Get Instant Help From 5000+ Experts For
question

Writing: Get your essay and assignment written from scratch by PhD expert

Rewriting: Paraphrase or rewrite your friend's essay with similar meaning at reduced cost

Editing:Proofread your work by experts and improve grade at Lowest cost

And Improve Your Grades
myassignmenthelp.com
loader
Phone no. Missing!

Enter phone no. to receive critical updates and urgent messages !

Attach file

Error goes here

Files Missing!

Please upload all relevant files for quick & complete assistance.

Guaranteed Higher Grade!
Free Quote
wave

Describe about the Power Systems For the Positive Sequence Impedance.

Background

The "remote area Microgrids" being developed in remote areas of far NW West Australia consist of basically a zone substation  supplying a local town supply, a long distribution feeder of poor reliability which supplies a small remote town some distance from the main zone substation supply.  The remote town system is arranged in a normally open local ring.  

MV:  Group A=11 kV, Group B=13.8 kV,  Group C=16.8 kV; balanced 3-phase.  

F1, F2, F3, F4 are fault locations.  Fault level on zone substation MV bus = Group A 200 MVA, Group B 250 MVA, Group C 300 MVA for balanced 3-phase; and Group A 200 MVA, Group B 250 MVA, Group C 350 MVA for single-phase to earth faults.

Line segments:  Line from Bus 1to Bus 2: 1 km of  impedance = 0.27 + j0.31 ohms/km;  Line from Bus 2 to Bus 3: 10 km of impedance = Group A 0.38 + j0.32,  Group B 0.475 + j0.40, Group C 0.57 + j0.48 ohms/km; each arm of remote microgrid ring (i.e. Bus 3 to Bus 4): 1 km of impedance = Group A 0.38 + j0.32,  Group B 0.475 + j0.40, Group C 0.57 + j0.48 ohms/km.  These are positive sequence impedances.  Assume negative sequence impedances equal positive sequence and zero sequence impedances are 2.5 times the positive sequence.

Line loading:  Line from Bus 1to Bus 2 (town distribution network) group A 1,000 kVA; group B 1,250 kVA; group C 1,500 kVA.  Line from Bus 2 to Bus 3: Group A 500kVA; Group B 625kVA; Group C 750kVA; each arm of remote microgrid ring (i.e. Bus 3 to Bus 4):group A 500kVA; Group B 625kVA; Group C 750kVA each, total 1,000/1250/1500 kVA.  All loads are at 0.9 PF lagging and evenly distributed along the line segments.

The distributed generators (DG's) in the remote microgrid are all solar PV types.  Peak output equals the peak load in the microgrid. The DG inverter fault levels are regulated by the inverter controls to 2.0 times full load current for both three-phase and single-phase to earth faults.  Internal impedance appears to the circuit to be highly resistive.

Main system protection: Overcurrent and earth fault relays are provided at CB3, CB4 and both CB1's.

Please undertake the following analyses:

Calculate the 3-phase fault levels on Bus 2, Bus 3 and Bus 4 with the main supply connected but DG's disconnected.  Then calculate the fault levels on Bus 3 and Bus 4 with the microgrid islanded (synch switch open).  To do this, assume that there is a total of (group A/B/C) 500/625/750 kW of DG output in the centre of each arm of the

  1.  Then calculate the fault levels on Bus 3 and Bus 4 with the microgrid DG's in service and the main grid also connected (parallel supply).

Zero sequence impedance:

Positive sequence impedance:

Negative sequence impedance:

Assume line to line fault between lines b and c:

The first step in solving questions 2-5 is to determine the zero, negative and positive sequence impedance when looked from bus 2. We then draw the sequence diagram which will assist us in calculating the equivalent Thevanin’s impedance. We then use the formulas in the lecture notes to determine the sequence currents.

Network Data

The voltage magnitude and phase currents for each bus and corresponding phase are shown below;

Number

Name

Phase Volt A

Phase Volt B

Phase Volt C

Phase Ang A

Phase Ang B

Phase Ang C

1

One

0.64022

0.64022

0.64022

0

-120

120

2

Two

0.51217

0.51217

0.51217

0

-120

120

3

Three

0.69145

0.69145

0.69145

0

-120

120

4

Four

0.20484

0.20484

0.20484

0

-120

120

5

Five

0.79389

0.79389

0.79389

0

-120

120

6

Six

0

0

0

0

0

0

7

Seven

0.78108

0.78108

0.78108

0

-120

120

Table 4: Per unit bus voltage magnitude

Number of Bus

Name of Bus

Phase Cur A

Phase Cur B

Phase Cur C

Phase Ang A

Phase Ang B

Phase Ang C

5

Five

0.64028

0.64028

0.64028

-90

150

30

6

Six

4.375

4.375

4.375

-90

150

30

7

Seven

0.89639

0.89639

0.89639

-90

150

30

Table 5: Per unit current values supplied by each generator

Number of Bus

Name of Bus

Phase Cur A

Phase Cur B

Phase Cur C

Phase Ang A

Phase Ang B

Phase Ang C

5

Five

26787.18

26787.18

26787.18

-90

150

30

6

Six

140328.2

140328.2

140328.2

-90

150

30

7

Seven

25876.41

25876.41

25876.41

-90

150

30

Table 6: Actual current values(A) supplied by each generator

Calculate the voltage drop at Bus 3 with the main supply connected but DG's disconnected at peak loading.  Assume even loading; the volt drop is then equal to (half total load at the end of the line) * (line impedance).  Then calculate the voltage drop at Bus3 and Bus 4 with the microgrid islanded, again at peak loading times.  Although microgrid loads and DG's are evenly spread around the whole microgrid ring, for the purposes of modelling assume that the total microgrid peak load of (group A/B/C) 1000/1250/1500 kW is met by the DG's modelled as before as two of (group A/B/C) 500/625/750 kW sources, one in the middle of each arm of the microgrid.

When a fault is placed at Bus 2

The voltage magnitude and phase currents for each bus and corresponding phase are shown below;

Number

Name

Phase Volt A

Phase Volt B

Phase Volt C

Phase Ang A

Phase Ang B

Phase Ang C

1

One

0.25

0.25

0.25

0

-120

120

2

Two

0

0

0

0

0

0

3

Three

0.35

0.35

0.35

0

-120

120

4

Four

0.3663

0.3663

0.3663

0

-120

120

5

Five

0.55

0.55

0.55

0

-120

120

6

Six

0.61044

0.61044

0.61044

0

-120

120

7

Seven

0.525

0.525

0.525

0

-120

120

Table 1: Per unit bus voltage magnitude

Number of Bus

Name of Bus

Phase Cur A

Phase Cur B

Phase Cur C

Phase Ang A

Phase Ang B

Phase Ang C

5

Five

1.25

1.25

1.25

-90

150

30

6

Six

1.8315

1.8315

1.8315

-90

150

30

7

Seven

1.75

1.75

1.75

-90

150

30

Table 2: Per unit current values supplied by each generator

Number of Bus

Name of Bus

Phase Cur A

Phase Cur B

Phase Cur C

Phase Ang A

Phase Ang B

Phase Ang C

5

Five

52296.22

52296.22

52296.22

-90

150

30

6

Six

58745.45

58745.45

58745.45

-90

150

30

7

Seven

50518.15

50518.15

50518.14

-90

150

30

Table 3:Actual current values(A) supplied by each generator

7.25: When the fault is placed at bus 6, the following results are obtained;

The voltage magnitude and phase currents for each bus and corresponding phase are shown below;

Number

Name

Phase Volt A

Phase Volt B

Phase Volt C

Phase Ang A

Phase Ang B

Phase Ang C

1

One

0.64022

0.64022

0.64022

0

-120

120

2

Two

0.51217

0.51217

0.51217

0

-120

120

3

Three

0.69145

0.69145

0.69145

0

-120

120

4

Four

0.20484

0.20484

0.20484

0

-120

120

5

Five

0.79389

0.79389

0.79389

0

-120

120

6

Six

0

0

0

0

0

0

7

Seven

0.78108

0.78108

0.78108

0

-120

120

Table 4: Per unit bus voltage magnitude

Number of Bus

Name of Bus

Phase Cur A

Phase Cur B

Phase Cur C

Phase Ang A

Phase Ang B

Phase Ang C

5

Five

0.64028

0.64028

0.64028

-90

150

30

6

Six

4.375

4.375

4.375

-90

150

30

7

Seven

0.89639

0.89639

0.89639

-90

150

30

Table 5: Per unit current values supplied by each generator

Number of Bus

Name of Bus

Phase Cur A

Phase Cur B

Phase Cur C

Phase Ang A

Phase Ang B

Phase Ang C

5

Five

26787.18

26787.18

26787.18

-90

150

30

6

Six

140328.2

140328.2

140328.2

-90

150

30

7

Seven

25876.41

25876.41

25876.41

-90

150

30

Table 6: Actual current values(A) supplied by each generator

7.26: When fault is placed midway between buses 2 and 3;

This is done by right setting the location at 50%

The voltage magnitude and phase currents for each bus and corresponding phase are shown below;

Number

Bus Name

Phase Volt A

Phase Volt B

Phase Volt C

Phase Ang A

Phase Ang B

Phase Ang C

1

One

0.43151

0.43151

0.43151

0

-120

120

2

Two

0.23823

0.23823

0.23823

0

-120

120

3

Three

0.21

0.21

0.21

0

-120

120

4

Four

0.52142

0.52142

0.52142

0

-120

120

5

Five

0.66344

0.66344

0.66344

0

-120

120

6

Six

0.71017

0.71017

0.71017

0

-120

120

7

Seven

0.42

0.21

0.21

0

180

180

8

Faultpt

0.00000

0.00000

0.00000

0.00

0.00

0.00

Table 7: Per unit bus voltage magnitude

Number of Bus

Name of Bus

Phase Cur A

Phase Cur B

Phase Cur C

Phase Ang A

Phase Ang B

Phase Ang C

5

Five

0.96639

0.96639

0.96639

-90

150

30

6

Six

1.41595

1.41595

1.41595

-90

150

30

7

Seven

2.1

2.1

2.1

-90

150

30

Table 8: Per unit current values supplied by each generator

Number of Bus

Name of Bus

Phase Cur A

Phase Cur B

Phase Cur C

Phase Ang A

Phase Ang B

Phase Ang C

5

Five

40430.75

40430.75

40430.75

-90

150

30

6

Six

45416.71

45416.71

45416.71

-90

150

30

7

Seven

60621.78

60621.78

60621.78

-90

150

30

Table 9: Actual current values(A) supplied by each generator

For protection purposes, overcurrent and earth-fault relays are installed at CB's CB3, CB4 and each CB1.  The overcurrent relays at each CB has a pick-up current, as reckoned in primary amps, of 200% of the peak load flowing at that point (the extra 100% is to prevent spurious tripping under peak load conditions).  The single-phase to earth pickup currents can be set at 20% of the overcurrent (three-phase) setting on the corresponding overcurrent relay.

Design the overcurrent and earth fault relay settings (pick-up currents and time dial settings) at CB's CB3, CB4 and each CB1 when the feeder is supplied by the main grid only.   To do the design, start at the relays at CB1 and work your way up the feeder to the relays at CB3and CB4.  Overcurrent relay pick-up currents will be set at 150% of load.  The CB1relay time setting will be 0.1.  Then, using time grading techniques (use 0.35 sec grading margin between relays) determine the settings for firstly overcurrent and then earth fault relays at CB2, CB3, and CB4.  The earth fault relay pick-up current settings will be 20% of the corresponding overcurrent relay setting.  Again, start off with a CB1 relay time setting of 0.1, and use the standard inverse (SI) characteristic.  You can calculate directly from the SI relay formula, or use the overcurrent relay setting spreadsheet "Ocurrent.xls" which is provided for your use (or alternatively, make you own one!).

In questions 9-12 we are required to calculate the phase voltages at bus 3. Since we had already calculated the voltages at bus 2 we take the voltage sequence at bus 2 and sequence currents we then apply current divider rule to obtain currents flowing to or from bus 3. We then apply KVL to obtain the sequence voltages at bus 3 from where we convert the sequence voltage components to phase voltages.

Then analyse the situation in the microgrid when it is islanded.  Will the overcurrent relays at each CB1 (Bus #3) detect microgrid faults when fed by the DG's only?  Suggest some solutions to the problem.  As one possible solution, consider back-to-back directional overcurrent and earth fault relays at each CB1 and recommend settings for the backward direction.

No it will not detect the faults. Bus coupler and bus section circuit-breakers, are required to make and interrupt the maximum fault current associated with short-circuit infeeds from all connected circuits when energising a section of busbar which is still inadvertently earthed. The resultant direction of current flow in the substation is also indicated. The making onto such a fault also imposes maximum duty on substation infrastructure equipment.

Cite This Work

To export a reference to this article please select a referencing stye below:

My Assignment Help. (2021). Essay: Power Systems' Positive Sequence Impedance." (49 Characters). Retrieved from https://myassignmenthelp.com/free-samples/eng767-power-systems/backward-direction.html.

"Essay: Power Systems' Positive Sequence Impedance." (49 Characters)." My Assignment Help, 2021, https://myassignmenthelp.com/free-samples/eng767-power-systems/backward-direction.html.

My Assignment Help (2021) Essay: Power Systems' Positive Sequence Impedance." (49 Characters) [Online]. Available from: https://myassignmenthelp.com/free-samples/eng767-power-systems/backward-direction.html
[Accessed 24 February 2024].

My Assignment Help. 'Essay: Power Systems' Positive Sequence Impedance." (49 Characters)' (My Assignment Help, 2021) <https://myassignmenthelp.com/free-samples/eng767-power-systems/backward-direction.html> accessed 24 February 2024.

My Assignment Help. Essay: Power Systems' Positive Sequence Impedance." (49 Characters) [Internet]. My Assignment Help. 2021 [cited 24 February 2024]. Available from: https://myassignmenthelp.com/free-samples/eng767-power-systems/backward-direction.html.

Get instant help from 5000+ experts for
question

Writing: Get your essay and assignment written from scratch by PhD expert

Rewriting: Paraphrase or rewrite your friend's essay with similar meaning at reduced cost

Editing: Proofread your work by experts and improve grade at Lowest cost

loader
250 words
Phone no. Missing!

Enter phone no. to receive critical updates and urgent messages !

Attach file

Error goes here

Files Missing!

Please upload all relevant files for quick & complete assistance.

Other Similar Samples

support
Whatsapp
callback
sales
sales chat
Whatsapp
callback
sales chat
close