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Question: Q5. a) A 6 kV RMS three-phase power supply is applied to a balanced Y-connected three-phase load consisting of three identical (same) impedances of 48236.870 0. Taking the phase to neutral voltage Van as a reference, calculate; (i) The phasor currents in each line. (ii) The total active and reactive power supplied to the load. b) Repeat the problem with the same three-phase impedances arranged in a Delta connection. Take Vab as a reference. -Free Course Hero Question Answer.

Question Description: Q5. a) A 6 kV RMS three-phase power supply is applied to a balanced Y-connected three-phase load consisting of three identical (same) impedances of 48236.870 0. Taking the phase to neutral voltage Van as a reference, calculate; (i) The phasor currents in each line. (ii) The total active and reactive power supplied to the load. b) Repeat the problem with the same three-phase impedances arranged in a Delta connection. Take Vab as a reference.

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a) Y- connected 3- $ Balanced Load a Ia a Ia 48 36. 872 Supply 6 K V $8 36. 87 – 12 48 36 . 87 52 C Ic Ic b b Ib Given data; V L = 6 KV coms ) Vph = 6x 10 = 3464. 10 V Van = Vph = 3464. 10 V By taking Van as reference: Van = 3464. 10 109 V Von = 3464. 10 1 – 120 0 V Ven = 34 64 . 10 | + 120 1 0+ 3464. 10 – 240 V 9 ). phasor currents: Ia = Van = 3 464. 10 0 48 36- 87 = 72.16 – 36-87A Ib – = 3 464 . 10 – 120 72. 16 -156- 87 A 48 36.87 Ic = yen = 3464.10 |120 72. 16| 83. 13 A 48 36 . 87 = 72.16 – 276. 87 A

1i ). Total active & Reactive power Supplied to Load 8 = 3 Van Ian S = 3 X 3 4 64. 10 10 X 72. 16 |+ 36. 87 S = 59 9. 92 + j 4 49. 94 KVA P = 59 9 . 92 Kul Q = 449 . 94 KVAR D). A – Connected 3- Balanced Load a Ia a Ias 6 KY omg 48 36. 87 48 36. 87 Ica A C Ibc b Ic C 48 36. 87 6 Ib " . Vph = VL = 6000 V . Vab = 6000y By taking Vab as Reference: Vab = 6000 10 v Vbe = 60001- 12. V V/ ca = 6000 -240 v


Image transcription text

i ) phasor currents Iab = Vab 125 -36 87 A 48 36. 87 Ibe = Vbc 600. |-120 125 – 156. 87′ A 2 48 36.81 Ica = Vca 6000 -240 = 125 -276. 87 A 48 36.87 From above figure . Ia = Tab- Ica = 125 -36.87- 125-276.87 Ia = 216. 50 / -66. 87 A Similarly. Ib = Ibc – Iab – 216.50 1-186. 87 A Ic = Ica – Ibe = 216. 50 +93.13. A ii). Total active & Reactive power supplies to Load. 8 = 3 * Vab + Iab 8 = 3 X 6000 X 125 |+36.87 8 = 179 9 . 99 + J 1350.00 KVA P= 179 9. 91 EW CO = 1350. 00 KVAR

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