EE 255 - Chapter 2c

# EE 255 - Chapter 2c - 45 S OLUTION(a The transformer is...

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Unformatted text preview: 45 S OLUTION (a) The transformer is connected Y-Y, so the primary and secondary phase voltages are the line voltages divided by 3 . The turns ratio of each autotransformer is given by SE 13.8 kV/ 3 13.2 kV/ 3 H C L C V N N V N + = = SE 13.2 13.2 13.8 C C N N N + = SE 13.2 0.6 C N N = Therefore, SE / C N N = 22. (b) The power advantage of this autotransformer is IO SE 22 23 C C C W C C S N N N N S N N + + = = = so 1/22 of the power in each transformer goes through the windings. Since 1/3 of the total power is associated with each phase, the windings in each autotransformer must handle ( )( ) 2000 kVA 30.3 kVA 3 22 W S = = (c) The voltages across each phase of the autotransformer are 13.8/ 3 = 7967 V and 13.2/ 3 = 7621 V. The voltage across the common winding ( C N ) is 7621 kV, and the voltage across the series winding ( SE N ) is 7967 kV – 7621 kV = 346 V. Therefore, a single phase of the autotransformer connected as an ordinary transformer would be rated at 7621/346 V and 30.3 kVA. 2-13. Two phases of a 13.8-kV three-phase distribution line serve a remote rural road (the neutral is also available). A farmer along the road has a 480 V feeder supplying 120 kW at 0.8 PF lagging of three-phase loads, plus 50 kW at 0.9 PF lagging of single-phase loads. The single-phase loads are distributed evenly among the three phases. Assuming that the open-Y—open- Δ connection is used to supply power to his farm, find the voltages and currents in each of the two transformers. Also find the real and reactive powers supplied by each transformer. Assume the transformers are ideal. S OLUTION The farmer’s power system is illustrated below: Load 1 Load 2 V LL,P V LL,S +- I L,P I L,S The loads on each phase are balanced, and the total load is found as: 1 120 kW P = 46 ( ) ( )-1 1 1 tan 120 kW tan cos 0.8 90 kvar Q P θ = = = 2 50 kW P = ( ) ( )-1 2 2 tan 50 kW tan cos 0.9 24.2 kvar Q P θ = = = TOT 170 kW P = TOT 114.2 kvar Q = 1 1 TOT TOT 114.2 kvar PF cos tan cos tan 0.830 lagging 170 kW Q P-- = = = The line current on the secondary side of the transformer bank is ( )( ) TOT 170 kW 246.4 A 3 PF 3 480 V 0.830 LS LS P I V = = = The open-Y—open Δ connection is shown below. From the figure, it is obvious that the secondary voltage across the transformer is 480 V, and the secondary current in each transformer is 246 A. The primary voltages and currents are given by the transformer turns ratios to be 7967 V and 14.8 A, respectively. If the voltage of phase A of the primary side is arbitrarily taken as an angle of 0°, then the voltage of phase B will be at an angle of –120°, and the voltages of phases A and B on the secondary side will be V 480 ° ∠ = AS V and V 120 480 °- ∠ = BS V respectively....
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EE 255 - Chapter 2c - 45 S OLUTION(a The transformer is...

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