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Homework Problems for ECE 821 Power Electronics
Problem 8: (a), (b), and (c) three parts.
(a)
The figure shows a voltage-source inverter based reactive & harmonic compensator for a
nonlinear load. The system and parameters are based on 240 V, 20 kVA, and 60 Hz. Assume
that the load is 3-phase balanced and generates 10 kVar lagging reactive power and 10 A 5th
harmonic current (negative-sequence), i.e., iLa = 24 2 sin(ωt −
π
) − 10 2 sin(5ωt ) referring to
2
the source voltage, vSa = 240 2 / 3 sin(ωt ) , design the compensator by doing the following
things:
1. Calculate the
vSa
iLa
ZS iSa
−
+
required dc voltage
for the inverter to
vSb
Nonlinear
−
+
compensate the
Load
reactive and
vSc
harmonic current;
−
+
2. Determine the
240V 1%
iCa
required dc
60Hz
capacitance of the
20KVA
Lr
vCa
inverter to keep the
vCb
dc voltage ripple
Vdc
vCc
within ±5% of the
Cd
required dc voltage;
6%
3. Draw a control block
diagram for the
Reactive and harmonic
compensator.
compensator
Solution:
Vll2− base (240V )
=
= 2.88 Ω . ω = 60 ⋅ 2π = 377 rad / sec .
Sbase
20kVA
The total inductance, L=LS+Lr=7%, or 0.07*2.88/377 = 535 µH. (Lr=6%, or 0.06*2.88/377 = 458
µH). Since the load is 3-phase balanced and generates 10 kVar lagging reactive power and 10 A
5th harmonic current (negative-sequence), therefore,
iLa = −24 2 cos(ωt ) + 10 2 sin( −5ωt ) ,
2π
2π
iLb = −24 2 cos(ωt −
),
) + 10 2 sin(−5ωt −
3
3
2π
2π
iLc = −24 2 cos(ωt +
).
) + 10 2 sin(−5ωt +
3
3
The compensator injects current to cancel the load reactive and harmonic current, thus iCa=iLa,
iCb=iLb, iCc=iLc. Consider phase “a” to determine the peak value of phase voltage, vCa.
di
d
vCa = vSa + ( L or Lr ) Ca = 240 2 / 3 sin ωt + 535µ
− 24 2 cos ωt + 10 2 sin( −5ωt )
dt
dt
= 196 sin ωt + 6.85 sin ωt − 14.3 cos(5ωt )
2
1. Z base =
(
)
Homework Problems for ECE 821 Power Electronics
Therefore, (vCa)peak ≈ 196+6.85=202.85 V. Because the system is balanced, the peak value
should be the same for all phases. The required dc voltage should be Vdc ≥ 2(vCa ) peak = 405.7 V
for the triangular-wave PWM or Vdc ≥ 3 (vCa ) peak = 351.35 V for the space vector PWM.
2. The instantaneous real power flowing out from the compensator, pC, can be expressed as:
pC = v S • i C = vSaiCa + vSbiCb + vSciCc = 240V ⋅ 10A ⋅ 3 ⋅ cos(6ωt ) . This real power causes the
dc voltage to fluctuate (Note that reactive power of a balanced three-phase system has no
effects on the dc voltage when assuming the PWM switching frequency is much higher than
60 Hz.). Thus we have
T6
1
2
2
∫0 pC dt = 2 Cd Vdc + p − Vdc − p or
(
Cd =
∫
T6
0
pC dt
V dc ⋅ ∆Vdc
=
)
∫
T6
0
pC dt
2
2 ⋅ V dc ⋅ ε
, where ε is the allowable voltage ripple.
⎧ 2400 3 62ω
= 223 µF for the triangular - wave PWM
⎪
2
⎪ 2 ⋅ (405.7 ) ⋅ 0.05
.
∴ Cd = ⎨
2
⎪ 2400 3 6ω
⎪ 2 ⋅ (351.35)2 ⋅ 0.05 = 298 µF for the space vector PWM
⎩
3.
vS
iS
iL
iC
iL
Compute
reactive &
harmonic
components
PWM
signals
PWM
Control
Nonlinear
Load
iC_ref
diC
iC
OR
diC
diC
vS
PI
vC_ref
Space Vector PWM
or
Triangular-wave PWM
Here two PWM methods are shown although others may be used.
Homework Problems for ECE 821 Power Electronics
(b)
The figure shows a voltage-source inverter based reactive
power compensator (STATCOM) rated at ±100 Mvar for
a power system with an infinite bus voltage of 161 kV
(i.e., V=161 kV). The total leakage inductance (X) of the
transformer is 10% of the rated (or base) value (100 Mvar,
161 kV). Determine the STATCOM’s output voltage
Vo’s magnitude, Vo, and phase angle, α, referred to the
bus voltage V when it produces a reactive power of 100
Mvar leading and has a total power loss of 1 MW. (Hint:
use phasors and/or power equations to calculate Vo).
V = V∠0o
± Q MVar
X
Vo = Vo ∠α
Inverter
Solution 1: The figure can be reduced into a phase-to-neutral circuit as follows:
o
o
X I C = I P ∠0 + I Q ∠90
Vp =
V
∠0 o
3
Vop =
Vo
∠α
3
Ploss
1 MW
Q
100 Mvar
=
= 3.59 A and I Q =
=
= 359 A . Therefore,
3 ⋅V
3 ⋅ 161 kV
3 ⋅V
3 ⋅ 161 kV
(161 kV )2 (3.59∠0o + 359∠90o )
161 kV o
Vop = Vp − jXI C =
∠0 − j 0.1 ⋅
100 Mvar
3
o
= (93 + 9.3) kV∠0 − 0.093 kV∠90o = 102.3 kV∠0o − 0.093 kV∠90o
where, I P =
= 102.3 kV∠ − 0.052o
∴ Vo = 3 ⋅ 102.3 kV = 177 kV and α = -0.052o .
Solution 2:
For a two voltage-source (bus) system, we have the following power (from
V1 to V2) equations:
V1V2
sin(α 2 − α1 )
X
V
Q12 = 2 (V1 cos(α 2 − α1 ) − V2 )
X
From the above equations, Vo and α can be obtained.
P12 =
Homework Problems for ECE 821 Power Electronics
(c)
Consider a UPFC in a two-machine power system. The UPFC can be represented a synchronous
voltage source, Vpq, with controllable magnitude Vpq and angle ρ. The real and reactive powers
VV
transmitted over the line prior the compensation (or when Vpq=0) are P0 = S R sin δ and
X
V
Q0 = R (VS cos δ − VR ) at a given angle δ. Determine the voltage, Vpq, (both the magnitude Vpq
X
and angle ρ) that should be ejected by the UPFC when the commands for the real and reactive
power to be transmitted are Pref and Qref.
VS = VS ∠0o
Sending End
V pq = V pq ∠ρ
VT = VT ∠α
VR = VR ∠δ
X
Receiving End
Solution:
According to the power equations, we have
V
VV
P0 = S R sin δ , and Q0 = R (VS cos δ − VR )
X
X
VV
V
Pref = T R sin (δ − α ) and Qref = R (VT cos(δ − α ) − VR ) . In addition, VT = Vpq + VS.
X
X
Solving the above equations, we can obtain Vpq.