Tutorial 1 Latest

8/17/2019 Tutorial 1 Latest

1/12

1. In the two-bus system shown in Figure 1, bus 1 is a slack bus with 01

01.0V pu. A

load of 100 MW and 50 MVAR is taken from bus 2. The line impedance is

z 12 = 0.12 + j 0.16 pu on a base of 100 MVA. Using Newton-Raphson method, obtain

the voltage magnitude and phase angle of bus 2. Start with an initial estimate of)0(

2V = 1.0 pu and (0)

2δ = 0. Perform two iterations.

Figure 1

Y =

j43 j43 j43 j43

Y

55

55 and θ =

00

00

53.13126.87

126.8753.13

In this problem 0.5QI

1.0PI

2

2

and unknown quantities =

2

2

V

δ

With initial value0

1 01.0V ;0

2 01.0V

iP =

ii

2

iGV +

N

in

1n

iV

nV

niY cos (

niθ +

nδ -

iδ )

iQ =

ii

2

iBV

N

in

1n

iV

nV

niY sin (

niθ +

nδ -

iδ )

2P = )δδθ(cosYVVGV

2112121222

2

2

= (1 x 1 x 3 ) + (1 x 1 x 0)0(126.87cos5 0 x ) = 0 pu

22

2

22BVQ

2V

1V

21Y sin (

21θ +

1δ -

2δ )

= -(1 x 1 x -4) - (1 x 1 x 5 x )00(126.87sin0

= 0 pu


2/12

1.001.0PPIΔP222

pu

0.500.5QQIΔQ222 pu

ii

Hii

2

iiBVQ

iiN iP + ii2

i GV

ii

M i

P -ii

2

iGV

ii

L i

Qii

2

iBV

22

2

2222BVQH 0 - ( 1 x 1 x -4 ) = 4

22

N 2

P +22

2

2GV = 0 + (1 x 1 x 3) = 3

22

M 2

P -22

2

2GV = 0 - (1 x 1 x 3) = -3

22

2

2222BVQL 0 - ( 1 x 1 x -4 ) = 4

43

34

2

2

2

V

VΔ

Δδ

=

0.5

1.0

2

2

2

V

VΔ

Δδ

=

0.2

0.1

0)1(

2.73rad.0.10.1)0δ 5(

pu0.80.2)1.0V)1(

2 (

0

101.0V

0

2.730.8V 5

This completes the first iteration.


3/12

Second iteration:

2P = )δδθ(cosYVVGV

2112121222

2

2

= (0.8 x 0.8 x 3 ) + (0.8 x 1 x 5.73))(-0(126.87cos50

x ) = -0.7875 pu

22

2

22BVQ

2V

1V

21Y sin (

21θ +

1δ -

2δ )

= -(0.8 x 0.8 x -4) - (0.8 x 1 x 5 x )5.73))(0(126.87sin0

= -0.3844 pu

0.2125-0.7875)1.0PPIΔP222

( pu

0.11560.3844)(0.5QQIΔQ222

pu

22

2

2222BVQH -(-0.38440) - (0.8 x 0.8 x -4 ) = 2.9444

22

N 2

P +22

2

2GV = -0.7875 + (0.8 x 0.8 x 3) = 1.1325

22

M 2

P -22

2

2GV = -0.7875 - (0.8 x 0.8 x 3) = -2.7075

22

2

2222BVQL -0.3844 - ( 0.8 x 0.8 x -4 ) = 2.1756

2.17562.7075

1.13252.9444

2

2

2

V

VΔ

Δδ

=

0.1156

0.2125

2

2

2

V

VΔ

Δδ

=

0.0967

0.0350

2

VΔ - 0.0967 x 0.8 = -0.0773

0)(2

27.73rad.0.1350.035)(0.1-δ

pu0.72270.0773)0.8V)2(

2 (

0

101.0V

0

27.730.7227V

This completes the second iteration.


4/12

2. Figure 2 shows the one-line diagram of a simple three-bus power system with

generation at buses 1 and 2. The voltage at bus 1 is 01

01.0V per unit. Voltage

magnitude at bus 2 is fixed at 1.05 pu with a real power generation of 400 MW. A

load consisting of 500 MW and 400 MVAR is taken from bus 3. Line admittances

are marked in per unit on a 100 MVA base. For the purpose of hand calculations,

line resistances and line charging susceptances are neglected. UsingNewton-Raphson method, start with the initial estimates of

)0(

2V = 1.05 + j 0 and

)0(

3V = 1.0 + j 0, and keeping

2V = 1.05 pu, determine the phasor values of

2V and

3V . Perform two iterations.

Figure 2

Y =

000

000

000

904090209020

902090609040

902090409060

Y

402020

206040

204060

and θ =

000

000

000

909090

909090

909090

In this problem

4QI

5PI

4PI

3

3

2

and unknown quantities =

3

3

2

V

δ

δ

With flat start0

101.0V ;

0

201.05V ;

0

301.0V


5/12

ii

2

iiii

iiiiii

ii

2

iiii

BVQL

PM;PN

BVQH

)δδ(sinYVVM

)δδ(sinYVVN

)δδ(cosYVVH

i j ji ji ji

i j ji ji ji

i j ji ji ji

iP =

ii

2

iGV +

N

in

1n

iV

nV

niY cos (

niθ +

nδ -

iδ )

iQ =

ii

2

iBV

N

in

1n

iV

nV

niY sin (

niθ +

nδ -

iδ )

2P = )δδθ(cosYVV)δδθ(cosYVVGV

233232322112121222

2

2

= 0 + 0))90(cosx20x1x(1.05))90(cosx40x1x1.05(00


323232233113131333

2

3

= 0 + 0))90(cosx20x1.05x1())90(cosx20x1x(100

22

2

22BVQ

2V

1V )δδθ(sinY

211212)δδ(θsinYVV

232332

23

= 60)-x1.05x(1.05- ))(90sinx40x1x(1.050

))(90sinx20x1x(1.05-0 = 3.15

33

2

33BVQ

3V

1V )δδθ(sinY


3232323

2

= 40)-x1x(1- ))(90sinx20x1x(10

))(90sinx20x1.05x(1-0 = -1

404PPIΔP222

505PPIΔP333

31)(4QQIΔQ333

Sinceii

G are zero and ji

θ are0

90

22

2

2222BVQH -3.15 + ( 1.05 x 1.05 x 60 ) = 63

33

2

3333BVQH = 1 + ( 1 x 1 x 40 ) = 41

0PM;0PN333333

33

2

3333BVQL -1 + ( 1 x 1 x 40 ) = 39

32

H2

V3

V )δδ(cosY2332

= 211x20x1x1.05 and23

H = -21

32

N2

V3

V 0)δδ(sinY2332

; 32

M3

V2

V 0)δδ(sinY3223


6/12

3900

04121

02163

3

3

3

2

V

VΔ

Δδ

Δδ

=

3

5

4

3

3

3

2

V

VΔ

Δδ

Δδ

=

0.0769-

0.1078-

0.0275

0)1(

21.5782rad.0.02750.02750δ

0)1(

36.179rad.0.10780.10780δ

0.92310.07691.0V)1(

3

0

101.0V

0

21.581.05V

0

36.180.9231V

This completes the first iteration.

After second iteration :

0)2(

21.61rad.0.02810.00060.0275δ

0)2(

3

6.898rad.0.12040.01260.1078δ

0.90560.01750.9231V)2(

3 pu

Thus at the end of second iteration :

0

101.0V

0

21.611.05V

0

36.90.9056V


7/12

3. Referring to question 2 above, perform 2 iterations in computing :

a) Power flow using Decouple Power Flow method.

b) Power flow using Fast Decouple Power Flow method.

Figure 2

a) Power flow using Decouple Power Flow method.

4121

2163

3

2

Δδ

Δδ

=

5

4

0)1(

2

1.5756rad.0.02750.02750δ 0)1(

36.1765rad.0.10780.10780δ

0

101.0V

0

21.581.05V

0

36.181.0V

33

2

33BVQ

3V

1V )δδθ(sinY


3232323

2

= 40)-x1x(1- )))6.18(0(90insx20x1x(100

))6.18)(1.58(90insx20x1.05x(1- 00 = -0.6915

3.3085-0.6915)(4QQIΔQ333

33

2

3333BVQL -0.6915 - ( 1 x 1 x -40 ) = 39.3085


8/12

ii

2

iiii

ii

2

iiii

BVQL

BVQH

)δδ(cosYVVHi j ji ji ji

39.3085

3

3

V

VΔ = 3.3085

3VΔ = - 3.3085 x 1.0 / 39.3085 = - 0.0842

0.91580.08421.0V)1(

3

After first iteration :

0

101.0V

0

21.581.05V

0

36.180.9158V

Second iteration:


233232322112121222

2

2

= 0 +

3.7548))90(cosx20x0.9158x(1.05))90(cosx40x1x1.05(00

000

58.118.658.1


323232233113131333

2

3

= 0 +

4.5685))90(cosx20x1.05x())90(cosx20x1x(0.915800

000

18.658.19158.018.6

22

2

22BVQ

2V

1V )δδθ(sinY


232332

23

=

))58.118.690sin(x20x9158.0x05.1())58.190sin(x40x1x05.1( 00000

60)-x1.05x(1.05-

33

2

33BVQ

3V

1V )δδθ(sinY


3232323

2

=

.658.190sin(x20x05.1x9158.0())18.690sin(x20x1x9158.0( 0000

40)-x0.9158x(0.9158-

= -3.7177

2452.07548.3 4PPIΔP222

4315.0)5685.4( 5PPIΔP333

0.2823)(4QQIΔQ333

7177.3

Sinceii

G are zero and ji

θ are0

90


9/12

22

2

2222BVQH -5.1103 + ( 1.05 x 1.05 x 60 ) = 61.0397

33

2

3333BVQH = 3.7177 + ( 0.9158 x 0.9158 x 40 ) = 37.2653

32

H2

V3

V )δδ(cosY2332

= -1.05 x 0.9158 x 20 x cos (-6.180-1.58

0)

= -19.0557

23H = -19.0557

37.2653.0557

.055761.0397

19

19

3

2

Δδ

Δδ

=

0.4315

0.2452

3

2

Δδ

Δδ

=

0.0113

0.0005

0)2(

21.60rad.0.02800.00050.0275δ

0)2(

36.82rad.0.11910.01130.1078δ

0

101.0V

0

21.05V 6.1

0

30.9158V 85.6

33

2

33BVQ

3V

1V )δδθ(sinY


3232323

2

=

8.66.190sin(x20x05.1x9158.0())85.690sin(x20x1x9158.0(

0000

40)-x0.9158x(0.9158- = -3.6607

0.3393-.6607)(4QQIΔQ333

3

33

2

3333BVQL -3.6607 + ( 0.9158 x 0.9158 x 40 ) = 29.8869

29.8869

3

3

V

VΔ = 0.3393

3VΔ = - 0.3393 x 0.9158 / 29.8869 = - 0.0104

0.90540.01040V)1(

3 9158.

Thus at the end of second iteration, bus voltages are

0

101.0V

0

21.V 6.105

0

36.850.9054V


10/12

B =

b) Power flow using Fast Decouple Power Flow method.

40

60

60

20203

20402

20401

321

The constant matrices are

40203

2602

32

0

Initial solution is

0

3

0

2

0

1

01.0V

01.05V

01.0V

4ΔP2 ; 5ΔP

3

5V

ΔP;3

V

ΔP

3

3

2

2 8095.

V

ΔPΔδB

'

40 20

2060

3

2

Δδ

Δδ

=

5

3.8095

On solving the above

3

2

Δδ

Δδ

=

0.1119

0.0262

0)1(

2rad.0.02620.02620δ 50.1

0)1(

341rad.0.11190.11190δ .6

0

3

0

2

0

1

411.0V

1.05V

01.0V

.6

50.1

'

B and "

B 40


11/12

33

2

33BVQ

3V

1V )δδθ(sinY


3232323

2

= ))41.65.190sin(x20x05.1x1())41.690sin(x20x1x1( 00000

40)-x1x(1-

= -0.6752

3.3248-)(4QQIΔQ 333 6752.0

V

ΔQVΔB

"

40 3

VΔ = - 3.3248 i.e.3

VΔ = - 0.0831

0.91690.08311.0V)1(

3

At the end of first iteration, bus voltages are

0

3

0

2

0

1

410.9169V

1.05V

01.0V

.6

50.1

Second iteration:


233232322112121222

2

2

= 0 +

3.7492))90(cosx20x0.9169x(1.05))90(cosx40x1x1.05(00

000

50.141.65.1


323232233113131333

2

3

= 0 +

6971.441.65.19169.041.6 000

))90(cosx20x1.05x())90(cosx20x1x(0.916900

2508.07492.3 4ΔP2

.30295ΔP3

06971.4

3304.02389.0 3

3

2

2

V

ΔP

;V

ΔP

V

ΔPΔδB

'

40 20

2060

3

2

Δδ

Δδ

=

3304.0

0.2389


12/12

3

2

Δδ

Δδ

=

0.0075

0.0015

0)1(

2rad.0.02770.001502620δ 59.1.

0)1(

3 6.84rad.0.11940.00750.1119δ

0

3

0

2

0

1

0.9169V

1.05V

01.0V

84.6

59.1

33

2

33BVQ

3V

1V )δδθ(sinY


3232323

2

=

.659.190sin(x20x05.1x9169.0())84.690sin(x20x1x9169.0( 0000

40)-x0.9169x(0.9169-

= -3.6261

0.3739-)(4QQIΔQ333

6261.3

V

ΔQVΔB

"

40 3

VΔ = - 0.4078 i.e.3

VΔ = - 0.0102

0.90670.01020.9169V)1(

3

At the end of second iteration, bus voltages are

0

3

0

2

0

1

0.9067V

1.05V

01.0V

84.6

59.1

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