Power Factor Correction is a technique that uses capacitors to reduce the reactive power component of an AC circuit to increase efficiency and reduce current.
When dealing with direct current (DC) circuits, the power expended by the connected load is calculated as the multiplication of the DC voltage of the DC current, that is, the V*I is given in watts (W). For a constant resistant load, the current is proportional to the applied voltage, so the electrical power distributed by the resistant load will be linear. However, in the alternating current (AC) circuit, the situation is somewhat different as the reassurance affects the behavior of the circuit.
For an AC circuit, the power spent in watts at any time of time is equal to the multiplication of volts and amps at exactly the same time, this is due to the fact that an AC voltage (and current) is sinusoidal, so both the size and the constant change.
The average power in a DC circuit is simply V*I, but the average power of an AC circuit is that where there is no current, there are coils, windings, transformers, etc. A degree of phase with voltage, resulting in less real power, voltage and current spent in watts than the product. Because in circuits containing both resistance and reacctance, the phase angle (Ε) between them must also be taken into account.
In the tutorial on sinusoidal waveforms, we found that the phase angle (∠Ε) is the angle in electrical degrees, where the current lags behind the voltage. For a fully resistant load, the voltage and current are "intra-phase" since there is no reassurance.
However, for an AC circuit containing an inductor, coil or solenoid or other inductive load, its inductive reactance (XL) creates a phase angle in which the current lags 90o behind the voltage. Therefore, both have resistance (R) and inductive reactance (XL) given in Ohm, and the combined effect is called Impedance. Therefore, the impedance represented by the capital letter Z is the result value given in Ohm due to the combined effect of a circuit resistance and reactance.
Consider the RL series circuit below.
RL Series Circuit
Since it is a serial circuit, the current must be common to both resistance and inductor, so that the voltage drops along the resistance, vr is "in the same phase" as the serial current, while the voltage drop in the inductor "leads" to the VL " current with 90o (ELI). As a result, since both vectors are in the same phase, the voltage falling along the resistance is placed on the current vector, while the voltage developed along the inductor coil is drawn in a vertical direction due to the voltage directing the current to 90o.
Thus, the vector diagram drawn for each component will have the current vector as a reference, with two voltage vectors drawn according to their position as shown.
R and L Vector Diagrams
The resistance voltage is drawn along the VR, horizontal or "real axis", and the inductor voltage is drawn on the VL, vertical or "imaginary axis". To find the resulting VS voltage developed during the series connected circuit, we must combine two separate vectors using the current as a reference. The resulting vector voltage can be easily found using pythagorean theorem, as the combination of VR and VL creates a right-angle triangle, as shown below.
Phaser Diagram for Serial RL Circuit
The vector sum of VR and VL not only gives us the amplitude of VS due to the Pythagorean equation: we can use not only the
V2 S =
V2 R + V2 L, but also the
resulting phase angle (∠Ε) between VS and i, so we can use any of the standard Trigonometry functions of sinus, cosine and tangent to find it.
Power Factor Correction Example
An RL series circuit consists of an inductor with a resistance of 15Ω and an inductive recess of 26Ω. If a 5-amp current passes around the circuit, calculate:
1) Supply voltage.
2) Phase angle between feed voltage and circuit current.
3) Draw the resulting phaser diagram.
1). supply voltage VS
Using the impedances of the circuit, we can double-check this 150Vrms response as follows:
2). Phase angle using triganometry functions Ε:
3). Resulting phaser diagram showing VS
The calculated voltage that fell along the resistance (real component) was 75 volts, while the voltage produced along the inductor (imaginary component) was 130 volts. Clearly, the sum of 75 volts plus 130 volts is equal to 205 volts, which is much larger than the calculated 150 volts. This is because the value of 150V represents the sum of the phaser. Knowing individual voltage drops and impedances, we can convert these values into values that represent the power consumed in the circuit, real or imaginary.
Power in Serial RL Circuits
In a current-containing circuit, depending on whether the reactance is capacitive or inductive, I direct or delay the i voltage somewhat. The power consumed by resistance in watts is called "real power", so the symbol "P" (or W) is given. Watts can also be calculated as I2R, where R is the total resistance of the circuit. However, in order to calculate the value of real power in rms voltage and rms current (Vrms*Irms), we must also multiply these values by the cosine of the phase angle, cosΕ:
As we can see above, since the voltage and current for a resistance are "in-phase", the phase angle is therefore zero (0), so it gives us the value cos(Ε) = 1. give us the same true power value as using I2R. Then, using our example of the coil above, the power distributed by 15Ω resistance:
PR = I2R = 52x 15 = 375 watts
which is the same as saying:
PR = VR*I cos(Ε) = 75 x 5 x cos(Ε) = 375 watts
When voltage and current are "out of phase" with each other because the circuit contains reactenses, the V I's product is called "visible power" when the volt-amper (VA) unit is given instead of wattage. The volt-amperage has the symbol "S". The current for a fully inductive circuit is 90o behind the voltage, so the actual power for the inductive load is given as follows: V I cos(+90o) and this is: VI0. Clearly then there is no power consumed by an inductant, so there is no loss of power, so PL = 0 watts. However, to indicate that this wattage power is present in an AC circuit, the volt-amperage is called reagent (VAR) and is indicated by the symbol "Q". Therefore, for an inductive circuit, the volt-amperage uses the symbol of reactive or simply "reactive power" QL.
Similarly, the current for a fully capacitive circuit directs the voltage to 90o, the actual power for a capacitive load is given asfollows: V I cos(-90o) and again: VI*0. Clearly then and as before, there is no power consumed by a capacitance, so there is no loss of power in PC = 0 watts. Therefore, to indicate that this watt-free power is present in a capacitive circuit, the volt-amperage is called reactive capacitive and the QC symbol is given. Note that here the reactive power of a capacitance is defined as negative, resulting in -QC.
Again, using our example above, the reactive power in and out of the inductor at a rate determined by the frequency is given as follows:
QL = I2XL = 52x 26 = 650 VAR
Since there is a 90o phase difference between voltage and current waveforms in a pure recess (inductive or capacitive), we multiply the V*I by sin(Ε) to give the vertical component that is 90o out of phase. But the sine of the angle (sin 90o) gives the result as "1", that is, we can find reactive power by simply multiplying the rms voltage and current values as in the figure.
QL = I2XL = VIsin(Ε) = 1305sin(90o) = 13051 = 650 VAR
Then we can see that the volt-amperage reagent or VAR part has a magnitude (the same as the actual power) but there is no phase angle associated with it. So reactive power is always on the 90o vertical axis. So if we know this:
PR = I2R = 375 Watts
QL = I2XL = 650 VAR (ind.)
as shown, we can create a power triangle to show the relationship between P, Q and S.
Inductive Power Triangle
Capacitive Power Triangle
Again, we can use the previous Pythagorean Theorem and trigonometry functions of Sinus, Cosine and Tangent to describe a power triangle.
Force Triangle Equations
Power Factor Correction
Power Factor Correction improves the phase angle between the supply voltage and the current while the actual power consumption in watts remains the same, since as we have seen, a pure reassurance does not consume any real power. Adding an impedance in the form of capacitive reacance in parallel with the above coil will reduce Ε, thereby increasing the power factor, which will reduce the rms current of the circuits drawn from the feed.
The power factor of an AC circuit can vary from 0 to 1 depending on the strength of the inductive load, but in reality it can never be less than about 0.2 for the heaviest inductive loads. As we see above, a power factor smaller than 1 means that there is increased (fully inductive) consumption of reactive power as we get closer to 0. Clearly, a power factor that is exactly "1" means that the circuit consumes zero reactive power (fully resistant), resulting in a power factor angle of 0o. It's called the unity power factor.
Adding a parallel capacitor to the coil will not only reduce this unwanted reactive power, but also reduce the total amount of current from the source. In theory, capacitors can provide 100% of the compensating reactive power required in one circuit, but in practice a power factor correction of between 95% and 98% (0.95 to 0.98% is usually sufficient). Therefore, using our coil in example 2 above, what capacitor value is required to increase the power factor from 0.5 to 0.95.
A power factor of 0.95 is equal to the following phase angle: cos(0.95) = 18.2o, hence the required amount of VAR:
That's why we need a reactive power value of 82.2VAR for a phase angle of 18.2o. If the original uncorrected VAR value is 433VAR and the newly calculated value is 82.2VAR, we need a decrease in var (capacitive) value of 433 – 82.2 = 350.8 VAR. Then:
The capacitor required to reduce reactive power to 82.2VAR must have a capacitive reassurance of 28.5Ω at the nominal feed frequency. Therefore, the capacitor's capacitance is calculated as follows:
New Volt-Amperage Value
We can also create a power triangle to show the previous and next values for VA (S) and VAR (Q), as shown.
If the apparent power of the circuit has been reduced from 500VA to only 263VA, we can calculate the rms current provided as follows:
S = V*I, therefore: I = S/V = 263/100 = 2.63 Amps
So not only does it increase the total power factor from 0.5 to 0.95 by connecting a capacitor to the coil, but it also reduces the feed current from 5 amps to 2.63 amps, a reduction of about 47%. That's what the last circuit will look like.
Ultimate Power Factor Correction Circuit
If you wish, you can increase the capacitor value from the calculated value above 93uF for our simple example to a maximum of 114.8uF by further improving the power factor from the required 0.95 to 1.0 (unit). In reality, a single standard 100uF non-polarized capacitor will suffice for this sample.
In this tutorial, we found that a delayed power factor due to inductive load increases power losses in an AC circuit. We can reduce the phase difference between voltage and current by adding an appropriate capacitive reactive component in the form of a capacitor in parallel with an inductive load.
This has the effect of reducing the power factor of the circuit, that is, the ratio of active power to visible power, as well as improving the power quality of the circuit and reducing the required amount of welding current. This technique is called "Power Factor Correction".