10 three phase system

8/6/2019 10 Three Phase System

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Three phase system



Flux densityB [T]

l

d

U

U

L

A

M

N

Generator for single phase

Current induces in the coil as thecoil moves in the magnetic field

Current produced at terminal

Note

Induction motor cannot start by itself. This problem is

solved by introducing three

phase system



Instead of using one coil only , three coils are used arranged in one

axis with orientation of 120o each other. The coils are R-R 1 , Y-Y1

and B-B1. The phases are measured in this sequence R-Y-B. I.e Y

lags R by 120o , B lags Y by 120o.



The three winding can be represented by the above circuit. In this

case we have six wires. The emf are represented by eR , eY, eB.

t e [! sinE m

)120-sin(E mQt e [!

)240-sin(EmQt e [!

Load

Fi is

Fi is

Fi is

start

start

start

R

R1

Y

Y1

B

B1

L1

L2

L3

eR

eY

eB



The circuit can be simplified as follows, where R 1 can be

connected to Y and Y1 can be connected to B. In this case

the circuit is reduced to 4 wires.

BYR 1RB eeee !

)]240sin()120sin([sinmQQ

[[[! t t t

QQ 120sintcos120costsint[sin ��! [[[m

E

]240sintcos240costsin QQ�� [[

]t866.0tsin5.0tcos866.0tsin50t[sin [[[[[ kos. E m !

0!Since the total emf is zero, R and B1 can be connected together, thus

we arrive with delta connection system.

Finish

Finish

Finish

start

start

start

Y

Y

e +eY+e

e

eY

e



R

R1

Y

Y1

B

B1

eB

eR

eY

L i e

c o n d u c t o r s

R

R1

Y

Y1

B

B1

PL

PM

PN

�Since the total emf is zero, R and B1 can be

connected together as in Fig.A , thus we arrivewith delta connection system as in Fig. C.

�The direction of the emf can be referred to the

emf waveform as in Fig. B where PL is +ve (R 1-

R), PM is ±ve (Y-Y1) and P N is ±ve (B-B1).

Fig.A

Fig.

Fig. C



�R 1, Y1 and B1 are connected together.

�As the e.m.f generated are assumed in

positive direction , therefore the current

directions are also considered as flowingin the positive direction.

�The current in the common wire (MN)

is equal to the sum of the generated

currents. i.e iR +iY+iB .

�This arrangement is called four ±wirestar-connected system. The point N

refers to star point or neutral point.

R

R1

Y

Y1

B

B1

ener ator Load

N

iR

iB

iY

1

iR +i

Y+i

B

M



tsini [m RI !

)120tsin(i

Q

- I mY [!

)240tsin(iQ

- I B

[!

The instantaneous current in loads L1 , L2 and L3 are

BY R N iiii !

0)]240tsin(

)120tsin(t[sin

!

!Q

Q

[

[[m I

R

R

1

Y Y1

B

B1 L i n e c

o n

d u c t o r s

R

N

iR

iB

iY

iR +i

Y+i

B

L3

L1L

2

B

Y



Three-wire star-connected system with

balanced load

For balanced loads, the fourth wire carries no current , so it can

be dispensed

R

R

1

Y Y1

B

B 1

N

L

L

L

a b c balanced

loadgener ator

coils

5

10

5

0

8.66

2.6

9.7

7.1

8.66



p h a s e a

n d l i n e c u r r e n t i n

a m p e r e

Instantaneous currents¶ waveform for iR , iY and iB in

a balanced three-phase system.



VR

VY

VB-VY

-VB

-VR

VRY

VYB

VBR

)( Y RY R RY V V V V V !!

R

Y

VRY

VY

IR

IB

IY

R

Y

V

VRYV R

VY

IR

IB

IY

R

VY

V

)( BY BY Y B V V V V V !!

)( R B R B B R V V V V V !!

�VRY,VYB and VBR are called line voltage

�VR ,VY and VB are called phase voltage

From Kirchoff voltage law we have

In phasor diagram



VR

VY

VB-VY

-VB

-VR

VRY

VYB

VBR

For balanced load VR , VY and

VB are equaled but out of phase

VR = VP�30r;

VY = VP�-90r;

VB = VP�150r;

P

o

R RY V V V 330cos2 !!

therefore

VRY = VL�30r;

VYB = VL�-90r;

VBR = VL�150r;

P

o

Y Y B V V V 330cos2 !!

P

o

B B R V V V 330cos2 !!



P L

V V 3!

P LI I !

then

and



I3

R

Y

BIB

IR

IY

I2

I1

�IR , I

Yand I

Bare called line current

�I1, I2 and I3 are called phase current

)(III3131R I I !!

)(III 1212YI I !!

)(III2323BI I !!

I2

I1

I3

-I3-I2

-I1

IR

IY

IB

From Kirchoff current law we have

In phasor diagram

VL

V



I2

I1

I3

-I3-I2

-I1

IR

IY

IB

P RI I I )3(30cos2 1 !! Q

P Y I I I )3(30cos2 2 !!Q

P B I I I 330cos2

3!! Q

Since the loads are balanced, the magnitude of currents are

equaled but 120o out of phase. i.e I1 =I2=I3 ,=IP Therefore:-

Thus IR =IY=IB = IL

Where IP is a phase current and IL

is a line current

I1 = VP�30r;

I2 = VP�-90r;

I3

= VP�150r;

IR = IL�30r;

IY = IL�-90r;

IB = IL�150r;



Hence

P LI I 3!

P LV V !



Unbalanced load

In a three-phase four-wire system the line voltage is 400Vand non-inductive loads of 5 k W, 8 k W and 10 k W are

connected between the three conductors and the neutral.

Calculate: (a) the current in each phase

(b) the current in the neutral conductor.



V

V

V

L

P 2303

400

3 !!!

AV

P I

P

B

B7.21230

105 3

!

v

!!

AV

P I

P

Y Y 8.34

230

108 3

!v

!!

AV

P I

P

R

R5.43

230

104

!!!

Voltage to neutral

Current in 8k W resistor





IR

IYIB

IYH

IYV

IBH

IBV

A. I I I o

BY H 3.117.218.34866030cos30cos !!! Q

A I I I I

o

BY RV 0.13)7.218.34(5.05.4360cos60cos !!!Q

I NV

I NH

I N

A. I I I V NH N 2.170.13311 2222!!!

Resolve the current components into horizontal and verticalcomponents.



R

B

Y

400

400

400

IR

IB

IY

I3

I2

I

1R1=100;

R2=20;

=30QF

X2=60;

A delta ±connected load is arranged as in Figure below.

The supply voltage is 400V at 50Hz. Calculate:(a)The phase currents;

(b)The line currents.

A R

V I

RY 4100

400

1

1!!!

I1 is in phase with VRYsince there is only resistor in the branch

(a)



2

2

2

2YXR Z ! 22 6020 !

¹¹

º

¸©©

ª

¨!U

2

21Y

R

Xtan ¹

º

¸©

ª

¨!

20

60tan 1

BR

3

X

VI ! Q90)1030502/(1

4006

�vvvT!

A77.3!

A Z

V I

Y

Y B32.6

6020

400

222 !

!!

o90�

In branch between YB , there are two components , R 2 and X2

In the branch RB , only capacitor in

it , so the XC is -90 out of phase.

'3471Q

!

VRY

IR

I3

I2

VY ¡

V¡ R

I1

71o3 '

o

3 o

-I3

3 o



(b)31R III !

2

331

2

1

2

cos2I I I I I

R ! 3.5677.330cos77.30.420.4

222 !! o

R I

A I R

5.7!

12Y III !

2

121

2

2

2 cos2 I I I I I Y !

5.1050.4'3411cos32.60.4232.6222!!

o

Y I

A I Y

3.10!

I1

-I1I

2

IY

120o

71o 3 '

0o

U

U = 71o 34¶ -60o= 11o 34¶

U=3

0

o



2

223

2

3

2cos2 I I I I I B !

5.1877

.3

'26138cos

32.6

77

.3

232.6

222

!!

o

B

I

A I B

3.4!

U = 180-30o-11o 34¶ = 138o 34¶

23B III !

U

-I2

I2

I3

I2

71o 3 '0o

30o11o 3 '



Power in three phase

Active power per phase = I P V P x power factor Total active power= 3V

P I P

x power factor

N cos3 P P I V P !

If IL andVLare rms values for line current and line voltage

respectively. Then for delta (() connection: VP = VL

and IP= IL/�3. therefore:

cos3 L L I V P !

For star connection (9) : VP = VL/�3 and IP = IL. therefore:

cos3 L L I V P !



A three-phase motor operating off a 400V system is developing

20k W at an efficiency of 0.87 p.u and a power factor of 0.82.Calculate:

(a)The line current;

(b)The phase current if the windings are delta-connected.

(a) Since watt sin power input

watt sin power out put Eff iciency !

f pV I

watt sin power out put

L L.3 v

!L

82.04003

10002087.0

vvv

v!

L I

And line current =IL=40.0A

(b) For a delta-connected winding

Acurrent l ine

current Pha se 1.233

0.40

3!!!



Three identical coils, each having a resistance of 20; and

an inductance of 0.5 H connected in (a) star and (b) deltato a three phase supply of 400 V; 50 Hz. Calculate the

current and the total power absorbed by both method of

connections.

;! 20R P;!vvT! 1575.0502X P

N �!!22

P P P P P X R j X R Z

Q8315820

157tan15720 122�!¹

º ¸©

ª¨�!

First of all calculating the impedance of the coils

¹¹ º

¸©©ª

¨!N

P

P1

R

Xtanwhere

1264.083coscos !! QN



20;

3/400400

400

400

N

0.5H

0.5H 0.5H

20; 20;

V2313

400VP !!

Since it is a balanced load

Star-connection

A46.1158

231

Z

VII

P

P

LP !!!!

N cos3 L L I V P ! W1281264.046.14003 !vvv!

Power absorbed



20;

400400

400 0.5H

0.5H0.5H20;

20;

Star connection

V400VVLP

!! A38.4158

400

Z

VI

P

P

P!!!

N cos3 L L I V P ! W3831264.038.44003 !vvv!



A balanced three phase load connected in star, each phase consistsof resistance of 100 ; paralleled with a capacitance of 31.8 QF.

The load is connected to a three phase supply of 415 V; 50 Hz.

Calculate: (a) the line current;

(b) the power absorbed;

(c) total k VA;(d) power factor .

415

V2403

415

3

VVL

P !!!



PP

PX

1

R

1Y !

j

1X P [

!

Admittance of the load

where

jR

1

P

[!6108.31502 j

100

1 vvvT! S)01.0 j01.0( !

)01.001.0(240 jY V I I P P P L

!!! Q4539.34.24.2 �!! j

PPVA IVP ! QQ 454.8144539.3240 �!�v!

57645cos4.814 !!Q

P A P

k W728.15763PA !v!

Line current

Volt-ampere per phase

Active power per phase

Total active power



k W728.1 j5763 jPR

!v!

R eactive power per phase 576 j45sin4.814 jPPR !!Q

Total reactive power

k VA44.24.8143 !v!Total volt-ampere

(b)

(c)

(d) Power Factor = cosN = cos 45r = 0.707 (leading)



A three phase star-connected system having a phase voltage of 230V and loads consist of non reactive resistance of 4 ;, 5 ;

and 6; respectively.

Calculate:(a) the current in each phase conductor

(b) the current in neutral conductor

and (c) total power absorbed.

A5.574

230I4 !!;

A465

230I5 !!;

A3.386

230I6 !!;

10 three phase system

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