Engineering Principles, Electrical Networks Assignment

Engineering Principles, Electrical Networks - Assignment Example Question 1 1.1 Calculate line and phase currents? Given: 3 x 80â„ ¦ 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 â„ ¦ R = 415*2/4080 R = 44.27 â„ ¦ 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 â„ ¦ 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