How does the power factor compensation work
First of all, detached from three-phase motors, we want to consider compensation for ohmic-inductive loads in the alternating current network.
The purpose of compensation is to reduce the inductive reactive power consumed by the network, which is necessary to e.g. B. to build up a magnetic field of a motor or in the case of a transformer to limit the fact that the energy is temporarily stored in a capacitor after the field has been removed.
The reactive currents then no longer burden the previous network with its lines, switching devices and transformers.
This has a number of advantages:
- Reduction in power dissipation. The higher the current-carrying capacity of a line, the more power loss is lost. Since power loss is proportional to the square of the current, you not only save energy, but you may also benefit from further savings. 10% less electricity means 19% less power.
- Reduction of the cross-section or, alternatively, higher loads can be connected if the network is available.
- Transformers deliver apparent power. If less reactive power is transformed, higher real powers can be connected. If these resources are not needed, the savings are made by choosing a smaller transformer.
- If the power loss behind your own meter is reduced, the costs for real work decrease.
- Above a certain amount, reactive power must also be paid for. Compensation may even save the entire cost of reactive power.
We consider a circuit diagram and the corresponding vector diagrams:
Using an example we want to calculate the capacitance C of the compensation capacitor. Our ohmic-inductive load R and L (e.g. a motor) has the following data:
U = 230 V
f = 50 Hz
I = 3 A (uncompensated)
cos ϕ = 0.9
We first calculate the apparent power of our small AC motor:
From this we calculate the reactive power
From the formulas for reactive power of the capacitor and reactance, we create a new formula to calculate the capacitance:
We add the two formulas together:
At this point we want to completely compensate for the reactive power. In practice, an inductive remainder is kept to avoid overcompensation. We therefore use the reactive power of the motor for the compensation reactive power and obtain the following capacitance for the capacitor:
Pointer image after compensation:
Let us now briefly see what relief this action has brought us in terms of electricity. If you look at our parallel connection from the network, it no longer has any reactive power, so cos ϕ = 1, the real power of the motor is the apparent power of the entire circuit and we get:
We have compensated 10% less current load on the line.
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