LESSON 23. Polyphase systems – delta in generation - three phase load connections

Delta connection

The three coils of Fig. 9.5 can be con­nected as shown in Fig. 9.10 (a), the diagram being simplified in Fig. 9.10 (b). The end of each coil, which was con­nected to the neutral in star (Y) earlier, is now connected to the outer end of the next coil, as shown in Fig. 9.10  (a). As points o and a are now connected directly together, the o's are now superfluous and are dropped. Fig. 9.11 (a) shows vectorially the three voltages Eab, Ebc and Eca, acting from a to b, b to c, and c to a, respectively.  At first sight Fig. 9.10 looks like a short circuit, the three coils, each containing a source of voltage, being short circuited on themselves. The actual conditions existing in this closed cir­cuit may be demonstrated by the use of the subscript notation. Assume that the coil bc is broken at c', Fig. 9.12 (a). The voltage Ebc = Eba + Eac. The vector sum of these two voltages, shown in Fig. 9.12  (b), lies along voltage Ebc’, and is equal to it.    Therefore, the voltage Ecc' = 0 and points c and c' can be connected without any resulting flow of current. This is the same condition which exists when two direct current generators having equal voltages are connected in parallel. No current flows between the two if the proper polarity is observed.

Figure 9.10

Fig. 9. 10 Delta connection of alternator coils.

Figure 9.11

Fig. 9.11 Relation of coil voltages and currents in a delta system at unity power factor.

Figure 9.12

Fig. 9.12 Showing that Delta connection is not a short circuit

The coil currents of Fig. 9.10 are shown in Fig. 9.11 in phase with their respective voltages, balanced conditions being assumed. The line current

Iac’ = Iba + Ica

This addition is made vectorially in Fig. 9.11 (a), giving Iac’ 30° from Eca. It will be observed that Iac ' is Ö3 times the coil current. Line currents Ibb’ and Icc’ may be found in a similar manner, with the result shown in Fig. 9.11 (b). Therefore, in the delta system there is a phase difference of 30° between the line currents and the line voltages at unity power factor, just as in the star (Y) system. It is obvious that the line voltage is equal to the coil voltage in a delta system. Moreover, the sum of the three voltages acting around the delta must be zero by Kirchhoff's second law.

In a balanced delta system, the line voltage is equal to the coil voltage, but the line current is \/3 times the coil current.

Figure 9.13 shows three lamp loads, each requiring 10 amp at 115 volts. They are first connected in star (Y) and then in delta. In order to supply the proper voltage in each case, there are 199 volts across lines in the Y system and 115 volts in the delta system. There are 10 amp. per line in the Y system and 17.3 amp per line in the delta system. The power supplied is the same in each system.

Power in Delta system The total power in a delta system is

P  =  3 Ecoil Icoil cosθcoil

Figure 9.13

Fig. 9.13 Lamp loads in Y and in delta

This power is equal to that in the line, as there is no interven­ing loss.   Also, the line current

Iline  = √3 Icoil and Eline =  Ecoil

Hence, substituting in the above equation P  =   √3  Eline Iline  cosθload

This equation is the same as that developed for the star (Y) system. This should be so, for the relations in a three phase line are the same whether the power originates in a delta  or in a Y connected generator. The power factor of the delta system is the same as that for a Y system.

P F.   = P / √3 EI, where P is the total power of the system, and E and I are the line voltage and line current respectively.

The denominator, √3 EI, gives the volt amperes of the three phase system.

Last modified: Tuesday, 1 October 2013, 7:18 AM