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and so the total average power is 1870.9 W. From the results of Problem 11.13, the wattmeter readings are: 208 8:01 p 48:938 817:1 W WB Re VBA eff I eff Re p 608 B 2 2 208 10:21 p 127:598 1052:8 W WC Re VCA eff IC eff Re p 24008 2 2 The total power read by the two wattmeters is 1869.9 W.

0) 2.21)

11.15 A three-phase, three-wire, balanced, -connected load yields wattmeter readings of 1154 W and 557 W. Obtain the load impedance, if the line voltage is 141.4 V.

0.46) 2.21)

4. Draw a DFD for the dental o ce system (see Problem 4.2). See Fig. 8-11.

tan and, using PT p W2 W1 p 577 3 0:577 3 1731 W2 W1 30:08

0.46 and 2.21 (Fig. 4-5). Therefore,

u2/2

p 3VL eff IL eff cos , p 2 3VL eff 3VL eff cos 3 100 2 cos 30:08 VLeff 15:0 Z IPh eff IL eff PT 1154 577

30 e

11.16 A balanced -connected load, with Z 30 308 , is connected to a three-phase, three-wire, 250-V system by conductors having impedances Zc 0:4 j0:3 . Obtain the line-to-line voltage at the load.

30 e

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The single-line equivalent circuit is shown in Fig. 11-34. By voltage division, the voltage across the substitute Y-load is 10 308 250 p 08 137:4 0:338 V VAN 0:4 j0:3 10 308 3 p whence VL 137:4 3 238:0 V.

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Date Check date Patient data Name Check patient data Avail. date Make appointment Generate schedules Appointment file Patient records Confirmation Daily schedule Weekly schedule

30 e

CHAP. 11]

u2/2

0.6636 (d) Required area (Fig. 4-6) (area between z (area between z 0.4738 This is the same as P(0.81 (e) Required area (Fig. 4-7) Z 0.2910 0 and z 0 and z 0.1828 1.94) 0.81)

Fig. 11-34 Considering the magnitudes only, the line voltage at the load, 238.0 V, represents a drop of 12.0 V. The wire size and total length control the resistance in Zc , while the enclosing conduit material (e.g., steel, aluminum, or ber), as well as the length, a ects the inductive reactance.

1.94). 1.28 and z 0) 0.8997 0)

In the following, the voltage-current phasor diagram will not be included in the answer, even though the problem may ask speci cally for one. As a general rule, a phasor diagram should be constructed for every polyphase problem. 11.17 Three impedances of 10:0 53:138 are connected in delta to a three-phase, CBA system with an a ective line voltage 240 V. Obtain the line currents. Ans: IA 58:8 143:138 A; IB 58:8 23:138 A; IC 58:8 96:878 A Three impedances of 4:20 358 are connected in delta to a three-phase, ABC system having VBC 495:0 08 V. Obtain the line currents. Ans: IA 20:41 1258 A; IB 20:41 58 A; IC 20:41 1158 A A three-phase, three-wire system, with an e ective line voltage 100 V, has currents IA 15:41 1608 A IB 15:41 408 A IC 15:41 808 A

1.28).

Fig 8-11.

Fig. 4-6

What is the sequence of the system and what are the impedances, if the connection is delta Ans: CBA; 15:9 708 11.20 A balanced Y-connected load, with impedances 6:0 458 , is connected to a three-phase, four-wire CBA system having e ective line voltage 208 V. Obtain the four line currents. Ans: IA 28:31 1358 A; IB 28:31 158 A; IC 28:31 1058 A; IN 0 A balanced Y-connected load, with impedances 65:0 208 , is connected to a three-phase, three-wire, CBA system, where VAB 678:8 1208 V. Obtain the three line currents. Ans: IA 6:03 708 A; IB 6:03 508 A; IC 6:03 1708 A A balanced -connected load, with Z 9:0 308 , and a balanced Y-connected load, with ZY 5:0 458 , are supplied by the same three-phase, ABC system, with e ective line voltage 480 V. Obtain the line currents, using the single-line equivalent method. Ans: IA 168:9 93:368 A; IB 168:9 26:648 A; IC 168:9 146:648 A A balanced -connected load having impedances 27:0 258 and a balanced Y-connected load having impedances 10:0 308 are supplied by the same three-phase, ABC system, with VCN 169:8 1508 V. Obtain the line currents. Ans: IA 35:8 117:368 A; IB 35:8 2:648 A; IC 35:8 122:648 A

Fig. 4-7

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