# Audel electrical course for apprentices and journeymen - Rosenberg P.

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P = IE23 = 17.32 X 1000 X 1.732 = 30,000 VA (6)

Figure 24-4 Relative voltages and currents delivered by a delta-connected alternator.

Power in Polyphase Circuits 267

If the PF is less than 100%, then P in watts = IE cos f 23

Note that P is equal to the total power in the system and is in VA, or apparent power. To get kilovolt-amperes, kVA = VA/1000. At 100% PF, P would be both volt-amperes and watts, but with other than 100% PF, P in VA has to be multiplied by cos f to get watts.

Figure 24-5 will be used to compare power loss in a three-phase circuit as compared to a one-phase two-wire circuit. The resistance (O) of each line times I2 (10 ) equals 200 VA lost in each line, or 200 X 3 = 600 VA lost in all three lines.

Total P = IE23 = 10 X 1000 X 1.732 = 17,320 VA

Figure 24-5 Power lost in three-phase transmission.

To carry this same power in a two-wire, one-phase line, I = P/E = 17,320/1000 = 17.32 A per line. The total loss in the three-phase system was 600 VA, so for the same power in the one-phase system, 600/2 = 300 VA lost per conductor. Now R = P/1 so R = 300/17.322 = 1 ohm. This means that the one-phase lines would have twice the cross-section of the three-phase lines, so each conductor will weigh twice as much for the one-phase line as for the three-phase lines. See Figure 24-6.

Figure 24-6 Copper required for a one-phase system.

268 Chapter 24

Economy and Phases

The following are comparisons of weights of copper required for single-phase and three-phase power systems for carrying equal amounts of power the same distances and with the same losses. Two-wire, one-phase will be used as the base for comparison. The amperes and watts or volt-amperes won’t be called out, as they are all equal. The base voltage will be 100 volts. See Figure 24-7.

Figure 24-7 Total copper required for a one-phase 100 LBS transmission.

TOTAL

It was just shown that the conductors of a three-phase system were half the cross-section of those for a one-phase system for the same power delivered; Figure 24-8 illustrates the comparison of a three-phase circuit to the one-phase circuit.

25 LBS

^7 17 IV 17 3 V 25 LBS

17 (V 25 LBS

Figure 24-8 Total copper required for a three-phase, three-wire circuit.

The following will illustrate the economy of a three-phase, four-wire system. The economy of copper in a transmission line varies as the square of the applied voltage, as we saw in a previous chapter. In Figure 24-9, there are still 100 volts between any phase and the neutral and 173 volts between phases. So 1732 = 29,929 and 1002 = 10,000, or a 3-to-1 ratio. So the conductors will weigh as shown in Figure 24-9.

Power in Polyphase Circuits 269

8 1/3 LBS

^ 173 V 17; nnnmnnn V 10 DV 8 1/3 LBS

T sxT IOC )V 17 17: V V 8 1/3 LBS

i 100 V + 8 1/3 LBS

33.3 LBS ► TOTAL

Figure 24-9 Total copper required for a three-phase, four-wire system.

Neutral Current in Three-Phase, Four-Wire Systems

On three-phase, four-wire systems used for discharge lighting, the neutral current with two phases will be considered as being the same value as the phase current. This is due to the third harmonic of the line frequency. (Harmonics are multiples of the fundamental frequency.) This is illustrated in Figure 24-10A, where only the second and third harmonics are shown.

Figure 24-10B shows the third harmonic with the three fundamental phases. By studying this illustration, it is plain that the third harmonic of each phase adds to the third harmonic of the other phases. This is the reason that the NEC won’t permit the derating of the neutral of a three-phase, four-wire circuit when discharge lighting is supplied.

For three-phase, four-wire circuits with other than discharge lighting, the following vector methods of calculation of the neutral current will be found to be useful.

Problem

Using Figure 24-11, draw vectors of the currents as in Figure 24-12, scaling the lines to represent the amperes per phase, and the angles of phase displacement: OA = 10 amperes in phase A, OB = 10 amperes in phase B, and OC = 10 amperes in phase C. Draw the line ON, which comes out 10 amperes and diametrically opposite

270 Chapter 24

Oo 60° 1200180° 240° 300° 360° Figure 24-10 Harmonics of the phase current.

to line OC. This is 10 amperes in the neutral for phases A and B, but OC is diametrically opposite to ON, so OC and ON will cancel and the resultant neutral current for a balanced system will be zero.

For a three-phase, four-wire system with two phases balanced and one phase current variable, as in Figure 24-13, draw a vector as in Figure 24-14 except OC will be marked in various current

Power in Polyphase Circuits 271

Figure 24-11 Three phases with balanced loads.

Figure 24-12 Vectors of currents in Figure 24-11.

B

lengths in proportion to the currents for OC as shown in Figure

24-14. For example, suppose OC takes the values 3, 6, and 10 amperes.

OC2= 3 amperes so neutral current = 7 amperes

OC1= 6 amperes so neutral current = 4 amperes

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