Shakespeare and Angela Carter and their works Macbeth and Blood Chamber Essay Example for Free

Shakespeare and Angela Carter and their works Macbeth and Blood Chamber Essay Shakespeare and Angela Carter present through their work of ‘Macbeth’ and ‘The Bloody Chamber’ the struggles of women rebelling against gender stereotypes and how they fall victim to the patriarchal society they are a part of. Firstly, Shakespeare shows the importance of gender boundaries in society and how women who do not conform to their gender stereotypes will be punished. In Macbeth Shakespeare uses the character of Lady Macbeth to show this. Lady Macbeth throughout the play influences and controls her husband Macbeth for personal gain, she does this by taunting her husband which is already the first sign that she doesn’t conform to the stereotypical 16th-century woman who would usually be submissive to their husband. â€Å"When you durst do it, then you were a man† here we see that the gender roles have switched because it’s the wife who has the power here, by saying you ‘were a man’ is telling of this, she is almost mocking him by inferring that if he doesn’t murder Duncan then he is a ‘woman’ in the sense that he was too weak to carry out the murder showing how society viewed women as almost ‘delicate’ however in this scene it’s the woman who has the power. Lady Macbeth teases Macbeth because she knows how if he murders Duncan that she will gain power, now as they’re in a highly patriarchal society Lady Macbeth would not have had any social hierarchy over men but if she becomes queen she will have that power and this would have been desirable for her, as Shakespeare has shown us through her not follow social stereotypes that she has more ‘masculine’ desires rather than women who conform to society by following their only purpose in having children which Lady Macbeth opposes through her description of infanticide she has no interest in having children. To a Jacobean audience, this would be absurd because women’s only purpose in society in the 16th century was to provide men with children. Shakespeare is showing us here how she is against the most feminine thing for a woman and is almost striving for her own personal gain of power which is more masculine in the sense that in most Jacobean plays/stories it is onl y men who strive for power and is a masculine trait traditionally. In The Bloody Chamber, we see a masculine trait again is given to a female character, Carter gives the role of the hero who saves the damsel in distress to the protagonist’s mother which again in traditional stories is a male stereotype where a man would come to save the female who has been captured. The mother is identified with power as she’s described as having â€Å"furious justice† which shows that Carter and Shakespeare have both given female characters that power or shown that like Lady Macbeth they are capable of wanting/having that power. However, a female craving this power is not punished in The Bloody Chamber like it is in Macbeth with her being haunted by her feeling of guilt with her having visions of blood on her hands â€Å"come out, damned spot! But I command you!† as we can see from the way Shakespeare wants this presented as she’s shouting it’s obviously driving her insane as she can see it but no one else can and thus Shak espeare shows how a woman is punished for rebelling against gender stereotypes by her seeking power and has gone insane due to a man’s actions and therefore takes the consequences. But The Bloody Chamber doesn’t punish women for trying to gain power like in Macbeth but through the use of fetishes and sexual desires, they are punished. In The Bloody Chamber, the main protagonist is seen as having sexual desires through the words â€Å"thrusting†, â€Å"ecstasy†, †burning cheek† and â€Å"tender† the main protagonist is usually described with sexual words like this throughout the story. This is a continuative theme throughout the story even by things such as the red necklace she is given by the count which can symbolize blood is a reference to sexual fetishes and desires. However, the same intrigue that allows her to explore sexuality also forces her to explore the secret room which allows her to discover the bodies of the counts old wives. Like Lady Macbeth who is punished for her husband’s actions, the main protagonist in this story would also have been punished for her husband’s actions that being that she discovered his murders and would have been killed by him for discovering them until she is saved by her empowered mother. Secondly, we also see the rebelling of gender stereotypes through the merging of the genders with female and masculine traits being given to opposite-sex characters in Macbeth and The Bloody Chamber. In Macbeth, we see the characters of the witches who cross the line between female and male gender stereotypes. â€Å"You should be a woman, yet your beards forbid me† showing that Shakespeare even blurred the lines of gender through their appearance and thus they rebel against the patriarchal society by not conforming to these stereotypes of looking like a woman and it shows through the confusion of Banquo who has been indoctrinated by the patriarchy and knows nothing else. However, even with the witches who have tried to rebel from gender stereotypes even through appearance they still fall victim to the patriarchy. They have received the punishment of being segregated from a society which is most likely due to the time being that in the 16th century England the fad for witch hun ting was at its peak. A critic states: ‘Of all Shakespeares female characters Lady Macbeth stands out far beyond the rest — remarkable for her ambition, strength of will, cruelty, and dissimulation.’ I agree with this statement that Lady Macbeth stands out and has traits such as ‘strength of will’ because of â€Å"That made you break your enterprise to me? When you durst do it, then you were a man† it would have been unheard of for someone like Lady Macbeth who is a representation of a woman in 16th century Elizabethan England to stand up and taunt their husband like she does which does show strength, it also shows her rebelling against gender stereotypes because it’s unfeminine for a woman to mock their husband for not being a man in considering if she wasn’t married she would have no power at all in society and therefore doing this could be seen as a risk to a woman and therefore making Lady Macbeth ‘strong’ and ‘ambitious’ As Lady Macbeth propels her husband toward committing Duncans murder, she indicates that she must take on masculine characteristics. Her most famous speech

Team Building Exercise And Purpose Management Essay

Team Building Exercise And Purpose Management Essay Team building is a wide range of activities that are presented to business, sports, schools, religions or even NPOs to ensure improvements performance in teams. Team building is usually done through several activities and practices, which can range from simple bonding exercise to complicated simulations along with multi-day team building retreats that are designed with the aim of developing a particular team, usually falling somewhere in between. This practice in most cases sits within theories along with practices of organizational development. However, it can also be applied in to sports teams, groups in schools, as well as other contexts. Team building practices need not to be confused with what is referred to as team recreation, which consists of activities for teams that are mainly recreational. In most cases, team building is considered as a very important factor in any environment (DeSimone Et al. 2002). Its concentration is to specialize in bringing out the best from a certain team, as to ensure self development, positive communication practices, leadership skills as well as the capability of working closely together as a team to problem solving. Different literatures have stated that, work environment in most cases concentrates on individual and personal objectives, with regard and recognition singling out the achievements that individual employees have attained. How to create effective teams is a challenge in every organization. In other words, team building also means the processes that are used in the selection and of teams from scratch. Team Building Exercise and Purpose The exercises involved in team building consists of a variety of tasks that are designed with the aim of developing group members as well as their capability of working together effectively and efficiently. There exist several types of team building activities that range from kid games to games involving novel complex tasks; however, they are designed for specific needs (Field Ford, 1995). There also exist complex team building activities that are composed of multiple exercises like rope courses, corporate drumming, as well as exercises that might last for several days. The main objective of team building exercises is for helping teams to become human units that are much cohesive for them to work together effectively to complete tasks. Types of Team Building Exercises The first type is communication exercise. This type of exercise is just as it sounds. Communication exercises involve problem solving practices that are geared towards upgrading communication skills. Most of the issues encountered in such practices are in most cases solved through effective communication among the members (Foley, 2000). The main objective of this activity is to come up with an activity that addresses the significance of good communication in team performance and /or potential problems with communication. Another type of team building activity includes problem solving, also known as decision making exercise. This exercise concentrates specifically on groups that are working together to deal with difficult, or come up with complex solutions. Such like exercises are the most common exercises as they appear to be having the most direct link, to what most of the employers are looking for to be incorporated in their working force (Noe, 2005). The main goal of this exercise is to provide the participants with a problem whose solution is not easily apparent, or it might require the team to with solutions that are much creative. The next exercise is planning also called adaptability exercise. These exercises concentrates on concepts that deal with planning as well as being able to adapt to different environmental climates, that are changing every now and then, due to political instability din the country. These are important things for teams to be able to do when they are assigned complex tasks. Or even complex circumstances. The main object of this type of exercise is to show the significance of planning before implementing any solution or recommendation. The last but not least type of team building exercise is trust exercise. This exercise on the other hand involves engagement of team members in a manner that induces trust between them. Some time, the practices are difficult to engage in, or even implement. This is based on the fact that there are varying levels of trust that exist between individuals, as well as varying degrees of personal comforts trusting others generally (Rigg Et al. 2007). The main goal of such like exercises of is the creation of trust between team members. Tips for Team Building It is true that, individuals at work place keep on talking about team building, working together as a team, however, it is only a few of them that really understands the creation as well as the experience gained from team work, or even how to come up and develop one that is much effective. Belonging to a team, looking at it in the broadest point of view, should be the essence of feeling being part and parcel of something that is much larger than oneself (Achua Lussier, 2007). It has a lot of things to do with somebodys understanding of the mission, vision and objectives of that particular group that one belongs in. Fit need to be noticed that, team oriented environment, every individual who happens to be a member, needs to contribute to the overall success of that particular group. But how many individuals do that, or how many groups allow that? If though individuals might be having specific job functions and belongs to specific departments, done needs to know that is unified with other team members for the accomplishment of the overall goal of the organization or group. The larger picture need to be the one that drives actions in an organization as someones function exists to serve the larger picture. In group building, it is much essential to differentiate between the overall sense of teamwork from the task of coming up with an effective and intact team, that is build to attain certain goals or objectives. As effect, on the ground, individuals tent to confused two team building objectives. As a result, this is the reasons as to why so many team building seminars, meetings, retreats as well as activities are deemed to be failures in the society by those who usually participate in them. It is true that leaders of such group activities fail to define the team that they desired to build (Larry, 1991). The process of developing an overall sense of team work is far much different from the process of building an effective, focused work team, when considering team building approaches. Different Cs for Team building Executives, managers as well as the organization staff members, have to universally engage in the process of exploring different ways of improving business results along with profitability. Many individuals look at team-based, horizontal organization structures as being the best design for involving all the employees in the creation of business successes. But the fact is, no matter the name given to ones team-based improvement effort; continuous improvement, total quality, lean manufacturing or even self directed work teams, every body need to strive at improving the results for the benefit of customers of ones organization (Heathfield, 2010). It is true but disappointing to note that, only few organizations are however totally pleased with the results of their team improvement efforts produce. It should be noted that, if ones team improvement efforts are not living to ones expectations, this self diagnosing list of check, may tell the reasons as to why. Some successful team building, that ends up creating effective, focused work teams, in most cases need attention to each one of the following: Clear expectations; the executive leadership needs to clearly communicate its expectations for the performance of the team, along with the expected outcomes. Team members on the other hand need to understand the reason as ton why the group was formed. In addition, the organization needs to demonstrate constancy of purpose in providing support to the team with resources of time, individuals, as well as finances. The work that is performed by the team, need to receive sufficient emphasis as a major priority in terms of time, attention, discussions well as interest directed its way by the executive leaders. Context; team members should have an understanding as to why they are taking part in the team, in such a way, they will have an understanding effects of team strategy in helping the organization in attaining it communicated business objectives and goals. As an effect, the team members ought to define their teams significance to the attainment of the corporate goals and objectives. To attain this, the team members need to understand where its work fits in the total context of different goals, principles, vision as well as values of the organization. Commitment; there is need for the members of the team to take part in the teams activities fully, as this will make team members to be more committed to the attainment of the mission and outcome expectations of the team. In so doing, the team members will perceive their services pas much valuable to the organization as well as their own careers. This in one way or the other will make the make the team members to anticipate for the recognition for their contributions (Heathfield, 2011). It should be noted by team members that, their participation will enhance the growth of their skills and experience on the team. After this realization, team members will then be excited and challenged by the team opportunity. Competence is another C; it will be good if the team feels that it has the appropriate individuals taking part in it, for instance, in the process improvement, each step of the operation need to represent in the team. The team should feel that, its members should have the necessary knowledge, skills and capabilities to address the issues for which the team was formed to address. In case this is not there, the team ought to have an access to the help it requires. The team should feel that, it has been given the necessary, strategies and support that it requires fin order to accomplish the assigned responsibility. Charter; fit is the responsibility of the team to take the assigned area of responsibility and design its own mission, vision as well as the strategies that will be much helpful in the attainment of the overall mission and objective. It is always good for the team to define and co9mmunicate its objectives, its anticipated results as well as its report contribution, its timelines, and the procedure it will employ in measuring its results of its work as well as the processes along with operations followed by the team to accomplish its tasks. The leadership team or any other group charged with the responsibility of coordinating group activities should at all times support all what the team has designed. Control; the team at all times need to be given the necessary freedom and as well as the empowerment so that it can feel the ownership that is much necessary for it to accomplish its charter. At the same juncture, the team members should clearly understand their limits or boundaries. This will in one way or the other determine the extent at which the team can go in pursuit of solutions, determine its limitations like in terms of time and monitory resources, that was defined at the initial stages of the project, before the team experiences barriers as well as the rework. The teams reporting relationship as well as accountability should be understood by all members of the organization. This is because; the organization defined the teams authority, to make recommendation, as well complimenting its plan. It holds more water if all there will be a defined process of reviewing both the organization and the team consistently and aligned to the purpose and in direction. Every team member sho uld hold each other accountable for project timelines, commitments and results. As an effect, the organization should have plans for increasing the opportunities for self management among organization members. Collaboration; the team should be in a position of understanding the team as well as the group process. Members need to be in a position of understanding group stages involved in group development. This will enable them to work interpersonally together effectively. All team members are acquainted with the duties and responsibilities [charged to each team member as well as to the group as a whole. They should know the responsibilities of team leaders as well as the team recorders to avoid confusion and mixing of responsibilities. The team ought to be in a position of solving problems, process improvement, and project measurement techniques. This will encourage cooperation among group members to accomplish the group charter. The team should establish team norms and rules of conduct that will be used for governing like conflict resolution, consensus decision making along with meeting management. This will enable the team to use appropriate strategy to accomplish its action plan. Communication; the team members should always be clear about the priority of their tasks. This will enable the team to establish proper methods that will be used during feedback giving, as well as receiving honest performance feedback. The organization on its part needs to provide important business data regularly, enabling the team to understand the complete context of its existence. The team members should also be in a position of communicating clearly and honestly with each other. This in one way or the other help members top provide diversified opinions to the table for discussions and conclusions, which will end up addressing necessary conflicts. Creative Innovations; the organization should in one way or the other show that it really want change. This will enable the team to realize that the company needs creative thinking, solutions that are unique and also new ideas. The organization as a result rewards individuals who reasonably risks coming up with improvements, other than victimizing them. It should reward these who fit in and maintain the status quo. Never the less, it should provide training education, as well as providing access to learning materials like books and films, along with field trips that will stimulate new thinking. Consequences; all team members need to feel responsible and accountable for team achievements. To enhance this attitude, rewards should be awarded whenever the team becomes successful], in addition, this will also be achieved if reasonable risks are respected and encouraged in the organization. Though the reward supply ought to be encouraged by the organization, the rewarding system ought to be in a position of recognizing both team as well as individual performances. Do team members fear reprisal? the team members need not to waste their time pointing fingers at each other instead of resolving problems. It will be encourage if the organization will be willing to share gains and the increased profitability wit team along with individual contributors, as for the contributors to be motivated, they ought to see their impact don the increased organization success. Coordination; though there are different systems of coordination in teamwork, but there need to be a central leadership team that will be assisting the group to get what they require for their success. This central body will ensure priorities and resource allocation have been planned across all departments. In support of this, teams ought to understand the aspect of internal customer serving- as the next process, anybody to whom they provide products or even services. To accelerates this, cross-functional along with multi-departmental groups need to work need to work together as one, in developing customer focused, process-focused orientation , as the move away from traditional departmental thinking. Cultural change is the last but not least C involved in team building. The organizations or even societies need to recognize that the team-focused, empowering, enabling organizational culture of the future is different than the traditional, hierarchical organization it may currently be. The organization need to be having plans on how to recruit its employees, along with how to reward, plan development, how it carries out motivation practices, and how it manages individuals they employ. It is good for the organization to determine if will be using failures for learning as well as supporting reasonable risks (Van Dijk Phoads, 2007). The organization should also look forward on how to change its climate towards supporting teams, as this will increase the payback from them. When more time is spend in looking at such tips to ensure effective and efficient teams contribution to business successes. Conclusion A team that is well managed forms the cornerstone of success in any organization. The success of a team appears if and only if team members are in good terms, able to accomplish something tangible, by working together as a group other than working as individuals. It should be taken that, the success and the progress of a team, is the responsibility that is charged to every team member, regardless of the hierarchy. Being in a group, has nothing to do with the dissolution of somebodys individualism, though helps in build in g the identity by addressing somebodys strength as well as potentialities. The basis of team building is based on effective communication. however, the co-existence can be achieved by having by engaging in team building practices, which are fun-filled with the aim of attaining goals by making work far much enjoyable as well as gratifying , though there has to be consistency, to ensure that individuals are not taking it as time and effort wastage (Cardy Et al. 2009). The aim of team building in most cases is to encourage members with expertise in certain areas contribute to the objectives of teams. Lack of harmony in a team makes individual opinions and ideas, which might lead to conflicts. In addition, the absence of co-ordination between diversified personalities is one of the biggest factors that might lead to failure at work place. For the effective team management, certain pointers ought to be in place, for instance, responsibility division between members and be much clear about roles; this effectively avoids confusion (Andrejev, 2006). The process of decision making should also not be left top management alone; the entire team should whenever possible. This in one was or the other ensures, better productivity, absence of trust and belonging among members. Team building helps in boosting self-confidence as well as moral of the involved individuals. The show of appreciation as well as giving timely feedback, also motivate employees. It is good to notice that, teams are never static; they grow and change with time. Effective team management recognizes each members strong qualities and mobilizes them to work together.

George Boole: The Genius Essay -- essays research papers

George Boole: The Genius   Ã‚  Ã‚  Ã‚  Ã‚  George Boole was a British mathematician, and he is known as the inventor of Boolean Algebra. His theories combined the concepts of logic and mathematics, and hence he is known as the father of mathematical logic. This combination of mathematics and logic came to be known as Boolean algebra, and is the basis of digital electronic design, which is used in fields ranging from telephone switching to computer engineering. Because of the utilization of the concepts of Boolean algebra in electronics and computers, George Boole is regarded by many as the father of computing also.   Ã‚  Ã‚  Ã‚  Ã‚  George was born on 2nd November, 1815 in Lincoln, England. His father, John Boole was a shoemaker, and his mother a housewife. John Boole proved to be a great influence in George’s life due to his keen interest in science and mathematics. He shared his passion with his son, and started teaching George at an early age. By the time he was seven, George was deeply in love with mathematics, and used to be lost in the world of mathematics. He acquired a reputation as a child genius, and one day, he was found spelling difficult words for people’s amusement after going missing from school.   Ã‚  Ã‚  Ã‚  Ã‚  George was from a poor family, and his parents could not afford to pay fees for grammar school, so the child genius ended up going to a small school called Mr. Bainbridge’s Academy. He made fast progress in studies, and was soon assisting teachers in teaching and grading. His exploits weren’t limited to just math and science either; he loved to read and learn, and was very well read in a lot of subjects. His father John also introduced him to literature and Latin, but George soon learned all his father had to offer. After that, John found George a tutor – bookseller William Brooke. Mr. Brooke turned out to be a great asset for George; he gave George access to all the books in his store, and also taught him. Mr. Brooke and George ended up being lifelong friends. However, just knowing Latin was not enough for George. He added Greek to his repertoire, and this was completely self-taught. He also went on to study French, German, and Italian. In Ma y 1930, the local paper published George’s translation of Greek poet Meleager’s work, and this got George his reputation as a boy genius.   Ã‚  &n... .... His wife Mary’s approach to trying to cure him was also one of the primary reasons of his death. She believed in the theory that the cause would also be the cure, so instead of keeping him warm, she regularly drenched him with water in bed, leading to severe complications. Ironically, Mary said she did it because it seemed ‘logical’ to her!   Ã‚  Ã‚  Ã‚  Ã‚  George’s works considered purely mathematical until the year 1937. In 1937, Claude Shannon, a graduate student at MIT, discovered the connection between electronic circuits and Boolean algebra. This connection is essential to the operation of computers and modern electronics circuits. Computers and circuits utilize Boolean algebra for all their decision making calculations, and without it they would be quite useless.   Ã‚  Ã‚  Ã‚  Ã‚  George Boole was well ahead of his time with his mathematical theories and the combination of mathematics and logic. His theories are in use today, a century after his time, and will be in use as the basis of one of the most important machines man has ever built. He was a true genius, and his work has gotten him the deserved title of the father of mathematical logic. George Boole: The Genius Essay -- essays research papers George Boole: The Genius   Ã‚  Ã‚  Ã‚  Ã‚  George Boole was a British mathematician, and he is known as the inventor of Boolean Algebra. His theories combined the concepts of logic and mathematics, and hence he is known as the father of mathematical logic. This combination of mathematics and logic came to be known as Boolean algebra, and is the basis of digital electronic design, which is used in fields ranging from telephone switching to computer engineering. Because of the utilization of the concepts of Boolean algebra in electronics and computers, George Boole is regarded by many as the father of computing also.   Ã‚  Ã‚  Ã‚  Ã‚  George was born on 2nd November, 1815 in Lincoln, England. His father, John Boole was a shoemaker, and his mother a housewife. John Boole proved to be a great influence in George’s life due to his keen interest in science and mathematics. He shared his passion with his son, and started teaching George at an early age. By the time he was seven, George was deeply in love with mathematics, and used to be lost in the world of mathematics. He acquired a reputation as a child genius, and one day, he was found spelling difficult words for people’s amusement after going missing from school.   Ã‚  Ã‚  Ã‚  Ã‚  George was from a poor family, and his parents could not afford to pay fees for grammar school, so the child genius ended up going to a small school called Mr. Bainbridge’s Academy. He made fast progress in studies, and was soon assisting teachers in teaching and grading. His exploits weren’t limited to just math and science either; he loved to read and learn, and was very well read in a lot of subjects. His father John also introduced him to literature and Latin, but George soon learned all his father had to offer. After that, John found George a tutor – bookseller William Brooke. Mr. Brooke turned out to be a great asset for George; he gave George access to all the books in his store, and also taught him. Mr. Brooke and George ended up being lifelong friends. However, just knowing Latin was not enough for George. He added Greek to his repertoire, and this was completely self-taught. He also went on to study French, German, and Italian. In Ma y 1930, the local paper published George’s translation of Greek poet Meleager’s work, and this got George his reputation as a boy genius.   Ã‚  &n... .... His wife Mary’s approach to trying to cure him was also one of the primary reasons of his death. She believed in the theory that the cause would also be the cure, so instead of keeping him warm, she regularly drenched him with water in bed, leading to severe complications. Ironically, Mary said she did it because it seemed ‘logical’ to her!   Ã‚  Ã‚  Ã‚  Ã‚  George’s works considered purely mathematical until the year 1937. In 1937, Claude Shannon, a graduate student at MIT, discovered the connection between electronic circuits and Boolean algebra. This connection is essential to the operation of computers and modern electronics circuits. Computers and circuits utilize Boolean algebra for all their decision making calculations, and without it they would be quite useless.   Ã‚  Ã‚  Ã‚  Ã‚  George Boole was well ahead of his time with his mathematical theories and the combination of mathematics and logic. His theories are in use today, a century after his time, and will be in use as the basis of one of the most important machines man has ever built. He was a true genius, and his work has gotten him the deserved title of the father of mathematical logic.

MDMA aka Ecstasy Essay -- Research Drugs Medicine Essays

MDMA aka Ecstasy Adam, Ecstasy, XTC, hug, beans, love drug, X and E are all street names for the drug that is known as MDMA in the medical and scientific world. MDMA is most commonly referred to as Ecstasy and is a "synthetic, psychoactive drug with both stimulant (amphetamine-like) and hallucinogenic (LSD-like) properties". MDMA is also a neurotoxin, which in high doses can raise body temperature and cause muscle and kidney breakdown that eventually leads to failure of the cardiovascular system. This said, the drug is still one of the most popular on the market, and demand is rising. (1) MDMA is made up of a chemical structure consisting of 3,4 methylenedioxymethamphetamine. It depletes serotonin-producing neurons after prompting "nerve cells to release a flood of serotonin". MDA, a parent drug of MDMA, destroys neurons that contain the neurotransmitter dopamine whose destruction is an underlying cause in the lack of coordination, tremors, and paralysis that come with Parkinson's disease. (1) Dopamine is also thought to be a chemical messenger to learning. (10) The use of MDMA not only reduces the presence of these two very important neurotransmitters, but can have serious side effects as well. Psychological side effects consist of confusion, depression, sleep problems, drug craving, severe anxiety and paranoia. Physical effects and signs of Ecstasy users are muscle tension, involuntary teeth clenching, nausea, blurred vision, rapid eye movement, faintness, chills or sweating, as well as increased heart rate and blood pressure. Long-term use of this drug is evident by an acne-like rash that develops all over the body and ca n lead to liver damage. This gives the drug many risks similar to those of cocaine. (1) ... ... http://www.dancesafe.org/slideshow/ 5) Ecstasy http://www.geocities.com/sunsetstrip/Limo/4325/ecstasy.html 6) Ecstasy's Legacy http://www.sciam.com/missing.cfm 7) Scientists Study Serotonin Markers for Suicide Prevention http://www.mhsource.com/pt/p950907.jhtml?_requestid=374901 8) Serotonin and Judgment http://web.sfn.org/briefings/serotonin.html 9) Suicide Facts http://www.nimh.nih.gov/research/suifact.htm 10) Breakthrough? Study finds Dopamine cannot be source of Pleasure in Brain http://www.sciencedaily.com/realeases/1999/03/990304052313.htm 11)Dextromethorphan http://www.nlm.nih.gov/medlineplus/druginfo/dextromethorphansystemic202187.html 12) Mixing Ecstasy with Other Drugs http://www.seeq.com/popupwrapper.jsp?referrer=http%3A%2F%2Fserendip.brynmawr.edu%2Fbiology%2Fb103%2Ff01%2Fweb3%2Fsterling.html&domain=wakeywakey.com

Franz Kafka :: essays research papers

â€Å"When he lifted his head a little, he saw his vaulted brown belly, sectioned by arch-shaped ribs, to whose dome the cover, about to slide off completely, could barely cling. His many legs, pitifully thin compared with the size of the rest of him, were waving helplessly before his eyes.† Gregor Samsa is the main character in this story to go through a metamorphosis. This change has turned Gregor into a â€Å"monstrous vermin†. Kafka expresses the anxieties, inner terrors, and cynicism of Gregor’s life throughout the novella, Metamorphosis. Gregor’s feelings towards his job, the effect his job has on his family, and the cruelty that his family displays show many of the changes that occur in the story. The novel opens with Gregor in his monstrous state, late for work. He surmises that his job as a traveling salesman is very important, since he has to pay off his father’s debts, yet he is growing extremely tired and frustrated about it. â€Å"The upset of doing business is much worse than the actual business in the home office, and, besides, I’ve got the torture of traveling, worrying about changing trains, eating miserable food at all hours, constantly seeing new faces, no relationships that last or get more intimate. To the devil with it all!† Gregor has a great amount of anger towards his job, which soon leads to his resentment towards society as a whole. The fact that his office manager showed up at Gregor’s house plays an immense role in creating apprehension and anxieties in Gregor’s mind. He feels strangled by his job and is too weak to tolerate the pressure. Along with the pressure created by his office manager and society, the Samsa’s, especially Gregor’s father, take advantage of him. Gregor earns the basic income to support his family. â€Å"But of course he actually could have paid off more of his father’s debt to the boss with this extra money, and the clay on wh ich he could have gotten rid of his job would have been much closer, but now things were undoubtedly better, the way his father had arranged them.† The bad taste of the Samsa’s has put Gregor in a difficult position, which I feel is one of the largest issues leading to Gregor’s metamorphosis. Gregor’s family in general had given him the negative attitude he has on life. They took advantage of him to the point where he was the means of the family’s survival.

Cengel Solutions

Chapter 4 Fluid Kinematics Solutions Manual for Fluid Mechanics: Fundamentals and Applications by Cengel & Cimbala CHAPTER 4 FLUID KINEMATICS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (â€Å"McGraw-Hill†) and protected by copyright and other state and federal laws.By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party.No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill. 4-1 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics Introductory Problems 4-1C Solution We are to define and explain kinematics and fluid kinematics. Analysis Kinematics means the study of motion.Fluid kinematics is the study of how fluids flow and how to describe fluid motion. Fluid kinematics deals with describing the motion of fluids without considering (or even understanding) the forces and moments that cause the motion. Discussion Fluid kinematics deals with such things as describing how a fluid particle translates, distorts, and rotates, and how to visualize flow fields. 4-2 Solution We are to write an equation for centerline speed through a nozzle, given that the flow speed increases parabolically. Assumptions 1 The flow is steady. 2 The flow is axisymmetri c. The water is incompressible. Analysis A general equation for a parabola in the x direction is u = a + b ( x ? c) General parabolic equation: 2 (1) We have two boundary conditions, namely at x = 0, u = uentrance and at x = L, u = uexit. By inspection, Eq. 1 is satisfied by setting c = 0, a = uentrance and b = (uexit – uentrance)/L2. Thus, Eq. 1 becomes u = uentrance + Parabolic speed: ( uexit ? uentrance ) L2 x2 (2) Discussion You can verify Eq. 2 by plugging in x = 0 and x = L. 4-3 Solution location. For a given velocity field we are to find out if there is a stagnation point.If so, we are to calculate its Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis The velocity field is V = ( u , v ) = ( 0. 5 + 1. 2 x ) i + ( ? 2. 0 ? 1. 2 y ) j (1) At a stagnation point, both u and v must equal zero. At any point (x,y) in the flow field, the velocity components u and v are obtained from Eq. 1, Velocity components: u = 0. 5 + 1. 2 x v = ? 2. 0 ? 1. 2 y (2) x = ? 0. 4167 y = ? 1. 667 (3) Setting these to zero yields Stagnation point: 0 = 0. 5 + 1. 2 x 0 = ? 2. 0 ? 1. 2 y So, yes there is a stagnation point; its location is x = -0. 17, y = -1. 67 (to 3 digits). Discussion If the flow were three-dimensional, we would have to set w = 0 as well to determine the location of the stagnation point. In some flow fields there is more than one stagnation point. 4-2 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-4 Solution location. For a given velocity field we are to find out if there is a stagnation point.If so, we are to calculate its Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis The velocity field is ( )( ) V = ( u, v ) = a 2 ? ( b ? cx ) i + ? 2cby + 2c 2 xy j 2 (1) At a stagna tion point, both u and v must equal zero. At any point (x,y) in the flow field, the velocity components u and v are obtained from Eq. 1, Velocity components: u = a 2 ? ( b ? cx ) 2 v = ? 2cby + 2c 2 xy (2) b? a c y=0 (3) Setting these to zero and solving simultaneously yields Stagnation point: 0 = a 2 ? ( b ? cx ) 2 x= v = ? 2cby + 2c xy So, yes there is a stagnation point; its location is x = (b – a)/c, y = 0. Discussion If the flow were three-dimensional, we would have to set w = 0 as well to determine the location of the stagnation point. In some flow fields there is more than one stagnation point. Lagrangian and Eulerian Descriptions 4-5C Solution We are to define the Lagrangian description of fluid motion. Analysis In the Lagrangian description of fluid motion, individual fluid particles (fluid elements composed of a fixed, identifiable mass of fluid) are followed. DiscussionThe Lagrangian method of studying fluid motion is similar to that of studying billiard balls and other solid objects in physics. 4-6C Solution We are to compare the Lagrangian method to the study of systems and control volumes and determine to which of these it is most similar. Analysis The Lagrangian method is more similar to system analysis (i. e. , closed system analysis). In both cases, we follow a mass of fixed identity as it moves in a flow. In a control volume analysis, on the other hand, mass moves into and out of the control volume, and we don’t follow any particular chunk of fluid.Instead we analyze whatever fluid happens to be inside the control volume at the time. Discussion to a point. In fact, the Lagrangian analysis is the same as a system analysis in the limit as the size of the system shrinks 4-3 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-7C Sol ution description. We are to define the Eulerian description of fluid motion, and explain how it differs from the LagrangianAnalysis In the Eulerian description of fluid motion, we are concerned with field variables, such as velocity, pressure, temperature, etc. , as functions of space and time within a flow domain or control volume. In contrast to the Lagrangian method, fluid flows into and out of the Eulerian flow domain, and we do not keep track of the motion of particular identifiable fluid particles. Discussion The Eulerian method of studying fluid motion is not as â€Å"natural† as the Lagrangian method since the fundamental conservation laws apply to moving particles, not to fields. -8C Solution We are to determine whether a measurement is Lagrangian or Eulerian. Analysis Since the probe is fixed in space and the fluid flows around it, we are not following individual fluid particles as they move. Instead, we are measuring a field variable at a particular location in sp ace. Thus this is an Eulerian measurement. Discussion If a neutrally buoyant probe were to move with the flow, its results would be Lagrangian measurements – following fluid particles. 4-9C Solution We are to determine whether a measurement is Lagrangian or Eulerian. AnalysisSince the probe moves with the flow and is neutrally buoyant, we are following individual fluid particles as they move through the pump. Thus this is a Lagrangian measurement. Discussion If the probe were instead fixed at one location in the flow, its results would be Eulerian measurements. 4-10C Solution We are to determine whether a measurement is Lagrangian or Eulerian. Analysis Since the weather balloon moves with the air and is neutrally buoyant, we are following individual â€Å"fluid particles† as they move through the atmosphere. Thus this is a Lagrangian measurement.Note that in this case the â€Å"fluid particle† is huge, and can follow gross features of the flow – the ballo on obviously cannot follow small scale turbulent fluctuations in the atmosphere. Discussion When weather monitoring instruments are mounted on the roof of a building, the results are Eulerian measurements. 4-11C Solution We are to determine whether a measurement is Lagrangian or Eulerian. Analysis Relative to the airplane, the probe is fixed and the air flows around it. We are not following individual fluid particles as they move. Instead, we are measuring a field variable at a particular location in space relative to the moving airplane.Thus this is an Eulerian measurement. Discussion The airplane is moving, but it is not moving with the flow. 4-4 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-12C Solution We are to compare the Eulerian method to the study of systems and contr ol volumes and determine to which of these it is most similar. Analysis The Eulerian method is more similar to control volume analysis.In both cases, mass moves into and out of the flow domain or control volume, and we don’t follow any particular chunk of fluid. Instead we analyze whatever fluid happens to be inside the control volume at the time. Discussion In fact, the Eulerian analysis is the same as a control volume analysis except that Eulerian analysis is usually applied to infinitesimal volumes and differential equations of fluid flow, whereas control volume analysis usually refers to finite volumes and integral equations of fluid flow. 4-13C Solution flow. We are to define a steady flow field in the Eulerian description, and discuss particle acceleration in such aAnalysis A flow field is defined as steady in the Eulerian frame of reference when properties at any point in the flow field do not change with respect to time. In such a flow field, individual fluid particle s may still experience non-zero acceleration – the answer to the question is yes. Discussion ( a = dV / dt ) Although velocity is not a function of time in a steady flow field, its total derivative with respect to time is not necessarily zero since the acceleration is composed of a local (unsteady) part which is zero and an advective part which is not necessarily zero. 4-14C SolutionWe are to list three alternate names for material derivative. Analysis The material derivative is also called total derivative, particle derivative, Eulerian derivative, Lagrangian derivative, and substantial derivative. â€Å"Total† is appropriate because the material derivative includes both local (unsteady) and convective parts. â€Å"Particle† is appropriate because it stresses that the material derivative is one following fluid particles as they move about in the flow field. â€Å"Eulerian† is appropriate since the material derivative is used to transform from Lagrangian to Eulerian reference frames. Lagrangian† is appropriate since the material derivative is used to transform from Lagrangian to Eulerian reference frames. Finally, â€Å"substantial† is not as clear of a term for the material derivative, and we are not sure of its origin. Discussion All of these names emphasize that we are following a fluid particle as it moves through a flow field. 4-5 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.Chapter 4 Fluid Kinematics 4-15 Solution We are to calculate the material acceleration for a given velocity field. Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Analysis The velocity field is V = ( u , v ) = (U 0 + bx ) i ? byj (1) The acceleration field components are obtained from its definition (the material acceleration) in Cartesian coordinates, ? u ?u ?u ?u +u +v +w = 0 + (U 0 + bx ) b + ( ? by ) 0 + 0 ?t ?x ?y ?z ?v ?v ?v ?v ay = + u + v + w = 0 + (U 0 + bx ) 0 + ( ? by )( ? b ) +0 ?t ?x ?y ?z ax = (2) here the unsteady terms are zero since this is a steady flow, and the terms with w are zero since the flow is twodimensional. Eq. 2 simplifies to ax = b (U 0 + bx ) ay = b2 y (3) a = b (U 0 + bx ) i + b 2 yj Material acceleration components: (4) In terms of a vector, Material acceleration vector: Discussion For positive x and b, fluid particles accelerate in the positive x direction. Even though this flow is steady, there is still a non-zero acceleration field. 4-16 Solution particle. For a given pressure and velocity field, we are to calculate the rate of change of pressure following a fluid Assumptions 1 The flow is steady. The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Analysis The pressure field is P = P0 ? Pressure field: 2U 0 bx + b 2 ( x 2 + y 2 ) ? 2? ? (1) By definition, the material derivative, when applied to pressure, produces the rate of change of pressure following a fluid particle. Using Eq. 1 and the velocity components from the previous problem, DP ? P ?P ?P = +u +v + Dt ?t ?x ?y Steady ( w ?P ?z (2) Two-dimensional ) ( = (U 0 + bx ) ? ?U 0 b ? ? b 2 x + ( ? by ) ? ? b 2 y ) where the unsteady term is zero since this is a steady flow, and the term with w is zero since the flow is two-dimensional.Eq. 2 simplifies to the following rate of change of pressure following a fluid particle: ( ) DP 2 = ? ? ? U 0 b ? 2U 0 b 2 x + b3 y 2 ? x 2 ? ? ? Dt (3) Discussion The material derivative can be applied to any flow property, scalar or vector. Here we apply it to the pressure, a scalar quantity. 4-6 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permi ssion. Chapter 4 Fluid Kinematics 4-17 SolutionFor a given velocity field we are to calculate the acceleration. Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis The velocity components are Velocity components: u = 1. 1 + 2. 8 x + 0. 65 y v = 0. 98 ? 2. 1x ? 2. 8 y (1) The acceleration field components are obtained from its definition (the material acceleration) in Cartesian coordinates, ? u ?u ?u ?u +u +v +w = 0 + (1. 1 + 2. 8 x + 0. 65 y )( 2. 8 ) + ( 0. 98 ? 2. 1x ? 2. 8 y )( 0. 65 ) + 0 ? t ?x ?y ?z ?v ?v ?v ?v + u + v + w = 0 + (1. 1 + 2. 8 x + 0. 65 y )( ? 2. 1) + ( 0. 98 ? 2. 1x ? 2. 8 y )( ? 2. ) +0 ay = ?t ?x ?y ?z ax = (2) where the unsteady terms are zero since this is a steady flow, and the terms with w are zero since the flow is twodimensional. Eq. 2 simplifies to Acceleration components: ax = 3. 717 + 6. 475 x a y = ? 5. 054 + 6. 475 y (3) At the point (x,y) = (-2,3), the acceleration components of Eq. 3 are Acceleration compone nts at (-2,3): ax = ? 9. 233 ? -9. 23 a y = 14. 371 ? 14. 4 Discussion The final answers are given to three significant digits. No units are given in either the problem statement or the answers. We assume that the coefficients have appropriate units. 4-18 SolutionFor a given velocity field we are to calculate the acceleration. Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis The velocity components are Velocity components: u = 0. 20 + 1. 3 x + 0. 85 y v = ? 0. 50 + 0. 95 x ? 1. 3 y (1) The acceleration field components are obtained from its definition (the material acceleration) in Cartesian coordinates, ? u ?u ?u ?u +u +v +w = 0 + ( 0. 20 + 1. 3 x + 0. 85 y )(1. 3) + ( ? 0. 50 + 0. 95 x ? 1. 3 y )( 0. 85 ) + 0 ? t ?x ?y ?z ?v ?v ?v ?v + u + v + w = 0 + ( 0. 20 + 1. 3 x + 0. 85 y )( 0. 95 ) + ( ? 0. 50 + 0. 95 x ? 1. y )( ? 1. 3 ) +0 ay = ?t ?x ?y ?z ax = (2) where the unsteady terms are zero since this is a steady flow, and the terms with w are zero since the flow is twodimensional. Eq. 2 simplifies to Acceleration components: ax = ? 0. 165 + 2. 4975 x a y = 0. 84 + 2. 4975 y (3) At the point (x,y) = (1,2), the acceleration components of Eq. 3 are Acceleration components at (1,2): ax = 2. 3325 ? 2. 33 a y = 5. 835 ? 5. 84 Discussion The final answers are given to three significant digits. No units are given in either the problem statement or the answers. We assume that the coefficients have appropriate units. -7 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-19 Solution We are to generate an expression for the fluid acceleration for a given velocity. Assumptions 1 The flow is steady. 2 The flow is axisymmetric. 3 The water is incompressible. Analysis In Problem 4-2 we found that along the centerline, u = uentranc e + Speed along centerline of nozzle: ( uexit ? uentrance ) x2 (1) ?u ?u ?u ?u +u +v +w ?t ?x y ?z (2) L2 To find the acceleration in the x-direction, we use the material acceleration, ax = Acceleration along centerline of nozzle: The first term in Eq. 2 is zero because the flow is steady. The last two terms are zero because the flow is axisymmetric, which means that along the centerline there can be no v or w velocity component. We substitute Eq. 1 for u to obtain Acceleration along centerline of nozzle: ax = u ( uexit ? uentrance ) 2 ? ( uexit ? uentrance ) ?u ? = ? uentrance + x ? ( 2) x ? ? ?x ? L2 L2 ? (3) or ax = 2uentrance Discussion ( uexit ? uentrance ) L2 x+2 ( uexit ? uentrance )L4 2 x3 (4) Fluid particles are accelerated along the centerline of the nozzle, even though the flow is steady. 4-20 Solution We are to write an equation for centerline speed through a diffuser, given that the flow speed decreases parabolically. Assumptions 1 The flow is steady. 2 The flow is axis ymmetric. Analysis A general equation for a parabola in x is General parabolic equation: u = a + b ( x ? c) 2 (1) We have two boundary conditions, namely at x = 0, u = uentrance and at x = L, u = uexit. By inspection, Eq. 1 is satisfied by setting c = 0, a = uentrance and b = (uexit – uentrance)/L2. Thus, Eq. becomes Parabolic speed: Discussion u = uentrance + ( uexit ? uentrance ) L2 x2 (2) You can verify Eq. 2 by plugging in x = 0 and x = L. 4-8 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-21 Solution We are to generate an expression for the fluid acceleration for a given velocity, and then calculate its value at two x locations. Assumptions 1 The flow is steady. 2 The flow is axisymmetric. AnalysisIn the previous problem, we found that along the centerline, u = uent rance + Speed along centerline of diffuser: ( uexit ? uentrance ) 2 L x2 (1) To find the acceleration in the x-direction, we use the material acceleration, Acceleration along centerline of diffuser: ax = ?u ?u ?u ?u +w +u +v ?z ?t ?x ?y (2) The first term in Eq. 2 is zero because the flow is steady. The last two terms are zero because the flow is axisymmetric, which means that along the centerline there can be no v or w velocity component. We substitute Eq. 1 for u to obtain Acceleration along centerline of diffuser: ( uexit ? uentrance ) x 2 ? ( uexit ? entrance ) x ?u ? = ? uentrance + ax = u ? ( 2) ? ?x ? L2 L2 ? ? or ax = 2uentrance ( uexit ? uentrance ) 2 L x+2 ( uexit ? uentrance ) 2 4 L x3 (3) At the given locations, we substitute the given values. At x = 0, Acceleration along centerline of diffuser at x = 0: ax ( x = 0 ) = 0 (4) At x = 1. 0 m, Acceleration along centerline of diffuser at x = 1. 0 m: ax ( x = 1. 0 m ) = 2 ( 30. 0 m/s ) ( ? 25. 0 m/s ) ( ? 25. 0 m/s ) 3 (1. 0 m ) + 2 (1. 0 m ) 2 4 ( 2. 0 m ) ( 2. 0 m ) 2 (5) = -297 m/s 2 Discussion ax is negative implying that fluid particles are decelerated along the centerline of the diffuser, even though the flow is steady.Because of the parabolic nature of the velocity field, the acceleration is zero at the entrance of the diffuser, but its magnitude increases rapidly downstream. 4-9 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics Flow Patterns and Flow Visualization 4-22C Solution We are to define streamline and discuss what streamlines indicate. Analysis A streamline is a curve that is everywhere tangent to the instantaneous local velocity vector.It indicates the instantaneous direction of fluid motion throughout the flow field. Discussion If a flow field is steady, streamlines, pathlines, and st reaklines are identical. 4-23 Solution For a given velocity field we are to generate an equation for the streamlines. Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. The steady, two-dimensional velocity field of Problem 4-15 is Analysis V = ( u , v ) = (U 0 + bx ) i ? byj Velocity field: (1) For two-dimensional flow in the x-y plane, streamlines are given by Streamlines in the x-y plane: dy ? v = dx ? along a streamline u (2) We substitute the u and v components of Eq. 1 into Eq. 2 and rearrange to get dy ?by = dx U 0 + bx We solve the above differential equation by separation of variables: dy dx = by ? U 0 + bx Integration yields 1 1 1 ? ln ( by ) = ln (U 0 + bx ) + ln C1 b b b (3) where we have set the constant of integration as the natural logarithm of some constant C1, with a constant in front in order to simplify the algebra (notice that the factor of 1/b can be removed from each term in Eq. 3). When we recall that ln(ab) = lna + lnb, and that –lna = ln(1/a), Eq. 3 simplifies to Equation for streamlines: y= CU 0 + bx ) ( (4) The new constant C is related to C1, and is introduced for simplicity. Discussion Each value of constant C yields a unique streamline of the flow. 4-10 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-24E Solution For a given velocity field we are to plot several streamlines for a given range of x and y values. 3 Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis From the solution to the previous problem, an equation for the streamlines is 1 Streamlines in the x-y plane: y= C (U 0 + bx ) (1) y0 (ft) Constant C is set to various values in order to plot the streamlines. Several streamlines in the given range of x and y are plotted in Fig. 1. The directi on of the flow is found by calculating u and v at some point in the flow field. We choose x = 1 ft, y = 1 ft. At this point u = 9. 6 ft/s and v = –4. 6 ft/s. The direction of the velocity at this point is obviously to the lower right. This sets the direction of all the streamlines. The arrows in Fig. indicate the direction of flow. Discussion -1 -2 -3 0 1 2 3 x (ft) 4 5 The flow is type of converging channel flow. FIGURE 1 Streamlines (solid blue curves) for the given velocity field; x and y are in units of ft. 4-25C Solution We are to determine what kind of flow visualization is seen in a photograph. Analysis Since the picture is a snapshot of dye streaks in water, each streak shows the time history of dye that was introduced earlier from a port in the body. Thus these are streaklines. Since the flow appears to be steady, these streaklines are the same as pathlines and streamlines. DiscussionIt is assumed that the dye follows the flow of the water. If the dye is of nearly th e same density as the water, this is a reasonable assumption. 4-26C Solution We are to define pathline and discuss what pathlines indicate. Analysis A pathline is the actual path traveled by an individual fluid particle over some time period. It indicates the exact route along which a fluid particle travels from its starting point to its ending point. Unlike streamlines, pathlines are not instantaneous, but involve a finite time period. Discussion If a flow field is steady, streamlines, pathlines, and streaklines are identical. -27C Solution We are to define streakline and discuss the difference between streaklines and streamlines. Analysis A streakline is the locus of fluid particles that have passed sequentially through a prescribed point in the flow. Streaklines are very different than streamlines. Streamlines are instantaneous curves, everywhere tangent to the local velocity, while streaklines are produced over a finite time period. In an unsteady flow, streaklines distort and t hen retain features of that distorted shape even as the flow field changes, whereas streamlines change instantaneously with the flow field.Discussion If a flow field is steady, streamlines and streaklines are identical. 4-11 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-28C Solution We are to determine what kind of flow visualization is seen in a photograph. Analysis Since the picture is a snapshot of dye streaks in water, each streak shows the time history of dye that was introduced earlier from a port in the body.Thus these are streaklines. Since the flow appears to be unsteady, these streaklines are not the same as pathlines or streamlines. Discussion It is assumed that the dye follows the flow of the water. If the dye is of nearly the same density as the water, this is a r easonable assumption. 4-29C Solution We are to determine what kind of flow visualization is seen in a photograph. Analysis Since the picture is a snapshot of smoke streaks in air, each streak shows the time history of smoke that was introduced earlier from the smoke wire. Thus these are streaklines.Since the flow appears to be unsteady, these streaklines are not the same as pathlines or streamlines. Discussion It is assumed that the smoke follows the flow of the air. If the smoke is neutrally buoyant, this is a reasonable assumption. In actuality, the smoke rises a bit since it is hot; however, the air speeds are high enough that this effect is negligible. 4-30C Solution We are to determine what kind of flow visualization is seen in a photograph. Analysis Since the picture is a time exposure of air bubbles in water, each white streak shows the path of an individual air bubble.Thus these are pathlines. Since the outer flow (top and bottom portions of the photograph) appears to be ste ady, these pathlines are the same as streaklines and streamlines. Discussion It is assumed that the air bubbles follow the flow of the water. If the bubbles are small enough, this is a reasonable assumption. 4-31C Solution We are to define timeline and discuss how timelines can be produced in a water channel. We are also to describe an application where timelines are more useful than streaklines. Analysis A timeline is a set of adjacent fluid particles that were marked at the same instant of time.Timelines can be produced in a water flow by using a hydrogen bubble wire. There are also techniques in which a chemical reaction is initiated by applying current to the wire, changing the fluid color along the wire. Timelines are more useful than streaklines when the uniformity of a flow is to be visualized. Another application is to visualize the velocity profile of a boundary layer or a channel flow. Discussion Timelines differ from streamlines, streaklines, and pathlines even if the flo w is steady. 4-32C Solution For each case we are to decide whether a vector plot or contour plot is most appropriate, and we are to explain our choice.Analysis In general, contour plots are most appropriate for scalars, while vector plots are necessary when vectors are to be visualized. (a) A contour plot of speed is most appropriate since fluid speed is a scalar. (b) A vector plot of velocity vectors would clearly show where the flow separates. Alternatively, a vorticity contour plot of vorticity normal to the plane would also show the separation region clearly. (c) A contour plot of temperature is most appropriate since temperature is a scalar. (d) A contour plot of this component of vorticity is most appropriate since one component of a vector is a scalar.Discussion There are other options for case (b) – temperature contours can also sometimes be used to identify a separation zone. 4-12 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution pe rmitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-33 Solution For a given velocity field we are to generate an equation for the streamlines and sketch several streamlines in the first quadrant. Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane.Analysis The velocity field is given by V = ( u , v ) = ( 0. 5 + 1. 2 x ) i + ( ? 2. 0 ? 1. 2 y ) j (1) For two-dimensional flow in the x-y plane, streamlines are given by dy ? v = ? dx ? along a streamline u Streamlines in the x-y plane: (2) We substitute the u and v components of Eq. 1 into Eq. 2 and rearrange to get dy ? 2. 0 ? 1. 2 y = dx 0. 5 + 1. 2 x We solve the above differential equation by separation of variables: dy dx = ?2. 0 ? 1. 2 y 0. 5 + 1. 2 x > dy dx ? ? 2. 0 ? 1. 2 y = ? 0. 5 + 1. 2 x Integration yields ? 1 1 1 ln ( ? 2. 0 ? 1. 2 y ) = ln ( 0. 5 + 1. 2 x ) ? ln C1 1. 2 1. 2 1. 2 here we have set the constant of integration as the natural logarithm of some constant C1, with a constant in front in order to simplify the algebra. When we recall that ln(ab) = lna + lnb, and that –lna = ln(1/a), Eq. 3 simplifies to Equation for streamlines: y= 5 y 4 3 2 C ? 1. 667 1. 2 ( 0. 5 + 1. 2 x ) 1 The new constant C is related to C1, and is introduced for simplicity. C can be set to various values in order to plot the streamlines. Several streamlines in the upper right quadrant of the given flow field are shown in Fig. 1. The direction of the flow is found by calculating u and v at some point in the flow field.We choose x = 3, y = 3. At this point u = 4. 1 and v = -5. 6. The direction of the velocity at this point is obviously to the lower right. This sets the direction of all the streamlines. The arrows in Fig. 1 indicate the direction of flow. Discussion 6 (3) 0 0 1 2 3 4 5 x FIGURE 1 Streamlines (solid black curves) for the given velocity field. The flow appea rs to be a counterclockwise turning flow in the upper right quadrant. 4-13 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-34 Solution For a given velocity field we are to generate a velocity vector plot in the first quadrant. Scale: 6 Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis 5 y4 The velocity field is given by V = ( u , v ) = ( 0. 5 + 1. 2 x ) i + ( ? 2. 0 ? 1. 2 y ) j 3 (1) 2 At any point (x,y) in the flow field, the velocity components u and v are obtained from Eq. 1, Velocity components: u = 0. 5 + 1. 2 x 10 m/s v = ? 2. 0 ? 1. 2 y 1 0 (2) 0To plot velocity vectors, we simply pick an (x,y) point, calculate u and v from Eq. 2, and plot an arrow with its tail at (x,y), and its tip at (x+Su,y+Sv) where S is some scale factor for the vector plot. For the vector plot shown in Fig. 1, we chose S = 0. 2, and plot velocity vectors at several locations in the first quadrant. 1 2 3 4 5 x FIGURE 1 Velocity vectors for the given velocity field. The scale is shown by the top arrow. Discussion The flow appears to be a counterclockwise turning flow in the upper right quadrant. 4-35 Solution For a given velocity field we are to generate an acceleration vector plot in the first quadrant.Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis The velocity field is given by V = ( u , v ) = ( 0. 5 + 1. 2 x ) i + ( ? 2. 0 ? 1. 2 y ) j (1) At any point (x,y) in the flow field, the velocity components u and v are obtained from Eq. 1, Velocity components: u = 0. 5 + 1. 2 x v = ? 2. 0 ? 1. 2 y Scale: (2) 6 The acceleration field is obtained from its definition (the material acceleration), Acceleration components: ?u ?u ?u ?u ax = +u +v +w = 0 + ( 0. 5 + 1. 2 x )(1. 2 ) + 0 + 0 ?t ?x ?y ?z ?v ?v ?v ?v ay = + u + v + w = 0 + 0 + ( ? 2. 0 ? 1. 2 y )( ? 1. 2 ) +0 t ?x ?y ?z 5 4 y 3 2 (3) 1 0 0 where the unsteady terms are zero since this is a steady flow, and the terms with w are zero since the flow is two-dimensional. Eq. 3 simplifies to Acceleration components: ax = 0. 6 + 1. 44 x a y = 2. 4 + 1. 44 y 10 m/s2 (4) 1 2 3 4 5 x FIGURE 1 Acceleration vectors for the velocity field. The scale is shown by the top arrow. To plot the acceleration vectors, we simply pick an (x,y) point, calculate ax and ay from Eq. 4, and plot an arrow with its tail at (x,y), and its tip at (x+Sax,y+Say) where S is some scale factor for the vector plot. For the vector plot shown in Fig. , we chose S = 0. 15, and plot acceleration vectors at several locations in the first quadrant. Discussion Since the flow is a counterclockwise turning flow in the upper right quadrant, the acceleration vectors point to the upper right (centripetal acceleration). 4-14 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill C ompanies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-36 For the given velocity field, the location(s) of stagnation point(s) are to be determined.Several velocity Solution vectors are to be sketched and the velocity field is to be described. Assumptions 1 The flow is steady and incompressible. 2 The flow is two-dimensional, implying no z-component of velocity and no variation of u or v with z. Analysis (a) The velocity field is Scale: V = ( u , v ) = (1 + 2. 5 x + y ) i + ( ? 0. 5 ? 1. 5 x ? 2. 5 y ) j (1) 5 Since V is a vector, all its components must equal zero in order for V itself to be zero. Setting each component of Eq. 1 to zero, Simultaneous equations: x = -0. 421 m 4 3 u = 1 + 2. 5 x + y = 0 v = ? 0. 5 ? 1. 5 x ? 2. y = 0 y 2 We can easily solve this set of two equations and two unknowns simultaneously. Yes, there is one s tagnation point, and it is located at Stagnation point: 10 m/s y = 0. 0526 m 1 0 (b) The x and y components of velocity are calculated from Eq. 1 for several (x,y) locations in the specified range. For example, at the point (x = 2 m, y = 3 m), u = 9. 00 m/s and v = -11 m/s. The magnitude of velocity (the speed) at that point is 14. 21 m/s. At this and at an array of other locations, the velocity vector is constructed from its two components, the results of which are shown in Fig. . The flow can be described as a counterclockwise turning, accelerating flow from the upper left to the lower right. The stagnation point of Part (a) does not lie in the upper right quadrant, and therefore does not appear on the sketch. -1 0 1 2 3 4 5 x FIGURE 1 Velocity vectors in the upper right quadrant for the given velocity field. Discussion The stagnation point location is given to three significant digits. It will be verified in Chap. 9 that this flow field is physically valid because it satisfies th e differential equation for conservation of mass. 4-15 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-37 For the given velocity field, the material acceleration is to be calculated at a particular point and plotted at Solution several locations in the upper right quadrant. Assumptions 1 The flow is steady and incompressible. 2 The flow is two-dimensional, implying no z-component of velocity and no variation of u or v with z. Analysis (a) The velocity field isV = ( u , v ) = (1 + 2. 5 x + y ) i + ( ? 0. 5 ? 1. 5 x ? 2. 5 y ) j (1) Using the velocity field of Eq. 1 and the equation for material acceleration in Cartesian coordinates, we write expressions for the two non-zero components of the acceleration vector: ax = ?u ?u +u ?t ?x +v ?u ?y +w ?u ?z Scale: = 0 + (1 + 2. 5 x + y )( 2. 5 ) + ( ? 0. 5 ? 1. 5 x ? 2. 5 y )(1) + 0 10 m/s2 5 4 and ay = ?v ?v +u ?t ?x +v ?v ?y +w ?v ?z = 0 + (1 + 2. 5 x + y )( ? 1. 5 ) + ( ? 0. 5 ? 1. 5 x ? 2. 5 y )( ? 2. 5 ) + 0 3 y 2 1 At (x = 2 m, y = 3 m), ax = 11. 5 m/s2 and ay = 14. 0 m/s2. b) The above equations are applied to an array of x and y values in the upper right quadrant, and the acceleration vectors are plotted in Fig. 1. Discussion The acceleration vectors plotted in Fig. 1 point to the upper right, increasing in magnitude away from the origin. This agrees qualitatively with the velocity vectors of Fig. 1 of the previous problem; namely, fluid particles are accelerated to the right and are turned in the counterclockwise direction due to centripetal acceleration towards the upper right. Note that the acceleration field is non-zero, even though the flow is steady. 0 -1 0 1 2 3 4 5 x

Benefits of Technology

Ever since it developed, there has been controversy over whether or not mechanical inventions benefit our lives. Doug Rennie’s article, â€Å"Faxed to the Max†, says â€Å"the timesaving devices we created give us more freedom to do what we want have instead imprisoned us in our own technology. † Although this may be true to some people, everyone has the power to resist technological devices. People can chose to use these tools to enhance their lifestyles, like they can chose to use a hammer or a saw for a project; they don’t necessarily need it, but it would save time and energy.It is the ability to mistreat these inventions that makes some people uneasy towards about using technology in their daily lives. Although it can be abused in many ways, technology used properly is a tool that aids in education, health and communication. Our physical health has benefited greatly from technology. Through inventions like the pacemaker and artificial limbs, a tremend ous amount or people have had better physical conditions. This biological technology is not favourable to all. Some say that it is better for nature to do its work and people should not interfere.The spread of new inventions can soon get out of hand. For example, if we have the ability to clone humans, our lifestyles will change. People will abuse this power by being careless about their health. They believe people think technology can save them in the end, so they can exercise less and eat more unhealthy foods. On the other hand, curative inventions have saved lives and allowed people to lead healthier ones. There can be ways to make technology accessible to only certain people, but there is no reason to stop the usage of these medical products if they can help save lives.An example of one of these products is the pacemaker, invented by Canadian electrical engineer, John Hopps. Because the heart stops beating when it cools, he found that mechanical or electric stimulation could mak e the heart start up again. Since then, many were given healthier lives. Prosthetic limbs have also played a great role in helping the lives of amputees. Artificial limbs with sensors and microchips have recently been designed so that these people can maintain an active lifestyle rather than being confined to wheelchairs. Technology plays a very important role in the communication of people  today.With the touch of a button, the Internet and the cellular telephone can allow us to transmit our message to someone on the other side of the planet. Certain people chose not to use any of these appliances because of the negative consequences. The Internet contains inappropriate sites and some parents don’t want to risk the chance of their child meeting a dangerous stranger in a chat room. People are so used to e-mailing that they don’t know how to send letters anymore. Other antagonists of technological communication believe that the invention of cellular phones is prone to car accidents.Contrariwise, internet has allowed to world to communicate and to easily get in touch with someone. There are parental controls on the Internet and the phone is simply a tool that people can chose not use. In a car accident, it is more of the driver’s responsibility than the cellular phone because the driver chose to use the cellular phone. The Internet is a cheaper and quicker way to communicate. Typing an e-mail and it getting it to the other person takes only a matter of seconds, whereas writing a letter and sending it could take days.Through instant messages and chat rooms on the Internet, a person can converse with another on the other side of the world and save money on their phone bills. Cell phones are convenient devices, especially in emergencies. For instance, if someone was lost in the woods, they can either phone for help or check the map on their cell phone. Cellular phones are also timesaving tools because they can fit in pockets and there is no n eed to find a payphone or carry coins on the street. We can easily abuse the technology around us, but we can just as easily find that the consequences are more beneficial when we use it properly.We should consider the negative effects of technology and find ways to limit them, but we cannot let these issues stop us from the possibilities that the devices can offer. Many of the appliances that scientists created have improved our communication with the world, easier and faster learning, and have even extended the life spans of numerous people. It is wrong to say that everyone has been trapped into being time-sick patients of modern devices. We are simply using these tools to our advantage so we can accomplish the numerous tasks in our lives easier. Without any technology at all, we will be taking a step back from mankind.