The Tomb Of Tutankhamen Essays - Egyptian Gods, Egyptian Mythology

The Tomb of Tutankhamen The Tomb of Tutankhamen What does the tomb of tutankhamen and its contents show about the Egyptian concern for the afterlife? Tutakhamen's tomb, and the artifacts inside are an indication of the concern the Ancient Egyptians held for the after-life of their king. In 26th Nov. 1922, the English archaeologist Howard Carter opened the virtually intact tomb of a largely unknown pharaoh: Tutankhamen. This was the first, and to date the finest royal tomb found virtually intact in the history of Egyptology. It took almost a decade of meticulous and painstaking work to empty the tomb of Tutankhamen. Around 3500 individual items were recovered. When the Burial Chamber of Tutankhamen was officially opened, on 17 February 1923, the Antechamber had been emptied. It had taken near fifty days to empty the Antechamber; the time required to dismantle and restore the contents of the Burial Chamber including the gilded wooden and the sarcophagus was to be greater, and the work was not completed until November 1930, eight years after the original discovery. One must examine both the tomb itself, and its contents, to see the connection between the tombs and burial rituals and the doctrine of eternal life. The royal tombs were not merely homes in the hereafter for the kings, as are the private tombs of commoners and nobility. Instead the tombs are cosmological vehicles of rebirth and deification as much as "houses of eternity." As the king is supposed to become Osiris in a far more intimate way than commoners, he is equipped with his very own Underworld. And as the king is supposed to become R in a way entirely unavailable to commoners, he is equipped with his very own passage of the sun, whether this is thought of as the way through the underworld or through the heavens. Tutankhamon's tomb, hurriedly prepared for the premature death of the king at the age of only about 18, is, as Romer says, a "hole in the ground," compared to a proper royal tomb. The theme of fours is conspicuous in Egyptian religious practice. Tutankhamon's tomb contains four chambers. The burial chamber, with a ritual if not an actual orientation towards the West, is the chamber of departure towards the funeral destinies. The internment of the body certainly is the beginning of the sojourn of the dead, and the Egyptians saw the dead as departing"into the West." The room called the "Treasury" is then interpreted to have a ritual orientation towards the North as the "chamber of reconstitution of the body." Since the most conspicuous object in the Treasury was a great gilt sledge holding the shrine containing the canopic chest, which holds the king's viscera, this could well suggest the problem of reassembling the king's living body. That task, indeed, has a very important place in Egyptian mythology. After the goddess Isis had retrieved her husband Osiris's murdered body from Byblos, their common brother, Seth, the original murderer, stole the body, cut it into pieces, and tossed them in the Nile. Isis then had to retrieve the parts of the body before Osiris could be restored to life. Her search through the Delta, which is in the North of Egypt, seems to parallel the "sacred pilgrimage" to cities of the Delta that Desroches-Noblecourt relates as one of ritual acts of the funeral, as many of the other objects in the Treasury seem to be accessories for that pilgrimage. For the sovereign to be reborn it was necessary that a symbolic pilgrimage be made to the holy cities of the delta. The principal halts of the journey corresponded almost exactly to the four cardinal points of the delta where these cities were situated. Sais, to the west, represented the necropolis where the body was buried; Buto to the north, with its famous canal, was an essential stage of the transformations within the aquatic world of the primordial abyss, evoking the water surrounding the unborn child; and Mendes to the east whose name could be written with the two pillars of Osiris, the djed pillars, evoking the concept of air. There, said the old texts, the gods Shu and Tefenet were reunited, or again, according to the 17th chapter of The Book of the Dead, that was where the souls of Osiris and Re had joined. Finally, the southern-most city which completed the cycle of Heliopolis, the city of the sun, symbolizing the fourth [sic] element, fire, where the heavenly body arose in youth glory between the two hills on the horizon. [Christiane Desroches-Noblecourt, 1963, p. 238-9] As these four cities parallel the four rooms

Free sample - Ethical Standards. translation missing

Ethical Standards. Ethical StandardsMoral and ethical aspects of human life are very important for the personal development and statement. These life values determine attitude towards people, God and family. They are not regulated by law or other governmental documents, and there is no determined punishment for their violation, but still breaking ethical rules means breaking own personality and acting against oneself. 1. Ethical standard in majority of cases is taken from the parents’ family, social life, religion and historical period. Also, some people think that social groups, mentality and corporate culture, make impact on ethical standards formation. In other words, it is possible to consider the social environment as a major factor, which influences the ethical and moral standards of the personality. 2. According to the respondents’ point of view, those standards of personal ethics, which were grounded by parents in the childhood, are completely impossible to change because personality makes all important life decisions and actions, according to these standards. In some cases, it is possible to make some impact for the personal ethics while changing environment and religious- person would have to adapt to the new life conditions and to the people, surrounding him or her with their life principles. 3. In most cases ethical standards consider God as the main judge of all our actions. The God is also represented for many people as the personal example on how people should live and act. Family values are the core principle of ethics and people think that it is impossible to be fair with him/herself without being fair in own family. Parents, kids and beloved are the sense of life. Belief in afterlife time – one more important issue in following all ethical norms and standards. Respondents consider the afterlife time as the remuneration for the fair and faithful life, for acting, according to the ethical norms. References: Auditing Practices Board (2010) Ethical Standards. [Online] (Updated 17 July 2010) World Health Organisation (2010) Ethical standards and procedures for research with human beings

Concept Analysis Research Paper Example | Topics and Well Written Essays - 2000 words

Concept Analysis - Research Paper Example Fortunately, these are steadfast and always find a place for the duty of care as this form the tenets of their profession. This is the place of nursing in our contemporary set up. The importance of those in the profession is largely ignored by the public and their value is only recognized when one is need of their expertise. It is this maligned nature of the profession that has strengthened the call for â€Å"the heart of nursing†. The heart of nursing is an apparent reference to a set of qualities that a nurse needs to discharge their duty in oblivion of the external factors that would otherwise affect successful discharge of commensurate duties. This is a key part of ensuring successful delivery of nursing care in light of pandemic ailments affecting the world populace (Schnatz, 2007). In trying to understand nursing and what it encapsulates we must try to look at the underlying enablers of nursing care. It should be noted that nursing as a practice is as strongly reliant on professional education as it is on human duty and obligation to care for others. Notably, in primitive societies where the decision to be a nurse (caregiver) was made on behalf of a person at birth the successful discharge of nursing duties was still possible. ... In the midst of these informal means of gathering nursing expertise the sole or the basic drive was the willingness to care given the plight of patients. The fact that nursing has been existent since antiquity and the knowledge that the first formal nurse training school was established in 1872, nurse training school of Women’s Hospital of Philadelphia, further underlines the duty for care or compassion as the most fundamental aspect in nursing (Egenes, 2010). Method of concept analysis and why this method was chosen In discussing compassion as nursing’s most precious asset or as the most dominant altruistic ideal in nursing care this study adopts Chinn & Kramer’s (1991) approach to concept analysis. Chinn & Kramer (1991) identify the purpose of their concept analysis approach as â€Å"producing tentative definition of the concept and a set of tentative criteria for determining if the concept ‘exists’ in a particular situation (88). Under this appro ach the first step is selecting the concept, clarifying the purpose, identifying the sources of data, exploring context and values and formulating criteria. Data sources Data to be used in the concept analysis will be derived from existent studies on nursing care and specific review of compassion and compassionate care as a key ideal in nursing. A host of this information is published in peer reviewed journals and professional websites which have sought to explain and research on nursing. Equally important will be historical data on the place of nursing in our society as well as the foundations and prerequisites of being a successful care giver. In essence, this analysis envisions that successful care giving as a factor is determined by the compassion of the care

Outline Essay Example | Topics and Well Written Essays - 750 words - 2

Outline - Essay Example In other words, I need to address a lifestyle change that is responsive to my desire to curtail irresponsible spending. This question is very important to me because I realize that in order to improve the quality of my life I need to be more budget conscious and more responsible with what I do with my earnings. This is all the more important because my irrational and irresponsible spending impacts not only my own comfort, but the comfort of my family. At this point I am able to admit that I am at worst a compulsive spender, at best, unable to manage my earnings responsibly. I want to make a lifestyle change that reverses this conduct and improves the quality of my life and the lives of my loved ones. In this regard, I have focused my attention on what it is that I need to do to invoke an effective lifestyle change to curtail my spending habits. Current research into the topic of lifestyle changes typically agree that lifestyle change commences with identifying future goals and acknowledging current personal obstacles that can be changed to achieve a successful lifestyle change. McGraw (1999) informs that lifestyle changes begin with a realization that everyone faces challenges but the individual must take sole responsibility for meeting their own challenges (pp. 11-12). In general there are ten life laws for lifestyle changes. They are: becoming one of those that get it; accept that you are responsible for your own life; do that which works; accept what needs to be changed; take positive action; identify your perceptions; take control of your life; dictate how you are treated; be merciful and identify what you want to achieve (McGraw 1999). Markway et al (2003) caution that making lifestyle changes are never easy as it requires commitment, vision and discipline (p. 92). Tracy (2009) maintains that success driven persons are those who think ahead to the future and do not

Philosophy of healthcare in USA Essay Example | Topics and Well Written Essays - 250 words

Philosophy of healthcare in USA - Essay Example The target population includes the elderly especially the chronically ill older patients, children, and unemployed or underemployed people. Much attention has been given to older people because of their poor health and it is estimated by 2050 thirty percent of the population in USA will consist of elderly people. It is estimated that over 45 million people in USA are either uninsured or underinsured (Papadimos, 2007). This means that access to quality healthcare is little or none. In some instances political agendas have resulted in creating healthcare programs that are ineffective. This has downplayed the importance of better healthcare services to the population. The quality and appropriateness of healthcare services impact heavily on costs. The insured have access to better healthcare services than those uninsured. This means the uninsured and those unemployed or underemployed access cheap healthcare services that may be poor in quality. The government should make sure that all people have access to quality healthcare regardless of their socioeconomic status. Health insurance policies should at least cover the most vulnerable people in the population. Programs proposed by the Congress should be workable and

Slack Bus And Slack Generator

Slack Bus And Slack Generator The Table below shows input data of each busbar in the system used to solve the power flow and the simulation result according to instruction described in question 1. BUS Input Data [Simulation Result] BUS 1 pu P (load) 100 MW Q (load) 0 Mvar BUS 2 P (load) 200 MW Q (load) 100 Mvar CB of Generation Open BUS 3 1 pu P (Gen) 200 MW P (load) 100 MW Q (load) 50 Mvar AVR On AGC Off Slack bus and slack generator In power flow calculation, unique numerical solution cannot be calculated without reference voltage magnitude and angle due to unequal number of unknown variables and independent equations. The slack bus is the reference bus where its voltage is considered to be fixed voltage magnitude and angle (1à ¢Ã‹â€ Ã‚  0 °), so that the various voltage angle difference among the buses can be calculated respect. In addition, the slack generator supplies as much real power and reactive power as needed for balancing the power flow considering power generation, load demand and losses in the system while keep the voltage constant as 1à ¢Ã‹â€ Ã‚  0 °. In real power system, when relatively weak system is linked to the larger system via a single bus, this bus can represent the large system with an equivalent generator keeping the voltage constant and generating any necessary power like slack bus. [1] Bus type (PQ bus or PV bus) BUS Bus type Comments BUS 2 PQ Bus Generator is disconnected to Bus 2 BUS 3 PV Bus Generator is connected to Bus 3 and the magnitude of voltage of generator keep constant by using AVR In general, each bus in the power system can be categorized into three bus 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 bus at which the magnitude of the bus voltage is kept constant by the generator. Even though the bus has several generators and load, if any generators connected to the bus regulate the bus voltage with AVR, then this bus is referred to PV Bus. For PV bus, the magnitude of the bus voltage and real power supplied to the system are specified, and reactive power and angle of the bus voltage are accordingly determined. If a preset maximum and minimum reactive power limit is reached, the reactive output of the generator remains at the limited values, so the bus can be considered as PQ Bus instead of PV Bus. [2] PQ Bus (Load Bus): It is a bus at which the voltage is changed depending on total net real power and reactive power of loads and generators without voltage regulator. Therefore, in the power simulation and calculation, the real power and reactive power of the loads are specified as input data and accordingly the voltage (magnitude and angle) is calculated based on the above input. The following table specifies input and output of each bus type in the power system simulation and calculation. Bus Type P Q (Magnitude) ÃŽÂ ´ (Angle) PQ Bus Input Input Output Output PV Bus Input Output Input Output Slack Bus Output Output Input Input System Balance Total Generation Load Demand BUS Real Power (MW) Imaginary Power (Mvar) Generation Load Generation Load BUS 1 204.093 100 56.240 0 BUS 2 0 200 0 100 BUS 3 200 100 107.404 50 Total 404.093 400 163.644 150 Difference Pgen Pdemand = 4.093 Qgen Qstored in load = 13.644 Reason: Real power loss due to resistance of transmission line and imaginary power storage due to reactance of transmission line are the reasons for the difference between power generation and load demand in the system. P (Losses) Q (Storage) over the transmission line BUS Real Power (MW) Imaginary Power (Mvar) Sending Receiving Losses 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 Total Plosses = 4.093 Qstored in load = 13.644 1) Imaginary power flows from Bus 3 to Bus 1. The summation of real power losses and imaginary power storage over the transmission line are exactly same with total difference between generation and load. Therefore, it is verified that the difference is shown over the transmission line. Kirchoff balance as each bus [4] Bus1 ÃŽÂ £ P1 = + Pgen1 Pload1 P12 P13 = 204.093 100 102.714 1.379 = 0 ÃŽÂ £ Q1 = + Qgen1 Qload1 Q12 Q13 = 56.24 0 56.653 + 0.413 = 0 Bus2 ÃŽÂ £ P2 = + Pgen2 Pload2 P21 P23 = 0 200 + 100.65 + 99.35 = 0 ÃŽÂ £ Q2 = + Qgen2 Qload2 Q21 Q23 = 0 100 + 49.773 + 50.227 = 0 BUS3 ÃŽÂ £ P3 = + Pgen3 Pload3 P31 P32 = 200 100 + 1.378 101.378 = 0 ÃŽÂ £ Q3 = + Qgen3 Qload3 Q31 Q32 = 107.404 50 0.414 56.99 = 0 According to the calculation above, as summation of incoming outgoing real power and imaginary power at each bus become zero, it is verified that each busbar obeys a Kirchoff balance. In addition, the total power system is completely balanced, because total generation power (real imaginary) are equal to summation of total load demand and real power loss stored imaginary power over the transmission (i.e. Pgen Pdemand = Plosses, Qgen Qstored in load = Q stored in system) as shown above. Voltage Angle and Angle Difference As a result of the Powerworld, the voltage angle and angle difference are shown in the table below. BUS Voltage Angle Voltage Angle Difference BUS1 ÃŽÂ ´1 = 0.00 ° BUS1- BUS2 ÃŽÂ ´1 ÃŽÂ ´2 = 0.00 ° (-2.5662 °) = 2.5662 ° BUS2 ÃŽÂ ´2 = -2.5662 ° BUS2- BUS3 ÃŽÂ ´2 ÃŽÂ ´3 = -2.5662 ° (-0.043 °) = -2.5232 ° BUS3 ÃŽÂ ´3 = -0.043 ° BUS3- BUS1 ÃŽÂ ´3 ÃŽÂ ´1 = -0.043 ° 0.00 ° = -0.043 ° Power System Analysis -1 The table below summarizes generation and voltage angle variation at each bus as generation at Bus 3 varies from 0 MW to 450 MW by 50MW. Simulation Results and 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 generation Q3(gen) decrease while real power generation P3(gen) increase because Bus 3 as a PV Bus regulates the constant bus voltage magnitude by controlling excitation of the generation through the AVR. Power Generation at Bus 1: It is found that P1(gen) decreases and Q1(gen) increases simultaneously, while P3(gen) increases and Q3(gen) decrease. As the total load demand in the system keeps constant (i.e. Ptotal(load) = 400 MW, Qtotal(load) = 150Mvar), any necessary real power and reactive power for the system balance need to be supplied by generator (slack generator) at Bus 1. Therefore, power generation P1(gen) and Q1(gen) at Bus 1 change reversely compared to power generation change at Bus 3. Voltage Angle Difference: In general, real power flow is influenced by voltage angle difference between sending bus and receiving bus according to PR =. Therefore, it is observed that as real power generation P3(gen) increases real power flow from Bus 3 to Bus2 increase, accordingly voltage angle difference (ÃŽÂ ´3 ÃŽÂ ´2) between Bus 3 and Bus 2 increases. However, decrease in real power from Bus 1 to Bus 2 due to increase of P3(gen) result in decrease of voltage angle difference (ÃŽÂ ´1 ÃŽÂ ´2). In addition, Real power between Bus 1 and Bus 3 flows from Bus 1 to Bus 3 until P3(gen) reach to 200 MW and as P3(gen) increase more than 200 MW the real power flows from Bus 3 to Bus 1. So, it is also observed that voltage angle difference (ÃŽÂ ´3 ÃŽÂ ´1) is negative angle when P3(gen) is less than 200MW and the difference increase while P3(gen) increase. Power System Analysis -2 The table below summarizes the variation of power generation and voltage angle difference at each bus when the load demand at Bus 3 varies by 50MW and 25Mvar. Simulation Results and 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 total load demand in the system increases due to increase of load demand P2(load) Q2(load) at Bus 2, any necessary real power for the system balance is supplied by generator (slack generator) at Bus 1 considering constant P3(gen), so P1(gen) increases. In addition, any necessary reactive power for the system balance is supplied from Bus 1 as well as Bus 3, so both Q1(gen) and Q3(gen) increase. Voltage Angle Difference: It is found that real power flow increase both from Bus 1 to Bus 2 and from Bus 3 to Bus 2 due to increase of load demand at Bus2. Accordingly, both voltage angle difference ÃŽÂ ´1 ÃŽÂ ´2 and ÃŽÂ ´3 ÃŽÂ ´2 increase when the power flow P12 and P32 increase. In addition, when P2(load) is less than 200 MW, P1gen is relatively low. Therefore real power between Bus 3 and Bus 1 flows from Bus 3 to Bus 1 at lower P2(load) (less than 200MW). On the other hand, while P2(load) increase more than 200 MW, the real power flow direction changes (Bus 1 to Bus 3) and the real power flow increases. Accordingly, the voltage angle difference ÃŽÂ ´1 ÃŽÂ ´3 change from negative to positive and increase. Voltage Magnitude at Bus 2: It is observed that magnitude of bus voltage at Bus2 drops due to increase of the load demand at Bus 2. Question 2 System Model Admittance Matrix In order to construct the admittance matrix of Powerworld B3 case, single phase equivalent circuit can be drawn as below; z = r + jx (r = 0, x = 0.05) z12 = z21= j0.05 pu, y12 = 1/ z12 = 1/j0.05 = -j20 pu = y12 z13 = z31= j0.05 pu, y13 = 1/ z13 = 1/j0.05 = -j20 pu = y31 z23 = z32= j0.05 pu, y23 = 1/ z23 = 1/j0.05 = -j20 pu = y32 Admittance matrix can be defined as follows; BUS = Diagonal elements Y(i,i) of the admittance matrix, called as the self-admittance [lecture slide] [6], are the summation of all admittance connected with BUS i. = y12 + y13 = -j20 j20 = -j40 pu = y21 + y23 = -j20 j20 = -j40 pu = y31 + y32 = -j20 j20 = -j40 pu Off diagonal elements Y(i,j) of the admittance matrix, called as the mutual admittance [lecture slide] [6], are negative admittance between BUS i and BUS j. = y12 = -(-j20) = j20 pu = y13 = -(-j20) = j20 pu = y21 = -(-j20) = j20 pu = y23 = -(-j20) = j20 pu = y31 = -(-j20) = j20 pu = y32 = -(-j20) = j20 pu Therefore, the final admittance matrix BUS is; BUS = = The following figure shows the BUS of the Powerworld B3 case and it is verified that the calculated admittance matrix is consistent with the result of the Powerworld. Power Flow Calculation Nodal equation with the admittance matrix can be used to calculate voltage at each bus if we know all the current (i.e. total generation power and load demand at each BUS) and finally the power flow can be calculated accordingly. , therefore, In this question, however, simulation results of the voltage at each bus from the Powerworld are used for the power flow calculation as follows; [Simulation result] Voltage at each Bus and Voltage Difference V1 = 1 à ¢Ã‹â€ Ã‚  0.00 ° pu (BUS1) V2 = 1 à ¢Ã‹â€ Ã‚  -0.48 ° pu (BUS2) V3 = 1 à ¢Ã‹â€ Ã‚  0.48 ° pu (BUS 3) Voltage difference between BUS 1 and BUS 2 V12 = V1 V2 = 1 à ¢Ã‹â€ Ã‚  0.00 ° 1 à ¢Ã‹â€ Ã‚  -0.48 ° = 3.5 x 10-5 + j 8.38 x 10-3 = 8.38 x 10-3 à ¢Ã‹â€ Ã‚  89.76 ° pu V21 = V2 V1 = V12 = 3.5 x 10-5 j 8.38 x 10-3 = 8.38 x 10-3 à ¢Ã‹â€ Ã‚  -90.24 ° pu Voltage difference between BUS 3 and BUS 2 V32 = V3 V2 = 1 à ¢Ã‹â€ Ã‚  0.48 ° 1 à ¢Ã‹â€ Ã‚  -0.48 ° = j 16.76 x 10-3 = 16.76 x 10-3 à ¢Ã‹â€ Ã‚  90 ° pu V23 = V2 V3 = V32 = j 16.76 x 10-3 = -16,76 x 10-3 à ¢Ã‹â€ Ã‚  -90 ° pu Voltage difference between BUS 3 and BUS 1 V31 = V3 V1 = 1 à ¢Ã‹â€ Ã‚  0.48 ° 1 à ¢Ã‹â€ Ã‚  0.00 ° = 3.5 x 10-5 + j 8.38 x 10-3 = 8.38 x 10-3 à ¢Ã‹â€ Ã‚  90.24 ° pu V13 = V1 V3 = V31 = 3.5 x 10-5 j 8.38 x 10-3 = 8.38 x 10-3 à ¢Ã‹â€ Ã‚  -89.76 ° pu Line Current Current flow from BUS i and BUS j can be calculated by using voltage difference and interconnected admittance of the line between buses. [ Iij = yij * (Vi Vj) ] Line current between BUS 1 and BUS 2 I12 = y12 x (V1 V2) = -j20 x 8.38 x 10-3 à ¢Ã‹â€ Ã‚  89.76 ° = 167.6 x 10-3 à ¢Ã‹â€ Ã‚  -0.24 ° pu (BUS 1 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ BUS 2) I21 = y21 x (V2 V1) = -j20 x 8.38 x 10-3 à ¢Ã‹â€ Ã‚  -90.24 ° = 167.6 x 10-3 à ¢Ã‹â€ Ã‚  -180.24 ° pu (BUS 2 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ BUS 1) Line current between BUS 3 and BUS 2 I32 = y32 x (V3 V2) = -j20 x 16.76 x 10-3 à ¢Ã‹â€ Ã‚  90 ° = 335.2 x 10-3 à ¢Ã‹â€ Ã‚  0.00 ° pu (BUS 3 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ BUS 2) I23 = y23 x (V2 V3) = -j20 x 16.76 x 10-3 à ¢Ã‹â€ Ã‚  -90 ° = 335.2 x 10-3 à ¢Ã‹â€ Ã‚  180 ° pu (BUS 2 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ BUS 3) Line current between BUS 3 and BUS 1 I31 = y31 x (V3 V1) = -j20 x 8.38 x 10-3 à ¢Ã‹â€ Ã‚  90.24 ° = 167.6 x 10-3 à ¢Ã‹â€ Ã‚  0.24 ° pu (BUS 3 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ BUS 1) I13 = y13 x (V1 V3) = -j20 x 8.38 x 10-3 à ¢Ã‹â€ Ã‚  -89.76 ° = 167.6 x 10-3 à ¢Ã‹â€ Ã‚  -179.76 ° pu (BUS 1 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ BUS 3) Apparent Power Flow Apparent flow from BUS i and BUS j can be calculated by voltage at the sending bus and line current. [ Sij = Vi * I*ij ] Apparent Power from BUS 1 to BUS 2 S12 = V1* I*12 = 1 à ¢Ã‹â€ Ã‚  0.00 ° x 167.6 x 10-3 à ¢Ã‹â€ Ã‚  0.24 ° = 167.6 x 10-3 à ¢Ã‹â€ Ã‚  0.24 ° = 0.1676 + j 7.02 x 10-4 pu Apparent Power from BUS 2 to BUS 1 S21=V2* I*21=1à ¢Ã‹â€ Ã‚  -0.48 ° x 167.6 x 10-3à ¢Ã‹â€ Ã‚  180.24 °=167.6 x 10-3à ¢Ã‹â€ Ã‚  179.76 ° = -0.1676 + j7.02 x 10-4 pu Apparent Power from BUS 3 to BUS 2 S32 = V3* I*32 = 1 à ¢Ã‹â€ Ã‚  0.48 ° x 335.2 x 10-3 à ¢Ã‹â€ Ã‚  0.00 ° = 335.2 x 10-3 à ¢Ã‹â€ Ã‚  0.48 ° = 0.3352 + j 2.81 x 10-3 pu Apparent Power from BUS 2 to BUS 3 S23=V2* I*23=1 à ¢Ã‹â€ Ã‚  -0.48 ° x 335.2 x 10-3 à ¢Ã‹â€ Ã‚  180 °= 335.2 x 10-3 à ¢Ã‹â€ Ã‚  179.76 ° = -0.3352 + j 2.81 x 10-3 pu Apparent Power from BUS 3 to BUS 1 S31 = V3* I*31 = 1à ¢Ã‹â€ Ã‚  0.48 ° x 167.6 x 10-3à ¢Ã‹â€ Ã‚  -0.24 ° = 167.6 x 10-3 à ¢Ã‹â€ Ã‚  0.24 ° = 0.1676 + j 7.02 x 10-4 pu Apparent Power from BUS 1 to BUS 3 S13=V1* I*13=1à ¢Ã‹â€ Ã‚  0.00 ° x 167.6 x 10-3à ¢Ã‹â€ Ã‚  179.76 °= 167.6 x 10-3à ¢Ã‹â€ Ã‚  179.76 ° = -0.1676 + j 7.02 x 10-4 pu Comparison with simulation results The unit of the above calculation results is pu value, so in order to compare the results with simulation results pu value of current and power flow need to be converted to actual values by using the following equation considering Sbase = 100MVA and Vline_base = 345kV. [3] Sactual = Sbase ÃÆ'- Spu = 100 MVA ÃÆ'- Spu Iactual = Ibase ÃÆ'- Ipu = ÃÆ'- Ipu = ÃÆ'- Ipu = 167.3479 A ÃÆ'- Ipu Calculation Result and Simulation Result Flow direction Value Calculation Result Simulation Result BUS 1 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ BUS 2 |S12| 0.1676 ÃÆ'- 100 = 16.76 MVA 16.67 MVA P12 16.76 MW 16.67 MW Q12 0.0702 Mvar 0.07 Mvar |I12| 0.1676 ÃÆ'- 167.3479 = 28.0475 A 27.89 A BUS 3 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ BUS 2 |S32| 0.3352 ÃÆ'- 100 = 33.52 MVA 33.33 MVA P32 33.52 MW 33.33 MW Q32 0.281 Mvar 0.28 Mvar |I32| 0.3352 ÃÆ'- 167.3479 = 56.0950 A 55.78 A BUS 3 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ BUS 1 |S31| 0.1676 ÃÆ'- 100 = 16.76 MVA 16.67 MVA P31 16.76 MW 16.67 MW Q31 0.0702 Mvar 0.07 Mvar |I31| 0.1676 ÃÆ'- 167.3479 = 28.0475 A 27.89 A BUS 2 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ BUS 1 |S21| 0.1676 ÃÆ'- 100 = 16.76 MVA 16.67 MVA P21 -16.76 MW -16.67 MW Q21 0.0702 Mvar 0.07 Mvar |I21| 0.1676 ÃÆ'- 167.3479 = 28.0475 A 27.89 A BUS 2 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ BUS 3 |S23| 0.3352 ÃÆ'- 100 = 33.52 MVA 33.33 MVA P23 -33.52 MW -33.33 MW Q23 0.281 Mvar 0.28 Mvar |I23| 0.3352 ÃÆ'- 167.3479 = 56.0950 A 55.78 A BUS 1 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ BUS 3 |S13| 0.1676 ÃÆ'- 100 = 16.76 MVA 16.67 MVA P13 -16.76 MW -16.67 MW Q13 0.0702 Mvar 0.07 Mvar |I13| 0.1676 ÃÆ'- 167.3479 = 28.0475 A 27.89 A It is found that calculation results of current flow and apparent power flows (i.e. 28.0475 A and 56.0950 A/ 33.52 MVA and 16.76MVA) are about 0.5 % higher than simulation result (i.e. 27.89 A and 55.78 A / 33.33 MVA and 16.67 MVA) which can be considered slightly different. Difference of the voltage angle at each bus between calculation (0.48 °) and simulation (0.4775 °) could be the reason for this minor difference. Question 3 Admittance Matrix and Nodal Equation Admittance between two buses y12 = y21 = -j8 pu y13 = y31 = -j4 pu y14 = y41 = -j2.5 pu y23 = y32 = -j4 pu y24 = y42 = -j5 pu y30 = -j0.8 pu (BUS3-Neutral BUS) y40 = -j0.8 pu (BUS4-Neutral BUS) Admittance Matrix Ybus (Admittance Matrix) = Diagonal elements Y(i,i) of the admittance matrix, called as the self-admittance [2] [4], are the summation of all admittance connected with BUS i. = y12 + y13 + y14 = -j8 -j4 j2.5 = -j14.5 = y21 + y23 + y24 = -j8 -j4 j5 = -j17 = y30 + y31 + y32 = -j08 -j4 j4 = -j8.8 = y40 + y41 + y42 = -j0.8 -j2.5 j5 = -j8.3 Off diagonal elements Y(i,j) of the admittance matrix, called as the mutual admittance [2] [4], are negative admittance between BUS i and BUS j. = y12 = -(-j8) = j8 pu = y13 = -(-j4) = j4 pu = y14 = -(-j2.5) = j2.5 pu = y21 = -(-j8) = j8 pu = y23 = -(-j4) = j4 pu = y24 = -(-j5) = j5 pu = y31 = -(-j4) = j4 pu = y32 = -(-j4) = j4 pu = y34 = 0 pu = y41 = -(-j2.5) = j2.5 pu = y42 = -(-j5) = j5 pu = y43 = 0 pu Therefore, admittance matrix Ybus is as follows; Ybus = = Power Flow Analysis Power flow ignoring transmission line capacitance Nodal Equation Current from the neutral bus to each bus are given and admittance matrix (Ybus) is calculated above. Therefore, final nodal equation is as follows; Ibus = Ybus * Vbus à ¢Ã¢â‚¬ ¡Ã¢â‚¬â„¢ Vbus = Y-1bus * Ibus = Ybus à ¢Ã¢â‚¬ ¡Ã¢â‚¬â„¢ = = Voltage Analysis Voltage at each bus can be derived from the equation (Vbus = Y-1bus * Ibus) and Matlab was used for calculate matrix division. (Source code is attached in Appendix-1) Vbus == V12 = 0.0034 + j 0.0031 pu V13 = -0.0277 j 0.0257 pu V14 = 0.0336 + j 0.0311 pu V21 = -0.0034 j 0.0031 pu V23 = -0.0311 j 0.0288 pu V24 = 0.0302 + j 0.0280 pu V31 = 0.0277 + j 0.0257 pu V32 = 0.0311 + j 0.0288 pu V41 = -0.0336 j 0.0311 pu V42 = -0.0302 j 0.0280 pu Current flow in the system Current flow from BUS i and BUS j can be calculated by using voltage difference and interconnected admittance of the line between buses. [Iij = yij * (Vi Vj) ] The calculation result from Matlab is as follows; I12 = 0.0249 j 0.0269 pu I13 = -0.1026 + j 0.1108 pu I14 = 0.0777 j 0.0840 pu I21 = -0.0249 + j 0.0269 pu I23 = -0.1151 + j 0.1243 pu I24 = 0.1399 j 0.1511 I31 = 0.1026 j 0.1108 pu I32 = 0.1151 j 0.1243 pu I34 = 0 pu I41 = -0.0777 + j 0.0840 pu I42 = -0.1399 + j 0.1511 pu I43 = 0 pu Power flow in the system Apparent flow from BUS i and BUS j can be calculated by voltage at the sending bus and line current. [ Sij (pu) = Vi * I*ij = Pij + jQij ] The calculation result from Matlab is as follows; S12 = 0.0311 + j 0.0175 pu S13 = -0.1283 j 0.0723 pu S14 = 0.0972 + j 0.0548 pu S21 = -0.0311 j 0.0174 pu S23 = -0.1438 j 0.0803 pu S24 = 0.1749 + j 0.0977 pu S31 = 0.1283 + j 0.0780 pu S32 = 0.1438 + j 0.0875 pu S34 = 0 pu S41 = -0.0972 j 0.0496 pu S42 = -0.1749 j 0.0892 pu S44 = 0 pu Admittance Matrix considering transmission line capacitance According to the instruction of the Question 3, power system model can be drawn by using à Ã¢â€š ¬ equivalent circuit of the lines with capacitive shunt admittance (yc) of 0.1 pu at each side as shown below. Admittance Matrix Contrary to equivalent model in Question 3-1, the current flow through the capacitor in the transmission line needs to be considered to find the admittance matrix. Therefore, considering the capacitors the current equation with Kirchhoffs current law at each bus is as follows; [2] [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 separate and group individual products by voltage. 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 admittance matrix are calculated as follows; = y12 + y13 + y14 + yc12 + yc13+ yc14 = -j8 -j4 j2.5 + j0.1 + j0.1 +0.1j = -j14.2 pu = y21 + y23 + y24 + yc21 + yc23+ yc24 = -j8 -j4 j5 + j0.1 + j0.1 +0.1j = -j16.7 pu = y30 + y31 + y32 + yc31+ yc32 = -j08 -j4 j4 + j0.1 +0.1j = -j8.6 pu = y40 + y41 + y42 + yc41+ yc42 = -j0.8 -j2.5 j5 + j0.1 +0.1j = -j8.1 pu = y12 = -(-j8) = j8 pu = y13 = -(-j4) = j4 pu = y14 = -(-j2.5) = j2.5 pu = y21 = -(-j8) = j8 pu = y23 = -(-j4) = j4 pu = y24 = -(-j5) = j5 pu = y31 = -(-j4) = j4 pu = y32 = -(-j4) = j4 pu = y34 = 0 pu = y41 = -(-j2.5) = j2.5 pu = y42 = -(-j5) = j5 pu = y43 = 0 pu Therefore, admittance matrix Ybus is as follows; Ybus = = Annex-1: Matlab source code and Calculation results with Matlab Matlab Source Code % define self admittance and mutual admittance by using admittace between % the buses (y12=y21=-j8, y13=y31=-j4, y14=y41=-j2.5, y23=y32=-j4, % y24=y42=-j5, y34=0, y43=0, y30=-j0.8, y40=-j0.8 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 not connected, so admittance Y34 and Y43 are equal to zero % define the 44 admittance matrix (Ybus) Ybus=[Y11 Y12 Y13 Y14; Y21 Y22 Y23 Y24; Y31 Y32 Y33 Y34; Y41 Y42 Y43 Y44]; % In order to define the nodal equation (I = Ybus*V), the given I needs to defined. i1=0; i2=0; i3=-i; i4=-0.4808-0.4808i; Ibus=[i1; i2; i3; i4]; % Each bus voltage can be calculated by using matrix division (V= YbusI) Vbus=YbusIbus; v1=Vbus(1,1); v2=Vbus(2,1); v3=Vbus(3,1); v4=Vbus(4,1); % Calculate voltage difference between buses 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 buses 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; % apparent power can be calculated by s12 = v1 * conj(i12) s12=v1*conj(i12); s13=v1*conj(i13); s14=v1*conj(i14); s21=v2*conj(i21); s23=v2*conj(i23); s24=v2*conj(i24); s31=v3*conj(i31); s32=v3*conj(i32); s34=v3*conj(i34); s41=v4*conj(i41); s42=v4*conj(i42); s43=v4*conj(i43); % Real power and Reactive power can be derived by 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); % end Matlab Calculation Results

Physics of the Acoustic Guitar Essays -- physics guitars

Stringed instruments are most likely the oldest type of musical instruments. The acoustic guitar has been around since the 1500's, and is the most common stringed instrument used today. The guitar is one of the most versatile instruments in existence, being used in virtually every style of music. It can produce a wide range of sounds depending upon the style of the guitar and the type of strings used on it. The two main styles of acoustic guitars are the nylon-stringed and the steel-stringed. Nylon-stringed guitars are used mainly for styles such as classical music and flamenco. Steel-stringed guitars are most commonly associated with country and folk music. Before getting into how the guitar works, it will be beneficial to take a look at the different parts that make up a guitar. A guitar has three main parts- a body, neck and headstock. There are various styles and designs for acoustic guitars, but no matter the style, all of them have essentially the same parts. The body acts as an anchor for the neck and bridge. The top of the body is called the sounding board. It is responsible for almost all of the guitars acoustics, and is what allows the ear to hear the sound created by playing the guitar. The sounding board generally has a large hole cut in it called the sound hole. The bridge, which is also located on the sounding board, is what anchors the strings to the body. The neck is the long, slender part of the guitar connecting the body to the head. the front of the neck is called the fret board. This is where the frets are spaced down the entire neck. The frets are thin bars that are set perpendicular to the strings. The frets make it possible to create a variety of tones by... ... pattern of these resonance waves in order to cut down on interference, thus creating a richer sound. Bibliography Askill, John. Physics of Musical Sounds. New York: D. Van Norstrand Company, 1979. Billington, Ian. The Physics of the Acoustic Guitar. 9 Apr. 2003. 211.web.stuff/billington/main.htm> Guitar Acoustics. 9 Apr. 2003. Hokin, Sam. The Physics of Everyday Stuff: The Guitar. 9 Apr. 2003. /physics/stuff/guitar.html> How Stuff Works: How Acoustic Guitars Work. 9 Apr. 2003. com/guitar1.htm> Phillips, Mark, and Jon Chappell. Guitar For Dummies. Foster City: IDG Books, 1998. Sumit's Acoustic Guitar Anatomy. 9 Apr. 2003. anatomy.html> Physics of the Acoustic Guitar Essays -- physics guitars Stringed instruments are most likely the oldest type of musical instruments. The acoustic guitar has been around since the 1500's, and is the most common stringed instrument used today. The guitar is one of the most versatile instruments in existence, being used in virtually every style of music. It can produce a wide range of sounds depending upon the style of the guitar and the type of strings used on it. The two main styles of acoustic guitars are the nylon-stringed and the steel-stringed. Nylon-stringed guitars are used mainly for styles such as classical music and flamenco. Steel-stringed guitars are most commonly associated with country and folk music. Before getting into how the guitar works, it will be beneficial to take a look at the different parts that make up a guitar. A guitar has three main parts- a body, neck and headstock. There are various styles and designs for acoustic guitars, but no matter the style, all of them have essentially the same parts. The body acts as an anchor for the neck and bridge. The top of the body is called the sounding board. It is responsible for almost all of the guitars acoustics, and is what allows the ear to hear the sound created by playing the guitar. The sounding board generally has a large hole cut in it called the sound hole. The bridge, which is also located on the sounding board, is what anchors the strings to the body. The neck is the long, slender part of the guitar connecting the body to the head. the front of the neck is called the fret board. This is where the frets are spaced down the entire neck. The frets are thin bars that are set perpendicular to the strings. The frets make it possible to create a variety of tones by... ... pattern of these resonance waves in order to cut down on interference, thus creating a richer sound. Bibliography Askill, John. Physics of Musical Sounds. New York: D. Van Norstrand Company, 1979. Billington, Ian. The Physics of the Acoustic Guitar. 9 Apr. 2003. 211.web.stuff/billington/main.htm> Guitar Acoustics. 9 Apr. 2003. Hokin, Sam. The Physics of Everyday Stuff: The Guitar. 9 Apr. 2003. /physics/stuff/guitar.html> How Stuff Works: How Acoustic Guitars Work. 9 Apr. 2003. com/guitar1.htm> Phillips, Mark, and Jon Chappell. Guitar For Dummies. Foster City: IDG Books, 1998. Sumit's Acoustic Guitar Anatomy. 9 Apr. 2003. anatomy.html>