Power System Analysis and Design 6th Edition Glover Test Bank

Power System Analysis and Design 6th Edition Glover Test Bank

richard@qwconsultancy.com

1|Pa ge

Name: ________________________

ECE 476 Exam #1 Tuesday , October 11, 2011

Closed book, closed notes One note sheet allowed

1. ________ / 25 2. ________ / 20 3. ________ / 25 4. ________ / 30

Total

________ / 100

1.

(25 points total)

A 3, 300 mile, 345-kV line has series impedance of z = 0.03 + j0.4 /mile and shunt admittance y = j5.0 x 10-6 mho/mile.

(9 pts) a.

Calculate the line's characteristic impedance Zc and the propagation constant 

(9 pts) b.

Initially assume that the receiving end of the line is open-circuited and has a line to line voltage of 345 kV. Calculate the sending end line to line voltage magnitude.

(7 pts) c.

Draw the long line  equivalent circuit for this line, showing the values of all parameters.

2.

(20 points total)

The circuit shown below is a balanced three phase system with a wye-connected generator producing 1 volt (phase to neutral). Assume that each inductor has an impendance ZL = j10 and each capacitor ZC = -j15. Determine Ia, Icap, and the total three phase complex power supplied by the y-connected voltage source.

3.

(25 points total)

A 3-phase, 60Hz, 50km long, completely transposed transmission line is built using Drake conductor. Drake conductor has an outside diameter of 1.108 inches; stranding of 26/7 (Al/St), which yields a GMR for the conductor of 0.0375 feet. Resistance at 60-Hz for this conductor is 0.117 /mile. A horizontal tower configuration is used, with a phase spacing of 25 feet (25 feet between left and center phases, 25 feet between center and right, and hence 50 feet between left and right). Bundling is used, with 2 conductors per phase, spaced 1 foot apart. For reference µ0 = 4π x 10-7 H/m and Ɛ0 = 8.854 x 10-12 F/m. a) Find the positive sequence inductance in H/m and inductive reactance in Ω/km. b) Find the capacitance to neutral in F/m and the admittance to neutral in S/km.

Neglect the effect of the earth plane. c) Since the line is 50km, we can make the short line approximation. Draw the short line π equivalent circuit, labeling appropriate values.

For reference  = 4 x 10-7 H/m and 0 = 8.854 x 10-12 F/m. There are 1609 meters per mile.

4.

Multiple Choice. Circle the most correct answer. Ten problems, three points each for a total of 30 points.

Note this Problem Continues on the Next Page A. As discussed in class, the most common load model for the power flow is A. Constant impedance B. Constant current C. Constant power B. We can only use per phase analysis when: A. All loads and sources are Y connected B. All loads and sources are delta connected C. There is no mutual inductance between phases D. A and C E. B and C C. Which statement about phase and line voltages (V) and currents (I) in Wye and Delta connections is correct?

A. Phase I = line I in Wye, and phase V = line V in Delta B. Phase I = line I in Delta, and phase V = line V in Wye C. Phase I = line I and phase V = line V in Wye D. Phase I = line I and phase V = line V in Delta D. Using Newton’s method to solve the equation x2 –sin(x) – 2 = 0, with an initial guess of x=1 (that is, its value at the zero iteration), select the value that is closest to the value of x after the second iteration: A. 1.6754 B. 1.7775 C. 1.8073 D. 2.3041 E. 2.3253 E. The impedance of a 10 MVA, 22/220 kV transformer is 0.2 per unit. What is this impedance in per unit for a power base of 10 MVA, and a voltage base of 11 kV on the low voltage side of the transformer?

A. 0.05 p.u.

B. 0.1 p.u.

C. 0.2 p.u.

D. 0.4 p.u.

E. 0.8 p.u.

F. Transformer open circuit (OC) and short circuit (SC) tests allow us to calculate certain equivalent circuit values. Which parameters are calculated from the OC test and which are from the SC test?

A. The OC test gives the resistances and the SC test gives the reactances B. The OC test gives the series impedance and the SC test gives the shunt admittance C. The OC test gives the shunt admittance and the SC test gives the series impedance D. The OC test gives the voltages and the SC test gives currents G. In a long line Pi model, we use series impedance Z’ and shunt admittance Y’/2. In the short line model, we use: A. Series impedance Z’ and shunt admittance Y’/2 B. Series impedance Z’ and shunt admittance Y/2 C. Series impedance Z and shunt admittance Y/2 D. Series impedance Z and shunt admittance 0 E. Series impedance 0 and shunt admittance 0 H. In the power flow at the slack (swing) bus, A. P and V are fixed B. P and Q are fixed C. P and ƟV are fixed D. V and Q are fixed E. None of the above I.

A solid conductor has a conductor area of 1113 kcmils. What is its outside diameter, in inches?

A. 1.055” B. 1.293” C. 1113” D. 1.044”

J.

If the diameter of a transmission line conductor is increased, then: A. Both the inductance and the capacitance increase B. The inductance increases and the capacitance decreases C. The inductance decreases and the capacitance increases D. Both the inductance and the capacitance decrease E. The inductance decreases and the capacitance remains unchanged

Name: _____Answers___________

ECE 476 Exam #1 Tuesday , October 6, 2015

Closed book, closed notes One note sheet allowed Scientific calculators allowed

1. ________ / 24 2. ________ / 24 3. ________ / 24 4. ________ / 28

Total

________ / 100

1.

(24 points total)

A new 3 phase, 60 Hz, 200 mile transmission line is to be built using Bluebird conductor.

Bluebird conductor has an outside diameter of 1.762 inches; stranding of 84/19 (Al/St) yields a GMR for the conductor of 0.0588 feet. Resistance for each conductor is 0.0505 /mile. Bundling is to be used, with the three conductors per phase spaced symmetrically 1 feet apart. A horizontal tower configuration is used, with a distance of 20 feet between adjacent phases (i.e., 20 feet between the left and center phases, 20 feet between the center and right phases, and hence 40 feet between the left and right phases).

Assuming that the line is uniformly transposed, draw the medium length -equivalent circuit for the line, showing the value of all parameters. For reference  = 4 x 10-7 H/m and 0 = 8.854 x 10-12 F/m. There are 1609 meters per mile.

0.0505  200  3.367  3 Rb  3 0.0588 11  0.389 ft

R

GMD  3 20  20  40  25.2 ft  25.2  7 L  2 107 ln   8.342  10 H /m   0.389  X L  0.506  /mile  200 mile =101  Rbc  3 0.0734 11  0.419 ft 2 0  1.358 1011 F/m  25.2  ln    0.419  YC 2 60  C 1609  200   0.00082 S 2 2

C

2.

(24 points total)

Two three-phase generators supply a three-phase load through separate three-phase lines.

The load absorbs 30 kW (three-phase) at unity (1.0) p.f. The line impedance is 0 + 1j ohms per phase between generator G1 and the load, and 0 + 2j ohms per phase between generator G2 and the load. If generator G1 supplies 15 kW (three-phase) at unity p.f., with a terminal voltage of 460 V line-to-line, determine: (12 pts) a.

The real power supplied by generator G2

(6 pts) b.

The line-to-line voltage at the load terminals

(6 pts) c.

The reactive power supplied by generator G2

(a)

15 kW

(b) *

 15000  I1     18.820 amps  460 3  460 Vload ,ln  0  j 18.82  265.58  j18.82  266.2  4.05 volts 3 Vload ,ll  3  266.2  461.1 volts (c) *

  10000 IL     37.46  j 2.65 amps  265.58  j18.82  I 2  I L  I1  18.64  j 2.65 amps, I 2  18.82 amps G2 must be supplying all the reactive losses, so its reactive power output is



3  1 I1  2  I 2 2

2

  3.18 kvars

3.

(24 points total)

True/False – Two points each. Circle T if statement is true, F if statement is False.

T

F

1.

Supplying reactive power locally leads to increased line current, thus increased line losses.

T

F

2.

When a three-phase system is delta-connected, the phase currents equal the line currents.

T

F

3.

Excepting a few small isolated systems, the electric grid in North America consists of a single large synchronous 60 Hz system.

T

F

4.

The total real power losses of an ideal transformer are always zero.

T

F

5.

Transposition of high voltage transmission lines, particularly those operating at over 345 kV, is practically always done on every transmission line to keep the system blanced.

T

F

6.

Per unit values are always dimensionless.

T

F

7.

With balanced three-phase circuits, per-phase analysis is commonly done after converting the Y-connected loads to Δconnected loads, thereby solving only one phase of the circuit

T

F

8.

LTC transformers can be used for controlling bus voltage magnitudes.

T

F

9.

In the PowerWorld example shown in class the use of phase shifting transformers was demonstrated for controlling the flow of power in New York, NY.

T

F

10.

The area control error (ACE) for an electric balancing authority can never be negative because transmission lines always have real power losses.

T

F

11.

The ballpark figure given in class for the real power losses on a high voltage transformer (e.g. 500 MVA) was about 5%.

T

F

12.

A load that consumes 100 kW and 50 kvar has a lagging power factor.

4.

(28 points total)

Short Answer, seven points each a)

For a three-phase system using a 100 MVA power base and a 138 kV voltage base, what is the per unit impedance of Z = 5.06 + j23.7 Ω?

0.0266 + j 0.1244 b)

As discussed in class, a common power flow load model is to assume the load is constant power. That is, with no voltage dependence. Why might this be an appropriate model for a resistive heater in which the instantaneous power consumed varies with the square of the voltage?

Electric heaters are often controlled by a thermostat, so in aggregate they behave as constant energy loads, hence a constant power model is appropriate for the power flow time frame; c)

Give the 3 by 3 Bus Admittance Matrix (Ybus) for the three bus power system shown below with the indicated short line model per unit line impedances Bus 2

Bus 1

j0.2

j0.1

j0.1 Bus 3

10   15 5 Ybus  j  5 15 10   10 10 20  d)

For the equation 2 x  x  3  0 with an initial guess x(0) = 1, do two Gauss iterations to determine x(2).

x(0) = 1, x(1) = 2.0, x(2) = 2.207

Name: ________________________

ECE 476 Exam #2 Tuesday, November 15, 2011 75 Minutes Closed book, closed notes One new note sheet allowed, one old note sheet allowed

1. ________ / 25 2. ________ / 25 3. ________ / 30 4. ________ / 20

Total

________ / 100

1.

(25 points total)

A generator bus (with a 1.0 per unit voltage) supplies a 150 MW, 50 Mvar load through a lossless transmission line with per unit (100 MVA base) impedance of j0.1 and no line charging. Starting with an initial voltage guess of 1.00, determine the first iteration value of the load bus voltage (magnitude and angle) using the Newton-Raphson power flow method.

2.

(25 points total)

For the balanced, three phase network shown below assume that all data is per unit on a 100 MVA base except for the transmission line reactance. Assume a 20 kV voltage base for the generator and motor, and a 138 kV voltage base for the transmission line.

(10pts) a)

If the system is initially operating unloaded with all voltages at 1.0 per unit, what is the magnitude of the fault current (in amps) if a balanced, three phase fault occurs on the terminal of the generator on the left. You should neglect the dc offset current.

(8 pts) b)

During the fault from part a, what is the per unit voltage magnitude on the terminal of the motor?

(7 pts) c)

Repeat part a, except now assume that the generator is supplying 100 MVA with a 0.8 lagging power factor and a terminal voltage of 1.0 per unit.

3.

Short Answer, ten problems, three points each 1. In the Fast Decoupled Power Flow formulation, we assume that: A. Shunt admittances (G) are zero and voltages are 1p.u. in the Jacobian B. sin(x)=0 and cos(x)=1 C. Reactive power flows (Q) are negligible D. All of the above 2. For a solid (no fault impedance) single-line-to-ground fault the sequence networks are connected A. In parallel B. In a star configuration C. In series D. Just the positive and negative networks are connected since there is no fault impedance 3. Congestion on transmission lines (ie flows reaching line limits) will tend to cause Local Marginal Prices (LMPs) to: A. Always decrease B. Stay the same C. Always increase D. Some may increase and some may decrease 4. An impedance (distance) relay on a particular line may also act as a backup for a relay on another line. True or false?

T. True F. False 5. A balanced three phase fault is the most common type of transmission line fault.

True or false?

A. True B. False 6. Generator penalty factors relate power generation to system losses. What values can the penalty factor for the slack bus (Lslack) have: A. 0 < Lslack < 1 B. Lslack = 1 C. Lslack > 1 D. Any value greater than 0

Continued on Next Page

7. For which fault types will there be a zero sequence current that is non-zero: A. 3 phase, L-L, SLG, DLG B. L-L, SLG, DLG C. SLG, DLG D. 3 phase, L-L 8. A wind turbine with just an induction generator (a Type 1 design) is usually modeled in the power flow as a PV bus. True or false?

T. True F. False 9. Pick the positive sequence set: A. a = 0.50°, b = 1.0+120°, c = 2.0-120° B. a = 0.50°, b = 1.0-120°, c = 2.0+120° C. a = 0.50°, b = 1.00°, c = 2.00° D. a = 1.030°, b = 1.0-90°, c = 1.0+150° 10. For the general minimization problem Lagrangian as . Then . True or false?

T. True F. False

, we define the is just a restatement of

4.

(Short Answer: 20 points total – five points each)

A.

Give two reasons why the slack (reference) bus is needed for the power flow problem..

B.

Explain how you could use power flow analysis to approximate the penalty factor for a generator?

C.

What is the purpose of power system economic dispatch, and what is a necessary condition for an economic dispatch of the generation?

D.

An ideal inductor with L = 1 H is connected in series with an ac voltage source (v(t) = sin(t) volts) and a switch. The switch, which is initially open, is closed at t = 0. Sketch the current through the circuit (as a function of time) for the first few cycles for t  0.

Name: ______Answers______

ECE 476 Exam #2 Thursday, November 12, 2015 75 Minutes Closed book, closed notes One new note sheet allowed, one old note sheet allowed

1. ________ / 25 2. ________ / 20 3. ________ / 20 4. ________ / 15 5. ________ / 20

Total

________ / 100

1.

(25 points total)

A generator bus (slack bus) supplies a 100 MW, 50 Mvar load through a transmission transmission line that can be modeled with a lossless short transmission line model with per unit impedance (100 MVA base) of j0.12. Also assume there is also a 100 Mvar capacitor (measured at 1.0 per unit voltage) at the load bus.

A)

Determine the bus admittance matrix (Ybus) for the system.

 8.33 8.33  Ybus  j    8.33 7.33 B)

Calculate the voltage magnitude and angle at the load bus using the NewtonRapson method. Use the flat-start. Provide the values after 2 iterations. The general power balance equations are given below.

P2 (x)  V2 (8.33sin  2 ) 1.0

 0

Q2 (x)  V2 (8.33cos 2 )  V2 (7.33)  0.5  0 2

J ( x)

8.33 V2 cos 2    8.33 V2 sin  2

8.33sin  2  8.33cos 2  14.67 V2 

1

x

(1)

0   1   0.12  0  8.33     0.5  1.0789  1 0 6.34        

x

(2)

1   0.076   0.113  0.12   8.93         1.0789   1.076 7.56   0.110  1.0653 

1

2.

(20 points total)

The fuel-cost curves for a two generator system are given as follows: C1(PG1) = 2000 + 45 * PG1 + 0.05 * (PG1)2 C2(PG2) = 500 + 50 * PG2 + 0.03 * (PG2)2 Generator limits are: 0  PG1  200 200  PG2  800 For a load of 600 MW, use the lambda iteration method to determine the values of M, PG1(M) and PG2(M) after two iterations. Show the values of all variables at each iteration. Use starting values of L = 60 and H = 80. Be sure to consider the generator limits; you may ignore any penalty factors.

P ( ) 

  45   50 0.1



0.06

-600 but limits need to be considered

P(60) = 150 + 200 – 600 = -300 (PG2 at low limit) P(80) = 200 +500 – 600 = 100 (PG1 at high limit) Initial values bound the solution Set m = (60+80)/2 = 70 P(70) = 200+333.3-600 = -66.7 Less than zero so L m; new m = (70+80)/2 = 75 P(75) = 200+416.7 - 600 = 16.7 Greater than zero so H m; new m = (70+75)/2 = 72.5 Note, after the first iteration it is clear that PG1 = 200 MW at the solution so PG2 = 400 meaning  will be $ 74/MWh

3.

(20 points total)

For the balanced, three phase network shown below assume that all data is per unit on a 100 MVA base except for the transmission line reactance. Assume a 13.8 kV voltage base for the generator and motor, and a 138 kV voltage base for the transmission line.

(10 pts)

a)

If the system is initially operating unloaded with all voltages at 1.0 per unit, what is the magnitude of the fault current (in amps) if a balanced, three phase fault occurs on the terminal of the motor.

You should neglect the dc offset current.

(10 pts)

b)

During the fault from part a, what is the per unit voltage magnitude on the terminal of the generator?

a)The per unit imedaance is j 0.452 j0.2  j 0.1386 1 100 MVA I f , pu    j 7.21, I base   4184 amps j 0.1386 3 13.8 kV I f  30, 200 amps b) The generator terminal voltage is 1 1.0 - (j0.15)   0.67 pu j 0.452

4.

(Short Answer: 15 points total – five points each)

1.

Explain how you could use power flow analysis to approximate the penalty factor for a generator?

Solve the power flow and get the losses. Change the generator output slightly and get the new losses. This gives a numerical approximation of how the losses change for a change in generation.

2.

As discussed in class, what is contingency analysis and why is it used?

Contingency analysis is the process of sequentially applying various “contingencies” and then solving the power flow and seeing if there are any limit violations. An example contingency would be opening a transmission line. It is used to determine if a power flow solution is n-1 reliable.

3.

Assume the per unit losses on a small system can be approximated as 0.15(PG1)2 + 0.1(PG2)2 - 0.05 PG1 PG2. If the per unit generator outputs are PG1 = 1 and PG2 = 2, what is the penalty factor for generator 1?.

PLosses  0.3PG1  0.05PG 2  0.2 PG1 L1 

1  1.25 PLosses 1 PG1

5.

(20 points total)

True/False – Two points each. Circle T if statement is true, F if statement is False.

T

F

1.

Nonlinear equations, such as those for the power flow, can have multiple solutions.

T

F

2.

Because of line resistance, the power flow Jacobian matrix is guaranteed to never be singular.

T

F

3.

In the power flow the voltage angle at the slack bus does not change.

T

F

4.

The dc power flow solves for the per unit voltage magnitudes, assuming that the voltage angles are constant at zero degrees.

T

F

5.

When solving a minimization problem adding contraints will result in a cost function value that is greater than or equal to that of the unconstrained problem.

T

F

6.

In the PJM and MISO LMP markets generators are never paid more than they offer into the market.

T

F

7.

In three-phase systems using symmetrical components, the negative sequence is used to represent non-zero neutral currents.

T

F

8.

Economic dispatch primarily is concerned with determining which units to turn on/off.

T

F

9.

In economic dispatch, the penalty factor at the slack bus is always unity.

T

F

10.

When considering faults on high voltage transmission lines, such as those operating at 500 or 765 kV, the most common type of fault is the balanced three-phase faults covered in Chapter 7.

Quiz 1 ECE 476 1. Two balanced three-phase loads are in parallel. a. Load 1 draws 10 kW at 0.8 PF lagging b. Load 2 draws 20 kVA at 0.6 PF leading The loads are supplied by a balanced three-phase 480 VLL source. (a) Draw the power triangle for the combined load.

PL = P1 + P2 = 22 kW 21.12°

QL = Q1 + Q2 = -8.5 kVAR

SL = 23.59 kVA

(b) Determine PF of the combined load. cos21.12° = 0.933 leading

(c) Determine the magnitude of the line current from the source.

𝐼𝐿 =

𝑆𝐿 √3𝑉𝐿𝐿

=

23.59×103 √3×480

= 28.37 𝐴

(d) Y-connected inductors are now installed in parallel with the combined load. What value of inductive reactance is needed in each leg of the Y to make the source power factor unity?

𝑄𝐼𝑛𝑑 = |𝑄𝐿 | = 8.5 × 103 𝑉𝐴𝑅 = 𝑋Y =

3(480/√3) 8.5×103

2

= 27.1 Ω

3(𝑉𝐿𝑁 )2 𝑋Y

Quiz 2 ECE 476 Name:‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗ 1. A 60-Hz single –phase, two-wire overhead line has solid cylindrical copper conductors with 2.4 cm diameter. The conductors are arranged in a horizontal configuration with 3 m spacing.

Calculate the inductance of each conductor due to both internal and external flux linkages in mH/km (40 points). 𝐷 𝐿𝑥 = 𝐿𝑦 = 2 × 10−7 𝐿𝑛 ( ) 𝐻/𝑚 𝑟′ 𝐷 = 3𝑚 0.024 𝑟 ′ = 0.7788 × 𝑟 = 0.7788 × ( ) = 9.346 × 10−3 2 3 𝐻 1000𝑚 1000𝑚𝐻 𝐿𝑥 = 𝐿𝑦 = 2 × 10−7 𝐿𝑛 ( ) ( )( ) = 1.154 𝑚𝐻/𝑘𝑚 −3 9.346 × 10 𝑚 𝑘𝑚 𝐻

2. The figure below is a completely transposed three-phase overhead transmission line with bundled phase conductors. All conductors have a radius of 2 cm.

0.4m

0.4m

0.4m

0.4m 12m

12m

a. Determine the inductance per phase in mH/km (40 points). 𝑟 ′ = 0.7788 ∗ 0.02 = 0.0156 4

𝑅𝑏 = √(𝑟 ′ )(0.4)(0.4)(√2 × 0.4) = 0.1938𝑚 𝐷𝑒𝑞 = 3√𝐷𝐴𝐵 𝐷𝐵𝐶 𝐷𝐶𝐴 = 3√(12)(12)(24) = 15.12𝑚 𝐿 = 0.2𝐿𝑛 (

𝐷𝑒𝑞 15.12 ) = 0.2𝐿𝑛 ( ) = 0.8714 𝑚𝐻/𝑘𝑚 𝑅𝑏 0.1938

b. Find the inductive line reactance per phase in Ω/km at 60 Hz (20 points). 𝑋 = 𝜔𝐿 = 2𝜋60 × 0.8714 × 10−3 = 0.3285 Ω/𝑘𝑚

Quiz 3 ECE 476 Name:‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗ A generator bus (slack bus) supplies a 100 MW, 50 Mvar load through a lossless transmission line with parameters on 100 MVA base below. Also, there is a 10 Mvar shunt capacitor at the load bus.

X per unit: 0.2 & B per unit: 0.2 1. Determine the bus admittance matrix (Ybus) for the system.

10 Mvar = 0.1 p.u. 𝑆 0.1 = 2 = 0.1 𝑝. 𝑢. 𝑉2 1 1 0.2 1 1 1 0.2 𝑌11 = 𝑗0.2 + 𝑗 2 , 𝑌12 = − 𝑗0.2, 𝑌21 = − 𝑗0.2, 𝑌22 = 𝑗0.2 + 𝑗 2 + 𝑗0.1

Assuming 1 pu bus voltage, 𝑌𝑐𝑎𝑝 =

𝑌𝑏𝑢𝑠 = [

−𝑗4.9 𝑗5 ] 𝑗5 −𝑗4.8

2. Calculate the voltage magnitude and angle at the load bus using the Newton-Rapson method. Use the flat-start. Provide the values after 2 iterations. General power balance equations are given below. 𝑛

𝑃𝑖 = ∑|𝑉𝑖 ||𝑉𝑘 |(𝐺𝑖𝑘 𝑐𝑜𝑠𝜃𝑖𝑘 + 𝐵𝑖𝑘 𝑠𝑖𝑛𝜃𝑖𝑘 ) = 𝑃𝐺𝑖 − 𝑃𝐷𝑖 𝑘=1 𝑛

𝑄𝑖 = ∑|𝑉𝑖 ||𝑉𝑘 |(𝐺𝑖𝑘 𝑠𝑖𝑛𝜃𝑖𝑘 − 𝐵𝑖𝑘 𝑐𝑜𝑠𝜃𝑖𝑘 ) = 𝑄𝐺𝑖 − 𝑄𝐷𝑖 𝑘=1

𝑃2 (𝑥) = |𝑉2 |(5𝑠𝑖𝑛𝜃2 ) + 1 = 0 𝑄2 (𝑥) = |𝑉2 |(−5𝑐𝑜𝑠𝜃2 ) + |𝑉2 |2 (4.8) + 0.5 = 0 𝐽(𝑥) = [

5|𝑉2 |𝑐𝑜𝑠𝜃2 5|𝑉2 |𝑠𝑖𝑛𝜃2

5𝑠𝑖𝑛𝜃2 ] −5𝑐𝑜𝑠𝜃2 + 9.6|𝑉2 |

0 1 5 0 𝑥 (0) = [ ], 𝑓(𝑥 (0) ) = [ ], 𝐽(𝑥 (0) ) = [ ] 1 0.3 0 4.6 0 5 𝑥 (1) = [ ] − [ 1 0 𝑓(𝑥 (1) ) = [

𝐽(𝑥 (1) ) = [

−0.2 0 −1 1 ] [ ]=[ ] 0.3 0.9348 4.6

0.9348 × 5 sin(−0.2) + 1 0.0714 ]=[ ] 0.1137 0.9348 × −5 cos(−0.2) + 0.93482 (4.8) + 0.5

5(0.9348) cos(−0.2) 5 sin(−0.2) 4.5808 −0.9933 ]=[ ] 5(0.9348) sin(−0.2) −5 cos(−0.2) + 9.6(0.9348) −0.9286 4.0737

−0.2 −0.2228 4.58 −0.9933 −1 0.0714 𝑥 (2) = [ ]−[ ] [ ]=[ ] 0.9348 0.1137 0.9017 −0.9286 4.0737 ∴ 𝑉2 = 0.9017∠ − 12.765°

Quiz 4 ECE 476 Name:‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗

1. A 1000-MVA 20-kV, 60-Hz three-phase generator is connected through a 1000-MVA 20-kV Δ/138-kV Y transformer to a 138-kV circuit breaker and a 138-kV transmission line. The generator reactances are 𝑋𝑑′′ =0.15, 𝑋𝑑′ =0.45 and 𝑋𝑑 =1.8 pu. The transformer series reactance is 0.1 pu; transformer losses and exciting current are neglected. A 3-phase short-circuit occurs on the line side of the circuit breaker when the generator is operated at rated terminal voltage and at no-load. Determine the substransient current through the breaker in pu and in kA rms ignoring any dc offset.

A. The base current on the high voltage side of the transformer is: 𝐼𝑏𝑎𝑠𝑒,𝐻𝑉 = 𝐼" =

𝑆3∅ √3𝑉𝐿𝐿

1000

=

√3 × 138

= 4.184 𝑘𝑉𝐴

𝐸𝑞 1 = = 4 𝑝. 𝑢. = 4 × 4.184 = 16.735 𝑘𝐴 𝑋𝑑′′ + 𝑋𝑇𝑅 0.15 + 0.1

2. Determine Ybus in per-unit for the circuit below using a 1000 MVA system base. Per unit values for the generators and transformers are given on their own base so you need to first convert some of them to the system MVA base.

T1 Δ

T2 Y

G1 500 MVA, 13.8 kV X”=0.2 pu

′′ Hint: 𝑋𝐺1 = 0.2 ×

1

500 kV Line X = 50 Ω

2

500

Δ

G2

500 MVA 13.8 Δ/500 Y kV X=0.12 pu

1000

Y

1000 MVA 20 Δ/500 Y kV X=0.10 pu 𝑉2

5002

𝑆𝑏𝑎𝑠𝑒

1000

= 0.4 𝑝𝑢, 𝑍𝑏𝑎𝑠𝑒 = 𝑏𝑎𝑠𝑒 =

= 250 Ω, 𝑋𝑙𝑖𝑛𝑒 =

1000

A. 𝑋𝑇1 = 0.12 × 500 = 0.24 𝑝𝑢 1

1

1

1

1

𝑌11 = −𝑗 (0.4+0.24 + 0.2), 𝑌12 = 𝑌21 = 𝑗 0.2, 𝑌22 = −𝑗 (0.1+0.17 + 0.2) −6.56 5 𝑌𝑏𝑢𝑠 = 𝑗 [ ] 5 −8.7

1000 MVA, 20 kV X”=0.17 pu 50 250

= 0.2 𝑝𝑢