Examples Examples

7/30/2019 Examples Examples

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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:

Field

Current

(A)

10 20 30 40

Terminal 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

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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

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

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

TerminalVolts (OC)

9800 13000 14900 15800

SC Current 195 291


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Example 3: A 3-phase, 800 kVA, 3000 V, 50 Hz alternator gave thefollowing results:

ExcitingCurrent (A)

30 35 40 50 60 65 70 75 77.5 80 85 90 100 110

O.C. volt(line)

_ _ _ 2560 3000

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 loadcurrent on short circuit of the alternator. The armature resistance perphase is 0.27225 . 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.


4/14

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.

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

. . . , .

also the vector diagram. Solution:IL = 30 kVA /(3) (650) = 26.6469355 AZS = [ 650 / (3) ] / 26.6469355 = 14.08333333 A

Ra=0; XS = ZSIXS = 375.2776749 V

Eph = Vph + IL ( Ra + j XL ) ; Vph + IXS < 53.13010235Eph = 622.7425899 < 28.8224976 VVR% = 622.7425899 (555/ 3)

(555/ 3)= 94.34627131 %


5/14

Solution:IL = 30 kVA /(3) (550) = 31.49183286 AVph = 550 / 3 = 317.5426481 V

IRa= 4.72377493 V/phase

Eph = Vph + IL ( Ra + j XL ) ; Vph + IRa < 36.86989765Eph = 321.3341678 < 0.50537273 VELL = 556.5671049 V

If = 20


6/14

:Solution

( )191188

144toslots

P

S ==

=

2

10x3cospk

== 1018

180

1. The following information is given in

connection with an alternator: slots = 144;

poles = 8; rpm = 900; conductors/slot = 6; fluxper 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

( ) conductorsZT 8641446 ==

864==

965925826.0=pk

6

3

8

144

==

phase

poles

slots

m

=

2

10sin6

2

10x6sin

dk 95614277.0=dk

3p

phaseturnsT /1442

288==

( )( )( )( )( )( )86 10x16014410x8.144.4 = pdg kkE VEg 7283753.637=

( )

V

E LLg

577947.1104

7283753.6373

=

=

1 pt

1 pt

1 pt

1 pt

8 pts

8 pts


7/14

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. Whendriven 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

VELL 6600=

VV

Eg 511777.38103

6600==

( )1311224

288toslots

P

S==

180 424

288

==poles

slots

m( )

polesN

fP 24

120==

==12

=

2

15x2cospk

965925826.0=pk 1 pt

=

2

15sin4

2

15x4sin

dk 957662196.0=dk

1 pt

slot

cond

slots

turns

coil

turnsx

slot

coils .6432162==

( ) conductorsZT 1843228864==

61443

18432==phZ

phaseturnsT /30722

6144

==

6 pts

6 pts


8/14

:2. Solutionnoofoncontinuati

( ) ( )( )( )50307244.4511777.3810 pd kk=

mWb040223388.6= 6 pts


9/14

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

( ) Hzf 50120

60010==

( ) conductorsZT 10801290 ==

3603

1080==phZ 1 pt

phaseturnsT /1802 ==

Wye:

( )( )( )( )( )180501.0144.4 dg kE =

VEg 16.3836= 6 pts

1 pt

( )

V

E LLg

424026.6644

16.38363

=

=

6 pts

Delta:

( )( )( )( )( )180501.0144.4 dg kE =

LLg EVE == 16.3836

6 pts


10/14

1. Three non-inductive resistances, each of100 , are connected in star to 3-phase, 440

< 0O V supply. Three equal choking coilseach of reactance 100 are also connectedin 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

303

440

1

== IIL

AIIL 30540341184.21 ==

For Delta load:

a)

100

0440

jI =

AI 904.4 =

AIL 120621023553.72 =

b) total line current

AIII LLT 5650512.101033264177.821 =+=

( ) ( ) WIP LT 3333331.64510021 ==

( ) ( ) VarsIQ LT 19361002

2 ==

VAUT 723183.2040=

)(316227765.0 laggingU

Ppf

T

T ==


11/14

2. A symmetrical 3-phase, 3-wire supply with aline voltage of 173 < 0OV supplies two balanced

3-phase loads; one Y-connected with eachbranch impedance equal to (6 + j8) ohm andthe other -connected with each branchimpedance 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 circuitd) total real power and apparent power

For Wye load:

86

303

173

1j

IIL+

==

AIIL 309881.91 ==

a)

:Solution

For Delta load:

2418 jI +=

AI 13010235.53766666667.5 =

AIL 13010235.83988159757.92 =

b) total line current

AIII LLT 13010235.8397631931.1921 =+=


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Solution continuation No.2:

( ) ( ) WIP LL 58.5986211 ==

VAUL 6333333.9971 =

( ) ( ) WIP LL 5800001.59862

22 ==

VAUL 6333334.9972 =

WPPP LLT 16.119721 =+=

VAUUU LLT 266667.199521 =+=

)(6.0 laggingU

Ppf

T

T ==


13/14

pfIVP LLT 3=

= 78833062.31

( )( )( )85.0554003=TP

WPT 28511.35628=

3. A 440-V, 50-Hz induction motor takes a linecurrent of 55 A at a power factor of 0.85

(lagging). Three -connected capacitors areinstalled to improve the power factor to 0.9(lagging). Calculate the kVA of the capacitorbank and the capacitance of each capacitor.

power factor of 0.85 (lagging).

TNEW PQ )tan(=

= 8419327.25

VARSQNEW 56604.17255=

TOLD PQ )tan(=

VARSQOLD 42799.22080=

power factor of 0.9 (lagging).


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NEWOLDCAP QQQ =

Continuation of No. 3 solution:

VARSQCAP 861954.4824=

VARS

VARS

QCAP 287318.16083

861954.4824

==

VARS65519745.3

287318.1608==

== 376501.120440

C

CI

VX

( )( )FC

44285915.26

376501.120502

1==

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