Three Phase Systems 95 9. MEASUEMENT OF POE Star-Connected Balanced Load with Neutral Point Power can be measured in this case by connecting a single wattmeter with its current coil in one line and the pressure coil between the line and the neutral point as shown in Fig. 9.6. The reading of the wattmeter thus connected, gives the power per phase. Total power = power per phase = wattmeter reading L -phase ac supply N -phase balanced load L L Y B Fig. 9.6 Measurement of power by single wattmeter Measurement of Power by Two attmeter Method Power in a -phase three wire system, with balanced or unbalanced load can be measured by using two wattmeters. The load may be star or delta connected. The current coils of the two wattmeters are connected in any of the two lines and the pressure coils are connected between these lines and the third line, as shown in Fig. 9.7. Let e N,, e YN and e BN be the voltages across the three phases of the load and i, i Y and i B the currents flowing in the three lines. Total instantaneous power in the load = e N i + e YN i Y + e BN i B Instantaneous current through the current coil of wattmeter = i Instantaneous voltage across the pressure coil of wattmeter e B = e N e BN Instantaneous power measured by wattmeter = i (e N e BN ) = i e B Instantaneous current through current coil of wattmeter = i Y
96 Basic Electrical Engineering i e N Z -phase ac supply N -phase load Z Z e YN e BN Y i Y B Fig. 9.7 i B Measurement of power in -phase, three wire system Instantaneous voltage across the pressure coil of wattmeter e YB = e YN e BN Instantaneous power measured by = i Y (e YN e BN ) = i Y e YB Total instantaneous power = e N i + e YN i Y + e BN i B (9.8) Moreover, for -phase, -wire system, i + i Y + i B = 0 \ i B = (i + i Y ) Substituting the value of i B in Eq. (9.8) Total instantaneous power = e N i + e YN i Y + e BN ( i i Y ) = i (e N e BN ) + i Y (e YN e BN ) = Power measured by + Power measured by Hence at any instant, the total power is equal to the sum of the two wattmeter readings. However, the inertia of the moving system of the wattmeters causes it to indicate the mean value of power taken over a cycle. Thus the sum of the readings of the two wattmeters gives the average value of the total power fed to the three phase load. No particular conditions have been assumed, while deriving the above relationship, hence, the result holds good for balanced as well as unbalanced load. Though a star-connected load has been assumed in the above derivation, but the same result will be obtained by taking a delta-connected load.
Three Phase Systems 97 Measurement of Power by Three attmeter Method In case the supply is -phase, 4-wire system, then for unbalanced loads the two wattmeter method cannot be used as there is a current flowing in the neutral. For such a system, three wattmeters are to be used to measure the total -phase power. Each wattmeter is connected with its current coil in series with the line and the pressure coil connected between the phase and neutral as shown in Fig. 9.8. Total Power = P + P Y + P B = + Y + B Thus, the addition of the readings of three wattmeters will give the total power consumed by the load. I -phase ac supply -f V N unbalanced load N N V BN Y Y I Y V YN B B I B Fig. 9.8 Measurement of power by three wattmeter method 9. MEASUEMENT OF POE AND POE FACTO IN THEE-PHASE SYSTEM ITH BALANCED LOAD USING TO ATTMETE METHOD Figure 9.7 shows -phase star-connected balanced load, supplied from a -phase supply system with two wattmeters properly connected in the circuit to measure the input power to the load. Let I, I Y and I B be the rms values of the currents in the lines and V N,V YN and V BN the rms values of voltages across the phases. The phasor diagram of such a circuit with an assumption of lagging current has been shown in Fig. 9.9. The load being balanced, currents I, I Y and I B are taken equal in magnitude and lagging by an angle f with respect to its own phase voltage. Similarly, phase voltage V N,V YN and V BN are equal in magnitude but displaced by 0, the phase sequence being YB. Current through the current coil of wattmeter = I
98 Basic Electrical Engineering V BN V N I V B f V YB I B f f 0 V BN V BN Fig. 9.9 I Y V YN Phasor diagram for measurement of power Voltage across the pressure coil of wattmeter = V N V BN = V B eferring to the phasor diagram, Fig. 9.9, phase difference between I and V B = (0 f) Thus, the reading of wattmeter = I V B cos (0 f) (9.9) Current through the current coil of wattmeter = I Y Voltage across the pressure coil of wattmeter = V YN V BN = V YB eferring to the phasor diagram (Fig. 9.9), the phase difference between the current I Y and voltage V YB = (0 + f). Thus, reading of wattmeter = I Y V YB cos (0 + f) (9.0) As the load is balanced, I = I Y = I B = I L V Y = V YB = V B = V L Substituting these in Eqs 9.9 and 9.0, = V L I L cos (0 f) (9.) = V L I L cos (0 + f) (9.) Adding Eqs. (9. and 9.) + = V L I L cos (0 f) + V L I L cos (0 + f) = V L I L [cos 0 cos f + sin 0 sin f + cos 0 cos f sin 0 sin f] = V L I L cos f = Total power input to a balanced load. Hence, the sum of the readings indicated by the two wattmeters connected as per Fig. 9.7 is equal to the total power drawn by a -phase balanced load. Now, + = V L I L cos f (9.)
Three Phase Systems 99 Subtracting Eq. (9.) from Eq. (9.), = V L I L cos (0 f) [V L I L cos (0 + f)] = V L I L [cos 0 cos f + sin 0 sin f cos 0 cos f + sin 0 sin f] = V L I L sin f (9.4) Dividing Eq. (9.4) by Eq. (9.), VL IL sinf - = + V I cosf tan f = Power factor, cos f = = sec f cos f = F HG L L - + sec f + tan f (9.5) = L - O + NM + QP (9.6) Hence, power factor can be calculated using Eq. (9.6) from the readings of two wattmeters. Equation (9.6) may also be put in the following form, cos f = F - I + + HG KJ cos f = (9.7) L - O + NM r + rqp where r = ratio of wattmeter readings = A graph of power factor vs the ratio of wattmeter reading using Eq. (9.7), has been drawn in Fig. 9.0 for r varying from + to. This graph can also be used for finding out the value of power factor, corresponding to a particular ratio of wattmeter readings. The following important conclusions can be drawn from the graph of power factor plotted in Fig. 9.0. (i) eading of wattmeter is zero, when the load power factor is 0.5 lagging, i.e. f = 60. I K J
00 Basic Electrical Engineering.0 0.8 0.6 atio of wattmeter reading 0.4 0. 0 0. 0.4 0.6 0.8.0 Fig. 9.0 0 0. 0. 0. 0.4 0.5 0.6 0.7 0.8 0.9.0 Power factor elation between pf and ratio of wattmeter readings (ii) eading of wattmeter is negative, for the load power factor less than 0.5 lagging, i.e. for f > 60. In such a case it is necessary to reverse the connections to either the current or pressure coil, in order to measure the power registered by wattmeter. However, the reading thus obtained must be taken as negative, while calculating the total power and the power factor. (iii) The reading of wattmeter will be positive, when the load power factor is greater than 0.5 lagging, i.e. f < 60. (iv) Both the wattmeters will indicate the same readings, when the power factor of the load is unity, i.e. f = 0. Examples on Power Measurement Example 9.8 Two wattmeters connected to measure the power input to a -phase circuit indicate 5 k and.5 k respectively, the latter reading being obtained after reversing the current coil connections. Calculate the power and power factor of the load. Solution: eading of first wattmeter, = 5 k eading of second wattmeter has been obtained after reversing the current coil connections. As such this reading is really negative, because of reversing the current coil connections. Thus reading of second wattmeter, =.5 k Total power fed to the load = + = 5 + (.5) =.5 k
Three Phase Systems 0 Power factor of the load can be calculated from the above reading by first finding out tan f, that is tangent of the power factor angle. tan f = - + = 5 -(-5. ) 5 + (-5. ) 6. 5 = =.7 5. Hence power factor angle, f = 64.7 Power factor of the load, cos f = 0.47 Example 9.9 Two wattmeters have been used to measure the power input to a 50 k, 440 V, -phase slip ring induction motor running at full load. The wattmeter readings are 5 k and 50 k. Calculate (i) the input to the motor, (ii) power factor of the motor, (iii) line current drawn by the motor and (iv) efficiency of the motor. Solution: (i) Power input to the motor = + = 5 + 50 = 65 k 5-50 (ii) tan f = - = = 0.68 + 5 + 50 Power factor angle, f = 4. Power factor, cos f = 0.86 (iii) Power input to the motor = V L I L cos f = 440 I L 0.86 65 0 Thus line current, I L = = 6. A 440 0. 86 (iv) Output of the motor = 50 k Input to the motor = 65 k output 50 Thus, the efficiency of the induction motor = = input 65 = 0.909 = 90.9% Example 9.0 A balanced star-connected load is supplied from a symmetrical, -phase, 440 V, 50 Hz supply system. The current in each phase is 0 A and lags behind its phase voltage by an angle 40. Calculate (i) phase voltage, (ii) load parameters, (iii) total power and (iv) readings of two wattmeters, connected in the load circuit to measure the total power. Solution: Figure 9.7 shows a balanced star-connected load connected across a -phase supply. (i) Line voltage, V L = 440 V V 440 Phase voltage in star-connected circuit, V ph = L = = 54 V
0 Basic Electrical Engineering (ii) Current in each phase = 0 A Vph 54 Impedance of the load per phase, Z ph = = =.7 I 0 Z ph = + X =.7 (i) Moreover, the current in each phase lags behind its voltage by 40, hence X tan 40 = or Inductive reactance of the load, X = 0.89 Substituting the value of X in Eq. (i) + (0.89) = (.7). 7 esistance of the load, = + ( 0. 89) = 9.7 Inductive reactance of the load, X = 0.89 9.7 = 8.6 (iii) Power consumed by each phase = V Ph I ph cos f = 54 0 cos 40 = 89.5 Total power = 89.5 = 674.5 =.6745 k (iv) Total power = + (reading of two wattmeters) Thus + =.6745 (ii) Also tan 40 = - +. 6745 0. 89 or = = 5.656 (iii) Solving Eqs (ii and iii), = 8.665 k =.009 k Example 9. Power input to a -phase 440 V, 7. k induction motor, whose efficiency and power factor are respectively 88 percent and 0.8, is to be measured by two wattmeter method. Find the reading of both the wattmeters and the full load line current drawn by the motor. Solution: Power output of -phase induction motor = 7. k Efficiency of the motor = 88 % output 7. Power input to the motor = = = 4.86 k efficiency 0.88 Power input when measured by two wattmeter method = + where, and are the readings of the two wattmeters. Hence + = 4.86 (i) Power factor of the motor, cos f = 0.8 Power factor angle, f = cos 0.8 = 4.9 ph
Three Phase Systems 0 In power measurement by two wattmeter method, tan f = tan 4.9 = - + - 4. 86 Thus, 4.86 0. 698 = = 7.08 k (ii) Solving Eqs (i and ii) = 4.86 + 7.08 59.467 = = 9.74 k = 4.86 9.74 =.65 k Hence the readings of two wattmeters are 9.74 and.65 k. Full load line current drawn by the motor = power input cos f 4.86 0 = 440 0. 8 = 67.8 A Example 9. Three inductive coils, each having a resistance of 0 and reactance of 5 are connected (i) in star (ii) in delta, across a -phase, 400 V, 50 Hz supply. Calculate in each case, the readings on two wattmeters connected in the circuit to measure the power input. Also determine the phase and line currents in each case. Solution: (i) Three coils with resistance 0 and reactance 5 each, are connected in star configuration. V 400 Phase voltage, V ph = L = = V Impedance of each coil, i.e. of each phase, Z ph = ( 0) + ( 5) = 5 Vph Phase current = = = 9.4 A Zph 5 Line current = phase current (for star connected circuits) = 9.4 A ph 0 Power factor of the circuit, cos f = = = 0.8 Z 5 Power input to the circuit, P = V L I L cos f = 400 9.4 0.8 = 5 hen two wattmeters are connected in the circuit to measure the power input to the circuit, then, ph V L
04 Basic Electrical Engineering Power input = + ( and are the readings of two wattmeters) Hence + = 5 (i) Power factor of the circuit as calculated from the parameters of the circuit, cos f = 0.8 or f = 6.87 However, when power is measured by two wattmeters tan f = tan 6.87 = - + - 5 or 0. 75 5 = = 7.5 (ii) Solving Eqs (i and ii) 5 + 7. 5 = = 669.5 = 45.75 Hence the readings of two wattmeters in case of star connection are 669.5 and 45.75. (ii) Now the same coils are connected in mesh across a -phase, 400 V, 50 Hz supply. Thus, phase voltage = 400 V (equals line voltage) Impedance per phase, Z ph = 5 400 Current per phase = 5 = 6 A Line current = phase current (in delta-connected circuits) = 6 = 7.7 A Power factor of the delta connected circuit = 0.8 (remains same) Power input to the circuit = V L I L cos f = 400 7.7 0.8 = 56 Thus, + = 56 (iii) tan f = 0.75 = - + - 56 0. 75 56 = = 665.5 (iv) Solving Eqs (iii and iv), = 007.75 = 45.5 eadings on two wattmeters are, 007.75 and 45.5.