Kurt Vonnegut Analysis - 825 Words

The attempts by scholars to define Vonnegut’s style of writing lean toward the belief that his work may be modern, postmodern and postmodern humanist (Davis). Accordingly, following the postmodern lean, Davis describes Vonnegut’s fiction as â€Å"lies that enable a humanism of practice.† (Davis). Decidedly anti-war, Vonnegut refused to glorify his most hurtful memories of World War II. His writings took on a common thread of sharp wit and satire. Hilariously, he made fun of his world and attempted to teach a lesson regarding society’s quirks and highlight what he thought about society. Born November 11, 1922 and raised by wealthy Germans in Indianapolis, Kurt Vonnegut, Jr’s family abandoned their German culture to prove their patriotism†¦show more content†¦After his release, Kurt received the Purple Heart award for what he referred to as â€Å"frost bite†. (Allen) Unfortunately, Dresden, well into German territory, built no bomb shelters for its citizens, let alone prisoners of war. As such, soon after Kurt found himself a POW, the Allied troops started bomb runs on the City. Vonnegut fictionalized his experience in his 1969 work, Slaughterhouse-Five. Being a pacifist, the number of pro-war movies spurred Vonnegut to write about his experiences in Dresden in a way that could not be construed to support war. This fictional work, heralded as one of his best books, has frequently been banned from schools for being unpatriotic. In Kurt Vonnegut and the American Novel, Tally struggles to define Kurt’s work as acclaimed postmodern revolutionary or marginal science fiction. Tally continues to describe â€Å"Vonnegut is a modern author tackling postmodern issues†. Vonnegut’s lessons in morality in â€Å"Cat’s Cradle† push the idea that big government is bad and please, add salt. These ideas are carried throughout history in movies and literature. For example, Demolition Man (Stallone 1993) and Animal Farm (Orwell 1945) both deal with the notion that too much government can be a bad thing for society. The commonShow MoreRelatedAnalysis Of Kurt Vonnegut s Slaughterhouse Five Essay2080 Words   |  9 PagesNatalie Lubben December 5, 2016 Rhetorical Analysis Essay Draft Slaughterhouse-five War is a virus, a plagues our world and has experienced since the early ages of time. Once a war is cured a new strain begins stronger and more unforgiving as the last. Humans are creatures of habit which continue the violence. Kurt Vonnegut’s novel, slaughterhouse-five, is a deliberate and well developed statement against war as expressed through the tone, rhetoric, and characters, making anti war a prominentRead MoreAnalysis Of Harrison Bergeron By Kurt Vonnegut905 Words   |  4 PagesIn â€Å"Harrison Bergeron† by Kurt Vonnegut, Vonnegut expresses the dangers that could be caused in total equality. He deliberates the pain his characters have to endure through their handicaps they received from the government to assure equality in society. Vonnegut explores the dangers that total equality brings to society. Harrison’s attempt to free people of their equality is accompanied by Harrison’s parents, sitting on the couch having to deal with their hand icaps while trying to focus on Harrison’sRead MoreAnalysis Of Kurt Vonnegut s Slaughterhouse Five 1453 Words   |  6 PagesTo Go or Not to Go? So it goes. Defining post-modern works, can be daunting, but the main traits of post-modernism are embracing skepticism and overturning conventions. With this in mind, Kurt Vonnegut explores war drawing parallels from his own past experience and depicts it through his character Billy Pilgrim allowing the reader to see the dichotomy in reality and fiction, separating his novel from the normal layout of a linear novel. Also, Slaughterhouse-Five discusses the controversial militaryRead MoreAnalysis Of Kurt Vonnegut s Slaughterhouse Five 1634 Words   |  7 PagesKurt Vonnegut once said, â€Å"So it goes† to describe the unavoidableness of fate. This aspect of seeing terrible things and being able to continue on would become a main theme in his novels. Vonnegut, as an author, received his essential voice by writing about his own experiences, using what would become his signature pessimistic yet humanist view. Vonnegut is described by Lindsay Clark as, â€Å"Worse than a pessimist†¦ he is an eternal optimist doomed to disappointment† (Clark, â€Å"Viewing Four Vonnegut NovelsRead MoreAnalysis Of The Book Harrison Bergeron By Kurt Vonnegut859 Words   |  4 Pages Kurt Vonnegut wrote a short story called â€Å"Harrison Bergeron.† This story takes place in a world where everyone is one hundred percent equal through restraining anyone with an above average natural anything. It start of with the George and Hazel Bergeron watching television. Then a strong and intelligent young man named Harrison Bergeron made the choice to appear on that broadcast and speak against the handicaps before getting shot by the Handicapper General. This story shows that individuality makesRead MoreAnalysis Of Kurt Vonnegut s Harrison Bergeron 935 Words   |  4 PagesIn â€Å"Harrison Bergeron† Kurt Vonnegut reveals the truth about world. Vonnegut farther explains how humankind is controlled by America’s first amendment of everyone being created equal. The main character of the story, also a protagonist Harrison Bergeron aims to let the world know what is truly happening to them. He is killed by an antagonist Dianna Moon Clampers who is a handicapper general. The futuristic short story is written in a third person omniscient. It’s told y a narrator who seems to knowRead MoreAnalysis Of Kurt Vonnegut s Harrison Bergeron 953 Words   |  4 PagesEquality. The most sought after desire in society. Each person has the felt the disease of envy for another’s talent, wisdom, or beauty. The heart, mind and soul are never replicated between two humans. This places one’s envy as a never ending cycle. Kurt Vonnegut’s story, Harrison Bergeron, focuses on the theme that society and government aspire to make all citizens equal by restricting them and making all handicapped for the purpose of obedience only to inadvertently achieve inequality. The shortRead MoreAnalysis Of Harrison Bergeron By Kurt Vonnegut Jr.1242 Words   |  5 Pages If you were to live in a society in which citizens are literally equal in every aspect of their lives, would you consider this kind of society a utopia or dystopia? The science-fiction short story, â€Å"Harrison Bergeron† by Kurt Vonnegut Jr. depicts the future of a world where the First Amendment of the Constitution of the United States of America is interpreted and executed literally, where every man is believed to be created equal. If you happened to have been born strong, beautiful, intelligentRead MoreAnalysis Of Kurt Vonnegut s Harrison Bergeron 1322 Words   |  6 Pagesdon’t stand out. This is life in Kurt Vonnegut, Jr.’s short story â€Å"Harrison Bergeron.† Harrison, a brilliant and strong 14 year old boy, decides he doesn’t want to follow the government s rules anymore. He sets out to overthrow the government, knowing he may not succeed. Throughout the story Vonnegut expresses the theme that standing up for what you believe in despite the dangers is the morally righteous and necessary thing to do. A major theme in Kurt Vonnegut, Jr’s short story, â€Å"Harrison BergeronRead MoreAnalysis Of Kurt Vonnegut s Harrison Bergeron 873 Words   |  4 Pagesstate of being equal, especially in status, rights and opprtunites, But in Kurt Vonnegut story â€Å"Harrison Bergeron† equality is far from being equal. In this story, though it tries to paint an imaginary picture of a future in American society where everybody is finally equal before God and the Law, and where nobody is smatter, better, looking, stronger or quicker than anybody else, but that is actually not the case. Vonnegut, indirectly in this story re-echo’s the popular Animal Farm saying that â€Å"All

Homeland Security Policy And Practice - 1482 Words

Homeland Security Response Paper Assignment 1 FREDDIE BURSE Florida State University Summer 2015 Dr. Audrey Heffron Casserleigh Homeland Security: Policy and Practice PAD 5376 AND 4935 - Tues. 5:30 Abstract This is an APA formatted essay of the Department of Homeland Security and its major role it play within and outside the borders of the United States of America. The essay is outlined to follow prompts requested by Professor Dr. Audrey Heffron Casserleigh. Does the United States’ involvement in International Organizations (The UN, NATO, WTO, etc) and international treaties negatively or positively affect our Homeland Security? Why is this the case? Chose a Homeland Security law, portion of law (ie a portion of the USA PATRIOT Act) or policy that we discussed in class. Briefly describe this, and explain if you believe it is a useful (or not useful) tool in homeland security. Homeland Security The United States Department of Homeland Security is known as an organization that ensures the safety of society in the nation (Homeland Security). The department primary purpose is to address threats, disasters, or hazards that may terrorize the nation and its borders. The department confronts and resolves conflicts between outside nations and the United States of America. For many years the United States Department of Homeland Security have assigned laws. As well as planned and assumed threatened, susceptible, and vulnerable events that may occur within the UnitedShow MoreRelatedU.s. Department Of Homeland Security1668 Words   |  7 Pages1. Purpose Among one of the missions of The U.S. Department of Homeland Security is to protect and preserve the security of the Cyberspace in the country. 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Nwatu Strayer University Critical Infrastructure Protection Introduction The Presidential Policy Directive (PPD) on Critical Infrastructure Security and Resilience advances a national unity of effort to strengthen and maintain secure, functioning, and resilient critical infrastructure (The White House Office of the Press Secretary, February 2013). It is imperative for every nationRead MoreReview Of The Literature : Failure Of Intelligence1476 Words   |  6 PagesReview of the Literature Due to failure of intelligence which led to September 11 terrorist attacks in the United States, President George W. Bush created the Office of Homeland Security in October 2001 and later converted it to a full cabinet department with a bill signed in November 2002 (Zimmerman, 2011). Creation of the DHS falls into the overall Federal government goal of reducing loss of life and property during emergency and her continuous efforts to prevent constant and evolving threatsRead MorePresident Donald J. 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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