Writing and Comprehensive Form

Description of Reading and Writing Measures Standardized Test Description KTEA II Reading comprehension and Written Expression The Reading comprehension and written expression subtests were given and scored. The Kaufman Test of Educational Achievement, Second Edition is an individually administered measure of academic achievement for ages 4 and a half through 25. The test is available in two versions: the Brief form which assesses the achievement of reading , mathematics, written expression; and the Comprehensive Form which covers a wide range of achievement domains and an analysis of students’ errors.The Kaufman Test of Educational Achievement, Comprehensive Form, Second Edition represents a revision of the Kaufman Test of Educational Achievement Comprehensive Form. The KTEA –II Comprehensive Form has an expanded age range and has retained the five subtests from the original KTEA and has modified to allow for testing of children and adults from preschool-age through co llege-age. Nine new subtests have been added to allow for assessment of a broad range of achievement domains and skills. KTEA-II Comprehensive Form age norms are provided for ages 4 and a half through 25, and grade norms are provided for Kindergarten through Grade 12.KTEA III Comprehensive Form is curriculum-based it provides norm-referenced and error analysis systems, criterion-referenced assessment in reading, mathematics, written language, and oral language. The KTEA-II Comprehensive Form has two independent , parallel forms (A and B) and the KTEA-II Brief Form norms at ages 4 and a half through 90. These three non-overlapping batteries make the KTEA II useful for measuring student progress. The KTEA II Comprehensive Form make it an important tool for assessing academic achievement.The KTEA II measures achievement in reading, mathematics, written language, and oral language and allows the examiner to administer a single subtest or a combination of subtests to assess achievement i n one or more domains. All seven specific learning disability areas identified in the Individuals with Disabilities Education Act Amendment of 1997 (IDEA,1997)are measured: basic reading skills, reading comprehension, mathematics calculation, mathematics, reasoning, oral expression, listening comprehension, and written expression.The KTEA Comprehensive Form like the KTEA was developed from a clinical model of assessment. Curriculum experts defined specific sub skills measured by each subtest and the different types of errors students are likely to make on each subtest. Standardization data guided the final error analysis System. KTEA-II Comprehensive Form content has undergone bias reviews to ensure that students of either sex and ethnic and socioeconomic backgrounds can be assessed. The KTEA II Comprehensive Form was normed using two separate representative, nationwide standardizations, one in the fall and one in the spring.The procedure accurately measures students’ perform ance both at the beginning and end of he year. The KTEA-II Comprehensive Form is conformed with the Kaufman Assessment Battery for Children, Second Edition. The KTEA II Comprehensive Form is a reliable, valid measure of academic achievement. The KTEA II allows the examiner to observe the student’s test taking behavior, motivation, and visual-motor coordination. The two parallel forms make it an ideal instrument for longitudinal studies. KTEA II Written Expression subtestDescription of writing task Students are administered an item set based on their grade. (3rd Grade) Following assessment directions from the KTEA II manual and easel, I provided the Level 3 booklet and a pencil to my student. The written expression booklet is titled Kyra’s Dragon. I explain to my student following the provided directions from the easel that this story is about a girl named Kyra and the dragon she has to find. As we go through the story, you’ll write some of the words and sentence s. This is similar to the â€Å"cloze† technique that was used in the informal assessment, The McLeod Assessment of Reading Comprehensions. ) I tell my student to write the best words and sentences he can and not to worry if he doesn’t know how to spell a word – spelling won’t count. The first item we starts with is #31 I say â€Å" Let’s start by writing your full name here† and I point to the to of the booklet. The next item #32, my student writes the sentence â€Å"The dragon carries people away. † That I dictate. Tets: Writing SkillsOn item #3 my student has to write one word to complete the sentence â€Å" The king says to Kyra, â€Å"Finding the dragon_____________ save us all. † For item #34, my student has to write one good sentence to complete a part of the story, â€Å" Kyra’s Dragon. This fill in the blank interactive story goes on with similar tasks inserting words, sentences, combining sentences, proper word usage of specific words and punctuation into the story booklet until my student gets item # 49. Item #49 is where my student must complete a timed retell of the entire story, pretending my student is the king’s scribe.My student must retell the story of Kyra’s dragon so that his grandchildren will know how people came to live in their new town. He is given 10 minutes to complete his retell. My task as administrator of this test is to follow the script on the easel, read the prompts, and point to the correct place for the student to write his answers. This took about 25 minutes to administer. I am allowed to repeat story segments and item instructions if necessary. I may also tell a student how to spell a word if they ask, since spelling is not scored in this subtest, but only if examinee asks for assistance.

Slack Bus And Slack Generator Engineering Essay

The Table below shows input informations of each busbar in the system used to work out the power flow and the simulation consequence harmonizing to direction described in inquiry 1.BusInput Data[ Simulation Result ] BUS 1 plutonium P ( burden ) 100 MW Q ( burden ) 0 Mvar BUS 2 P ( burden ) 200 MW Q ( burden ) 100 Mvar CB of Generation Open BUS 3 1 plutonium P ( Gen ) 200 MW P ( burden ) 100 MW Q ( burden ) 50 Mvar AVR On AGC OffSlack coach and slack generatorIn power flow computation, alone numerical solution can non be calculated without mention electromotive force magnitude and angle due to unequal figure of unknown variables and independent equations. The slack coach is the mention coach where its electromotive force is considered to be fixed voltage magnitude and angle ( 1a? 0A ° ) , so that the assorted electromotive force angle difference among the coachs can be calculated regard. In add-on, the slack generator supplies as much existent power and reactive power as needed for equilibrating the power flow sing power coevals, load demand and losingss in the system while maintain the electromotive force changeless as 1a? 0A ° . In existent power system, when comparatively weak system is linked to the larger system via a individual coach, this coach can stand for the big system with an tantamount generator maintaining the electromotive force changeless and bring forthing any necessary power like sla ck coach. [ 1 ]Bus type ( PQ coach or PV coach )BusBus typeRemarksBUS 2 PQ Bus Generator is disconnected to Bus 2 BUS 3 PV Bus Generator is connected to Bus 3 and the magnitude of electromotive force of generator support invariable by utilizing AVR In general, each coach in the power system can be categorized into three coach types such as Slack Bus, Load ( PQ ) Bus, and Voltage Controlled ( PV ) Bus. The definition and difference between PQ Bus and PV Bus are described as follows ; [ 2 ] PV Bus ( Generator Bus or Voltage Controlled Bus ) : It is a coach at which the magnitude of the coach electromotive force is kept changeless by the generator. Even though the coach has several generators and burden, if any generators connected to the coach modulate the coach electromotive force with AVR, so this coach is referred to PV Bus. For PV coach, the magnitude of the coach electromotive force and existent power supplied to the system are specified, and reactive power and angle of the coach electromotive force are consequently determined. If a preset upper limit and minimal reactive power bound is reached, the reactive end product of the generator remains at the limited values, so the coach can be considered as PQ Bus alternatively of PV Bus. [ 2 ] PQ Bus ( Load Bus ) : It is a coach at which the electromotive force is changed depending on entire net existent power and reactive power of tonss and generators without electromotive force regulator. Therefore, in the power simulation and computation, the existent power and reactive power of the tonss are specified as input informations and consequently the electromotive force ( magnitude and angle ) is calculated based on the above input. The following table specifies input and end product of each coach type in the power system simulation and computation. Bus Type Phosphorus Q ( Magnitude ) I? ( Angle ) PQ Bus Input signal Input signal End product End product PV Bus Input signal End product Input signal End product Slack Bus End product End product Input signal Input signalSystem BalanceEntire Generation & A ; Load DemandBusReal Power ( MW )Fanciful Power ( Mvar )CoevalsLoadCoevalsLoadBUS 1 204.093 100 56.240 0 BUS 2 0 200 0 100 BUS 3 200 100 107.404 50 Entire 404.093 400 163.644 150DifferencePgen – Pdemand = 4.093Qgen – Qstored in burden = 13.644Reason: Real power loss due to opposition of transmittal line and fanciful power storage due to reactance of transmittal line are the grounds for the difference between power coevals and load demand in the system.P ( Losses ) & A ; Q ( Storage ) over the transmittal lineBusReal Power ( MW )Fanciful Power ( Mvar )SendingReceivingLosingssSendingReceivingStoredBUS 1 – Bus 2 102.714 100.650 2.064 56.653 49.773 6.88 BUS 1 – Bus 3 1.379 1.378 0.001 0.4141 ) 0.4131 ) 0.001 BUS 3 – Bus 2 101.378 99.350 2.028 56.990 50.227 6.763 Entire Palestine liberation organizations =4.093Qstored 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 same with entire difference between coevals and burden. Therefore, it is verified that the difference is shown over the transmittal line. ‘Kirchoff ‘ balance as each coach [ 4 ] Bus1 I? P1 = + Pgen1 – Pload1 – P12 – P13 = 204.093 – 100 – 102.714 – 1.379 = 0 I? Q1 = + Qgen1 – Qload1 – Q12 – Q13 = 56.24 – 0 – 56.653 + 0.413 = 0 Bus2 I? P2 = + Pgen2 – Pload2 – P21 – P23 = 0 – 200 + 100.65 + 99.35 = 0 I? Q2 = + Qgen2 – Qload2 – Q21 – Q23 = 0 – 100 + 49.773 + 50.227 = 0 BUS3 I? P3 = + Pgen3 – Pload3 – P31 – P32 = 200 – 100 + 1.378 – 101.378 = 0 I? Q3 = + Qgen3 – Qload3 – Q31 – Q32 = 107.404 – 50 – 0.414 – 56.99 = 0 Harmonizing to the computation supra, as summing up of incoming & A ; surpassing existent power and fanciful power at each coach become zero, it is verified that each busbar obeys a ‘Kirchoff ‘ balance. In add-on, the entire power system is wholly balanced, because entire coevals power ( existent & A ; fanciful ) are equal to summing up of entire load demand and existent power loss & A ; stored fanciful power over the transmittal ( i.e. Pgen – Pdemand = Plosses, Qgen – Qstored in burden = Q stored in system ) as shown above.Voltage Angle and Angle DifferenceAs a consequence of the Powerworld, the electromotive force angle and angle difference are shown in the tabular array below.BusVoltage AngleVoltage Angle DifferenceBUS1 I?1 = 0.00A ° BUS1- BUS2 I?1 – I?2 = 0.00A ° – ( -2.5662A ° ) = 2.5662A ° BUS2 I?2 = -2.5662A ° BUS2- BUS3 I?2 – I?3 = -2.5662A ° – ( -0.043A ° ) = -2.5232A ° BUS3 I?3 = -0.043A ° BUS3- BUS1 I?3 – I?1 = -0.043A ° – 0.00A ° = -0.043A °Power System Analysis -1The tabular array below summarizes coevals and electromotive force angle fluctuation at each coach as coevals at Bus 3 varies from 0 MW to 450 MW by 50MW.Simulation Consequences and ObservationP3 = 0 MW P3 = 50 MW P3 = 100 MW P3 = 150 MW P3 = 250 MW P3 = 300 MW P3 = 350 MW P3 = 400 MW P3 = 450 MW Reactive Power Generation at Bus 3: It is found that reactive power coevals Q3 ( gen ) lessening while existent power coevals P3 ( gen ) addition because Bus 3 as a PV Bus regulates the changeless coach electromotive force magnitude by commanding excitement of the coevals through the AVR. Power Generation at Bus 1: It is found that P1 ( gen ) decreases and Q1 ( gen ) increases at the same time, while P3 ( gen ) additions and Q3 ( gen ) lessening. As the entire load demand in the system keeps changeless ( i.e. Ptotal ( burden ) = 400 MW, Qtotal ( burden ) = 150Mvar ) , any necessary existent power and reactive power for the system balance demand to be supplied by generator ( loose generator ) at Bus 1. Therefore, power coevals P1 ( gen ) and Q1 ( gen ) at Bus 1 alteration reversely compared to power coevals alteration at Bus 3. Voltage Angle Difference: In general, existent power flow is influenced by electromotive force angle difference between directing coach and having coach harmonizing to PR = . Therefore, it is observed that every bit existent power coevals P3 ( gen ) increases existent power flow from Bus 3 to Bus2 addition, consequently voltage angle difference ( I?3 – I?2 ) between Bus 3 and Bus 2 additions. However, lessening in existent power from Bus 1 to Bus 2 due to increase of P3 ( gen ) consequence in lessening of electromotive force angle difference ( I?1 – I?2 ) . In add-on, Real power between Bus 1 and Bus 3 flows from Bus 1 to Bus 3 until P3 ( gen ) range to 200 MW and as P3 ( gen ) addition more than 200 MW the existent power flows from Bus 3 to Bus 1. So, it is besides observed that electromotive force angle difference ( I?3 – I?1 ) is negative angle when P3 ( gen ) is less than 200MW and the difference addition while P3 ( gen ) addition.Power System Analysis -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 MW P2 = 50 MW Q2 = 25 MW P2 = 100 MW Q2 = 50 MW P2 = 150 MW Q2 = 75 MW P2 = 250 MW Q2 = 125 MW P2 = 300 MW Q2 = 150 MW P2 = 350 MW Q2 = 175 MW P2 = 400 MW Q2 = 200 MW P2 = 450 MW Q2 = 225 MW Power Generation at Bus 1 and Bus 3: It is observed that as the entire load demand in the system increases due to increase of load demand P2 ( burden ) & A ; Q2 ( burden ) at Bus 2, any necessary existent power for the system balance is supplied by generator ( loose generator ) at Bus 1 sing changeless P3 ( gen ) , so P1 ( gen ) increases. In add-on, any necessary reactive power for the system balance is supplied from Bus 1 every bit good as Bus 3, so both Q1 ( gen ) and Q3 ( gen ) addition. Voltage Angle Difference: It is found that existent power flow addition both from Bus 1 to Bus 2 and from Bus 3 to Bus 2 due to increase of load demand at Bus2. Consequently, both electromotive force angle difference I?1 – I?2 and I?3 – I?2 addition when the power flow P12 and P32 addition. In add-on, when P2 ( burden ) is less than 200 MW, P1gen is comparatively low. Therefore existent power between Bus 3 and Bus 1 flows from Bus 3 to Bus 1 at lower P2 ( burden ) ( less than 200MW ) . On the other manus, while P2 ( burden ) addition more than 200 MW, the existent power flow way alterations ( Bus 1 to Bus 3 ) and the existent power flow additions. Consequently, the electromotive force angle difference I?1 – I?3 alteration from negative to positive and addition. Voltage Magnitude at Bus 2: It is observed that magnitude of coach electromotive force at Bus2 beads due to increase of the load demand at Bus 2.Question 2System Model & A ; Admittance MatrixIn order to build the entree matrix of Powerworld B3 instance, individual stage tantamount circuit can be drawn as below ;omega = R + jx ( r = 0, x = 0.05 )z12 = z21= j0.05 plutonium, y12 = 1/ z12 = 1/j0.05 = -j20 plutonium = y12 z13 = z31= j0.05 plutonium, y13 = 1/ z13 = 1/j0.05 = -j20 plutonium = y31 z23 = z32= j0.05 plutonium, y23 = 1/ z23 = 1/j0.05 = -j20 plutonium = y32 Admittance matrix can be defined as follows ; BUS = Diagonal elements Y ( I, I ) of the entree matrix, called as the self-admittance [ talk slide ] [ 6 ] , are the summing up of all entree connected with BUS I. = y12 + y13 = -j20 – j20 = -j40 plutonium = y21 + y23 = -j20 – j20 = -j40 plutonium = y31 + y32 = -j20 – j20 = -j40 plutonium Off diagonal elements Y ( I, J ) of the entree matrix, called as the common entree [ talk slide ] [ 6 ] , are negative entree between BUS I and BUS J. = – y12 = – ( -j20 ) = j20 plutonium = – y13 = – ( -j20 ) = j20 plutonium = – y21 = – ( -j20 ) = j20 plutonium = – y23 = – ( -j20 ) = j20 plutonium = – y31 = – ( -j20 ) = j20 plutonium = – y32 = – ( -j20 ) = j20 plutonium Therefore, the concluding entree matrix BUS is ; BUS = = The undermentioned figure shows the BUS of the Powerworld B3 instance and it is verified that the deliberate entree matrix is consistent with the consequence of the Powerworld.Power Flow CalculationNodal equation with the entree matrix can be used to cipher electromotive force at each coach if we know all the current ( i.e. entire coevals power and load demand at each BUS ) and eventually the power flow can be calculated consequently. , hence, In this inquiry, nevertheless, simulation consequences of the electromotive force at each coach from the Powerworld are used for the power flow computation as follows ; [ Simulation consequence ]Voltage at each Bus and Voltage 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 ° plutonium V21 = 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 ° plutonium V23 = 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 ° plutonium V13 = 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 current 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† Ã¢â‚¬â„¢ 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† Ã¢â‚¬â„¢ 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† Ã¢â‚¬â„¢ 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† Ã¢â‚¬â„¢ 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† Ã¢â‚¬â„¢ 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† Ã¢â‚¬â„¢ 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 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 the above computation consequences is pu value, so in order to compare the consequences with simulation consequences pu value of current and power flow demand to be converted to existent values by utilizing the undermentioned equation sing Sbase = 100MVA and Vline_base = 345kV. [ 3 ] Sactual = Sbase A- Spu = 100 MVA A- Spu Iactual = Ibase A- Ipu = A- Ipu = A- Ipu = 167.3479 A A- IpuCalculation Result and Simulation ResultFlow way & A ; ValueCalculation ConsequenceSimulation ConsequenceBUS 1 a† Ã¢â‚¬â„¢ BUS 2|S12| 0.1676 A- 100 = 16.76 MVA 16.67 MVA P12 16.76 MW 16.67 MW Q12 0.0702 Mvar 0.07 Mvar |I12| 0.1676 A- 167.3479 = 28.0475 A 27.89 ABUS 3 a† Ã¢â‚¬â„¢ BUS 2|S32| 0.3352 A- 100 = 33.52 MVA 33.33 MVA P32 33.52 MW 33.33 MW Q32 0.281 Mvar 0.28 Mvar |I32| 0.3352 A- 167.3479 = 56.0950 A 55.78 ABUS 3 a† Ã¢â‚¬â„¢ BUS 1|S31| 0.1676 A- 100 = 16.76 MVA 16.67 MVA P31 16.76 MW 16.67 MW Q31 0.0702 Mvar 0.07 Mvar |I31| 0.1676 A- 167.3479 = 28.0475 A 27.89 ABUS 2 a† Ã¢â‚¬â„¢ BUS 1|S21| 0.1676 A- 100 = 16.76 MVA 16.67 MVA P21 -16.76 MW -16.67 MW Q21 0.0702 Mvar 0.07 Mvar |I21| 0.1676 A- 167.3479 = 28.0475 A 27.89 ABUS 2 a† Ã¢â‚¬â„¢ BUS 3|S23| 0.3352 A- 100 = 33.52 MVA 33.33 MVA P23 -33.52 MW -33.33 MW Q23 0.281 Mvar 0.28 Mvar |I23| 0.3352 A- 167.3479 = 56.0950 A 55.78 ABUS 1 a† Ã¢â‚¬â„¢ BUS 3|S13| 0.1676 A- 100 = 16.76 MVA 16.67 MVA P13 -16.76 MW -16.67 MW Q13 0.0702 Mvar 0.07 Mvar |I13| 0.1676 A- 167.3479 = 28.0475 A 27.89 A It is found that computation consequences of current flow and evident power flows ( i.e. 28.0475 A and 56.0950 A/ 33.52 MVA and 16.76MVA ) are about 0.5 % higher than simulation consequence ( i.e. 27.89 A and 55.78 A / 33.33 MVA and 16.67 MVA ) which can be considered somewhat different. Difference of the electromotive force angle at each coach between computation ( 0.48A ° ) and simulation ( 0.4775A ° ) could be the ground for this minor difference.Question 3Admittance Matrix and Nodal EquationEntree between two coachsy12 = y21 = -j8 plutonium y13 = y31 = -j4 plutonium y14 = y41 = -j2.5 plutonium y23 = y32 = -j4 plutonium y24 = y42 = -j5 plutonium y30 = -j0.8 plutonium ( BUS3-Neutral BUS ) y40 = -j0.8 plutonium ( BUS4-Neutral BUS )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 = -j8.3 Off diagonal elements Y ( I, J ) of the entree matrix, called as the common entree [ 2 ] [ 4 ] , are negative entree between BUS I and BUS J. = – y12 = – ( -j8 ) = j8 plutonium = – y13 = – ( -j4 ) = j4 plutonium = – y14 = – ( -j2.5 ) = j2.5 plutonium = – y21 = – ( -j8 ) = j8 plutonium = – y23 = – ( -j4 ) = j4 plutonium = – y24 = – ( -j5 ) = j5 plutonium = – y31 = – ( -j4 ) = j4 plutonium = – y32 = – ( -j4 ) = j4 plutonium = – y34 = 0 plutonium = – y41 = – ( -j2.5 ) = j2.5 plutonium = – y42 = – ( -j5 ) = j5 plutonium = – y43 = 0 plutonium Therefore, entree matrix Ybus is as follows ;Ybus = =Power Flow AnalysisPower flow disregarding transmittal 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†¡Ã¢â‚¬â„¢ Vbus = Y-1bus * Ibus = Ybus a†¡Ã¢â‚¬â„¢ ==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 plutonium V21 = -0.0034 – J 0.0031 plutonium V23 = -0.0311 – J 0.0288 plutonium V24 = 0.0302 + J 0.0280 plutonium V31 = 0.0277 + J 0.0257 plutonium V32 = 0.0311 + J 0.0288 plutonium V41 = -0.0336 – J 0.0311 plutonium V42 = -0.0302 – J 0.0280 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 plutonium I21 = -0.0249 + J 0.0269 plutonium I23 = -0.1151 + J 0.1243 plutonium I24 = 0.1399 – J 0.1511 I31 = 0.1026 – J 0.1108 plutonium I32 = 0.1151 – J 0.1243 plutonium I34 = 0 plutonium I41 = -0.0777 + J 0.0840 plutonium I42 = -0.1399 + J 0.1511 plutonium I43 = 0 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 plutonium S21 = -0.0311 – J 0.0174 plutonium S23 = -0.1438 – J 0.0803 plutonium S24 = 0.1749 + J 0.0977 plutonium S31 = 0.1283 + J 0.0780 plutonium S32 = 0.1438 + J 0.0875 plutonium S34 = 0 plutonium S41 = -0.0972 – J 0.0496 plutonium S42 = -0.1749 – J 0.0892 plutonium S44 = 0 plutoniumAdmittance Matrix sing transmittal line electrical capacityHarmonizing to the direction of the Question 3, power system theoretical account can be drawn by utilizing Iˆ tantamount circuit of the lines with capacitive shunt entree ( yc ) of 0.1 plutonium at each side as shown below.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 + yc14V1 Bus 2: I2 = I21 + I23 + I24 + Ic21 + Ic23 + Ic24 I2 = y21 ( V2-V1 ) + y23 ( V2-V3 ) + y24 ( V2-V4 ) + yc21V2 + yc23V2 + yc24V2 Bus 3: I3 = I30 + I31 + I32 + Ic31 + Ic32 I3 = y30V3 + y31 ( V3-V1 ) + y32 ( V3-V2 ) + yc31V3 + yc32V3 Bus 4: I4 = I40 + I41 + I42 + Ic41 + Ic42 I4 = y40V4 + y41 ( V4-V1 ) + y42 ( V4-V2 ) + yc41V4 + yc42V4 Equation above can be rearranged to divide and group single merchandises by electromotive force. Bus 1: I1 = ( y12 + y13 + y14 + yc12 + yc13+ yc14 ) V1 – y12V2 – y13V3 – y14V4 = Y11V1 + Y12V2 + Y13V3 + Y14V4 Bus 2: I2 = ( y21 + y23 + y24 + yc21 + yc23+ yc24 ) V2- y21V1 – y23V3 – y24V4 = Y21V1 + Y22V2 + Y23V3 + Y24V4 Bus 3: I3 = ( y30 + y31 + y32 + yc31+ yc32 ) V3 – y31V1 – y32V2 = Y31V1 + Y32V2 + Y33V3 + Y34V4 Bus 4: I4 = ( y40 + y41 + y42 + yc41+ yc42 ) V4 – y41V1 – y42V2 = Y41V1 + Y42V2 + Y43V3 + Y44V4 Finally, Diagonal elements Y ( I, I ) and off diagonal elements Y ( I, J ) of the entree matrix are calculated as follows ; = y12 + y13 + y14 + yc12 + yc13+ yc14 = -j8 -j4 – j2.5 + j0.1 + j0.1 +0.1j = -j14.2 plutonium = y21 + y23 + y24 + yc21 + yc23+ yc24 = -j8 -j4 – j5 + j0.1 + j0.1 +0.1j = -j16.7 plutonium = y30 + y31 + y32 + yc31+ yc32 = -j08 -j4 – j4 + j0.1 +0.1j = -j8.6 plutonium = y40 + y41 + y42 + yc41+ yc42 = -j0.8 -j2.5 – j5 + j0.1 +0.1j = -j8.1 plutonium = – y12 = – ( -j8 ) = j8 plutonium = – y13 = – ( -j4 ) = j4 plutonium = – y14 = – ( -j2.5 ) = j2.5 plutonium = – y21 = – ( -j8 ) = j8 plutonium = – y23 = – ( -j4 ) = j4 plutonium = – y24 = – ( -j5 ) = j5 plutonium = – y31 = – ( -j4 ) = j4 plutonium = – y32 = – ( -j4 ) = j4 plutonium = – y34 = 0 plutonium = – y41 = – ( -j2.5 ) = j2.5 plutonium = – y42 = – ( -j5 ) = j5 plutonium = – y43 = 0 plutonium Therefore, entree matrix Ybus is as follows ;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=-j0.8 y12=-8i ; y21=-8i ; y13=-4i ; y31=-4i ; y14=-2.5i ; y41=-2.5i ; y23=-4i ; y32=-4i ; y24=-5i ; y42=-5i ; y34=0 ; y43=0 ; y30=-0.8i ; y40=-0.8i ; Y11=-8i-4i-2.5i ; Y12=8i ; Y13=4i ; Y14=2.5i ; Y21=8i ; Y22=-8i-4i-5i ; Y23=4i ; Y24=5i ; Y31=4i ; Y32=4i ; Y33=-0.8i-4i-4i ; Y34=0 ; Y41=2.5i ; Y42=5i ; Y43=0 ; Y44=-5i-2.5i-0.8i ; % Bus 3 and Bus 4 is non connected, so admittance Y34 and Y43 are equal to zero % define the 4Ãâ€"4 entree matrix ( Ybus ) Ybus= [ Y11 Y12 Y13 Y14 ; Y21 Y22 Y23 Y24 ; Y31 Y32 Y33 Y34 ; Y41 Y42 Y43 Y44 ] ; % In order to specify the nodal equation ( I = Ybus*V ) , the given I needs to specify. i1=0 ; i2=0 ; i3=-i ; i4=-0.4808-0.4808i ; Ibus= [ i1 ; i2 ; i3 ; i4 ] ; % Each coach electromotive force can be calculated by utilizing matrix division ( V= YbusI ) Vbus=YbusIbus ; v1=Vbus ( 1,1 ) ; v2=Vbus ( 2,1 ) ; v3=Vbus ( 3,1 ) ; v4=Vbus ( 4,1 ) ; % Calculate electromotive force difference between coachs v12=v1-v2 ; v13=v1-v3 ; v14=v1-v4 ; v21=v2-v1 ; v23=v2-v3 ; v24=v2-v4 ; v31=v3-v1 ; v32=v3-v2 ; v34=v3-v4 ; v41=v4-v1 ; v42=v4-v2 ; v43=v4-v3 ; % current flow between coachs can be calculated by i12 = y12* ( v1-v2 ) i12=y12*v12 ; i13=y13*v13 ; i14=y14*v14 ; i21=y21*v21 ; i23=y23*v23 ; i24=y24*v24 ; i31=y31*v31 ; i32=y32*v32 ; i34=y34*v34 ; i41=y41*v41 ; i42=y42*v42 ; i43=y43*v43 ; % evident power can be calculated by s12 = v1 * conj ( i12 ) s12=v1*conj ( i12 ) ; s13=v1*conj ( i13 ) ; s14=v1*conj ( i14 ) ; s21=v2*conj ( i21 ) ; s23=v2*conj ( i23 ) ; s24=v2*conj ( i24 ) ; s31=v3*conj ( i31 ) ; s32=v3*conj ( i32 ) ; s34=v3*conj ( i34 ) ; s41=v4*conj ( i41 ) ; s42=v4*conj ( i42 ) ; s43=v4*conj ( i43 ) ; % Real power and Reactive power can be derived by following p12=real ( s12 ) ; p13=real ( s13 ) ; p14=real ( s14 ) ; q12=imag ( s12 ) ; q13=imag ( s13 ) ; q14=imag ( s14 ) ; p21=real ( s21 ) ; p23=real ( s23 ) ; p24=real ( s24 ) ; q21=imag ( s21 ) ; q23=imag ( s23 ) ; q24=imag ( s24 ) ; p31=real ( s31 ) ; p32=real ( s32 ) ; p34=real ( s34 ) ; q31=imag ( s31 ) ; q32=real ( s32 ) ; q34=imag ( s34 ) ; p41=real ( s41 ) ; p42=real ( s42 ) ; p43=real ( s43 ) ; q41=imag ( s41 ) ; q42=real ( s42 ) ; q43=imag ( s43 ) ; % terminalMatlab Calculation Results

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