Engineering Principles, Electrical Networks - Assignment Example

Engineering Principals, Electrical Networks, Three-Phase Supplies Facility Question Paper Three-phase supplies Question 1 Calculateline and phase currents Given: 3 x 80 resistors star connected 415 v three phase supply 50 Hz frequency IL= V/R IL = 415 / 80 IL = 5.18 A In a star connected load IL = Iph Iph = 5.18 A 1.2 Total power dissipated by load P = 3 x VL x IL x Cos Taking Power Factor (Cos ) as 0.85, P will be: P = 3 x 415 x 5.18 x 0.85 P = 3164.88 Watt 1.3 The power dissipated if, due to fault, one phase of the load went open Circuit P = 2 x VL/3 x IL cos P = 2 x 414/3 x 5.18 x 0.85 P = 2109.

92 Watt Question 2 Given: Delta connected load Power Factor 0.8 lagging P = 4080 Watt 425 V supply 50 Hz frequency 2.1 Calculate resistance and reactance of each phase In Delta Connected load Vph = VL This P = 3 x VL x IL x Cos 4080 = 3 x425 x IL x 0.8 Iph = 4080/(3 x425 x IL x 0.8) Iph = 6.93 A R = V*2 / P R = 415*2/4080 R = 44.27 Apparent Power S = P/pf S = 4080 / 0.8 S = 5100 Watt Reactive Power Q = (S*2 - P*2) Q = (5100*2 - 4080*2) Q = 3060 Watt Inductance XL = V*2/Q XL = 425*2/3060 XL = 59.02 2.1 Calculate Line current if same load was connected to star Delta IL = P/V Delta IL = 4080/425 Delta IL = 9.

6 A In Star IL = Iph Star IL = 9.6A Question 3 Given: Balanced 10 kW three phase load 415 V three wire supply 18 A line current 3.1 determine Load Power Factor Power Factor = True Power/Apparent Power = P/Q This: P = V x I Watts P = 415 x 18 P = 7470 Watts S (apparent Power) = 10000 Watt PF = 7470 / 10000 PF = 0.75 Question 4 Given: Star connected induction motor Supply 415 V Frequency 50 Hz Line current 20A Power factor 0.75 4.1 Output power of motor if efficiency is 0,85% P in = 3 x V x I x pf Watts P in = 3 x 415 x 20 x 0.

75 P in = 10782 Watt P out = P in x P out = 10782 x 0.85 P out = 9164.71 Watt 4.2 The value of the capacitor required to improve power factor to 0.9 S = 3 x 415 x 20 S = 14376 Watt P = 10789 Watt 0.75 = Cos 41.4 Q = S sin Q = 14376 x Sin 41.4 Q = 9507 Watt Induction motor will have a lagging power factor this Load KVA = 14376 -41.4 = 10789 - j9507 Resultant KVAr = 14376 -25.8 = 10789 - j11987.58 KVAr of Capacitor = KVAr - Load KVA = (10789 - j11987) - (10789 - j9507) = 0 + j2480 = 2480 90 KVAr Question 5 Given: Star connected generator Vph = 240 V IL = 10A Delta connected balanced load 5.

1 From diagram draw to scale, phasor diagrams for the generator and load to show the relative magnitudes and phase relationship of the line and phase voltage currents. In a Star connected generator VL = 3 x Vph V VL = 3 x 240 VL = 415.69 V IL = Iph IL = 10 A This: VRph = 240 0 VBph = 240 120 VYph = 240 -120 but VL leads Vph by 30 This for Delta connected load: Vph = VL IL = Iph / 3 IL = 10 / 3 IL = 5.77 A But in Load IL leads Iph by 30 Question 6 6.1 One-watt meter method of measuring power in 3 phase system A one-watt power meter can only be used effectively in a balanced load situation. 6.2 Two-watt meter method of measuring power in 3 phase system 6.

2 Two-watt meter method of measuring power in 3 phase system The three-watt method can be used on balanced and unbalanced loads.