Slack Bus And Slack Generator

at large(p) motor heap And Slack GeneratorThe Table below shows input data of apiece messbar in the organization use to solve the author come down and the simulation result concord to instruction described in straits 1. good deal foreplay DataSimulation Result mess 1puP ( fill) snow MWQ ( encumbrance)0 Mvar passel 2P ( hitch)cc MWQ (load) coulomb MvarCB of GenerationOpen passenger vehicle 31 puP (Gen)200 MWP (load) cytosine MWQ (load)50 MvarAVROnAGCOffSlack handler and cliff referenceIn force out give calculation, unique numerical solution shagnot be calculated without germ potential difference order and angle ascribable to unequal number of unknown variables and independent equalitys. The delaying agglomerate is the reference bus where its emf is considered to be fixed potentiality magnitude and angle (10), so that the various potential drop angle difference among the buses cornerstone be calculated respect. In addition, the slack generator supplies as much corporeal world personnel and reactive super role as needed for balancing the mightiness flow considering spring multiplication, load demand and losses in the governance tour keep the potentiality constant as 10. In real military unit strategy, 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(prenominal) prerequisite power like slack bus. 1 jitney type (PQ bus or PV bus) slew wad typeComments pile 2PQ batchGenerator is disconnected to great deal 2BUS 3PV motorcoachGenerator is connected to lot 3 and the magnitude of voltage of generator keep constant by using AVRIn general, apiece bus in the power system can be categorized into three bus types such as Slack mountain, Load (PQ) flock, and Voltage Controlled (PV) Bus. The definition and difference amongst PQ Bus and PV Bus ar described as follows 2PV 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 argon 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 take of the generator remains at the limited values, so the bus can be considered as PQ Bus instead of PV Bus. 2PQ Bus (Load Bus) It is a bus at which the voltage is changed depending on fare 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 higher up input.The pursuance board specifies input and output of each bus type in the power system simulation and calculation.Bus TypePQ(Magnitude) ( rake)PQ BusInputInputOutputOutputPV BusInputOutputInputOutputSlack BusOutputOutputInputInput placement BalanceTotal Generation Load commandBUS realistic berth (MW)complex quantity Power (Mvar)GenerationLoadGenerationLoadBUS 1204.093 cytosine56.2400BUS 202000100BUS 3200100107.40450Total404.093400163.644150 discrepancyPgen Pdemand = 4.093Qgen Qstored in load = 13.644Reason Real power loss due to resistance of transmission preeminence and complex number power storage due to reactance of transmission line are the reasons for the difference amidst power generation and load demand in the system.P (Losses) Q (Storage) over the transmission lineBUSReal Power (MW) conceptional Power (Mvar)SendingReceivingLossesSendingReceivingStoredBUS 1 BUS 2102.714100.6502.06456.65349.7736.88BUS 1 BUS 31.3791.3780.0010.4 141)0.4131)0.001BUS 3 BUS 2101.37899.3502.02856.99050.2276.763TotalPlosses =4.093Qstored in load =13.6441) 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 4Bus1 P1 = + Pgen1 Pload1 P12 P13 = 204.093 100 102.714 1.379 = 0 Q1 = + Qgen1 Qload1 Q12 Q13 = 56.24 0 56.653 + 0.413 = 0Bus2 P2 = + Pgen2 Pload2 P21 P23 = 0 200 + 100.65 + 99.35 = 0 Q2 = + Qgen2 Qload2 Q21 Q23 = 0 100 + 49.773 + 50.227 = 0BUS3 P3 = + Pgen3 Pload3 P31 P32 = 200 100 + 1.378 101.378 = 0 Q3 = + Qgen3 Qload3 Q31 Q32 = 107.404 50 0.414 56.99 = 0 correspond to the calculation above, as summation of incoming outgo 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 goingAs a result of the Powerworld, the voltage angle and angle difference are shown in the table below.BUSVoltage AngleVoltage Angle DifferenceBUS11 = 0.00BUS1- BUS21 2 = 0.00 (-2.5662) = 2.5662BUS22 = -2.5662BUS2- BUS32 3 = -2.5662 (-0.043) = -2.5232BUS33 = -0.043BUS3- BUS13 1 = -0.043 0.00 = -0.043Power System Analysis -1The 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 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 entrap 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 imperative excitation of the generation by means of 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 volt age angle difference (3 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 (1 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 (3 1) is detrimental angle when P3(gen) is less(prenominal) than 200MW and the difference increase while P3(gen) increase.Power System Analysis -2The 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 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 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(prenominal) 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 1 2 and 3 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 different hand, while P2(load) increase more than 200 MW, the real power flow direction cha nges (Bus 1 to Bus 3) and the real power flow increases. Accordingly, the voltage angle difference 1 3 change from disallow 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.Question 2System Model entree ground substanceIn order to construct the entranceway hyaloplasm of Powerworld B3 case, single chassis equivalent circuit can be drawn as belowz = r + jx (r = 0, x = 0.05)z12 = z21= j0.05 pu, y12 = 1/ z12 = 1/j0.05 = -j20 pu = y12z13 = z31= j0.05 pu, y13 = 1/ z13 = 1/j0.05 = -j20 pu = y31z23 = z32= j0.05 pu, y23 = 1/ z23 = 1/j0.05 = -j20 pu = y32Admittance ground substance can be defined as followsBUS =Diagonal elements Y(i,i) of the addition hyaloplasm, called as the self-admittance lecture slide 6, 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 puOff bezzant e lements Y(i,j) of the admittance matrix, called as the mutual admittance lecture slide 6, 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 puTherefore, the final admittance matrix BUS isBUS = =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 CalculationNodal equation with the admittance matrix can be employ to calculate voltage at each bus if we know all the underway (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 followsSimulation resultVoltage at each Bus and Voltage DifferenceV1 = 1 0.00 pu (BU S1) V2 = 1 -0.48 pu (BUS2) V3 = 1 0.48 pu (BUS 3)Voltage difference between BUS 1 and BUS 2V12 = V1 V2 = 1 0.00 1 -0.48 = 3.5 x 10-5 + j 8.38 x 10-3 = 8.38 x 10-3 89.76 puV21 = V2 V1 = V12 = 3.5 x 10-5 j 8.38 x 10-3 = 8.38 x 10-3 -90.24 puVoltage difference between BUS 3 and BUS 2V32 = V3 V2 = 1 0.48 1 -0.48 = j 16.76 x 10-3 = 16.76 x 10-3 90 puV23 = V2 V3 = V32 = j 16.76 x 10-3 = -16,76 x 10-3 -90 puVoltage difference between BUS 3 and BUS 1V31 = V3 V1 = 1 0.48 1 0.00 = 3.5 x 10-5 + j 8.38 x 10-3 = 8.38 x 10-3 90.24 puV13 = V1 V3 = V31 = 3.5 x 10-5 j 8.38 x 10-3 = 8.38 x 10-3 -89.76 puLine flowingCurrent 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 2I12 = y12 x (V1 V2) = -j20 x 8.38 x 10-3 89.76 = 167.6 x 10-3 -0.24 pu (BUS 1 BUS 2)I21 = y21 x (V2 V1) = -j20 x 8.38 x 10-3 -90.24 = 167.6 x 10-3 -180.24 pu (BUS 2 BUS 1)Line current between BUS 3 and BUS 2I32 = y32 x (V3 V2) = -j20 x 16.76 x 10-3 90 = 335.2 x 10-3 0.00 pu (BUS 3 BUS 2)I23 = y23 x (V2 V3) = -j20 x 16.76 x 10-3 -90 = 335.2 x 10-3 180 pu (BUS 2 BUS 3)Line current between BUS 3 and BUS 1I31 = y31 x (V3 V1) = -j20 x 8.38 x 10-3 90.24 = 167.6 x 10-3 0.24 pu (BUS 3 BUS 1)I13 = y13 x (V1 V3) = -j20 x 8.38 x 10-3 -89.76 = 167.6 x 10-3 -179.76 pu (BUS 1 BUS 3) unvarnished Power FlowApparent 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 2S12 = V1* I*12 = 1 0.00 x 167.6 x 10-3 0.24 = 167.6 x 10-3 0.24 = 0.1676 + j 7.02 x 10-4 puApparent Power from BUS 2 to BUS 1S21=V2* I*21=1-0.48 x 167.6 x 10-3180.24=167.6 x 10-3179.76 = -0.1676 + j7.02 x 10-4 puApparent Power from BUS 3 to BUS 2S32 = V3* I*32 = 1 0.48 x 335.2 x 10-3 0.00 = 335.2 x 10-3 0.48 = 0.3352 + j 2.81 x 10-3 puApparent Power from BUS 2 to BUS 3S23=V2* I*23=1 -0.48 x 335.2 x 10-3 180= 33 5.2 x 10-3 179.76 = -0.3352 + j 2.81 x 10-3 puApparent Power from BUS 3 to BUS 1S31 = V3* I*31 = 10.48 x 167.6 x 10-3-0.24 = 167.6 x 10-3 0.24 = 0.1676 + j 7.02 x 10-4 puApparent Power from BUS 1 to BUS 3S13=V1* I*13=10.00 x 167.6 x 10-3179.76= 167.6 x 10-3179.76 = -0.1676 + j 7.02 x 10-4 puComparison with simulation resultsThe 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. 3Sactual = Sbase - Spu = 100 MVA - SpuIactual = Ibase - Ipu = - Ipu = - Ipu = 167.3479 A - IpuCalculation Result and Simulation ResultFlow direction placeCalculation ResultSimulation ResultBUS 1 BUS 2S120.1676 - 100 = 16.76 MVA16.67 MVAP1216.76 MW16.67 MWQ120.0702 Mvar0.07 MvarI120.1676 - 167.3479 = 28.0475 A27.89 ABUS 3 BUS 2S320.3352 - 100 = 33.52 MVA33.33 MVAP3233.52 MW33.33 MWQ320.281 Mvar0 .28 MvarI320.3352 - 167.3479 = 56.0950 A55.78 ABUS 3 BUS 1S310.1676 - 100 = 16.76 MVA16.67 MVAP3116.76 MW16.67 MWQ310.0702 Mvar0.07 MvarI310.1676 - 167.3479 = 28.0475 A27.89 ABUS 2 BUS 1S210.1676 - 100 = 16.76 MVA16.67 MVAP21-16.76 MW-16.67 MWQ210.0702 Mvar0.07 MvarI210.1676 - 167.3479 = 28.0475 A27.89 ABUS 2 BUS 3S230.3352 - 100 = 33.52 MVA33.33 MVAP23-33.52 MW-33.33 MWQ230.281 Mvar0.28 MvarI230.3352 - 167.3479 = 56.0950 A55.78 ABUS 1 BUS 3S130.1676 - 100 = 16.76 MVA16.67 MVAP13-16.76 MW-16.67 MWQ130.0702 Mvar0.07 MvarI130.1676 - 167.3479 = 28.0475 A27.89 AIt 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.Question 3Ad mittance Matrix and Nodal EquationAdmittance between two busesy12 = y21 = -j8 pu y13 = y31 = -j4 pu y14 = y41 = -j2.5 puy23 = y32 = -j4 pu y24 = y42 = -j5 puy30 = -j0.8 pu (BUS3-Neutral BUS) y40 = -j0.8 pu (BUS4-Neutral BUS)Admittance MatrixYbus (Admittance Matrix) =Diagonal elements Y(i,i) of the admittance matrix, called as the self-admittance 2 4, 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 = -j8.3Off diagonal elements Y(i,j) of the admittance matrix, called as the mutual admittance 2 4, 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 puTherefore, a dmittance matrix Ybus is as followsYbus = =Power Flow AnalysisPower flow ignoring transmission line contentNodal EquationCurrent from the neutral bus to each bus are given and admittance matrix (Ybus) is calculated above. Therefore, final nodal equation is as followsIbus = Ybus * Vbus Vbus = Y-1bus * Ibus= Ybus ==Voltage AnalysisVoltage 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 puV21 = -0.0034 j 0.0031 pu V23 = -0.0311 j 0.0288 pu V24 = 0.0302 + j 0.0280 puV31 = 0.0277 + j 0.0257 pu V32 = 0.0311 + j 0.0288 puV41 = -0.0336 j 0.0311 pu V42 = -0.0302 j 0.0280 puCurrent flow in the systemCurrent 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 a s followsI12 = 0.0249 j 0.0269 pu I13 = -0.1026 + j 0.1108 pu I14 = 0.0777 j 0.0840 puI21 = -0.0249 + j 0.0269 pu I23 = -0.1151 + j 0.1243 pu I24 = 0.1399 j 0.1511I31 = 0.1026 j 0.1108 pu I32 = 0.1151 j 0.1243 pu I34 = 0 puI41 = -0.0777 + j 0.0840 pu I42 = -0.1399 + j 0.1511 pu I43 = 0 puPower flow in the systemApparent 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 followsS12 = 0.0311 + j 0.0175 pu S13 = -0.1283 j 0.0723 pu S14 = 0.0972 + j 0.0548 puS21 = -0.0311 j 0.0174 pu S23 = -0.1438 j 0.0803 pu S24 = 0.1749 + j 0.0977 puS31 = 0.1283 + j 0.0780 pu S32 = 0.1438 + j 0.0875 pu S34 = 0 puS41 = -0.0972 j 0.0496 pu S42 = -0.1749 j 0.0892 pu S44 = 0 puAdmittance Matrix considering transmission line capacitanceAccording to the instruction of the Question 3, power system model can be drawn by using equivalent circuit of the lines with capacitive shunt adm ittance (yc) of 0.1 pu at each side as shown below.Admittance MatrixContrary 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 Kirchhoffs current law at each bus is as follows 2 5Bus 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 + I41 + I42 + Ic41 + Ic42 I4 = y40V4 + y41(V4-V1) + y42(V4-V2) + yc41V4 + yc42V4Equation above can be rearranged to separate and assort individual products by voltage.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 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 puTherefore, admittance matrix Ybus is as followsYbus = =Annex-1 Matlab kickoff code and Calculation results with MatlabMatlab 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=-j0.8y12=-8i y21=-8i y13=-4i y31=-4i y14=-2.5i y41=-2.5i y23=-4i y32=-4iy24=-5i y42=-5i y34=0 y43=0 y30=-0.8i y40=-0.8iY11=-8i-4i-2.5i Y12=8i Y13=4i Y14=2.5iY21=8i Y22=-8i-4i-5i Y23=4i Y24=5iY31=4i Y32=4i Y33=-0.8i-4i-4i Y34=0Y41=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 44 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=-0.4808-0.4808iIbus=i1 i2 i3 i4% separately bus voltage can be calculated by using matrix division (V= YbusI)Vbus=YbusIbusv1=Vbus(1,1) v2=Vbus(2,1) v3=Vbus(3,1) v4=Vbus(4,1)% a ugur voltage difference between busesv12=v1-v2 v13=v1-v3 v14=v1-v4v21=v2-v1 v23=v2-v3 v24=v2-v4v31=v3-v1 v32=v3-v2 v34=v3-v4v41=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*v14i21=y21*v21 i23=y23*v23 i24=y24*v24i31=y31*v31 i32=y32*v32 i34=y34*v34i41=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 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=i mag(s43) % endMatlab Calculation Results