Download IGEE 402 – Power System Analysis FINAL EXAMINATION - SAMPLE

IGEE 402 – Power System Analysis
FINAL EXAMINATION - SAMPLE
Fall 2004
Special instructions:
-
Duration: 180 minutes.
-
Material allowed: a crib sheet (double sided 8.5 x 11), calculator.
-
Attempt 5 out of 7 questions.
-
Make any reasonable assumption.
-
All questions and sub-questions carry equal weight.
QUESTION 1 (20 points)
A single phase 550 V, 60 Hz feeder supplies 2 loads in parallel: (a) a 20 kW heating load; (b) a 50
kW motor load, with an efficiency of 0.90 and a power factor of 0.80 (lagging).
(a) Draw the instantaneous current and voltage waveforms for the total load, indicating peak and
rms values. Give the equation for the total instantaneous power and draw the waveform. Indicate
the value of the total real power.
(b) Draw the V-I vector diagram and the power diagram. Compute total reactive power and power
factor.
(b) A capacitor is connected in parallel with the loads to increase the power factor to 1.0. Compute
the reactive power required from the capacitor. Redraw the power diagram.
(c) Explain the benefits in having a unity power factor from (i) the utility point of view; (ii) the
consumer point of view.
QUESTION 2 (20 points)
A single phase 24 kV/2.4 kV, 60 Hz transformer, with a series equivalent impedance of 0.2 + j1.0 Ω
referred to the low-voltage side, is connected at the end of a transmission line of impedance 50 +
j400 Ω. The transformer feeds a load drawing 200 kW, at a 0.8 power factor lagging, and 2300 V.
(a) The feeder voltage is adjusted so that the load voltage is 2300 V. Assume the load voltage has
an angle of 0 deg. Find the amplitude and angle of the voltage at the sending end of the
transmission line. (Hint: refer all quantities to the low voltage side).
(b) Compute the current in the transmission line, the losses in the transmission line and transformer,
and the overall system efficiency.
(c) Calculate the no-load voltage and the voltage regulation.
(d) Explain how the voltage regulation can be reduced. (Hint: consider shunt compensation).
QUESTION 3 (20 points)
A three-phase feeder supplies three identical single-phase 135 kV/6.9 kV, 60 Hz, 100 MVA
transformers, each with a leakage reactance of 0.15 pu on the transformer base, and connected in a
IGEE 402 – Power System Analysis – FINAL EXAMINATION – SAMPLE
Joós, G.
∆-Y connection. The load is a three-phase Y-connected 12 kV load, with per phase parameters of 60
MW, 0.85 power factor (lagging). A capacitor bank of three elements of impedance (-j5 Ω) connected
in ∆ is used for power factor correction. The secondary voltage is assumed to be 12 kV.
(a) Compute the total power and total reactive power drawn from the feeder.
(b) Draw the single line pu circuit, using a 300 MVA (three-phase), 12 kV base.
(c) Compute the total line current, the total power and total reactive power drawn from the feeder, in
pu. Compute the voltage at the feeder. Compare with results in (a).
(d) Phase c of the load becomes an open circuit. In the original V-A three-phase circuit, compute the
current in the lines feeding the load, when (i) the load neutral is solidly grounded; (ii) the load
neutral is floating.
QUESTION 4 (20 points)
A 400 km, 138 kV, 60 Hz transmission line has the following distributed parameters: r = 0.106 Ω/km,
x = 0.493 Ω/km, y = j3.36 x 10-6 S/km. Losses are neglected.
(a) Compute the nominal π equivalent circuit parameters and draw the circuit. Compute the
corresponding ABCD parameters.
(b) Find the surge impedance and surge impedance loading.
(c) The line delivers 40 MW at 132 kV with a power factor of 0.95 lagging. Using the ABCD
parameters, compute the sending end voltage, current and δ angle. Confirm using the nominal π
equivalent circuit, and the short line equivalent.
(d) Draw the approximate voltage profile of this line for the following power delivered: (i) 0 MW, 20
MW, 50 MW, and surge impedance loading. Indicate the methods available to maintain the
voltages within the range of 0.95 and 1.05.
QUESTION 5 (20 points)
A two-bus transmission system consists of the following: (i) a voltage regulated bus (1.0 pu), the
swing bus; (ii) a load bus, with a load of power (0.3 = j1.0), and a fixed capacitor, supplying a reactive
power of (j1.1); (iii) a transmission line of impedance (j0.4).
(a) Draw the single line diagram. Assuming the load bus voltage magnitude is 1 pu, compute the
load angle δ. Compute the Ybus matrix.
(b) Formulate the Gauss Seidel solution. Indicate the unknowns. Compute the first iteration.
(c) Formulate the Newton Raphson solution. Compute the first iteration.
(d) Comment on the features of the 2 methods described above. Indicate approaches to accelerate
the convergence to the solution. Indicate the changes required to the formulation above if a
voltage-controlled bus replaces the load bus.
QUESTION 6 (20 points)
A three-phase power system is fed from the secondary of a ∆-Y connected transformer, with a
neutral grounded through an impedance of j1 pu. The transformer series impedance is j0.15 pu. The
transmission line impedance is j0.2 pu, with no zero sequence component. The sending end voltage
is assumed to be 1 pu. The load is a Y-connected impedance of 1 pu (resistive).
(a) Draw the sequence component networks, indicating all the impedance values, and voltage
sources.
(b) A three-phase symmetrical short circuit occurs at the load end, with a 0.1 pu impedance. Give
the sequence impedance matrix for the fault current. Draw the sequence networks
corresponding to the fault. Compute the sequence components of the fault currents. Convert into
phase quantities.
(c) A single line-to-ground fault occurs on phase a with an impedance of 0.1 pu. Calculate the
sequence components of the fault current. Draw the sequence networks corresponding to the
Page 2 of 3
IGEE 402 – Power System Analysis – FINAL EXAMINATION – SAMPLE
Joós, G.
fault. Compute the value of the sequence components of the fault currents. Convert into phase
quantities.
(d) For the symmetrical three-phase fault and the single line-to-ground fault, compute the
symmetrical components of the voltage at the fault, and convert to abc values. Draw conclusions
as to the severity and impact of those two types of faults on the voltages at the load..
QUESTION 7 (20 points)
A three-phase. 60 Hz, round rotor hydroelectric generator has an H constant of 5 s. It delivers Pm =
1.0 pu power at a power factor of 0.90 lagging to an infinite bus through a transmission line of
reactance X = 0.6. The voltage at the infinite bus is 1.0 pu with an angle of 0 deg. The machine
transient reactance X’d = 0.3 pu.
(a) Compute the transient machine internal voltage and angle δ. Give the corresponding equation
for the electric power.
(b) Write the pu swing equation. A three-phase bolted short occurs midway along the transmission
line. Determine the power angle 4 cycles after the initiation of the short circuit. Assume the
mechanical input power remains constant at the initial value.
(c) Draw the P- δ curve for the above operating conditions. Indicate the initial conditions, and the
fault clearing point. Explain the application of the equal area criterion to this case.
(d) Discuss methods to ensure that the machine does not loose synchronism (breaker operation,
series compensation, …)
APPENDIX A – SYMMETRICAL COMPONENT TRANSFORMATION MATRICES
A=
-1
A =
1
1
1
1
2
a
a
1
a
a2
1
1
1
1
a
a2
1
a2
a
Page 3 of 3
ECSE 464 - Power Systems Analysis
Final Exam - Solutions
QUESTION 1 (18 points)
A single phase 550 V, 60 Hz feeder supplies 2 loads in parallel: (a) a 20 kW heating load; (b) a 50
kW motor load, with an efficiency of 0.90 and a power factor of 0.80 (lagging).
1. Draw the instantaneous current and voltage waveforms for the total load, indicating
peak and rms values. Give the equation for the total instantaneous power and draw the
waveform. Indicate the value of the total real power.
2. Draw the V-I vector diagram and the power diagram. Compute total reactive power and power
factor.
3. A capacitor is connected in parallel with the loads to increase the power factor to 1.0.
Compute the reactive power required from the capacitor. Redraw the power diagram.
4. Explain the benefits in having a unity power factor from (i) the utility point of view; (ii) the
consumer point of view.
Vl := 550
Pr := 20000
Pm := 50000
pf_m := 0.80
ω := 2 ⋅ π ⋅ 60
1. Instantaneous voltage, current, and power waveforms
Heating Load:
Ir :=
Pr
Vl
Motor Load:
Sm :=
Pm
pf_m
Imotor_mag :=
Sm
Vl
φ := acos( pf_m)
Imotor := Imotor_mag⋅ ( cos( φ) − j ⋅ sin( φ) )
Combined Load:
Is := Ir + Imotor
mag( Is) = 144.385
a_deg( Is) = −28.179
Plot Waveforms:
t := 0 , 0.0001 .. 0.04
2 ⋅ mag( Is) ⋅ cos( ω ⋅ t + arg( Is) )
i( t) :=
Ip :=
2 ⋅ mag( Is)
Vlp :=
v ( t) :=
2 ⋅ Vl⋅ cos( ω ⋅ t)
2 ⋅ Vl
1000
800
Vlp = 777.817
600
Vl = 550
400
i( t)
v( t)
Ip = 204.192
200
mag( Is) = 144.385
0
200
400
600
800
1000
0
0.004 0.008 0.012 0.016 0.02 0.024 0.028 0.032 0.036 0.04
t
Instantaneous Power:
p ( t) := v ( t) ⋅ i( t)
Total Real Power:
Ptotal := Pm + Pr
5
1.49 .10
5
1.18 .10
4
p( t)
8.59 .10
4
Ptotal = 7 × 10
Ptotal
4
5.41 .10
4
2.24 .10
9407.23
0
0.008
0.016
0.024
t
0.032
0.04
2. VI vector diagram, total reactive power, power factor, and power triangle
Qt := Sm⋅ sin( φ)
−cos( pf ) ⋅
180
π
Stotal := ( Pm + Pr ) + j ⋅ Qt
pf :=
( Pm + Pr )
mag( Stotal)
= −36.441
pf = 0.881
Qt = 3.75 × 10
4
mag( Stotal) = 7.941 × 10
4
3. Calculate the reactive power required from the capacitor
The reactive power required from the capacitor is simply the negative of the total reactive power
from part 2. This way the load power factor becomes unity and the feeder apparent power has
only a real component. The new power triangle looks as follow:
4. Explain the benefits of having unity power factor from (i) the utility point of view and
(ii) the customer's point of view.
The utility benefits from unity power factor since in this way the reactive power flows on the
line become zero. Although this doesn't directly contribute to losses, reactive power flows
contribute to increased currents and the associated losses involved. In addition, the
increased current may also play a role in reducing the lifespan of the lines and other
equipment. Also, providing reactive compensation at the load can help to improve voltage
stability
In terms of the customer, unity power factor is desirable since it indirectly affects the power
quality through the regulation of the voltage. Large inductive loads can cause undervoltages
which can be undesirable for sensitive loads. Also, the customer will likely inherit some of
the cost due to the losses mentioned above and therefore saving from the point of view of the
utility also result in savings for the customer as well.
QUESTION 2 (18 points)
A single phase 24 kV/2.4 kV, 60 Hz transformer, with a series equivalent impedance of 0.2 +
j1.0 Ω referred to the low-voltage side, is connected at the end of a transmission line of
impedance 50 + j400 . The transformer feeds a load drawing 200 kW, at a 0.8 power factor
lagging, and 2300 V.
1.
2.
3.
4.
The feeder voltage is adjusted so that the load voltage is 2300 V. Assume the load voltage
has an angle of 0 deg. Find the amplitude and angle of the voltage at the sending end of the
transmission line. (Hint: refer all quantities to the low voltage side).
Compute the current in the transmission line, the losses in the transmission line and
transformer, and the overall system efficiency.
Calculate the no-load voltage and the voltage regulation.
Explain how the voltage regulation can be reduced. (Hint: consider shunt compensation).
1. Calculate the feeder voltage and angle
Define Load:
Pload := 200000
Zloadmag :=
Vload
pfload := 0.8
Sload :=
Pload
pfload
Vload := 2300
2
Zload := Zloadmag⋅ ( pfload + j ⋅ sin( acos( pfload) ) )
Sload
Zload = 16.928 + 12.696i
mag( Zload) = 21.16
a_deg( Zload) = 36.87
Calculate Feeder Voltage:
a :=
24000
2400
Ztrans := 0.2 + j ⋅ 1.0
Zline :=
( 50 + j ⋅ 400 )
a
Vs := Vload⋅
( Ztrans + Zline + Zload)
Zload
2
Zline = 0.5 + 4i
mag( Vs) = 2.715 × 10
3
a_deg( Vs) = 8.24
2. Calculate the transmission line current, all losses, and efficiency
Current:
Iline_low :=
Iline_high :=
( Vs − Vload)
( Zline + Ztrans)
Iline_low
mag( Iline_high ) = 10.87
a
a_deg( Iline_high ) = −36.87
Losses:
2
Transmission Line
Pline := mag( Iline_low) ⋅ Re( Zline )
Transformer
Ptrans := mag( Iline_low) ⋅ Re( Ztrans)
2
Efficiency:
Eff :=
Pload
Pload + Pline + Ptrans
⋅ 100
3
Pline = 5.907 × 10
3
Ptrans = 2.363 × 10
Eff = 96.029
3. No-load voltage and voltage regulation
Since the line contains no shunt components the no-load voltage will equal the
magnitude of the source voltage.
Vno_load := Vs
VR :=
( Vno_load − Vload)
Vload
Vno_load = 2.715 × 10
⋅ 100
3
VR = 18.043 %
4. Improvement of Voltage Regulation
The receiving end voltage varies as a function of the current on line which is dependent on the
load connected. We can compensate for the change in voltage by adding shunt
compensation at the load. In this way the voltage can be regulated to a given reference, say
1 pu. In this way the receiving end voltage may change by only a very small amount between
no-load an full-load. Therefore, a decrease in the VR is possible which will depend on the
rating of the reactive power source, in the case where the rating of the compensation is
infinite the voltage regulation becomes 0%.
VR will also depends upon the impedance of the line and therefore, series compensation or
installation of a parallel line could also help to limit VR.
QUESTION 3 (18 points)
A three-phase feeder supplies three identical single-phase 135 kV / 6.9 kV, 60 Hz, 100 MVA
transformers, each with a leakage reactance of 0.15 pu on the transformer base, and connected
in a -Y connection. The load is a three-phase Y-connected 12 kV load, with per phase
parameters of 60 MW, 0.85 power factor (lagging). A capacitor bank of three elements of
impedance (-j 5 ) connected in ∆ is used for power factor correction. The secondary voltage is
assumed to be 12 kV.
1. Compute the total power and total reactive power drawn from the feeder.
2. Draw the single line pu circuit, using a 300 MVA (three-phase), 12 kV base.
3. Compute the total line current, the total power and total reactive power drawn from the feeder,
in pu. Compute the voltage at the feeder. Compare with results in (1).
4. Phase c of the load becomes an open circuit. In the original V-A three-phase circuit,
compute the current in the lines feeding the load, when (i) the load neutral is solidly
grounded; (ii) the load neutral is floating.
1. Compute the total real and reactive power drawn from the feeder
6
Sbase := 300 × 10
Vbase := 12000
Zbase :=
Vbase
2
Zbase = 0.48
Sbase
Load:
pf := 0.85
Zload :=
Pload := 3 ⋅ 60⋅ 10
1
Sload
6
⋅ ( pf + j ⋅ sin( acos( pf ) ) )
Sload :=
Pload
pf ⋅ Sbase
Ibase :=
Zload = 1.204 + 0.746i
Capacitor Bank:
Xc := −j ⋅
5
3
Xcpu :=
Xc
Zbase
Xcpu = −3.472i
1
Qc :=
Xcpu
Total Real and reactive power:
Ptotal := Pload
Ptotal = 1.8 × 10
8
Qtotal := Sbase⋅ ( Sload⋅ sin( acos( pf ) ) − Qc )
7
Qtotal = 2.515 × 10
Sbase
3 ⋅ Vbase
2. Single Line Diagram
3. Total line current, total real and reactive power and power factor in pu. Feeder voltage.
Iline :=
1
Zload
+
1
Iline = 0.606
Xcpu
a_deg( Iline) = −7.955
Sline := 1 ⋅ conj( Iline)
Pline := Re( Sline)
Pline = 0.6
Qline := Im( Sline)
Pline
pfline :=
Qline = 0.084
Sline
8
7
Pline⋅ Sbase = 1.8 × 10
Compare with (1):
Ptotal = 1.8 × 10
Vfeeder := 1 + Iline⋅ 0.1⋅ Zbase
Vf_high := Vfeeder⋅ 135 ⋅ 10
3
Qline⋅ Sbase = 2.515 × 10
8
Vfeeder = 1.029
pfline = 0.99
7
Qtotal = 2.515 × 10
pu
Vf_high = 1.389 × 10
5
V
4. Currents in the line feeding the load with phase c open circuited
(i) neutral is solidly grounded
We simply refer to the below figure and solve the two loop equations, we disregard the
capacitors since we are interested only in the current feeding the load.
Va := 1




Vb := 1 ⋅  cos −120 ⋅
Z := j ⋅ 0.15 + 1.2 + j ⋅ 0.75
 + j ⋅ sin −120 ⋅ π  

180 
180  

π
Vab := Va − Vb
ia :=
Va
ib :=
Z
ia ⋅ Ibase = 9.623 × 10
3
Vb
io := ia + ib
Z
3
3
ib ⋅ Ibase = 9.623 × 10
io ⋅ Ibase = 9.623 × 10
a_deg( ib) = −156.87
a_deg( io) = −96.87
a_deg( ia) = −36.87
(i) neutral is floating
In this case, the problem is trivial since I 0 = 0, since there is no path for zero sequence
current. Therefore, we can say that Ia = -Ib and the circuit looks as follows:
Ia :=
Vab
Ib := −Ia
2Z
3
Ia ⋅ Ibase = 8.333 × 10
a_deg( Ia) = −6.87
Ib ⋅ Ibase = 8.333 × 10
3
a_deg( Ib) = 173.13
QUESTION 4 (18 points)
A 400 km, 138 kV, 60 Hz transmission line has the following distributed parameters:
r = 0.106 /km, x = 0.493 /km, y = j3.36 x 10-6 S/km. Losses are neglected.
(a) Compute the nominal equivalent circuit parameters and draw the circuit. Compute
the corresponding ABCD parameters.
(b) Find the surge impedance and surge impedance loading.
(c) The line delivers 40 MW at 132 kV with a power factor of 0.95 lagging. Using the
ABCD parameters, compute the sending end voltage, current and angle. Confirm
using the nominal equivalent circuit, and the short line equivalent.
(d) Draw the approximate voltage profile of this line for the following power delivered: (i)
0 MW, 20 MW, 50 MW, and surge impedance loading. Indicate the methods
available to maintain the voltages within the range of 0.95 and 1.05.
(a) Compute the nominal π equivalent circuit parameters, ignore losses
−6
rl := 0.106
xl := 0.493
yl := 3.36⋅ 10
X := ixl⋅ l
X = 197.2i
Y := iyl⋅ l
Y = 1.344i × 10
l := 400
Vl := 138000
Vl
−3
3
Api := 1 + Y⋅
X
= 7.967 × 10
Api = 0.867
2
Bpi := X
Bpi = 197.2i
Cpi := Y⋅  1 + Y⋅

X
4
Cpi = 1.255i × 10
(b) Find the surge impedance and the SIL
Zc :=
β :=
 xl 

 yl 
Zc = 383.049
xl⋅ yl
SIL :=
( Vl⋅ Vl)
SIL = 4.972 × 10
Zc
β = 1.287 × 10
7
−3
(c) Calculate sending end voltage and angle using ABCD, nominal pi, and short line
equivalent
Lossless ABCD
A := cos( β ⋅ l)
A = 0.87
B := i⋅ Zc⋅ sin( β ⋅ l)
B = 188.604i
C := i⋅
sin( β ⋅ l)
Zc
C = 1.285i × 10
−3
4
−3
pf := 0.95
Prec := 40⋅ 10
6
Srated :=
Prec
3
Vrl := 132 ⋅
Ir :=
10
3
pf
4
Vrl = 7.621 × 10
 Srated   1  ⋅ ( pf + j ⋅ sin( acos( pf ) ) )


 3   Vrl 
Ir = 184.163
a_deg( Ir) = 18.195
Sending end voltage and angle,using ABCD, nominal pi, and short line
Vsl := A⋅ Vrl + B⋅ Ir
4
Vsl = 5.549 × 10 + 3.3i × 10
4
4
4
4
4
Vsl := Api ⋅ Vrl + Bpi⋅ Ir Vsl = 5.477 × 10 + 3.45i × 10
Vsl := Vrl + X⋅ Ir
Vsl = 6.487 × 10 + 3.45i × 10
Vsl = 6.456 × 10
4
a_deg( Vsl) = 30.739
Vsl = 6.473 × 10
4
a_deg( Vsl) = 32.207
Vsl = 7.347 × 10
4
a_deg( Vsl) = 28.006
Sending end current using ABCD, nominal pi, and short line
Isl := C⋅ Vrl + A⋅ Ir
Isl = 152.277 + 148.013i
Isl = 212.359
a_deg( Isl) = 44.186
Isl := Cpi⋅ Vrl + Api ⋅ Ir
Isl = 151.77 + 145.524i
Isl = 210.265
a_deg( Isl) = 43.796
Isl := Ir
Isl = 174.955 + 57.505i
Isl = 184.163
a_deg( Isl) = 18.195
(d) Approximate voltage profile for 0 MW, 20 MW, 50 MW, and SIL (49.7 MW)
It is possible to keep the receiving end voltage between 0.95 and 1.05 using shunt
compensation in the form of switched capacitors or inductors. Static compensators which
are more elegant can be used as well which provide for better adjustment of the reactive
power injected or consumed, and can respond much quicker. Examples include Static var
compensators (SVCs) and static shunt compensators (STATCOM).
QUESTION 5 (18 points)
A two-bus transmission system consists of the following: (i) a voltage regulated bus (1.0
pu), the swing bus; (ii) a load bus, with a load of power (0.3 + j1.0), and a fixed capacitor,
supplying a reactive power of (j1.1); (iii) a transmission line of impedance (j0.4).
(a) Draw the single line diagram. Assuming the load bus voltage magnitude is 1 pu,
compute the load angle . Compute the Ybus matrix.
(b) Formulate the Gauss Seidel solution. Indicate the unknowns. Compute the first
iteration.
(c) Formulate the Newton Raphson solution. Compute the first iteration.
(d) Comment on the features of the 2 methods described above. Indicate approaches to
accelerate the convergence to the solution. Indicate the changes required to the
formulation above if a voltage-controlled bus replaces the load bus.
(a) Draw the single line diagram, calculate load angle, δ assuming V2 = 1.0, calculate the
admittance matrix.
Use: P = V1V2/ X sin(δ1-δ2)
δ 2 := −asin( P⋅ X)
 y −y 
Ybus := −j ⋅ 
 −y y 
P := 0.3
δ 2⋅
180
π
X := 0.4
= −6.892
y :=
1
X
Q2 := 0.1
P2 := −P
Ybus =
 −2.5i 2.5i 

 2.5i −2.5i 
(b) Formulate the Gauss-Siedel problem: state the unknows and calculate the 1st iteration
We apply the Guass-Siedel formula to the voltage at the load bus:
V2 (k+1) = 1/Y22 [ (P2 - jQ 2 )/V 2 - Y21*V1 (k) ]
The unknowns are the voltage and angle at bus 2. Once the solution has converged we are
able to obtain the unknown flows which are the real and reactive power at bus 1.
So, we calculate the first iteration assuming a flat voltage profile as our initial guess.
V2 :=
1
Ybus
 ( P2 − j ⋅ Q2)

− Ybus ⋅ 1
1
,
0
1


⋅
1, 1
a_deg( V2) = −6.582
V2 = 1.047
(c) Formulate the Newton-Raphson problem: state the unknows and calculate the 1st
iteration
In order to calculate the first iteration we need to determine the Jacobian, which will define
our search direction. The vector of unknows, x is given by the angle and magnitude of the
voltage at the load bus:
x :=
Initial guess:
V2 := x
0

1
The unknown flows again are the real and reactive
power at bus 1. Which are calculated once the
voltage at the load bus has been determined.
V1 := V2
1
∆y :=
Calculate Jacobian:
dP2dδ2 := −V2⋅ Ybus
1, 0
(
 P2 

 Q2 
(
⋅ V1⋅ sin arg ( V2) − arg( V1) − arg Ybus
1, 0
∆y =
 −0.3 

 0.1 
))
( (
)) + Ybus ⋅V1⋅cos(arg(V2) − arg(V1) − arg(Ybus ))
dQ2dδ2 := V2⋅ Ybus
⋅ V1⋅ cos( arg( V2) − arg ( V1) − arg( Ybus ) )
dQ2dV2 := −2 V2⋅ Ybus
⋅ sin( arg ( Ybus ) ) + Ybus
⋅ V1⋅ sin( arg ( V2) − arg( V1) − arg ( Ybus ) )
dP2dV2 := 2V2⋅ Ybus
1, 1
⋅ cos arg Ybus
1, 1
1, 0
1, 0
1, 0
1, 0
1, 1
J :=
1, 1
1, 0
 dP2dδ2 dP2dV2 

 dQ2dδ2 dQ2dV2 
Calculate the new voltage:
J=
 2.5 0 

 0 2.5 
−1
∆x := J
x=
⋅ ∆y
 −0.12 

 1.04 
1, 0
∆x =
x ⋅
180
0 π
 −0.12 

 0.04 
x := x + ∆x
= −6.875
(d) Comments
Newton Raphson is more well suited to solve the load flow equations since it is more often
used to solve non-linear equations while Gauss-Siedel is more well suited for solving linear
equations. Methods to improve speed of convergence are to use fast-decoupled load flow
where the Jacobian is only calculated once, and only inverted once as well. Sparse matrices
methods are also implemented for larger systems. If bus 2 becomes a PV bus, the
formulation is changed slightly in that x becomes the angle δ2 and whereas y becomes the
known flow P2. In Gauss-Siedel we iterate w.r.t Q 2 and not V 2
QUESTION 6 (18 points)
A three-phase power system is fed from the secondary of a -Y connected transformer, with
a neutral grounded through an impedance of j 1 pu. The transformer series impedance is
j0.15 pu. The transmission line impedance is j 0.2 pu, with no zero sequence component.
The sending end voltage is assumed to be 1 pu. The load is a Y-connected impedance of 1
pu (resistive).
(a) Draw the sequence component networks, indicating all the impedance values, and
voltage sources.
(b) A three-phase symmetrical short circuit occurs at the load end, with a 0.1 pu
impedance. Give the sequence impedance matrix for the fault current. Draw the
sequence networks corresponding to the fault. Compute the sequence components of
the fault currents. Convert into phase quantities.
(c) A single line-to-ground fault occurs on phase a with an impedance of 0.1 pu.
Calculate the sequence components of the fault current. Draw the sequence
networks corresponding to the fault. Compute the value of the sequence components
of the fault currents. Convert into phase quantities.
(d) For the symmetrical three-phase fault and the single line-to-ground fault, compute the
symmetrical components of the voltage at the fault, and convert to abc values. Draw
conclusions as to the severity and impact of those two types of faults on the voltages
at the load.
(a) Draw the sequence component networks (Note: grounding impedance should be
3Zn)
(b) Three phase Symmetrical Short Circuit
Zo :=  1 +


j ⋅ 1.15 
1
−
Zo = 0.569 + 0.495i
Z1 :=  1 +


j ⋅ 0.35 
1
−
Z1 = 0.109 + 0.312i
E1 :=
1
1 + j ⋅ 0.35
E1 = 0.891 − 0.312i
E1 = 0.944
a_deg( E1 ) = −19.29
Zf := 0.1
Z2 := Z1
a := −0.5 + j ⋅
I1 :=
E1
Z1 + Zf
I1 = 2.514
 Io 

Iph := Aseq⋅  I1 
I
 2
I2 := 0
Io := 0
a_deg( I1) = −75.44
 Iph0 
 2.514 
 Iph  =  2.514
1  

 2.514 
 Iph2


3
2
1 1 1 
 2
Aseq :=  1 a a 

2
1 a a 
 a_deg( Iph0) 
 −75.44 
 a_deg( Iph )  =  164.56
1  

 44.56 
 a_deg( Iph2)


(c) Single Line to Ground Fault
I1 :=
E1
I2 := I1
( Z1 + Z2 + Zo + 3 ⋅ Zf )
I1 = 0.605
Io := I1
a_deg( I1) = −65.097
 Io 

Iph := Aseq⋅  I1 
I
 2
 Iph0

 Iph
 1
 Iph
 2

 1.815 
= 0
 
 0 

( ) = −65.097
a_deg Iph
0
(d) Calculate fault voltages
Three phase fault:
V2 := 0
Vo := 0
V1 := E1 ⋅
Zf
Zf + Z1
V1 = 0.063 − 0.243i
V1 = 0.251
a_deg( V1) = −75.44
 Vo 

Vph := Aseq⋅  V1 
V
 2
 Vph0

 Vph
1

 Vph
2


 0.251 
 =  0.251
 
 0.251 

(
a_deg Vph
0
) = −75.44
Line to ground fault:
Vo := Io⋅ Zf
V1 := Vo
 Vo 

Vph := Aseq⋅  V1 
V
 2
V2 := V1
 Vph0

 Vph
1

 Vph
2

Vo = 0.06
a_deg( Vo) = −65.097

 0.181 
= 0
 
 0 

(
a_deg Vph
0
) = −65.097
QUESTION 7 (18 points)
A three-phase. 60 Hz, round rotor hydroelectric generator has an H constant of 5 s. It
delivers Pm = 1.0 pu power at a power factor of 0.90 lagging to an infinite bus through a
transmission line of reactance X = 0.6. The voltage at the infinite bus is 1.0 pu with an
angle of 0 deg. The machine transient reactance X'd = 0.3 pu.
(a) Compute the transient machine internal voltage and angle . Give the corresponding
equation for the electric power.
(b) Write the pu swing equation. A three-phase bolted short occurs midway along the
transmission line. Determine the power angle 4 cycles after the initiation of the short
circuit. Assume the mechanical input power remains constant at the initial value.
(c) Draw the P- curve for the above operating conditions. Indicate the initial conditions,
and the fault clearing point. Explain the application of the equal area criterion to this
case.
(d) Discuss methods to ensure that the machine does not loose synchronism (breaker
operation, series compensation, …)
(a) Compute the transient machine internal voltage and angle
X := 0.6
I :=
P
pf
Xd := 0.3
pf := 0.9
⋅ ( pf − j ⋅ sin( acos( pf ) ) )
E := 1 + j ⋅ ( X + Xd) ⋅ I
H := 5
I = 1.111
P := 1
a_deg( I) = −25.842
E = 1.695
a_deg( E) = 32.079
(b) Provide and use the swing equation
( 2H)
ωsyn
ω pu( t) ⋅
d
2
dt
2
δ ( t)
= pm,pu(t) -p e,pu (t)
ωsyn := 2 ⋅ π ⋅ 60
δ := arg( E)
Integrating swing equation twice,assuming p m remains constant we obtain
δ ( t)
=
ωsyn 2
⋅t + δ
4H
For four cycles following the fault we determine the new δ
δ new :=
⋅ 
2
 +δ
4H  60 
ωsyn
4
δ new = 0.644
δ new⋅
180
π
= 36.879
(c) Draw the P-δ curve for the fault operation
p m( δ ) =
E ⋅
1
( X + Xd)
δ1
⌠
A1 := 
⌡δ
δ2
⌠
A2 := 
⌡δ
1
( X + Xd)
= 1.883
δ1
p m dδ
0
E
⋅ sin( δ ) = 1.18⋅ sin( δ )
⌠

⌡δ
1.0 dδ =
δ new − δ = 0.084
0
(1.18sinδ − pm) dδ
( ( )
( )) − (δ2 − δ1) =
1.18 cos δ 1 − cos δ 2
( )
1.46
δ 2 := 2.13
δ 2⋅
1.18 cos δ 2 + δ 2 =
180
π
A1 := 0.168
= 122.04
We calculate A1 using the swing equation and the fault duration to determine the angle at which
the fault is cleared, δ1 . The machine continues to accelerate until A2 = A1. We can determine
whether the machine retains stability determining whether there in fact does exist and angle δ2 for
which A2 does in fact equal A1. Following the above calculations we see that this is true for
approximately 122 degrees, and therefore stability is maintained.
(d) Discuss methods to maintain stability
There exist various methods to help maintain stability. Many of the most common are listed
here:
•
•
•
•
•
•
Smaller equivalent line impedance (parallel lines, different conductors, smaller transformer
leakage reactances, series compensation...)
Higher voltages
FACTS devices
Faster clearing of faults
Larger machine inertia
Define function
mag( x ) :=
2
( Re( x ) ) + ( Im( x ) )
conj( x ) := Re( x ) − Im( x ) ⋅ j
a_deg( x ) := arg ( x ) ⋅
180
π
2
n
1
1
r