Download Sinusoidal Steady-state Analysis

BALANCED THREE-PHASE CIRCUITS
Osman Parlaktuna
Osmangazi University
Eskisehir, TURKEY
BALANCED THREE-PHASE VOLTAGES
A set of balanced three-phase voltages consists of three sinusoidal
voltages that have identical amplitudes and frequencies but are out of
phase with each other by exactly 1200. The phases are referred to as
a, b, and c, and usually the a-phase is taken as the reference.
abc (positive) phase sequence: b-phase lags a-phase by
1200,
and Vc-phase leads a-phase by 1200.
c
Va  Vm 00
Va
Vb  Vm -1200
Vc  Vm + 1200
Vb
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acb (negative) phase sequence: b-phase leads a-phase by 1200,
and c-phase lags a-phase by 1200.
Vb
Va  Vm 00
Va
Vb  Vm + 1200
Vc  Vm -1200
Vc
Another important characteristic of a set of balanced three-phase
voltages is that the sum of the voltages is zero.
Va  Vb  Vc  0
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THREE-PHASE VOLTAGE SOURCES
A three-phase voltage source is a generator with three separate
windings distributed around the periphery of the stator.
The rotor of the generator is an electromagnet driven at synchronous speed by
a prime mover. Rotation of the electromagnet induces a sinusoidal voltage
in each winding.
The phase windings are designed so that
the sinusoidal voltages induced in them
are equal in amplitude and out of phase
with each other by 1200. The phase
windings are stationary with respect to
the rotating electromagnet.
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There are two ways of interconnecting the separate phase windings
to form a 3-phase source: in either a wye (Y) or a delta ().
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ANALYSIS OF THE Y-Y CIRCUIT
VN
VN  Van
VN  Vbn
VN  Vcn



0
Z0 Z A  Z1a  Zga ZB  Z1b  Zgb ZC  Z1c  Zgc
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BALANCED THREE-PHASE CIRCUIT





The voltage sources form a set of balanced threephase voltages, this means Va,n, Vb,n, Vc,n are a set of
balanced three-phase voltages.
The impedance of each phase of the voltage source
is the same: Zga=Zgb=Zgc.
The impedance of each line conductor is the same:
Z1a=Z1b=Z1c.
The impedance of each phase of the load is the
same: ZA=ZB=ZC
There is no restriction on the impedance of the
neutral line, its value has no effect on the system.
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If the circuit is balanced, the equation can be written as
VN (
1
1
V  Vbn  Vcn
 )  a n
Z0 Z
Z
Z  Z A  Z1a  Z ga  ZB  Z1b  Z gb  ZC  Z1c  Z gc
Since the system is balanced,
Va n  Vbn  Vcn  0 and VN  0
This is a very important result. If VN=0, there is no potential difference
between n and N, consequently the current in the neutral line is zero.
Hence we may either remove the neutral line from a balanced Y-Y
circuit (I0=0) or replace it with a perfect short circuit between n and N
(VN=0). Both are convenient to use when modeling balanced 3-phase
circuits.
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When the system is balanced, the three line currents are
Van
Vbn
Vcn
IaA 
, IbB 
, IcC 
Z
Z
Z
The three line currents from a balanced set of three-phase currents,
that is, the current in each line is equal in amplitude and frequency
and is 1200 out of phase with the other two line currents. Thus, if
we calculate IaA and know the phase sequence, we can easily write
IbB and IcC.
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SINGLE-PHASE EQUIVALENT CIRCUIT
We can use line current equations to construct an equivalent circuit
for the a-phase of the balanced Y-Y circuit. Once we solve this circuit,
we can easily write the voltages and currents of the other two phases.
Caution: The current in the neutral conductor in this figure is IaA,
which is not the same as the current in the neutral conductor of the
balanced three-phase circuit, which is I0=IaA+IbB+IcC
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LINE-TO-LINE AND LINE-TONEUTRAL VOLTAGES
The line-to-line voltages are VAB, VBC, and
VCA. The line-to-neutral voltages are VAN,
VBN, and VCN. Using KVL
VAB  VAN  VBN
VBC  VBN  VCN
VCA  VCN  VAN
For a positive (abc) phase sequence where
a-phase taken as reference
VAN  V 00
VBN  V -1200
VCN  V + 1200
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VAB  V 00  V -1200  3V 300
VBC  V -1200  V 1200  3V - 900
VAB  V 1200  V 00  3V 1500
• The magnitude of the line-to-line voltage is
3 times the magnitude of the line-to-neutral
voltage.
• The line-to-line voltages form a balanced
three-phase set of voltages.
•The set of line-to-line voltages leads the set
of line-to-neutral voltages by 300
•For a negative sequence, the set of line-to-line
voltages lags the set of line-to-neutral voltages
by 300
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EXAMPLE
A balanced three-phase Y-connected generator with positive sequence
has an impedance of 0.2+j0.5 Ω/ and an internal voltage of 120V/ .
The generator feeds a balanced three-phase Y-connected load having
an impedance of 39+j28 Ω/. The line connecting the generator to the
load is 0.8+j1.5 Ω/. The a-phase internal voltage of the generator is
specified as the reference phasor.
Calculate the line currents IaA, IbB, IcC
120 00
IaA 
 2.4  36.870 A
40 + j30
IbB  2.4  156.870 A
IcC  2.4 83.130 A
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Calculate the three phase voltages at the load VAN, VBN, VCN
VAN  (39  j 28)( 2.4  36.87 0 )  115.22  1.190 V
VBN  115.22  121.190 V
VCN  115.22 118.810 V
Calculate the line voltages VAB, VBC, VCA
VAB  ( 3 300 )VAN  199.58 28.810V
VBC  199.58 - 91.190V
VCA  199.58 148.810V
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Calculate the phase voltages at the terminals of the generator
Van, Vbn, Vcn
Van  120  (0.2  j 0.5)( 2.4  36.87 0 )  118.9  0.320 V
Vbn  118.9  120.320 V
Vcn  118.9 119.680 V
Calculate the line voltages at the terminals of the generator
Vab, Vbc, Vca
Vab  ( 3 300 )Van  20594
.
29.680V
Vbc  20594
.
 90.320V
Vca  20594
. 149.680V
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Backfill: Resistive Delta/Wye
Conversion
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Delta/Wye continued...
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Other way,
Wye to Delta...
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ANALYSIS OF THE Y- CIRCUIT
If the load in a three-phase circuit is connected in a delta, it can be
transformed into a Y by using the  to Y transformation. When the load
is balanced, the impedance of each leg of the Y is ZY=Z/3. After this
transformation, the a-phase can be modeled by the previous method.
When a load is connected in a delta, the current in each leg of the delta
is the phase current, and the voltage across each leg is the phase voltage.
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To demonstrate the relationship between the phase currents and line
currents, assume a positive phase sequence and
IAB  I 00 , IBC  I -1200 , ICA  I 1200
IaA  I AB  ICA  I 00  I 1200  3I - 300
IbB  IBC  I AB  I -1200  I 00  3I -1500
IcC  ICA  IBC  I 1200  I -1200  3I 900
The magnitude of the line currents is 3 times the magnitude of the
phase currents and the set of line currents lags the set of phase
currents by 300. For the negative sequence, line currents lead the
phase currents by 300.
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Positive sequence
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Negative sequence
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EXAMPLE
The Y-connected load of the previous example feeds a -connected load
through a distribution line having an impedance of 0.3+j0.9/. The load
impedance is 118.5+j85.8 /.
Transforming  load into Y and drawing a-phase of the circuit gives
118.5  j85.8
 39.5  j 28.6 / 
3
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Calculate the line currents IaA, IbB, and IcC
120 00
IaA 
 2.4  36.870 A
40  j30
IbB  2.4  156.870 A IcC  2.4 8313
. 0A
Calculate the phase voltages at the load terminals:
Because the load is  connected, the phase voltages are the same as
the line voltages.
VAN  (39.5  j 28.6)( 2.4  36.870 )  117.04  0.960V
VAB  3 300 VAN  202.72 29.040V
VBC  202.72  90.960V VCA  202.72 149.040V
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Calculate the phase currents of the load.
I AB
IBC
1

300 IaA  139
.
 6.870 A
3
= 139
.
 126.870 A ICA  139
. 11313
. 0A
Calculate the line voltages at the source terminals
Van  (39.8  j 29.5)(2.4  36.870 )  118.9 - 0.320V
Vab  3 300 Van  20594
. 29.680V
Vbc  20594
.
 90.320V Vca  20594
. 149.680V
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AVERAGE POWER IN
A BALANCED Y LOAD
PA  VAN IaA cos(vA  iA )
PB  VBN IbB cos(vB  iB )
PC  VCN IcC cos(vC  iC )
In a balanced three-phase system, the
magnitude of each line-to-neutral voltage
is the same as is the magnitude of each phase
current. The argument of the cosine functions
is also the same for all three phases.
V  VAN  VBN  VCN
I  IaA  IbB  IcC
  vA  iA  vB  iB  vC  iC
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For a balanced system, the power delivered to each phase of the
load is the same
PA  PB  PC  P  V I cos 
PT  3P  3V I cos 
PT  3(
VL
) I L cos   3VL I L cos 
3
Where PT is the total power delivered to the load.
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COMPLEX POWER IN
A BALANCED Y LOAD
Reactive power in a balanced system
Q  V I sin 
QT  3Q  3V I sin 
Complex power in a balanced system
S  VAN I*aA  VBN I*bB  VCN I*cC  V I*
S  P  jQ  V I*
ST  3S  3VL I L 
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POWER CALCULATIONS IN A
BALANCED DELTA LOAD
PAB  VAB I AB cos(vAB  ivAB )
PBC  VBC IBC cos(vBC  ivBC )
PCA  VCA ICA cos(vCA  ivCA )
For a balanced system
VAB  VBC  VCA  V
I AB  IBC  ICA  I
vAB  iAB  vBC  iBC  vCA  iCA  
PAB  PBC  PCA  P  V I cos 
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PT  3P  3V I cos 
IL
PT  3( )VL cos   3VL I L cos 
3
Q  V I sin 
QT  3Q  3V I sin 
S  P  jQ  V I*
ST  3S  3VL I L 
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INSTANTANEOUS POWER IN
THREE-PHASE CIRCUITS
In a balanced three-phase circuit, the total instantaneous power is
invariant with time. Thus the torque developed at the shaft of a
three-phase motor is constant, which means less vibration in machinery
powered by three-phase motors.
Considering a-phase as the reference, for a positive phase
sequence, the instantaneous power in each phase becomes
pA  v AN iaA  Vm Im cos t cos(t   )
pB  v BN ibB  Vm Im cos(t  1200 )cos(t    1200 )
pC  vCN icC  Vm I m cos(t  1200 )cos(t    1200 )
pT  pA  pB  pC  15
. Vm I m cos 
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EXAMPLE
A balanced three-phase load requires 480 kW at a lagging power factor
of 0.8. The line has an impedance of 0.005+j0.025/. The line voltage
at the terminals of the load is 600V
Calculate the magnitude of the
line current.
600 *
IaA  (160  j120)103
3
I*aA  577.35 36.870 A
Single phase equivalent
Circuit Analysis II
IaA  577.35 - 36.870 A
I L  577.35 A
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Calculate the magnitude of the line voltage at the sending end of the line.
600
Van  VAN  ZLIaA 
 ( 0.005  j 0.025)(577.35 - 36.870 )
3
 357.51 157
. 0V
VL  3 Van  619.23V
Calculate the power factor at the sending end of the line
pf  cos[157
. 0  ( 36.870 )]
= cos 38.440  0.783 Lagging
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EXAMPLE
Determine all the load
currents:
Transforming  to Y as
60/3=20, and drawing aphase of the circuit gives:
120 00
IaA 
 7.5 00 Arms
4  20 // 30
VAN  (7.5 00 )(12)  90 00Vrms
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In the original Y connected load
I AN
IBN
90 00

 3 00 Arms
30
 3 - 1200 Arms ICN  3 1200 Arms
For the original delta-connected load
VAB  90 3 0 0  30 0  155.88 300 Vrms
I AB
IBC
155.88 30 0

 2.6 300 Arms
60
 2.6  90 0 Arms ICA  2.6 1500 Arms
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Unbalanced Three Phase Systems
 An unbalanced system is due to unbalanced voltage sources or unbalanced load.
In a unbalanced system the neutral current is NOT zero.
Unbalanced three phase Y connected load.
Line currents DO NOT add up to zero.
In= -(Ia+ Ib+ Ic) ≠ 0
Eeng 224
‹#›
Eeng 224
‹#›
Three Phase Power Measurement
 Two-meter method for measuring three-phase power
Eeng 224
‹#›
Residential Wiring
Single phase three-wire residential wiring
Eeng 224
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