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Contents Introduction 3 Three-phase power 4 Phase power 4 Total true power 7 The power triangle Power measurement 13 23 One wattmeter 23 Two wattmeter method 25 Three wattmeters 27 Power factor measurement 30 Summary 34 Answers 39 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 1 2 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 Introduction When a load is connected to a three phase supply, it draws power from that supply. The value of the load power is important in determining the actual size of the load which the source is capable of supplying, the size of the cables connecting the source to the load and the type and size of the protection used to protect both the load and the source. In this section we will examine some of the concepts relating to three phase power, the methods used to minimise reactive power at the source and the means of measuring true three phase power and determining the power factor of a three phase load. At the end of this section you should be able to: explain the purposes for measuring power, energy, power factor and maximum demand of a.c. power systems and loads explain the difference between true power, apparent power and reactive power and the units in which these quantities are measured in a three phase system draw the power triangle to show the relationships between true power, apparent power and reactive power in a three phase system describe the methods used to measure three phase power , energy, power factor and demand determine how the power factor of a three phase installation can be improved using manufacturers catalogues, select measurement equipment for a particular installation. EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 3 Three-phase power Power in a three-phase circuit is calculated in much the same way as singlephase power. The main difference is that three-phase power refers to the total power of a three-phase load, whereas single-phase power refers to only one part of a three-phase load, or to the total load if it is supplied from a single-phase source. Phase power The power developed, used and/or consumed in a phase is determined using the equations already covered in this module. Look at the basic three-phase circuit in Figure 1. Figure 1: Basic three-phase circuit In the circuit drawn in Figure 1 we can see that there are three independent loads, supplied from a three-phase source. The power used by each section of the load is calculated by using one of the following equations. Pphase Vphase I phase cos phase 2 Pphase I phase Rphase Pphase 4 2 Vphase Rphase EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 The equations for each phase are therefore the same, except that when we are dealing with a specific phase subscripts identifying that phase are used. For example, the equations for A phase would be written as: PA VA I A cos A PA I A2 RA PA VA2 RA Example 1 Determine the phase powers consumed by a three-phase star connected load consisting of inductors having an impedance of 24 with a power factor of 0.8 lagging, when supplied from a three-phase 415 V, 50 Hz source. Solution Step 1 Determine the phase voltage. As the load is star connected, use the following equation: Vline 3 Vphase Rearrange to find Vphase: V phase VLine 3 415 3 240 V VA Similarly, VB and VC are also 240 V. Determine the phase current using Ohm’s law. Step 2 I phase IA V phase Z phase VA ZA 240 24 10 A By similar calculation, IB and IC are also 10 A. EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 5 Step 3 Determine the phase power using the equation. Pphase V phase I phase cos phase 240 10 0.8 1920 W Therefore: PA PB PC 1920 W P C VC I C cos C 240 10 0. 8 1920 W As you can see from Example 1, if the load is balanced the values for voltage, current and power have the same magnitude for each phase. Student exercise 1 1 Determine the phase powers consumed by a three-phase star connected load consisting of heaters having an impedance of 50 if it is supplied from a three-phase 415 V, 50 Hz source. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 2 Determine the phase powers consumed by a three-phase delta connected motor, whose winding impedance is 12 with a power factor of 0.7 lagging, if it is supplied from a three-phase 415 V, 50 Hz source. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ Check your answers with those given at the end of the section. 6 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 Total true power Total power in any circuit is determined by simply adding together the powers used by the load. With true power there is no phase separation. This means that total true power for a three-phase load is the numerical sum of the three individual phase powers. This can be written as: Ptotal PA PB PC where PA, PB and PC are the three-phase powers. This equation is valid for any type of load, balanced or unbalanced. How the equation works for balanced loads is demonstrated in Example 2, while Example 3 deals with unbalanced loads. Example 2 Determine the total power consumed by a three-phase star connected load where each phase of the load consists of inductors having an impedance of 10 with a power factor of 0.6 lagging, if it is supplied from a three-phase 415 V, 50 Hz source. Step 1 Determine the phase voltage. As the load is star connected, use the following equation: Vline 3 Vphase Rearrange to find Vphase V phase VLine 3 415 3 240 V V phase Therefore: VA VB VC 240 V Step 2 Determine the phase current using Ohm’s law. EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 7 I phase V phase Z phase 240 10 24 A I A I B IC 12A Step 3 Determine the phase power using the equation. Pphase V phase I phase cos phase PA VA I A cos A 240 24 0.6 3456 W PB VB I B cos B 240 24 0.6 3456 W PC VC I C cos C 240 24 0.6 3456 W Step 4 Determine the total power. Pt PA PB PC 3456 3456 3456 10 368 W 10.37 kW Example 3 Determine the total power consumed by a three-phase star connected load supplied from a three-phase 415 V, 50 Hz source. The load consists of a 20 resistor (A phase), a 15 impedance with a power factor of 0.7 (B phase) and a 25 impedance with a power factor of 0.9 (C phase). Solution Step 1 Determine the phase voltage. As the load is star connected, use the following equation: Vline 3 Vphase 8 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 Rearrange to find Vphase VLine 3 V phase 415 3 240 V VA 415 3 240 V VB 415 3 240 V VC Step 2 Determine the phase current using Ohm’s law. I phase IA V phase Z phase VA ZA 240 20 12 A IB VB ZB 240 15 16 A IC VC ZC 240 25 9.6 A Step 3 Determine the phase power using the equation. Pphase V phase I phase cos phase EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 9 PA VA I A cos A 240 12 1 2880 W PB VB I B cos B 240 16 0.7 2688 W PC VC I C cos C 240 9.6 0.9 2074 W Step 4 Determine the total power. Pt PA PB PC 2880 2688 2074 7642 W 7.64 kW Balanced loads As Example 1 showed, if the load is balanced the phase powers are equal, and the simplest way of determining the total power is to calculate one phase power and multiply it by three. That is: Pt PA PB PC but PA PB PC so Pt 3PA 3PB 3PC that is, Pt 3Pphase In star connected loads the following equations apply: I phase I line V phase Vline 3 If we substitute the star equations into the total power equation we can write the often-used equation for total power in a balanced load: Pt 3 V phase I phase cos phase 3 Vline 3 I line cos phase 3 Vline I line cos phase 10 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 In delta connected loads the following equations apply: V phase Vline I phase I line 3 If we substitute the delta equations into the total power equation we can write the often-used equation for total power in a balanced load: Pt 3 V phase I phase cos phase 3 Vline I line 3 cos phase 3 Vline I line cos phase As you can see, the total power equation is identical for both types of balanced loads, either star connected or delta connected. As the phase values are typically unobtainable, being within the machine or equipment, it is very important to be able to use line values of voltage and current—these are the only values that can be measured at the point of connection. Example 4 A three-phase delta connected induction motor draws 20 A at a power factor of 0.8 lagging from a three-phase 415 V, 50 Hz supply. Determine the electrical input to the motor. Solution Pt 3 Vline I line cos phase 3 415 20 0.8 11 500 W 11.5 kW EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 11 Student exercise 2 1 A three-phase delta connected induction motor draws 20 A at a power factor of 0.8 lagging from a three-phase 415 V, 50 Hz supply. Determine the electrical input to the motor. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 2 An 11 kV transformer supplies a 300 kW balanced load operating at a 0.9 power factor lagging. Determine the line current supplied by the transformer. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 3 Determine the total power consumed by a three-phase delta connected load where each phase of the load consists of inductors having an impedance of 25 with a power factor of 0.8 lagging, if it is supplied from a three-phase 415 V, 50 Hz source. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ Check your answers with those given at the end of the section. 12 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 The power triangle As you know there are three types of power in ac circuits: true power, apparent power and reactive power. You will remember that these powers can be linked together by means of a power triangle. As power is the same for three-phase power as for singlephase, the triangle is the same (Figure 2). Figure 2: Power triangle for three-phase system In Figure 2 the horizontal line represents the true power (P), the vertical line represents the reactive power (Q) and the diagonal line represents the apparent power (S). The triangle is always drawn this way, regardless of the load type. Remember that the triangle must be scaled if it is to be used for solving problems. Calculations using the power triangle use the trigonometric relationships of sine, cosine and tangent, and Pythagoras’ theorem. The equations commonly used for calculating each power are: St Pt 2 Qt2 Pt St cos Qt St sin These equations apply both to the individual phase powers and to the total power of the three-phase load. Take care when performing calculations with three-phase power that you keep phase power values separate from the three-phase values. EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 13 Example 5 A three-phase transformer is rated at 50 kVA. If the transformer is fully loaded, determine the true power and the reactive power for the transformer if the load it is supplying has a power factor of 0.6 lagging. Solution Step 1 Draw the circuit (Figure 3). Figure 3: Circuit for example 5 Step 2 Convert the power factor to an angle. pf = cos ° cos –1pf cos –1 0.6 53.13 Since the power factor was lagging the phase angle is a negative value with respect to the voltage. = –53.13° Step 3 Determine the true power. Pt St cos 50 103 cos (–53.13) 30 103 30 kW Step 4 Determine the reactive power. Qt St sin 50 103 sin (–53.13) 40 103 40 kW 14 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 The power factor We have seen that keeping the power factor as close as practicable to unity has a number of advantages, the most important being that the line current drawn from the supply will have a minimum value. The power triangle can also be used to illustrate and/or solve problems dealing with power factor correction. (If the triangle is used to solve a problem, remember that it must be drawn to scale). Example 6 gives the two methods of solving a problem dealing with power factor correction, one graphical and one analytical. Example 6 A three-phase balanced load requires 180 kW of power when operating at a power factor of 0.6 lagging and connected to a 415 V, 50 Hz three-phase supply. Determine the kVAr rating and capacitance of a star connected capacitor bank that would improve the power factor to 0.9 lagging. Solution 1 - Graphical method Step 1 Draw the circuit (Figure 4). Figure 4: Circuit for example 6 Step 2 Determine the original circuit phase angle. pf = cos cos –1 (pf) cos –1 0.6 53.13 Step 3 Determine the true, apparent and reactive powers for the original load. P = 180 kW (given) EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 15 S P cos 180 103 cos(53.13) 180 103 0.6 300 kVA Q S sin 300 103 sin 53.13 300 103 0.8 240 kVAr Note that you only need P and the angle to construct the power triangle. Step 4 Select a suitable scale and convert the values determined in Step 3 to scaled values. We will use a scale of 5 kW (or kVA, or kVAr) to 1 mm. 300 1 5 60 mm S 180 1 5 36 mm P 240 1 5 48 mm Q Step 5 Draw the triangle (Figure 5). Figure 5: Phasors for example 6 Step 6 16 Determine the new circuit phase angle. EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 cos –1 (pf) cos –1 0.9 25.8 Step 7 Draw a line at the new phase angle of the circuit onto the original triangle representing the original load as in Figure 6. Figure 6: Line for new phase angle for example 6 Step 8 Using Figure 6, measure the distance between the points at which the original and the new lines representing the apparent power touch the reactive power side of the triangle. This distance represents the capacitive VAr added to the circuit. Qcap distance between Qold and Qnew 30.6 mm Convert this value to a real unit using the scale. 30.6 5 1 153 kVAr Qcap Step 9 Determine the phase current in the capacitor bank. In star, we have: I line 3 I phase Qt 3 Vphase I phase sin phase I phase Qt 3 V phase sin phase 153 103 3 240 sin 90 212.5 A EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 17 or, as the capacitor bank is star connected, and in star I Line I phase , the phase current may be calculated using Qt 3 Iline Iline sin phase Rearrange to find Iline Qt I line 3 Vline sin phase 153 103 3 415 sin 90 212.5 A I phase I line 212.5 A Step 10 Determine the capacitive reactance of each capacitor. Xc Vcap I cap 240 212.8 1.13 Step 11 Determine the capacitance in each phase of the load. C 1 2 fX cap 1 2 50 1.13 0.00282 F =2820 F Solution 2 - Analytical method Step 1 Draw the circuit for the example as in Figure 4. Step 2 Determine the original circuit phase angle. pf = cos ° cos –1 (pf) cos –1 0.6 53.13 18 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 Step 3 Determine the true, apparent and reactive powers for the original load. P = 180 kW (given) S P cos 180 103 cos (53.13) 180 103 0.6 =300 kVA Q S sin 300 103 sin (53.13) 300 103 0.8 240 kVAR Step 4 Determine the new circuit phase angle. cos –1pf cos –1 0.9 25.8 Step 5 Determine the true, apparent and reactive powers for the new load including the capacitor bank. P = 180 kW (given) S new P cos 180 103 cos (–25.8) 180 103 0.9 200 kVA Qnew S sin 200 10 3 sin (25.8) 200 10 3 0.435 87 kVAr Step 6 Determine the capacitor bank rating by finding the difference between the original reactive power and the new reactive power. EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 19 Qcap Qold – Qnew 240 – 87 153 kVAr Step 7 Determine the phase current in the capacitor bank. Remember that in star VLine 3 Vphase Qt 3 Vphase I phase sin phase I phase Qt 3 V phase sin phase 153 103 3 240 sin 90 212.5 A Or, as the capacitor bank is star connected, and in star I Line I phase , the phase current may be calculated using Qt 3 Vline Iline sin phase Rearrange to find Iline I line Qt 3 Vline sin phase 153 103 3 415 sin 90 212.5 A I phase I line 212.5 A Step 8 Determine the phase impedance of each capacitor using Ohm’s law. Z cap Vcap I cap 240 212.8 1.13 As a capacitor is a pure component, its impedance equals the capacitive reactance: ie Z cap X cap X cap 1.13 20 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 Step 9 Determine the capacitance in each phase of the load. C 1 2 fX cap 1 2 50 1.13 0.00282 F 2820 F The analytical method looks easier, as there are fewer steps in solving the problem. The graphical solution, however, allows the principles of power factor correction to be more easily understood. The most common three-phase arrangement for connecting capacitor banks is delta, as this allows for maximum effect on the line current with minimum current rating of the actual capacitors. The capacitors may be arranged in delta whether the load is star connected or delta connected. This is shown in Figures 7(a) for a star load and 7(b) for a delta load. (a) star load (b) delta load Figure 7: Connection of three-phase capacitor banks in delta EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 21 Student exercise 3 A three-phase balanced load requires 750 kW of power when operating at a power factor of 0.75 lagging. It is connected to a 415 V, 50 Hz three-phase supply. Determine the kVAr rating and capacitance of a delta connected capacitor bank that would improve the power factor to (a) 0.85 lagging ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ (b) 0.95 lagging ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ Check your answers with those given at the end of the section. 22 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 Power measurement There are a number of methods to measure the power used and/or developed by a load. They all involve a meter called a wattmeter. This meter was discussed in other resource modules. Three-phase power is measured using one, two or three wattmeters. The method which is used, is determined by three factors: • whether the load is balanced or unbalanced • whether the supply system is three wire or four wire • the availability and cost of the wattmeters. One wattmeter The one wattmeter method of connection can be used on both three wire and four wire three-phase systems. It is easiest to look first at the three-phase four wire system. This is illustrated in Figure 8. Figure 8: Connection of single analog wattmeter In this method the voltage coil of the wattmeter is connected between one phase and neutral. The power measured is phase power. If the load is balanced, the total power used by the load is three times the measured value. This is written as Pt 3 wattmeter reading EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 23 Example 7 A single wattmeter is used to measure the power consumed by a balanced load. If the wattmeter is connected to measure the power in the A phase of the load and the reading is 1500 W, determine the total power in the load. Solution Pt 3 wattmeter reading =3 1500 =4500 W =4.5 kW If the load is unbalanced, the technique is to measure each phase power and add the three phase values together. This is written as Pt PA PB PC This method suffers the disadvantage that the connection process is more tedious than for a balanced load, as the actual connections of the wattmeter must be changed for each reading. Example 8 A single wattmeter is used to measure the power used by an unbalanced three-phase load. If the power measured in A phase, B phase and C phase is 1200 W, 3245 W and 2790 W respectively, determine the total power in the load. Solution Pt PA PB PC 1200 3245 2790 7235 W 7.24 kW A disadvantage of this type of wattmeter connection is that it is not accurate for varying unbalanced loads. The three-phase three wire system of supply can also be monitored using a single wattmeter. Remember that to measure phase power the wattmeter must be connected between one line and neutral. In the three wire system there is no neutral. To overcome this, an artificial star point (neutral point) is created by connecting one end of the voltage coil of the wattmeter to one line and the other end to two impedances which are in turn connected to the other two lines. This is illustrated in Figure 9. 24 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 Figure 9: Single analog wattmeter with artificial star point It is very important that the impedance values of the two impedances used with the voltage coil to create the artificial star point are the same as that of the voltage coil of the wattmeter, otherwise inaccurate readings will result. The easiest way to accomplish this is to use two coils identical to the voltage coil of the wattmeter. These are obtainable from the manufacturers and distributors of these meters. As for the four-wire connection of a single wattmeter, the load may be balanced or unbalanced. The total power for the load is determined in the same way as for the four wire connection. That is, for balanced loads Pt 3 wattmeter reading . and for unbalanced loads Pt PA PB PC Two wattmeter method This method may only be used on three wire systems. The connection diagram is shown in Figure 10. EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 25 Figure 10: Connection of analog wattmeters for two wattmeter method The total power is found by adding the readings of the two meters. This is written as: Pt W1 W2 This can be proven mathematically by considering the instantaneous values of current and voltage to be e1 and i1 for A, e2 and i2 for B with e3 and i3 for C. Voltage across W1 = e3 – e2 Voltage across W2 = e1 - e2 Average power indicated by W1 = i3 (e3 – e2) Average power indicated by W2 = i1 (e1 – e2) W1 +W2 i3 (e3 e2 ) i1 (e1 e2 ) i3e3 i3e2 i1e1 i1e2 i1e1 i3e3 e2 (i3 i1 ) But (i3 i1 ) i2 total zero at node W1 +W i1e1 i3e3 i2 e2 This method of connection is accurate for balanced and unbalanced loads, but it does give inaccurate readings for loads with low power factors. A further disadvantage is that phase power cannot be measured. Example 9 The two wattmeter method of measuring total power is used to measure the power used by an unbalanced three phase load. If W1 reads 10 kW and W2 reads 7 kW, determine the total power in the load. Solution Pt W1 W2 10 kW 7 kW 17 kW As most analog meters deflect in one direction only, for some loads, the meter will try to deflect backwards due to the phase relationship between current and voltage within the meter. When this occurs, the voltage coil connections must be reversed. Some meters provide a reversing switch that will reverse the connections of the voltage coil in the wattmeter. This will then allow the meter to indicate the power that the load is consuming. For the total power calculation, this reversal of the coil will mean that the value 26 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 measured by that wattmeter should be regarded as negative. Don’t forget to alter the polarity of the reading in the total power equation. Example 10 The two wattmeter method of measuring total power is used to measure the power used by an unbalanced three phase load. If W1 reads 4.5 kW after the reversing switch is closed and W2 reads 13 kW, determine the total power in the load. Solution Pt W1 W2 – 4.5 kW 13 kW 8.5 kW Three wattmeters Like the single wattmeter, this may be used with either three phase three wire system or three phase four wire systems. We will look first at the four wire system. The current coil of each wattmeter is connected in a line and the voltage coil is connected between one line and neutral. This is shown in Figure 11. Figure 11: Connection of three analog wattmeters With this method each phase power is measured, and the total power is the sum of the three readings. This is written as Pt W1 W2 W3 The main disadvantage of this method is the expense of installing and using three meters. EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 27 Example 11 Three wattmeters are used to measure the power consumed in an unbalanced three phase load supplied from a three phase four wire source. If the power measured by W1, W2 and W3 is 4700 W, 5200 W and 3750 W respectively, determine the total power in the load. Solution Pt W1 W2 W3 4700 5200 3750 =13.65 kW The connection diagram for using three wattmeters in a three-wire system is given in Figure 12. Figure 12: Connection of three analog wattmeters using artificial star point Figure 12 shows how an artificial star point is created by connecting one end of each of the voltage coils together. It must be stressed that this artificial star point can only be created with the voltage coils if they are identical. If they are not identical, they will not develop symmetrical phase voltages, and the wattmeters will give incorrect readings. The total power is again the sum of the phase powers measured by the three wattmeters. 28 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 Student exercise 4 1 A single wattmeter is used to measure the power consumed by a balanced load. If the wattmeter is connected to measure the power in the B phase of the load and the reading is 4200 W, determine the total power in the load. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 2 The two wattmeter method of measuring total power is used to measure the power used by an unbalanced three phase load. If W1 reads 40 kW and W2 reads 27 kW, determine the total power in the load. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 3 The two wattmeter method of measuring total power is used to measure the power used by an unbalanced three phase load. If W1 reads 6 kW after the reversing switch is closed and W2 reads 9.2 kW, determine the total power in the load. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 4 Three wattmeters are used to measure the power consumed in an balanced three phase load supplied from a three phase three wire source. If the power measured by W1, W2 and W3 is 1500 W, 3650 W and 2740 W respectively, determine the total power in the load. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ Check your answers with those given at the end of the section. EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 29 Power factor measurement An advantage of the two wattmeter method of measuring three phase power is that it allows the power factor of the load to be determined. The circuit for the two wattmeter connection is shown in Figure 13. Figure 13: Two wattmeter connection The equation for determining the power factor angle or phase angle of the load is W – W1 tan 3 2 W2 W1 Note: For this equation to work, W2 must be in the phase immediately before W1 as shown below: Phase sequence : A B C A B C A W2 W1 Recall that the cosine of the phase angle is the power factor (cos = pf). Example 13 The two wattmeter method of measuring total power is used to measure the power used by an unbalanced three phase load. If W1 reads 5 kW and W2 reads 7 kW, determine the power factor of the load. tan 3 30 W1 – W2 W1 W2 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 7–5 tan –1 3 75 tan –1 3 0.167 tan –1 0.289 16.1 pf cos cos16.1 0.96 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 31 As mentioned earlier, most analogue meters deflect in one direction only, and in some meters a reversing switch is provided which will reverse the connections of the voltage coil in the wattmeter. This allows the meter to indicate the power the load is consuming. For the power factor calculation, this reversal of the coil means that the value measured by that wattmeter will be a negative value. Don’t forget to alter the polarity of the reading in the power factor equation. If the wattmeter reading of W2 is greater than that of W1 the power factor is lagging, while if the reading of W1 is greater than that of W2, the power factor is leading. Typically, however, the nature of the load is used to determine whether the power factor is leading or lagging. Inductive loads have a lagging power factor and capacitive loads have a leading power factor. As capacitors and capacitive loads are mainly used for power factor correction, for normal distribution and utilisation purposes where wattmeters are connected the usual power factor for a load is lagging. If you have Jenneson, refer to section 9.8 ‘Three phase power’. See also Appendix 1 for a description of a power factor meter. Note particularly Figure A1.4 which clearly shows the same configuration of potential and current coils used for the two wattmeter method. If you have Hampson, refer to ‘Energy and Power requirements of ac systems, on page 204. 32 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 Student exercise 5 1 The two wattmeter method of measuring total power is used to measure the power used by an unbalanced three phase load. If W1 reads 2.5 kW and W2 reads 12 kW, determine the power factor of the load. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 2 The two wattmeter method of measuring total power is used to measure the power used by an unbalanced three phase load. If W1 reads 0 kW and W2 reads 10 kW, determine the power factor of the load. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 3 The two wattmeter method of measuring total power is used to measure the power used by an unbalanced three phase load. If W1 reads 10 kW and W2 reads 10 kW, determine the power factor of the load. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ Check your answers with those given at the end of the section. EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 33 Summary Phase power is determined using one of the following: Pphase Vphase I phase cos phase 2 Pphase I phase Rphase Pphase 2 Vphase Rphase Total power is the power used by the total load. For three phase loads it is the sum of the three phase powers. Pt PA BB PC For balanced loads the equations that can be used are Pt 3 Pphase Pt 3 PA 3 PB 3 PC Pt 3 Vline I line cos phase There are three types of powers, apparent power (S), reactive power (Q) and true power (P). The three types can be linked by a power triangle in which the horizontal line represents true power, the vertical line reactive power and the diagonal line apparent power. Capacitor banks are used to improve the power factor of an inductive load. These capacitors are normally connected in parallel with the load at the supply terminals of the load. • When connecting capacitor banks in three phase systems, the capacitors are usually connected in delta. To measure power, a wattmeter is used. To measure three phase power either one, two or three wattmeters are used, depending upon the load, the type of supply and the availability and cost of the wattmeters. The power factor of the load can be determined when using the two wattmeter method by the equations W – W1 tan 3 2 and cos pf W2 W1 34 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 Check your progress 15 In questions 1 to 9, place the letter matching your answer in the brackets provided. 1 The minimum number of fixed wattmeters required to measure total power drawn by a three-phase four-wire unbalanced load is: (a) one (b) three (c) two (d) four. 2 ( ) A maximum demand indicator measures maximum: (a) volt amperes for a given time period (b) volt amperes irrespective of time (c) power used for a given time period (d) power used irrespective of a time period. 3 ( ) The total power in a three-phase system may be measured using a single fixed wattmeter, provided the: (a) load is not balanced (b) load is balanced (c) neutral is not connected (d) load is star connected. 4 ( ) When using a wattmeter to measure power in a three-phase star connected system, the current coil is connected in: (a) series with the load (b) parallel with the supply (c) series with the neutral (d) parallel with the load. 5 ( ) When using a wattmeter to measure power in a three-phase star connected system, the voltage coil is connected in: (a) series with the neutral (b) series with the load (c) parallel with the load (d) series with the supply. EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 ( ) 35 6 When measuring balanced three-phase power with the two wattmeter method and both readings are equal, the power factor is: (a) unity (b) zero (c) 0.5 leading (d) 0.5 lagging. 7 ( ) The power factor of a three-phase load may be determined by the two wattmeter method provided: (a) the load is balanced (b) the power factor is greater than 0.5 (c) the neutral is connected (d) there is no current in the middle phase. 8 ( ) A three-phase wattmeter can be constructed using the concepts of the two wattmeter method using: (a) two current coils on a single shaft (b) two potential coils on a single shaft (c) fixed potential coils (d) a current coil and potential coil on a single shaft. 9 ( ) Three-phase power factor can be measured using a: (a) crossed coil meter with coils at 60° (b) VAr modified wattmeter (c) maximum demand indicator (d) crossed coil meter with coils at 90°. ( ) 10 A star connected ac generator develops 11 000 volts per phase. Determine the MVA rating of the machine if the current per phase is 50 A. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 36 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 11 A three-phase alternator delivers a full load of 140 A at a power factor of 0.9 lagging. If the terminal voltage is 415 V, calculate the: (a) apparent power (b) true power. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 12 A factory draws a line current of 130 A from a 415V supply at a power factor of 0.6 lag. Determine the: (a) kVAr rating of a capacitor bank to improve the power factor to 0.866 lag (b) line current with the capacitor bank connected. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 37 13 The power in a three-phase 415 V system is measured by the two wattmeter method where W1 indicates 30 kW and W2 indicates 20 kW. Calculate the: (a) total power (b) power factor (c) line current. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 14 The power input to a three-phase, 415 V induction motor is measured by the two wattmeter method where W1 indicates 10 kW and W2 indicates -6 kW (note the minus sign). Calculate the: (a) power input (b) power factor of the motor. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 38 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 Answers Student exercise 1 1 1152 W 2 10.05 kW Student exercise 2 1 11.5 kW 2 17.5 A 3 16.53 kW Student exercise 3 (a) 196 kVAr, 2290 µF per phase (b) 414 kVAr, 8160 µF per phase Student exercise 4 1 12.6 kW 2 67 kW 3 3.2 kW 4 7.89 kW Student exercise 5 1 0.6 3 1 (unity) 2 0.5 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980 39 Check your progress 1 (b) 2 (c) 3 (b) 4 (a) 5 (c) 6 (a) 7 (a) 8 (b) 9 (d) 10 1.65 MVA 11 (a) 100.632 kVA (b) 90.75 kW (a) 42.38 kVAR (b) 90 A (a) 50 kW (b) 0.945 lag (c) 73.6 A (a) 4 kW (b) 0.143 12 13 14 2 EGG202A: 14 Calculate and measure power in three phase systems NSW DET 2017 2006/060/04/2017 LRR 4980