Power System Analysis And Design SI Edition 6th Edition By Glover – Test Bank

Complete Test Bank With Answers

Sample Questions Posted Below

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 ________ / 1001. (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

Y

bus j

  

8.33 7.33

  

B) 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. The

general power balance equations are given below.

P ( ) (8.33sin ) 1.0 0

x

  

V



2 2 2

2

Q V V

( ) ( 8.33cos ) (7.33) 0.5 0

x

    



2 2 2 2

8.33 cos 8.33sin

 

V

 

2 2 2

J x

( ) 8.33 sin 8.33cos 14.67

      

 

V V

2 2 2 2



1

0 8.33 0 1 0.12

        

(1)

x

          

1 0 6.34 0.5 1.0789

        





1 8.93 1 0.076 0.113

0.12

         

(2)



x



         



1.0789

1.076 7.56 0.110 1.0653

       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.

45 50 ( ) P  

 



 

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/MWh3. (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 0.452 j0.2 0.1386

j j

1 100 MVA 7.21, 4184 amps

    

f pu base

I j I j

,

0.1386 3 13.8 kV





I 30,200 amps

f

b) The generator terminal voltage is

 1 1.0 – (j0.15) 0.67 pu 0.452

j4. 1. 2. 3. (Short Answer: 15 points total – five points each)

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.

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.

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?.



P P

Losses

  

0.3P 0.05 0.2

G G

1 2



P

G

1

L P

 

1 1.25

1



1

Losses



P

G

1T F T F 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.

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

change.

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.

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.

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.

T F T F