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Contents Introduction 2 Low power factor 3 Effects of low power factor 3 Power factor measurement 4 Low power factor circuits 4 Power factor improvement 6 Power factor improvement capacitors 10 Power factor correction methods 14 Summary 16 Answers 19 EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 1 Introduction In this section, we will look at power factor, which is very important in electrical theory. In all domestic, industrial and commercial applications, power is the prime consideration of the installation. Power must be supplied as efficiently as possible, and this is where the power factor comes in. Power factor control is an effective way of conserving power, and it also determines the cable size of the conductors in both the final sub-circuit and the main supply line. We will consider power factor, looking at the relevant terms and the methods used to accomplish power factor control. At the end of this section you should be able to: 2 describe the effects of low power factor. describe the requirements for power factor improvement. list the methods used to improve low power factor of a installation. state local supply authority and AS/NZS 3000 requirements regarding the power factor of an installation and power factor improvement equipment. describe the methods used to measure single phase power factor. using manufacturers catalogues, select power factor equipment for a particular installation. EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 Low power factor Consider a parallel circuit that begins as a resistive circuit, but has more and more inductance added to it. The phase angle of this circuit will start at 0 degrees, but will increase as we add more inductance. Figure 1: The power triangle As a circuit becomes more reactive, the phase angle increases. Looking at the power triangle in figure 1, you can see the effect this has on the power relations. As φ grows, the magnitude of the reactive power Q grows relative to P. This means that a smaller and smaller component of the apparent power S is devoted to supplying the true power of the circuit. This situation is known as ‘low power factor’, because as φ grows from 0 toward 90° the power factor (cos φ) decreases from 1 towards 0. Thus if the voltage is constant, and the power factor decreases, the current must increase to maintain the same real power in the circuit. Effects of low power factor There are definite drawbacks to a low power factor. It results in the following disadvantages: There is an increased voltage drop in the supply conductors. This can only be compensated for by increasing the cable size, which will lead to higher costs for the larger cable and its accessories. Alternator efficiency is reduced. This is due to the internal power losses of the alternator I2R having a higher value. Real power EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 3 delivered to the load is also decreased in proportion to the power factor where apparent power is constant, since P = S cos Again this is due to the fact that the load current is high on low power factor loads. To overcome this drop in efficiency and decreased output of the alternator, a larger unit must be installed; again, the cost of the machine and its accessories increases. To handle the increased current, larger control devices are required. These control devices comprise the switch gear and transformers and their associated protection devices. Again, the cost of the installation escalates. There are larger power losses in the supply cables, due to the increased current. This is minimised by increasing the cable size (higher cost), but as P = I 2 Rcable, the cable size would have to vary at an exponential rate to keep the loss in the cable constant. Power factor measurement Consumers try to ensure that the installation is not creating excess cost through having low power factor. To this end they need to be aware of the actual power factor at all times. A simple solution is to compare energy meter readings against voltmeter and ammeter readings to determine the power factor. An installation may also justify the extra expense of installing specifically designed meters that measure actual power factor. Low power factor circuits The types of circuit which cause or produce a low circuit power factor are given below. 4 Large fluorescent lighting loads. Inside the lamp circuit of each fitting, a very inductive ballast is used to reduce the lamp current so as to avoid wasting true power by using resistors. This is one of the most common forms of artificial lighting. Lightly loaded motors. The motor inductive losses are relatively constant, but the output from the shaft, in mechanical watts, varies with the load. Normally, in the fully loaded motor, the mechanical power far exceeds the inductive losses within the motor, and the power factor is as high as 0.85, but on light loads this ratio decreases, and so the power factor falls. In some installations this condition occurs as a consequence of the duty cycle of the motor. Lightly loaded transformers. Since a transformer operates on the same basis as a motor, the considerations outlined above apply. EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 Capacitance in long transmission lines. Remember how a capacitor is constructed: two conductors separated by an insulator. This is what a transmission line is. There are three separate phases (conductors) separated by air (an insulator) from each other. Each phase is also separated from the ground (also a conductor) by the same insulator. Thus a three-phase system can be thought of as having six capacitors, one between each phase and the ground and one between each of the phase combinations. This is illustrated in Figure 2. Figure 2: Transmission line capacitance EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 5 Power factor improvement The local supply authority often requires consumers to maintain the power factor of loads at a nominal value. In any case, the power factor of the load should generally be around 0.8 (unless there is a particular reason for it), as this is the most cost effective value. A lower power factor will lead to higher losses, which will reduce the efficiency of the circuit. But if the power factor was forced higher, the rise in efficiency may not outweigh the increased cost of the installation. AS/NZS 3000:2000 does not quote a minimum value for power factor but does however give guidelines for the installation of power factor correction equipment and this is found in clause 4.6 ‘Capacitors’. To find what the minimum value of power factor required for your area is, consult your local Supply Authority. By way of example, the NSW Service and Installation Rules require a power factor of 0.9 lagging. To overcome the disadvantages of a lower power factor being generated, a technique is used whereby the reactive components of the circuit cancel each other out. You will remember that a practical circuit generally contains either a resistive or a resistive/inductive load. As power factor improvement is not required for resistive circuits, the examples dealing with power factor improvement refer to resistive/inductive circuits only. This is illustrated in Figure 3. Figure 3: Single phase circuit with PF improvement 6 EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 So, as we are looking at a circuit containing inductance, the power factor of the circuit will be lagging. The power triangle for the circuit in Figure 3 is given in Figure 4, showing apparent power, true power, reactive power and power factor angle. Figure 4: Power triangle for PF improvement To improve the power factor of the circuit, the circuit phase angle must be altered. It can be seen from the power triangle that a change in either the true power or the reactive power would have this effect. The true power cannot be altered, however; it is the normal output or consumption of the load. This means that the only way is to alter the reactive power of the load. As the two types of reactive power, inductive and capacitive, are 180° out of phase, they may be directly added to each other. This simplifies the calculation considerably. Remember that the total reactive power is always less than the largest type of reactive power if power factor improvement is used. Example A 240 V, 50 Hz single phase installation draws a current of 40 A from a supply at a power factor of 0.4 lagging. Determine the VAR rating of a capacitor bank to be connected in parallel with the load to achieve an installation power factor of: (a) 0.866 lagging (b) unity Solution Draw the circuit (Figure 5). EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 7 Figure 5 cos 0.4 66.4 S VI 240 40 9.6 kVA P S cos 9600 0.4 3.84 kW Q S sin 9600 sin 66.4 8.8 kVAr (a) New cos φ is 0.866: φ = 30° Now, as P is constant in the load Q P tan 30 3840 tan 30 2.22 kVAr Q new Qold Qcapacitor Qcapacitor Qold Q new 8.88 2.22 6.58 kVAr (b) New cos φ is 1: φ = 0° 8 EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 Now, as P is constant in the load: Q = P tan 0 = 3.84 tan 0 = 0 VAr Q capacitor = Qold – Q new = 8.8 – 0 = 8.8 kVAr Student exercise 1 1 A 240 V, 50 Hz single phase installation draws a current of 16 A from a supply at a power factor of 0.52 lagging. Determine the VAR rating of a capacitor bank to be connected in parallel with the load to achieve an installation power factor of: (a) 0.8 lagging ___________________________________________________________________ ___________________________________________________________________ (b) unity. ___________________________________________________________________ ___________________________________________________________________ 2 A 240 V, 50 Hz single phase installation comprises of an inductive load which draws a current of 32 A from a supply at a power factor of 0.6 lagging. Determine the VAR rating of a capacitor bank to be connected in parallel with the load to achieve an installation power factor of: (a) 0.75 lagging ___________________________________________________________________ ___________________________________________________________________ (b) 0.9 lagging. ___________________________________________________________________ ___________________________________________________________________ Check your answers with those given at the end of the section. EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 9 Power factor improvement capacitors To determine the actual value of the capacitance used to correct the power factor of the circuit, Ohm’s law and the equation for capacitance are required. Both of these were covered in previous modules, but we will revise them here. To determine the capacitive reactance, first use the Ohm’s law equation for reactive power to determine the capacitor current. I capacitor capacitor VAr supply voltage This can be written as: IC QC VC Next, calculate the value of capacitive reactance: XC VC IC Finally, determine the capacitance of the capacitor: C= 1 2fXC Example A 240 V, 50 Hz single phase motor draws a current of 10 A from a supply at a power factor of 0.6 lagging. Determine the capacitance required to be connected in parallel with the load in order to achieve an installation power factor of: (a) 0.8 lagging (b) 0.9 lagging 10 EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 Solution Draw a circuit diagram (Figure 6). Figure 6 cos 0.6 53.13 S VI 240 10 2.4 kVA P S cos 2400 cos 53.15 1.44 kW Q S sin 2400 sin 53.15 1.92 kVAr (a) New cos is 0.8 = 36.86° Now, as P is constant in the load Q P tan = 1440 tan 36.86° = 1.08 kVAR Qcapacitor = Qold Qnew 1.92 kVAr -1.08 kVAr = 840 VAr EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 11 IC QC VC 840 240 3.5 A XC VC IC 240 3.5 68.6 1 2 fX C C 1 2 50 68.6 46.4 μF (b) New cos is 0.9 Now, as P is constant in the load Q P tan 1.44 tan 25.84 697 VAr Q capacitor Qold Qnew 1920 697 1.22 kVAr IC QC VC 1220 240 5.08 A XC VC IC 240 5.08 47 12 EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 C 1 2 fX C 1 2 50 68.6 46.4 μF Student exercise 2 1 A 240 V, 50 Hz single phase motor draws a current of 8 A from a supply at a power factor of 0.7 lagging. Determine the capacitance required to be connected in parallel with the load to achieve an installation power factor of: (a) 0.8 lagging ___________________________________________________________________ ___________________________________________________________________ (b) 0.85 lagging. ___________________________________________________________________ ___________________________________________________________________ 2 A 240 V, 50 Hz single supply is connected to fluorescent lighting circuit which draws a current of 13.3 A at a power factor of 0.55 lagging. Determine the capacitance required to be connected in parallel with the load to achieve an installation power factor of: (a) 0.75 lagging ___________________________________________________________________ ___________________________________________________________________ (b) 0.85 lagging. ___________________________________________________________________ ___________________________________________________________________ Check your answers with those given at the end of the section. EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 13 Power factor correction methods The power factor correction device used is dependent upon the size of the installation. Once you have determined what device is most appropriate and the size required, you need to refer to manufacturer’s catalogues for the most cost effective solution. Capacitors For most discharge lighting applications, even though the total load is large it is comprised of small individual loads, and a small value capacitor connected in every second (or in every fitting) is the typical circuit arrangement. The total capacitance for capacitors in parallel is the numerical sum of the capacitance of each capacitor. C t C1 C 2 C3 .....Cn Example A single phase fluorescent lighting circuit comprises 14 twin 36 W fittings. If a 30 F capacitor is connected in every second fitting, as illustrated in Figure 7, determine the total capacitance of the circuit. Figure 7 14 EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 Solution The total capacitance of the circuit: C t C1 + C 2 + C3 + C 4 + C 5 + C6 + C7 30 F +F +F + 30 F +F +F+F = 210 F We can see that instead of using seven 30 F capacitors, we could have used a single 210 F capacitor. In some applications this is what is actually done, but in most cases it is too expensive, both in materials and installation costs. For large single function loads either a large capacitor or a capacitor bank is installed. Synchronous motors Another way of improving the power factor of a load involves the use of a synchronous motor. Basically the synchronous motor consists of two windings, a stator winding (stationary winding) supplied with alternating current and a rotor winding (rotating winding) supplied with direct current. If the dc supply to the rotor winding is increased beyond the normal full load dc value, over-excitation occurs and the motor supply current changes its phase angle from a lagging angle to a leading angle. Thus the power factor of the motor goes from a lagging value to a leading value. The reactive VARs produced by the motor are now capacitive and may be used to correct or cancel the inductive VAR in the installation. In this application, the motor is called a synchronous capacitor or synchronous condenser. It will be covered in greater detail in dealing with ac machines and synchronous motors. The single drawback to this method is the cost, both very high initial costs and high ongoing maintenance costs. Thus this method only finds application in the largest types of loads—typically in the MVA range—in large factories and at power stations. An advantage of course is that this method may incorporate a synchronous motor driving a load such as a blower which results in a machine serving more than one purpose. EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 15 Summary The power factor is the cosine of the phase angle (φ) between the current and the voltage in a circuit. The power factor is described in terms of: – a number ranging from 0 to 1, plus a statement that the current leads or lags the voltage (unless the power factor is 1, where the two are in phase. The disadvantages of a low power factor are: – – – – – – reduced overall efficiency increased power losses larger alternators and transformers larger transmission conductors higher rated control and protection equipment increased voltage drop – overall increased costs To improve the power factor of a load, the inductive VARs are cancelled by the application of capacitive VARs. The equations used to determine the value of the capacitor required are: XC The capacitor VARs are implemented by using: – – – 16 VC 1 and C = IC 2fX C capacitors individually in a small load capacitors arranged in a bank (parallel and parallel/series) for larger single loads synchronous capacitors (motors) for very large installations. EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 Check your progress In questions 1–5, place the letter matching your answer in the brackets provided. Part A 1 The power factor of an ac circuit cannot exceed: (a) 0.5 (b) 0.707 (c) 1.0 (d) 2.0 . 2 ( ) Reducing the value of inductance in parallel with a resistor connected to an ac supply will cause the power factor to: (a) remain the same (b) reach unity (c) increase (d) decrease. 3 ( ) A main disadvantage of a low power factor in a supply system is that: (a) larger currents are necessary for the same power (b) the waveshape of the EMF changes from a sine waveform (c) the true power increases (d) smaller conductors must be used. 4 ( ) Connecting a capacitor in parallel to a fluorescent lamp circuit will increase the: (a) circuit power factor. (b) apparent power (c) true power (d) reactive power. EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 ( ) 17 5 When capacitors improve the power factor of an inductive load the total power will: (a) remain the same (b) fluctuate (c) increase (d) decrease. ( ) Part B 1 The fluorescent lighting load in a shopping centre is measured at 6000 W at 0.26 power factor. Calculate the rating of a capacitor bank to improve the power factor to 0.8. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 2 A 1.5 kW single phase load is to be connected to a portable alternator. Determine the rating of the alternator and the current carrying capacity of the connecting leads if the load is supplied at 240 V and has a power factor of: (a) unity ___________________________________________________________________ ___________________________________________________________________ (b) 0.8 lag ___________________________________________________________________ ___________________________________________________________________ (c) 0.6 lag. ___________________________________________________________________ ___________________________________________________________________ Answers to Check your progress are at the end of the module. 18 EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 Answers Student exercise 1 1 (a) 1783 VAR (1.78 kVAR) (b) 3280 VAR (3.28 kVAR) 2 (a) 2.08 kVAR (b) 3.91 kVAR Student exercise 2 1 (a) 20.1 µF (b) 35.2 µF 2 (a) (b) 61 µF 87.2 µF Check your progress Part A 1 (c) 4 (a) 2 (d) 5 (a) 3 (a) Part B 1 17.8 kVAR 2 (a) (b) 1 5 kVA, 62.5 A 18.75 kVA, 78.125 A EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817 19 (c) 2 25 kVA, 104.17 A EGG202A: 8 Improve the power factor of ac circuits NSW DET 2017 2006/060/04/2017 LRR 3817