Leadership And Ethics Program At The Jepson School Of The...

There is an age old question that goes like this: Are individuals made to be a leader or are individuals born to be leaders? JoAnne Ciculla, a professor in leadership and ethics program at the Jepson School of the University of Richmond states the question should not be, what is leadership, but what is good leadership? (Giampetro-Meyer, Brown, Browne, 1998). Ciculla referenced three types of leadership: transformational, transitional, and servant leadership. While these are just a few, with leadership, you have to be willing to take charge and direct others. At times people may not be receptive because people have a tendency to want to do their own thing and not be controlled or directed by others. As we observe what is going on in society today with the Black Lives Movement, and with police brutality, some have asked the following question. Is there a leader who could bring unity by coming together to allow different groups to talk and move toward a sensible resolution? For an organization to be successful, the economical, organizational, and political systems depends on how effective and efficient the leader is (Parris Peachey, 2013). When it comes to being able to influence an organization, you would like to have someone that is able to do this well, however, in some cases a person may not necessarily have the proper skills needed to be a good leader, who could move the organization to be productive. I will compare, and contrast at least three leadership

The Mexican Revolution By Porfirio Diaz - 887 Words

Verdin, Ramon Khan, Kamal Santillan, Karen Research Paper The Mexican Revolution was brought on by tremendous disagreement among the Mexican people over the dictatorship of President Porfirio I. Dà ­az. Diaz stayed in office for 34 years. During that time, power was in the hands of only a few people. The people had no power to express their opinions or select their public officials. Wealth was also in the hands of a few people. Porfirio pushed peasants off of their lands and had business men take their land, they called the land haciendas. The dictator-like rule of Porfirio I. Diaz and the want of land were the reason for fighting in the Mexican Revolution. The main â€Å"antagonist† of the Mexican revolution was Porfirio Diaz. Diaz became president of Mexico in 1876. He was the autocrat who took full control of Mexico for 34 years. The revolutionist wanted a democracy so they fought against his power. The political instability during this time was also a major question in Mexico. Porfirio Diaz was very corrupt. He used his power as the president to his advantage. The Mexican Constitution called for public election but Dà ­az and his supporters used their political and economic resources to stay in power. He would often imprison his political opponents or assassinate them just to keep his power over Mexico. This caused the citizens of Mexico to increasingly oppose Porfirio Diaz and his regime. Francisco I. Madero, a wealthy landowner and industrialist, was one of the strongestShow MoreRelatedThe Mexican Revolution By Porfirio Diaz2015 Words   |  9 PagesThe Mexican Revolution began in 1910, when after decades of authoritarian rule, the Mexican leader, Porfirio Dà ­az, was challenged by Francisco I. Madero. Dà ­az was both the official and unofficial ruler of Mexico from 1876 to 1911. During the Porfiriato, Dà ­az managed to modernise the country through the development of mines, plantations, railways and telegraph lines. 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In the 1860’s Diaz was important to Mexican politics and then was elected president in 1877. Diaz said that he would only be president for one year and then would resign, but after four years he was re-elected as the President of Mexico. Porfirio Diaz and the Mexican revolution had a huge impact on the country of Mexico that isRead MoreMexico Post Colonial1566 Words   |  7 PagesThe history of Post colonial Mexico included many successful and influential leaders: Porfirio Diaz, Francisco Madero, Francisco (Pancho) Villa, and Emiliano Zapata. Disparities in classification of the revolution arise from the numerous factions and ideological assumptions advocated for the overthrow of Diaz’s rule, hence one can argue that it was a political, social, or economical revolution. A social revolution advocates a complete t ransformation of all characteristic aspects of society, encompassingRead MoreTaking a Look at the Mexican Revolution982 Words   |  4 Pagessuccess didn’t included â€Å"all† Mexicans which its majority were lower class citizens, on this equation the majority of people were never the ones to gain; wealth and land inequalities and abuse of power will part of the daily struggle, as result of this Mexican revolution will began. In 1908, after a publication of an interview of Porfirio Diaz by James Creelman, sentiments of Revolution began to spark in Mexico, but it wouldn’t be until November 1910 when the Mexican Revolution started. The revolutionaryRead MoreThe Mexican Revolution Essay1272 Words   |  6 PagesThe Mexican Revolution The Mexican Revolution was the culmination of a mass of political, economic, and social tension that accompanied the regime of the dictator Porfirio Diaz. 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Slack Bus And Slack Generator Engineering Essay Free Essays

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. Bus Input 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 Off Slack coach and slack generator In 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. We will write a custom essay sample on Slack Bus And Slack Generator Engineering Essay or any similar topic only for you Order Now 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 slack coach. [ 1 ] Bus type ( PQ coach or PV coach ) Bus Bus type Remarks BUS 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 signal System Balance Entire Generation A ; Load Demand Bus Real Power ( MW ) Fanciful Power ( Mvar ) Coevals Load Coevals Load BUS 1 204.093 100 56.240 0 BUS 2 0 200 0 100 BUS 3 200 100 107.404 50 Entire 404.093 400 163.644 150 Difference Pgen – Pdemand = 4.093 Qgen – Qstored in burden = 13.644 Reason: 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 line Bus Real Power ( MW ) Fanciful Power ( Mvar ) Sending Receiving Losingss Sending Receiving Stored BUS 1 – Bus 2 102.714 100.650 2.064 56.653 49.773 6.88 BUS 1 – Bus 3 1.379 1.378 0.001 0.4141 ) 0.4131 ) 0.001 BUS 3 – Bus 2 101.378 99.350 2.028 56.990 50.227 6.763 Entire Palestine liberation organizations = 4.093 Qstored in burden = 13.644 1 ) 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 Difference As a consequence of the Powerworld, the electromotive force angle and angle difference are shown in the tabular array below. Bus Voltage Angle Voltage Angle Difference BUS1 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 -1 The 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 Observation P3 = 0 MW P3 = 50 MW P3 = 100 MW P3 = 150 MW P3 = 250 MW P3 = 300 MW P3 = 350 MW P3 = 400 MW P3 = 450 MW Reactive Power Generation at Bus 3: It is found that reactive power 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 -2 The 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 Observation P2 = 0 MW Q2 = 0 MW P2 = 50 MW Q2 = 25 MW P2 = 100 MW Q2 = 50 MW P2 = 150 MW Q2 = 75 MW P2 = 250 MW Q2 = 125 MW P2 = 300 MW Q2 = 150 MW P2 = 350 MW Q2 = 175 MW P2 = 400 MW Q2 = 200 MW P2 = 450 MW Q2 = 225 MW Power Generation at Bus 1 and Bus 3: It is observed that as the 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 2 System Model A ; Admittance Matrix In 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 Calculation Nodal 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 Difference V1 = 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 2 V12 = 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 ° plutonium Voltage difference between BUS 3 and BUS 2 V32 = 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 ° plutonium Voltage difference between BUS 3 and BUS 1 V31 = 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 ° plutonium Line Current Current 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 2 I12 = 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 2 I32 = 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 1 I31 = 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 Flow Apparent 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 2 S12 = 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 plutonium Apparent Power from BUS 2 to BUS 1 S21=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 plutonium Apparent Power from BUS 3 to BUS 2 S32 = 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 plutonium Apparent Power from BUS 2 to BUS 3 S23=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 plutonium Apparent Power from BUS 3 to BUS 1 S31 = 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 plutonium Apparent 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 plutonium Comparison with simulation consequences The 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- Ipu Calculation Result and Simulation Result Flow way A ; Value Calculation Consequence Simulation Consequence BUS 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 A BUS 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 A BUS 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 A BUS 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 A BUS 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 A BUS 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 3 Admittance Matrix and Nodal Equation Entree between two coachs y12 = 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 Matrix Ybus ( 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 Analysis Power flow disregarding transmittal line electrical capacity Nodal Equation Current 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 Analysis Voltage 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 plutonium Current flow in the system Current 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 plutonium Power flow in the system Apparent 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 plutonium Admittance Matrix sing transmittal line electrical capacity Harmonizing 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 Matrix Contrary 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 Matlab Matlab 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 ) ; % terminal Matlab Calculation Results How to cite Slack Bus And Slack Generator Engineering Essay, Essay examples

DBQ Great Depression free essay sample

During the time of the Great Depression there were major changes throughout America, it helped change the role of the government to help the people. When looking at the Great Depression one must look at the effects of the New Deal policy it helped the American people and changed the role of the of the government during this time period. As a result of these successful policies it assisted the American people by giving them jobs and it expanded the government into a bureaucracy. There were many people affected by the Great Depression the worst off were the farmers because of overproduction of what they were harvesting they were not getting enough money to pay their farms (Doc J) but because of the Agricultural Adjustment Act it helped pay for the their farms but the draw-back about this that the benefits were distributed unevenly to farmers. Women and minorities had it not quite as bad but it was pretty bad for them. We will write a custom essay sample on DBQ Great Depression or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page Most African Amerithingscan people where on the street (Doc I) because there where people who where so desperae for a job they would be hired for the normal pay because of this invisble scar it was so bd people would go out and steal food to sell to try and bring something home.. So to slove this problem Congress with the approval of Roosevelt passed the Cilvilian Conservation Corps to provide work relief for young men from unployed families by having them dredge out rivers to clran and buld bridges over them along wit other nature perseving things like making hiking trails so people dont destroy the natural area around them. The Civil Works Administration to work within a short amount of time by setting up sewer systems through out cities but the effectiveness was limited due to poor leadership (Doc A). The next thing that was affecected was the banks Congress with the approval of Roosevelt passed the Emergency Banking Act that ressaured that the banks were safe and soon more deposits were made than withdrawls because of this it helped the bsnks go back into business. The people where not the only thing that was helped soon after Roosevelt made some major changes to the government. Some of the changes had to do with the size of the government Roosevelt had made the government so big it because a bureaucracy (Doc C) because Roosevelt helped so many people with his changes a lot of people voted for his political party because of this he was voted into office many times but many people beleved that the president was a socialist because of the New Deal (Doc D) that explained the gobernment. Many people civilized the New Deal saying that the programs will cause problems in the future. As will as the NRA (Doc B) taking over the work place in fact it was proven to be socialism in the court case Schecter V.U. S. the Supreme Court had sided with Schecter and said that the NRA was socialism and should back away and let compaanies fail of they wanted to. The New Deal had helped expand the government into a bureaucracy it helped millions of people but it did have its limitations. Successful the New Deal helped both Americans and helped expanded the government. Because of t hese policies it helped America come out of the Great Depression. It is intersting to note even though there were many success there were a few limits to these important policies.