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# Slack Bus And Slack Generator

 ✅ Paper Type: Free Essay ✅ Subject: Engineering ✅ Wordcount: 4500 words ✅ Published: 1st May 2017

The Table below shows input data of each busbar in the system used to solve the power flow and the simulation result according to instruction described in question 1.

## Input Data

[Simulation Result]

BUS 1

pu

100 MW

0 Mvar

BUS 2

200 MW

100 Mvar

CB of Generation

Open

BUS 3

1 pu

P (Gen)

200 MW

100 MW

50 Mvar

AVR

On

AGC

Off

## Slack bus and slack generator

In power flow calculation, unique numerical solution cannot be calculated without reference voltage magnitude and angle due to unequal number of unknown variables and independent equations. The slack bus is the reference bus where its voltage is considered to be fixed voltage magnitude and angle (1âˆ 0°), so that the various voltage angle difference among the buses can be calculated respect. In addition, the slack generator supplies as much real power and reactive power as needed for balancing the power flow considering power generation, load demand and losses in the system while keep the voltage constant as 1âˆ 0°. In real power system, when relatively weak system is linked to the larger system via a single bus, this bus can represent the large system with an equivalent generator keeping the voltage constant and generating any necessary power like slack bus. 

## Bus type

BUS 2

PQ Bus

Generator is disconnected to Bus 2

BUS 3

PV Bus

Generator is connected to Bus 3 and the magnitude of voltage of generator keep constant by using AVR

In general, each bus in the power system can be categorized into three bus 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; 

PV Bus (Generator Bus or Voltage Controlled Bus): It is a bus at which the magnitude of the bus voltage is kept constant by the generator. Even though the bus has several generators and load, if any generators connected to the bus regulate the bus voltage with AVR, then this bus is referred to PV Bus. For PV bus, the magnitude of the bus voltage and real power supplied to the system are specified, and reactive power and angle of the bus voltage are accordingly determined. If a preset maximum and minimum reactive power limit is reached, the reactive output of the generator remains at the limited values, so the bus can be considered as PQ Bus instead of PV Bus. 

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PQ Bus (Load Bus): It is a bus at which the voltage is changed depending on total net real power and reactive power of loads and generators without voltage regulator. Therefore, in the power simulation and calculation, the real power and reactive power of the loads are specified as input data and accordingly the voltage (magnitude and angle) is calculated based on the above input.

The following table specifies input and output of each bus type in the power system simulation and calculation.

Bus Type

P

Q

(Magnitude)

Î´ (Angle)

PQ Bus

Input

Input

Output

Output

PV Bus

Input

Output

Input

Output

Slack Bus

Output

Output

Input

Input

BUS 1

204.093

100

56.240

0

BUS 2

0

200

0

100

BUS 3

200

100

107.404

50

Total

404.093

400

163.644

150

## Qgen – Qstored in load = 13.644

Reason: Real power loss due to resistance of transmission line and imaginary power storage due to reactance of transmission line are the reasons for the difference between power generation and load demand in the system.

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

Total

Plosses =

## 4.093

Qstored in load =

## 13.644

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

The summation of real power losses and imaginary power storage over the transmission line are exactly same with total difference between generation and load. Therefore, it is verified that the difference is shown over the transmission line.

‘Kirchoff’ balance as each bus 

Bus1

Î£ P1 = + Pgen1 – Pload1 – P12 – P13 = 204.093 – 100 – 102.714 – 1.379 = 0

Î£ Q1 = + Qgen1 – Qload1 – Q12 – Q13 = 56.24 – 0 – 56.653 + 0.413 = 0

Bus2

Î£ P2 = + Pgen2 – Pload2 – P21 – P23 = 0 – 200 + 100.65 + 99.35 = 0

Î£ Q2 = + Qgen2 – Qload2 – Q21 – Q23 = 0 – 100 + 49.773 + 50.227 = 0

BUS3

Î£ P3 = + Pgen3 – Pload3 – P31 – P32 = 200 – 100 + 1.378 – 101.378 = 0

Î£ Q3 = + Qgen3 – Qload3 – Q31 – Q32 = 107.404 – 50 – 0.414 – 56.99 = 0

According to the calculation above, as summation of incoming & outgoing real power and imaginary power at each bus become zero, it is verified that each busbar obeys a ‘Kirchoff’ balance. In addition, the total power system is completely balanced, because total generation power (real & imaginary) are equal to summation of total load demand and real power loss & stored imaginary power over the transmission (i.e. Pgen – Pdemand = Plosses, Qgen – Qstored in load = Q stored in system) as shown above.

## Voltage Angle and Angle Difference

As a result of the Powerworld, the voltage angle and angle difference are shown in the table below.

## Voltage Angle Difference

BUS1

Î´1 = 0.00°

BUS1- BUS2

Î´1 – Î´2 = 0.00° – (-2.5662°) = 2.5662°

BUS2

Î´2 = -2.5662°

BUS2- BUS3

Î´2 – Î´3 = -2.5662° – (-0.043°) = -2.5232°

BUS3

Î´3 = -0.043°

BUS3- BUS1

Î´3 – Î´1 = -0.043° – 0.00° = -0.043°

## Power System Analysis -1

The table below summarizes generation and voltage angle variation at each bus as generation at Bus 3 varies from 0 MW to 450 MW by 50MW.

## Simulation Results 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 generation Q3(gen) decrease while real power generation P3(gen) increase because Bus 3 as a PV Bus regulates the constant bus voltage magnitude by controlling excitation of the generation through the AVR.

Power Generation at Bus 1: It is found that P1(gen) decreases and Q1(gen) increases simultaneously, while P3(gen) increases and Q3(gen) decrease. As the total load demand in the system keeps constant (i.e. Ptotal(load) = 400 MW, Qtotal(load) = 150Mvar), any necessary real power and reactive power for the system balance need to be supplied by generator (slack generator) at Bus 1. Therefore, power generation P1(gen) and Q1(gen) at Bus 1 change reversely compared to power generation change at Bus 3.

Voltage Angle Difference: In general, real power flow is influenced by voltage angle difference between sending bus and receiving bus according to PR =. Therefore, it is observed that as real power generation P3(gen) increases real power flow from Bus 3 to Bus2 increase, accordingly voltage angle difference (Î´3 – Î´2) between Bus 3 and Bus 2 increases. However, decrease in real power from Bus 1 to Bus 2 due to increase of P3(gen) result in decrease of voltage angle difference (Î´1 – Î´2). In addition, Real power between Bus 1 and Bus 3 flows from Bus 1 to Bus 3 until P3(gen) reach to 200 MW and as P3(gen) increase more than 200 MW the real power flows from Bus 3 to Bus 1. So, it is also observed that voltage angle difference (Î´3 – Î´1) is negative angle when P3(gen) is less than 200MW and the difference increase while P3(gen) increase.

## Power System Analysis -2

The table below summarizes the variation of power generation and voltage angle difference at each bus when the load demand at Bus 3 varies by 50MW and 25Mvar.

## Simulation Results 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 total load demand in the system increases due to increase of load demand P2(load) & Q2(load) at Bus 2, any necessary real power for the system balance is supplied by generator (slack generator) at Bus 1 considering constant P3(gen), so P1(gen) increases. In addition, any necessary reactive power for the system balance is supplied from Bus 1 as well as Bus 3, so both Q1(gen) and Q3(gen) increase.

Voltage Angle Difference: It is found that real power flow increase both from Bus 1 to Bus 2 and from Bus 3 to Bus 2 due to increase of load demand at Bus2. Accordingly, both voltage angle difference Î´1 – Î´2 and Î´3 – Î´2 increase when the power flow P12 and P32 increase. In addition, when P2(load) is less than 200 MW, P1gen is relatively low. Therefore real power between Bus 3 and Bus 1 flows from Bus 3 to Bus 1 at lower P2(load) (less than 200MW). On the other hand, while P2(load) increase more than 200 MW, the real power flow direction changes (Bus 1 to Bus 3) and the real power flow increases. Accordingly, the voltage angle difference Î´1 – Î´3 change from negative to positive and increase.

Voltage Magnitude at Bus 2: It is observed that magnitude of bus voltage at Bus2 drops due to increase of the load demand at Bus 2.

## System Model & Admittance Matrix

In order to construct the admittance matrix of Powerworld B3 case, single phase equivalent circuit can be drawn as below;

## z = r + jx (r = 0, x = 0.05)

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

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

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

Admittance matrix can be defined as follows;

BUS =

Diagonal elements Y(i,i) of the admittance matrix, called as the self-admittance [lecture slide] , are the summation of all admittance connected with BUS i.

= y12 + y13 = -j20 – j20 = -j40 pu

= y21 + y23 = -j20 – j20 = -j40 pu

= y31 + y32 = -j20 – j20 = -j40 pu

Off diagonal elements Y(i,j) of the admittance matrix, called as the mutual admittance [lecture slide] , are negative admittance between BUS i and BUS j.

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

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

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

Therefore, the final admittance matrix BUS is;

BUS = =

The following figure shows the BUS of the Powerworld B3 case and it is verified that the calculated admittance matrix is consistent with the result of the Powerworld.

## Power Flow Calculation

Nodal equation with the admittance matrix can be used to calculate voltage at each bus if we know all the current (i.e. total generation power and load demand at each BUS) and finally the power flow can be calculated accordingly.

, therefore,

In this question, however, simulation results of the voltage at each bus from the Powerworld are used for the power flow calculation as follows;

[Simulation result]

## Voltage at each Bus and Voltage Difference

V1 = 1 âˆ 0.00° pu (BUS1) V2 = 1 âˆ -0.48° pu (BUS2) V3 = 1 âˆ 0.48° pu (BUS 3)

## Voltage difference between BUS 1 and BUS 2

V12 = V1 – V2 = 1 âˆ 0.00° – 1 âˆ -0.48° = 3.5 x 10-5 + j 8.38 x 10-3 = 8.38 x 10-3 âˆ 89.76° pu

V21 = V2 – V1 = – V12 = – 3.5 x 10-5 – j 8.38 x 10-3 = 8.38 x 10-3 âˆ -90.24° pu

## Voltage difference between BUS 3 and BUS 2

V32 = V3 – V2 = 1 âˆ 0.48° – 1 âˆ -0.48° = j 16.76 x 10-3 = 16.76 x 10-3 âˆ 90° pu

V23 = V2 – V3 = – V32 = – j 16.76 x 10-3 = -16,76 x 10-3 âˆ -90° pu

## Voltage difference between BUS 3 and BUS 1

V31 = V3 – V1 = 1 âˆ 0.48° – 1 âˆ 0.00° = – 3.5 x 10-5 + j 8.38 x 10-3 = 8.38 x 10-3 âˆ 90.24° pu

V13 = V1 – V3 = – V31 = 3.5 x 10-5 – j 8.38 x 10-3 = 8.38 x 10-3 âˆ -89.76° pu

## Line Current

Current flow from BUS i and BUS j can be calculated by using voltage difference and interconnected admittance of the line between buses. [ Iij = yij * (Vi – Vj) ]

## Line current between BUS 1 and BUS 2

I12 = y12 x (V1 – V2) = -j20 x 8.38 x 10-3 âˆ 89.76° = 167.6 x 10-3 âˆ -0.24° pu (BUS 1 â†’ BUS 2)

I21 = y21 x (V2 – V1) = -j20 x 8.38 x 10-3 âˆ -90.24° = 167.6 x 10-3 âˆ -180.24° pu (BUS 2 â†’ BUS 1)

## Line current between BUS 3 and BUS 2

I32 = y32 x (V3 – V2) = -j20 x 16.76 x 10-3 âˆ 90° = 335.2 x 10-3 âˆ 0.00° pu (BUS 3 â†’ BUS 2)

I23 = y23 x (V2 – V3) = -j20 x 16.76 x 10-3 âˆ -90° = 335.2 x 10-3 âˆ 180° pu (BUS 2 â†’ BUS 3)

## Line current between BUS 3 and BUS 1

I31 = y31 x (V3 – V1) = -j20 x 8.38 x 10-3 âˆ 90.24° = 167.6 x 10-3 âˆ 0.24° pu (BUS 3 â†’ BUS 1)

I13 = y13 x (V1 – V3) = -j20 x 8.38 x 10-3 âˆ -89.76° = 167.6 x 10-3 âˆ -179.76° pu (BUS 1 â†’ BUS 3)

## Apparent Power Flow

Apparent flow from BUS i and BUS j can be calculated by voltage at the sending bus and line current. [ Sij = Vi * I*ij ]

## Apparent Power from BUS 1 to BUS 2

S12 = V1* I*12 = 1 âˆ 0.00° x 167.6 x 10-3 âˆ 0.24° = 167.6 x 10-3 âˆ 0.24° = 0.1676 + j 7.02 x 10-4 pu

## Apparent Power from BUS 2 to BUS 1

S21=V2* I*21=1âˆ -0.48° x 167.6 x 10-3âˆ 180.24°=167.6 x 10-3âˆ 179.76° = -0.1676 + j7.02 x 10-4 pu

## Apparent Power from BUS 3 to BUS 2

S32 = V3* I*32 = 1 âˆ 0.48° x 335.2 x 10-3 âˆ 0.00° = 335.2 x 10-3 âˆ 0.48° = 0.3352 + j 2.81 x 10-3 pu

## Apparent Power from BUS 2 to BUS 3

S23=V2* I*23=1 âˆ -0.48° x 335.2 x 10-3 âˆ 180°= 335.2 x 10-3 âˆ 179.76° = -0.3352 + j 2.81 x 10-3 pu

## Apparent Power from BUS 3 to BUS 1

S31 = V3* I*31 = 1âˆ 0.48° x 167.6 x 10-3âˆ -0.24° = 167.6 x 10-3 âˆ 0.24° = 0.1676 + j 7.02 x 10-4 pu

## Apparent Power from BUS 1 to BUS 3

S13=V1* I*13=1âˆ 0.00° x 167.6 x 10-3âˆ 179.76°= 167.6 x 10-3âˆ 179.76° = -0.1676 + j 7.02 x 10-4 pu

## Comparison with simulation results

The unit of the above calculation results is pu value, so in order to compare the results with simulation results pu value of current and power flow need to be converted to actual values by using the following equation considering Sbase = 100MVA and Vline_base = 345kV. 

Sactual = Sbase Ã- Spu = 100 MVA Ã- Spu

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

## BUS 1 â†’ BUS 2

|S12|

0.1676 Ã- 100 = 16.76 MVA

16.67 MVA

P12

16.76 MW

16.67 MW

Q12

0.0702 Mvar

0.07 Mvar

|I12|

0.1676 Ã- 167.3479 = 28.0475 A

27.89 A

## BUS 3 â†’ BUS 2

|S32|

0.3352 Ã- 100 = 33.52 MVA

33.33 MVA

P32

33.52 MW

33.33 MW

Q32

0.281 Mvar

0.28 Mvar

|I32|

0.3352 Ã- 167.3479 = 56.0950 A

55.78 A

## BUS 3 â†’ BUS 1

|S31|

0.1676 Ã- 100 = 16.76 MVA

16.67 MVA

P31

16.76 MW

16.67 MW

Q31

0.0702 Mvar

0.07 Mvar

|I31|

0.1676 Ã- 167.3479 = 28.0475 A

27.89 A

## BUS 2 â†’ BUS 1

|S21|

0.1676 Ã- 100 = 16.76 MVA

16.67 MVA

P21

-16.76 MW

-16.67 MW

Q21

0.0702 Mvar

0.07 Mvar

|I21|

0.1676 Ã- 167.3479 = 28.0475 A

27.89 A

## BUS 2 â†’ BUS 3

|S23|

0.3352 Ã- 100 = 33.52 MVA

33.33 MVA

P23

-33.52 MW

-33.33 MW

Q23

0.281 Mvar

0.28 Mvar

|I23|

0.3352 Ã- 167.3479 = 56.0950 A

55.78 A

## BUS 1 â†’ BUS 3

|S13|

0.1676 Ã- 100 = 16.76 MVA

16.67 MVA

P13

-16.76 MW

-16.67 MW

Q13

0.0702 Mvar

0.07 Mvar

|I13|

0.1676 Ã- 167.3479 = 28.0475 A

27.89 A

It is found that calculation results of current flow and apparent power flows (i.e. 28.0475 A and 56.0950 A/ 33.52 MVA and 16.76MVA) are about 0.5 % higher than simulation result (i.e. 27.89 A and 55.78 A / 33.33 MVA and 16.67 MVA) which can be considered slightly different. Difference of the voltage angle at each bus between calculation (0.48°) and simulation (0.4775°) could be the reason for this minor difference.

## Admittance between two buses

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

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

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

Ybus (Admittance Matrix) =

Diagonal elements Y(i,i) of the admittance matrix, called as the self-admittance  , are the summation of all admittance 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 admittance matrix, called as the mutual admittance  , are negative admittance between BUS i and BUS j.

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

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

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

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

Therefore, admittance matrix Ybus is as follows;

## Nodal Equation

Current from the neutral bus to each bus are given and admittance matrix (Ybus) is calculated above. Therefore, final nodal equation is as follows;

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

= Ybus â‡’ =

## Voltage Analysis

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

Vbus ==

V12 = 0.0034 + j 0.0031 pu V13 = -0.0277 – j 0.0257 pu V14 = 0.0336 + j 0.0311 pu

V21 = -0.0034 – j 0.0031 pu V23 = -0.0311 – j 0.0288 pu V24 = 0.0302 + j 0.0280 pu

V31 = 0.0277 + j 0.0257 pu V32 = 0.0311 + j 0.0288 pu

V41 = -0.0336 – j 0.0311 pu V42 = -0.0302 – j 0.0280 pu

## Current flow in the system

Current flow from BUS i and BUS j can be calculated by using voltage difference and interconnected admittance of the line between buses. [Iij = yij * (Vi – Vj) ] The calculation result from Matlab is as follows;

I12 = 0.0249 – j 0.0269 pu I13 = -0.1026 + j 0.1108 pu I14 = 0.0777 – j 0.0840 pu

I21 = -0.0249 + j 0.0269 pu I23 = -0.1151 + j 0.1243 pu I24 = 0.1399 – j 0.1511

I31 = 0.1026 – j 0.1108 pu I32 = 0.1151 – j 0.1243 pu I34 = 0 pu

I41 = -0.0777 + j 0.0840 pu I42 = -0.1399 + j 0.1511 pu I43 = 0 pu

## Power flow in the system

Apparent flow from BUS i and BUS j can be calculated by voltage at the sending bus and line current. [ Sij (pu) = Vi * I*ij = Pij + jQij ] The calculation result from Matlab is as follows;

S12 = 0.0311 + j 0.0175 pu S13 = -0.1283 – j 0.0723 pu S14 = 0.0972 + j 0.0548 pu

S21 = -0.0311 – j 0.0174 pu S23 = -0.1438 – j 0.0803 pu S24 = 0.1749 + j 0.0977 pu

S31 = 0.1283 + j 0.0780 pu S32 = 0.1438 + j 0.0875 pu S34 = 0 pu

S41 = -0.0972 – j 0.0496 pu S42 = -0.1749 – j 0.0892 pu S44 = 0 pu

## Admittance Matrix considering transmission line capacitance

According to the instruction of the Question 3, power system model can be drawn by using Ï€ equivalent circuit of the lines with capacitive shunt admittance (yc) of 0.1 pu at each side as shown below.

Contrary to equivalent model in Question 3-1, the current flow through the capacitor in the transmission line needs to be considered to find the admittance matrix. Therefore, considering the capacitors the current equation with Kirchhoff’s current law at each bus is as follows;  

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 separate and group individual products by voltage.

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 admittance matrix are calculated as follows;

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

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

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

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

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

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

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

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

Therefore, admittance matrix Ybus is as follows;

## Matlab Source Code

% define self admittance and mutual admittance by using admittace between

% the buses (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 not connected, so admittance Y34 and Y43 are equal to zero

% define the 4×4 admittance matrix (Ybus)

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

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

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

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

% Each bus voltage can be calculated by using matrix division (V= YbusI)

Vbus=YbusIbus;

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

% Calculate voltage difference between buses

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 buses 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;

% apparent 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); % end

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