Download 11. Three phase system

Three phase system
d
 L
M
Flux density
B [T]
N
l
A

Generator for single phase
Note
Induction motor cannot start
by itself. This problem is
solved by introducing three
phase system
Current induces in the coil as the
coil moves in the magnetic field
Current produced at terminal
Instead of using one coil only , three coils are used arranged in one
axis with orientation of 120o each other. The coils are R-R1 , 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.
Finish R
start
eR
L1
eY
L2
eB
L3
R1
Finish Y
start
Load
Y1
Finish B
start
B1
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.
e R  E m sin t
eY  E m sin( t-120  )
eB  E m sin( t-240  )
The circuit can be simplified as follows, where R1 can be
connected to Y and Y1 can be connected to B. In this case
the circuit is reduced to 4 wires.
Finish R
eRB1  eR  eY  eB
eR
start
 E m [sin t  sin( t  120 )  sin( t  240 )]


 Em [sin t  sin t  cos120  cos t  sin 120

Finish Y
eY
start

 sin t  cos 240  cos t  sin 240 ]
R1
eR+eY+eB
Y1
Finish B
eB
start
B1
 Em [sin t  0.5 sin t  0.866 cos t  0.5 sin t  0.866kost ]
0
Since the total emf is zero, R and B1 can be connected together, thus
we arrive with delta connection system.
Fig. B
R
Fig.A
PL
Y
PM
Y1
B
Line
conductors
R1
PN
B1
•Since the total emf is zero, R and B1 can be
connected together as in Fig.A , thus we arrive
with 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 (R1R), PM is –ve (Y-Y1) and PN is –ve (B-B1).
Fig. C
R
eY
Y1
Y
eR
B1
R1
B
eB
•R1, 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 flowing
in 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 –wire
star-connected system. The point N
refers to star point or neutral point.
R
iR
R1
Y
iY
Y1
B
N
M
iR+iY+iB
iB
B1
Generator
1
Load
The instantaneous current in loads L1 , L2 and L3 are
R
R1
iY  I m sin( t-120 )

N
Y1
B1
i B  I m sin( t-240 )

B
i N  i R  iY  i B
 I m [sin t  sin( t  120 )
 sin( t  240 )]  0
Y
Y
iY
iB
B
iR+iY+iB
L3
L2
L1
Line conductors
i R  I m sin t
R
iR
Three-wire star-connected system with
balanced load
For balanced loads, the fourth wire carries no current , so it can
be dispensed
8.66A
2.6A
5A
R
L
R1
10A
N
Y1
0
9.7A
Y
L
B1
B
L
7.1A
5A
8.66A
generator
coils
a
b
c
balanced
load
phase and line current in ampere
Instantaneous currents’ waveform for iR, iY and iB in
a balanced three-phase system.
•VRY, VYB and VBR are called line voltage
•VR, VY and VB are called phase voltage
R
VRY
VBR
N
Y
From Kirchoff voltage law we have
VB
VRY  VR  VY  VR  (VY )
VYB  VY  VB  VY  (VB )
VBR  VB  VR  VB  (VR )
IR
VR
B
VBR
IY
VY
VYB
IB
-VY
VB
VRY
VR
-VR
In phasor diagram
VY
VYB
-VB
For balanced load VR , VY and
VB are equaled but out of phase
VR = VP30;
VRY = VL30;
VY = VP-90;
VYB = VL-90;
VB = VP150;
VBR = VL150;
VRY  2VR cos 30o 
 3 V
 3 V
  3 V
VBR  2VB cos 30o 
P
P
P
VRY
VR
-VR
VY
therefore
VYB  2VY cos 30o
VBR
-VY
VB
VYB
-VB
then
VL 
 3 V
and
IL  IP
P
IR
•IR, IY and IB are called line current
•I1, I2 and I3 are called phase current
From Kirchoff current law we have
I1
IY
I3
VP
Y
VL
I R  I1  I3  I1  ( I 3 )
I2
IY  I2  I1  I 2  (I1 )
IB
IR
I B  I 3  I 2  I 3  ( I 2 )
In phasor diagram
R
I1
-I3
-I2
IB
I3
I2
-I1
IY
B
Since the loads are balanced, the magnitude of currents are
equaled but 120o out of phase. i.e I1 =I2=I3 ,=IP Therefore:-
IR = IL30;
I1 = VP30;
IY = IL-90;
I2 = VP-90;
IB = IL150;
I3 = VP150;
Where IP is a phase current and IL is a line current
IR
I R  2 I1 cos 30  ( 3 ) I P

I1
I Y  2 I 2 cos 30  ( 3 ) I P
I B  2I 3 cos 30 

Thus IR=IY=IB = IL
 3I
-I3
-I2
IB
P
I3
I2
-I1
IY
Hence
IL 
 3 I
P
VL  VP
Unbalanced load
In a three-phase four-wire system the line voltage is 400V
and non-inductive loads of 5 kW, 8 kW and 10 kW are
connected between the three conductors and the neutral.
Calculate:
(a) the current in each phase
(b) the current in the neutral conductor.
Voltage to neutral
VP 
VL 400

 230V
3
3
Current in 10kW resistor
PR 10 4
IR 

 43.5 A
VP 230
Current in 8kW resistor
PY 8 103
IY 

 34.8 A
VP
230
Current in 5kW resistor
PB 5 103
IB 

 21.7 A
VP
230
IR
INV
IBH
IN
IYH
IBV
IYV
IB
IY
INH
Resolve the current components into horizontal and vertical
components.
I H  IY cos 30  I B cos 30o  0.86634.8  21.7  11.3 A
IV  I R  I Y cos 60  I B cos 60o  43.5  0.5(34.8  21.7)  13.0 A
I N  I NH  I NV  11.32  13.02  17.2 A
2
2
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.
IR
R
400V 400V
Y
400V
R1=100
I1
IY
R2=20 I2
IB
X2=60
C=30F
I3
B
(a)
I1 
VRY 400

 4A
R1 100
I1 is in phase with VRY since there is only resistor in the branch
In branch between YB , there are two components , R2 and X2
I2 
VYB
400

 6.32 A
2
2
ZY
20  60
ZY  R 2  X 2
2
2
X
 Y  tan 1  2
 R2
 20  60
2
IR

400
1 /(2  50  30 10  6 )  90 
o

90
 3.77 A
I1
-I3

 60 
  tan 1    71 34'
 20 

In the branch RB , only capacitor in
it , so the XC is -90 out of phase.
V
I 3  BR
XC
VRY
2
30o
90o
71o34'
VBR
30o
I3
I2
VYB
(b)
I R  I1  I 3
I R2  I12  2 I1 I 3 cos   I 32
=30o
I1
I R2  4.0  24.03.77  cos 30o  3.77   56.3
2
2
I R  7.5 A
120o
 = 71o 34’ -60o= 11o 34’

I Y  I 2  I1
2
2
-I1
I2
I  I  2I1I 2 cos  I
2
Y
2
1
I Y2  6.32  24.06.32 cos11o34'4.0  105.5
2
IY  10.3 A
60o
71o 34'
2
IY
 = 180-30o-11o 34’ = 138o 34’
I B  I3  I 2
I B2  I 32  2 I 3 I 2 cos   I 22
I B2  6.32  23.77 6.32 cos 138o 26'3.77   18.5
2
2
-I2
I B  4.3 A
I2

90o
11o 34'
71o 34'
30o
I3
I2
Power in three phase
Active power per phase = IPVP x power factor
Total active power= 3VPIP x power factor
P  3VP I P cos 
If IL and VL are rms values for line current and line voltage
respectively. Then for delta () connection: VP = VL and IP
= IL/3. therefore:
P  3VL I L cos 
For star connection () : VP = VL/3 and IP = IL. therefore:
P  3VL I L cos 
A three-phase motor operating off a 400V system is developing
20kW 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 Efficiency  output power in watts
input power in watts

output power in watts
3 I LVL  p. f
0.87 
20 1000
3  I L  400  0.82
And line current =IL=40.0A
(b) For a delta-connected winding
line current 40.0
Phase current 

 23.1A
3
3
Three identical coils, each having a resistance of 20 and
an inductance of 0.5 H connected in (a) star and (b) delta
to a three phase supply of 400 V; 50 Hz. Calculate the
current and the total power absorbed by both method of
connections.
First of all calculating the impedance of the coils
X P  2  50  0.5  157 
R P  20
Z P  RP  jX P  RP  X P 
2
2
XP
where   tan 
 RP
 157 

 20  157  tan 
  15883
 20 
2
2
1
cos   cos 83  0.1264
1 



Star-connection
20
400V
400V
400 / 3
0.5H
N
0.5H
400V
20
0.5H
20
Since it is a balanced load
VP 
400
 231V
3
IP  IL 
VP 231

 1.46A
Z P 158
Power absorbed
P  3VL I L cos 
 3  400  1.46  0.1264  128W
Star connection
400V
20
400V
400V
VP  VL  400 V
0.5H
20
0.5H
20
0.5H
VP 400
IP 

 4.38A
Z P 158
P  3VL I L cos   3  400  4.38  0.1264  383W
A balanced three phase load connected in star, each phase consists
of resistance of 100  paralleled with a capacitance of 31.8 F.
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 kVA;
(d) power factor .
415
VP 
VL
3

415
3
 240 V
Admittance of the load
1
1
YP 

R P XP

where
XP 
1
jC
1
1
 jC 
 j2  50  31.8  10 6  (0.01  j0.01)S
RP
100
Line current
I L  I P  VPYP  240(0.01  j0.01)  2.4  j 2.4  3.3945
Volt-ampere per phase



240

3
.
39

45

814
.
4

45
PVA  VP I P
Active power per phase
PPA  814.4 cos 45  576
Total active power
PA  3  576  1.728 kW
(b)

P

j
814
.
4
sin
45
 j576
Reactive power per phase PR
Total reactive power
(c)
(d)
Total volt-ampere
PR  j3  576  j1.728 kW
 3  814 .4  2.44 kVA
Power Factor = cos = cos 45 = 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.
230
I 4 
 57.5A
4
230
I 5 
 46A
5
230
I 6 
 38.3A
6
38.3 A
57.5 A
(b)
46 A
X-component = 46 cos 30 + 38.3 cos 30 - 57.5 = 15.5 A
Y-component = 46 sin 30 - 38.3 sin 30 = 3.9 A
Therefore
(c)
I N  15.5 2  3.9 2  16A
P  230 57.5  46  38.3  32.61kW