Power Factor Improvement

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Salman bin AbdulazizUniversity College of Engineering Electrical Engineering Department EE 2050Electrical Circuit Laboratory Power Factor Improvement Experiment # 4 Objectives: 1.

1 Salman bin AbdulazizUniversity College of Engineering Electrical Engineering Department EE 2050Electrical Circuit Laboratory Power Factor Improvement Experiment # 4 Objectives: 1. To introduce the concept of power factor and method of power factor improvement for a single phase inductive loads. Equipments: 1. One Fluorescent lamp. (110 V, 20 W) 2. Two multi-meters. 3. One Wattmeter. 4. One decade capacitor box. Note: Background: Please record any damage, if exists any while performing the experiment. You can also write the difficulties you are confronted with when using the equipment, and your suggestions and critics related with the equipment you used. 1. Power in resistive and reactive AC circuits Consider a circuit for a single-phase AC system shown in Fig. 1, where a 120 volt, 60 Hz AC voltage source is delivering power to a resistive load. Fig. 1. Ac source drives a purely resistive load. Experiment # 4 Page 1

2 In this example, the RMS current to the load would be 2 A. Therefore power dissipated at the load (active power) would be (2) 2 60 = 240 W. Because this load is purely resistive, the current is in phase with the voltage. Fig. 2, shows the voltage, current and power waveforms. Fig. 2. Current is in phase with voltage in a resistive circuit. Note that the power waveform is always positive, this means that power is always being dissipated by the resistive load, and never returned to the source. For comparison, let s consider a simple AC circuit with a purely reactive load as shown in Fig. 3. Fig. 3. AC circuit with a purely reactive (inductive) load. Experiment # 4 Page 2

3 Fig. 4 Current lags voltage with 90º in a pure inductive circuit. As seen from Fig. 4, power (p) alternates equally between cycles of positive and negative. This means that power is being alternately absorbed from and returned to the source. Now, let s consider an AC circuit with a load consisting of both inductance and resistance as shown in Fig. 5. Fig. 5. AC circuit with both reactance and resistance. Experiment # 4 Page 3

4 Fig. 6. Current, voltage and power waveforms As with any reactive circuit, the power alternates instantaneously between positive and negative values over time. In a purely reactive circuit, alternation between positive and negative power is equally divided, resulting in a net power dissipation of zero. However, in circuits with mixed resistance and reactance like this one, the power waveform will still alternate between positive and negative, but the amount of positive power will exceed the amount of negative power. In other words, the combined inductive/resistive load will consume more power than it returns back to the source. 2. True, Reactive, and Apparent power We know that reactive loads such as inductors and capacitors dissipate zero power, yet the fact that they drop voltage and draw current gives the deceptive impression that they actually do dissipate power. This ghost power is called reactive power, and it is measured in a unit called Volt-Amps-Reactive (VAR), rather than watts. The mathematical symbol for reactive power is the capital letter Q. The actual amount of power being used, or dissipated, in a circuit is called true (active) power, and it is measured in watts (symbolized by the capital letter P). The combination of reactive power and true power is called apparent power, and it is the product of a circuit s voltage and current, without reference to phase angle. Apparent power is measured in the unit of Volt-Amps (VA) and is symbolized by the capital letter S. There are several power equations relating the three types of power to resistance, reactance, and impedance: Experiment # 4 Page 4

5 The three types of power can be represented by a triangle shown in Fig. 7. Fig. 7. Power triangle relating apparent power to true power and reactive power. From Fig. 7, the angle of this power triangle graphically indicates the ratio between the amount of dissipated (or consumed) power and the amount of absorbed/returned power. It also happens to be the same angle as that of the circuit s impedance in polar form. When expressed as a fraction, this ratio between true power and apparent power is called the power factor for this circuit. It should be noted that power factor, like all ratio measurements, is a unitless quantity. For the purely resistive circuit, the power factor is unity (perfect), because the reactive power equals zero. Here, the power triangle would look like a horizontal line, because the opposite (reactive power) side would have zero length. For the purely inductive circuit, the power factor is zero, because true power equals zero. Here, the power triangle would look like a vertical line, because the adjacent (true power) side would have zero length. 3. Improving the System Power Factor Consider an RL circuit whose circuit impedance can be expressed as ZL = R+j XL = ZL θl. Suppose the expression for the supply voltage is v = 2 V sin ω t, the expression of current will be 2 I sin (ω t θ L ), where I = V / Z L, is the rms value of current and θ L is its phase angle. The instantaneous power P = vi = 2 VI sin (ωt) sin (ω t θ L ) can be expressed as: Experiment # 4 Page 5

6 From the above equation, it is clear that the first term is constant while the second term is a cosine wave which has a frequency of 2ω and a peak value of VI. The average value of this term is zero. Thus the average value of power delivered to load, P av, is The term cos θ L in the above equation is called power factor (PF) of load. If the load PF is less than unity it may be either capacitive (i.e. leading) PF or inductive (i.e. lagging) PF. However, in practice almost all the loads are inductive with lagging PF. When the PF of a load is less than unity, it takes more current for the same power from the supply. This large current requirement will cause more I 2 R losses and IR drop in the supply line. Moreover, large current will demand greater current capacity for the generator and associated systems. Therefore, it is desired to correct the PF to unity. Lagging PF can be improved by connecting capacitors across the inductive load as shown in Fig.8. The shunt capacitor will not absorb real power but instead it will take leading current from the line to neutralize the lagging current component of the load, thereby making the net supply current less and hence improving the system PF. Fig. 8. Inductive load with a compensating capacitor For the circuit of Fig.8, if the switch S is off, load current I L is given as Where Z L 2 = R 2 + X L 2. The first component of the current I L given above is called active component while the second component is called reactive component of current. This reactive component which is lagging the supply voltage by 90 can be neutralized if a capacitor is connected across the load, in case of the switch S is turned on, which draws a reactive current of similar magnitude but leading the supply voltage by 90 (i.e. I C = j V X L / Z L 2 ). Thus, with the capacitor connected parallel to the load, the supply current I S is given as Since I C = j V /X C = j V (2 π f C). Thus, to get a PF of unity, we must put C such that Experiment # 4 Page 6

7 If C is less or more than that given by the above equation, the resulting power factor will be less than unity and will be lagging or leading respectively. Figure.9 (a) depicts the phasor diagram of voltage and current whereas Fig.9 (b) shows the power diagram. Both of these diagrams show how the leading current or leading VA of the capacitor neutralizes the lagging reactive current or lagging reactive volt ampere of the load, thereby, improving the PF. Fig. 9. Improving PF using capacitor In both of these diagrams we see that initial or load phase angle, θ L (before compensation) is more than the final or supply phase angle, θ S (after compensation), i.e. Procedure 1. Connect the circuit as shown in Fig.10. Fig. 10. Circuit used in procedure Experiment # 4 Page 7

8 2. When the switch S being OFF, record the reading of voltmeter (V), ammeter (I) and wattmeter before connecting the capacitor C in Table For the fluorescent lamp, without capacitor, estimate the load resistance R, inductive reactance X L, impedance Z L, phase angle θ L and PF. Record the value of PF in table 1. Calculate the value of C required to make PF = 1. C = (µf) Table 1. Reading of voltmeter, ammeter and wattmeter at different values of C. Capacitor C (µf) Not connected Voltage (V) Current (A) Power (W) Power Factor PF 1.0 (µf) 2.0 (µf) 3.0 (µf) 3.2 (µf) 3.4 (µf) 3.6 (µf) 3.8 (µf) 4.0 (µf) 4.2 (µf) 4.4 (µf) 4.6 (µf) 4.8 (µf) 5.0 (µf) 5.2 (µf) 5.4 (µf) 5.6 (µf) 5.8 (µf) 6.0 (µf) Experiment # 4 Page 8

9 4. Switch ON the switch S to connect the capacitor C parallel to the load with a capacitance value of 1.0 µf, and record the reading of voltmeter, ammeter and wattmeter again. Then calculate the load PF then record in Table Increase the value of the capacitor in steps given in Table 1, and record the required readings until you get a point where the current is minimum. 6. Plot the relationship between the load power factor PF and the capacitor value. From this curve, Find the value of C at which the PF is maximum. 7. Compare this value with that calculated in step 3. Load PF Fig. 11. Power factor vs capacitor values C (µf) Experiment # 4 Page 9

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