Exercises of course ELEC0029 Power flow computation and...

Exercises of course ELEC0029

Power flow computation and sensitivity analysisin a simple system

T. Van Cutsem

February 2018

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System data

Lines data

UN (kV) X (Ω/km) X/R ωC/2 (µS/km) SN (MVA) km

380 0.3 10 1.5 1350 50150 0.4 5 1.2 300 50

Transformers data

orig extr R (pu) X (pu) B (pu) SN (MVA) n (“extr” side)

A B 0 0.14 0 500 1.08F B 0 0.14 0 250 1.08D B 0 0.14 0 300 1.00E C 0 0.14 0 600 1.00

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Operating point data

I generators:I voltage at bus A: 15 kVI voltage at bus F: 15 kVI production of F: 150 MW

I loads:I at bus E: 400 MW / 120 MvarI at bus D: 100 MW / 30 Mvar

3 / 27

I Specify the data at all buses and run a power flow computation todetermine the electrical state of the system. In this computationassume that transformers D-B and E-C are of Yd11 type, and takeinto account the phase angle of their transformer ratios.

The data are available in the file 6bus.dat, while the results can bedisplayed on the one-line diagram by using the file 6bus.svg

I Detail the active and reactive power balances of the system.

I Check that line D-E and transformer C-E operate below theirthermal ratings.

4 / 27

Transformers D-B et E-C of type Yd11; complex ratios

150 kV

380 kVpuissances réactives (Mvar)

puissances actives (MW)

D E

A

B

C

modules tensions (pu)15 kV phases tensions (deg)

F

0.9912

1.0000

−106

−14

−53

14

54

352

163

348

128−347

−137

154

56

1.0363

1.0134

1.0194

−347

−5.9 −7.8

20.2

0.0

100

30

1.0000

150

79

−0.9

400

120

17.6

5 / 27

I Simulate the misuse of a transformer E-C of type Yy0, whiletransformer D-B is of type Yd11. Show that the resulting electricalstate is unacceptable.

6 / 27

Transformer D-B of type Yd11 and E-C of type Yy0

150 kV



D E

A

B

C


F

0.9205

1.0000

−207

87

−310

20

332

372

296

90

206−90

−221

432

147

0.9967

0.9495

0.9732

−90

−6.5 −6.9

11.2

0.0

100

30

1.0000

150

145

−1.3

400

120

−8.3

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I Getting back to the correct configuration of the transformers,perform a new power flow computation in which the phase angles ofthe transformer ratios are set to zero. Comment on the resultsobtained.

Unless otherwise specified, the phase angles are neglected in thesequel.

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Phase angles of transformer ratios neglected

150 kV



D E

A

B

C


F

−12.4

0.9912

1.0000

−106

−14

−53

14

54

352

163

348

128−347

−137

154

56

1.0363

1.0134

1.0194

−347

−5.9 −7.8

−9.8

0.0

100

30

1.0000

150

79

−0.9

400

120

9 / 27

I Assuming that transformer D-B is equipped with a phase shiftingdevice allowing to vary progressively the phase angle of itstransformer ratio, study the impact of this phase angle on branchpower flows and bus voltage magnitudes.

10 / 27

Varying the phase angle of the ratio of transformer D-B

1∠0 for E-C and 1∠φ for D-B. Example for φ = −10.

150 kV



D E

A

B

C


F

−10.6

0.9815

1.0000

−132

12

−153

7

157

355

179

248

136−247

−151

257

68

1.0315

1.0073

1.0145

−247

−6.0 −7.3

−2.6

0.0

100

30

1.0000

150

87

−0.9

400

120

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Variation of φ

-100

0

100

200

300

400

500

600

-15 -10 -5 0 5 10 15

phi (deg)

P B-C (MW)P D-E (MW)P B-D (MW)

0

20

40

60

80

100

120

140

160

-15 -10 -5 0 5 10 15

phi (deg)

Q B-C (MVar)Q D-E (MVar)Q B-D (MVar)

0.97

0.975

0.98

0.985

0.99

0.995

1

1.005

1.01

1.015

1.02

-15 -10 -5 0 5 10 15

phi (deg)

V C (pu)V D (pu)V E (pu)

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I Assuming that transformer D-B is equipped with a tap changerallowing to vary progressively (the magnitude of) its transformerratio, study the impact of this ratio on branch power flows and busvoltage magnitudes.

I Same question for an identical variation of the ratios of bothtransformers D-B an E-C.

13 / 27

Varying the (magnitude) of the ratio of transformer D-B

1∠0 for E-C and n∠0 for D-B. Example for n = 0.95.

150 kV



D E

A

B

C


F

−12.2

1.0029

1.0000

−74

−46

−61

48

62

352

162

340

91−339

−102

162

91

1.0364

1.0541

1.0232

−339

−5.9 −7.8

−9.7

0.0

100

30

1.0000

150

79

−0.9

400

120

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Variation of n

0

50

100

150

200

250

300

350

400

90 95 100 105 110

n (pu)


-50

0

50

100

150

200

90 95 100 105 110

n (pu)


0.94

0.96

0.98

1

1.02

1.04

1.06

1.08

1.1

90 95 100 105 110

n (pu)


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Same variation of the ratios of transformers D-B et E-C

n∠0 for both transformers. Example for n = 0.95.

150 kV



D E

A

B

C


F

−11.8

1.0487

1.0000

−105

−15

−55

15

56

352

159

346

123−345

−133

156

55

1.0374

1.0705

1.0210

−345

−5.9 −7.8

−9.5

0.0

100

30

1.0000

150

77

−0.9

400

120

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Same variation of n in both transformers

50

100

150

200

250

300

350

400

90 95 100 105 110

n (pu)


0

20

40

60

80

100

120

140

90 95 100 105 110

n (pu)


0.85

0.9

0.95

1

1.05

1.1

1.15

90 95 100 105 110

n (pu)


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I Let ARTERE (the power flow computation software) adjustautomatically the ratios of both transformers D-B and E-C so as tobring the voltages at buses D and E in the dead-band [1.02 1.04] pu

18 / 27

Automatic adjustment of the ratios of both transformersto bring the voltages at buses D and E in [1.02 1.04] pu

iter max mismatches : MW Mvar phi(MVA)

1 395.2 407.1 605.1

2 27.4 62.4 78.0

3 1.0 1.2 2.0

4 0.1 0.1 0.1

in-phase trfo taps : 2 down 0 up

4 4.4 58.8 90.6

5 0.2 0.8 1.0

6 0.0 0.1 0.1


6 4.5 61.1 83.7

7 0.1 0.7 0.7

8 0.0 0.0 0.0


8 4.5 63.5 86.9

9 0.1 0.7 0.7

10 0.0 0.0 0.0

transfo ctld bus volt set-pt tol ratio tap max

E-C E 1.0275 1.0300 0.0100 0.963 14 25

D-B D 1.0350 1.0300 0.0100 0.988 16 2519 / 27

Situation after the transformers have been adjusted

150 kV



D E

A

B

C


F

−12.1

1.0275

1.0000

−121

1

−51

−2

51

352

161

351

143−349

−151

151

38

1.0368

1.0350

1.0184

−349

−5.9 −7.8

−9.6

0.0

100

30

1.0000

150

78

−0.9

400

120

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I Study the effect of varying the voltage of generator F on thereactive power productions of both generators, on the branch powerflows and on the bus voltage magnitudes.

I Same question for an identical variation of the voltages of bothgenerators A and F.

21 / 27

Varying the voltage of generator F

Example: VF = 0.98 pu

150 kV



D E

A

B

C


F

−12.5

0.9828

1.0000

−106

−14

−53

14

54

352

189

348

129−347

−137

154

56

1.0283

1.0052

1.0112

−347

−5.9 −7.9

−9.9

0.0

100

30

0.9800

150

55

−0.8

400

120

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Variation of VF

0

50

100

150

200

250

0.96 0.98 1 1.02 1.04

V F (pu)

Q Gen A (MVar)Q Gen F (MVar)

0.97

0.98

0.99

1

1.01

1.02

1.03

1.04

0.96 0.98 1 1.02 1.04

V F (pu)


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Variation of VF

50

100

150

200

250

300

350

0.96 0.98 1 1.02 1.04

V F (pu)


0

20

40

60

80

100

120

140

0.96 0.98 1 1.02 1.04

V F (pu)


24 / 27

Same variation of voltages of generators A and F

Example: VF = VA = 0.98 pu

150 kV



D E

A

B

C


F

−13.0

0.9658

0.9800

−106

−14

−53

15

54

352

168

348

131−347

−139

154

57

1.0124

0.9888

0.9949

−347

−6.2 −8.1

−10.3

0.0

100

30

0.9800

150

81

−0.9

400

120

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Variation of VF = VA

60

80

100

120

140

160

180

0.96 0.98 1 1.02 1.04

V Gen (pu)

Q Gen A (MVar)Q Gen F (MVar)

0.92

0.94

0.96

0.98

1

1.02

1.04

1.06

1.08

1.1

0.96 0.98 1 1.02 1.04

V Gen (pu)

V C (pu)V D (pu)V E (pu)V B (pu)

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Variation of VF = VA

50

100

150

200

250

300

350

0.96 0.98 1 1.02 1.04

V Gen (pu)


0

20

40

60

80

100

120

140

0.96 0.98 1 1.02 1.04

V Gen (pu)


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Exercises of course ELEC0029 Power flow computation and...

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