Download Three Phase System Generation of three phase system: N

Three Phase System
Generation of three phase system: Consider following 3 phase simple loop generator shown in
figure 1a. Its generated voltage will be as shown in figure 1b.
N
c2
2400
b1
1.5
1
a1
ω
1200
a2
0.5
ea1a2
eb1b2
0
c1
b2
0
2
4
6
8
10
ec1c2
-0.5
S
-1
1200
2400
-1.5
Fig.1b: 3-Phase wave from
Fig.1a
ea1a 2  Em Sin(t )
-----(1)
ec1c 2  Em Sin (t  1200 )
-----(2)
eb1b 2  Em Sin(t  2400 )
-----(3)
Eb1b2
1200
1200
1200
Ea1a2, Eb1b2 & Ec1c2 are RMS values.
Ea1a2
Ec1c2
Fig.1c: Phasor Diagram
Double Subscript notation:
-VAB=+VBA
VAB
A
B
IAB
IBA=-IAB
VAB=VA-VB
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Phase Sequence: It is the sequence in which current or voltages in different phases attain their
maximum values.
Phase sequence may be R-Y-B or R-B-Y as shown bellow
1.5
1
0.5
ea1a2
eb1b2
0
0
2
4
6
8
10
ec1c2
Fig.2a: Phase Sequence R-Y-B
-0.5
-1
1200
2400
-1.5
1.5
1
0.5
ea1a2
eb1b2
0
0
2
4
6
8
10
ec1c2
Fig.2b: Phase Sequence R-B-Y
-0.5
-1
1200
2400
-1.5
Interconnection of 3-phase supply: The six terminals of three phase winding can be connected
to form any of the below.
1. Star or WYE (Y) connected 3-ϕ System
2. Mesh or Delta(∆) connected 3- ϕ System
Star or WYE (Y) connected 3-ϕ System:
Line Current
R
IR
Phase
Voltage
ER
Line Voltage
(ERY)
N
EY
IB
EB
IY
EBR
Y
EYB
B
Fig.3a
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ER, EY & EB are called phase voltages.
IR, IY & IB are called phase currents.
ERY, EYB & EBR are called line voltages.
From figure it is clear that
IR+IY+IB=0
Line current (IL) = Phase Current (IP)
EB
- EY
ERY
EBR
600
-ER 1200
1200 ER
EY
-EB
EYB
Fig.3b:
Line voltage from above phasor diagram
ERY  ER  EY  ER  ( EY )
Its magnitude
ERY  ER  EY  2 ER EY cos 600
2
2
For balanced 3-phase system
ER = EY = EB =EP
So
(Phase Voltage)
E RY  E P  E P  2 E P E P cos 60 0  E P  E P  2 E P 
2
2
2
2
2
1
2
E RY  3E P
Similarly
EYB  EY  ( E B )  EY  E B  2 EY E B cos 600  3E P
2
2
E BR  E B  ( E R )  E B  E R  2 E B E R cos 600  3E P
2
3
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Hence for balanced system
ERY = EYB = EBR = √3EL = EL
If ϕ is the angle between phase voltage and phase current then
EL
I L Cos ϕ = √3ELIL Cos ϕ
3
Active power of 3 phase =3 EPIPCos ϕ =3
W
Reactive power of 3 phase =3 EPIPSin ϕ =3
EL
I L Sin ϕ = √3ELIL Sin ϕ
3
VAR
Apparent power of 3 phase =3 EPIP
EL
IL
3
VA
=3
= √3ELIL
Mesh or Delta (∆) connected 3- ϕ System
IR
R
IRY
ERY
Line Voltage
Phase
Voltage
EBR
IBR
IY
IYB
Y
Line Voltage
EYB
IB
B
Fig.4a
ERY, EYB & EBR are called phase voltages.
IRY, IYB & IBR are called phase currents.
From figure it is clear that
ERY+EBR+EBY=0
Line Voltage (EL) = Phase Voltage (EP)
IBR - IYB
I
IY
R
600
-IRY
1200
1200
IYB
IRY
-IBR
IB
Fig.4b
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Apply KCL at node R
I R  I BR  I RY  I BR  ( I RY )
Its magnitude from above phasor diagram
I R  I BR  I RY  2 I BR I RY cos 60 0
2
2
For balanced 3-phase system
IRY = IYB = IBR =IP
So
(Phase Current)
I R  I P  I P  2 I P I P cos 60 0  I P  I P  2 I P 
2
2
2
2
2
1
2
I R  3I P
Similarly
I Y  I RY  ( I YB )  I p  I P  2 I p I P cos 60 0  3I P
2
2
I Y  I YB  ( I BR )  I p  I P  2 I p I P cos 60 0  3I P
2
2
Hence for balanced system
IR = IY = IB = IL = √3IP
If ϕ is the angle between phase voltage and phase current then
Active power of 3 phase =3 EPIPCos ϕ =3 E L
IL
Cos ϕ = √3ELIL Cos ϕ
3
Reactive power of 3 phase =3 EPIPSin ϕ =3 E L
Apparent power of 3 phase =3 EPIP
=3 E L
W
IL
Sin ϕ = √3ELIL Sin ϕ
3
VAR
IL
3
VA
= √3ELIL
Example 1: If the phase voltage of a three phase star connected alternator is EP. What will be the
line voltage?
i.
ii.
When the phases are correctly connected?
When the connection to one phase is reversed?
Solution:
i.
Line voltage (EL) = Phase Voltage (EP)
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ii.
Let connation of phase R is reversed
EB
EB=ERY
- EY
- EY=EBR
ERY
EBR
600
-ER 1200
ER
1200 ER
EY
-ER
EY
-EB
-EB
EYB
Fig.5b: Phasor Diagram when connection of
phase R is reversed
EYB
Fig.5a: Phasor Diagram when phases
are correctly connected
Line voltage from above phasor diagram
ERY  ER  EY  ER  ( EY )
Its magnitude
E RY  E R  EY  2 E R EY cos 120 0
2
2
E RY  E P  E P  2 E P E P cos120 0  E P  E P  2 E P  
2
2
2
2
2
1
2
ERY  EP
Similarly
EYB  EY  ( E B )  EY  E B  2 EY E B cos 600  3E P
2
2
E BR  E B  ( E R )  E B  E R  2 E B E R cos 120 0  E P
2
2
Example 2: Three identical of 20 ohm are connected in star to a 415 V, 3-phase, 50 Hz supply.
Calculate
i.
Total power taken by load
ii.
Power consumed in the resistance if they are connected in delta to same supply.
iii.
If one of the resistance is open circuited in each case calculated the power consumed.
Solution:
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i.
Power in star
VL=415 V,
VL=√3VP=415 V ⇒VP=415/√3 Volts
IL=IP= VP/RP =415/20√3
Amp
Total power consumed
(Resistive load⇒ PF=1)
P=3VPIPCos ɸ = 3*(415/√3)*(415/20√3)*1=8611.25 Watts
ii.
Power in Delta
VL=415 V,
VL=VP=415 V
IL=IP= VP/RP =415/20
Amp
Total power consumed
P=3VPIPCos ɸ = 3*(415)*(415/20)*1=25833.75 Watts
iii.
Power when one resistance is open circuit
Star Connected
Delta Connected
20Ω
415V
415 V
RP=20Ω
20Ω
RP=20Ω
415V
415 V
Fig.6a
Fig.6a
V 2 415 2
P

 4305.625 Watt
Req
40
P
V 2 V 2 415 2 415 2



Rp Rp
20
20
 17222.5 Watt
Measurement of power in 3-phase load: There are three methods
1. One wattmeter method
2. Two wattmeter method
3. Three wattmeter method
1. One wattmeter method :
It is used only for balanced load
Total power consumed by 3-ϕ load
M
COM
CC
PC
L
R
V
RP
N
RP
P=3x Power consumed by one phase load
=3x Reading of one wattmeter (W)
RP
Y
B
Fig.7: 3-ϕ Star Connected Balanced Load
Note: if load is delta connected, voltage terminal (V) of PC will be connected to ground.
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2. Three wattmeter method:
For balanced or unbalanced load total power
P=W1+W2+W3
W1
CC
M
COM
L
R
V
PC
RP
N
W2
CC
M
COM
COM
RP
L
Y
B
V
PC
W3
CC
M
RP
L
V
PC
Fig.7
Note: If load is delta connected, voltage terminal (V) of PC of all wattmeters will be connected
to ground.
3. Two Wattmeter method:
a. When load is star connected:
M
W1
CC
L
i1
COM
1
V
PC
v1
v3
i2
3
M
COM
W2
CC
PC
i3
v2
2
L
V
Fig.7
Instantaneous power given by wattmeter 1
p1=i1.(v1-v3) ---(1)
Instantaneous power given by wattmeter 2
p2=i2.(v2-v3) ---(2)
Adding (1) & (2)
p1+p2= i1.(v1-v3)+ i2.(v2-v3)
=v1i1-v3i1+v2i2-v3i2
=v1i1+v2i2-v3 (i1+i2)
 i1  i2  i3  0  i1  i2  i3 
=v1i1+v2i2+v3i3
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= Total instantaneous power in 3-phase load
b. When load is delta connected:
M
COM
W1
CC
i1-i3
L
i1
V
PC
v1
v3
i3
i2
v2
M
COM
W2
CC
PC
i2-i1
L
V
Fig.7
Instantaneous power given by wattmeter 1
p1=-v3.(i1-i3) ---(1)
Instantaneous power given by wattmeter 2
p2=v2.(i2-i1) ---(2)
Adding (1) & (2)
p1+p2= -v3.(i1-i3)+ v2.(i2-i1)
=-v3i1-v3i3+v2i2-v2i1
=-(v2+v3)i1+v2i2+v3i3
 v1  v2  v3  0  v2  v3  v1
=v1i1+v2i2+v3i3
= Total instantaneous power in 3-phase load
Hence a load may be balanced or unbalanced, star connected or delta
connected, total instantaneous power of 3-ϕ load will be the sum of the
instantaneous power given by the two wattmeters.
Since wattmeter measures active power so total active power of 3-ϕ load will be
P= W1+W2
W1= Active power measured by wattmeter 1
W2= Active power measured by wattmeter 2
Determination of power factor: Consider a star connected balanced load
Let
V1= V2 = V3=Vp
Phase voltages (RMS)
I1= I2 = I3=Ip
Phase currents (RMS)
Reading of wattmeter 1
W1=I1V13Cos(30-ϕ)
=IP√3VP Cos(30-ϕ)
= √3VP IP Cos(30-ϕ) ---(1)
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V3 I3
ϕ
ϕ
ϕ
I2
300
30
V1
I1
0
V2
-V3
V13
V23
Fig.3b
Reading of wattmeter 2
W2=I2V23Cos(30+ϕ)
=IP√3VP Cos(30+ϕ)
= √3VP IP Cos(30+ϕ)
W1+W2 = 3VP IP Cosϕ=Total active power of 3-ϕload
W1-W2 =√3VP IP Sinϕ
(4)/(3)
---(2)
---(3)
---(4)
W1  W2
3VP I P Sin

W1  W2
3VP I P Cos
 W  W2 
 W  W2 
    tan 1 3  1

tan   3  1
 W1  W2 
 W1  W2 
Power factor
 W  W2 

Cos  tan 1 3  1
 W1  W2 
Effect of power factor on wattmeter readings:
We know
W1=VLILCos(30-ɸ)
=√3VPIP Cos (30-ɸ)
--- (1)
W2=VLILCos(30-ɸ)
=√3VPIP Cos (30+ɸ)
--- (2)
 When ɸ=0 i.e PF Cosɸ=1:
W1= VLILCos(30)= VLIL(√3/2)
W2= VLILCos(30)= VLIL(√3/2)
Both wattmeters will show equal readings
 When ɸ=600 i.e PF Cosɸ=0.5:
W1= VLILCos(30-60)= VLIL(√3/2)
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W2= VLILCos(30+60)= 0
One Wattmeter will show zero reading and total power consumed = W1
 When ɸ=90 i.e PF Cosɸ=0:
W1= VLILCos(30-90)= VLIL(1/2)
W2= VLILCos(30+90)= -VLIL(1/2)
Hence we can say when power factor angle is greater than 600 one
wattmeter will give negative reading. For obtaining the reading of that wattmeter either
the connection of current coil(CC) or pressure coil(PC) should be changed and reading
will be taken as negative.
Example 3: For a certain load one wattmeter reads 20 KW and other 5 KW after the voltage of
this wattmeter has been reversed. Calculate power and power factor of the load.
Solution:
Power P=W1+W2=20+ (-5)= 15 KW
 W  W2 
 20   5 
 25 
  Cos tan 1 3 
  Cos tan 1 3    Cos 70.89 0  0.327
Cos  Cos tan 1 3  1
 15 
 20   5 
 W1  W2 
Q 1. Draw connection diagram fro measurement of power in 3-phase Y connected load using two
wattmeter method. In one such experiment the load supplied was 30 KW at 0.7 power factor lagging.
Find the reading of each wattmeter.
Hint:
W1+W2=P= 30 KW
 W  W2 
  0.7
Cos  tan 1 3  1
 W1  W2 
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