CHAPTER 20 THREE PHASE SYSTEMS Exercise 112, Page 327 1. Three loads, each of resistance 50 are connected in star to a 400 V, 3-phase supply. Determine (a) the phase voltage, (b) the phase current and (c) the line current. 400 V, 3-phase supply means that 400 V is the line voltage. (a) For a star connection, VL 3 VP Hence, phase voltage, VP (b) Phase current, I P VL 400 = 231 V 3 3 VP 231 = 4.62 A R P 50 (c) For a star connection, IP IL Hence, line current, IL = 4.62 A 2. A star-connected load consists of three identical coils, each of inductance 159.2 mH and resistance 50 . If the supply frequency is 50 Hz and the line current is 3 A determine (a) the phase voltage and (b) the line voltage. Inductive reactance, XL = 2πfL = 2π(50)(159.2 × 10- 3) = 50 Impedance of each phase, Zp = For a star connection, IL = Ip = R 2 X2L = 502 502 = 70.71 Vp Zp Hence, phase voltage, Vp = Ip Zp = (3)(70.71) = 212 V Line voltage, VL = 3 Vp = 3 (212) = 367 V 3. Three identical capacitors are connected in star to a 400 V, 50 Hz, 3-phase supply. If the line current is 12 A determine the capacitance of each of the capacitors. © John Bird Published by Taylor and Francis 251 For a star connection, IL 12A IP VL 3 VP hence, VP XC VL 400 = 231 V 3 3 VP 231 19.25 IP 12 and capacitance, C = thus, 1 1 19.25 2 f C 2 50 19.25 = 165.4 F 4. Three coils each having resistance 6 and inductance L H are connected in star to a 415 V, 50 Hz, 3-phase supply. If the line current is 30 A, find the value of L. For a star connection, IL 30A IP VL 3 VP hence, VP ZL VL 415 = 239.6 V 3 3 VP 239.6 7.987 IP 30 X L 7.987 2 R 2 7.987 2 62 = 5.272 Ω from which, Hence, and R 2 XL 2 7.987 thus, 5.272 = 2π f L inductance, L = 5.272 = 16.78 mH 2 50 5. A 400 V, 3-phase, 4 wire, star-connected system supplies three resistive loads of 15 kW, 20 kW and 25 kW in the red, yellow and blue phases respectively. Determine the current flowing in each of the four conductors. For a star connected system, VL 3 VP from which, VP Power, P = VI for a resistive load, hence I Thus, IR VL 400 = 230.94 V 3 3 P V PR 15000 = 64.95 A, VR 230.94 IY PY 20000 = 86.60 A VY 230.94 © John Bird Published by Taylor and Francis 252 and IB PB 25000 = 108.25 A VB 230.94 The phasor diagram of the three currents is shown in (i) below. Adding phasorially gives diagram (ii) below, where I N is the neutral current. (i) (ii) Total horizontal component = 64.95 cos 90 + 108.25 cos 210 + 86.60 cos 330 = - 18.75 Total vertical component = 64.95 sin 90 + 108.25 sin 210 + 86.60 sin 330 = - 32.475 Hence, magnitude of neutral current, I N = 18.752 32.4752 = 37.50 A © John Bird Published by Taylor and Francis 253 Exercise 113, Page 329 1. Three loads, each of resistance 50 are connected in delta to a 400 V, 3-phase supply. Determine (a) the phase voltage, (b) the phase current and (c) the line current. (a) For a delta connection, VL VP Since VL = 400 V, then phase voltage, VP = 400 V (b) Phase current, I P VP 400 =8A RP 50 (c) For a delta connection, line current, IL 3 IP 3 8 = 13.86 A 2. Three inductive loads each of resistance 75 and inductance 318.4 mH are connected in delta to a 415 V, 50 Hz, 3-phase supply. Determine (a) the phase voltage, (b) the phase current, and (c) the line current. (a) For a delta connection, VL VP Since VL = 410 V, then phase voltage, VP = 415 V (b) Phase impedance, ZP R 2 X L 2 = Phase current, I P 75 2 2 50 318.4 103 2 752 1002 125 VP 415 = 3.32 A ZP 125 (c) For a delta connection, line current, IL 3 IP 3 3.32 = 5.75 A 3. Three identical capacitors are connected in delta to a 400 V, 50 Hz, 3-phase supply. If the line current is 12 A determine the capacitance of each of the capacitors. IL 12 6.93 A 3 3 © John Bird Published by Taylor and Francis For a delta connection, IL 3 IP hence IP 254 VL VP = 400 V XC VP 400 57.74 I P 6.93 and capacitance, C = thus, 1 2 50 57.74 1 57.74 2 f C = 55.13 F 4. Three coils each having resistance 6 and inductance L H are connected in delta, to a 415 V, 50 Hz, 3-phase supply. If the line current is 30 A, find the value of L. For a delta connection, IL 3 IP hence IP IL 30 17.32 A 3 3 VL VP = 415 V ZP VP 415 23.96 I P 17.32 from which, thus, 23.96 = X L 23.962 62 23.197 Hence, inductance, L = R 2 XL2 62 XL2 i.e. 2π f L = 23.197 Ω 23.197 = 73.84 mH 2 50 5. A 3-phase, star-connected alternator delivers a line current of 65 A to a balanced delta-connected load at a line voltage of 380 V. Calculate (a) the phase voltage of the alternator, (b) the alternator phase current and (c) the load phase current. (a) In star, VL 3 VP from which, phase voltage of alternator, VP VL 380 = 219.4 V 3 3 (b) In star, IP IL hence, alternator phase current = 65 A (c) In delta, IL 3 IP from which, load phase angle, IP IL 65 = 37.53 A 3 3 © John Bird Published by Taylor and Francis 255 6. Three 24 F capacitors are connected in star across a 400 V, 50 Hz, 3-phase supply. What value of capacitance must be connected in delta in order to take the same line current? In star, VL 3 VP hence, VP XC VL 400 = 230.94 V 3 3 1 1 = 132.63 2 f C 2 50 24 106 Hence, IP VP 230.94 = 1.741 A = line current for star connection. X C 132.63 In delta, if IL 1.741 3 IP hence, IP 1.741 = 1.00517 A 3 VP VL 400 = 397.94 IP I P 1.00517 397.94 = XC from which, i.e. capacitance in delta, C = 1 2 f C 1 = 8 F 2 50 397.94 © John Bird Published by Taylor and Francis 256 Exercise 114, Page 331 1. Determine the total power dissipated by three 20 resistors when connected (a) in star and (b) in delta to a 440 V, 3-phase supply. (a) In star, VL 440V 3 VP hence, VP and IP VP 254 12.70 A I L RP 20 Hence in star, power, P = or 440 254 V 3 3 VL IL cos 3 440 12.70 cos 0 = 9.68 kW P = 3I P 2 R P 3 12.70 20 = 9.68 kW 2 (b) In delta, VL VP 440 V and IP VP 440 22 A RP 20 IL 3 IP 3 22 = 38.11 A Hence in delta, power, P = 3 VL IL cos 3 440 38.11 cos 0 = 29.04 kW P = 3I P 2 R P 3 22 20 = 1350 W = 29.04 kW 2 or 2. Determine the power dissipated in the circuit of Problem 2, Exercise 112, page 327. X 50 Circuit phase angle, ϕ = tan 1 L tan 1 tan 1 1 45 50 R Power, P = or 3 VL IL cos 3 367 3 cos 45 = 1348 W = 1.35 kW P = 3I P 2 R P 3 3 50 = 1.35 kW 2 3. A balanced delta-connected load has a line voltage of 400 V, a line current of 8 A and a lagging power factor of 0.94. Draw a complete phasor diagram of the load. What is the total power dissipated by the load ? © John Bird Published by Taylor and Francis 257 If the power factor is 0.94, then 0.94 = cos ϕ from which, ϕ = cos 1 0.94 = 19.95º Line voltage = phase voltage = 400 V Line current = 8 A Phase current = 8 = 4.62 A 3 From the diagram above, IR IRY IBR , IY IYB IRY and IB IBR IYB The complete phasor diagram is shown below, with the line current lagging the line voltage by 19.95º Power, P = 3 VL IL cos 3 400 8 0.94 since power factor = cos ϕ = 5210 W = 5.21 kW © John Bird Published by Taylor and Francis 258 4. Three inductive loads, each of resistance 4 and reactance 9 are connected in delta. When connected to a 3-phase supply the loads consume 1.2 kW. Calculate (a) the power factor of the load, (b) the phase current, (c) the line current and (d) the supply voltage. (a) Phase impedance, ZP 42 92 9.849 X 9 and phase angle, tan 1 L tan 1 66.04 4 R Hence, power factor of load = cos = cos 66.04 = 0.406 i.e. 1.2 103 3IP 2 4 (b) Power, P = 3I P 2 R P from which, phase current, I P 1200 = 10 A 3(4) (c) In delta, line current, IL 3 (10) = 17.32 A (d) Power, P = from which, 3 VL IL cos i.e. 1200 = supply voltage, VL 3 VL (17.32)(0.406) 1200 = 98.53 V 3 (17.32)(0.406) 5. The input voltage, current and power to a motor is measured as 415 V, 16.4 A and 6 kW respectively, Determine the power factor of the system. Power, P = from which, 3 VL IL cos i.e. 6000 = 3 (415) (16.4) cos cos = power factor of system = 6000 = 0.509 3 (415)(16.4) 6. A 440 V, 3-phase a.c. motor has a power output of 11.25 kW and operates at a power facror of 0.8 lagging and with an efficiency of 84%. If the motor is delta connected determine (a) the power input, (b) the line current and (c) the phase current. © John Bird Published by Taylor and Francis 259 (a) Efficiency = from which, (b) Power, P = power output power input i.e. power input = 0.84 = 11250 power input 11250 = 13393 W or 13.39 kW 0.84 3 VL IL cos hence, line current, IL (c) In delta, IL 3 IP from which, phase current, I P P 13393 = 21.97 A 3 VL cos 3 440 0.80 I L 21.97 = 12.68 A 3 3 © John Bird Published by Taylor and Francis 260 Exercise 115, Page 336 1. Two wattmeters are connected to measure the input power to a balanced three-phase load. If the wattmeter readings are 9.3 kW and 5.4 kW determine (a) the total output power, and (b) the load power factor (a)Total input power, P = P1 + P2 = 9.3 + 5.4 = 14.7 kW (b) tan = P P 3 1 2 = P1 P2 9.3 5.4 3 = 9.3 5.4 3.9 3 = 0.459524 14.7 Hence, = tan 1 (0.459524) = 24.68o Power factor = cos = cos 24.68o = 0.909 2. 8 kW is found by the two-wattmeter method to be the power input to a 3-phase motor. Determine the reading of each wattmeter if the power factor of the system is 0.85 Working in kilowatts, 8 = P1 P2 If power factor = 0.85, then cos = 0.85 and tan = tan 31.79 = 0.6197 Hence, tan = 0.6197 = from which, and phase angle, = cos 1 0.85 = 31.79 P P P P 3 1 2 3 1 2 8 P1 P2 P1 P2 Adding equations (1) and (2) gives: from which, (1) 0.6197 8 3 = 2.862 (2) 2 P1 10.862 P1 10.862 = 5.431 kW 2 and from equation (1), P2 8 5.431 = 2.569 kW Thus the readings of the two wattmeters are: 5.431 kW and 2.569 kW © John Bird Published by Taylor and Francis 261 3. When the two-wattmeter method is used to measure the input power of a balanced load, the readings on the wattmeters are 7.5 kW and 2.5 kW, the connections to one of the coils on the meter reading 2.5 kW having to be reversed. Determine (a) the total input power, and (b) the load power factor Since the reversing switch on the wattmeter had to be operated the 2.5 kW reading is taken as – 2.5 kW (a) Total input power, P = P1 + P2 = 7.5 + (- 2.5) = 5 kW (b) tan = P P 3 1 2 = P1 P2 7.5 ( 2.5) 3 = 7.5 ( 2.5) 10 3 2 3 5 Angle = tan 1 (2 3 ) = 73.90o Power factor = cos = cos 73.90o = 0.277 4. Three similar coils, each having a resistance of 4.0 and an inductive reactance of 3.46 are connected (a) in star and (b) in delta across a 400 V, 3-phase supply. Calculate for each connection the readings on each of two wattmeters connected to measure the power by the twowattmeter method. (a) Star connection: VL 3 VP Hence, VP IL IP and VL 400 = 230.94 V 3 3 Phase impedance, ZP 4.02 3.462 = 5.289 Phase current, I P VP 230.94 = 43.664 A ZP 5.289 Total power, P = 3I P 2 R P 3 43.664 4.0 = 22.879 kW 2 If the wattmeter readings are P1 and P2 then: © John Bird Published by Taylor and Francis 262 P1 + P2 = 22.879 (1) X 3.46 Phase angle, = tan 1 L tan 1 40.86 4 R and tan 40 86 = P P P P 3 1 2 3 1 2 from (1) 22.879 P1 P2 22.879 tan 40.86 3 from which, P1 - P2 = i.e. P1 - P2 = 11.426 (2) 2 P1 = 22.879 + 11.426 = 34.305 Adding equations (1) and (2) gives: 34.305 = 17.15 kW 2 and P1 = Substituting in (1) gives: P2 = 22.879 – 17.15 = 5.73 kW Hence, in star, the wattmeter readings are 17.15 kW and 5.73 kW (b) Delta connection: VL VP Phase current, I P IL 3 IP and VP 400 = 75.629 A ZP 5.289 Total power, P = 3I P 2 R P 3 75.629 4.0 = 68.637 kW 2 P1 + P2 = 68.637 Hence tan 40 86 = (3) P P P P 3 1 2 3 1 2 from (3) 68.637 P1 P2 68.637 tan 40.86 3 from which, P1 - P2 = i.e. P1 - P2 = 34.278 Adding equations (3) and (4) gives: (4) 2 P1 = 68.637 + 34.278 = 102.915 102.915 = 51.46 kW 2 and P1 = Substituting in (3) gives: P2 = 68.637 – 51.46 = 17.18 kW © John Bird Published by Taylor and Francis 263 Hence, in delta, the wattmeter readings are 51.46 kW and 17.18 kW 5. A 3-phase, star-connected alternator supplies a delta connected load, each phase of which has a resistance of 15 and inductive reactance 20 . If the line voltage is 400 V, calculate (a) the current supplied by the alternator and (b) the output power and kVA rating of the alternator, neglecting any losses in the line between the alternator and the load. A circuit diagram of the alternator and load is shown below. Vp (a) Considering the load: Phase current, Ip = Zp Vp = VL for a delta connection, hence, Vp = 400 V Phase impedance, Zp = Hence, Ip = Vp Zp = R 2p X 2L = 152 202 = 25 400 = 16 A 25 For a delta-connection, line current, IL = 3 Ip = 3 (16) = 27.71 A Hence, 27.71 A is the current supplied by the alternator (b) Alternator output power is equal to the power dissipated by the load i.e. P = 3 VL IL cos , where cos = Rp Zp = 15 = 0.6 25 Hence, P = 3 (400)(27.71)(0.6) = 11518.8 W = 11.52 kW © John Bird Published by Taylor and Francis 264 Alternator output kVA, S = 3 VL IL = 3 (400)(27.71) = 19198 VA = 19.20 kVA 6. Each phase of a delta-connected load comprises a resistance of 40 and a 40 F capacitor in series. Determine, when connected to a 415 V, 50 Hz, 3-phase supply (a) the phase current, (b) the line current, (c) the total power dissipated, and (d) the kVA rating of the load. (a) Capacitive reactance, X C 1 1 79.58 2 f C 2 50 40 106 Phase impedance, ZP R P 2 XC2 402 79.582 89.067 Phase current, I P VP VL 415 = 4.66 A ZP ZP 89.067 (since VP VL in delta) (b) Line current, IL 3 IP 3 (4.66) = 8.07 A (c) From the impedance triangle, cos = Hence, total power dissipated, P = (d) The kVA rating of the load, S = RP 40 = 0.449 ZP 89.067 3 VL IL cos 3 4158.07 0.449 = 2.605 kW 3 VL IL 3 4158.07 = 5.80 kVA © John Bird Published by Taylor and Francis 265