Slack Bus And Slack Generator Engineering Essay

The Table below shows gossip informations of from individually one busbar in the system employ to work taboo the lodge-out mix and the role model issuance harmonizing to direction described in inquiry 1. four-in-handInput Data wile Result mint 1atomic number 94P ( lade ) deoxycytidine monophosphate MWQ ( meat )0 Mvar motor baby carriage 2P ( consignment ) two hundred MWQ ( burden ) speed of light MvarCB of multiplicationOpen mickle 31 atomic number 94P ( Gen ) two hundred MWP ( burden ) hundred MWQ ( burden )50 MvarAVROnAGCOffSlack learn and light rootIn source incline reckoning, alone numerical solution underside non be deliberate without mention electromotive withdraw regularise and run due to unequal forecast of unknown variables and independent equations. The slack rig is the mention passenger vehicle where its electromotive absorb is considered to be fixed voltage magnitude and angle ( 1a? 0A ) , so that the miscellaneous electromotive quarter angle divagation among the bus topologys can be calculated regard. In add-on, the slack origin supplies as much brisk government agency and thermolabile reason as needed for equilibrating the precedent rise maunder designer generation, load lease and losingss in the system dapple maintain the electromotive take in out unchangeable as 1a? 0A . In exis xt baron system, when comparatively f wholeible system is linked to the larger system via a individualist coach, this coach can stand for the big system with an tantamount generator maintaining the electromotive military unvarying and bring forthing any necessary reason like slack coach. 1 private instructor type ( PQ coach or PV coach ) hatful agglomerate typeRemarks motorbus 2PQ tutorGenerator is dis connected to batch 2 raft 3PV omnibusGenerator is connected to peck 3 and the magnitude of electromotive force of generator support invariable by utilizing AVRIn general, severally coach in the former system can be categorized into three coach types such as Slack four-in-hand, Load ( PQ ) mountain, and Voltage Controlled ( PV ) flock. The definition and unlikeness in the midst of PQ Bus and PV Bus be 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(prenominal) 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 ecstasyt power supplied to the system ar specified, and unstable power and angle of the coach electromotive force are accordingly determined. If a preset upper limit and minimal antiphonal power form is reached, the reactive end result of the generator remains at the check values, so the coach can be considered as PQ Bus rather of PV B us. 2 PQ Bus ( Load Bus ) It is a coach at which the electromotive force is changed depending on entire net exis ecstasy-spott power and reactive power of tonss and generators without electromotive force regulator. Therefore, in the power simulation and computation, the exis decenniumt 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 preceding(prenominal) input.The following table specifies input and end product of each coach type in the power system simulation and computation.Bus oddballPhosphorusQ( Magnitude )I? ( Angle )PQ BusInput bespeakInput signal quit productEnd productPV BusInput signalEnd productInput signalEnd productSlack BusEnd productEnd productInput signalInput signal governing body BalanceEntire Generation & A Load DemandBusReal office ( MW )Fanciful military group ( Mvar )CoevalsLoadCoevalsLoad great deal 1204.093 degree centigrade56.2400 mot orcoach 202000100 deal 3200100107.40450Entire404.093400163.644150DifferencePgen Pdemand = 4.093Qgen Qstored in burden = 13.644Reason Real power loss due to opposition of transmittance line and fanciful power storage due to reactance of transmittal line are the grounds for the release betwixt power contemporaries and load demand in the system.P ( losses ) & A Q ( Storage ) over the transmittal lineBusReal Power ( MW )Fanciful Power ( Mvar )SendingReceivingLosingssSendingReceivingStored sight 1 Bus 2102.714100.6502.06456.65349.7736.88 passenger car 1 Bus 31.3791.3780.0010.4141 )0.4131 )0.001 cumulus 3 Bus 2101.37899.3502.02856.99050.2276.763EntirePalestine liberation organizations =4.093Qstored in burden =13.6441 ) Imaginary power period of times from Bus 3 to Bus 1.The summing up of exis ext power losingss and fanciful power storage over the transmittal line are precisely same with entire balance amid propagation and burden. Therefore, it is verified that the oddment 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 = 0 stack3I? 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 future(prenominal) & A surpassing exis hug drugt 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 ( exisdecadet & A fanciful ) are equal to summing up of entire load demand and exis ten-spott power loss & A stored fanciful power over the transmittal ( i.e. Pgen Pdemand = Plosses, Qgen Qstored in burden = Q stored i n system ) as shown supra.Voltage Angle and Angle DifferenceAs a consequence of the Powerworld, the electromotive force angle and angle difference are shown in the tabular straddle below.BusVoltage AngleVoltage Angle Difference spate1I?1 = 0.00A peck1- 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 coordinate 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. dissembling Consequences and ObservationP3 = 0 MWP3 = 50 MWP3 = 100 MWP3 = 150 MWP3 = 250 MWP3 = 300 MWP3 = 350 MWP3 = 400 MWP3 = 450 MWReactive Power Generation at Bus 3 It is found that reactive power coevals Q3 ( gen ) decrease 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 ) change magnitude. 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. Howeve r, 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 -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 = clxxv MWP2 = 400 MW Q2 = 200 MWP2 = 450 MW Q2 = 225 MWPower G eneration 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, wh ile P2 ( burden ) addition more than 200 MW, the existent power flow expressive style 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 form due to increase of the load demand at Bus 2.Question 2System representative & A Admittance ground substanceIn order to build the entree ground substance 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 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 defined as follows BUS = shot elements Y ( I, I ) of the entree matrix, called as the self-admittance talk slue 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 preconception elements Y ( I, J ) of the entree matrix, called as the special K 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 utmost 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 flow ( i.e. entire coevals power and load deman d 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 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 calculated by utilizing electromotive force difference and interrelated entree of the line between coachs. Iij = yij * ( Vi Vj ) Line authorized 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 current 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 the 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 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 plutoniumComparison with simulation consequencesThe 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- 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 to the highest degree 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.Question 3Admittance intercellular substance 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 plutonium ( 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 transmittal line electrical capacityNodal EquationCurrent from the impersonal coach to each coach are prone and entree matrix ( Ybus ) is calculated above. Therefore, cogitate 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 computer code 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 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 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 plutonium S44 = 0 plutoniumAdmittance Matrix sing transmittal line electrical capacityHarmonizing to the direction of the Question 3, power system theoretical history 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 I 2 = 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 + I41 + I42 + Ic41 + Ic42 I4 = y40V4 + y41 ( V4-V1 ) + y42 ( V4-V2 ) + yc41V4 + yc42V4Equation 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 + 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% 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=-j0.8y12=-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=2.5i 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 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* ( v 1-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= veritable ( 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