7/28/2019 2012-ss1

1/7

1

EE 521 Analysis of Power Systems

Fall 2012

C. C. Liu

Problem Set 1

Problem 1: The following is a one-line diagram of a 3-bus power system.

The bus and line input data are shown in Tables 1 and 2.

Table1 - Bus Input Data for Problem 1Bus No. Type V Angle PG QG PL QL

per unit degrees per unit per unit per unit per unit1 swing 1.000 0.000 --- --- 0.000 0.000

2 load --- --- 0.000 0.000 2.000 0.2003 generator 1.010 --- 1.300 --- 0.200 0.600

Table 2 - Line Input Data for Problem 1Line No. From Bus To Bus r x g/2 b/2

per unit per unit per unit per unit1 2 3 0.036 0.400 0.000 0.4302 2 1 0.018 0.200 0.000 0.2203 1 3 0.009 0.100 0.000 0.110

The power flow results are shown in Table 3.

Table 3 - Power Flow Output Data for Problem 1Bus No. V Angle PG QG PL QL

per unit Degrees per unit per unit per unit per unit1 1.000 0.000 --- --- 0.000 0.0002 0.951 -15.642 0.000 0.000 2.000 0.2003 1.010 1.980 1.300 0.595 0.200 0.600

2

1

3

7/28/2019 2012-ss1

2/7

2

a) Calculate the reactive power generation at bus 1 in p.u. (You do NOT need thecomplete Ybus matrix.)

b) Calculate the real power flow on line 1-2, at bus 1 end.c) Calculate the total reactive power generated by the shunt capacitors of the 3 lines

equivalents.

Solutions:

0.951-15.64

Qg1

2

1

3

10

1.011.98O

a) Y12 =- 1/ (r12+jx12) =- 1/(0.018+j0.2) =G12 +jB12=-0.4464 +j 4.9598Y13 =- 1/ (r13+jx13) =- 1/(0.009+j0.1) =G13 +jB13=-0.8928 +j 9.9197Y11 =- y12 -y13+jb12/2+jb13/2=-(-0.4464 +j 4.9598) - (-0.8928 +j 9.9197)

+j0.220+j0.110 =1.3392 j 14.5495

Qg1= -V12B11 +V1V2[G12sin12 - B12cos12] +V1V3[G13sin13 - B13cos13]

=-0.0888 p.u.(Note that12,13are Voltage angle difference)

b) I12 = (V1-V2) 1/( R12+jX12) =(10o- 0.951-15.64o) (0.4464 j4.9598)=1.3093 j 0.3032 p.u.

S12=V1I12* =1.3093 +j0.3032P12=Re( S

12) =1.3093 p.u.

c) Qshunt =V12(b13/2 +b12/2) +V2

2 (b21/2 +b23/2) +V32(b31/2 +b32/2) =

=1.02(0.110 +0.220) +0.9512 (0.220 +0.430) +1.012(0.110 +0.430)=1.47 p.u.

7/28/2019 2012-ss1

3/7

3

Problem 2:This problem is about the Newton Raphson power flow algorithm. A 3-buspower system is shown below. Bus 1 is the slack bus of this system. All values are in perunit.

Now suppose the generating unit at bus 2 is to be completely shut down due to technicalproblems. In other words, the power plant is to be disconnected from the system. Followthe next steps to find the resulting system operating condition.

a) What equations are needed for the power flow solution after the generating plant atbus 2 is completely disconnected? You should minimize the number of equations youneed for the N-R iteration to find the complex bus voltages.

( ) ( )

( )[ ] ( )332333231331133331331133

cos11.111.1cos)(

sin11.1sin)(

VVBVBVVQ

VBVVP

=+=

==

x

x

=

11.111.1

11.111.1

jj

jjYbus

b) Find the Jacobian matrix for the system after the generator unit at bus 2 isdisconnected. You need to write down the definition (of Jacobian) first, and thenformulas and their values for the flat start condition.

V2 2

V3 3

V1=001 p.u.

8.113 jz =

9.023 jz =

9.012 jz =

2.01.03 jSD +=

7/28/2019 2012-ss1

4/7

4

( )

=

=

3

3

3

3

3

3

3

3

3

3

VQQ

V

PP

J

V

x

x

Definition

( )( ) ( )( ) ( )

=

3313333313

3313313

cos2sin

sincos

BVBBV

BBVJ x Formulas

( )

=

11.10

011.1xJ Values for the flat start condition

c) Now calculate the voltages at V2 and V3 based on one iteration of the NewtonRaphson algorithm.

( )( )

=

=

=

2.0

1.0

0

0

2.0

1.00

3

03

3

3

0

3

3

x

x

Q

P

Q

P

Q

P

=

11.10

011.10J

=

9009.00

09009.01

0J

=

=

=

=

=

1802.01623.5

1802.00901.0

2.01.0

01

0

3

310

0

3

30 JQPJ

Vx

=

+

=+=

8195.0

1623.5

1802.0

1623.5

1

0 00001xxx

Therefore after the first iteration 03 1623.58195.0 =V

Using the calculated values for the voltage magnitude and angle at bus 3, the

voltage at bus 2 can be found.

( ) ( ) 00003112 3249.29088.01623.58195.0012

101

2

1=== VVVV

7/28/2019 2012-ss1

5/7

5

Problem 3: Consider the following power system. All quantities are in per unit. All linesare purely inductive. The reactance of all transmission lines is 0.5 p.u. Ignore shuntcapacitances of transmission lines. Bus 1 is the slack bus.

Follow the steps below to perform one iteration of thefast-decoupled power flow. Useflat start to initialize the power flow algorithm. a) Write the 3 equations needed to

calculate the power flow solution. b) Calculate one iteration of the P- equations.Compute the new voltage angles. Calculate one iteration of the Q-V equation. Computethe new voltage magnitude.

Solutions:

The bus admittance matrix is:

=

422

242

224

jjj

jjj

jjj

Y bus

The power flow equations are:

0coscos

0sinsin

0sinsin

31311332322333

2

3

313113323223

2232332212112

===

D

D

g

QBVVBVVBV

PBVVBVV

PBVVBVV

Calculate one iteration of the P- equations. Compute the new voltage angles.

For a flat start: V1(0) =V3

(0) =1.0 , V2(0) = 1.05 p.u. and1

(0) =2(0) =3

(0) =0.0 rad.

P2(0) =1.05*1.0*2*0.0 +1.05*1.0*2*0.0 =0

P3(0) =1.0*1.05*2*0 +1.0*1.0*2*0 =0

LoadPD+jQD=1+j1 p.u.

V3

1

3

2

V1=1/0PG2=0.5 p.u.V2=1.05

7/28/2019 2012-ss1

6/7

6

P2(0) =0.5 - 0 =0.5 p.u.

P3(0) =-1.0 - 0 =-1 p.u.

=

=

=

254.0

10*937.7

0.1

476.0

42

24

12

1

0.1/0.1

05.1/5.0

42

24

3

3

2

3

2

3

3

2

2

3

2

3332

2322

VP

V

P

BB

BB

2(1) =0 - 7.937*10-3 =- 7.937*10-3 rad. =-0.455

3(1) =0 - 0.254 =-0.254 rad. =-14.55

Calculate one iteration of the Q-V equation. Compute the new voltage magnitude.

Q3(0) =-1.0 2*(-4) - 1.0*1.05*2*cos32

(1) - 1.0*1.02*cos31(1) =-0.09996 -0.1 p.u.

Q3(0) =-1.0 - (-0.1) =-0.9 p.u.

[ ] [ ]225.0

0.1

9.0

4

13

3

3333

=

V

V

QVB

V3(1) =1.0 - 0.225 =0.775 p.u.

7/28/2019 2012-ss1

7/7

7

Problem 4: DC load flow

A three-bus power system has a double circuit between bus 2 and bus 3 as shown inFigure P1. Suppose one of the double lines is to be scheduled for maintenance.

PG2

PG1

PLoad

To bemaintained

B=2

B=2

B=4(total)

PG1=0.05 p.u.

Pload=0.2 p.u.

V1,V2,V3=1 p.u.

03

=(for reference)

Bus1 Bus2

Bus3

a) Find the B matrix and Z=B-1

matrix for the system condition before the dotted line isde-energized for maintenance.

B old =

62

24, Zold =j

2.01.0

1.03.0

b) Use the DC load flow method to find the nodal angles 21, after

the dotted line is de-

energized for maintenance.

B new=

42

24 , inverse(B new)=

33.017.017.033.0

2

1

=

33.017.0

17.033.0

2.0

05.0=

0575.0

0175.0rad=

29.3

1