PARALLEL-CONNECTED NETWORKS in Software

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PARALLEL-CONNECTED NETWORKS
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The complex power S is also useful in analyzing practical networks, for example, the collection of households drawing on the same power lines. Referring to Fig. 10-8, ST Veff I Veff I I I eff 1;eff 2;eff n;eff S1 S2 Sn from which PT P1 P2 Pn QT Q1 Q2 Qn q ST P2 Q2 T T pf T PT ST
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These results (which also hold for series-connected networks) imply that the power triangle for the network may be obtained by joining the power triangles for the branches vertex to vertex. In the example shown in Fig. 10-9, n 3, with branches 1 and 3 assumed inductive and branch 2 capacitive. In such diagrams, some of the triangles may degenerate into straight-line segments if the corresponding R or X is zero. If the power data for the individual branches are not important, the network may be replaced by its equivalent admittance, and this used directly to compute ST .
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EXAMPLE 10.12 Three loads are connected in parallel to a 6-kVeff ac line, as shown in Fig. 10-8. P1 10 kW; pf 1 1; P2 20 kW; pf2 0:5 lagging; Given
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P3 15 kW; pf3 0:6 lagging
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Find PT , QT , ST , pfT , and the current Ieff .
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AC POWER
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Fig. 10-8 We rst nd the reactive power for each load: pf1 cos 1 1 pf2 cos 2 0:5 pf3 cos 3 0:6 Then PT , QT , ST , and pfT , are PT P1 P2 P3 10 20 15 45 kW QT Q1 Q2 Q3 0 34:6 20 54:6 kvar q p ST P2 Q2 452 54:62 70:75 kVA pf T PT =ST 0:64 cos ;  50:58 lagging Ieff S=Veff 70:75 kVA = 6 kV 11:8 A Ieff 11:8 50:58 A The current could also be found from I I1 I2 I3 . However, this approach is more time-consuming. tan 1 0 tan 2 1:73 tan 3 1:33 Q1 P1 tan 1 0 kvar Q2 P2 tan 2 34:6 kvar Q3 P3 tan 3 20 kvar
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Fig. 10-9
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POWER FACTOR IMPROVEMENT
Electrical service to industrial customers is three-phase, as opposed to the single-phase power supplied to residential and small commercial customers. While metering and billing practices vary among the utilities, the large consumers will always nd it advantageous to reduce the quadrature component of their power triangle; this is called improving the power factor. Industrial systems generally have an overall inductive component because of the large number of motors. Each individual load tends to be either pure resistance, with unity power factor, or resistance and inductive reactance, with a lagging power factor. All of the loads are parallel-connected, and the equivalent impedance results in a lagging current and a corresponding inductive quadrature power Q. To improve the power factor, capacitors, in three-phase banks, are connected to the system either on the primary or secondary side of the main transformer, such that the combination of the plant load and the capacitor banks presents a load to the serving utility which is nearer to unit power factor.
AC POWER
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EXAMPLE 10.13 How much capacitive Q must be provided by the capacitor bank in Fig. 10-10 to improve the power factor to 0.95 lagging
Fig. 10-10 Before addition of the capacitor bank, pf cos 258C 0:906 lagging, and I1 240 08 68:6 258 A 3:5 258    240 68:6 p 258 8232 258 7461 j3479 S Veff I p 08 eff 2 2 Then (see Fig. 10-11),
After the improvement, the triangle has the same P, but its angle is cos 1 0:95 18:198. 3479 Qc tan 18:198 7461 or Qc 1027 var (capacitive)
The new value of apparent power is S 0 7854 VA, as compared to the original S 8232 VA. The decrease, 378 VA, amounts to 4.6 percent.
Fig. 10-11
The transformers, the distribution systems, and the utility company alternators are all rated in kVA or MVA. Consequently, an improvement in the power factor, with its corresponding reduction in kVA, releases some of this generation and transmission capability so that it can be used to serve other customers. This is the reason behind the rate structures which, in one way or another, make it more costly for an industrial customer to operate with a lower power factor. An economic study comparing the cost of the capacitor bank to the savings realized is frequently made. The results of such a study will show whether the improvement should be made and also what nal power factor should be attained.
EXAMPLE 10.14 A load of P 1000 kW with pf 0:5 lagging is fed by a 5-kV source. A capacitor is added in parallel such that the power factor is improved to 0.8. Find the reduction in current drawn from the generator. Before improvement: P 1000 kW; cos  0:5; S P= cos  2000 kVA; I 400 A After improvement: P 1000 kW; cos  0:8; S P= cos  1250 kVA; I 250 A
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