By Anthony J. Pansini
This paintings describes undefined, mechanical and financial issues linked to the winning making plans, layout, development, upkeep and operation of electric transmission and distribution of strength.
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Additional resources for Power Transmission And Distribution, 2nd Edition
Single-Phase Transformer Connections Distribution System Electrical Design Figure 2-20. Connections for Two-Phase Transformers 49 50 Power Transmission and Distribution Y- Y Connection With the Y-Y connection, the secondary circuit is in phase with the primary circuit and the ratio of primary to secondary voltage is the same as the ratio of turns in the phases. The secondary may be either threewire or four-wire as desired. Unbalanced currents on the secondary circuit are transmitted through the transformers to the primary unchanged in phase relation, although reduced in magnitude according to the ratio of the windings.
In the field of reliability, emergency connections with other circuits, loop circuits (open and closed), primary network, and throwover arrangements for individual consumers may be considered. Permissible voltage drop is affected by regulators (both at substations and on lines), capacitors, boosters. The choice of primary voltage is dependent not only on the availability of standard methods and equipment, but also on operating and maintenance procedures, and its impact on other utilities. And all are subject to further modification to provide for load growth and environmental requirements, and above all, for the safety of workers and the general public.
The structures and facilities associated with the secondary conductors are also assumed to be the same for any size of the secondary conductor and will be neglected in the determination. In the determination, the length of secondary is divided into two parts, one each way from the transformer, of length l/2 and load D . 2000 For uniformly distributed load, the loss is equal to what it would be if the load were all concentrated at one-third the distance from the transformer. Hence D × 1000 D 2000 I= = 2E × pf E × pf 1× 3 2 R = A and where I D l E pf Rl r A = load current = = = = = load density in kW per 1000 feet total length of secondary in feet circuit voltage power factor of load resistance of circuit of length twice one-third of the total length l/2 = resistivity of conductor in ohms per mil foot = cross sectional area of conductor in circular mils Distribution System Electrical Design 23 2 D2 r × 4E 2pf 2 3A I 2R = then Annual cost of losses =2 I 2R × 365tC 1000 2 = 2 × 365 and rD 2 tC 3 × 4 × 1000AE 2pf 2 t = equivalent hours (see above) C = cost of conductor loss per kW hour kc = annual carrying charges of three conductors in place, including the neutral The factor 2 includes the total losses on the secondary on both sides of the transformer.