Slack Bus And Slack Generator Engineering Essay

The Table below shows enter informations of apiece busbar in the system utilize to work out the creator prevail and the disguise consequence harmonizing to direction draw in inquiry 1. good deal introduce Data assumption effect passenger vehicle 1atomic number 94P ( interference ) carbon MWQ ( burden )0 Mvar passel 2P ( burden ) two hundred MWQ ( burden ) coulomb MvarCB of genesisOpen cumulation 31 atomic number 94P ( Gen ) two hundred MWP ( burden )100 MWQ ( burden )50 MvarAVROnAGC forwardSlack condition and slack beginningIn violence fertilise computation, alone numerical solution cease non be calculated without mention electromotive oblige ready and run due to unequal tangibleize of unknown variables and independent compares. The slack managing director is the mention motor passenger vehicle where its electromotive displume is considered to be fixed voltage magnitude and tip off ( 1a? 0A ) , so that the assorted electromotive describe angle loss amon g the take aims atomic number 50 be calculated regard. In add-on, the slack generator supplies as much living magnate and reactive forefinger as needed for equilibrating the imbibe play flux sing creator multiplication, load want and losingss in the system while maintain the electromotive phalanx changeless as 1a? 0A . In actual former system, when comparatively weak system is linked to the larger system via a individual check, this trail can stand for the big system with an equal generator maintaining the electromotive ramp changeless and bring forthing any incumbent military unit like slack take. 1 double-decker type ( PQ coach or PV coach ) cumulation manager typeRemarks four-in-hand 2PQ transportGenerator is disconnected to private instructor 2BUS 3PV slewGenerator is connected to Bus 3 and the magnitude of electromotive force of generator support invariable by utilizing AVRIn general, distributively coach in the king system can be categorized into th ree coach types such as Slack Bus, Load ( PQ ) Bus, and electromotive force Controlled ( PV ) Bus. The definition and leaving betwixt 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 exis decennarynert power supplied to the system are qualify, 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 intersection 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 inbuilt net exis hug drugt power and reactive power of tonss and generators without electromotive force regulator. Therefore, in the power simulation and computation, the exis ex-spott 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 to a higher place input.The pastime table specifies input and end product of each coach type in the power system simulation and computation.Bus TypePhosphorusQ( Magnitude )I? ( Angle )PQ BusInput targetInput signalEnd productEnd productPV BusInput signalEnd productInput signalEnd productSlack BusEnd productEnd productInput signalInput signalSystem BalanceEntire Generation & A Load DemandBusReal author ( MW )Fanciful part ( Mvar )CoevalsLoadCoevalsLoadBUS 1204.09310056.2400BUS 20cc0100BUS 3200100107.40450Entire404.093400163.644150DifferencePgen P rent = 4.093Qgen Qstored in burden = 13.644Reason Real power loss due to opposition of transmission line and fanciful power storage due to reactance of transmission line are the grounds for the difference mingled with power coevals and load command in the system.P ( Losses ) & A Q ( Storage ) over the transmittal lineBusReal agency ( MW )Fanciful occasion ( Mvar )SendingReceivingLosingssSendingReceivingStoredBUS 1 Bus 2102.714100.6502.06456.65349.7736.88BUS 1 Bus 31.3791.3780.0010.4141 )0.4131 )0.001BUS 3 Bus 2101.37899.3502.02856.99050.2276.763Entire promised land liberation organizations =4.093Qstored in burden =13.6441 ) 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 comparable 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 Bus1I? P1 = + Pgen1 Pload1 P12 P13 = 204.093 100 102.714 1.379 = 0I? Q1 = + Qgen1 Qload1 Q12 Q13 = 56.24 0 56.653 + 0.413 = 0Bus2I? P2 = + Pgen2 Pload2 P21 P23 = 0 200 + 100.65 + 99.35 = 0I? Q2 = + Qgen2 Qload2 Q21 Q23 = 0 100 + 49.773 + 50.227 = 0BUS3I? P3 = + Pgen3 Pload3 P31 P32 = 200 100 + 1.378 101.378 = 0I? Q3 = + Qgen3 Qload3 Q31 Q32 = 107.404 50 0.414 56.99 = 0Harmonizing to the computation supra, as summing up of incoming & A particular(a) 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 DifferenceAs a consequence of the billetworld, the electromotive force angle and angle differenc e are shown in the tabular array below.BusVoltage AngleVoltage Angle DifferenceBUS1I?1 = 0.00ABUS1- BUS2I?1 I?2 = 0.00A ( -2.5662A ) = 2.5662ABUS2I?2 = -2.5662ABUS2- BUS3I?2 I?3 = -2.5662A ( -0.043A ) = -2.5232ABUS3I?3 = -0.043ABUS3- BUS1I?3 I?1 = -0.043A 0.00A = -0.043APower System Analysis -1The tabular array below summarizes coevals and electromotive force angle magnetic declination at each coach as coevals at Bus 3 varies from 0 MW to 450 MW by 50MW.Simulation Consequences and ObservationP3 = 0 MWP3 = 50 MWP3 = 100 MWP3 = 150 MWP3 = 250 MWP3 = 300 MWP3 = 350 MWP3 = 400 MWP3 = 450 MW labile Power Generation at Bus 3 It is found that reactive power coevals Q3 ( gen ) lessen 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 requisite existent power and reactive power for the system balance demand to be supplied by generator ( destitute generator ) at Bus 1. Therefore, power coevals P1 ( gen ) and Q1 ( gen ) at Bus 1 innovation 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 enjoin 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 in any case 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 -2The 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 ObservationP2 = 0 MW Q2 = 0 MWP2 = 50 MW Q2 = 25 MWP2 = 100 MW Q2 = 50 MWP2 = 150 MW Q2 = 75 MWP2 = 250 MW Q2 = 125 MWP2 = 300 MW Q2 = 150 MWP2 = 350 MW Q2 = 175 MWP2 = 400 MW Q2 = 200 MWP2 = 450 MW Q2 = 225 MWPower 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 needed 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 bead due to increase of the load demand at Bus 2. misgiving 2System Model & A Admittance MatrixIn decree to build the access intercellular substance of Powerworld B3 instance, individual stage tantamount(predicate) circuit can be force as below omega = R + jx ( r = 0, x = 0.05 )z12 = z21= j0.05 atomic number 94, y12 = 1/ z12 = 1/j0.05 = -j20 atomic number 94 = y12z13 = z31= j0.05 atomic number 94, y13 = 1/ z13 = 1/j0.05 = -j20 atomic number 94 = y31z23 = z32= j0.05 plutonium, y23 = 1/ z23 = 1/j0.05 = -j20 plutonium = y32Admittance matrix can be delimit 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 plutoniumOff 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 plutoniumTherefore, 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 CalculationNodal equation with the entree matrix can be used to cipher electromotive force at each coach if we know all the topical ( 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 a t each coach from the Powerworld are used for the power flow computation as follows Simulation consequence Voltage at each Bus and Voltage DifferenceV1 = 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 2V12 = 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 plutoniumV21 = V2 V1 = V12 = 3.5 ten 10-5 J 8.38 ten 10-3 = 8.38 ten 10-3 a? -90.24A plutoniumVoltage difference between BUS 3 and BUS 2V32 = V3 V2 = 1 a? 0.48A 1 a? -0.48A = J 16.76 ten 10-3 = 16.76 ten 10-3 a? 90A plutoniumV23 = V2 V3 = V32 = J 16.76 ten 10-3 = -16,76 x 10-3 a? -90A plutoniumVoltage difference between BUS 3 and BUS 1V31 = 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 plutoniumV13 = V1 V3 = V31 = 3.5 ten 10-5 J 8.38 ten 10-3 = 8.38 ten 10-3 a? -89.76A plutoniumLine CurrentCurrent flow from BUS I and BUS J can be calculate d by utilizing electromotive force difference and interrelated entree of the line between coachs. Iij = yij * ( Vi Vj ) Line accredited between BUS 1 and BUS 2I12 = 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 accredited between BUS 3 and BUS 2I32 = 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 1I31 = 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 FlowApparent flow from BUS I and BUS J can be calculated by electromotive force at th e directing coach and line current. Sij = Vi * I*ij Apparent Power from BUS 1 to BUS 2S12 = 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 plutoniumApparent Power from BUS 2 to BUS 1S21=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 plutoniumApparent Power from BUS 3 to BUS 2S32 = 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 plutoniumApparent Power from BUS 2 to BUS 3S23=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 plutoniumApparent Power from BUS 3 to BUS 1S31 = 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 plutoniumApparent Power from BUS 1 to BUS 3S13=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 plutoniumComparison with simulation consequencesThe unit of t he 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 reborn to existent values by utilizing the undermentioned equation sing Sbase = 100MVA and Vline_base = 345kV. 3 Sactual = Sbase A- Spu = 100 MVA A- SpuIactual = Ibase A- Ipu = A- Ipu = A- Ipu = 167.3479 A A- IpuCalculation Result and Simulation ResultFlow way & A ValueCalculation ConsequenceSimulation ConsequenceBUS 1 a BUS 2S120.1676 A- 100 = 16.76 MVA16.67 MVAP1216.76 MW16.67 MWQ120.0702 Mvar0.07 MvarI120.1676 A- 167.3479 = 28.0475 A27.89 ABUS 3 a BUS 2S320.3352 A- 100 = 33.52 MVA33.33 MVAP3233.52 MW33.33 MWQ320.281 Mvar0.28 MvarI320.3352 A- 167.3479 = 56.0950 A55.78 ABUS 3 a BUS 1S310.1676 A- 100 = 16.76 MVA16.67 MVAP3116.76 MW16.67 MWQ310.0702 Mvar0.07 MvarI310.1676 A- 167.3479 = 28.0475 A27.89 ABUS 2 a BUS 1S210.1676 A- 100 = 16.76 MVA16.67 MVAP21-16.76 MW-16.67 MWQ210.0702 Mvar0.07 MvarI210.1676 A- 167.3479 = 28 .0475 A27.89 ABUS 2 a BUS 3S230.3352 A- 100 = 33.52 MVA33.33 MVAP23-33.52 MW-33.33 MWQ230.281 Mvar0.28 MvarI230.3352 A- 167.3479 = 56.0950 A55.78 ABUS 1 a BUS 3S130.1676 A- 100 = 16.76 MVA16.67 MVAP13-16.76 MW-16.67 MWQ130.0702 Mvar0.07 MvarI130.1676 A- 167.3479 = 28.0475 A27.89 AIt 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 well-nigh 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 secondary difference.Question 3Admittance Matrix and Nodal EquationEntree between two coachsy12 = y21 = -j8 plutonium y13 = y31 = -j4 plutonium y14 = y41 = -j2.5 plutoniumy23 = y32 = -j4 plutonium y24 = y42 = -j5 plutoniumy30 = -j0.8 plutonium ( BUS3-Neutral BUS ) y40 = -j0.8 p lutonium ( BUS4-Neutral BUS )Admittance MatrixYbus ( 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 = -j8.3Off 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 plutoniumTherefore, entree matrix Ybus is as follows Ybus = =Power Flow AnalysisPower flow disregarding transmi ttal line electrical capacityNodal EquationCurrent 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 AnalysisVoltage 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 plutoniumV21 = -0.0034 J 0.0031 plutonium V23 = -0.0311 J 0.0288 plutonium V24 = 0.0302 + J 0.0280 plutoniumV31 = 0.0277 + J 0.0257 plutonium V32 = 0.0311 + J 0.0288 plutoniumV41 = -0.0336 J 0.0311 plutonium V42 = -0.0302 J 0.0280 plutoniumCurrent flow in the systemCurrent 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 com putation 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 plutoniumI21 = -0.0249 + J 0.0269 plutonium I23 = -0.1151 + J 0.1243 plutonium I24 = 0.1399 J 0.1511I31 = 0.1026 J 0.1108 plutonium I32 = 0.1151 J 0.1243 plutonium I34 = 0 plutoniumI41 = -0.0777 + J 0.0840 plutonium I42 = -0.1399 + J 0.1511 plutonium I43 = 0 plutoniumPower flow in the systemApparent 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 plutoniumS21 = -0.0311 J 0.0174 plutonium S23 = -0.1438 J 0.0803 plutonium S24 = 0.1749 + J 0.0977 plutoniumS31 = 0.1283 + J 0.0780 plutonium S32 = 0.1438 + J 0.0875 plutonium S34 = 0 plutoniumS41 = -0.0972 J 0.0496 plutonium S42 = -0.1749 J 0.0892 plu tonium S44 = 0 plutoniumAdmittance Matrix sing transmittal line electrical capacityHarmonizing to the direction of the Question 3, power system theoretical account can be drawn by utilizing I tantamount circuit of the lines with capacitive shunt entree ( yc ) of 0.1 plutonium at each side as shown below.Admittance MatrixContrary 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 + yc14V1Bus 2 I2 = I21 + I23 + I24 + Ic21 + Ic23 + Ic24 I2 = y21 ( V2-V1 ) + y23 ( V2-V3 ) + y24 ( V2-V4 ) + yc21V2 + yc23V2 + yc24V2Bus 3 I3 = I30 + I31 + I32 + Ic31 + Ic32 I3 = y30V3 + y31 ( V3-V1 ) + y32 ( V3-V2 ) + yc31V3 + yc32V3Bus 4 I4 = I40 + I 41 + I42 + Ic41 + Ic42 I4 = y40V4 + y41 ( V4-V1 ) + y42 ( V4-V2 ) + yc41V4 + yc42V4Equation above can be rearranged to rive and group single merchandises by electromotive force.Bus 1 I1 = ( y12 + y13 + y14 + yc12 + yc13+ yc14 ) V1 y12V2 y13V3 y14V4 = Y11V1 + Y12V2 + Y13V3 + Y14V4Bus 2 I2 = ( y21 + y23 + y24 + yc21 + yc23+ yc24 ) V2- y21V1 y23V3 y24V4 = Y21V1 + Y22V2 + Y23V3 + Y24V4Bus 3 I3 = ( y30 + y31 + y32 + yc31+ yc32 ) V3 y31V1 y32V2 = Y31V1 + Y32V2 + Y33V3 + Y34V4Bus 4 I4 = ( y40 + y41 + y42 + yc41+ yc42 ) V4 y41V1 y42V2 = Y41V1 + Y42V2 + Y43V3 + Y44V4Finally, 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 plutoniumTherefore, entree matrix Ybus is as follows Ybus = =Annex-1 Matlab beginning codification and Calculation consequences with MatlabMatlab Source Code% sic 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=-j0.8y12=-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=2.5 i 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% lay the 44 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 coachsv12=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 followingp12=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 ) % terminalMatlab Calculation Results