Examples Examples

Field 10 20 30 Current (A) Terminal 1200 2100 2830 Volts (OC) SC Current

40

3460

13.2 26.0

Calculate the synchronous impedance and the synchronous reactance per phase for this machine, using the highest point given on the saturation or open circuit voltage curve to obtain the values.

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Example 1 . A three-phase, Y-connected generator is rated at 100 kVa, 60 cycles, 2300 volts. The effective resistance of the armature is 1.5 ohms per leg. The test data are given below:

Example 2 . A three-phase, slow speed, Y-connected alternator is rated at 5000 kVA and 13,200 volts. The resistance of the armature between terminals is 0.192 ohm at 75 C. The effective resistance is 1.6 times the dc-value at 75 C.. The test data on this machine is given below. Field 90 Current (A)

135

Terminal Volts (OC)

9800 13000

SC Current

195

180

225

14900

15800

291

a) Calculate the regulation at a pf of 0.8 lagging. b) Calculate the regulation for a load of unity pf.

Example 3: A 3-phase, 800 kVA, 3000 V, 50 Hz alternator gave the following results: Exciting Current (A)

30

35

40

50

60

65

70

75

77.5

80

85

90

100

110

O.C. volt (line)

_

_

_

2560

300 0

3250

3300

3450

3500

3600

3700

3800

3960

4050

S.C. current

140

150

170

190

_

_

_

_

_

_

_

_

_

_

a) A field current of 110 A is found necessary to circulate a full load current on short circuit of the alternator. The armature resistance per phase is 0.27225 Ω. Calculate the voltage regulation at 0.8 p.f. lagging and 0.9 p.f leading, using synchronous impedance method. Show also the vector diagram.

Example 4 . A 30-kVA, Y-connected alternator rated at 555 volts at 50 Hz has the open circuit characteristics given by the following data: Field Current (A) Terminal Volts

2

4

7

9

12

15

20

22

24

25

155

287

395

440

475

530

555

560

610

650

A field current of 25 A is found necessary to circulate a full load current on short circuit of the alternator. Calculate the voltage regulation at 0.8 p.f. lagging and 0.8 p.f. leading, using synchronous impedance method. Show Solution: also the vector diagram. IL = 30 kVA /(√3) (650) = 26.6469355 A ZS = [ 650 / (√3) ] / 26.6469355 = 14.08333333 A

Ra=0; XS = ZS IXS = 375.2776749 V Eph = Vph + IL ( Ra + j XL ) ; Vph + IXS < 53.13010235 Eph = 622.7425899 < 28.8224976 V VR% = 622.7425899 – (555/ √3) (555/ √3) = 94.34627131 %

Solution: IL = 30 kVA /(√3) (550) = 31.49183286 A Vph = 550 / √3 = 317.5426481 V IRa= 4.72377493 V/phase Eph = Vph + IL ( Ra + j XL ) ; Vph + IRa < 36.86989765 Eph = 321.3341678 < 0.50537273 V ELL = 556.5671049 V If = 20