacitor bank C appears to be supplying all the reactive current of the load. For this reason, capacitors are sometimes referred to as “generators of leading vars”. | ||
In diagram '''(c)''' of '''Figure L9''', the active-power current component has been added, and shows that the (fully-compensated) load appears to the power system as having a power factor of 1.<br>In general, it is not economical to fully compensate an installation. | |||
'''Figure L9''' uses the power diagram ('''see Fig. L2''') to illustrate the principle of compensation by reducing a large reactive power Q to a smaller value Q’ by means of a bank of capacitors having a reactive power Qc. <br>In doing so, the magnitude of the apparent power S is seen reduced to S’. | |||
Qc can be calculated by the following formula deduced from figure L9: | |||
Qc=P.(tan(ϕ)-tan(ϕ')) | |||
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'''Example:<br>'''A motor consumes 100 kW at a power factor of 0.75 (i.e. tan ϕ = 0.88). To improve the power factor to 0.93 (i.e. tan ϕ = 0.4), the reactive power of the capacitor bank must be : Qc = 100 (0.88 - 0.4) = 48 kvar<br>The selected level of compensation and the calculation of rating for the capacitor bank depend on the particular installation. The factors requiring attention are explained in a general way in clause 5, and in clauses 6 and 7 for transformers and motors. | '''Example:<br>'''A motor consumes 100 kW at a power factor of 0.75 (i.e. tan ϕ = 0.88). To improve the power factor to 0.93 (i.e. tan ϕ = 0.4), the reactive power of the capacitor bank must be : Qc = 100 (0.88 - 0.4) = 48 kvar<br>The selected level of compensation and the calculation of rating for the capacitor bank depend on the particular installation. The factors requiring attention are explained in a general way in clause 5, and in clauses 6 and 7 for transformers and motors. |
Revision as of 16:11, 23 August 2013
Improving the power factor of an installation requires a bank of capacitors which acts as a source of reactive energy. This arrangement is said to provide reactive energy compensation |
An inductive load having a low power factor requires the generators and transmission/distribution systems to pass reactive current (lagging the system voltage by 90 degrees) with associated power losses and exaggerated voltage drops, as noted in sub-clause 1.3. If a bank of shunt capacitors is added to the load, its (capacitive) reactive current will take the same path through the power system as that of the load reactive current. Since, as pointed out in sub-clause 1.3, this capacitive current Ic (which leads the system voltage by 90 degrees) is in direct phase opposition to the load reactive current (IL). The two components flowing through the same path will cancel each other, such that if the capacitor bank is sufficiently large and Ic = IL, there will be no reactive current flow in the system upstream of the capacitors.
This is indicated in Figure L8 (a) and (b) which show the flow of the reactive components of current only.
In this figure:
R represents the active-power elements of the load
L represents the (inductive) reactive-power elements of the load
C represents the (capacitive) reactive-power elements of the power-factor correction equipment (i.e. capacitors).
a) Reactive current components only flow pattern
b) When IC = IL, all reactive power is supplied from the capacitor bank
c) With load current added to case (b)
Fig. L8: Showing the essential features of power-factor correction
It will be seen from diagram (b) of Figure L9, that the capacitor bank C appears to be supplying all the reactive current of the load. For this reason, capacitors are sometimes referred to as “generators of leading vars”.
In diagram (c) of Figure L9, the active-power current component has been added, and shows that the (fully-compensated) load appears to the power system as having a power factor of 1.
In general, it is not economical to fully compensate an installation.
Figure L9 uses the power diagram (see Fig. L2) to illustrate the principle of compensation by reducing a large reactive power Q to a smaller value Q’ by means of a bank of capacitors having a reactive power Qc.
In doing so, the magnitude of the apparent power S is seen reduced to S’.
Qc can be calculated by the following formula deduced from figure L9:
Qc=P.(tan(ϕ)-tan(ϕ'))
Fig. L9: Diagram showing the principle of compensation: Qc = P (tan ϕ - tan ϕ’)
Example:
A motor consumes 100 kW at a power factor of 0.75 (i.e. tan ϕ = 0.88). To improve the power factor to 0.93 (i.e. tan ϕ = 0.4), the reactive power of the capacitor bank must be : Qc = 100 (0.88 - 0.4) = 48 kvar
The selected level of compensation and the calculation of rating for the capacitor bank depend on the particular installation. The factors requiring attention are explained in a general way in clause 5, and in clauses 6 and 7 for transformers and motors.
Note: Before starting a compensation project, a number of precautions should be observed. In particular, oversizing of motors should be avoided, as well as the no-load running of motors. In this latter condition, the reactive energy consumed by a motor results in a very low power factor (≈ 0.17); this is because the kW taken by the motor (when it is unloaded) are very small.
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