Examples Examples

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:

Field Current (A) 10 20 30 40 Terminal Volts (OC) 1200 2100 2830 3460

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.

Volts (OC) SC

Current

13.2 26.0

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

Current (A)

90 135 180 225

Terminal 9800 13000 14900 15800

a) Calculate the regulation at a pf of 0.8 lagging.

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

Terminal Volts (OC)

9800 13000 14900 15800 SC Current 195 291

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:

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

Field Current (A) 2 4 7 9 12 15 20 22 24 25 Terminal Volts 155 287 395 440 475 530 555 560 610 650

lagging and 0.8 p.f. leading, using synchronous impedance method. Show

also the vector diagram.

Solution:

I_L= 30 kVA /(√3) (650) = 26.6469355 A Z_S= [ 650 / (√3) ] / 26.6469355 = 14.08333333 A Ra=0; X_S= Z_S IX_S= 375.2776749 V E_ph = V_ph + I_L ( R_a + j X_L) ; V_ph + IX_S< 53.13010235 E_ph = 622.7425899 < 28.8224976 V VR% = 622.7425899 – (555/ √3) (555/ √3) = 94.34627131 %

Solution: I_L= 30 kVA /(√3) (550) = 31.49183286 A V_ph= 550 / √3 = 317.5426481 V IRa= 4.72377493 V/phase E_ph = V_ph + I_L ( R_a + j X_L ) ; V_ph + IRa < 36.86989765 E_ph = 321.3341678 < 0.50537273 V E_LL= 556.5671049 V If = 20<90.50537273 + 7<216.8698976 A = 16.82207726 < 110.0830856 A

:

Solution

(

1

19

)

18

8

144

to

slots

P

S

=

=













°

=

2

10

x

3

cos

p

k

β

=

°

= 10

18

180

1. The following information is given in connection with an alternator: slots = 144; poles = 8; rpm = 900; conductors/slot = 6; flux per pole = 1.8 x 106 _{maxwells; coil span = slots}

1 to 16; winding connection = star. Calculate: (a) the voltage per phase; (b) the voltage between terminals. 20 PTS

( )

conductors

Z

_T

=

6

144

=

864

288

864 =

=

ph

Z

1 pt

965925826

.

0

=

p

k

6

3

8

144

=

=

phase

poles

slots

m

























°

=

2

10

sin

6

2

10

x

6

sin

d

k

₌

₀

_.

_95614277

d

k

288

3

=

=

ph

Z

phase

turns

T

144

/

2

288 =

=

( )

( )

(

6

)

( )( )

(

8

)

10

x

1

60

144

10

x

8

.

1

44

.

4

−

=

_{d p} g

k

k

E

_φ

V

E

_gφ

=

637

.

7283753

(

)

V

E

_{g LL}

577947

.

1104

7283753

.

637

3

=

=

1 pt 1 pt 1 pt 1 pt 8 pts 8 pts

2. In a 3 phase, start connected alternator, there are 2 coil per slot and 16 turns per coil. Armature has 288 slots on its periphery. When driven at 250 rpm it produces 6600 V between the lines at 50 Hz. The pitch of the coil is 2 slots less than the full pitch. Calculate the flux per pole, total number of conductors and turns per phase. 20 PTS

:

Solution

V

E

_LL

=

6600

V

V

E

_g

3810

.

511777

3

6600

=

=

φ

(

1

13

)

12

24

288

to

slots

P

S

=

=

β

=

°

= 15

180

₃

4

24

288

=

=

phase

poles

slots

m

( )

poles

N

f

P

=

120

=

24

β

=

°

= 15

12

180













°

=

2

15

x

2

cos

p

k

965925826

.

0

=

p

k

1 pt

























°

=

2

15

sin

4

2

15

x

4

sin

d

k

₌

₀

_.

_957662196

d

k

1 pt

3 phase

slot

cond

slots

turns

coil

turns

x

slot

coils

16

32

64

.

2

=

=

(

)

conductors

Z

_T

=

64

288

=

18432

6144

3

18432 =

=

ph

Z

phase

turns

T

3072

/

2

6144 =

=

6 pts 6 pts

:

2

.

Solution

no

of

on

continuati

( )

( )

( )(

3072

)( )

50

44

.

4

511777

.

3810

=

k

_d

k

_p

φ

mWb

040223388

.

6

=

φ

_{6 pts}

3. A 3-phase, 10-pole alternator has 90 slots, each containing 12 conductors. If the speed is 600 r.p.m. and the flux per pole is 0.1 Wb, calculate the line e.m.f. and voltage per phase when the phases are (i) star connected (ii) delta connected. Assume the winding factor to be 0.96 and the flux sinusoidally distributed.

20 PTS

:

Solution

(

)

Hz

f

50

120

600

10

=

=

( )

conductors

Z

_T

=

90

12

=

1080

360

3

1080 =

=

ph

Z

360

1 pt

phase

turns

T

180

/

2

360 =

=

Wye:

( )( )( )( )( )

1

0

.

1

50

180

44

.

4

_d g

k

E

_φ

=

V

E

_gφ

=

3836

.

16

6 pts 1 pt

(

)

V

E

_{g LL}

424026

.

6644

16

.

3836

3

=

=

6 pts Delta:

( )( )( )( )( )

1

0

.

1

50

180

44

.

4

_d g

k

E

_φ

=

LL g

V

E

E

_φ

=

3836

.

16

=

6 pts

1. Three non-inductive resistances, each of 100 Ω, are connected in star to 3-phase, 440 < 0O _{V supply. Three equal choking coils}

each of reactance 100 Ω are also connected in delta to the same supply.

Calculate:

a) line current of each 3-phase load (in polar form)

b) the total line current (in polar form) c) power factor of the system

For Wye load:

:

Solution

100

30

3

440

1

−

∠

=

=

I

_φ

I

_L

A

I

I

_L1

=

_φ

=

2

.

540341184

∠

−

30

For Delta load: a) c) pf of the system

100

0

440

j

I

_φ

=

∠

A

I

_φ

=

4

.

4

∠

−

90

A

I

_L2

=

7

.

621023553

∠

−

120

b) total line current

A

I

I

I

_T

=

_L1

+

_L2

=

8

.

033264177

∠

−

101

.

5650512

( ) ( )

I

W

P

_T

=

_L1 2

100

=

645

.

3333331

c) pf of the system

( ) ( )

I

Vars

Q

_T

=

_L2 2

100

=

1936

VA

U

_T

=

2040

.

723183

)

(

316227765

.

0

lagging

U

P

pf

T T

=

=

2. A symmetrical 3-phase, 3-wire supply with a line voltage of 173 < 0O _{V supplies two balanced}

3-phase loads; one Y-connected with each branch impedance equal to (6 + j8) ohm and the other Δ-connected with each branch impedance equal to (18 + j24) ohm. Calculate:

a) line current taken by each 3-phase load (in polar form)

b) the total line current (in polar form) c) power factor of the entire load circuit d) total real power and apparent power For Wye load:

8

6

30

3

173

1

j

I

I

_L

+

−

∠

=

=

_φ

A

I

I

_L1

=

_φ

=

9

.

9881

∠

−

30

a)

:

Solution

For Delta load:

24

18

j

I

_φ

=

+

A

I

_φ

=

5

.

766666667

∠

−

53

.

13010235

A

I

_L2

=

9

.

988159757

∠

−

83

.

13010235

b) total line current

A

I

I

Solution continuation No.2:

( ) ( )

I

W

P

_L1

=

_L1 2

6

=

598

.

58

VA

U

_L1

=

997

.

6333333

( ) ( )

I

W

P

_L2

=

_L2 2

6

=

598

.

5800001

VA

U

_L2

=

997

.

6333334

W

P

P

P

_T

=

_L1

+

_L2

=

1197

.

16

VA

U

U

U

_T

=

_L1

+

_L2

=

1995

.

266667

)

(

6

.

0

lagging

U

P

pf

T T

=

=

pf

I

V

P

_T

=

3

_{L L}

°

=

31

.

78833062

θ

(

400

)( )(

55

0

.

85

)

3

=

T

P

W

P

_T

=

35628

.

28511

3. A 440-V, 50-Hz induction motor takes a line current of 55 A at a power factor of 0.85 (lagging). Three Δ-connected capacitors are installed to improve the power factor to 0.9 (lagging). Calculate the kVA of the capacitor bank and the capacitance of each capacitor.

power factor of 0.85 (lagging).

T NEW

P

Q

=

tan(

β

)

°

=

25

.

8419327

β

VARS

Q

_NEW

=

17255

.

56604

T OLD

P

Q

=

tan(

θ

)

VARS

Q

_OLD

=

22080

.

42799

NEW OLD

CAP

Q

Q

Q

=

−

Continuation of No. 3 solution:

VARS

Q

_CAP

=

4824

.

861954

VARS

VARS

Q

_CAP

1608

.

287318

3

861954

.

4824

=

=

φ

A

V

VARS

I

_C

3

.

65519745

400

287318

.

1608

=

=

φ

A

V

I

_C

3

.

65519745

400

=

=

φ

Ω

=

=

440

120

.

376501

φ φ C C

I

V

X

( )(

)

F

C

µ

π

50

120

.

376501

26

.

44285915

2

1

=

=