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10.7 Power Factor Correction

Question What loads do we normally find in industrial facilities?

Industrial facilities use to employ inductive receivers such as motors, transformers, gas discharge lamps, etc.

Question How do inductive receivers generate the magnetic fields?

These inductive receivers work with considerable reactive power to generate the magnetic fields they need to operate.

Question What effects does the reactive power consumption have?

The consumption of reactive power causes that the power supply company has to provide extra energy that does not make effective work. Because the reactive power is fed back into the grid every ¼ cycle, the electric current in the grid rises. Having a low power factor has the following drawbacks:

For the same useful power Pu, the current increases as the power factor decreases. As a result, the Joule effect in the wires increases, and then the losses increase as well. Therefore, the efficiency of machines and distribution systems decreases. Recalling that:

$P = U \cdot I \cdot cos φ$ (10.20)

$P = R \cdot I^{2}$ (10.21)

A low power factor results in greater voltage drops. This requires more frequent regulation of generators, transformers and lines than would be necessary with a higher power factor. There will be a waste of the generating plant's available energy.

Question Then, what do the companies require from with respect to the power factor?

Because of these disadvantages, electricity supply companies require that their subscribers consume energy with a power factor ≥ 0.55. Distribution companies apply an extra fee for the reactive energy that inductive loads consume in the billing of electrical energy to ensure that users avoid a low power factor.

This extra charge consists of a surcharge of €/kVAr that they add to the billing, whose value depends on the average power factor of the facility. Both the supplying company and the subscribers must work with a power factor as high as possible. However, in some cases it is not always practical to correct the power factor for high values (for example cos φ > 0.95) because the costs are lower than the benefits.

练习题

Which of the following are examples of inductive receivers commonly found in industrial facilities?

A. Resistive heaters and incandescent lamps

B. Motors, transformers, and gas discharge lamps

C. Capacitors and batteries

D. Solar panels and fuel cells

According to the formula $P = U \cdot I \cdot cos φ$ , what happens to the current $I$ if the useful power $P$ remains constant while the power factor $cos φ$ decreases?

A. The current decreases proportionally

B. The current remains unchanged

C. The current increases

D. The current becomes zero

What is the minimum power factor that electricity supply companies typically require from their subscribers?

cos φ \geq 0.95

cos φ \geq 0.55

cos φ \geq 0.80

cos φ \geq 1.00

Which of the following are disadvantages caused by a low power factor in an electrical system?

A. Increased current for the same useful power

B. Increased Joule effect losses in conductors

C. Greater voltage drops in the system

D. Waste of generating plant's available energy

E. Reduced efficiency of machines and distribution systems

How do distribution companies penalize low power factor in industrial facilities?

A. They disconnect the facility from the grid

B. They apply an extra fee for reactive energy consumed by inductive loads

C. The surcharge is calculated in €/kVAr

D. The surcharge value depends on the average power factor of the facility

Reactive power consumed by inductive receivers performs effective work in the electrical system.

According to the formula $P = R \cdot I^{2}$ , the Joule losses in conductors are directly proportional to the square of the current.

It is always practical and economical to correct the power factor to values above $cos φ > 0.95$ .

Inductive receivers such as motors and transformers require reactive power to generate the magnetic fields needed for their operation.

The active power in an AC circuit is calculated using the formula: $P = U \cdot I \cdot$ ___

A low power factor causes greater voltage drops, which requires more frequent ___ of generators, transformers, and lines.

Explain why reactive power consumption increases the electric current in the grid even though it does not perform effective work.

Why must synchronous generators be regulated more frequently when supplying loads with low power factor?

A synchronous generator supplies power to an industrial facility with several induction motors operating at a power factor of 0.7. Compared to operating at unity power factor with the same useful power output, how does the low power factor affect the generator's terminal voltage and the system losses?

A. Terminal voltage drop decreases and system losses decrease

B. Terminal voltage drop increases and system losses increase due to higher current

C. Terminal voltage drop increases but system losses remain unchanged

D. Terminal voltage drop remains the same but system losses increase

Which of the following statements correctly describe the relationship between inductive loads in industrial facilities and the magnetic fields they require? Select all that apply.

A. Induction motors require reactive power to generate their magnetic fields

B. The magnetic field in transformers is produced without any reactive power consumption

C. Similar to synchronous generators that need excitation to produce magnetic flux, inductive loads need reactive power for their magnetic fields

D. The reactive power consumed by inductive loads does effective work in the load

E. Low power factor caused by inductive loads results in higher grid current

When a synchronous generator supplies a load with a low power factor, the voltage regulation coefficient ε will be larger compared to supplying the same useful power at a higher power factor, because the increased current causes greater armature reaction and voltage drop.

For a synchronous generator supplying a constant useful power P to an industrial load, if the power factor decreases from 0.9 to 0.6, the current in the armature windings will ___ according to the relationship $P = U \cdot I \cdot cos φ$ , resulting in increased Joule losses calculated as $P = R \cdot I^{2}$ .

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10.7 Power Factor Correction