Download 24 Three phase systems

Three-phase systems
Power stations generate so-called three phase voltages
A three-phase voltage source is a
generator with three separate
windings distributed around the
periphery of the stator. Each winding
comprises one phase of the generator.
The rotor of the generator is an
electromagnet driven at synchronous
speed by a prime mover, such as a
steam or gas turbine. 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 120°.
Three phase generator and waveforms
Three-phase generator is equivalent
to three AC voltage sources synchronously
generating three voltages with the same
frequency f and different phases Θ (shifted
by 120O each)
Y-connected three-phase voltage sources
Equivalent circuit of Y- connected ideal threephase voltage sources
Equivalent circuit of Y-connected threephase voltage sources with winding
impedances
∆-connected three-phase voltage sources
Equivalent circuit of ∆- connected ideal threephase voltage sources
Equivalent circuit of ∆-connected threephase voltage sources with winding
impedances
Advantages and importance of three-phase systems
1.
2.
3.
The instantaneous power in three-phase system can be constant (not
pulsating as in a single phase circuit).
For the same transmitted power, three-phase systems consume less
wires.
For the above two reasons, most of the power distribution systems in
US and over the world, use three-phase systems.
Y-Y three-phase system
Source
Power line
Load
Y-Y three-phase system
Let us find the voltage of the node N.
This is a single-node circuit.
Assume:
Zga =Zgb = Zgc;
Z1a = Z1b = Z1c;
ZA = ZB = ZC;
The circuits that meet these conditions,
are called balanced three-phase circuits
Let us define the total impedances:
ZAT = Zga+Z1a + ZA
ZBT = Zgb+Z1b + ZB
ZCT = Zgc+Z1c + ZC
Y-matrix for the node N:
Y = 1/ZAT +1/ZBT +1/ZCT + 1/Z0 = 3/ZAT +1/Z0
Current source matrix:
IS = Va/ZAT +Vb/ZBT +Vc/ZCT = (Va +Vb +Vc) /ZAT
Node potential:
VN = Y-1*Is = ZATZ0/(3Z0+ZAT) *(Va +Vb +Vc) /ZAT =
Z0/(3Z0+ZAT) *(Va +Vb +Vc)
Y-Y three-phase system (cont.)
Node potential:
VN = Z0/(3Z0+ZAT) *(Va +Vb +Vc)
Consider the sum of three phase voltages:
(Va +Vb +Vc)
Assuming equal amplitudes and 120o phase shifts:
Phase a: Va ∠ 0
Phase b Vb = Va ∠ 1200
Phase c Vc = Va ∠ -1200
Using Euler’s formula for the complex amplitudes Vb and Vc:
Vb = Va [cos(1200) + i sin (1200)]
Vc = Va [cos(1200) - i sin (1200)]
Vb+ Vc = Va [cos(1200) + i sin (1200)] + Va [cos(1200) - i sin (1200)] =
= Va *2cos(1200)
Note that cos(1200) = -0.5;
Hence, Vb+Vc = Va *2*(-0.5) = -Va;
Hence,
(Va +Vb +Vc) = 0 and VN = 0
Voltages and currents in Y-Y three-phase system
VN = 0:
in balanced three-phase circuits,
the voltage across the neutral wire
equals zero
In principal, the neutral wire can be
eliminated or at least made less
conducting.
Load currents: IA = (Va-VN) /ZT = Va /ZT, as in a single phase circuit.
Similarly, IB = Vb /ZT
IC = Vc /ZT
However, unlike a single-phase circuit, the current through the neutral wire N
IN = 0 as the voltage VN = 0
Using the KCL, we conclude that IA + IB + IC = 0
Instantaneous Load Powers in Y-Y system
Assuming equal amplitudes VM and 120o phase shifts:
va = VM cos(ωt)
vb = VM cos(ωt +120o)
vc = VM cos(ωt-120o)
In balanced circuit, the load currents have the same amplitude and same phase
shift with respect to the corresponding phase voltages:
ia = IM cos(ωt - Θ)
ib = IM cos(ωt - Θ +120o)
ic = IM cos(ωt - Θ -120o)
Instantaneous powers: p(t) = i(t) * v(t):
pa = IM VM cos(ωt) cos(ωt - Θ)
pb = IM VM cos(ωt+120o) cos(ωt – Θ+120o)
pc = IM VM cos(ωt -120o) cos(ωt – Θ -120o)
Instantaneous Load Powers in Y-Y system
Total instantaneous power: pT = pa + pb + pc:
pT = IM VM [cos(ωt) cos(ωt - Θ) +
+ cos(ωt+120o) cos(ωt – Θ+120o) + cos(ωt -120o) cos(ωt – Θ -120o)]
Applying the trigonometric identity:
cos(ωt) cos(ωt - Θ) = ½ [cos(2ωt – Θ) +cos( Θ)]
cos(ωt+120o) cos(ωt – Θ+120o) = ½ [cos(2ωt – Θ+240o) +cos( Θ)]
cos(ωt-120o) cos(ωt – Θ-120o) = ½ [cos(2ωt – Θ-240o) +cos( Θ)]
Applying the trigonometric identity:
½ [cos(2ωt – Θ+240o) + cos(2ωt – Θ-240o)] =
½ * 2cos(2ωt – Θ)cos(240o) = - ½ cos(2ωt – Θ); Note, cos(240o) = -1/2;
The total power pT is now given by:
pT = ½ VM IM [3 cos(Θ)]; The total power does not have time dependence.
Using effective voltages and currents: Vrms = VM/S2 and Irms = IM/S2”
pT = Vrms Irms 3 cos(Θ); The total power equals to the sum of the phase average powers
Instantaneous Load Powers in Y-Y system
2.0
1.0
V1+V2+V3
V1, V2, V3
1.5
0.5
0.0
0
2
4
6
ωt
8
10
1.0
0.5
0.0
0
2
4
6
ωt
8
10
Complex Powers in Y-Y system
PA = VArms IArms cos (θA),
PB = VBrms IBrms cos (θB),
PA = VCrms ICrms cos (θC),
θ is the phase shift between the voltage
and the current.
In balanced circuits, all the amplitudes
and phase shifts are equal, hence:
PT = 3 VrmsIrms cos(θ )
Delta to Y transformation
See ch. 11 for more details on Delta – Y connections in three-phase systems