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.

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∆-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

IGEE 402 – Power System Analysis – FINAL EXAMINATION – SAMPLE Joós, G.

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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

1 1 1

A = 1 a2 a

1 a a2

1 1 1

A-1 = 1 a a2

1 a2 a

a_deg Is( ) 28.179−=

mag Is( ) 144.385=Is Ir Imotor+:=

Combined Load:

Imotor Imotor_mag cos φ( ) j sin φ( )⋅−( )⋅:=

φ acos pf_m( ):=Imotor_magSmVl

:=SmPmpf_m

:=

Motor Load:

IrPrVl

:=

Heating Load:

1. Instantaneous voltage, current, and power waveforms

ω 2 π⋅ 60⋅:=pf_m 0.80:=Pm 50000:=Pr 20000:=Vl 550:=

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 50kW motor load, with an efficiency of 0.90 and a power factor of 0.80 (lagging).

Draw the instantaneous current and voltage waveforms for the total load, indicating1.peak and rms values. Give the equation for the total instantaneous power and draw thewaveform. Indicate the value of the total real power.Draw the V-I vector diagram and the power diagram. Compute total reactive power and power2.factor.A capacitor is connected in parallel with the loads to increase the power factor to 1.0.3.Compute the reactive power required from the capacitor. Redraw the power diagram.Explain the benefits in having a unity power factor from (i) the utility point of view; (ii) the4.consumer point of view.

ECSE 464 - Power Systems Analysis

Final Exam - Solutions

Plot Waveforms:

t 0 0.0001, 0.04..:=

i t( ) 2 mag Is( )⋅ cos ω t⋅ arg Is( )+( )⋅:= v t( ) 2 Vl⋅ cos ω t⋅( )⋅:=

Ip 2 mag Is( )⋅:= Vlp 2 Vl⋅:=

0 0.004 0.008 0.012 0.016 0.02 0.024 0.028 0.032 0.036 0.041000

800

600

400

200

0

200

400

600

800

1000

i t( )

v t( )

t

Vlp 777.817=

Vl 550=

Ip 204.192=

mag Is( ) 144.385=

Instantaneous Power:

Total Real Power:

p t( ) v t( ) i t( )⋅:=

Ptotal Pm Pr+:=

0 0.008 0.016 0.024 0.032 0.049407.23

2.24 .104

5.41 .104

8.59 .104

1.18 .105

1.49 .105

p t( )

Ptotal

t

Ptotal 7 104×=

2. VI vector diagram, total reactive power, power factor, and power triangle

Qt Sm sin φ( )⋅:= Stotal Pm Pr+( ) j Qt⋅+:= pfPm Pr+( )mag Stotal( )

:=

cos pf( )−180π

⋅ 36.441−=

pf 0.881=

Qt 3.75 104×=

mag Stotal( ) 7.941 104×=

3. Calculate the reactive power required from the capacitor

The reactive power required from the capacitor is simply the negative of the total reactive powerfrom part 2. This way the load power factor becomes unity and the feeder apparent power hasonly 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 theline become zero. Although this doesn't directly contribute to losses, reactive power flowscontribute to increased currents and the associated losses involved. In addition, theincreased current may also play a role in reducing the lifespan of the lines and otherequipment. Also, providing reactive compensation at the load can help to improve voltagestability

In terms of the customer, unity power factor is desirable since it indirectly affects the powerquality through the regulation of the voltage. Large inductive loads can cause undervoltageswhich can be undesirable for sensitive loads. Also, the customer will likely inherit some ofthe cost due to the losses mentioned above and therefore saving from the point of view of theutility also result in savings for the customer as well.

a_deg Vs( ) 8.24=mag Vs( ) 2.715 103×=Vs VloadZtrans Zline+ Zload+( )

Zload⋅:=

Zline 0.5 4i+=Zline50 j 400⋅+( )

a2:=

Ztrans 0.2 j 1.0⋅+:=a240002400

:=

Calculate Feeder Voltage:

a_deg Zload( ) 36.87=mag Zload( ) 21.16=Zload 16.928 12.696i+=

Zload Zloadmag pfload j sin acos pfload( )( )⋅+( )⋅:=Zloadmag

Vload2

Sload:=

Vload 2300:=Sload

Ploadpfload

:=pfload 0.8:=Pload 200000:=

Define Load:

1. Calculate the feeder voltage and angle

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 ofimpedance 50 + j400 . The transformer feeds a load drawing 200 kW, at a 0.8 power factorlagging, and 2300 V.

The feeder voltage is adjusted so that the load voltage is 2300 V. Assume the load voltage1.has an angle of 0 deg. Find the amplitude and angle of the voltage at the sending end of thetransmission line. (Hint: refer all quantities to the low voltage side).Compute the current in the transmission line, the losses in the transmission line and2.transformer, and the overall system efficiency. Calculate the no-load voltage and the voltage regulation.3.Explain how the voltage regulation can be reduced. (Hint: consider shunt compensation).4.

4. Improvement of Voltage Regulation

The receiving end voltage varies as a function of the current on line which is dependent on theload connected. We can compensate for the change in voltage by adding shuntcompensation at the load. In this way the voltage can be regulated to a given reference, say1 pu. In this way the receiving end voltage may change by only a very small amount betweenno-load an full-load. Therefore, a decrease in the VR is possible which will depend on therating of the reactive power source, in the case where the rating of the compensation isinfinite the voltage regulation becomes 0%.

VR will also depends upon the impedance of the line and therefore, series compensation orinstallation of a parallel line could also help to limit VR.

%VR 18.043=VR

Vno_load Vload−( )Vload

100⋅:=

Vno_load 2.715 103×=Vno_load Vs:=

Since the line contains no shunt components the no-load voltage will equal themagnitude of the source voltage.

3. No-load voltage and voltage regulation

Eff 96.029=EffPload

Pload Pline+ Ptrans+100⋅:=

Efficiency:

Ptrans 2.363 103×=Ptrans mag Iline_low( )2 Re Ztrans( )⋅:=

Pline 5.907 103×=Pline mag Iline_low( )2 Re Zline( )⋅:=

Losses:

Transmission Line

Transformer

a_deg Iline_high( ) 36.87−=mag Iline_high( ) 10.87=Iline_high

Iline_lowa

:=

Iline_lowVs Vload−( )Zline Ztrans+( )

:=

Current:

2. Calculate the transmission line current, all losses, and efficiency

Qtotal 2.515 107×=Qtotal Sbase Sload sin acos pf( )( )⋅ Qc−( )⋅:=

Ptotal 1.8 108×=Ptotal Pload:=

Total Real and reactive power:

Qc1Xcpu

:=Xcpu 3.472i−=XcpuXcZbase

:=Xc j−53⋅:=

Capacitor Bank:

Zload 1.204 0.746i+=Zload1

Sloadpf j sin acos pf( )( )⋅+( )⋅:=

IbaseSbase

3 Vbase⋅:=

SloadPloadpf Sbase⋅

:=Pload 3 60⋅ 106⋅:=pf 0.85:=

Load:

Zbase 0.48=ZbaseVbase2

Sbase:=

Vbase 12000:=Sbase 300 106×:=

1. Compute the total real and reactive power drawn from the feeder

QUESTION 3 (18 points)A three-phase feeder supplies three identical single-phase 135 kV / 6.9 kV, 60 Hz, 100 MVAtransformers, each with a leakage reactance of 0.15 pu on the transformer base, and connectedin a -Y connection. The load is a three-phase Y-connected 12 kV load, with per phaseparameters of 60 MW, 0.85 power factor (lagging). A capacitor bank of three elements ofimpedance (-j 5 ) connected in ∆ is used for power factor correction. The secondary voltage isassumed to be 12 kV.

Compute the total power and total reactive power drawn from the feeder.1.Draw the single line pu circuit, using a 300 MVA (three-phase), 12 kV base.2.Compute the total line current, the total power and total reactive power drawn from the feeder,3.in pu. Compute the voltage at the feeder. Compare with results in (1).Phase c of the load becomes an open circuit. In the original V-A three-phase circuit,4.compute the current in the lines feeding the load, when (i) the load neutral is solidlygrounded; (ii) the load neutral is floating.

Z j 0.15⋅ 1.2+ j 0.75⋅+:=

Vab Va Vb−:=Vb 1 cos 120−

π

180⋅

j sin 120−π

180⋅

⋅+

⋅:=Va 1:=

(i) neutral is solidly groundedWe simply refer to the below figure and solve the two loop equations, we disregard thecapacitors since we are interested only in the current feeding the load.

4. Currents in the line feeding the load with phase c open circuited

VVf_high 1.389 105×=Vf_high Vfeeder 135⋅ 103⋅:=

puVfeeder 1.029=Vfeeder 1 Iline 0.1⋅ Zbase⋅+:=

Qtotal 2.515 107×=Ptotal 1.8 108×=

Compare with (1): Qline Sbase⋅ 2.515 107×=Pline Sbase⋅ 1.8 108×=

Qline 0.084=Qline Im Sline( ):=pfline 0.99=pfline

PlineSline

:=Pline 0.6=Pline Re Sline( ):=

Sline 1 conj Iline( )⋅:=

a_deg Iline( ) 7.955−=Iline 0.606=Iline1

Zload1Xcpu

+:=

3. Total line current, total real and reactive power and power factor in pu. Feeder voltage.

2. Single Line Diagram

a_deg Ib( ) 173.13=Ib Ibase⋅ 8.333 103×=a_deg Ia( ) 6.87−=Ia Ibase⋅ 8.333 103×=

Ib Ia−:=IaVab2Z

:=

(i) neutral is floating

In this case, the problem is trivial since I0 = 0, since there is no path for zero sequencecurrent. Therefore, we can say that Ia = -Ib and the circuit looks as follows:

a_deg io( ) 96.87−=a_deg ib( ) 156.87−=a_deg ia( ) 36.87−=

io Ibase⋅ 9.623 103×=ib Ibase⋅ 9.623 103×=ia Ibase⋅ 9.623 103×=

io ia ib+:=ibVbZ

:=iaVaZ

:=

Cpi Y 1 YX4⋅+

⋅:= Cpi 1.255i 10 3−×=

(b) Find the surge impedance and the SIL

Zcxlyl

:= Zc 383.049= SILVl Vl⋅( )Zc

:= SIL 4.972 107×=

β xl yl⋅:= β 1.287 10 3−×=

(c) Calculate sending end voltage and angle using ABCD, nominal pi, and short lineequivalent

Lossless ABCDA cos β l⋅( ):= A 0.87=

B i Zc⋅ sin β l⋅( )⋅:= B 188.604i=

C isin β l⋅( )Zc

⋅:= C 1.285i 10 3−×=

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. Computethe 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. Confirmusing 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 methodsavailable to maintain the voltages within the range of 0.95 and 1.05.

(a) Compute the nominal π equivalent circuit parameters, ignore losses

rl 0.106:= xl 0.493:= yl 3.36 10 6−⋅:= l 400:= Vl 138000:=

X ixl l⋅:= X 197.2i= Vl

37.967 104×=Y iyl l⋅:= Y 1.344i 10 3−

×=

Api 1 YX2⋅+:= Api 0.867=

Bpi X:= Bpi 197.2i=

Vsl 7.347 104×= 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 shuntcompensation in the form of switched capacitors or inductors. Static compensators whichare more elegant can be used as well which provide for better adjustment of the reactivepower injected or consumed, and can respond much quicker. Examples include Static varcompensators (SVCs) and static shunt compensators (STATCOM).

pf 0.95:= Prec 40 106⋅:= SratedPrecpf

:=

Vrl 132103

3⋅:= Vrl 7.621 104×=

IrSrated3

1Vrl

pf j sin acos pf( )( )⋅+( )⋅:= 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⋅+:= Vsl 5.549 104× 3.3i 104×+= Vsl 6.456 104×= a_deg Vsl( ) 30.739=

Vsl Api Vrl⋅ Bpi Ir⋅+:= Vsl 5.477 104× 3.45i 104×+= Vsl 6.473 104×= a_deg Vsl( ) 32.207=

Vsl Vrl X Ir⋅+:= Vsl 6.487 104× 3.45i 104×+=

QUESTION 5 (18 points)A two-bus transmission system consists of the following: (i) a voltage regulated bus (1.0pu), 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 theformulation 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 theadmittance matrix.

Use: P = V1V2/ X sin(δ1-δ2) P 0.3:= X 0.4:= y1X

:=

δ2 asin P X⋅( )−:= Q2 0.1:=δ2180π

⋅ 6.892−=

P2 P−:=

Ybus j−y

y−

y−

y

⋅:= Ybus2.5i−

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 - jQ2)/V2 - Y21*V1(k) ]

The unknowns are the voltage and angle at bus 2. Once the solution has converged we areable to obtain the unknown flows which are the real and reactive power at bus 1.

(d) CommentsNewton Raphson is more well suited to solve the load flow equations since it is more oftenused to solve non-linear equations while Gauss-Siedel is more well suited for solving linearequations. Methods to improve speed of convergence are to use fast-decoupled load flowwhere the Jacobian is only calculated once, and only inverted once as well. Sparse matricesmethods are also implemented for larger systems. If bus 2 becomes a PV bus, theformulation is changed slightly in that x becomes the angle δ2 and whereas y becomes theknown flow P2. In Gauss-Siedel we iterate w.r.t Q2 and not V2

x0180π

⋅ 6.875−=x0.12−

1.04

=

x x ∆x+:=∆x0.12−

0.04

=∆x J 1−∆y⋅:=Calculate the new voltage:

J2.5

0

0

2.5

=J

dP2dδ2

dQ2dδ2

dP2dV2

dQ2dV2

:=

dQ2dV2 2− V2 Ybus1 1,⋅ sin arg Ybus1 1,( )( )⋅ Ybus1 0,

V1⋅ sin arg V2( ) arg V1( )− arg Ybus1 0,( )−( )⋅+:=

dQ2dδ2 V2 Ybus1 0,⋅ V1⋅ cos arg V2( ) arg V1( )− arg Ybus1 0,( )−( )⋅:=

dP2dV2 2V2 Ybus1 1,⋅ cos arg Ybus1 1,( )( )⋅ Ybus1 0,

V1⋅ cos arg V2( ) arg V1( )− arg Ybus1 0,( )−( )⋅+:=

dP2dδ2 V2− Ybus1 0,⋅ V1⋅ sin arg V2( ) arg V1( )− arg Ybus1 0,( )−( )⋅:=

Calculate Jacobian:

∆y0.3−

0.1

=∆yP2

Q2

:=

V1 V2:=V2 x1:=

x0

1

:=Initial guess: The unknown flows again are the real and reactivepower at bus 1. Which are calculated once thevoltage at the load bus has been determined.

In order to calculate the first iteration we need to determine the Jacobian, which will defineour search direction. The vector of unknows, x is given by the angle and magnitude of thevoltage at the load bus:

(c) Formulate the Newton-Raphson problem: state the unknows and calculate the 1stiteration

a_deg V2( ) 6.582−=V2 1.047=V2

1Ybus1 1,

P2 j Q2⋅−( )1

Ybus1 0,1⋅−

⋅:=

So, we calculate the first iteration assuming a flat voltage profile as our initial guess.

QUESTION 6 (18 points)A three-phase power system is fed from the secondary of a -Y connected transformer, witha neutral grounded through an impedance of j 1 pu. The transformer series impedance isj0.15 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 1pu (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 thesequence networks corresponding to the fault. Compute the sequence components ofthe 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 sequencenetworks corresponding to the fault. Compute the value of the sequence componentsof the fault currents. Convert into phase quantities.

(d) For the symmetrical three-phase fault and the single line-to-ground fault, compute thesymmetrical components of the voltage at the fault, and convert to abc values. Drawconclusions as to the severity and impact of those two types of faults on the voltagesat the load.

(a) Draw the sequence component networks (Note: grounding impedance should be3Zn)

a_deg Iph0( )a_deg Iph1( )a_deg Iph2( )

75.44−

164.56

44.56

=

Iph0

Iph1

Iph2

2.514

2.514

2.514

=Iph Aseq

Io

I1

I2

⋅:=

a_deg I1( ) 75.44−=I1 2.514=Aseq

1

1

1

1

a2

a

1

a

a2

:=

Io 0:=I2 0:=I1E1

Z1 Zf+:=

a 0.5− j32

⋅+:=

Z2 Z1:=

Zf 0.1:=

a_deg E1( ) 19.29−=

E1 0.944=

E1 0.891 0.312i−=

E11

1 j 0.35⋅+:=

Z1 0.109 0.312i+=

Z1 11

j 0.35⋅+

−:=

Zo 0.569 0.495i+=

Zo 11

j 1.15⋅+

−:=

(b) Three phase Symmetrical Short Circuit

a_deg V1( ) 75.44−=

V1 0.251=V1 0.063 0.243i−=V1 E1Zf

Zf Z1+⋅:=

Vo 0:=V2 0:=

Three phase fault:

(d) Calculate fault voltages

a_deg Iph0( ) 65.097−=

Iph0Iph1Iph2

1.815

0

0

=

Iph Aseq

Io

I1

I2

⋅:=

a_deg I1( ) 65.097−=I1 0.605=

Io I1:=I2 I1:=I1

E1Z1 Z2+ Zo+ 3 Zf⋅+( )

:=

(c) Single Line to Ground Fault

Vph Aseq

Vo

V1

V2

⋅:=

Vph0Vph1Vph2

0.251

0.251

0.251

= a_deg Vph0( ) 75.44−=

Line to ground fault:

Vo Io Zf⋅:= V1 Vo:= V2 V1:= Vo 0.06= a_deg Vo( ) 65.097−=

Vph Aseq

Vo

V1

V2

⋅:=

Vph0Vph1Vph2

0.181

0

0

= a_deg Vph0( ) 65.097−=

δnew180π

⋅ 36.879=δnew 0.644=δnewωsyn4H

460

2⋅ δ+:=

For four cycles following the fault we determine the new δ

ωsyn4H

t2⋅ δ+=δ t( )

Integrating swing equation twice,assuming pm remains constant we obtain

δ arg E( ):=

ωsyn 2 π⋅ 60⋅:== pm,pu(t) -p e,pu (t)2H( )ωsyn

ωpu t( ) 2tδ t( )d

d

2⋅

(b) Provide and use the swing equation

a_deg E( ) 32.079=E 1.695=E 1 j X Xd+( )⋅ I⋅+:=

a_deg I( ) 25.842−=I 1.111=IPpf

pf j sin acos pf( )( )⋅−( )⋅:=

P 1:=H 5:=pf 0.9:=Xd 0.3:=X 0.6:=

(a) Compute the transient machine internal voltage and angle

QUESTION 7 (18 points)A three-phase. 60 Hz, round rotor hydroelectric generator has an H constant of 5 s. Itdelivers Pm = 1.0 pu power at a power factor of 0.90 lagging to an infinite bus through atransmission line of reactance X = 0.6. The voltage at the infinite bus is 1.0 pu with anangle 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 shortcircuit. 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 thiscase.

(d) Discuss methods to ensure that the machine does not loose synchronism (breakeroperation, series compensation, …)

We calculate A1 using the swing equation and the fault duration to determine the angle at whichthe fault is cleared, δ1 . The machine continues to accelerate until A2 = A1. We can determinewhether the machine retains stability determining whether there in fact does exist and angle δ2 forwhich A2 does in fact equal A1. Following the above calculations we see that this is true forapproximately 122 degrees, and therefore stability is maintained.

δ2180π

⋅ 122.04=δ2 2.13:=

1.461.18 cos δ2( ) δ2+ =

A1 0.168:=1.18 cos δ1( ) cos δ2( )−( ) δ2 δ1−( )− =A2δ1

δ2δ1.18sinδ pm−( )

⌠⌡

d:=

δnew δ− 0.084=δ0

δ1δ1.0

⌠⌡

d =A1δ0

δ1δpm

⌠⌡

d:=

EX Xd+( )

1.883=1.18 sin δ( )⋅E1

X Xd+( )⋅ sin δ( )⋅ =pm δ( ) =

(c) Draw the P-δ curve for the fault operation

(d) Discuss methods to maintain stability

There exist various methods to help maintain stability. Many of the most common are listedhere:•

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( ) Re x( )( )2 Im x( )( )2+:=

conj x( ) Re x( ) Im x( ) j⋅−:=

a_deg x( ) arg x( )180π

⋅:=

n

1

1

r