Jesus Christ

Jesus Christ was born about 6 B. C. and was crucified about 29 A. D. He was the founder of Christianity. Jesus was born in Bethlehem, a village in Palestine five miles (8km) south of Jerusalem. Palestine was under Roman domination with Herod the Great as its puppet ruler. There is no way of reckoning Jesus’ exact birth date. The Christian calendar, devised in the sixth century, places the Nativity in the year 1 A. D. Most calendars, however, believe that Herod died in 4 B. C. and that Jesus was born two or three years earlier.The nativity has been celebrated on December 25 since the fourth century A. D. Luke tells of the decree of Emperor Augustus that required everyone to go to his native village to be taxed. Mary and her husband Joseph had to go from their home in Nazareth, Galilee, to Bethlehem, Judea. Jesus was born in a stable because there was no room in the inn. A manger (a feeding trough for livestock) served as a crib. Luke reports that shepherds came to worship him, and Matthew tells of the wise men from the east, who saw a brilliant star in the sky and came to pay homage to him.Herod, alarmed by the birth of a child described by the wise men as â€Å"King of the Jews,† ordered the massacre of all young children in or near Bethlehem in an attempt to kill Jesus. Joseph and his family escaped by fleeting to Egypt. After Herod died Joseph took his family back to Nazareth. The parents were devout Jews. Jesus received careful religious instruction and spent much time in studying the Scriptures (Old Testament). His language was Aramaic, a Semitic tongue. Joseph was a carpenter, and Jesus probably learned the trade. Mark (6:3) reports that Jesus had four brother and these were James, Joseph, Judas, and Simon.

Locating Tetrahedral and Octahedral Voids Essay

The close packed structures have both octahedral and tetrahedral voids. In a ccp structure, there is 1 octahedral void in the centre of the body and 12 octahedral void on the edges. Each one of which is common to four other unit cells. Thus, in cubic close packed structure. Octahedral voids in the centre of the cube =1 Effective number of octahedral voids located at the 12 edge of = 12 Ãâ€" 1/4 = 3∠´ Total number of octahedral voids = 4 In ccp structure, there are 8 tetrahedral voids. In close packed structure, there are eight spheres in the corners of the unit cell and each sphere is in contact with three groups giving rise to eight tetrahedral voids Circles labelled T represent the centers of the tetrahedral interstices in the ccp arrangement of anions. The unit cell â€Å"owns† 8 tetrahedral sites. Circles labelled O represent centers of the octahedral interstices in the ccp arrangement of anions (fcc unit cell). The cell â€Å"owns† 4 octahedral sites. Illustration 16. In a solid, oxide ions are arranged in ccp. Cations A occupy one – sixth of the tetrahedral voids and cations B occupy one third of the octahedral voids. What is the formula of the compound? Solution: In ccp with each oxide there would be 2 tetrahedral voids and one octahedral voids 1/3rd octahedral voids is occupied by B and 1/6th tetrahedral void by A. Therefore the compound can be Illustration 17. In a crystalline solid, having formula AB2O4, oxide ions are arranged in cubic close packed lattice while cations A are present in tetrahedral voids and cations B are present in octahedral voids .

5 Strategies for Tackling SAT Subject Tests

You’ve probably heard plenty about the SAT. But SAT subject tests may be less familiar. What exactly are they? Should you take them? Want tips on how to prepare? Read on to learn what you can expect from the tests and our advice for how to do achieve high scores. Formerly known as the SAT II, an SAT subject test covers your specific knowledge of a given subject area, as opposed to the SAT, which covers general knowledge of only English and math. There are 20 individual subject tests across math, science, English, history, and foreign languages, and you may choose whichever tests you wish to take based on your individual strengths and knowledge areas. However, do keep in mind that if you are applying for certain majors or programs at some schools, you may be required to submit results from specific subject tests. SAT subject tests are administered six times per year, but not all subjects are offered in each sitting. Visit the CollegeBoard’s website to find out when a given subject is offered. You may take 1-3 subjects in a single test sitting, and can add or opt out of tests on the day of with some limitations . However, you may not take subject tests on the same day you take the SAT. The most important reason why you should take SAT subject tests is that many colleges require them as part of the admissions process. To find out which schools do, check out our list . Even colleges that don’t require them may still consider your results if you choose to take them, so they could give you an edge if you perform well. This is a good opportunity to demonstrate your strengths, since you’re able to choose your subjects, and could even compensate for weaker SAT scores, since you can show your aptitude in your strongest areas. If you’re self-taught or have studied certain subjects independently, SAT subjects tests can demonstrate that you’re a natural learner and seek out opportunities outside the classroom. Plus, your results can be used to determine placement or can be applied for credit at some colleges. Our students see an average increase of 250 points on their SAT scores. Remember: this is a chance to demonstrate your strengths in a particular area. Unlike the SAT, subject tests are more closely linked to the content you’ve studied in school. Therefore, if you’ve studied and done well in a particular subject, then chances are good that you’ll also do well on that particular SAT subject test. Of course, make sure to choose the subjects in which you perform best. For example if you’re a math whiz, opt for Math II. If you want another chance to prove your English prowess, take Literature. Also be sure to look into any requirements for the colleges to which you’re applying, since some programs may require specific tests. Make note of when specific tests are offered before you plan your schedule, and remember that you can’t take subject tests on the same day as the SAT. It’s a good idea to take the subject test around the same time as the AP for a given subject, since you’ll effectively be studying for both tests at the same time and will only have to do it once. That way, the material will still be fresh in your mind. You should also visit the CollegeBoard page for each test and consult the recommended preparation section. For instance, for Math II , CollegeBoard recommends â€Å"more than three years of college-preparatory mathematics, including two years of algebra, one year of geometry, and elementary functions (precalculus) or trigonometry or both.† You will also see a list of topics covered and anticipated skills you will need on this page, so make sure you’re familiar with these before you sit for the test. The good news is that there are plenty of resources available for practicing. It’s important to make use of them, because these tests are based on your knowledge of given subjects as opposed to rote skills. Here are a few resources: For every standardized test you take, you should come prepared with certain materials, such as pencils and your ID. But some SAT subject tests require additional materials. For instance, if you take a language with listening test, you will need a CD player. You are also allowed to use a calculator for both math tests, but be sure to make sure your brand and type is permitted. Also pay attention to materials you’re not allowed to bring. For example, you can’t use a calculator for Physics. Like the SAT, subject tests are scored on a 200-800 scale, with your percentile corresponding to a score between 200-800. You will also see your subscores, which are scaled between 20-80, for language with listening tests. All questions are multiple choice, and you lose a portion of a point for an incorrect answer. You do not lose credit for questions you do not answer. To learn more about scoring protocols, read CollegeBoard’s page, How SAT Subject Tests Are Scored . SAT subject tests are a great way to demonstrate your knowledge of certain areas and your strengths in the classroom. Even if the schools to which you are applying don’t require subject tests, they can still strengthen your application, and may help you receive credit or higher placement at some colleges. If you choose to take SAT subject tests, make sure to do your research and become familiar with the materials you need to know beforehand. Be sure to use the resources available to you to ensure you do your best. Looking for some more help for acing the SAT? The SAT Tutoring Program will help you achieve top scores on your test. We’ll pair you with two private tutors, one for reading, English, and writing, and one for math. All of our tutors have scored in the 99th percentile on the section they are teaching and are chosen based on teaching skills and ability to relate to their students.

Lessons learned from public health Essay Example | Topics and Well Written Essays - 750 words

Lessons learned from public health - Essay Example A major reason for my enriching and fulfilling experience with these modules was the instructor. The instructor gave us enough materials to guide both instructed and self study. During my spare time, I revised and studied the course materials to gain a deeper comprehension of what the course demands and balance it with what the instructor required. I also engaged other students in discussions and debates on the subject so that we could learn new concepts, eliminate our weaknesses and polish areas of strength. I found it particularly refreshing that during group discussions, each student had a fresh insight to offer. This has taught me that no matter the deficiencies, all individuals have a contribution to make to the development of knowledge. As I progress to other modules, I will bear this lesson in mind during academic and nonacademic interactions with students, teachers, and in my social life. I must say that the learning style models were the most exciting part of the modules. I found the Myers-Briggs Type Indicator (MBTI) and the Herrmann Brain Dominance Instrument to be very insightful and revealing. Using these models, teachers can develop effective ways of interacting with students and maximizing the potential of each student. These models can also be used by instructors to create blueprints for supporting below-average students in specific subjects. It is worth noting that each model has unique attributes.

Psychology (Clinical depression) Essay Example | Topics and Well Written Essays - 1000 words

Psychology (Clinical depression) - Essay Example (Robinson, Berman, Neimeyer, 1990) Depression can appear at any age, but it is prevalent now among children and adolescents, especially when they are the offspring of adults with depression. It can also strike late in life and its symptoms of impaired memory, slow speech and slowed movement can be mistaken for those of senility or stroke. Psychological treatment of depression assists the depressed individual in several ways. First of all, it helps to ease the pain of depression with addressing the feelings of hopelessness that go along with depression. That supportive counseling changes the pessimistic ideas, unrealistic expectations, and critical self-evaluations that create depression and sustain it. A qualified therapist is to help the depressed person to recognize which life problems are critical, and which are minor, and to develop positive life goals, and a more positive self-assessment. At last, the problem solving therapy changes the areas of the person's life that are creating significant stress, and contributing to the depression. Here can be a good impact of behavioral therapy that is aimed to develop coping skills, or an impact of interpersonal therapy that assist in solving relationship problems. (Weissman M.M., Markowitz J.C., 1994) Unfortunately, in practice, many poorly trained therapists can't provide supportive counseling, that leads to little improvements in treating depression. Psychotherapy is said to be helpful while being a part of the overall treatment plan, because, as researches show that in many cases unless the depressed person with the assistance of a therapist makes critical life changes, the depression still continues. This is explained by the fact of existence of internal and external changes, when internal changes are connected with problem assessment, self-evaluation, the evaluation of others, and the expectations the depressed person has for himherself, others and about life. It is an internal factor of hisher personality where there is a need to be fastidious with oneself. External changes in this case may be needed in problem solving skills, stress management, communication skills, life management skills, and the skills needed to develop and sustain relationships. (Carlson, Martin and Buskwist, 2004) They are connected with the system 'person - environment', and development of one's interaction skills with the surrounding. The length of treatment will vary due to the severity of depression, the number and kind of life problems the person has to be solved. Many people tend to experience some relief within 6 or 10 sessions, while approximately 70-80% of the treated notice improvement only within 20-30 sessions. (Simon G.E., VonKorff M., Wagner E.H., Barlow W., 1993) Treatment sessions are usually scheduled once per week, that can be a rather unsatisfactory point for those who suffer from constant problems and need to be assisted more often. Many psychiatrists now doubt if therapy and cognitive training are as effective as dugs in the treatment of depression.

Costing and Information Systems of the Worplestrop Essay - 1

Costing and Information Systems of the Worplestrop - Essay Example Product costs are identified with goods produced or purchased for resale in an organization. In marginal costing, the only variable cost is charged as a cost of sale and we get the contribution is which normally is given by sales revenue less the variable cost of product sold. The closing stock of goods which are work in progress or finished goods in the store is valued at variable production cost. Fixed cost are normally treated as period cost and are charged in full to the profit and loss accounts of the accounting period in which they are incurred. The principle of marginal costing of a product is that in a particular period, fixed cost will always be the same, for any volume of sales and production provided that the level of activity is within the relevant range or the budgeted range, therefore, by selling an extra item of product or service the following will happen. Similar if the volume of sale, if the volume of sales falls by one item, profit will fall by the same amount of contribution earned from the sale of items. Profit measurement should, therefore, base on analysis of total contribution. Since fixed cost relates to a period of time, and do not change with either increase or decrease in the sales volume, it is misleading to charge units of sales with a share of the fixed cost from the total contribution of the period to get the profit figure. When one unit of a commodity is produced then an extra cost must be incurred in its production as a variable cost but the fixed cost will always remain constant. Then it is true to say that it is in order to value the closing stock as a variable cost. It is sometimes called full costing, in this costing, all the manufacturing cost incurred in any particular period is accounted for, and also until the product is sold fixed overheads will remain a product cost.

Five approaches to qualitative research Coursework

Five approaches to qualitative research - Coursework Example This research problem focuses on the way of life of the members of the skin bleaching community, a community wherein a White complexion is almost an obsession. The purpose of this research is to observe and interview several members of this community and identify the practices that are distinctive to them. The ethnographer will observe the daily activities of this community and determine the underlying reasons for their practices. The ethnographic method enables a researcher to have a direct and personal look with the culture s/he wants to study. It puts the researcher at the center of the study, normally letting them to take part in the community or culture they aim to understand. The array of topics for ethnographic study is vast. Ethnographers can observe or examine the extraordinary or the ordinary. Since ethnography gives room for the researcher’s subjectivity, it puts the researcher in an exclusive, distinctive position where expressive interpretations and expression of emotions are allowed. The purpose of field research is to observe, understand, and interact with individuals in their normal environment. Field research involves observation in the natural or actual setting and of the daily lives of the individual under study. Field research may be described as a blanket term that involves the numerous tasks that field researchers carry out when they gather data—participation, observation, interview, and analysis of artifacts or documents produced by the individuals they study. An example of a field research question is: what are the different factors that influence the motivation and job satisfaction of employees in financial institutions? This research problem relates to the various factors (e.g. salary, benefits, interpersonal relations, etc.) that affect the job performance of employees in financial institutions, such as banking organizations. The purpose of this research is to directly

Fundamentals of Communication Assignment Example | Topics and Well Written Essays - 500 words

Fundamentals of Communication - Assignment Example The ethical issues should be analyzed well in advance when one understands the ideology behind the communication understandings because ethics is something that plays a very tacit role within any society of the world. This concept of the ethical issues within the book has been covered very appropriately. This is the reason why I believe it should be mentioned duly within this paper. Also this ethical issue construct needs to be given emphasis because it is necessary to speak in such terms to achieve what the eventual goals are for the organization and how it wishes to expand its network across the different domains and dimensions of the society. The need is to reach the masses and let the message be known easily so that there are no problems encountered by the people who remain a very significant part of the society nonetheless (Alberts, Nakayama & Martin, 2009). When the communication regimes are based on the basic element of dealing in an ethical manner with one another, the success is indeed guaranteed. This will be highlighted in the different zones where ethics has been employed and where it should be made use of to make sure that communication remains significant at the end of the day. It i s in essence a fundamental part of any communication setting which has come about with the passage of time, and which shall remain more important in the changing eras. What is required here is an understanding of the different ethical issues which come about in the discussion of communication. This is because communication feels that it is incomplete if ethical understandings are not reached upon quite clearly. This will set the basis for finding out other issues and concerns within the communication mechanisms. But if the ethical issues are properly gauged and realized, then the communication concerns will be dealt with in a proper manner. The need of the hour

Self-Evaluation Paper Essay Example | Topics and Well Written Essays - 3250 words

Self-Evaluation Paper - Essay Example However, this way of solving dispute has been criticized for being expensive, time-consuming, and in some cases leading to unfair and unjust outcomes regarding particular cases (Frenkel and James 22). As a result, alternative dispute resolution mechanisms have been devised in order to provide an economical, quicker and a more just and fair outcomes to parties in dispute. Mediation is one of the main alternative dispute resolution mechanisms. Mediation is a confidential and voluntary type of alternative dispute resolution. This method involves an impartial and independent person(s) who helps two or more parties to reach a solution over a dispute which is acceptable to all the parties. It may involve talking to the parties separately or together. Often, mediators do not make judgments, instead they ask questions in an attempt to unravel underlying problems relating to the dispute, help the parties in understanding issues and clarifying the alternatives for resolution of their dispute ( Goodman 51). Principles of Mediation Being a critical alternative dispute resolution mechanism, mediation is guided by a number of principles that ensures that it achieves its objectives effectively. There are five principles of mediation which are generally- recognized that guides the mediation process. However, it should be noted that there are more other elements or principles that have been incorporated, but in a broader sense they still fall under the five which are generally- recognizable (Frenkel and James 46). The first principle is the principle of informed consent that requires that prior to consenting participation in mediation, the parties in dispute should be informed about their legal options and rights, as well about the mediation process (Wall and Dunne 217-8). The second principle is that of voluntariness which requires that mediation process should be voluntary and that parties should enter into it freely and in any outcome resulting from the process. The parties a re also free to end their participation in the process at any time. The third principle is that of confidentiality which seeks to make parties aware that the process is confidential. This enables them to freely explore options and speak without fear that their communications in the process might be used against them. Also, confidentiality enables a party in dispute to request that particular information disclosed to the mediator not to be shared with the other parties (Frenkel and James 49). In addition, impartiality is a mediation principle which dictates that the parties are entitled to impartial and fair process involving an impartial and neutral mediator. The final principle is that of self-determination that requires that the parties and not the mediator to define the scope of issues to be addressed at mediation, and that the parties will determine the mediation’s outcome or results (Goodman 105-6). In order to understand the concept of mediation clearly and better, this self- evaluation paper will; consider the evaluation of my performance as a mediator. The paper will include the agreement reached by the parties and discuss the lessons learnt from the process that are deemed to be helpful for the future. I acted as a mediator between these two parties: Sammy Atwater (Plaintiff) and James

Strategic management Essay Example | Topics and Well Written Essays - 4000 words - 4

Strategic management - Essay Example Henry Ford founded the company with many innovations introduced to the ways products were manufactured. One of these is the best-known model T of mass production, the moving assembly line – which is actually composed of conveyor belts – where time of work was reduced (Goh & Garg, 2008, p. 57). Ford revolutionized the car industry, paid higher wages to factory workers, and made cars affordable to anyone. (Purvis, 1997) Ford became a multinational corporation in 1970 but was predominantly operating in North America with subsidiaries in major markets in countries like Britain, Germany or Australia. These subsidiaries however have their own manufacturing plants. In Britain, the most popular Ford in the Cortina. With intense globalisation, the Ford organisation started to restructure internationally. In Europe, Ford was consolidated with further product development and designs which were originally European. Ford manufactured cars of different designs. (Grant, 2005, p. 440) In the UK, Ford is the biggest in the automobile industry with over 550 dealers across the United Kingdom and employing about 35,000 people. In its website, it says that it currently sells 440,000 vehicles annually, with Ford Focus as the car most Britons love. (Ford Motor Company, 2010) In the United States, Ford is now best known for its cars, trucks, crossovers and SUVs (Ford, 2010). In 2008, Ford was adjudged by Standard and Poor’s as the world’s second largest motor vehicle manufacturer, producing cars and trucks, including plastic and glass parts of the cars they make, and including replacement parts. Ford has a 33% stake in Mazda Motor Corp. Financial services include Ford Motor Credit (automotive financing and insurance) and American Road Insurance. Ford has a big share in the world market. It has ventured in many countries, trying to feel its presence even in China, which is the fastest growing market in the automobile

The Game of Pig Portfolio Essay Example for Free

The Game of Pig Portfolio Essay To play the game you need a regular 6-sided die. Each turn of the game consists of one or more rolls of the die. You keep rolling until you get a 1 or decide to stop. You may choose to stop rolling at any time. If you stop rolling before you get a 1 your score for that turn, your score is the sum of all of the numbers you rolled that turn (e. g. if you roll a 4, 6, 3, and 2 your score is 15 pts. ). But, if you roll a 1 in that turn your turn is over and you receive 0 pts. for that turn. The central problem in the unit is finding strategies for several different probability games. The key ideas were developed in the unit through the presentation of many games and variations of those games to us. The key ideas helped us to solve the central problem by giving us many opportunities to learn how to solve strategies for probability games. In each different game there was a new game or variation in which we could find a new way to find optimal game strategies for probability games. Why I chose each item: Homework 27: The Pig and I: This piece of work begins to explains the basics of what we learned in the unit. Homework 28: Beginning Portfolio Selection: The pig and I: This explains the main points of the unit (Theoretical and experimental probability) and also begins to show the reasons to choose certain pieces of work. Spinner Give and Take: This was chosen because it shows the basics of theoretical probability in great detail. The Game of Little Pig: This was chosen to show the many variations of games we learned to play, and the beginning steps of finding a strategy. POW 7: Make a Game: This was chosen to show how to make a game based off of the basic principles of probability games, it also shows how to make strategies for new games. Homework 7: Portraits of Probabilities: This was chosen as a basic block that began the whole unit and how to show probability in many different models. Homework 9: Rollin’ Rollin’ Rollin’: This was chosen to show the basics of dice probability, a very important section of this unit. Homework 11: Two Dice Sums and Products: This was chosen to show the many types of probabilities in dice probability, it also shows how to find what is most likely to happen. POW #4: A Sticky Gum Problem: This was chosen to show in depth theoretical probability. POW #6: Linear Nim: This was chosen to show in depth experimental probability, as well as strategies to find the best strategies to win probability games. It was also chosen to display my best work. Homework #25: Should I go on? : This was chosen to show how to make decisions in probability games as well as to display some of my better very thorough work. Personal Growth: For me I feel that the core ideas of this unit will make me think twice on my strategies in probability games before playing using them. I learned a lot about probability in this unit; it’s not as simple as I thought. There are many calculations you must learn how to do and there is obviously so much more behind the strategies and probabilities that I am yet to discover, for example, statistics class. Best Strategy: The best strategy is to get to 20 points and end your turn. This should take roughly 5 rolls of the dice seeing as the expected value per roll is 4. Four multiplied by five is 20. You should take 5 rolls because after five rolls the expected value of points begins to go up by smaller and smaller margins.

Week 5 â€Finances Essay Example for Free

Week 5 –Finances Essay Debate on profit maximization ethics and corporate social responsibility: Traditionally, the duty of company’s management is to improve the financial welfare of the shareholders of the company by maximum of profits provided it is under the law. This is the canonical law and economics account (Elhauge: 2004) Companies are therefore liable for any act(s) that causes more that usual harm under an independent law. However if the operations do not cause any undesired effect, then, it is socially acceptable that the company maximizes profits. The management of companies is therefore required to consider the interests of other stakeholders in their daily operations. This is as a result of the law that was enacted at the height of corporate takeovers during the 1980s. This law could however be construed to mean that the management of corporations only consider the interests of others only if doing so enhances the company’s profits. Shareholders and management have no legal duty to maximize profit although they have a legal discretion to forego on profits in the public interest. (Elhauge: 2004) According to Manuel Costello Branco and Lucia Lima Rodriquez companies only engage in corporate social responsibility if they are set to gain from such an undertaking. (Brancho Rodriguez: 2007) Corporate social responsibility may include environmental protection, human resource management, health and safety at work, relation with community and with suppliers and customers. (Branco Rodriguez: 2007) CSR boarders on ethics and management should therefore consider the impacts of their activities on the various stakeholders. It is also considered to be a competition strategy which can give a company competitive advantage. (Branco Rodriguez: 2007) This debate on CSR is on shareholders-stakeholders point of view where shareholders view is that management should maximize their wealth while the stakeholders view is towards all stakeholders (Friedman 1998, Jensen 2001) The classical view of business encompasses purely economic basis and constrained profit making views. This is the shareholders view. The stakeholders view on the other hand is of socially aware business where corporations are sensitive to needs if other stakeholders (Lantos: 2001) Companies should therefore, not ignore the interests of other stakeholders if doing so could impact negatively on the company’s intention of maximizing shareholders wealth. (Stern berg 1997, Jensen 2001) Ethics basically is what constitutes right or wrong behavior in business in terms of operations and situations happening in companies. In the daily operations of companies many unethical actions and decisions are made. Corporate citizenship concept is propagated by the society where businesses promote goals that they view as important while at the same time solving social problems thus rejecting the idea of profit maximization and law compliance. The results of an activity rather than the activity itself are what determine whether an action is ethical or not. Clarkson further elaborates that an actions is morally right if it generates the greatest amount of good to many people. (Utilitarian theory) (Clarkson: 1995) It is evident therefore that the concept of ethics is controversial in the sense that there are conflicting positions as to what constitutes what is morally right or wrong as shown by Kantian ethics and Utilitarian ethics. (Hymson 2007) In business ethics, because of competition, actions of one company e.g. adoption of lower prices leads to the other companies adopting the same pricing strategy and hence business ethics tend to be uniform (Hymson 2007) Monopolistic businesses where there is no competition can elect to apply personal ethics. But the cost of following personal ethics is borne by the employees. Hence argument gives credence to the idea that businesses only social responsibility is t maximize profits (Friedman: 1998). Outside business ethics, CSR is usually all about making profits. Government regulations The government should take charge and ensure that companies do not undertake business practice that cause undesirable effects on the community and therefore it should pass laws and regulations that guide corporations in the business practice (Reich 2007). The stiff competition that many corporations in the world today face makes them focus more on ways of making more profits and therefore the need of government regulations to protect the environment, consumers and even the employees. Robert Reich further alludes to the fact that corporations cannot be moral or immoral and can only be responsible if publicly held by their shareholders. The shareholders interest is to maximize their profits and therefore companies should do public good in their quest to maximize their profits (Reich 2007) Reasons for government regulations An example of company that could justify the government regulations is Wal-Mart. This corporation on one hand spends money on CSR projects but on the other it is against employees union, pays low wages with minimum benefits. The company also is against living wage initiatives. Wal-Mart has also been accused of forcing employees to perform overtime duties, sex and race discrimination and a whole lot of other things. The World com and Enron scandals are other examples that call for government regulations on CSR and business ethics. The list is endless, Adelphi, Tyco, Computer associates. All these cases touch on the subject of business ethics. Business ethics violations can also lead to illegalities as exemplified by the Enron and World com cases (Hymson 2007) References Hymson, E.B (2007) Law: The force that harmonizes business ethics with profit maximization. Retrieved on 4/3/2008 from http://www.salsb.org/slj/vol-xv/14hymson.pdf Elhauge, E. (2004) Sacrificing corporate profits in the public interest. Retrieved on 4/3/08 from http://www.law.harvard.edu/programs/olin_center/corporate_governance/papers/04.Elhauge.sacrificing-corporate-profits.pdf. Branco, M.C and Rodriguez L. L (2007). Positioning stakeholders’ theory within the debate on corporate social responsibility (Vol 12. No. 1). Retrieved on 4/3/2008 from http://www.salsb.org/slj/vol-xv/14hymson.pdf Friedman. (1998), â€Å"The social responsibility of business is to increase its profits†, in Pincus, L.B. (Ed.), Perspectives in business ethics, McGraw-Hill, Singapore Jensen, M.C. (2001) â€Å"Value Maximization, Stakeholder Theory, and the Corporate Objective Function†, Journal of Applied Corporate Finance, Vol.14 No.3 Sternberg. (1997) â€Å"The Defects of Stakeholder Theory†, Corporate Governance, Vol 5 No.1. Lantos, G.P. (2001) â€Å"The boundaries of strategic corporate social responsibility†, Journal of Consumer Marketing, Vol.18 No.7 Clarkson, M.B.E. (1995) â€Å"A Stakeholder Framework for Analyzing and Evaluating Corporate Social Performance†, Academy of Management Review, Vol.20 No.1 Robert Reich, R (2007). Super capitalism. The Transformation of Business, Democracy and Everyday Life.

A Traditional Definition Of Leadership Management Essay

A Traditional Definition Of Leadership Management Essay A traditional definition of leadership: Leadership is an interpersonal influence directed toward the achievement of a goal or goals. Three important parts of this definition are the terms interpersonal, influence, and goal. Â · Interpersonal means between persons. Thus, a leader has more than one person (group) to lead. Â · Influence is the power to affect others. Â · Goal is the end one strives to attain. Basically, this traditional definition of leadership says that a leader influences more than one person toward a goal. The definition of leadership used in this course follows. LEADERSHIP is a dynamic relationship based on mutual influence and common purpose between leaders and collaborators in which both are moved to higher levels of motivation and moral development as they affect real, intended change. (Kevin Freiberg and Jackie Freiberg, NUTS! Southwest Airlines Crazy Recipe for Business and Personal Success, Bard Press, 1996, p. 298) Three important parts of this definition are the terms relationship, mutual, and collaborators. Relationship is the connection between people. Mutual means shared in common. Collaborators cooperate or work together. This definition of leadership says that the leader is influenced by the collaborators while they work together to achieve an important goal. Leadership versus Management A leader can be a manager, but a manager is not necessarily a leader. The leader of the work group may emerge informally as the choice of the group. If a manager is able to influence people to achieve the goals of the organization, without using his or her formal authority to do so, then the manager is demonstrating leadership. According to John P. Kotter in his book, A Force for Change: How Leadership Differs From Management (The Free Press, 1990), managers must know how to lead as well as manage. Without leading as well as managing, todays organizations face the threat of extinction. Management is the process of setting and achieving the goals of the organization through the functions of management: planning, organizing, directing (or leading), and controlling. A manager is hired by the organization and is given formal authority to direct the activity of others in fulfilling organization goals. Thus, leading is a major part of a managers job. Yet a manager must also plan, organize, and control. Generally speaking, leadership deals with the interpersonal aspects of a managers job, whereas planning, organizing, and controlling deal with the administrative aspects. Leadership deals with change, inspiration, motivation, and influence. Management deals more with carrying out the organizations goals and maintai ning equilibrium. The key point in differentiating between leadership and management is the idea that employees willingly follow leaders because they want to, not because they have to. Leaders may not possess the formal power to reward or sanction performance. However, employees give the leader power by complying with what he or she requests. On the other hand, managers may have to rely on formal authority to get employees to accomplish goals. Trait Theories In the 1920s and 1930s, leadership research focused on trying to identify the traits that differentiated leaders from non-leaders. These early leadership theories were content theories, focusing on what an effective leader is, not on how to effectively lead. The trait approach to understanding leadership assumes that certain physical, social, and personal characteristics are inherent in leaders. Sets of traits and characteristics were identified to assist in selecting the right people to become leaders. Physical traits include being young to middle-aged, energetic, tall, and handsome. Social background traits include being educated at the right schools and being socially prominent or upwardly mobile. Social characteristics include being charismatic, charming, tactful, popular, cooperative, and diplomatic. Personality traits include being self-confident, adaptable, assertive, and emotionally stable. Task-related characteristics include being driven to excel, accepting of responsibilit y, having initiative, and being results-oriented. Trait theories intended to identify traits to assist in selecting leaders since traits are related to leadership effectiveness in many situations. The trait approach to understanding leadership supports the use of tests and interviews in the selection of managers. The interviewer is typically attempting to match the traits and characteristics of the applicant to the position. For example, most interviewers attempt to evaluate how well the applicant can work with people. Trait theory has not been able to identify a set of traits that will consistently distinguish leaders from followers. Trait theory posits key traits for successful leadership (drive, desire to lead, integrity, self-confidence, intelligence, and job-relevant knowledge) yet does not make a judgment as to whether these traits are inherent to individuals or whether they can be developed through training and education. No two leaders are alike. Furthermore, no leader possesses all of the traits. Comparing leaders in different situations suggests that the traits of leaders depend on the situation. Thus, traits were de-emphasized to take into account situational conditions (contingency perspective). Behavioral Theories The behavioral theorists identified determinants of leadership so that people could be trained to be leaders. They developed training programs to change managers leadership behaviors and assumed that the best styles of leadership could be learned. Theory X and Theory Y Douglas McGregor described Theory X and Theory Y in his book, The Human Side of Enterprise. Theory X and Theory Y each represent different ways in which leaders view employees. Theory X managers believe that employees are motivated mainly by money, are lazy, uncooperative, and have poor work habits. Theory Y managers believe that subordinates work hard, are cooperative, and have positive attitudes. Theory X is the traditional view of direction and control by managers. 1. The average human being has an inherent dislike of work and will avoid if he or she can. 2. Because of this human characteristic of dislike of work, most people must be controlled, directed, and threatened with punishment to get them to put forth adequate effort toward the achievement of organizational objectives. 3. The average human being prefers to be directed, wishes to avoid responsibility, has relatively little ambition, wants security above all. Theory X leads naturally to an emphasis on the tactics of control to procedures and techniques for telling people what to do, for determining whether they are doing it, and for administering rewards and punishment. Theory X explains the consequences of a particular managerial strategy. Because its assumptions are so unnecessarily limiting, it prevents managers from seeing the possibilities inherent in other managerial strategies. As long as the assumptions of Theory X influence managerial strategy, organizations will fail to discover, let alone utilize, the potentialities of the average human being. Theory Y is the view that individual and organizational goals can be integrated. 1. The expenditures of physical and mental effort in work are as natural as play or rest. 2. External control and the threat of punishment are not the only means for bringing out effort toward organizational objectives. 3. Commitment to objectives is a function of the rewards associated with their achievement. 4. The average human being learns, under proper conditions, not only to accept but also to seek responsibility. 5. The capacity to exercise a relatively high degree of imagination, ingenuity, and creativity in the solution of organizational problems in widely, not narrowly, distributed in the population. 6. Under the condition of modern industrial life, the intellectual potentialities of the average human being are only partially utilized. Theory Ys purpose is to encourage integration, to create a situation in which an employee can achieve his or her own goals best by directing his or her efforts toward the objectives of the organization. It is a deliberate attempt to link improvement in managerial competence with the satisfaction of higher-level ego and self-actualization needs. Theory Y leads to a preoccupation with the nature of relationships, with the creation of an environment which will encourage commitment to organizational objectives and which will provide opportunities for the maximum exercise of initiative, ingenuity, and self-direction in achieving them. Ohio State and University of Michigan Studies conducted at the Ohio State University and the University of Michigan identified two leadership styles and two types of leader behaviors. The Ohio State study identified two leadership styles: considerate and initiating structure. The University of Michigan study classified leaders behaviors as being production- or employee-centered. The primary concern of leaders with considerate and employee-centered style is the employees welfare. The primary concern of leaders with initiating-structure and production-centered styles is achieving goals. Research findings on which dimension is most important for satisfaction and productivity are inconclusive. However, employee oriented leaders appear to be associated with high group productivity and job satisfaction. University of Iowa Another approach to leader behavior focused on identifying the best leadership styles. Work at the University of Iowa identified democratic (participation and delegation), autocratic (dictating and centralized) and laissez-faire styles (group freedom in decision making). Research findings were also inconclusive. The Managerial Grid The dimensions identified at the University of Michigan provided the basis for the development of the managerial grid model developed by Robert Blake and Jane Mouton. It identifies five various leadership styles that represent different combinations of concern for people and concern for production. Managers who scored high on both these dimensions simultaneously (labeled team management) performed best. The five leadership styles of the managerial grid include impoverished, country club, produce or perish, middle-of-the road, and team. The impoverished style is located at the lower left-hand corner of the grid, point (1, 1). It is characterized by low concern for both people and production. The primary objective of the impoverished style is for managers to stay out of trouble. The country club style is located at the upper left-hand corner of the grid, point (1, 9). It is characterized as a high concern for people and a low concern for production. The primary objective of the country club style is to create a secure and comfortable atmosphere and trust that subordinates will respond positively. The produce or perish style is located at the lower right-hand corner of the grid, point (9,1). A high concern for production and a low concern for people characterize it. The primary objective of the produce or perish style is to achieve the organizations goals. To accomplish the organizatio ns goals, it is not necessary to consider employees needs as relevant. The middle-of-the-road style is located at the middle of the grid, point (5, 5). A balance between workers needs and the organizations productivity goals characterize it. The primary objective of the middle-of-the-road style is to maintain employee morale at a level sufficient to get the organizations work done. The team style is located at the upper right-hand of the grid, point (9, 9). It is characterized by a high concern for people and production. The primary objective of the team style is to establish cohesion and foster a feeling of commitment among workers. Contingency Theories Successful leaders must be able to identify clues in an environment and adapt their leader behavior to meet the needs of their followers and of the particular situation. Even with good diagnostic skills, leaders may not be effective unless they can adapt their leadership style to meet the demands of their environment. Fiedlers Contingency Model Leadership Theory and Research: Perspectives and Directions (Academic Press Inc (HBJ), 1993) was a tribute to Fred Fiedlers 40 year study of leadership and organizational effectiveness. The editors, Martin M. Chemers and Roya Ayman, write of Fiedlers contribution: The realization that leadership effectiveness depends on the interaction of qualities of the leader with demands of the situation in which the leader functions, made the simplistic one best way approach of earlier eras obsolete. Fred E. Fiedlers contingency theory postulates that there is no best way for managers to lead. Situations will create different leadership style requirements for a manager. The solution to a managerial situation is contingent on the factors that impinge on the situation. For example, in a highly routinized (mechanistic) environment where repetitive tasks are the norm, a certain leadership style may result in the best performance. The same leadership style may not work in a very dynamic environment. Fiedler looked at three situations that could define the condition of a managerial task: 1. Leader member relations: How well do the manager and the employees get along? 2. The task structure: Is the job highly structured, fairly unstructured, or somewhere in between? 3. Position power: How much authority does the manager possess? Managers were rated as to whether they were relationship oriented or task oriented. Task oriented managers tend to do better in situations that have good leader-member relationships, structured tasks, and either weak or strong position power. They do well when the task is unstructured but position power is strong. Also, they did well at the other end of the spectrum when the leader member relations were moderate to poor and the task was unstructured. Relationship oriented managers do better in all other situations. Thus, a given situation might call for a manager with a different style or a manager who could take on a different style for a different situation. These environmental variables are combined in a weighted sum that is termed Favorable at one end and unfavorable at the other. Task oriented style is preferable at the clearly defined extremes of favorable and unfavorable environments, but relationship orientation excels in the middle ground. Managers could attempt to reshape the environment variables to match their style. Another aspect of the contingency model theory is that the leader-member relations, task structure, and position power dictate a leaders situational control. Leader-member relations are the amount of loyalty, dependability, and support that the leader receives from employees. It is a measure of how the manager perceives he or she and the group of employees is getting along together. In a favorable relationship the manager has a high task structure and is able to reward and or punish employees without any problems. In an unfavorable relationship the task is usually unstructured and the leader possesses limited authority. The spelling out in detail (favorable) of what is required of subordinates affects task structure. Positioning power measures the amount of power or authority the manager perceives the organization has given him or her for the purpose of directing, rewarding, and punishing subordinates. Positioning power of managers depends on the taking away (favorable) or increasing (unfavorable) the decision-making power of employees. The task-motivated style leader experiences pride and satisfaction in the task accomplishment for the organization, while the relationship-motivated style seeks to build interpersonal relations and extend extra help for the team development in the organization. There is no good or bad leadership style. Each person has his or her own preferences for leadership. Task-motivated leaders are at their best when the group performs successfully such as achieving a new sales record or outperforming the major competitor. Relationship-oriented leaders are at their best when greater customer satisfaction is gained and a positive company image is established. Hersey-Blanchard Situational Leadership The Hersey-Blanchard Situational Leadership theory is based on the amount of direction (task behavior) and amount of socio-emotional support (relationship behavior) a leader must provide given the situation and the level of maturity of the followers. Task behavior is the extent to which the leader engages in spelling out the duties and responsibilities to an individual or group. This behavior includes telling people what to do, how to do it, when to do it, where to do it, and whos to do it. In task behavior the leader engages in one-way communication. Relationship behavior is the extent to which the leader engages in two-way or multi-way communications. This includes listening, facilitating, and supportive behaviors. In relationship behavior the leader engages in two-way communication by providing socio-emotional support. Maturity is the willingness and ability of a person to take responsibility for directing his or her own behavior. People tend to have varying degrees of maturity, d epending on the specific task, function, or objective that a leader is attempting to accomplish through their efforts. To determine the appropriate leadership style to use in a given situation, the leader must first determine the maturity level of the followers in relation to the specific task that the leader is attempting to accomplish through the effort of the followers. As the level of followers maturity increases, the leader should begin to reduce his or her task behavior and increase relationship behavior until the followers reach a moderate level of maturity. As the followers begin to move into an above average level of maturity, the leader should decrease not only task behavior but also relationship behavior. Once the maturity level is identified, the appropriate leadership style can be determined. The four leadership styles are telling, selling, participating, and delegating. High task/low relationship behavior (S1) is referred to as telling. The leader provides clear instructions and specific direction. Telling style is best matched with a low follower readiness level. High task/high relationship behavior (S2) is referred to as selling. The leader encourages two-way communication and helps build confidence and motivation on the part of the employee, although the leader still has responsibility and controls decision making. Selling style is best matched with a moderate follower readiness level. High relationship/low task behavior (S3) is referred to as participating. With this style, the leader and followers share decision making and no longer need or expect the relationship to be directive. Participating style is best matched with a moderate follower readiness level. Low relationship/lo w task behavior (S4) is labeled delegating. This style is appropriate for leaders whose followers are ready to accomplish a particular task and are both competent and motivated to take full responsibility. Delegating style is best matched with a high follower readiness level. Houses Path-Goal Model The path-goal theory developed by Robert House is based on the expectancy theory of motivation. The managers job is viewed as coaching or guiding workers to choose the best paths for reaching their goals. Best is judged by the accompanying achievement of organizational goals. It is based on the precepts of goal setting theory and argues that leaders will have to engage in different types of leadership behavior depending on the nature and demands of the particular situation. Its the leaders job to assist followers in attaining goals and to provide direction and support needed to ensure that their goals are compatible with the organizations. A leaders behavior is acceptable to subordinates when viewed as a source of satisfaction, and motivational when need satisfaction is contingent on performance, and the leader facilitates, coaches and rewards effective performance. Path goal theory identifies achievement-oriented, directive, participative and supportive leadership styles. In achievement-oriented leadership, the leader sets challenging goals for followers, expects them to perform at their highest level, and shows confidence in their ability to meet this expectation. This style is appropriate when the follower suffers from a lack of job challenge. In directive leadership, the leader lets followers know what is expected of them and tells them how to perform their tasks. This style is appropriate when the follower has an ambiguous job. Participative leadership involves leaders consulting with followers and asking for their suggestions before making a decision. This style is appropriate when the follower is using improper procedures or is making poor decisions. In supportive leadership, the leader is friendly and approachable. He or she shows concern for followers psychological well being. This style is appropriate when the followers lack confidence. Path-Goal theory assumes that leaders are flexible and that they can change their style, as situations require. The theory proposes two contingency variables (environment and follower characteristics) that moderate the leader behavior-outcome relationship. Environment is outside the control of followers-task structure, authority system, and work group. Environmental factors determine the type of leader behavior required if follower outcomes are to be maximized. Follower characteristics are the locus of control, experience, and perceived ability. Personal characteristics of subordinates determine how the environment and leader are interpreted. Effective leaders clarify the path to help their followers achieve their goals and make the journey easier by reducing roadblocks and pitfalls. Research demonstrates that employee performance and satisfaction are positively influenced when the leader compensates for the shortcomings in either the employee or the work setting. Vroom, Yetton, Jago Leader-Participation Model The Vroom, Yetton, Jago leader-participation model relates leadership behavior and participation to decision making. The model provides a set of sequential rules to determine the form and amount of participative decision making in different situations. It is a decision tree, requiring yes and no answers incorporating contingencies about task structure and alternative styles. The following contingency questions must be answered to determine the appropriate leadership style in the leader-participation model. Â · Quality Requirement: How important is the technical quality of this decision? Â · Commitment Requirement: How important is subordinate commitment to the decision? Â · Leaders Information: Do you have sufficient information to make a high-quality decision? Â · Problem Structure: Is the problem well structured? Â · Commitment Probability: If you were to make the decision yourself, are you reasonably certain that your subordinates would be committed to the decision? Â · Goal Congruence: Do subordinates share the organizational goals to be attained in solving this problem? Â · Subordinate Conflict: Is conflict among subordinates over preferred solutions likely? Â · Subordinate Information: Do subordinates have sufficient information to make a high-quality decision? Transformational Leadership Transformational leadership blends the behavioral theories with a little dab of trait theories. Transactional leaders, such as those identified in contingency theories, guide followers in the direction of established goals by clarifying role and task requirements. However, transformational leaders, who are charismatic and visionary, can inspire followers to transcend their own self-interest for the good of the organization. Transformational leaders appeal to followers ideals and moral values and inspire them to think about problems in new or different ways. Leader behaviors used to influence followers include vision, framing, and impression management. Vision is the ability of the leader to bind people together with an idea. Framing is the process whereby leaders define the purpose of their movement in highly meaningful terms. Impression management is a leaders attempt to control the impressions that others form about the leader by practicing behaviors that make the leader more attra ctive and appealing to others. Research indicates that transformational, as compared to transactional, leadership is more strongly correlated with lower turnover rates, higher productivity, and higher employee satisfaction. A transformational leader instills feelings of confidence, admiration and commitment in the followers. He or she is charismatic, creating a special bond with followers, articulating a vision with which the followers identify and for which they are willing to work. Each follower is coached, advised, and delegated some authority. The transformational leader stimulates followers intellectually, arousing them to develop new ways to think about problems. The leader uses contingent rewards to positively reinforce performances that are consistent with the leaders wishes. Management is by exception. The leader takes initiative only when there are problems and is not actively involved when things are going well. The transformational leader commits people to action and converts followers into leaders. Transformational leaders are relevant to todays workplace because they are flexible and innovative. While it is important to have leaders with the appropriate orientation defining tasks and managing interrelationships, it is even more important to have leaders who can bring organizations into futures they have not yet imagined. Transformational leadership is the essence of creating and sustaining competitive advantage.

marketing consulting :: essays research papers

Marketing Departments in mid-sized businesses face lots of challenges such as how to target customers more efficiently, how to increase market share, how to compete in the market-place and be ahead of competition, and how to increase one-to-one communications with customers. The unsteady economy has pressured businesses to become as streamlined as possible, leaving marketing departments short on budget and staff. Today, our role as Marketing Consultants is to help companies not only retain current customers but also aggressively grow market-share, open new market potentials and add new customers. Our Marketing offerings can help companies conquer these challenges. We can help developing, supervising and executing your Marketing Strategies, whether through dealing with your Marketing Department or through assigning persons from our company. Why we execute, or supervise for you? Because today’s economic realities have put a tremendous amount of pressure on the Marketing Departments to do more with less staff and budget. Marketing is the core of the company’s activities and strategies in today’s high competitive market. In the coming years, due to globalization, competition will boost, professional companies will achieve additional success and economical pressure will become very high. Please note here that cutbacks put companies at risk for losing customers through a lack of new initiatives and competition monitoring. The only possible response is to increase efficiency through outsourcing your Marketing Planning to experts in the field. Many Business Owners or Managers think that by simply placing an Ad in a newspaper or a commercial on a radio or a television station, customers will automatically come to purchase their product or service. This could bring some customers out of curiosity but hundreds of other potential customers may never learn of your business. Just think of the money you’ll lose simply because you didn’t develop an adequate Marketing Plan. As mentioned previously, Marketing is the core of your Business Operations and it determines how successful your business will be. What you, as a Business Owner or Manager, must do is maintain a thorough understanding of the Marketing Plan, and use it to extract advantages from the marketplace. Remember, your aim is not only to attract and keep a steady group of loyal customers but also to expand your customer base by identifying and attracting new customers and to reduce risks by anticipating Market Shifts that can affect your bottom line. †¦we visualize success to your company and help you reach it

Humorous Best Man Speech -- Wedding Toasts Roasts Speeches

Humorous Best Man Speech Good afternoon ladies and gentlemen. Firstly on behalf of the bridesmaids, I would like to thank the groom for his kind words, and may I also say that they have done a fantastic job today and all of them look absolutely beautiful. I'd also like to say that the bride looks absolutely stunning today as I’m sure you will all agree. Unfortunately for the wedding photographs, the groom just looks stunned. When I was asked to be best man I consulted the Internet for help. I must confess I was perplexed by some of the things I was expected to do: Help the groom dress. Thanks, but no thanks. If he hasn't learned by nor then he never will. That his shoes are tied. That his face and hair are â€Å"in order† (God didn't put them in order first time round, I'm not convinced that I stood a chance) . That he has nothing between his teeth (or is that his ears?) That his trouser flies are done up. I came to the conclusion that best man is just a fancy title for nanny. I also found out some other interesting things on the Internet, but now is definitely not the time to tell you what they were. However, I have taken the job of being best man seriously and have made sure: That he got to the service on time. That he was well dressed and looking smart, which I’m sure you will agree he is. And that he got a good nights sleep last night, and I’m pleased to report he slept like a baby. He woke up every half hour like clockwork, crying for his mom. So what can I say about the groom. Well, he’s witty, intelligent, charming, successful, han†¦ han... Sorry. I'm having trouble reading your writing. I've been racking my brain for the last couple of weeks trying to think of suitable stories to be said about... ...Now. if you could join me in a toast to some very important people, without whom today just wouldn't be the same. I'm sure all of us at some point will shuffle past them and exchange a few kind words. Ladies and gentlemen, I'd like you to raise your glasses and say a toast. To the bar staff. And of course we should not forget the bride and groom. I would like to say to you both: "May your love be modern enough to survive the times, and old-fashioned enough to last forever." Today is a day when each one of us wishes the happy couple well. Being human they will have their disagreements. Life being what it is there will be sad moments as well as glad. Yet I know that today we are all wishing them happiness and health in those years to come, and I am sure that the love between them will be strong enough to last forever. Ladies and gentlemen. the bride and groom.

Myths of the American Dream Exposed in Arthur Millers Death of a Sales

Myths of the American Dream Exposed in Arthur Miller's Death of a Salesman    Willy Loman, the lead character of Miller’s play, Death of a Salesman, believes in "the myths of the capitalistic society"(DiYanni 412). This essay will examine the impact of the capitalistic myths on Willy Lowman.   Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   Willy believes in the myth that popularity and physical appearance are the keys that unlock the door to the â€Å"American Dream†. We are first introduced to the importance of popularity and physical appearance when Willy is speaking to his wife, Linda, about their son Biff.   â€Å"Biff Loman is lost,† says Willy.   â€Å"In the greatest country in the world, a young man with such personal attractiveness gets lost.†Ã‚   In this quote, not only is Willy confused about how Biff’s good looks can’t help him get a job, but also because his son can’t get a job in a country like America.   Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   Willy believes in appearance, in phoniness, in popularity with those he regards as important in the capitalistic machine. An example of how Willy depends on popularity to help achieve the dream is seen when Willy is having a flashback in which he’s speaking to both Biff and Happy about having his own business. The boys ask their father if his business will be like their Uncle Charley’s.   Willy responds by saying that he’ll be, â€Å"Bigger than Uncle Charley!   Because Charley is not- liked.   He’s liked, but he’s not- well liked.†Ã‚     Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   The most significant example of Willy’s belief in the popularity myth also takes place in one of Willy’s flashbacks.   Again, he is speaking to his sons about becoming successful.   He tells them, â€Å"...the man who makes an appearance in the business world, the man who creates personal interest, is the man who gets ... ... slogans for his own beliefs: â€Å"Chevrolet, Linda, is the greatest car ever built.† But his blind faith cannot sustain him: â€Å"That goddam Chevrolet, they ought to prohibit the manufacture of that car†.     Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   Each day Willy must run faster and stretch his arms out further in his attempt to catch the dream. When he is too tired to run, Willy is spewed out of the capitalistic machine as a worn-out and useless part. Willy then gives all that he has remaining so that his son can collect the insurance money and thereby pay his entrance fee to the capitalistic machine. The same machine that destroyed Willy. Works Cited DiYanni, Robert. Literature: Reading Fiction, Poetry, and Drama. Compact Edition. McGraw Hill, 2000. 395-530. Miller, Arthur. "Death of a Salesman" in Literature, Reading, Reacting, Writing, Compact Fourth Edition. Harcourt, Inc. 2000.

Men Essay Question

You are sitting in a small, not very well ventilated room that has no air conditioning. You are sitting alongside eleven other overheated, uncomfortable, impatient men. That sounds rather unpleasant, right? Perhaps the last place you would like to be. Imagine how the Jurors may have been compelled to hurry and reach a conclusion for the young boys life they were debating on. What If, these men were seated In a well-lit, cool, comfortable room, with decorations and armchairs†¦ Would they have reached a different conclusion?Or maybe the same decision, but In much fewer hours. In the movie, one or two Jurors had plans to go too baseball game. A few others have plans with family and friends. Most everyone was ready get out of there as soon as possible. When being rushed, your mind tends to blur around the facts as well as your temper being shortened. This caused many of the Jurors to lose their temper, sending the group Into outraged arguments. Let's say the Jurors were not Irritabl e, but Instead had clear heads, and relaxed emotions.I'm sure the first response would be to think about it reasonably. The questionable trial would have been debated smoother with much less arguments. Having fewer disagreements, the trial's outcome may have turned out differently. The boy was let go innocent, but only after hours of debating the matter. If the conditions in which they were debating the trial were changed for the better, the outcome could have easily been different. They could have all agreed at once that he was guilty; they could have all agreed much quicker that he was innocent.Better conditions may have allowed for different first mime impressions of the trial, and led to a better conversation on what side to vote for. I believe that if the Jurors were placed in better conditions to decide, they could have come to a conclusion much quicker than that of which they were placed in. If it were me being shoved in a room with other inhospitable, rushed, people; I too, even if I wasn't at first, would become grumpy and more argumentative. This only proves that the conditions in which people interact and converse in, do affect the attitudes and outcomes of the debated situation.

Solution Manual for Fluid Mech Cengel Book

Chapter 6 nervous impulse abstr doing of consort Systems Chapter 6 MOMENTUM depth psychology OF FLOW SYSTEMS Newtons Laws and conservation of neural impulse 6-1C Newtons proscribedset law states that a clay at rest hang ins at rest, and a body in operation ashes in doubtfulness at the similar hurrying in a refined path when the net soldiers performing on it is cryptograph. therefore, a body tends to conserve its state or inertia. Newtons succor law states that the reviveup of a body is sex act to the net twitch performing on it and is mutually proportionate to its visual modality. Newtons third law states when a body exerts a thread on a second body, the second body exerts an compeer and adversary personnel on the first. r 6-2C Since nerve impulse ( mV ) is the product of a vector ( speeding) and a scalar ( kettle of fish), neural impulse must(prenominal)iness be a vector that points in the identical flush as the speeding vector. 6-3C The conservation of nerve impulse principle is de nonative as the urge of a dodge trunk incessant when the net wildness acting on it is slide fastener, and and so the nervous impulse of much(prenominal) forms is conserved.The impulse of a body remains cons burnt if the net bosom acting on it is zero. 6-4C Newtons second law of app atomic number 18nt motion, as well c either(a)ed the angular impetus equating, is permit show uped as the compute of flip-flop of the angular impulse of a body is tally to the net tortuousness acting it. For a non-rigid body with zero net torque, the angular neural impulse remains constant, but the angular s realize number changes in accordance with I? = constant where I is the moment of inertia of the body. 6-5C No.Two rigid bodies having the equalize aggregative and angular speed pull up s labors have different angular pulsations unless they too have the equal moment of inertia I. Linear pulsation Equation 6-6C The kinship in the midst of the time stations of change of an extensive property for a system and for a keep derriere flock is verbalized by the Reynolds transport theorem, which volunteers the link between the r system and nurse script concepts. The zephyrar nervous impulse comparison is obtained by setting b = V and whencely r B = mV in the Reynolds transport theorem. -7C The big businessmans acting on the discover book of account consist of body repels that act by means of t come on ensembleow on the blameless body of the go tight majority ( such as gravity, electric car, and magnetic repulses) and erupt staff offices that act on the halt near (such as the coerce hugs and reply make ups at points of contact). The net crash acting on a decl ar muckle is the promenade field of wholly body and excavate soldierys. eloquent lading is a body thread, and wedge is a spring up twitch (acting per unit bea). -8C All of these uprise str engths arise as the ope grade deal is isolated from its milieu for analysis, and the incumbrance of any detached object is accounted for by a powerfulness at that location. We atomic number 50 minimize the number of cake forces overt by choosing the comptroller deal such that the forces that we atomic number 18 non fire in remain internal, and then they do non complicate the analysis. A well-c irrigaten confine volume exposes solitary(prenominal) the forces that atomic number 18 to be de terminal figureined (such as reply forces) and a minimum number of other forces. 6-9C The nervous impulse- integ assess discipline figure break forth ? nables us to express the pulsing flux in scathe of the r r r r & ? V (V ? n )dAc = ? mV avg . The regard as of ? is congruity for furnish mass go down grade and mean meld hurrying as ? Ac incline, such as a thousand prey, more or less unity for turbulent bunk (between 1. 01 and 1. 04), but just around 1. 3 for laminar fall. So it should be con facial expressionred in laminar prevail. 6-1 proprietary MATERIAL. cc6 The McGraw-Hill Companies, Inc. break dispersion permitted unaccompanied to t severallyers and educators for production line preparation. If you ar a school-age child utilize this Manual, you argon utilize it with come on permission.Chapter 6 pulse epitome of track down Systems 6-10C The momentum par for unconstipated running(a) stream for the graphic symbol of no external forces is r r r & & F= ? mV ? ? mV ? ? out ? in where the left hand side is the net force acting on the harbour volume, and first term on the right hand side is the elect(postnominal) momentum flux and the second term is the outperform momentum flux by mass. 6-11C In the application of the momentum comparability, we move disregard the atmosphericalalalal push and locomote with wad hales alone since the atmospheric oblige acts in all delegations, and its effect female ge nitaliacels out in every pleader. -12C The fireman who softens the hose backwards so that the piddle system makes a U-turn in front macrocosm comp aldepressione al paltry experience a greater reply force since the numerical determine of momentum fluxes a cover up the snoot ar added in this case instead of be subshargoned. 6-13C No, V is not the upper limit to the go ups ultimate f number. Without abrasion the projectile s whirligig number leave alone continue to outgrowth as more gas yields the bird of night. 6-14C A pearly hovers beca exercise the strong downdraft of send out, caused by the bash impeller blades, manifests a momentum in the bloodline pelt.This momentum must be countered by the helicopter go on force. 6-15C As the transmit niggardness decreases, it requires more vitality for a helicopter to hover, because more business line must be forced into the downdraft by the helicopter blades to provide the same lift force. Therefore, it as aut horizeds more index for a helicopter to hover on the steer of a senior high potentiometer than it does at ocean direct. 6-16C In winter the send outmanship is in the main colder, and frankincense denser. Therefore, less air must be driven by the blades to provide the same helicopter lift, requiring less motive. 6-2 proprietary MATERIAL. 2006 The McGraw-Hill Companies, Inc. contain diffusion permitted just to teachers and educators for job preparation. If you argon a learner utilise this Manual, you argon development it without permission. Chapter 6 urge compendium of liquefy Systems 6-17C The force mandatory to check out the carapace against the plain pissing stream testament growth by a element of 4 when the focal ratio is double since & F = mV = ( ? AV )V = ? AV 2 and hence the force is relative to the squ atomic number 18 of the fastness. 6-18C The quickening will not be constant since the force is not constant. The impulse force exerted by & we ewee on the weighing machine is F = mV = ( ? AV )V = ?AV 2 , where V is the congress speeding between the pissing system and the rest home, which is pitiable. The musical scale quickening will be a = F/m. precisely as the coat begins to move, V decreases, so the quickening must overly decrease. 6-19C The upper limit upper possible for the headquarters is the velocity of the pissing kilobyte. As long as the home headquarters is moving dumber than the kibibyte, the pissing supply will exert a force on the photographic dwelling, which will cause it to accele regularise, until terminal viridity-propelled plane velocity is reached. 6-20 It is to be shown that the force exerted by a fluid squirt of velocity V on a stationary nozzle is & proportional to V2, or alternatively, to m 2 . Assumptions 1 The accrue is calm down and incompressible. 2 The nozzle is give up to be stationary. 3 The nozzle involves a 90 turn and then the incoming and out dismissal tra ck down streams argon universal to each other. 4 The weewee supply is dismissed to the atmosphere, and and then the forage blackjack at the wall socket is zero. depth psychology We sire the nozzle as the tally volume, and the feed committee at the expiration as the x axis. observe that the nozzle makes a 90 turn, and thence it does not contribute to any hug force or momentum flux & term at the gateway in the x stress. Noting that m = ?AV where A is the nozzle passing atomic number 18a and V is the sightly nozzle event velocity, the momentum comparability for sweetheart additive consort in the x complaint reduces to r r r & & & & F= ? mV ? ? mV FRx = ? m out V out = ? mV ? ? out ? in where FRx is the re litigate force on the nozzle due to liquid small fry at the nozzle sales outlet. because, & m = ? AV & FRx = ? mV = AVV = AV 2 & & or FRx = ? mV = ? m & & m m2 =? ?A ? A Therefore, the force exerted by a liquid leafy vegetable of velocity V on this & stationary nozzle is proportional to V2, or alternatively, to m 2 . Liquid bird of night V FR 6-3 patented MATERIAL. 2006 The McGraw-Hill Companies, Inc. hit in dispersal permitted exclusively to teachers and educators for category preparation. If you be a savant exploitation this Manual, you be employ it without permission. Chapter 6 pulsation abbreviation of electric current Systems 6-21 A urine viridity of velocity V impinges on a central office moving toward the urine supply kibibyte with velocity ? V. The force essential to move the casing towards the spout is to be resolute in terms of F acting on the stationary exfoliation. Assumptions 1 The scat is still and incompressible. 2 The graduated table is tumid and the natural spring is commonplace to menage. 3 The wedge on both sides of the home office is atmospheric extort (and consequentlyly its effect cancels out). Fiction during motion is negligible. 5 There is no quickening of the denta l house necessitate. 6 The piss splashes wrap up the sides of the base in a plane normal to the small fry. 6 squirt liquefy is to the highest degree render and thus the effect of the momentum-flux subject grammatical constituent is negligible, ? ? 1. abstract We sh argon the headquarters as the control volume. The relative velocity between the carapace and the chiliad is V when the rest home is stationary, and 1. 5V when the plate is moving with a velocity ? V towards the plate. accordingly the momentum equating for smashed running(a) hunt in the plain boot reduces to r r r & & & & F= ? mV ? ? mV ? FR = ? mi Vi FR = miVi ? out ? in nonmoving plate ( Vi = V and moving plate ( Vi = 1. 5V and & mi = ? AVi = ? AV ) FR = ? AV 2 = F & mi = ? AVi = ? A(1. 5V ) ) FR = ? A(1. 5V ) 2 = 2. 25 ? AV 2 = 2. 25 F Therefore, the force necessitate to rent the plate stationary against the oncoming urine commons dies 2. 25 times when the commons velocity get goings 1. 5 times. discussion origin that when the plate is stationary, V is too the rave velocity. But if the plate moves toward the stream with velocity ? V, then the relative velocity is 1. 5V, and the amount of mass striking the plate (and falling eat up its sides) per unit time as well as increases by 50%. 1/2V V pissing resinous 6-4 proprietorship MATERIAL. 2006 The McGraw-Hill Companies, Inc. trammel scattering permitted moreover to teachers and educators for data track preparation. If you are a educatee development this Manual, you are exploitation it without permission. Chapter 6 impulse compend of melt down Systems 6-22 A 90 jostle deflects piss upward and dribbles it to the atmosphere at a contract rate. The kitty shove at the adit of the shove and the anchoring force fateed to remove the shove in place are to be determined. v Assumptions 1 The combine is plastered, frictionless, incompressible, and irrotational (so that the Bernoulli equiva lence is applicable). The clog of the human human human elbow and the urine supply in it is negligible. 3 The peeing is discharged to the atmosphere, and thus the bet pressure come at the outlet is zero. 4 The momentum-flux fudge factor in factor for each opening and outlet is attached to be ? = 1. 03. Properties We rent the immersion of pissing to be deoxyguanosine monophosphate kg/m3. analysis (a) We incorporate the elbow as the control volume, and fix the trance by 1 and the outlet by 2. We besides pin down the crosswise co-ordinate by x (with the program line of range as being the overbearing education) and the tumid aline by z.The persistence equality for this one- door one-outlet starchy die hard system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = ? AV , the mean inlet and outlet velocities of water are & & 25 kg/s m m = = = 3. 18 m/s 2 ? A ? (? D / 4) ( kibibyte kg/m 3 )? (0. 1 m) 2 / 4 Noting that V1 = V2 and P2 = Patm, the Bern oulli comparison for a streamline going finished the conc repose of the bring down elbow is express as V1 = V 2 = V = P V12 P V2 1 + + z1 = 2 + 2 + z2 P ? P2 = ? g ( z2 ? z1 ) P , post = ? g ( z2 ? z1 ) 1 1 ? g 2 g ? g 2 g substituting, ? ? 1 kN 2 ? P , second = ( kelvin kg/m3 )(9. 81 m/s 2 )(0. 35 m)? 1 ? vitamin C0 kg ? /s2 ? = 3. 434 kN/m = 3. 434 kPa ? ? r r r & & (b) The momentum compare for chill out one-dimensional be precondition is F= ? mV ? ? mV . We let the x- ? ? out ? in and z- components of the anchoring force of the elbow be FRx and FRz, and swallow them to be in the verificatory snaps. We also use gage pressures to avoid traffic with the atmospheric pressure which acts on all surfaces. and so the momentum pars on the x and y axes become & & FRx + P1,gage A1 = 0 ? ?m(+V1 ) = ? ?mV & & FRz = ? m(+V 2 ) = ? mV z x FRz 2 35 cm work out for FRx and FRz, and subbing the effrontery values, & FRx = ? ?mV ? P1, gage A1 ? N = ? 1. 03(25 kg/s)(3. 18 m/s )? ? 1 kg ? m/s 2 ? = ? 109 N ? ? ? (3434 N/m 2 )? (0. 1 m) 2 / 4 ? ? ? ? = 81. 9 N ? ? heat FRx = tan -1 water 25 kg/s FRx 1 ? 1N & FRy = ? mV = 1. 03(25 kg/s)(3. 18 m/s)? ? 1 kg ? m/s 2 ? and 2 2 FR = FRx + FRy = (? 109) 2 + 81. 9 2 = 136 N, ? = tan -1 81. 9 = ? 37 = 143 ? 109 tidings comment that the order of order of the anchoring force is 136 N, and its line of action makes 143 from the controlling(p) x way. Also, a proscribe value for FRx indicates the clutch caution is wrongly, and should be converse. 6-5 patented MATERIAL. 2006 The McGraw-Hill Companies, Inc. curb distribution permitted only to teachers and educators for run for preparation. If you are a learner victimization this Manual, you are use it without permission. Chapter 6 Momentum analytic thinking of Flow Systems 6-23 An clxxx elbow forces the feed to make a U-turn and discharges it to the atmosphere at a condition rate. The gage pressure at the inlet of the elbow and the anchoring force nee ded to see the elbow in place are to be determined. v Assumptions 1 The precipitate is potent, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli compare is applicable). The weight of the elbow and the water in it is negligible. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 4 The momentumflux rectification factor for each inlet and outlet is inclined to be ? = 1. 03. Properties We take the niggardness of water to be grand kg/m3. compendium (a) We take the elbow as the control volume, and de theatreate the approach by 1 and the outlet by 2. We also de signali oceante the swimming align by x (with the direction of operate as being the positive direction) and the erect array by z.The continuity equation for this one-inlet one-outlet starchy stream system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = ? AV , the mean inlet and outlet velocities of water are & & 25 kg/s m m = = = 3. 18 m/s 2 ? A ? (? D / 4) ( pace kg/m 3 )? (0. 1 m) 2 / 4 Noting that V1 = V2 and P2 = Patm, the Bernoulli equation for a streamline going by the center of the reducing elbow is evince as V1 = V 2 = V = P V12 P V2 1 + + z1 = 2 + 2 + z2 P ? P2 = ? g ( z2 ? z1 ) P , gage = ? g ( z2 ? z1 ) 1 1 ? g 2 g ? g 2 g substituting, ? ? 1 kN 2 ? P , gage = ( molar concentration kg/m3 )(9. 81 m/s2 )(0. 70 m)? 1 ? 1000 kg ? m/s2 ? 6. 867 kN/m = 6. 867 kPa ? ? r r r & & (b) The momentum equation for steady one-dimensional emanate is F= ? mV ? ? mV . We let the x- ? ? out ? in and z- components of the anchoring force of the elbow be FRx and FRz, and deport them to be in the positive directions. We also use gage pressures to avoid traffic with the atmospheric pressure which acts on all surfaces. Then the momentum equations on the x and z axes become & & & FRx + P1,gage A1 = ? m(? V 2 ) ? ? m(+V1 ) = ? 2 ? mV FRz = 0 Solving for FRx and exchange the given values, & FRx = ? 2 ? mV ? P1, gage A 1 ? 1N = ? 2 ? 1. 03(25 kg/s)(3. 18 m/s)? 1 kg ? m/s 2 ? = ? 218 N ? ? ? (6867 N/m 2 )? (0. 1 m) 2 / 4 ? ? 2 z x FRz Water 25 kg/s 35 cm and FR = FRx = 218 N since the y-component of the anchoring force is zero. Therefore, the anchoring force has a magnitude of 218 N and it acts in the veto x direction. preaching tick off that a contradict value for FRx indicates the expect direction is wrong, and should be reversed. FRx 1 6-6 copyrighted MATERIAL. 2006 The McGraw-Hill Companies, Inc. express distribution permitted only to teachers and educators for bunk preparation. If you are a scholar utilise this Manual, you are utilise it without permission.Chapter 6 Momentum psychoanalysis of Flow Systems 6-24E A horizontal water jet strikes a upended stationary plate normally at a specified velocity. For a given anchoring force needed to birth the plate in place, the feed in rate of water is to be determined. Assumptions 1 The full point is steady and incompressible. 2 The w ater splatters off the sides of the plate in a plane normal to the jet. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the displace water is the atmospheric pressure which is cut since it acts on the constitutional control surface. The steep forces and momentum fluxes are not considered since they have no effect on the horizontal answer force. 5 run ascend is close constant and thus the effect of the momentum-flux field of study factor is negligible, ? ? 1. Properties We take the absorption of water to be 62. 4 lbm/ft3. depth psychology We take the plate as the control volume such that it contains the spotless plate and cuts through the water jet and the support bar normally, and the direction of endure as the positive direction of x axis. The momentum equation for steady one-dimensional fertilise in the x ( liquefy) direction reduces in this case o r r r & & & & F= ? mV ? ? mV ? FRx = ? mV1 FR = mV1 ? ? out ? in We note tha t the reaction force acts in the opposite direction to track down, and we should not impart the invalidating & sign for forces and velocities in the ostracise x-direction. Solving for m and substitute the given values, & m= FRx 350 lbf = V1 30 ft/s ? 32. 2 lbm ? ft/s 2 ? ? 1 lbf ? ? ? = 376 lbm/s ? ? Then the volume flow rate becomes V& = & m ? = 376 lbm/s 62. 4 lbm/ft 3 = 6. 02 ft 3 /s Therefore, the volume flow rate of water infra stated self-assertions must be 6. 02 ft3/s.Discussion In reality, some water will be scattered back, and this will add to the reaction force of water. The flow rate in that case will be less. m 1 FRx = 350 lbf Waterjet 6-7 trademarked MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for move preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-25 A reducing elbow deflects water upwards and discharges it to the atmosphere at a specified rate.The anchoring force needed to batch the elbow in place is to be determined. v Assumptions 1 The flow is steady, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The weight of the elbow and the water in it is considered. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 4 The momentumflux correction factor for each inlet and outlet is given to be ? = 1. 03. Properties We take the density of water to be 1000 kg/m3. Analysis The weight of the elbow and the water in it is W = mg = (50 kg)(9. 1 m/s 2 ) = 490. 5 N = 0. 4905 kN We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction) and the just coordinate by z. The continuity equation for this one-inlet one-outlet steady flow system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = ? AV , the inlet and outlet velocities of water are & 30 kg/s m V1 = = = 2. 0 m/s ? A1 (1000 kg/m 3 )(0. 0 one hundred fifty m 2 ) & 30 kg/s m V2 = = = 12 m/s ? A2 (1000 kg/m 3 )(0. 025 m 2 ) Taking the center of the inlet cross sectionalisation as the annex level (z1 = 0) and noting that P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as ? V 2 ? V12 ? ? V22 ? V12 ? P V12 P V2 1 ? ? ? + + z1 = 2 + 2 + z2 P ? P2 = ? g ? 2 1 1 ? 2 g + z2 ? z1 ? P , gage = ? g ? 2 g + z2 ? ?g 2 g ? g 2 g ? ? ? ? Substituting, ? (12 m/s) 2 ? (2 m/s) 2 ? 1 kN ? = 73. 9 kN/m 2 = 73. 9 kPa P , gage = (1000 kg/m3 )(9. 81 m/s 2 )? + 0. 4 1 2 ? 1000 kg ? m/s 2 ? 2(9. 81 m/s ) ? ? The momentum equation for steady one-dimensional flow is & & ? F = ? mV ? ? ? mV . We let the x- and out in r r r z- components of the anchoring force of the elbow be FRx and FRz, and buy up them to be in the positive directi ons. We also use gage pressures to avoid dealing with the atmospheric pressure which acts on all surfaces. Then the momentum equations on the x and z axes become & & & FRx + P1,gage A1 = ? mV 2 cos ? ? ? mV1 and FRz ? W = ? mV 2 sin ? 2 25 cm2 Solving for FRx and FRz, and substituting the given values, & FRx = ? m(V 2 cos ? ? V1 ) ? P1, gage A1 ? 1 kN = 1. 03(30 kg/s)(12cos45 2) m/s? ? 1000 kg ? m/s 2 ? ? (73. 9 kN/m 2 )(0. 0150 m 2 ) = ? 0. 908 kN ? ? ? Water 30 kg/s 45 FRz FRx 150 m2 W 1 ? ? 1 kN ? & FRz = ? mV 2 sin ? + W = 1. 03(30 kg/s)(12sin45 m/s)? ? 1000 kg ? m/s 2 ? + 0. 4905 kN = 0. 753 kN ? ? 0. 753 2 2 2 2 -1 FRz FR = FRx + FRz = (? 0. 908) + (0. 753) = 1. 18 kN, ? = tan = tan -1 = ? 39. 7 FRx ? 0. 908 Discussion Note that the magnitude of the anchoring force is 1. 18 kN, and its line of action makes 39. 7 from +x direction. Negative value for FRx indicates the put one acrossd direction is wrong. 6-8 patented MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited di stribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-26 A reducing elbow deflects water upwards and discharges it to the atmosphere at a specified rate. The anchoring force needed to hold the elbow in place is to be determined. v Assumptions 1 The flow is steady, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The weight of the elbow and the water in it is considered. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. The momentumflux correction factor for each inlet and outlet is given to be ? = 1. 03. Properties We take the density of water to be 1000 kg/m3. Analysis The weight of the elbow and the water in it is W = mg = (50 kg)(9. 81 m/s 2 ) = 490. 5 N = 0. 4905 kN We take the elbow as the control volume, and designate the entra nce by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction) and the straight coordinate by z. The continuity equation for this one-inlet one-outlet steady flow system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = ?AV , the inlet and outlet velocities of water are & 30 kg/s m = = 2. 0 m/s V1 = ? A1 (1000 kg/m 3 )(0. 0150 m 2 ) & 30 kg/s m V2 = = = 12 m/s ? A2 (1000 kg/m 3 )(0. 0025 m 2 ) Taking the center of the inlet cross section as the reference level (z1 = 0) and noting that P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as ? V 2 ? V12 ? ? V22 ? V12 ? P V12 P V2 1 ? ? ? + + z1 = 2 + 2 + z2 P ? P2 = ? g ? 2 1 1 ? 2 g + z2 ? z1 ? P , gage = ? g ? 2 g + z2 ? ?g 2 g ? g 2 g ? ? ? ? or, P , gage = (1000 kg/m3 )(9. 81 m/s2 )? 1 ? ? ? (12 m/s)2 ? (2 m/s)2 2(9. 81 m/s ) ? 1 kN ? = 73. 9 kN/m 2 = 73. 9 kPa + 0. 4 1000 kg ? m/s 2 ? ? Th e momentum equation for steady one-dimensional flow is & & ? F = ? ?mV ? ? ? mV . We let the xout in r r r and y- components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive directions. We also use gage pressures to avoid dealing with the atmospheric pressure which acts on all surfaces. Then the momentum equations along the x and z axes become & & FRx + P1,gage A1 = ? mV 2 cos ? ? ? mV1 and & FRy ? W = ? mV 2 sin ? Solving for FRx and FRz, and substituting the given values, & FRx = ? m(V 2 cos ? V1 ) ? P1, gage A1 ? 1 kN = 1. 03(30 kg/s)(12cosone hundred ten 2) m/s? ? 1000 kg ? m/s 2 ? FRz ? ? ? (73. 9 kN/m 2 )(0. 0150 m 2 ) = ? 1. 297 kN ? ? ? ? 1 kN ? + 0. 4905 kN = 0. 8389 kN & = ? mV 2 sin ? + W = 1. 03(30 kg/s)(12sin110 m/s)? 2 ? ? 1000 kg ? m/s ? ? 2 25 cm2 110 2 2 FR = FRx + FRz = (? 1. 297) 2 + 0. 8389 2 = 1. 54 kN and FRz 0. 8389 = tan -1 = ? 32. 9 FRx ? 1. 297 Discussion Note that the magnitude of the anchoring force is 1. 54 kN, a nd its line of action makes 32. 9 from +x direction. Negative value for FRx indicates assumed direction is wrong, and should be reversed. ? = tan -1 FRz FRx Water 1 30 kg/s 50 m2 W 6-9 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 6 Momentum Analysis of Flow Systems 6-27 Water deepen by a nozzle strikes the back surface of a baby buggy moving horizontally at a constant velocity. The braking force and the proponent wasted by the brake are to be determined. . Assumptions 1 The flow is steady and incompressible. 2 The water splatters off the sides of the plate in all directions in the plane of the back surface. The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the move water is the atmospheric pressure which is disregarded since it acts on all surfaces. 4 Ficti on during motion is negligible. 5 There is no acceleration of the hang back. 7 The motions of the water jet and the drag are horizontal. 6 Jet flow is most resembling and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Analysis We take the squeeze as the control volume, and the direction of flow as the positive direction of x axis. The relative velocity between the cart and the jet is V r = V jet ?Vcart = 15 ? 10 = 10 m/s 15 m/s 5 m/s Therefore, we can assume the cart to be stationary and the jet to move Waterjet with a velocity of 10 m/s. The momentum equation for steady onedimensional flow in the x (flow) direction reduces in this case to r r r & & & & F= ? mV ? ? mV FRx = ? mi Vi Fbrake = ? mV r FRx ? ? out ? in We note that the brake force acts in the opposite direction to flow, and we should not allow the detrimental sign for forces and velocities in the blackball x-direction. Substituting the given values, ? 1N & Fbrake = ? mV r = ? (25 k g/s)(+10 m/s)? ? 1 kg ? m/s 2 ? ? ? = ? 250 N ? ?The negative sign indicates that the braking force acts in the opposite direction to motion, as expected. Noting that work is force times quad and the distance traveled by the cart per unit time is the cart velocity, the proponent wasted by the brakes is 1 kW ? ? & W = FbrakeV cart = (250 N)(5 m/s)? ? = 1. 25 kW ? 1000 N ? m/s ? Discussion Note that the ability wasted is equivalent to the maximum might that can be generated as the cart velocity is maintained constant. 6-10 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 6 Momentum Analysis of Flow Systems 6-28 Water accelerated by a nozzle strikes the back surface of a cart moving horizontally. The acceleration of the cart if the brakes fall apart is to be determined. Analysis The braking force was de termined in previous problem to be 250 N. When the brakes fail, this force will propel the cart forward, and the accelerating will be a= F 250 N ? 1 kg ? m/s 2 ? = m cart 300 kg ? 1N ? ? ? = 0. 833 m/s 2 ? ? Discussion This is the acceleration at the moment the brakes fail.The acceleration will decrease as the relative velocity between the water jet and the cart (and thus the force) decreases. 5 m/s 15 m/s 300 kg Waterjet FRx 6-11 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 6 Momentum Analysis of Flow Systems 6-29E A water jet hits a stationary rail-splitter, such that one-half of the flow is diverted upward at 45, and the other half is say down.The force essential to hold the splitter in place is to be determined. vEES Assumptions 1 The flow is steady and incompressible. 2 The water jet is expo sed to the atmosphere, and thus the pressure of the water jet before and subsequently the split is the atmospheric pressure which is disregarded since it acts on all surfaces. 3 The gravitative make are disregarded. 4 Jet flow is more or less uniform and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Properties We take the density of water to be 62. 4 lbm/ft3. Analysis The mass flow rate of water jet is & & m = ? V = (62. lbm/ft 3 )(100 ft 3 /s) = 6240 lbm/s We take the splitting section of water jet, including the splitter as the control volume, and designate the entrance by 1 and the outlet of either arm by 2 (both arms have the same velocity and mass flow rate). We also designate the horizontal coordinate by x with the direction of flow as being the positive direction and the vertical coordinate by z. r r r & & The momentum equation for steady one-dimensional flow is F= ? mV ? ? mV . We let ? ? out ? in the x- and y- components of the anchoring f orce of the splitter be FRx and FRz, and assume them to be in the & & positive directions.Noting that V2 = V1 = V and m 2 = 1 m , the momentum equations along the x and z 2 axes become & & & FRx = 2( 1 m)V 2 cos ? ? mV1 = mV (cos ? ? 1) 2 & & FRz = 1 m(+V 2 sin ? ) + 1 m(? V 2 sin ? ) ? 0 = 0 2 2 Substituting the given values, 1 lbf ? ? FRx = (6240 lbm/s)(20 ft/s)(cos45 1)? ? = ? 1cxxxv lbf 32. 2 lbm ? ft/s 2 ? ? FRz = 0 The negative value for FRx indicates the assumed direction is wrong, and should be reversed. Therefore, a force of 1 one hundred thirty-five lbf must be utilise to the splitter in the opposite direction to flow to hold it in place. No belongings force is necessary in the vertical direction.This can also be concluded from the symmetry. Discussion In reality, the gravitative personal effect will cause the upper stream to slow down and the lower stream to speed up after the split. But for minuscule distances, these effects are indeed negligible. 20 ft/s 100 ft/s FRz 45 45 FRx 6-12 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 6 Momentum Analysis of Flow Systems 6-30E Problem 6-29E is reconsidered.The effect of splitter angle on the force exerted on the splitter as the half splitter angle varies from 0 to one hundred eighty in increments of 10 is to be investigated. g=32. 2 ft/s2 rho=62. 4 lbm/ft3 V_dot=100 ft3/s V=20 ft/s m_dot=rho*V_dot F_R=-m_dot*V*(cos(theta)-1)/g lbf ?, 0 10 20 30 40 50 60 70 80 90 100 110 one hundred twenty 130 cxl 150 clx 170 180 8000 7000 6000 5000 & m , lbm/s 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 FR, lbf 0 59 234 519 907 1384 1938 2550 3203 3876 4549 5201 5814 6367 6845 7232 7518 7693 7752 FR, lbf 000 3000 2000 1000 0 0 20 40 60 80 100 120 one hundred forty 160 180 ?, 6-13 P ROPRIETARY 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 6 Momentum Analysis of Flow Systems 6-31 A horizontal water jet impinges normally upon a vertical plate which is held on a frictionless track and is initially stationary. The initial acceleration of the plate, the time it takes to reach a certain velocity, and the velocity at a given time are to be determined.Assumptions 1 The flow is steady and incompressible. 2 The water always splatters in the plane of the retreating plate. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure which is disregarded since it acts on all surfaces. 4 The tract is nearly frictionless, and thus fiction during motion is negligible. 5 The motions of the water jet and the cart are horizontal. 6 Th e velocity of the jet relative to the plate remains constant, Vr = Vjet = V. 7 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is egligible, ? ? 1. Properties We take the density of water to be 1000 kg/m3. Analysis (a) We take the vertical plate on the frictionless track as the control volume, and the direction of flow as the positive direction of x axis. The mass flow rate of water in the jet is & m = ? VA = (1000 kg/m 3 )(18 m/s)? (0. 05 m) 2 / 4 = 35. 34 kg/s The momentum equation for steady one-dimensional flow in the x (flow) direction reduces in this case to r r r & & & & F= ? mV ? ? mV FRx = ? mi Vi FRx = ? mV ? ? out ? in where FRx is the reaction force required to hold the plate in place.When the plate is released, an equal and opposite impulse force acts on the plate, which is determined to ? 1N & Fplate = ? FRx = mV = (35. 34 kg/s)(18 m/s)? ? 1 kg ? m/s 2 ? ? ? = 636 N ? ? Then the initial acceleration of the plate becomes a= Fplate m plate = 636 N ? 1 kg ? m/s 2 ? 1000 kg ? 1 N ? ? ? = 0. 636 m/s 2 ? ? 18 m/s 1000 kg Waterjet Frictionless track This acceleration will remain constant during motion since the force acting on the plate remains constant. (b) Noting that a = dV/dt = ? V/? t since the acceleration a is constant, the time it takes for the plate to reach a velocity of 9 m/s is ? t = ? V plate a = (9 ? ) m/s 0. 636 m/s 2 FRx = 14. 2 s (c) Noting that a = dV/dt and thus dV = adt and that the acceleration a is constant, the plate velocity in 20 s becomes V plate = V0, plate + a? t = 0 + (0. 636 m/s 2 )(20 s) = 12. 7 m/s Discussion The assumption that the relative velocity between the water jet and the plate remains constant is valid only for the initial moments of motion when the plate velocity is low unless the water jet is moving with the plate at the same velocity as the plate. 6-14 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educator s for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-32 A 90 reducer elbow deflects water downward into a smaller diam pipe. The concomitant force exerted on the reducer by water is to be determined. Assumptions 1 The flow is steady, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The weight of the elbow and the water in it is disregarded since the gravitative effects are negligible. 3 The momentum-flux correction factor for each inlet and outlet is given to be ? 1. 04. Properties We take the density of water to be 1000 kg/m3. Analysis We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction) and the vertical coordinate by z. The continuity equation for this one-inlet one-outle t steady flow system is & & & & m1 = m 2 = m = 353. 4 kg/s. Noting that m = ? AV , the mass flow rate of water and its outlet velocity are 2 & m = ? V1 A1 = ? V1 (? D1 / 4) = (1000 kg/m 3 )(5 m/s)? (0. 3 m) 2 / 4 = 353. 4 kg/s & & 353. kg/s m m = = = 20 m/s 2 ? A2 D 2 / 4 (1000 kg/m 3 )? (0. 15 m) 2 / 4 The Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as V2 = P V12 P V2 1 + + z1 = 2 + 2 + z2 ? g 2 g ? g 2 g ? V 2 ? V22 ? ? P2 = P + ? g ? 1 1 ? 2 g + z1 ? z2 ? ? ? Substituting, the gage pressure at the outlet becomes ? (5 m/s)2 ? (20 m/s)2 1 kPa ? 1 kN ? P2 = (300 kPa) + (1000 kg/m 3 )(9. 81 m/s 2 )? + 0. 5 = 117. 4 kPa 2 ? 1000 kg ? m/s 2 1 kN/m 2 ? 2(9. 81 m/s ) ? ? The momentum equation for steady one-dimensional flow is & & ? F = ? ?mV ? ? ? mV . We let the xout in r r and z- components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive directions. Then the momentum equations along the x and z axes become & FRx + P1,gage A1 = 0 ? ? mV1 & FRz ? P2,gage A2 = ? m(? V 2 ) ? 0 Note that we should not forget the negative sign for forces and velocities in the negative x or z direction. Solving for FRx and FRz, and substituting the given values, ? 1 kN & FRx = ? ?mV1 ? P1, gage A1 = ? 1. 04(353. 4 kg/s)(5 m/s)? ? 1000 kg ? m/s 2 ? ? ? (0. 3 m) 2 ? ? (300 kN/m 2 ) = ? 23. 0 kN ? 4 ? ? ? (0. 15 m) 2 ? + (117. 4 kN/m 2 ) = ? 5. 28 kN ? ? FRz ? 1 kN & FRz = ? ? mV 2 + P2, gage A1 = ? 1. 04(353. 4 kg/s)(20 m/s)? ? 1000 kg ? m/s 2 ? and 2 2 FR = FRx + FRz = (? 23. 0) 2 + (? 5. 28) 2 = 23. 6 kN FRx 30 cm Water 5 m/s ? = tan -1 FRz ? 5. 28 = tan -1 = 12. 9 FRx ? 23. 0 Discussion The magnitude of the anchoring force is 23. 6 kN, and its line of action makes 12. 9 from +x direction. Negative values for FRx and FRy indicate that the assumed directions are wrong, and should be reversed. 15 cm 6-15 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 6 Momentum Analysis of Flow Systems 6-33 A cheat on turbine with a given get across diameter and efficiency is subjected to steady principals. The power generated and the horizontal force on the supporting mast of the turbine are to be determined. vEES Assumptions 1 The wind flow is steady and incompressible. 2 The efficiency of the turbine-generator is self-sustaining of wind speed. 3 The frictional effects are negligible, and thus none of the incoming energising energy is reborn to thermic energy. summit flow is uniform and thus the momentum-flux correction factor is nearly unity, ? ? 1. Properties The density of air is given to be 1. 25 kg/m3. Analysis (a) The power say-so of the wind is its energising energy, & which is V2/2 per unit mass, and mV 2 / 2 for a given mass flow rate ? 1 m/s ? V1 = (25 km/h)? ? = 6. 94 m/s ? 3. 6 km/h ? & m = ? 1V1 A1 = ? 1V1 Wind V1 1 2 D V2 ?D 2 4 2 = (1. 25 kg/m 3 )(6. 94 m/s) ? (90 m) 2 4 2 = 55,200 kg/s V (6. 94 m/s) & & & W max = mke1 = m 1 = (55,200 kg/s) 2 2 ? 1 kN ? ? 1000 kg ? m/s 2 ? 1 kW ? 1 kN ? m/s ? = 1330 kW ? ? FR Then the actual power produced becomes & Wact = ? wind turbineW max = (0. 32)(1330 kW) = 426 kW (b) The frictional effects are assumed to be negligible, and thus the portion of incoming kinetic energy not reborn to electric power leaves the wind turbine as outgoing kinetic energy. Therefore, V2 V2 & & & & mke 2 = mke1 (1 ? ? wind turbine ) m 2 = m 1 (1 ? ? wind turbine ) 2 2 or V 2 = V1 1 ? ? wind turbine = (6. 94 m/s) 1 0. 32 = 5. 72 m/s We contain the control volume around the wind turbine such that the wind is normal to the control surface at the inlet and the outlet, and the entire control surface is at the atmospheric pressure.The momentum r r r & & equation for steady one-dimensional flow is F= ? mV ? ? mV . Writing it along the x-dir ection ? ? out ? in (without forgetting the negative sign for forces and velocities in the negative x-direction) and assuming the flow velocity through the turbine to be equal to the wind velocity give ? 1 kN & & & FR = mV 2 ? mV1 = m(V 2 ? V1 ) = (55,200 kg/s)(5. 72 6. 94 m/s)? ? 1000 kg ? m/s 2 ? ? ? = ? 67. 3 kN ? ? The negative sign indicates that the reaction force acts in the negative x direction, as expected.Discussion This force acts on top of the hulk where the wind turbine is installed, and the digression moment it generates at the substructure of the hover is obtained by multiplying this force by the tower height. 6-16 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 6 Momentum Analysis of Flow Systems 6-34E A horizontal water jet strikes a sheer plate, which deflects the water back to i ts headmaster direction.The force required to hold the plate against the water stream is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure, which is disregarded since it acts on all surfaces. 3 Friction between the plate and the surface it is on is negligible (or the friction force can be include in the required force to hold the plate). 4 There is no sputter of water or the deformation of the jet, and the reversed jet leaves horizontally at the same velocity and flow rate. Jet flow is nearly uniform and thus the momentum-flux correction factor is nearly unity, ? ? 1. Properties We take the density of water to be 62. 4 lbm/ft3. Analysis We take the plate together with the curved water jet as the control volume, and designate the jet inlet by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the di rection of incoming flow as being the positive direction). The continuity equation for this one-inlet one-outlet steady & & & flow system is m1 = m 2 = m where & m = ? VA = ? V ? D 2 / 4 = (62. 4 lbm/ft 3 )(one hundred forty ft/s)? (3 / 12 ft) 2 / 4 = 428. lbm/s r r r & & The momentum equation for steady one-dimensional flow is F= ? mV ? ? mV . Letting the ? ? out ? in reaction force to hold the plate be FRx and assuming it to be in the positive direction, the momentum equation along the x axis becomes & & & FRx = m(? V 2 ) ? m(+V1 ) = ? 2mV Substituting, 1 lbf ? ? FRx = ? 2(428. 8 lbm/s)(140 ft/s)? ? = ? 3729 lbf 2 ? 32. 2 lbm ? ft/s ? Therefore, a force of 3729 lbm must be apply on the plate in the negative x direction to hold it in place. Discussion Note that a negative value for FRx indicates the assumed direction is wrong (as expected), and should be reversed.Also, there is no need for an analysis in the vertical direction since the fluid streams are horizontal. 2 140 ft/s Wat erjet FRx 1 140 ft/s 3 in 6-17 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 6 Momentum Analysis of Flow Systems 6-35E A horizontal water jet strikes a bent plate, which deflects the water by 135 from its original direction. The force required to hold the plate against the water stream is to be determined.Assumptions 1 The flow is steady and incompressible. 2 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure, which is disregarded since it acts on all surfaces. 3 Frictional and gravitational effects are negligible. 4 There is no splattering of water or the deformation of the jet, and the reversed jet leaves horizontally at the same velocity and flow rate. 5 Jet flow is nearly uniform and thus the momentum-flux correct ion factor is nearly unity, ? ? 1. Properties We take the density of water to be 62. 4 lbm/ft3.Analysis We take the plate together with the curved water jet as the control volume, and designate the jet inlet by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of incoming flow as being the positive direction), and the vertical coordinate by z. The continuity equation for & & & this one-inlet one-outlet steady flow system is m1 = m 2 = m where & m = ? VA = ? V ? D 2 / 4 = (62. 4 lbm/ft 3 )(140 ft/s)? (3 / 12 ft) 2 / 4 = 428. 8 lbm/s r r r & & The momentum equation for steady one-dimensional flow is F= ? mV ? ? mV . We let the x- ? ? out ? in nd z- components of the anchoring force of the plate be FRx and FRz, and assume them to be in the positive directions. Then the momentum equations along the x and y axes become & & & FRx = m(? V 2 ) cos 45 ? m(+V1 ) = ? mV (1 + cos 45) & (+V 2 ) sin 45 = mV sin 45 & FRz = m Substituting the given values, 1 lbf ? ? FRx = ? 2(428. 8 lbm/s)(140 ft/s)(1 + cos45)? 2 ? ? 32. 2 lbm ? ft/s ? = ? 6365 lbf 1 lbf ? ? FRz = (428. 8 lbm/s)(140 ft/s)sin45? = 1318 lbf 2 ? ? 32. 2 lbm ? ft/s ? 2 140 ft/s Waterjet 135 FRz FRx 3 in 1 and 2 2 FR = FRx + FRz = (? 6365) 2 + 1318 2 = 6500 lbf , ? = tan -1 FRy FRx = tan -1 1318 = ? 1. 7 = 168. 3 ? 6365 Discussion Note that the magnitude of the anchoring force is 6500 lbf, and its line of action makes 168. 3 from the positive x direction. Also, a negative value for FRx indicates the assumed direction is wrong, and should be reversed. 6-18 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 6 Momentum Analysis of Flow Systems 6-36 Firemen are holding a nozzle at the end of a hose speckle trying to extinguish a fire.The fair(a) water outlet velocity and the escapeance force requ ired of the firemen to hold the nozzle are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water jet is exposed to the atmosphere, and thus the pressure of the water jet is the atmospheric pressure, which is disregarded since it acts on all surfaces. 3 Gravitational effects and vertical forces are disregarded since the horizontal resistance force is to be determined. 5 Jet flow is nearly uniform and thus the momentum-flux correction factor can be interpreted to be unity, ? ? 1. Properties We take the density of water to be 1000 kg/m3.Analysis (a) We take the nozzle and the horizontal portion of the hose as the system such that water enters the control volume vertically and outlets horizontally (this way the pressure force and the momentum flux at the inlet are in the vertical direction, with no contribution to the force dimension in the horizontal direction), and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinat e by x (with the direction of flow as being the positive direction). The sightly outlet velocity and the mass flow rate of water are determined from V= V& A = V& ? D / 4 2 = 5 m 3 /min ? (0. 06 m) 2 / 4 1768 m/min = 29. 5 m/s & m = ? V& = (1000 kg/m 3 )(5 m 3 /min) = 5000 kg/min = 83. 3 kg/s (b) The momentum equation for steady one-dimensional flow is & & ? F = ? ?mV ? ? ? mV . We let out in r r r horizontal force employ by the firemen to the nozzle to hold it be FRx, and assume it to be in the positive x direction. Then the momentum equation along the x direction gives ? ? 1N ? = 2457 N & & FRx = mVe ? 0 = mV = (83. 3 kg/s)(29. 5 m/s)? ? 1kg ? m/s 2 ? ? ? Therefore, the firemen must be able to resist a force of 2457 N to hold the nozzle in place. Discussion The force of 2457 N is equivalent to the weight of about 250 kg.That is, holding the nozzle requires the strength of holding a weight of 250 kg, which cannot be through with(p) by a single person. This demonstrates why severa l firemen are used to hold a hose with a high flow rate. FRz FRx 5 m3/min 6-19 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 6 Momentum Analysis of Flow Systems 6-37 A horizontal jet of water with a given velocity strikes a flat plate that is moving in the same direction at a specified velocity.The force that the water stream exerts against the plate is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water splatters in all directions in the plane of the plate. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure, which is disregarded since it acts on all surfaces. 4 The vertical forces and momentum fluxes are not considered since they have no effect on the horizontal force exerted on the p late. 5 The velocity of the plate, and the velocity of the water jet relative to the plate, are constant. Jet flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Properties We take the density of water to be 1000 kg/m3. Analysis We take the plate as the control volume, and the flow direction as the positive direction of x axis. The mass flow rate of water in the jet is & m = ? V jet A = ? V jet 10 m/s 30 m/s FRx 5 cm Waterjet ?D 4 2 = (1000 kg/m 3 )(30 m/s) ? (0. 05 m) 2 4 = 58. 9 kg/s The relative velocity between the plate and the jet is V r = V jet ? V plate = 30 ? 10 = 20 m/s Therefore, we can assume the plate to be stationary and the jet to move with a velocity of 20 m/s.The r r r & & F= ? mV ? ? mV . We let the horizontal momentum equation for steady one-dimensional flow is ? ? out ? in reaction force applied to the plate in the negative x direction to counteract the impulse of the water jet be FRx. Then the momentum equation al ong the x direction gives ? ? 1N ? & & ? FRx = 0 ? mVi FRx = mV r = (58. 9 kg/s)(20 m/s)? ? 1kg ? m/s 2 ? = 1178 N ? ? Therefore, the water jet applies a force of 1178 N on the plate in the direction of motion, and an equal and opposite force must be applied on the plate if its velocity is to remain constant.Discussion Note that we used the relative velocity in the determination of the mass flow rate of water in the momentum analysis since water will enter the control volume at this rate. (In the trammel case of the plate and the water jet moving at the same velocity, the mass flow rate of water relative to the plate will be zero since no water will be able to strike the plate). 6-20 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 6 Momentum Analysis of Flow Systems 6-38 Problem 6-37 is reconsidered. T he effect of the plate velocity on the force exerted on the plate as the plate velocity varies from 0 to 30 m/s in increments of 3 m/s is to be investigated. rho=1000 kg/m3 D=0. 05 m V_jet=30 m/s Ac=pi*D2/4 V_r=V_jet-V_plate m_dot=rho*Ac*V_jet F_R=m_dot*V_r N Vplate, m/s 0 3 6 9 12 15 18 21 24 27 30 Vr, m/s 30 27 24 21 18 15 12 9 6 3 0 FR, N 1767 1590 1414 1237 1060 883. 6 706. 9 530. 1 353. 4 176. 7 0 1800 1600 1400 1200 1000 FR, N 800 600 400 200 0 0 5 10 15 20 25 30 Vplate, m/s 6-21PROPRIETARY 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 6 Momentum Analysis of Flow Systems 6-39E A devotee moves air at sea level at a specified rate. The force required to hold the lover and the minimum power introduce required for the raw sienna are to be determined. v Assumptions 1 The flow of air is steady and incompressib le. 2 Standard atmospheric conditions exist so that the pressure at sea level is 1 atm. song leaves the fan at a uniform velocity at atmospheric pressure. 4 Air approaches the fan through a humongous area at atmospheric pressure with negligible velocity. 5 The frictional effects are negligible, and thus the entire mechanical power excitant is converted to kinetic energy of air (no modulation to thermal energy through frictional effects). 6 Wind flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Properties The gas constant of air is R = 0. 3704 psi? ft3/lbm? R. The standard atmospheric pressure at sea level is 1 atm = 14. 7 psi.Analysis (a) We take the control volume to be a horizontal high-flown cylinder bounded by streamlines on the sides with air go into through the large cross-sectional (section 1) and the fan hardened at the narrow cross-section at the end (section 2), and let its center line be the x axis. The density, ma ss flow rate, and discharge velocity of air are 14. 7 psi P ? = = = 0. 0749 lbm/ft 3 RT (0. 3704 psi ? ft 3 /lbm ? R)(530 R) & m = ? V& = (0. 0749 lbm/ft 3 )(2000 ft 3/min) = 149. 8 lbm/min = 2. 50 lbm/s V2 = V& A2 = V& 2 ? D 2 / 4 = 2000 ft 3 /min ? (2 ft) 2 / 4 = 636. 6 ft/min = 10. ft/s & & ? F = ? ?mV ? ? ? mV . Letting the out in The momentum equation for steady one-dimensional flow is r r r reaction force to hold the fan be FRx and assuming it to be in the positive x (i. e. , the flow) direction, the momentum equation along the x axis becomes 1 lbf ? ? & & FRx = m(V 2 ) ? 0 = mV = (2. 50 lbm/s)(10. 6 ft/s)? ? = 0. 82 lbf 2 ? 32. 2 lbm ? ft/s ? Therefore, a force of 0. 82 lbf must be applied (through friction at the base, for example) to prevent the fan from moving in the horizontal direction under the influence of this force. (b) Noting that P1 = P2 = Patm and V1 ? , the energy equation for the selected control volume reduces to ?P V2 ? ?P V2 ? & & & & & m? 1 + 1 + gz1 ? + W p ump, u = m? 2 + 2 + gz 2 ? + W turbine + E mech,loss ? ? ? ? 2 2 ? ? ? ? Substituting, V & & Wfan, u = m 2 2 2 V2 (10. 6 ft/s) 2 ? 1 lbf 1W ? & & Wfan,u = m 2 = (2. 50 lbm/s) ? ? = 5. 91 W 2 2 2 ? 32. 2 lbm ? ft/s 0. 73756 lbf ? ft/s ? Therefore, a useful mechanical power of 5. 91 W must be supplied to 2000 cfm air. This is the minimum required power stimulation required for the fan. Discussion The actual power commentary to the fan will be larger than 5. 1 W because of the fan inefficiency in converting mechanical power to kinetic energy. Fan 1 2 24 in 6-22 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 6 Momentum Analysis of Flow Systems 6-40 A helicopter hovers at sea level while being ludicrous. The volumetric air flow rate and the required power stimulus during un sealed hover, and the r evolutions per minute and the required power arousal during mingy hover are to be determined. Assumptions 1 The flow of air is steady and incompressible. 2 Air leaves the blades at a uniform velocity at atmospheric pressure. 3 Air approaches the blades from the top through a large area at atmospheric pressure with negligible velocity. 4 The frictional effects are negligible, and thus the entire mechanical power input is converted to kinetic energy of air (no diversity to thermal energy through frictional effects). 5 The change in air pressure with blossom is negligible because of the low density of air. 6 There is no acceleration of the helicopter, and thus the lift generated is equal to the thoroughgoing weight. Air flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Properties The density of air is given to be 1. 18 kg/m3. Analysis (a) We take the control volume to be a vertical hyperbolic cylinder bounded by streamlines on the sides with air entering through the large cross-section (section 1) at the top and the fan located at the narrow cross-section at the pervade (section 2), and let its centerline be the z axis with upwards being the positive direction. r r r & & F= ? mV ? ? mV . Noting The momentum equation for steady one-dimensional flow is ? out ? in that the only force acting on the control volume is the total weight W and it acts in the negative z direction, the momentum equation along the z axis gives W & & ? W = m(? V 2 ) ? 0 W = mV 2 = ( ? AV 2 )V 2 = ? AV 22 V2 = ? A 1 where A is the blade bitstock area, 15 m A = ? D / 4 = ? (15 m) / 4 = 176. 7 m 2 2 2 Then the discharge velocity, volume flow rate, and the mass flow rate of air in the un laden mode become V 2,un affluent = m un prankish g = ? A (10,000 kg)(9. 81 m/s 2 ) (1. 18 kg/m 3 )(176. 7 m 2 ) = 21. 7 m/s ocean level 2 V&unloaded = AV 2,unloaded = (176. 7 m 2 )(21. m/s) = 3834 m 3 /s & munloaded = ? V&unloaded = (1. 18 kg/m 3 )(3834 m 3/s) = 4524 kg/s Load 15,000 kg Noting that P1 = P2 = Patm, V1 ? 0, the elevation effects are negligible, and the frictional effects are disregarded, the energy equation for the selected control volume reduces to ? P V2 ? ?P V2 ? V2 & & & & & & & m? 1 + 1 + gz1 ? + W pump, u = m? 2 + 2 + gz 2 ? + W turbine + E mech,loss Wfan, u = m 2 ? ? ? ? 2 2 2 ? ? ? ? Substituting, ? V2 ? 1 kW ? (21. 7 m/s) 2 ? 1 kN & ? ? & = (4524 kg/s) W unloaded fan,u = ? m 2 ? ? = 1065 kW 2 ? 1 kN ? m/s ? 1000 kg ? m/s ? ? 2 ? 2 ? ? ? ? nloaded (b) We now relieve the calculations for the loaded helicopter, whose mass is 10,000+15,000 = 25,000 kg V 2,loaded = m loaded g = ? A (25,000 kg)(9. 81 m/s 2 ) (1. 18 kg/m 3 )(176. 7 m 2 ) = 34. 3 m/s & mloaded = ? V&loaded = ? AV2, loaded = (1. 18 kg/m 3 )(176. 7 m 2 )(34. 3 m/s) = 7152 kg/s ? V2 ? (34. 3 m/s)2 & & = (7152 kg/s) Wloaded fan,u = ? m 2 ? ? 2 ? 2 ? ?loaded ? 1 kW ? 1 kN ? ? ? 1000 kg ? m/s 2 1 kN ? m/s ? = 4207 kW ? ? 6-23 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 6 Momentum Analysis of Flow Systems Noting that the average flow velocity is proportional to the command overhead blade rotational velocity, the revolutions per minute of the loaded helicopter blades becomes & V 2 = kn V 2,loaded V 2, unloaded = & n loaded & n unloaded & n loaded = V 2,loaded V 2, unloaded & n unloaded = 34. 3 (400 rev) = 632 rpm 21. 7 Discussion The actual power input to the helicopter blades will be considerably larger than the calculated power input because of the fan inefficiency in converting mechanical power to kinetic energy. -24 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 6 Momentum Analysis of Flow Systems 6-41 A helicopter hovers on top of a high locoweed where the air density considerably lower than that at sea level. The blade rotational velocity to hover at the higher top and the percent increase in the required power input to hover at high altitude relative to that at sea level are to be determined. Assumptions 1 The flow of air is steady and incompressible. 2 The air leaves the blades at a uniform velocity at atmospheric pressure. 3 Air approaches the blades from the top through a large area at atmospheric pressure with negligible velocity. 4 The frictional effects are negligible, and thus the entire mechanical power input is converted to kinetic energy of air. 5 The change in air pressure with elevation while hovering at a given location is negligible because of the low density of air. 6 There is no acceleration of the helicopter, and thus the lift generated is equal to the total weight. Air flow is nearly uniform and thus the momen tum-flux correction factor can be taken to be unity, ? ? 1. Properties The density of air is given to be 1. 18 kg/m3 at sea level, and 0. 79 kg/m3 on top of the fate. Analysis (a) We take the control volume to be a vertical hyperbolic cylinder bounded by streamlines on the sides with air entering through the large cross-section (section 1) at the top and the fan located at the narrow cross-section at the bottom (section 2), and let its centerline be the z axis with upwards being the positive direction. r r & & F= ? mV ? ? mV . Noting The momentum equation for steady one-dimensional flow is ? ? out ? in that the only force acting on the control volume is the total weight W and it acts in the negative z direction, the momentum equation along the z axis gives W & & ? W = m(? V 2 ) ? 0 W = mV 2 = ( ? AV 2 )V 2 = ? AV 22 V2 = ? A where A is the blade span area. Then for a given weight W, the ratio of discharge velocities becomes V 2,mountain V 2,sea = W / ? mountain A W / ? sea A = ? sea ? mountain = 1. 18 kg/m 3 0. 79 kg/m 3 = 1. 222Noting that the average flow velocity is proportional to the overhead blade rotational velocity, the rpm of the helicopter blades on top of the mountain becomes & n = kV 2 & n mountain V 2, mountain = & n sea V 2,sea & n mountain = V 2, mountain V 2,sea & nsea = 1. 222(400 rpm) = 489 rpm Noting that P1 = P2 = Patm, V1 ? 0, the elevation effect are negligible, and the frictional effects are disregarded, the energy equation for the selected control volume reduces to ? P V2 ? ?P V2 ? V2 & & & & & & & m? 1 + 1 + gz1 ? + W pump, u = m? 2 + 2 + gz 2 ? W turbine + E mech,loss Wfan, u = m 2 ? ? ? ? 1 2 2 2 ? ? ? ? or V2 V2 V3 & & Wfan,u = m 2 = ? AV2 2 = ? A 2 = 2 2 2 1 2 ?A? ? ? W ? ? = ? ? ? A ? 3 1 2 ?A? ? ?W ? ? ? ? ? A ? 1 . 5 = W 1 . 5 2 ? A 15 m Then the ratio of the required power input on top of the mountain to that at sea level becomes & Wmountain fan,u 0. 5W 1. 5 / ? mountain A = & Wsea fan,u 0. 5W 1. 5 / ? sea A 2 ? mountai n ?sea = 1. 18 kg/m3 = 1. 222 0. 79 kg/m3 Sea level Load 15,000 kg Therefore, the required power input will increase by 22. 2% on top of the mountain relative to the sea level.Discussion Note that both the rpm and the required power input to the helicopter are inversely proportional to the square root of air density. Therefore, more power is required at higher elevations for the helicopter to operate because air is less dense, and more air must be forced by the blades into the downdraft. 6-25 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 6 Momentum Analysis of Flow Systems 6-42 The flow rate in a channel is controlled by a scour gate by raising or lowering a vertical plate. A semblance for the force acting on a sluice gate of width w for steady and uniform flow is to be developed. Assumptions 1 The f low is steady, incompressible, frictionless, and uniform (and thus the Bernoulli equation is applicable. ) 2 Wall shear forces at surfaces are negligible. 3 The channel is exposed to the atmosphere, and thus the pressure at rationalize surfaces is the atmospheric pressure. 4 The flow is horizontal. Water flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Analysis We take point 1 at the bountiful surface of the upstream flow before the gate and point 2 at the free surface of the downstream flow after the gate. We also take the bottom surface of the channel as the reference level so that the elevations of points 1 and 2 are y1 and y2, respectively. The application of the Bernoulli equation between points 1 and 2 gives P1 V12 P V2 + + y1 = 2 + 2 + y 2 ? g 2 g ? g 2 g V 22 ? V12 = 2 g( y1 ? y 2 ) (1)The flow is assumed to be incompressible and thus the density is constant. Then the conservation of mass relation for this single stream steady flow device can be expressed as V&1 = V&2 = V& A1V1 = A2V 2 = V& V1 = V& A1 = V& wy1 and V2 = V& A2 = V& wy 2 (2) Substituting into Eq. (1), ? V& ? ? wy ? 2 ? ? V& ? 2 g ( y1 ? y 2 ) & ? ? ? ? wy ? = 2 g ( y1 ? y 2 ) V = w 1 / y 2 ? 1 / y 2 ? ? 1? 2 1 2 2 2 g ( y1 ? y 2 ) V& = wy 2 2 2 1 ? y 2 / y1 (3) Substituting Eq. (3) into Eqs. (2) gives the following transaction for velocities, V1 = y2 y1 2 g ( y1 ? y 2 ) 1? y2 / 2 y1 and V2 = 2 g ( y1 ? y 2 ) 2 2 1 ? y 2 / y1 (4) We choose the control volume as the water body surrounded by the vertical cross-sections of the upstream and downstream flows, free surfaces of water, the interior(a) surface of the sluice gate, and the bottom surface of r r r & & F= ? mV ? ? mV . The the channel. The momentum equation for steady one-dimensional flow is ? ? out ? in force acting on the sluice gate FRx is horizontal