# Slack Bus And Slack Generator Engineering Essay

### Slack Bus And Slack Generator Engineering Essay

The Table below shows input informations of each busbar in the system used to work out the power flow and the simulation consequence harmonizing to direction described in inquiry 1.

## Input Data

[ Simulation Result ]

BUS 1

plutonium

P ( burden )

100 MW

Q ( burden )

0 Mvar

BUS 2

P ( burden )

200 MW

Q ( burden )

100 Mvar

CB of Generation

Open

BUS 3

1 plutonium

P ( Gen )

200 MW

P ( burden )

100 MW

Q ( burden )

50 Mvar

AVR

On

AGC

Off

## Slack coach and slack generator

In power flow computation, alone numerical solution can non be calculated without mention electromotive force magnitude and angle due to unequal figure of unknown variables and independent equations. The slack coach is the mention coach where its electromotive force is considered to be fixed voltage magnitude and angle ( 1a? 0A° ) , so that the assorted electromotive force angle difference among the coachs can be calculated regard. In add-on, the slack generator supplies as much existent power and reactive power as needed for equilibrating the power flow sing power coevals, load demand and losingss in the system while maintain the electromotive force changeless as 1a? 0A° . In existent power system, when comparatively weak system is linked to the larger system via a individual coach, this coach can stand for the big system with an tantamount generator maintaining the electromotive force changeless and bring forthing any necessary power like slack coach. [ 1 ]

## Remarks

BUS 2

PQ Bus

Generator is disconnected to Bus 2

BUS 3

PV Bus

Generator is connected to Bus 3 and the magnitude of electromotive force of generator support invariable by utilizing AVR

In general, each coach in the power system can be categorized into three coach types such as Slack Bus, Load ( PQ ) Bus, and Voltage Controlled ( PV ) Bus. The definition and difference between PQ Bus and PV Bus are described as follows ; [ 2 ]

PV Bus ( Generator Bus or Voltage Controlled Bus ) : It is a coach at which the magnitude of the coach electromotive force is kept changeless by the generator. Even though the coach has several generators and burden, if any generators connected to the coach modulate the coach electromotive force with AVR, so this coach is referred to PV Bus. For PV coach, the magnitude of the coach electromotive force and existent power supplied to the system are specified, and reactive power and angle of the coach electromotive force are consequently determined. If a preset upper limit and minimal reactive power bound is reached, the reactive end product of the generator remains at the limited values, so the coach can be considered as PQ Bus alternatively of PV Bus. [ 2 ]

PQ Bus ( Load Bus ) : It is a coach at which the electromotive force is changed depending on entire net existent power and reactive power of tonss and generators without electromotive force regulator. Therefore, in the power simulation and computation, the existent power and reactive power of the tonss are specified as input informations and consequently the electromotive force ( magnitude and angle ) is calculated based on the above input.

The following table specifies input and end product of each coach type in the power system simulation and computation.

Bus Type

Phosphorus

Q

( Magnitude )

I? ( Angle )

PQ Bus

Input signal

Input signal

End product

End product

PV Bus

Input signal

End product

Input signal

End product

Slack Bus

End product

End product

Input signal

Input signal

BUS 1

204.093

100

56.240

0

BUS 2

0

200

0

100

BUS 3

200

100

107.404

50

Entire

404.093

400

163.644

150

## Qgen – Qstored in burden = 13.644

Reason: Real power loss due to opposition of transmittal line and fanciful power storage due to reactance of transmittal line are the grounds for the difference between power coevals and load demand in the system.

## Stored

BUS 1 – Bus 2

102.714

100.650

2.064

56.653

49.773

6.88

BUS 1 – Bus 3

1.379

1.378

0.001

0.4141 )

0.4131 )

0.001

BUS 3 – Bus 2

101.378

99.350

2.028

56.990

50.227

6.763

Entire

Palestine liberation organizations =

## 4.093

Qstored in burden =

## 13.644

1 ) Imaginary power flows from Bus 3 to Bus 1.

The summing up of existent power losingss and fanciful power storage over the transmittal line are precisely same with entire difference between coevals and burden. Therefore, it is verified that the difference is shown over the transmittal line.

‘Kirchoff ‘ balance as each coach [ 4 ]

Bus1

I? P1 = + Pgen1 – Pload1 – P12 – P13 = 204.093 – 100 – 102.714 – 1.379 = 0

I? Q1 = + Qgen1 – Qload1 – Q12 – Q13 = 56.24 – 0 – 56.653 + 0.413 = 0

Bus2

I? P2 = + Pgen2 – Pload2 – P21 – P23 = 0 – 200 + 100.65 + 99.35 = 0

I? Q2 = + Qgen2 – Qload2 – Q21 – Q23 = 0 – 100 + 49.773 + 50.227 = 0

BUS3

I? P3 = + Pgen3 – Pload3 – P31 – P32 = 200 – 100 + 1.378 – 101.378 = 0

I? Q3 = + Qgen3 – Qload3 – Q31 – Q32 = 107.404 – 50 – 0.414 – 56.99 = 0

Harmonizing to the computation supra, as summing up of incoming & A ; surpassing existent power and fanciful power at each coach become zero, it is verified that each busbar obeys a ‘Kirchoff ‘ balance. In add-on, the entire power system is wholly balanced, because entire coevals power ( existent & A ; fanciful ) are equal to summing up of entire load demand and existent power loss & A ; stored fanciful power over the transmittal ( i.e. Pgen – Pdemand = Plosses, Qgen – Qstored in burden = Q stored in system ) as shown above.

## Voltage Angle and Angle Difference

As a consequence of the Powerworld, the electromotive force angle and angle difference are shown in the tabular array below.

## Voltage Angle Difference

BUS1

I?1 = 0.00A°

BUS1- BUS2

I?1 – I?2 = 0.00A° – ( -2.5662A° ) = 2.5662A°

BUS2

I?2 = -2.5662A°

BUS2- BUS3

I?2 – I?3 = -2.5662A° – ( -0.043A° ) = -2.5232A°

BUS3

I?3 = -0.043A°

BUS3- BUS1

I?3 – I?1 = -0.043A° – 0.00A° = -0.043A°

## Power System Analysis -1

The tabular array below summarizes coevals and electromotive force angle fluctuation at each coach as coevals at Bus 3 varies from 0 MW to 450 MW by 50MW.

## Simulation Consequences and Observation

P3 = 0 MW

P3 = 50 MW

P3 = 100 MW

P3 = 150 MW

P3 = 250 MW

P3 = 300 MW

P3 = 350 MW

P3 = 400 MW

P3 = 450 MW

Reactive Power Generation at Bus 3: It is found that reactive power coevals Q3 ( gen ) lessening while existent power coevals P3 ( gen ) addition because Bus 3 as a PV Bus regulates the changeless coach electromotive force magnitude by commanding excitement of the coevals through the AVR.

Power Generation at Bus 1: It is found that P1 ( gen ) decreases and Q1 ( gen ) increases at the same time, while P3 ( gen ) additions and Q3 ( gen ) lessening. As the entire load demand in the system keeps changeless ( i.e. Ptotal ( burden ) = 400 MW, Qtotal ( burden ) = 150Mvar ) , any necessary existent power and reactive power for the system balance demand to be supplied by generator ( loose generator ) at Bus 1. Therefore, power coevals P1 ( gen ) and Q1 ( gen ) at Bus 1 alteration reversely compared to power coevals alteration at Bus 3.

Voltage Angle Difference: In general, existent power flow is influenced by electromotive force angle difference between directing coach and having coach harmonizing to PR = . Therefore, it is observed that every bit existent power coevals P3 ( gen ) increases existent power flow from Bus 3 to Bus2 addition, consequently voltage angle difference ( I?3 – I?2 ) between Bus 3 and Bus 2 additions. However, lessening in existent power from Bus 1 to Bus 2 due to increase of P3 ( gen ) consequence in lessening of electromotive force angle difference ( I?1 – I?2 ) . In add-on, Real power between Bus 1 and Bus 3 flows from Bus 1 to Bus 3 until P3 ( gen ) range to 200 MW and as P3 ( gen ) addition more than 200 MW the existent power flows from Bus 3 to Bus 1. So, it is besides observed that electromotive force angle difference ( I?3 – I?1 ) is negative angle when P3 ( gen ) is less than 200MW and the difference addition while P3 ( gen ) addition.

## Power System Analysis -2

The tabular array below summarizes the fluctuation of power coevals and electromotive force angle difference at each coach when the burden demand at Bus 3 varies by 50MW and 25Mvar.

## Simulation Consequences and Observation

P2 = 0 MW Q2 = 0 MW

P2 = 50 MW Q2 = 25 MW

P2 = 100 MW Q2 = 50 MW

P2 = 150 MW Q2 = 75 MW

P2 = 250 MW Q2 = 125 MW

P2 = 300 MW Q2 = 150 MW

P2 = 350 MW Q2 = 175 MW

P2 = 400 MW Q2 = 200 MW

P2 = 450 MW Q2 = 225 MW

Power Generation at Bus 1 and Bus 3: It is observed that as the entire load demand in the system increases due to increase of load demand P2 ( burden ) & A ; Q2 ( burden ) at Bus 2, any necessary existent power for the system balance is supplied by generator ( loose generator ) at Bus 1 sing changeless P3 ( gen ) , so P1 ( gen ) increases. In add-on, any necessary reactive power for the system balance is supplied from Bus 1 every bit good as Bus 3, so both Q1 ( gen ) and Q3 ( gen ) addition.

Voltage Angle Difference: It is found that existent power flow addition both from Bus 1 to Bus 2 and from Bus 3 to Bus 2 due to increase of load demand at Bus2. Consequently, both electromotive force angle difference I?1 – I?2 and I?3 – I?2 addition when the power flow P12 and P32 addition. In add-on, when P2 ( burden ) is less than 200 MW, P1gen is comparatively low. Therefore existent power between Bus 3 and Bus 1 flows from Bus 3 to Bus 1 at lower P2 ( burden ) ( less than 200MW ) . On the other manus, while P2 ( burden ) addition more than 200 MW, the existent power flow way alterations ( Bus 1 to Bus 3 ) and the existent power flow additions. Consequently, the electromotive force angle difference I?1 – I?3 alteration from negative to positive and addition.

Voltage Magnitude at Bus 2: It is observed that magnitude of coach electromotive force at Bus2 beads due to increase of the load demand at Bus 2.

## System Model & A ; Admittance Matrix

In order to build the entree matrix of Powerworld B3 instance, individual stage tantamount circuit can be drawn as below ;

## omega = R + jx ( r = 0, x = 0.05 )

z12 = z21= j0.05 plutonium, y12 = 1/ z12 = 1/j0.05 = -j20 plutonium = y12

z13 = z31= j0.05 plutonium, y13 = 1/ z13 = 1/j0.05 = -j20 plutonium = y31

z23 = z32= j0.05 plutonium, y23 = 1/ z23 = 1/j0.05 = -j20 plutonium = y32

Admittance matrix can be defined as follows ;

BUS =

Diagonal elements Y ( I, I ) of the entree matrix, called as the self-admittance [ talk slide ] [ 6 ] , are the summing up of all entree connected with BUS I.

= y12 + y13 = -j20 – j20 = -j40 plutonium

= y21 + y23 = -j20 – j20 = -j40 plutonium

= y31 + y32 = -j20 – j20 = -j40 plutonium

Off diagonal elements Y ( I, J ) of the entree matrix, called as the common entree [ talk slide ] [ 6 ] , are negative entree between BUS I and BUS J.

= – y12 = – ( -j20 ) = j20 plutonium = – y13 = – ( -j20 ) = j20 plutonium

= – y21 = – ( -j20 ) = j20 plutonium = – y23 = – ( -j20 ) = j20 plutonium

= – y31 = – ( -j20 ) = j20 plutonium = – y32 = – ( -j20 ) = j20 plutonium

Therefore, the concluding entree matrix BUS is ;

BUS = =

The undermentioned figure shows the BUS of the Powerworld B3 instance and it is verified that the deliberate entree matrix is consistent with the consequence of the Powerworld.

## Power Flow Calculation

Nodal equation with the entree matrix can be used to cipher electromotive force at each coach if we know all the current ( i.e. entire coevals power and load demand at each BUS ) and eventually the power flow can be calculated consequently.

, hence,

In this inquiry, nevertheless, simulation consequences of the electromotive force at each coach from the Powerworld are used for the power flow computation as follows ;

[ Simulation consequence ]

## Voltage at each Bus and Voltage Difference

V1 = 1 a? 0.00A° plutonium ( BUS1 ) V2 = 1 a? -0.48A° plutonium ( BUS2 ) V3 = 1 a? 0.48A° plutonium ( BUS 3 )

## Voltage difference between BUS 1 and BUS 2

V12 = V1 – V2 = 1 a? 0.00A° – 1 a? -0.48A° = 3.5 x 10-5 + J 8.38 ten 10-3 = 8.38 ten 10-3 a? 89.76A° plutonium

V21 = V2 – V1 = – V12 = – 3.5 ten 10-5 – J 8.38 ten 10-3 = 8.38 ten 10-3 a? -90.24A° plutonium

## Voltage difference between BUS 3 and BUS 2

V32 = V3 – V2 = 1 a? 0.48A° – 1 a? -0.48A° = J 16.76 ten 10-3 = 16.76 ten 10-3 a? 90A° plutonium

V23 = V2 – V3 = – V32 = – J 16.76 ten 10-3 = -16,76 x 10-3 a? -90A° plutonium

## Voltage difference between BUS 3 and BUS 1

V31 = V3 – V1 = 1 a? 0.48A° – 1 a? 0.00A° = – 3.5 ten 10-5 + J 8.38 ten 10-3 = 8.38 ten 10-3 a? 90.24A° plutonium

V13 = V1 – V3 = – V31 = 3.5 ten 10-5 – J 8.38 ten 10-3 = 8.38 ten 10-3 a? -89.76A° plutonium

## Line Current

Current flow from BUS I and BUS J can be calculated by utilizing electromotive force difference and interrelated entree of the line between coachs. [ Iij = yij * ( Vi – Vj ) ]

## Line current between BUS 1 and BUS 2

I12 = y12 x ( V1 – V2 ) = -j20 x 8.38 ten 10-3 a? 89.76A° = 167.6 ten 10-3 a? -0.24A° plutonium ( BUS 1 a†’ BUS 2 )

I21 = y21 x ( V2 – V1 ) = -j20 x 8.38 ten 10-3 a? -90.24A° = 167.6 ten 10-3 a? -180.24A° plutonium ( BUS 2 a†’ BUS 1 )

## Line current between BUS 3 and BUS 2

I32 = y32 x ( V3 – V2 ) = -j20 x 16.76 ten 10-3 a? 90A° = 335.2 ten 10-3 a? 0.00A° plutonium ( BUS 3 a†’ BUS 2 )

I23 = y23 x ( V2 – V3 ) = -j20 x 16.76 ten 10-3 a? -90A° = 335.2 ten 10-3 a? 180A° plutonium ( BUS 2 a†’ BUS 3 )

## Line current between BUS 3 and BUS 1

I31 = y31 x ( V3 – V1 ) = -j20 x 8.38 ten 10-3 a? 90.24A° = 167.6 ten 10-3 a? 0.24A° plutonium ( BUS 3 a†’ BUS 1 )

I13 = y13 x ( V1 – V3 ) = -j20 x 8.38 ten 10-3 a? -89.76A° = 167.6 ten 10-3 a? -179.76A° plutonium ( BUS 1 a†’ BUS 3 )

## Apparent Power Flow

Apparent flow from BUS I and BUS J can be calculated by electromotive force at the directing coach and line current. [ Sij = Vi * I*ij ]

## Apparent Power from BUS 1 to BUS 2

S12 = V1* I*12 = 1 a? 0.00A° ten 167.6 ten 10-3 a? 0.24A° = 167.6 ten 10-3 a? 0.24A° = 0.1676 + J 7.02 ten 10-4 plutonium

## Apparent Power from BUS 2 to BUS 1

S21=V2* I*21=1a? -0.48A° x 167.6 ten 10-3a? 180.24A°=167.6 ten 10-3a? 179.76A° = -0.1676 + j7.02 x 10-4 plutonium

## Apparent Power from BUS 3 to BUS 2

S32 = V3* I*32 = 1 a? 0.48A° ten 335.2 ten 10-3 a? 0.00A° = 335.2 ten 10-3 a? 0.48A° = 0.3352 + J 2.81 ten 10-3 plutonium

## Apparent Power from BUS 2 to BUS 3

S23=V2* I*23=1 a? -0.48A° x 335.2 ten 10-3 a? 180A°= 335.2 ten 10-3 a? 179.76A° = -0.3352 + J 2.81 ten 10-3 plutonium

## Apparent Power from BUS 3 to BUS 1

S31 = V3* I*31 = 1a? 0.48A° ten 167.6 ten 10-3a? -0.24A° = 167.6 x 10-3 a? 0.24A° = 0.1676 + J 7.02 ten 10-4 plutonium

## Apparent Power from BUS 1 to BUS 3

S13=V1* I*13=1a? 0.00A° x 167.6 ten 10-3a? 179.76A°= 167.6 ten 10-3a? 179.76A° = -0.1676 + J 7.02 ten 10-4 plutonium

## Comparison with simulation consequences

The unit of the above computation consequences is pu value, so in order to compare the consequences with simulation consequences pu value of current and power flow demand to be converted to existent values by utilizing the undermentioned equation sing Sbase = 100MVA and Vline_base = 345kV. [ 3 ]

Sactual = Sbase A- Spu = 100 MVA A- Spu

Iactual = Ibase A- Ipu = A- Ipu = A- Ipu = 167.3479 A A- Ipu

## BUS 1 a†’ BUS 2

|S12|

0.1676 A- 100 = 16.76 MVA

16.67 MVA

P12

16.76 MW

16.67 MW

Q12

0.0702 Mvar

0.07 Mvar

|I12|

0.1676 A- 167.3479 = 28.0475 A

27.89 A

## BUS 3 a†’ BUS 2

|S32|

0.3352 A- 100 = 33.52 MVA

33.33 MVA

P32

33.52 MW

33.33 MW

Q32

0.281 Mvar

0.28 Mvar

|I32|

0.3352 A- 167.3479 = 56.0950 A

55.78 A

## BUS 3 a†’ BUS 1

|S31|

0.1676 A- 100 = 16.76 MVA

16.67 MVA

P31

16.76 MW

16.67 MW

Q31

0.0702 Mvar

0.07 Mvar

|I31|

0.1676 A- 167.3479 = 28.0475 A

27.89 A

## BUS 2 a†’ BUS 1

|S21|

0.1676 A- 100 = 16.76 MVA

16.67 MVA

P21

-16.76 MW

-16.67 MW

Q21

0.0702 Mvar

0.07 Mvar

|I21|

0.1676 A- 167.3479 = 28.0475 A

27.89 A

## BUS 2 a†’ BUS 3

|S23|

0.3352 A- 100 = 33.52 MVA

33.33 MVA

P23

-33.52 MW

-33.33 MW

Q23

0.281 Mvar

0.28 Mvar

|I23|

0.3352 A- 167.3479 = 56.0950 A

55.78 A

## BUS 1 a†’ BUS 3

|S13|

0.1676 A- 100 = 16.76 MVA

16.67 MVA

P13

-16.76 MW

-16.67 MW

Q13

0.0702 Mvar

0.07 Mvar

|I13|

0.1676 A- 167.3479 = 28.0475 A

27.89 A

It is found that computation consequences of current flow and evident power flows ( i.e. 28.0475 A and 56.0950 A/ 33.52 MVA and 16.76MVA ) are about 0.5 % higher than simulation consequence ( i.e. 27.89 A and 55.78 A / 33.33 MVA and 16.67 MVA ) which can be considered somewhat different. Difference of the electromotive force angle at each coach between computation ( 0.48A° ) and simulation ( 0.4775A° ) could be the ground for this minor difference.

## Entree between two coachs

y12 = y21 = -j8 plutonium y13 = y31 = -j4 plutonium y14 = y41 = -j2.5 plutonium

y23 = y32 = -j4 plutonium y24 = y42 = -j5 plutonium

y30 = -j0.8 plutonium ( BUS3-Neutral BUS ) y40 = -j0.8 plutonium ( BUS4-Neutral BUS )

Ybus ( Admittance Matrix ) =

Diagonal elements Y ( I, I ) of the entree matrix, called as the self-admittance [ 2 ] [ 4 ] , are the summing up of all entree connected with BUS I.

= y12 + y13 + y14 = -j8 -j4 – j2.5 = -j14.5

= y21 + y23 + y24 = -j8 -j4 – j5 = -j17

= y30 + y31 + y32 = -j08 -j4 – j4 = -j8.8

= y40 + y41 + y42 = -j0.8 -j2.5 – j5 = -j8.3

Off diagonal elements Y ( I, J ) of the entree matrix, called as the common entree [ 2 ] [ 4 ] , are negative entree between BUS I and BUS J.

= – y12 = – ( -j8 ) = j8 plutonium = – y13 = – ( -j4 ) = j4 plutonium = – y14 = – ( -j2.5 ) = j2.5 plutonium

= – y21 = – ( -j8 ) = j8 plutonium = – y23 = – ( -j4 ) = j4 plutonium = – y24 = – ( -j5 ) = j5 plutonium

= – y31 = – ( -j4 ) = j4 plutonium = – y32 = – ( -j4 ) = j4 plutonium = – y34 = 0 plutonium

= – y41 = – ( -j2.5 ) = j2.5 plutonium = – y42 = – ( -j5 ) = j5 plutonium = – y43 = 0 plutonium

Therefore, entree matrix Ybus is as follows ;

## Nodal Equation

Current from the impersonal coach to each coach are given and entree matrix ( Ybus ) is calculated above. Therefore, concluding nodal equation is as follows ;

Ibus = Ybus * Vbus a‡’ Vbus = Y-1bus * Ibus

= Ybus a‡’ =

## Voltage Analysis

Voltage at each coach can be derived from the equation ( Vbus = Y-1bus * Ibus ) and Matlab was used for calculate matrix division. ( Source codification is attached in Appendix-1 )

Vbus ==

V12 = 0.0034 + J 0.0031 plutonium V13 = -0.0277 – J 0.0257 plutonium V14 = 0.0336 + J 0.0311 plutonium

V21 = -0.0034 – J 0.0031 plutonium V23 = -0.0311 – J 0.0288 plutonium V24 = 0.0302 + J 0.0280 plutonium

V31 = 0.0277 + J 0.0257 plutonium V32 = 0.0311 + J 0.0288 plutonium

V41 = -0.0336 – J 0.0311 plutonium V42 = -0.0302 – J 0.0280 plutonium

## Current flow in the system

Current flow from BUS I and BUS J can be calculated by utilizing electromotive force difference and interrelated entree of the line between coachs. [ Iij = yij * ( Vi – Vj ) ] The computation consequence from Matlab is as follows ;

I12 = 0.0249 – J 0.0269 plutonium I13 = -0.1026 + J 0.1108 plutonium I14 = 0.0777 – J 0.0840 plutonium

I21 = -0.0249 + J 0.0269 plutonium I23 = -0.1151 + J 0.1243 plutonium I24 = 0.1399 – J 0.1511

I31 = 0.1026 – J 0.1108 plutonium I32 = 0.1151 – J 0.1243 plutonium I34 = 0 plutonium

I41 = -0.0777 + J 0.0840 plutonium I42 = -0.1399 + J 0.1511 plutonium I43 = 0 plutonium

## Power flow in the system

Apparent flow from BUS I and BUS J can be calculated by electromotive force at the directing coach and line current. [ Sij ( plutonium ) = Vi * I*ij = Pij + jQij ] The computation consequence from Matlab is as follows ;

S12 = 0.0311 + J 0.0175 plutonium S13 = -0.1283 – J 0.0723 plutonium S14 = 0.0972 + J 0.0548 plutonium

S21 = -0.0311 – J 0.0174 plutonium S23 = -0.1438 – J 0.0803 plutonium S24 = 0.1749 + J 0.0977 plutonium

S31 = 0.1283 + J 0.0780 plutonium S32 = 0.1438 + J 0.0875 plutonium S34 = 0 plutonium

S41 = -0.0972 – J 0.0496 plutonium S42 = -0.1749 – J 0.0892 plutonium S44 = 0 plutonium

## Admittance Matrix sing transmittal line electrical capacity

Harmonizing to the direction of the Question 3, power system theoretical account can be drawn by utilizing Iˆ tantamount circuit of the lines with capacitive shunt entree ( yc ) of 0.1 plutonium at each side as shown below.

Contrary to tantamount theoretical account in Question 3-1, the current flow through the capacitance in the transmittal line needs to be considered to happen the entree matrix. Therefore, sing the capacitances the current equation with Kirchhoff ‘s current jurisprudence at each coach is as follows ; [ 2 ] [ 5 ]

Bus 1: I1 = I12 + I13 + I14 + Ic12 + Ic13 + Ic14 I1 = y12 ( V1-V2 ) + y13 ( V1-V3 ) + y14 ( V1-V4 ) + yc12V1 + yc13V1 + yc14V1

Bus 2: I2 = I21 + I23 + I24 + Ic21 + Ic23 + Ic24 I2 = y21 ( V2-V1 ) + y23 ( V2-V3 ) + y24 ( V2-V4 ) + yc21V2 + yc23V2 + yc24V2

Bus 3: I3 = I30 + I31 + I32 + Ic31 + Ic32 I3 = y30V3 + y31 ( V3-V1 ) + y32 ( V3-V2 ) + yc31V3 + yc32V3

Bus 4: I4 = I40 + I41 + I42 + Ic41 + Ic42 I4 = y40V4 + y41 ( V4-V1 ) + y42 ( V4-V2 ) + yc41V4 + yc42V4

Equation above can be rearranged to divide and group single merchandises by electromotive force.

Bus 1: I1 = ( y12 + y13 + y14 + yc12 + yc13+ yc14 ) V1 – y12V2 – y13V3 – y14V4 = Y11V1 + Y12V2 + Y13V3 + Y14V4

Bus 2: I2 = ( y21 + y23 + y24 + yc21 + yc23+ yc24 ) V2- y21V1 – y23V3 – y24V4 = Y21V1 + Y22V2 + Y23V3 + Y24V4

Bus 3: I3 = ( y30 + y31 + y32 + yc31+ yc32 ) V3 – y31V1 – y32V2 = Y31V1 + Y32V2 + Y33V3 + Y34V4

Bus 4: I4 = ( y40 + y41 + y42 + yc41+ yc42 ) V4 – y41V1 – y42V2 = Y41V1 + Y42V2 + Y43V3 + Y44V4

Finally, Diagonal elements Y ( I, I ) and off diagonal elements Y ( I, J ) of the entree matrix are calculated as follows ;

= y12 + y13 + y14 + yc12 + yc13+ yc14 = -j8 -j4 – j2.5 + j0.1 + j0.1 +0.1j = -j14.2 plutonium

= y21 + y23 + y24 + yc21 + yc23+ yc24 = -j8 -j4 – j5 + j0.1 + j0.1 +0.1j = -j16.7 plutonium

= y30 + y31 + y32 + yc31+ yc32 = -j08 -j4 – j4 + j0.1 +0.1j = -j8.6 plutonium

= y40 + y41 + y42 + yc41+ yc42 = -j0.8 -j2.5 – j5 + j0.1 +0.1j = -j8.1 plutonium

= – y12 = – ( -j8 ) = j8 plutonium = – y13 = – ( -j4 ) = j4 plutonium = – y14 = – ( -j2.5 ) = j2.5 plutonium

= – y21 = – ( -j8 ) = j8 plutonium = – y23 = – ( -j4 ) = j4 plutonium = – y24 = – ( -j5 ) = j5 plutonium

= – y31 = – ( -j4 ) = j4 plutonium = – y32 = – ( -j4 ) = j4 plutonium = – y34 = 0 plutonium

= – y41 = – ( -j2.5 ) = j2.5 plutonium = – y42 = – ( -j5 ) = j5 plutonium = – y43 = 0 plutonium

Therefore, entree matrix Ybus is as follows ;

## Matlab Source Code

% define ego entree and common entree by utilizing admittace between

% the coachs ( y12=y21=-j8, y13=y31=-j4, y14=y41=-j2.5, y23=y32=-j4,

% y24=y42=-j5, y34=0, y43=0, y30=-j0.8, y40=-j0.8

y12=-8i ; y21=-8i ; y13=-4i ; y31=-4i ; y14=-2.5i ; y41=-2.5i ; y23=-4i ; y32=-4i ;

y24=-5i ; y42=-5i ; y34=0 ; y43=0 ; y30=-0.8i ; y40=-0.8i ;

Y11=-8i-4i-2.5i ; Y12=8i ; Y13=4i ; Y14=2.5i ;

Y21=8i ; Y22=-8i-4i-5i ; Y23=4i ; Y24=5i ;

Y31=4i ; Y32=4i ; Y33=-0.8i-4i-4i ; Y34=0 ;

Y41=2.5i ; Y42=5i ; Y43=0 ; Y44=-5i-2.5i-0.8i ;

% Bus 3 and Bus 4 is non connected, so admittance Y34 and Y43 are equal to zero

% define the 4×4 entree matrix ( Ybus )

Ybus= [ Y11 Y12 Y13 Y14 ; Y21 Y22 Y23 Y24 ; Y31 Y32 Y33 Y34 ; Y41 Y42 Y43 Y44 ] ;

% In order to specify the nodal equation ( I = Ybus*V ) , the given I needs to specify.

i1=0 ; i2=0 ; i3=-i ; i4=-0.4808-0.4808i ;

Ibus= [ i1 ; i2 ; i3 ; i4 ] ;

% Each coach electromotive force can be calculated by utilizing matrix division ( V= YbusI )

Vbus=YbusIbus ;

v1=Vbus ( 1,1 ) ; v2=Vbus ( 2,1 ) ; v3=Vbus ( 3,1 ) ; v4=Vbus ( 4,1 ) ;

% Calculate electromotive force difference between coachs

v12=v1-v2 ; v13=v1-v3 ; v14=v1-v4 ;

v21=v2-v1 ; v23=v2-v3 ; v24=v2-v4 ;

v31=v3-v1 ; v32=v3-v2 ; v34=v3-v4 ;

v41=v4-v1 ; v42=v4-v2 ; v43=v4-v3 ;

% current flow between coachs can be calculated by i12 = y12* ( v1-v2 )

i12=y12*v12 ; i13=y13*v13 ; i14=y14*v14 ;

i21=y21*v21 ; i23=y23*v23 ; i24=y24*v24 ;

i31=y31*v31 ; i32=y32*v32 ; i34=y34*v34 ;

i41=y41*v41 ; i42=y42*v42 ; i43=y43*v43 ;

% evident power can be calculated by s12 = v1 * conj ( i12 )

s12=v1*conj ( i12 ) ; s13=v1*conj ( i13 ) ; s14=v1*conj ( i14 ) ;

s21=v2*conj ( i21 ) ; s23=v2*conj ( i23 ) ; s24=v2*conj ( i24 ) ;

s31=v3*conj ( i31 ) ; s32=v3*conj ( i32 ) ; s34=v3*conj ( i34 ) ;

s41=v4*conj ( i41 ) ; s42=v4*conj ( i42 ) ; s43=v4*conj ( i43 ) ;

% Real power and Reactive power can be derived by following

p12=real ( s12 ) ; p13=real ( s13 ) ; p14=real ( s14 ) ;

q12=imag ( s12 ) ; q13=imag ( s13 ) ; q14=imag ( s14 ) ;

p21=real ( s21 ) ; p23=real ( s23 ) ; p24=real ( s24 ) ;

q21=imag ( s21 ) ; q23=imag ( s23 ) ; q24=imag ( s24 ) ;

p31=real ( s31 ) ; p32=real ( s32 ) ; p34=real ( s34 ) ;

q31=imag ( s31 ) ; q32=real ( s32 ) ; q34=imag ( s34 ) ;

p41=real ( s41 ) ; p42=real ( s42 ) ; p43=real ( s43 ) ;

q41=imag ( s41 ) ; q42=real ( s42 ) ; q43=imag ( s43 ) ; % terminal