Regression Model

IntroductionA regression model with one explanatory variable is called a Simple linear regression, that is it involves 2 points: single explanatory variable and the response variable which is the x and y, coordinates in a Cartesian plane and finds a linear function a non-vertical straight line that, as precisely as possible it explains the dependent variable values as a function of the independent variables. The term simple refers to the fact that the response variable y is related to one predictor x. The regression model is given as Y=?0+?1 + ? and they are two parameters that are used estimate the slope of the line ?1 and the y- intercept of the line ?0. ? is the random error term.BackgroundRegression analysis is a vital statistical method for the analysis of medical data. It makes it possible for the recognition and grouping of relationships among multiple factors. It also enables the recognition of prognostically relevant risk factors and the calculation of risk scores for individual prognostication, this was made possible by English scientist Sir Francis Galton (1822–1911), a cousin of Charles Darwin, made significant contributions to both genetics and psychology. He is the one that came with regression and a pioneer in using statistics in a study of living organism. In his study the data sets that he considered consisted was the heights of fathers and first sons. He wanted to find out whether he can predict the height of a son based on the father height. Looking at the scatterplots of these heights, Galton saw that the was relationship which was linear and increasing. After fitting a line to these data using the statistical techniques, he observed that for fathers whose heights were taller than the average, the regression line predicted that taller fathers tended to have shorter sons and shorter fathers tended to have taller sons.PurposesSimple linear regression could be for example be purposefully when we Consider a relationship between weight Y (in kilograms) and height X(in centimeters), where the mean weight at a given height is ?(X) = 2X/4 – 45 for X > 100. Because of biological variability, the weight will vary for example, it might be normally distributed with a fixed ? = 4. The difference between an observed weight and mean weight at a given height is referred to as the error for that weight. To discover the relationship which is linear, we could take the weight of three individuals at each height and apply linear regression to model the mean weight as a function of height using a straight line, ?(X) = ?0 + ?1X . The most popular way to estimate the parameters, intercept ?0 and slope ?1 is the least squares estimator, which is derived by differentiating the regression with respect to ?0 and ?1 and solving, Let (xi , y i ) be the Ith pair of X and Y values. The least squares estimator, estimates ?0 and ?1 by minimizing the residual sum of squared errors, SSE = ?(y i – ? i)2, where y i are the observed value and ?i = b0 + b1xi are the estimated regression line points and are called the fitted, predicted or â€Å"hat† values. The estimates are given by b0 = ¯y – b1  ¯x and b1 = SSXX / SSYY, and where  ¯Xand  ¯Y are the means of samples X and Y, SSXX and SSYY being their standard deviation values and r = r(X,Y) being their Pearson correlation coefficient. It is also referred to as Pearson's r, the Pearson product-moment correlation coefficient, is a measure of the linear between two variables X and Y Where X is the independent variable and Y being the Dependant variable as stated above. The Pearson correlation coefficient, r can take a range of values from -1 to +1. A value of 0 suggests that there is no association between the two variables X and Y. A value greater than 0 indicates a positive association that is, as the value of one variable increases, so does the value of the other variable. Before using simple linear regression analysis it is always vital to follow these few steps: Choose an independent variable that is likely to cause the change in the dependent variable Be certain that the past amounts for the independent variable occur in the exact same period as the amount of the dependent variable. Plot the observations on a graph using the y-axis for the dependant variable and the x-axis for the independent variable review the plotted observations for a linear pattern and for any outliers keep in mind that there can be correlation without cause and effect.ImportancesSimple linear regression is considered to be extensively useful in many practical applications and methodologies. Simple linear regression functions by assuming that the variables x and y have a relationship which is linear within the given set of data. As assumptions are and results are interpreted, persons handling the analysing role in a such data will have to be more critical because it has been studied before that there are some variables which inhibit marginal changes to occur while others will not consider being held at a fixed point. Although the concept of linear regression is one complex subject, it still remains to be one of the most vital statistical approaches being used till date. Simple linear regression is important because it has be wildly being used in many biological, behavioural , environmental as well as social sciences. Because of its ability to describe possible relationships between identified variables independent and dependent , it has assisted the fields of epidemiology, finance, economics and trend line in describing significant data that proves to be of essence in the identified fields. More so, simple linear regression is important because it provides an idea of what needs to be anticipated, more specially in controlling and regulating functions involved on some disciplines. Despite the complexity of simple linear aggression, it has proven to be adequately useful in many daily applications of life.

Jacksonian Democracy Dbq

Kathy Dai M. Galvin AP USH Period 1 Jacksonian Democracy DBQ The Jacksonian democracy of the 1820s-1830s is often associated with an expansion of the political influence, economic opportunities, and social equality available to â€Å"the common man,† a concept of the masses which President Andrew Jackson and his newly founded Democratic party came to represent. The new administration certainly saw gains for the majority; namely, public participation in government increased to unprecedented levels, and several economic decisions were made to favor the people over monopolies.Beginning with their exaggerated portrayal of the â€Å"corrupt† 1824 election however, the Jacksonian democrats also left a legacy of substantial miscalculations in policies and acts of hypocrisy that conflicted with their claimed intents to promote and protect popular democracy. In particular, the dangerous implications of various political and economic policies, along with the deliberate disregard of social inequality, are aspects of the Jacksonian age that most clearly demonstrate discrepancies between Jacksonian ideals and realities.The political field saw the first advances accredited to the Jacksonian democracy in the forms of extended suffrage and increased government participation, but it also involved many questionable federal acts that conflicted with the vision of political democracy. With Andrew Jackson’s election in 1828 introducing the first president from West of the Appalachians, the common men that Jackson championed naturally arose to the political stage as well.States all across the country adopted universal suffrage for white males on their own in the 1820s, but Jackson indeed bolstered the democratic trend through influence in newspapers, popular campaigning, and even a huge inauguration party at the White House open to the masses. In terms of campaigning however, the election of 1828 was the first in which the political parties directly attacked eac h other’s candidates through the press.The increase in voter participation led to a negative pattern of smear campaigning that aimed more to sway the masses than convey the truth that a healthy democracy needs. Furthermore, Jackson’s presidency was characterized by use of the spoils system and the systematic rotation of officeholders. These stipulated that federal jobs were strictly given to loyal Democrats and that federal offices could be held for only one term. While these practices were meant to emphasize equal political opportunities and build party loyalty, they inherently promoted government corruption.In fact, the power that Jackson wielded by trading federal positions for party loyalty both overextended his executive power and practiced the same corrupt bargaining of office that the Democrats accused John Quincy Adams of in the election of 1824. Thus, the Jacksonian democrats dealt clear detriments and hypocrisies to the system of popular democracy that they s o strongly advocated, despite their encouragement of universal white male suffrage and participation in office.Similarly, the Jacksonian age affected the economy both in accordance with the Jacksonian ideal of equal economic opportunity and against it; an executive branch act and a judicial branch decision were made with the intent of favoring the people, but substantial opposition highlighted the negative side effects that undermined the Jacksonian goal. President Jackson represented the executive branch with his bold move of vetoing a bill which proposed a rechartering of the Second Bank of the United States.As conveyed by Jackson’s dramatic veto message on July 10, 1832 the democrats maintained that the national bank’s monopoly on trade catered too frequently to foreign and wealthy stockowners, thus posing a threat to the ideal of equal economic opportunity that they claimed to protect (B). The Jacksonians stuck with their vision of themselves in this sense, but opp osing reactions to the veto pointed out that the attack on the bank was unnecessary and dangerous.Daniel Webster’s reply to the veto correctly asserted that by raising the alarm about an encroachment of economic freedoms, the Democrats were really harming the stability of the economy needlessly (C). Webster’s analysis was proven accurate by the Panic of 1837, during which a bubble of inflation caused by the end of the national bank was abruptly burst, and several years of depression followed. The recession and unemployment caused indirectly by Jackson’s cancelation of the national bank did more harm to public economic opportunities than good, despite the Jacksonians’ passionate belief in the threat that the Bank posed.Also in 1837 however, Chief Justice Roger Taney’s Supreme Court decision of Charles River Bridge v. Warren Bridge was a decisive victory for the Jacksonian ideal of equal economic opportunity. Taney interpreted a 1785 charter for a br idge on the Charles River loosely so that a new bridge could be erected across the same river, thus dispelling a monopoly and financially benefitting the people (H).The Jacksonians evidently believed in their roles as the protectors of economic equality, but the results of the changes their administration made were again varied in agreement with their ideals. Finally, the Jacksonians most clearly drifted from their claimed ideals in the social sphere, as they actively neglected to guard the individual liberties of minority groups and women. The Jacksonian’s rosy call for extended suffrage only applied to white males, and the issue of slavery was deliberately avoided to prevent unwanted conflicts between the states.In fact, the Jacksonian administration even put in place a â€Å"gag rule† in 1836 that allowed Congressmen to file away abolition petitions without discussion because the Acts and Resolutions of South Carolina threatened independent state action if SC did no t receive national and sectional support in controlling its slaves (F). The slaves quickly lost any support from the proclaimed Jacksonian ideal of individual liberty when pitted against the preservation of the Union.Likewise, the administration did not hesitate to pass the Indian Removal Act of 1830, which revealed that grandiose Jacksonian ideals yielded to the American desire for new land as well. The Act forced thousands of Native Americans to resettle in the West, with no regard for their personal liberties either. Even President Jackson outright denied to protect the ideal when he refused to enforce the Supreme Court’s decision on Worcester v. Georgia in 1832; John Marshall had ruled that the Cherokee had a right to their land, but Jackson would not stop the army from pushing the Cherokee out of Georgia regardless.The only evidence of any agreement with the Jacksonians’ vision of guarding liberties is a romanticized painting of the Cherokee migration. The painted Cherokees appear comfortable, unified, and still dignified, implying that the painter must have either imagined this as the reality of the situation or painted an ideal version of the scene (G). The painting actually contrasts sharply with the chaos and tragedy of the Cherokees’ â€Å"trail of tears,† but it is important that the Jacksonian intent is present. Although the mixtures of realized and neglected Jacksonian ideals in the political and economic ields were more even, the Jacksonians’ goal to preserve individual liberty was not entirely lost in the social issues of the age. In conclusion, the Jacksonian democrats certainly believed in their roles as guardians of political democracy, equality of economic opportunity, and individual liberty, but their intentions were often misguided or secondary in the face of greater challenges. The few clear strides made by the Jacksonian age were interspersed with instances of failure in realizing its democratic ideals, particularly in the social sphere.

Factors of Conflict in the Workplace Research Paper

Factors of Conflict in the Workplace - Research Paper Example Conflict in the workplace is the result of certain factors. For instance, according to Fiore, the most important cause could be in the case where one feels taken advantage of. The employer/boss/manager may take advantage of the employees by overworking them but ultimately paying them close to nothing sometimes even not paying them at all. It could also be that the employee feels misunderstood in the workplace not only by the employer but also by colleagues. In other instances yet, the company may not clearly have interpreted its goals and values well to its employees. Conversely, the employee may have goals and values that are not in tandem with those of the organization that they work for. Thus, there are four main conflict resolution steps that employers and managers may take so as to reduce workplace conflict. (2008)  Sample the following scenario that has the potential for conflict in the workplace. A janitor is going about his daily duties in the organization that involve clea ning of not only the halls but also the restroom. He is about 45 years of age and has been a janitor for the last 20 years. On one occasion as he goes about his duties cleaning the restroom, a man in a business suit who forms part of the management approaches him and says to him, 'you should listen to me, you appear to be an intelligent fellow. For more than 15 years you have been cleaning the toilets. Why don't you try something else Do something with your life and get another job' In response, the janitor smiles and says to him, 'what And leave show business' the man in the suit cannot help but walk away and shake his head in disbelief? Few weeks go by and the janitor and the man in the suit meet gain. Just like before, the man in the suit says the same thing as before. This time the janitors seeks to question him, why he often speaks to him as he does.  Ã‚  Janitor: 'I often feel like you are putting me down when you constantly ask me to do something with my life and get anothe r job. Man in suit: I am only trying to help.  Janitor: While I understand that, I would appreciate it if you didn't ask me to get another job. I try not to let it bother me but it makes me feel inadequate. Man in suit: I apologize. Janitor apology accepted: Apology accepted. I enjoy interacting with the people I come into contact with as I do my job which is why I made fun of your comment last time. Man in suit: It was very smart and caught me off guard. I will try and be more appreciative of the job you do for us on a daily basis. Janitor: That will be very nice of you. According to Tillett, the first and the most important thing that the managers need to look into is communication skills. They need to understand how communication is carried out in their organization. Of particular significance is the way in which they communicate and how they could be teaching their employees to communicate with each other. (1999) For instance, it is important that the use of 'I' as opposed to you is encouraged.

Macroeconomics Creative Project Essay Example | Topics and Well Written Essays - 750 words

Macroeconomics Creative Project - Essay Example There exists a big distinction of social issues and economic issues within our society; but, what we may tend to consider more social can best be explained using the analytical tools of economics. Attending exhibitions such as art exhibitions or museum exhibitions is part of social life and fun. From a classroom perspective, attending an exhibition, particularly in a museum is often part of history, art, literature, and architecture courses, but acquiring any economics knowledge form an exhibition in a museum is quite novel. Attending an exhibit would be more beneficial to art, history and architecture student than it would be for an economics student. It would be perceived by many as having fun. Imagine of an exhibition relating to identity politics and issues of race in America in the 21st century. This is more history and political science. Race is a social issue in the society that has been on the forefront since the emergence of civil rights movements in the U.S. Such an exhibit can easily be integrated into a history and art class and help students understand class theory better; but, for a macroeconomics class, it would seem less beneficial. The typical introductory course in macroeconomics covers measurement of national incomes, theories of income determination, market systems, inflation and unemployment, and the effect of fiscal and monetary policies. Usually, there is no inclusion of a discussion on the economics of race. Taking a close reflection of the literature of race issues in the America, including some articles from reliable writers on issues of racial discrimination; for example, the issue of African American not being included in Social Security, it is possible to integrate race in economics. Issues of employment discrimination on the basis of race have been reported all over in the society since time in history. From classroom skills acquired in the macroeconomics course, I believe outside classroom activities

Business Ethics Essay Example | Topics and Well Written Essays - 1500 words - 1

Business Ethics - Essay Example Business ethics requires the view of all business activities through the moral values lens to determine their acceptability in the society. For smooth business operations, all business actors must understand various issues such as the personal aspects of business including family, sex, marriage and friendship, individual rights, and the moral values ascribed by society within which the business operate. This paper entails discussion of chapters covered between week 4 and 8 and identification of different sources of information with information relevant to the readings. â€Å"It’s Not My Problem†: The Concept of Responsibility (Chapter 8) The concept of responsibility is applied in business in different ways in which every actor in the business world has a role to play. The different forms of responsibilities in the business field include consumer responsibility, corporate responsibility and shareholder responsibility. The view of these responsibilities through the busin ess ethics lens converts them into moral responsibilities and prima facie obligations. In this case, all business actors are obliged to ensure fulfillment of such responsibilities in orders for the business undertakings to have constructive ends or positive impact on the society. Lange and Washburn (2012) have explored the meaning irresponsible behavior in the business world to divert attention from what corporate social responsibility implies, to what business actors do that is against the expectation of responsible behavior. The information covered by the two authors closely relates to information covered in this chapter in which much emphasis has been put on the different responsibilities and the misunderstanding of the responsibility concept in business. According to Lange and Washburn (2012), business responsibility is deeply rooted in the external expectations and the perception of the people experiencing the business activities. The survival and success of a business organiza tion or venture greatly depends on the widespread external perception as to whether the organization acts in socially responsible manner and the ability to meet external expectations. Nevertheless, it is important to understand that the responsibility concept in business does not imply going against self-interests in the business venture but the need to strike a balance between self-interests and societal interests. The greatest dilemma in business ethics revolves around the conflict of responsibilities, where individual’s power may be limited by external expectations or responsibilities. Social Responsibility and the Stakeholder (Chapter 9) Social responsibility in the business world refers to the different roles that businesses and actors in the business world have by virtue of operating within the society. The concept of corporate social responsibility encompasses the social concerns of stakeholders and the economic interests of business owners and shareholders. The corpor ate world and the society directly depend on each other, where the society cannot function in absence of economic and social roles of corporations while at the same time corporations cannot exist without the society. The stakeholder approach in business social responsibilities ensures that business owners commit to serve broader interests, in addition to the business economic and

Critically analyse the extent to which the classical doctrine on Essay

Critically analyse the extent to which the classical doctrine on communication of acceptance is relevant in modern contractual relations - Essay Example eement.1 The acceptance of an offer is crucial to formation of contract and question arises how and why an offer must be accepted to become a contract. Because there should be meeting of minds. An agreement can be verbal or in writing and in both cases how the acceptance reaches the offeror is important. The doctrine of communication of acceptance requires that it should be made within a reasonable time. In the following pages, the position relating to communication of acceptance is discussed. While dealing with acceptance, it is necessary to discuss the other ingredients of agreement. The objective test employed by the courts for an agreement having been reached is the presence of â€Å"offer and acceptance. 2 As is well-known, an offer is said to be made when one person proposes to another to do or abstain form doing some thing in order to get the consent of the other to the act or abstinence. An offer can be verbal, in writing, by conduct or by combination of these forms.3 An offer also can be made to an individual, a group of persons or to the world at large.4 An acceptance is made when the person to whom the proposal is made signifies his consent thereto and it should be absolute and unqualified. The acceptance must be communicated in the usual and reasonable manner except when the offeror stipulates a particular manner of acceptance.5 In other words, the acceptance of the offeree implies that the offeree has accepted the offer in the same terms and conditions of th e offeror without any reservation.6 Thus, it follows that an agreement comes into existence only when acceptance is received by the offeror so as to ensure that the offeror is not bound by his offer without his knowledge. The place of receipt of acceptance is generally regarded as the place of conclusion of contract. This applies to instantaneous methods such as verbal, telephone, telex and facsimile.7 In such forms of messages, the sender is notified as to its receipt or otherwise by the

Health insurance in UAE Essay Example | Topics and Well Written Essays - 750 words

Health insurance in UAE - Essay Example This paper discusses health insurance in United Arab Emirates (UAE). The United Arab Emirates provides high quality medical care to its citizens (Oxford Business Group, 2012). Health care services are available for the Emiratis in every public hospital and other primary health centres. This means that all citizens have good access to medical care. The efforts to reduce the expenditures in the UAE’s Ministry of Health have led to introduction of a compulsory health insurance scheme. This will enable the citizens to access medical services in both private and public hospitals. Hence it is a boost for the UAE economy as it enhances the quality of medical care for the UAE residents. Provision of healthcare insurance is very relevant and important for the economy of the United Arab Emirates. This is because it cuts on the government’s yearly medical budget as well as generates revenue from the taxes paid by the insurance providers. Health insurance promotes quality medical care hence improved living standards. This leads to increased productivity which is relevant for the economy. Expansion in the health insurance sector is a relevant economic factor because of increased premiums for the insurance industries. Borscheild and Haueter (2012) assert that the premiums from every employee are subsidized by the government to cater for the occurrence of a future need. Health insurance has a bearing in the social and health needs of the UAE. Medical history reveals that the UAE had only 7 hospitals in 1979, but these have increased to over 68 today (IBP USA, 2009). This shows expansion in the provision of social amenities and health care. It is revealed that the absence of private health care brought about a foul cry, which forced the UAE government to roll out free care for all the residents, both local and foreigners (Borscheild & Haueter, 2012). Hence incredible developments had to be adopted with

Explains the political changes the United States underwent in the Essay

Explains the political changes the United States underwent in the 2008-2011 period in regards to political philosophy - Essay Example This revealed that racism did not guide the elections, as people wanted change in governance. Citizens of the United States expressed dissatisfaction with the republican governance, following George W. Bush’s two terms in office. There have also been changes in the voting structure, with many young people and minority groups participating in the polls. As a result, the political philosophy in the United States from 2008 seems more different than any other period in the history of the United States. Preference for young contestants increased; as such older candidates faced enormous challenges and opposition from younger ones. John McCain could not have defeated the democrat candidate because most Americans did not prefer people of his age (Lasser, 2011). In addition, dissatisfaction with the administration of George Bush made the electorate opt for a difference. An increased number of casualties in Afghanistan and Iraq war changed the mood of American electorate, following deaths of many American soldiers while fighting terrorists. During the Bush administration, economy of the United States had started to experience reduced growth. The banking industry and the stock market performed poorly in the start of 2008. Therefore, the political preference shifted to a Democratic candidate rather than a Republican (Katzman, 2011). During his presidential term, Barrack Obama has publicly declared his stand on various issues regarding American politics. Several legislative measures have been undertaken between 2008 and 2011. Politics have mainly been centered on revival of the United States economy; and provision of heath care that all Americans can access and manage to pay for. Obama administration has also strengthened social security and system of education in America. Political ideals have also been centered on withdrawing American troops in Iraq and Afghanistan, as well as stopping

Representational view of the simpsons Essay Example for Free

Representational view of the simpsons Essay The Simpson began as a short series of cartoons in the late eighties. It was in 1989, when the Fox Broadcasting Company employed Matt Groening, a cartoonist, to make the Simpsons into 13 half hour episodes that began its path to fame. No sooner had the show aired than the Simpsons became the highest rated show on Fox. However along with the high rating came huge criticism about the content of the show. The Simpsons seemed to satire the educational system, religious system, the American political system and all the American institutions that the American people believed strongly in. It seems to be one of the most controversial programs on TV today because of this. The Simpsons consists of a family of seven. Homer and Marge as the parents, Bart, Lisa and Maggie as the children, a cat Snowball II and one dog Santas Little Helper. The characters of the show were named after Matts real family. The Simpsons are a stereotypical view of an American family and this is another thing that the show has been criticised for. They live in a small town called Springfield. There are many similarities between the Simpsons family and a traditional sitcom family, however they are far from being an ideal family. The Simpsons have been called a dysfunctional family, for not sticking to the familiar family traits that other sitcom families do, for example, the Cosby show or the Waltons. The shows executive producer, Mike Scully, said critics have blamed the Simpsons for being dysfunctional family but they forgot the part that they are still a family and a lot of other families dont survive marriages. They have been called dysfunctional as the Simpsons are not always happy, they always seem to have money problems, Homer is a lazy father and Bart is always naughty. However to me the Simpsons show a more true-to-life family one that all people of any age can relate too. We all know that all children are not prefect or that not all families get on all the time but the Simpsons family sticks together whatever happens. HOMER. J. SIMPSON. Homer is one of the many characters in the show that is stereotyped. He is a stereotype of a typical male middle class, white, American man, as he has a beer belly, is not very bright and is always thinking of food. He is the father of the family, but has very little control over them. Homer works in a nuclear power plant as a safety inspector. His very rich boss, Mr. Burns, owns the plant that Homer works in. Mr. Burns does not think much of Homer, he thinks Homer is a stupid lay about because he does no work. Homer hardly ever thinks before he acts, for example, he bought Lisa a pony because she said she did not love him anymore, but ended up giving it up as they could not afford it. Compared with Ned Flanders, as a father, Homer is not the happiest of the kindest, but like Ned, Homer loves his family. Homer does not like Ned very much but Ned considers Homer as a friend. Ned is always happy and very religious; this is exaggerated a lot on the show. He is famous foe saying howdy-doddlely . Ned also out stages Homer, not on purpose, for example, in the episode called Simpsons Roosting on a Open Fire. Homer puts up Christmas lights but they are not very good then Ned put his on and they are excellent. The Flanders role is to highlight the negative points of the Simpsons family life. Each character has a particular characteristic for them, Homers is that he only has two strands of hair and he loves his food, especially donnuts, umm Donnuts. MARGE SIMPSON. Marge is the mother of the family. She too is a stereotyped as a housewife. Marge does not get out much but does most of the running around for the family and has hardly any time for herself. Marge is the more dominant one of the marriage. She is a very kind, loving mother. When you compare Marge with Mrs Lovejoy, the vicars wife, Marge is not a gossip or as religious as Mrs. Lovejoy but is a considerate and sensitive lady. She loves all her children very much no matter what they do. For example, when Bart ruined Thanksgiving by setting Lisas centrepiece on fire, Marge still forgave him afterwards. Marge keeps her family together, and although more tired and stressed out than usual sitcom mothers her role is actually quite similar. Marges main characteristics that all people associate with Marge is her tall, blue hair and she always says ummmm BART SIMPSON. Bart is the oldest son of the family and the most mischievous. Barts character was created to be a typical naughty schoolboy. He is constantly getting into trouble and is always playing pranks on people, especially Moe. Bart always rings up Moes Cabin asking to speak to someone, for example, Can I speak to Daily please? First initials, I. P. then Moe says hey everyone listen I. P Daily. and then everyone laughs at him. If you compare Bart to Tod Flanders, his neighbour, Bart looks like the devil. Tod is exactly the opposite to Bart, he is extremely religious, happy, constantly singing hymns in his sleep and he never says anything bad to his parents or friends. Overall Tod seems to be the prefect child. Barts main characteristic that all people associate with his character is his sayings. For example, Eat my shorts! or Dont have a cow man! but his most famous one is hay rumba! He is also well know for the comical lines he writes at the beginning of each episode, for example, I will not sell land in Florida. And I will not teach others to fly. As he writes this we know that these are all things he has done in school and has got into trouble for. This also further increases the shows satirisation of the educational system. LISA SIMPSON. Lisa is the most intelligent member of the family. She enjoys various sorts of activities to do in her spare time. One of which is to play the saxophone. She loves to play it even though Homer hates the noise it makes. Lisa also loves to read and write essays, she has written a number of competition essays and one of which won her and her family a free trip to Washington DC. As well as being an intelligent young girl, Lisa is an A grade student and she hates getting anything lower than an A. She also has a good sense of morals, for example, when Homer was stealing cable or as he put it getting free cable. Lisa was totally against it because it was unethical. Lisas character is not stereotyped as much as other characters in the show, for example, Apu; he is a typical Indian character in the show. He owns a Quickie-mart with an Indian accent. This is a one-dimensional view of an Indian man. Lisa is an eight-year-old girl who has the metal mind of a person twice her age. Lisa has been told that sometimes she is too clever for her own good. In some ways however Lisa plays the part of a typical sitcom daughter, well behaved, and more intelligent than her older brother. MAGGIE SIMPSON. Maggie is the baby of the family and loves them all very much. Even though Maggie is only one year old she has achieved so much. However in many ways she is a stereotypical view of a baby, always sucking on her dummy, and playing. However Maggie is one of the few clever Simpsons. She has learned how to spell her own name on an Etch-A-Sketch, she has wandered around town by herself, shot the richest man in Springfield, and has survived living with bears in the wilderness. However Maggie has still not said her first words. Her main characteristic is the noise she makes when she sucks her dummy, suck suck suck suck. Even though the whole family are very different they all love each other a great deal, which is what gets them though all the difficulties they have come across. The end of each episode has the family resolving their problems and although they are criticised for being dysfunctional I believe that they actually portray a more realistic family unit. A solid family that works through problems together is certainly not something to be criticised especially in this day and age.

An Internal Analysis Of General Motors

An Internal Analysis Of General Motors General Motors is the international company; the global market share is estimated at 12.5 percent as of 2008 (General Motor, no date), therefore, base on the global market coverage, it can benefit from many regional countries, to allow better access to resources, talents and knowledge.. According to Boston Consulting Group, General Motors is one of the 25 most innovative companies in 2008, it arises from the concept cars like the electric Volt and a renewed focus on design. The workforce is value driven and understands the business mission and vision. 1.2 Intangible resources The intangible resources of the organization are its brand image and its reputation. General Motors has its strong reputation for innovation and new products in the automotives market. This means that customers will often associate the brand with new technology and products. 1.3 Core competencies General Motors core competencies including innovation. They constantly strive to deliver new and exciting products to the market, and human resources. General Motors have a well prepared knowledge and flexible workforce who is concentrated on reaching business goals. General Motors starts recommending vehicle installed OnStar satellite in case of an emergency or theft.(Onstar, no date) 2 Factors affecting the Automotive Industry (PESTEL Analysis) 2.1 Political The safer vehicle and the environment in the American market is not restricted at a great disadvantage. Government and authorities in the region made efforts to regulate, eventually start the implement and set the laws and regulation since the 1960s (L.S. Robertson, 2006) 2.2 Economic According the number of studies (Catherine Rampell, 2008), is the major user of steel, iron, lead, plastics, vinyl, rubber, aluminum, textiles, and computer chips. The study also mentioned that for each autoworker there are seven other jobs created in other industries. 2.3 Sociocultural The consumer automotive market, especially the vehicles market, used to be dominated by men but nowadays is not true. Marketing strategies are also focusing on emotions. The presence of women in the market shall turn out as fallback to cushion declining sales On the other hand, young consumers increase their spending power by working in a part-time job or by receiving larger allowance from their increasingly financially successful parents. 2.4 Technology the internet has a great impact on every industry in the anywhere and has also affect the automobile industry. According to the J.D. Power and associates 2006 New Autoshopper.com Study, 59% of the buyers referred to the internet before making their purchases and out of that 59%, 85% of the serious buyers visit the manufactures website before going to the dealer for a test drive. 2.5 Environmental On 30th November 2007, the business leaders of 150 global companies published a communiquà © to world leaders calling for a comprehensive, legally binding United Nations framework to tackle climate change. In fact, there is a programme in which WWF and businesses collaborate to show leadership in addressing climate change. 3. Porters Five Force Anaylsis 3.1 Threat of new entrants The threat of new entrants is relatively low. This is mainly due to the fact that it will cost a new company a lot of investment cost and know how to enter the automotive market. 3.2 Threat of substitute products Tthe coverage of destination is limited, besides, bus, metro, taxi are not products that can replace the vehicle. Therefore, the threat of substitute products is relatively low to moderate 3.3 Bargaining power of suppliers Due to the fact that suppliers are united, they are a very strong threat in the automotive market, nevertheless, the United Auto Worker, the only supplier of labor, has exerted a great deal of the leverage over the benefits and wages provided by the big three, therefore we can conclude that the bargaining power of supplier is strong. 3.4 Bargaining power of buyers As we look at consumer behavior when buying vehicles the conclusion can be made that their power is relatively low. Mainly the reason for this low power is because individual consumer has some power over price within a given dealership, but low power over manufacturers. 3.5 Competitive rivalry within the industry Due to the fact that there is a intense completion going on in the auto industry the rivalry within the automotive market is high. There are a lot of choice and intense competition in internet sales that can damage a company when they dont keep up with their competitors. See appendix 1 for more information. Summary Porters Five Forces Framework Porter (1980) http://dpj.typepad.co.uk/entrepreneur_zest/images/2008/03/26/porters_five_forces.gif 4. SWOT Analysis 4.1 Strengths 4.1.1 Large Market Share Although GMs market share in the US has declined, it still retains a competitive 21.5 percent in 2008. They also start to penetrate the Chinese customer automotives market and gain an increasing share. 4.1.2 Global Experience They have already been a global company for almost 100 years now and have established themselves as the global leader for most of them, they already have global experience. 4.1.3 Variety of Brand Names GM has been the automotive leader for the majority of the last century. The current GM brands included: Chevrolet, GMC, Cadillac, Buick, Pontiac, Saturn, Saab, Daewoo, Opel, and Holden. 4.1.4 GMAC Customer Financing Program Since its establishment in 1919 it has proven to be GMs most reliable way of revenue. 4.1.5 OnStar Satellite Technology This technology allows the vehicles to be tracked in the event of an emergency or theft, it also allows the driver or passengers the ability to communicate with OnStar personnel at the click of a button. 4.2 Weaknesses 4.2.1 Behind on Alternative Energy Movement The alternative hybrid trend has begun to take place in the automotive industry and GM has been one step behind the competition It is a reason that lead them to lose the market share. 4.2.2 Organizational Structure below standard We can see the organization structure above, it seems to be too vertically integrated. The shortcomings are that it is a lack of communication between employees from top to bottom and may have played a part in GM lag on the hybrid technology. 4.2.3Stagnant Profitability To review GMs profit we can see that they are struggling with respect to the size of the company. It is disappointed that shareholder would not happy about. 4.2.4 Too dependent on domestic market GM overly become dependent on the US market The competition will become tough if they only focus on just one country. 4.2.5 Too dependent on GMAC GM overly become dependent on its financing company. Maybe it is a very great advantage for GM, but they cannot rely on financing company for gain the profit, if they want to compete with competitors 4.2.6 Poor Credit Rating GMs credit status has been declining obviously. According to Standard Poors, they declared billion of dollars of debt owed by GM to be junk, it will increase borrowing costs and limit fund-raising options for them.(Sharon Silke Carty, 2005) 4.3 Opportunities 4.3.1 Alternative energy movement hybrid technology is giving to GM a opportunity to once again become the automotive industry leader in innovation and technology 4.3.2 Continuing to expand globally Recently, they found that there is a dramatically increase in China automotive market, that proves themselves to focus and emphasis on Chinese huge potential market. 4.3.3 Low interest rates It is a opportunity that they can gain immediate revenue in sales. 4.3.4 Develop new vehicle styles and models As you know the development of the vehicle model never be satisfied, GM should keep attempting the most popular model in the world. 4.4 Threats 4.4.1 Rising fuel prices GM the major business pieces is the sales of truck and SUV, but it is gradually declined because of the lack of fuel efficiency, for rising fuel prices this reason, 4.4.2 Growth of competitors Toyota gains the first mover advantage of hybrid system and become the leader of automotive industry, GM faces the reality they should develop more fuel efficiency vehicle and small car model. 4.4.3 Legacy cost GM provide a great welfare to their retired employees, so they need to experience it is a serious financial problem more and more retired employees begin to be benefited. 5 Key strategic issues firm face GM is facing several strategic issues. It will be shown by a number of aspects. The economic recession, Competition, and New entrants. 5.1 Economic Recession The economy of world is currently in the downturn, or even recession in which will impact a lot of industries and consumers in the world. As the economy slowing down, doing business may find it hard to increase the revenue because banks are not willing to lend money to companies, especially bad credit status companies. As a result, GM may suffer from decrease in sales of automotives. However, they may benefit from hybrid technology, since more people are buying new fuel efficiency car model if consumer intend to purchase a new car. 5.2 Competitor in automotive industry The competition between the main competitors for instance, Toyota, Honda, Ford, has expanded beyond that of the traditional energy car. These companies are now competing on value adding products, fuel efficiency, design and service of maintenance. In order to achieve competitive advantage companies try to merge or acquire other companies in gain experience and knowledge. 5.3 New entrants The fast moving and constantly changing industry gives opportunities to new entrants who have not previously specialized in the automotive market as technology can often be easily imitated. In addition an alliance formed by different companies, such as Magna and Sberbank, trying to seek a takeover of Opel and Vauxhall, parts of GM European division. There has also a growing number of new entrants in China due to relatively low investment costs. 6 Strategic Options The strategic options that GM might realistically pursue had been discussed as below such as product development, market development,, and market penetration as their strategic options. (Ansoff, 1987) 6.1 Product development Product development is a feasible option because it can be capable both in present and the prospect. This option is also suitable to GM since it is able to exploit the needs of customers. GM is facing the threats of too much competitors in the market and the improvement of technology is increasing rapidly. Doing more researches on GMs products and investigate and understanding more about the favour of customers are significant. To improve the quality of services in their technology should not be neglected in order to catch up the expect of consumers and the externally needs so the company can be enhanced and stable in the market. . Suitability: Due to the furious competition between the three consoles company, this strategy is very important. GM have to develop more innovative new products to catch up the pace of the technological advancement. This could ensure the company to gain a step ahead of its competitiors. The technology industry requires swiftness in developing new and innovative products. Feasibility: The research and development of the company is doing very well in the past few years. They had been rated the top 25 most innovative company in the world (businessweek, 2008). In skills term, they have talent in which had leaded and developed a lot of innovative ideas which attracted a lot of consumers from youths, adult and even elders!. Acceptability: Shareholders and stakeholders will agree wholly with GMs decision as technology investors, they realize that the technology development or innovation is very important to maintain themselves into the market. 6.2 Market penetration According to (Lynch, R., 2006), market penetration uses the existing range of products or services to attract potential customers or consumers from competitor to expand their own market share. It basically targets the existing customers in the present market of competitors and gain profit from it. The cost is lower, and the chance of attracting consumers is higher. GM uses price competitive advantage to deal with their competitors, which are Toyota, Honda, Fordà ¢Ã¢â€š ¬Ã‚ ¦etc. When GM introduces their brand, it is priced lower than others significantly. GM offer lower prices than competitors and it attract competitors consumers to have an attempt on the product due to its low price Suitability: This strategy is suitable to be implemented because this allows GM to grab more market shares from their numerous of competitors in the US market, Feasibility: GM is capable to compete with competitors as they are financially competent with them. Since GM stated that they retain the GM Europe section because they posses good health of liquidity again. This also shows that GM is financially secured and there is no problem for them to compete with their competitors. Acceptability: Stakeholders of GM will accept the strategy as long it is profitable to the organization and to them. 6.3 Market development From this strategy options, it involves the strategy of introducing both new and existing products to their existing customers and not overlooking the new customers. From here, both new and existing customers have the options and more information about those products. From here, they would be introduce their product to both side. They would be able to saturated new opportunities for geographical spread, entering new segments or new users in the environment. Such a change will enable GM in better turnover at medium risk by exploiting current strength or market knowledge. (Lynch, R., 2006) Suitability: It is suitable in market development as they tend to target new groups of people, and other new geographic places. (Johnson, Scholes Whittington, 2008). GM try to targets elders, adult or basically all adult group. Feasibility: It is feasible for GM as they have enough resources to develop into the new market. According to the U.S. Treasury Department, they has received $1 billion repayment from GM. This shows that they totally have enough fund to research and develop proper strategy in order to venture into the new market or exploring into a new market segment. Acceptability: Stakeholders and shareholders would surely agree with this strategy as it gain profit for the company and stakeholders or shareholders themselves. Recommendation of Strategy options The best strategic option for GM in their current situation is to develop a new vehicle which is alternative energy SUV; this strategy would significantly lead to a growth in sales. In addition, GM can gain a new business partner from Canada-Russia consortium because of their contractual relationship and this would be a great competitive advantage for GM. By using this strategy, which is an external method, it can help GM to solve their problems quickly. An alliance with Magna-Sberbank will also benefit GM in terms of saving expenses on research for the small car technology. This will aid GM in solving the tribulations that they are facing. 7 Implementation Issues When GM wants to withdraw or retain Europe operation, some obstacles will appear. They are called implementation issues. These implementation issues will be discussed how Government involved in its decision making. 7.1 Russian Government German Gref, the Sberbank boss stated that he acquires GM Europe section to enter new markets abroad and confirmed it is a really low price to takeover Opel, it can get the permission to get its company technology. Sberbank is a famous creditors of automotive industry in Russia in which inform to turn around its carmaker industry. The Russian Trade and Industry Minister stated that if not the takeover, they will not good enough to develop the international car company Since the Russian government plans to acquire international technology, all of the international autos keep continue to build some facilities such as manufacturing plants in there to reach domestic market demand. The global companies can avoid Russian import taxes on autos, besides, they can benefit from the domestic low salaries. Nowadays, to sell the GM Europe, GM will struggle in the Russian market and GM technology. 7.2 US Government According to American negotiating circle, the US government is not welcomed to work with the Magna as long as Russian companies get extensive access to GM patents. As Magna keen to get access to all developments from the GM patent pool. US government concerns about not only car technology, they concern about Russian getting the military items. On the other hands, GM concerned that the consortium will use GM technology to contribute a potential threat of competitor to GM another brand in the Russian market. (Reuters, 2009) In addition, China Beijing automotive offer and promised job cut, and required financial support are much better than the other two, Magna also design Opel the best way for future development, but the GM is welcomed RHJ. According the analyst, GM expects to be able to make a comeback again in the future control of Opel, RHJ operation mode exactly GM wants, RHJ can give a hand to GM achieve a deep restructuring.(Wall street journal, 2009) 7.3 Germany Government According to Reuters(2009), the Germany Chancellor Merkel played an important role in the incident that Opel will be sold to Magna. Why? Because she takes the mission to save German unemployment rate, this is a political activity, she is preparing the coming election. How Merkel convinces Magna, allowed to retain as many jobs as Germany can be affordable? This is simple word money. The Germany Government wait for the European Commission approved a loan of 4.5 billion Euros to the Opel. Furthermore, the plants will not be closed in Germany, this seems very favorable for Merkel, and she is expected to easily win the election. On the other hands, it will damage their relationship if there is any failure of Russian involvement in Opel since Germany is apparently work well with Russia about trading. Besides, German election is almost can be said to the future destiny of the Opel that become a political hot potato. Government has invested a lot of money into the Opel, which is an economic decision making, but also the reason of political consideration. To unresolved Opel future, Merkel will be faced more political pressure. On the other hands, Germany government will not give any financial support to GM Europe section if GM remains it. It makes sense that Government will not help a outsider in which Germany has big three carmakers (Audi, BMW, Volkswagen),to survive its operation. As GM has already damaged the GM Europe section operation and put Opel on the transfer list. Germany cannot build any trust on GM, but by the way they have a good relationship with Russia as I mentioned above. Conclusion As a conclusion, GM sales have been affected by the economic downturn and in order to increase their sales the best option is to launch a alternative energy SUV. Besides that, GM wants to protect themselves on the market, and for this reason GM needs to be more innovative and be more sensitive to the current trends of the market. Knowing the needs of consumer will retain GMs position in the market as the worlds leading producer of mobile phone. 8. Reflective Statement on Leadership Leadership is defined as entails some degree of voluntarism in followship (Pardey, D., 2007). However, Kotter (1990) argues that leadership is about coping with the changes by giving direction. Kotters (1990) leadership framework states that the role of leadership is to set a direction, develop a vision of where the organization should be going and the strategy for change to achieve that vision. Leaders align people to the direction being set, communicating it to people and building commitment to it. Motivating and inspiring people so that they work to achieve the vision drawing on their needs, values and emotions. Our group were using democratic style of leadership where all of our members were involved in the decision making process. Ryan, our group leader discussed with all of us and decides which part we need to work on step by step. This helps a lot as we were very unclear about the direction were heading to. Communications In my group, there is some communication issues as the all four of us come from different country and background. Sometimes, there will be conflict of ideas and difficulty in presenting the ideas. We got delays in meeting because some of our group members were busy or there were some lack of communication as in setting up our meet up time got messed up and caused some group members turned up and some did not. I learnt that working in a group is not easy as it needs more effort to discuss and accomplish it. Responsibility We were each given task to search for the information from external analysis to internal analysis. All of our group mates were very responsible that they all searched for the information on their own part and some of them even helped to search for others part of work in detail. This helped our group to have the sufficient information for us to do our work. I learnt that without time management, everything would be in a total mess as we might have hand our assignment late. Problem solving We overcome our communication problem by spending more time during our meetings and we make sure everyone understand what are the things that we need to accomplish and understand what our group members were discussing during the meeting. I had learnt that developing a vision is important as it helps our group to do our work on time. Reference list Ansoff, H.I. (1987) Corporate Strategy Revised Edition. London: Penguin Books. BusinessWeek,(2008),The 50 Most Innovative Companies. [online] Available at: http://bwnt.businessweek.com/interactive_reports/innovative_companies/ (Access: 9th January 2010) Catherine Rampell, (2008), How many jobs depend on the big three?, Economix, [online] Available at http://economix.blogs.nytimes.com/2008/11/17/how-many-jobs-depend-on-the-big-three/ (Accessed: 9th January 2010) General Motors,(no date) GM Global Vehicle Sales and Market Share-2008,December [Online]. Available at http://www.gm.com/europe/corporate/sales/global/ (Accessed:9th January 2010) J.D.Power (2006), Research a car online[online] Available at: http://www.jdpower.com/autos/articles/research-a-car-online (Access:9th January 2010) Johnson, G., Scholes, K. Whittington, R. (2008) Exploring Corporate Strategy. 8th edn. England: Prentice Hall. Kotter, J. (1990) what leaders really do Harvard Business Review. L.S. Robertson, (2006), Motor Vehicle Dealth: Failed Policy Analysis and Neglected Policy, Vol. 27,p. 182-189, Journal of Public Health Policy Lynch, R. (2006) Corporate Strategy. 4th edn. Essex : Pearson. Pardey, D.(2007) Introducing Leadership. Oxford: Butterworth-Heinemann. Onstar,(no date) [online] Available at: http://www.onstar.com/us_english/jsp/index.jsp (Access:9th January 2010) Reuters (2009) GMs U-turn on Opel sale angers Germany Russia. Available at: http://www.reuters.com/article/idUSTRE5a25RL20091104 (Accessed:: 5/01/2009) Sharon Silke Carty, (2005) SP cut GM, Ford debt rating to junk status, [online] Available at http://www.usatoday.com/money/autos/2005-05-05-gm-junk_x.htm ,USA Today(Accessed:9th January 2010) Vincent J. Galifi. (2009) Magna Confirms Offer for Opel, [Online]. Available at: http://www.magna.com/magna/de/media/pressreleases/?i=218 (Accessed:5/01/2010) Vincent J. Galifi. (2009) MAGNA AND SBERBANK OFFER SELECTED AS THE PREFERRED SOLUTION FOR OPEL [Online]. Available at: http://www.sbrf.ru/en/news/index.php?id114=11000080 (Accessed:5/01/2010) Wall Street Journal (2009) RHJ improve the terms of its bid for GMs Opel.[Online]. Available at:http://online.wsj.com/article/SB125182274337876583.html

Analysis of Changes in Wage Rates in India

Analysis of Changes in Wage Rates in India CACP – The data source which is an important source for wages in rural India unfortunately does not publish the wage data from these studies. The cost of cultivation scheme collects data from the selected household at regular intervals on all aspects of farm business. The valuation of human labor has always been a problematic issue for the CACP. There is however a number of limitations associated with this data such as lack of data for many crops, problem in aggregating the data from the state level, unavailability of data at the state level etc. The four reference points have been used to look at a change in the wage rate. However, the choice of reference points in the study have been limited to the years 1983 and 1987-88 in the 80’s and 1993-94 in the 90’s. AWI Wage Rate: It is the most widely used source for analyzing trends in wage rate for rural India. However, one given problem noticed by the researcher with respect to AWI is the time lag in the data and the method of aggregating all the data for th states and all India level. There were many studies performed to check the average wage rate of labourer in different parts of India by many previous researchers. However in all these studies used simple average of wage rates in different months to arrive at the annual figures. Also, all these studies used population of agricultural labourers’ from the census as weights to arrive at the wage rate. The average AWI wage rate is 30-40 percent higher as compared to the RLE/NSS estimates of wage rates. It was also studied that the AWI wage rates were found to be marginally upwards biased as compared to the FMS estimates of wages. A look at the wage trends from AWI since 1980-81 suggests that the wage rates have generally been higher in Punjab, Haryana, and Kerala as compared to the other states in the 80’s. However, by the end of 1990, Gujarat and West Bengal have seen to have made a significant contribution and is closer to the traditional high wage rates. Rajasthan, on the other hand which was closer to the traditional high wage rates has fallen behind and is considerably lower than Kerala, Punjab and Haryana. The states on the lower ends of the wage rate are Orissa, Bihar, Madhya Pradesh, Maharashtra, and Andhra Pradesh. Kerala’s wage rate is found to be three times those in Orissa and in Punjab and Haryana they are twice the wage rate in Orissa. Wages have continued to grow in all three time periods but there was a significant slowdown in the wage rate during the period 1987-88 and 93-94. For the period 1987-88 to 1993-94, which also included the year of the financial crisis and the conseque nt economic reforms, wage rates show a deceleration in almost all the states as well as the all India level. This decline in the growth of wages is sharper for Andhra Pradesh, Assam, Bihar, Gujarat, and Karnataka. Wage trends from CACP – In the study it was found out that this wage rate had been created by aggregating over crops using crop specific weights for individual states. Wage rates from CACP are generally found to correlate well with the AWI series in spite of the differences in methodology of collection of data. In fact, for both 1983 and 1987-88, wage rates from CACP show a very high correlation coefficient of 0.95 with the AWI wage rates. It was also analyzed in the study that wage rates from CACP show better growth rates in 1987-88 and 1993-94 as compared to the AWI series. Rajasthan turns out to be an exception which showed markedly improved wage rates as showed by AWI wage series. Even for other states, CACP does not show as sharp a deceleration as AWI. Maharashtra on the other hand shows sharp deceleration in the wage rate during 1987-88 and 1993-94. NSS – These wage rates were found to be statistically more significant and reliable than that of AWI or CACP because of their consistent and superior sampling framework but they also allow for a much higher level of disaggregation. The wage rates reported by NSS for males for agricultural occupations are considerably lower than the wage rates reported by AWI. It was further studied that the NSS rates were very well correlated with the rates of AWI series with a correlation coefficient of around 0.9 in the 1980’s, 0.87 in 1993-94 and 0.95 in 1999-00. However, again for most states there was a deceleration in the growth rate during the next sub period that is, between 1987-88 and 1993-94. Except for the state of Gujarat, deceleration was seen for almost all states. Gaps between agricultural and non agricultural wages narrowed down considerably in Punjab and Rajasthan. West Bengal and Karnataka were found to be the states having the lowest divergence between agricultural a nd non agricultural activities. Comparison between agricultural and non agricultural activities gender wise was also performed. It was analyzed that the growth rates in these activities in males suggested that in 1983-84 and 87-88 agricultural wages grew faster than non agricultural wages. For females, however non agricultural wages grew faster than agricultural wages. RLE/ALE wage trends – RLE uses a subset of households from the NSS employment and unemployment quinquennial surveys. The household types are either self employed in agriculture, non agriculture, agriculture labour, other labour and others. A comparison of the wage rate from the NSS and the RLE reveal that there is a high degree of correlation between them especially in the 1990. This is true not only at the all India level but also at the state level. For the period from 1983-1987, growth rates from RLE were shown to be 70-80 percent higher than the ones shown by the NSS series in the study. The growth rate of the NS were similar to those studied by AWI or CACP for many states, however the RLE suggests growth rates higher than any of these. Further it was also studied that the growth rate of wages in agriculture are much higher than those suggested by NSS or CACP or AWI. There was also an inter range comparison done in this study to check the difference in wages under the RLE scheme. It was analyzed that comparing the 1977 RLE scheme to that of 1983, real wage rates declined by almost 10 percent for the latter. Such a decline in wage rate is not accompanied by any other wage estimates including those from NSS. WRRI Wage rates from eleven agricultural operations and seven non agricultural operations. For wage rates for agricultural operations, simple average of sowing, transplanting, weeding, harvesting, winnowing and threshing was taken as the representative wages. The analysis in the study was based for the years 1990-2000 and 2002-03. It was studied that the wage rates from WRRI for agriculture than those reported by NSS and RLE for males and almost 60 percent higher for females. WRRI estimates of wage rates are found to be closer to the CACP or AWI estimates because of similarity in methodology and sources. WRRI is the only estimate that is available after 1999-00. It was further analyzed in the study that the wage rates between male and female for agricultural occupations as well as non agricultural occupations have grown in real terms. Coming to the state wise analysis, Bihar and Orissa are the states that saw the highest growth rates of wages for both males and females. The other st ate that achieved close to 10 percent per annum growth rate of wages is Kerala. Apart from Kerala, the other two states – Andhra Pradesh and Tamil Nadu saw slower growth rates of wages. Uttar Pradesh having a large concentration of poor and rural labourers witnessed growth rates of less than two percent in both agricultural and non agricultural operations.

Personality Essay -- essays research papers

  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Psychology covers a huge field and one interesting aspect of it is personality. Personality by itself involves various issues. Some aspects are Psychoanalytic, Ego, Biological, Behaviorist, Cognitive, Trait, and Humanistic. Different types of behaviors are amazing to learn about, mainly the behavior therapy, collective behavior, crime and punishment, and Social behavior and peer acceptance in children. I chose Behaviorism over the other aspects because I believe behavior determines human personality and is very interesting. You can tell what one is by his behavior, and one behaves according to what place he has in society. By doing this paper on Behavior, I hope to get a better understanding of, if behavior develops a personality or if personality guides behavior. I also see behaviorism helping me in the future with my personal and professional career by understanding human personality and behavior better than I do. No matter what your major is , if you can determine one's personality by his behavior you can really get your work done from that person and understand the better than you would otherwise. This person could be your employee or your employer. Behavior Therapy Behavior therapy is the application of experimentally derived principles of learning to the treatment of psychological disorders. The concept derives primarily from work of Russian psychologist Ivan Pavlov. Behavior-therapy techniques differ from psychiatric methods, particularly psychoanalysis, in that they are predominately symptom (behavior) oriented and shows little or no concern for unconscious processes, achieving new insight, or effecting fundamental personality change. The U.S. psychologist B.F. Skinner, who worked with mental patients in a Massachusetts State hospital, popularized behavior therapy. From his work in animal learning, Skinner found that the establishment and extinction of responses can be determined by the way reinforces, or rewards, are given. The pattern of reward giving, both in time and frequency, is known as a schedule of reinforcement. The gradual change in behavior in approximation of the desired result is known as shaping. More recent developments in behavior therapy emphasize the adaptive nature of cognitive processes. Behavior-therapy techniques have been applied with some success to such disturbances as enuresis (bed-wetting), tics, phobias, stutteri... ...nvolving children to learn social acceptance showed us clearly how one behaves makes him what he is. I believe the same for adults. I believe if one behaves in a certain way for a long time, not only society with believe you are what you are behaving as but he himself will start believing he is what he is behaving as. Also I have learned to be more patient with people because I take a step in the further and think why a person would behave in a particular way. I now can see a clear difference between normal and abnormal behaviors. Benjamin, L. S. (1982). Use of structural analysis of social behavior (SASB) to guide intervention in psychotherapy. In J. C. Anchin & D. J. Kiesler (Eds.), Handbook of interpersonal psychotherapy (pp. 190-212). New York: Pergamon. Benis A.M. Toward Self & Sanity: On the Genetic Origins of the Human Character. Psychological Dimensions Publishers, New York, 1985 http://pmc.psych.nwu.edu/personality.html Carson, R. C. (1969). Interaction concepts of personality. Chicago: Aldine Gurtman, M. B. (1992b). Trust, distrust, and interpersonal problems: A circumplex analysis. Journal of Personality and Social Psychology, 62, 989-1002.

Expeditions to Antarctica and Annapurna Essay -- Comparison, Shackleto

When taking a quick look at the two expeditions, one led by Ernest Shackleton to Antarctica and the other led by Arlene Blum to climb Annapurna, a quick summation can be made that Blum succeeded in her expedition and Shackleton failed. But this is a shallow view, not considering the nuances and actual experiences of the trips. Ernest Shackleton set out with his crew in 1915 to be the first expedition to cross Antarctica, but in fact, he never set foot on the continent. While failing at his initial goal, he was a highly successful leader and kept his 28 men safe for close to two years, while they were trapped on the ice floe and then, after the ice gave way, when they were paddling hundreds of miles across open seas in small, wooden lifeboats. He then completed the treacherous journey across South Georgia Island to reach a town and resources necessary to rescue his other men. All of his men were rescued alive and safe and all were able to return home to their families. In compar ison, Arlene Blum set out from the United States in 1978 with 9 of her teammates as the American Women's Himalayan Expedition – the first all-woman group to attempt to climb Annapurna. Though their expedition was too often hazardous, it didn’t have the same level of urgency, because at any point during their climb, the group was at liberty to descend. Blum struggled with her role as the leader of this group of highly independent women. She too often hesitated and showed her lack of confidence, which in turn led to many problems within the group and with the hired Sherpas. On October 15th, 1978, two members of the expedition, along with two Sherpas, reached the summit of Annapurna, fulfilling the goal of the group. However, two days later, on Octobe... ...ew to the path of being trapped by the ice, he excelled at his decision making. He didn’t believe that one should look back at past mistakes and waste time on regrets, indeed, â€Å"Shackleton never wasted time or energy lamenting things that had passed or that he couldn’t change (Morell, pp. 145).† As Shackleton himself said, â€Å"A man must shape himself to a new mark directly the old one goes to ground (Morell, pp. 145).† This ability to shift quickly and react to new circumstances served him and his crew members well and allowed all of them to save themselves from their icy trap. As Dennis Perkins says in his book, Leading at the Edge, in reference to the last leg of Shackleton’s expedition, â€Å"Their heroic journey across South Georgia Island had saved their shipmates. It remains a tribute to unremitting effort—and to the tenacious creativity at The Edge (pp. 148).† Expeditions to Antarctica and Annapurna Essay -- Comparison, Shackleto When taking a quick look at the two expeditions, one led by Ernest Shackleton to Antarctica and the other led by Arlene Blum to climb Annapurna, a quick summation can be made that Blum succeeded in her expedition and Shackleton failed. But this is a shallow view, not considering the nuances and actual experiences of the trips. Ernest Shackleton set out with his crew in 1915 to be the first expedition to cross Antarctica, but in fact, he never set foot on the continent. While failing at his initial goal, he was a highly successful leader and kept his 28 men safe for close to two years, while they were trapped on the ice floe and then, after the ice gave way, when they were paddling hundreds of miles across open seas in small, wooden lifeboats. He then completed the treacherous journey across South Georgia Island to reach a town and resources necessary to rescue his other men. All of his men were rescued alive and safe and all were able to return home to their families. In compar ison, Arlene Blum set out from the United States in 1978 with 9 of her teammates as the American Women's Himalayan Expedition – the first all-woman group to attempt to climb Annapurna. Though their expedition was too often hazardous, it didn’t have the same level of urgency, because at any point during their climb, the group was at liberty to descend. Blum struggled with her role as the leader of this group of highly independent women. She too often hesitated and showed her lack of confidence, which in turn led to many problems within the group and with the hired Sherpas. On October 15th, 1978, two members of the expedition, along with two Sherpas, reached the summit of Annapurna, fulfilling the goal of the group. However, two days later, on Octobe... ...ew to the path of being trapped by the ice, he excelled at his decision making. He didn’t believe that one should look back at past mistakes and waste time on regrets, indeed, â€Å"Shackleton never wasted time or energy lamenting things that had passed or that he couldn’t change (Morell, pp. 145).† As Shackleton himself said, â€Å"A man must shape himself to a new mark directly the old one goes to ground (Morell, pp. 145).† This ability to shift quickly and react to new circumstances served him and his crew members well and allowed all of them to save themselves from their icy trap. As Dennis Perkins says in his book, Leading at the Edge, in reference to the last leg of Shackleton’s expedition, â€Å"Their heroic journey across South Georgia Island had saved their shipmates. It remains a tribute to unremitting effort—and to the tenacious creativity at The Edge (pp. 148).†

African American Essay

In the year of 1870, it was the re invention of slavery. America could not be built without economic. The south was still a negative place and they failed to accept blacks. After decades of discrimination, the voting rights act of 1965 aimed to overcome legal barriers at the state and local levels that denied blacks to vote under the 15th amendment. The 15th amendment in 1870 gave African Americans the right to vote. The constitutional amendment passed after the civil war that it guaranteed blacks the right to vote. It affected not only freed slaves in the south but the blacks that were living in the north who was not allowed to vote(3). The amendment was favored by the Republican Party; since the votes of the slaves helped the party dominates national politics in the years after the war. During the same year, Hiram Rhodes Revels, who was a republican from Mississippi, became the first African American to sit in the United States congress when he was elected to the United States Senate. Millions of black men served in congress during reconstruction but more than 600 served in the states legislatures and many more held local offices(3). The Jim Crow laws were the era of struggle. The state and local laws in the United States enacted between 1876 and 1965. In 1890, there was a â€Å"separate but equal† status for African Americans. Jim Crow laws followed the Black codes which restricted the civil rights and civil liberties of African Americans with no equality. During the reconstruction period, the federal law provided civil rights protection in the United States for the African Americans who had formally been slaves(1). In 1890, Louisiana required by law that blacks ride in separate railroad cars. The state of Louisiana passed a law that required separate accommodations for black and whites on railroads, including separate railway cars. Plessy attempted to sit in an all-white railroad car. After refusing to sit in the black railway carriage car, Plessy was arrested for violating in1890. Louisiana statute that provided for segregated â€Å"separate but equal† railroad accommodations. Those using facilities not designated for their race were criminally liable under the statute(4). Plessy was found guilty on the grounds that the law was a reasonable exercise of the state’s police powers based upon custom, usage, and tradition in the state. Plessy filed a petition in the Supreme Court of Louisiana against Ferguson, asserting that segregation stigmatized blacks and stamped them with a badge of inferiority in violation of the Thirteenth and Fourteenth amendments(4). The case of Plessy vs. Ferguson was one of a combination of rulings passed by the U. S and the state Supreme Courts after reconstruction. Many of these decisions allowed and required Jim Crow segregation laws in southern states. At the highest level, the case was decided on May 18th in 1896, in favor of Ferguson and the state of Louisiana. The Supreme Court had given southern states all the permission they needed to let any remaining equality between the races fade away and be replaced by the Jim Crow laws standing(5). By the 1870s, many southern whites had resorted to intimidation and violence to keep blacks from voting and restore white supremacy in the region. Beginning in 1873, a series of Supreme Court decisions limited the scope of Reconstruction-era laws and federal support for the Reconstruction Amendments, particularly the 14th and 15th, which gave African Americans the status of citizenship and protection. The Compromise of 1877 occurred after the Presidential Election of 1876, when Congress formed the Electoral Commission to resolve disputed Democratic Electoral votes from the South. The republicans agreed to enact Federal legislation that would spur industrialization in the south. They agreed to withdraw federal soldiers from their remaining positions in the south(5). They did this to appoint democrats to positions in the south and to appoint a democrat to the president’s cabinet. The Compromise of 1877 effectively ended the Reconstruction era. The Southern Democrats promised to protect but the political rights of blacks were not kept. The end of federal interference in southern affairs led to widespread disenfranchisement of blacks voters(4). From the late 1870s, southern legislatures passed a series of laws requiring the separation of whites from â€Å"persons of color† on public transportation, in schools, parks, restaurants, theaters and other locations. These segregationist statutes governed life in the South through the middle of the next century, ending after the success of the civil rights movement in the 1960’s. The migration was a watershed in the history of African Americans. It lessened their overwhelming concentration in the South, opened up industrial jobs to people who had up to then been mostly farmers, and gave the first significant impetus to their urbanization. The black migration began in the 1890s as African Americans left for cities such as Chicago, Detroit, Cleveland, Philadelphia, and New York(8). The single largest movement of African Americans occurred during World War I, when people moved from rural areas and small towns in the South to cities in the North and the East. Even in the North, blacks encountered violence at the hands of whites, who resented competition for jobs and black economic success. Segregation and discrimination in housing, education, and jobs was pervasive in the North as well. From 1916, more than six million blacks left the South for other regions of the United States. Over the next fifteen years, more than one tenth of the country’s black population would voluntarily move north. The Great Migration lasted until 1930. This was the first step in the full nationalization of the African American population(2). The Klu Klux Klan is the oldest organization. During this time 1920’s, there were still 85 percent of African Americans in the south. The Klan was created in 1871 by the Democratic Party to prevent African Americans from voting the 15th Amendment. The Klan also became Americans 1st terrorist group and became an institutional part of American life and political colt. African Americans were intimidated and had fear towards the KKK. They hated blacks, republicans, Jews, etc(6). In the 1920s, many blacks had been brought to the south against their own will after the success of the northern states during the Civil War. Also, after the freeing of the blacks from slavery in 1865, a group was established which was designed to spread fear throughout the black population that still lived in the southern states. Most of the hatred was directed against the poor black families in the south who were very vulnerable to attack(6). The white hooded KKK burnt churches of the black population, murdered, raped, castrated etc. They were rarely caught as most senior law officers in the South were high ranking KKK men. White people who were in contact with any blacks had a reason to fear the KKK because they would be after them as well for being what they called â€Å"nigga lovers†(6). The Black Americans tried to fight back using non-violent methods. The NAACP asked Washington for new laws to help combat the KKK violence but received very little help. In the 1920’s Black Americans started to turn to the â€Å"Back to Africa† movement which told blacks that they should return to their native America. This was started by Marcus Garvey but the whole movement faltered when he was arrested for fraud and sent to prison. If African Americans were to move back to Africa, they would be giving the â€Å"white America† exactly what they asked for(6). African Americans drew to church. Going to church brought everyone closer to God despite everything that was going on in the 1920’s. In 1865, blacks started to create independent black churches. The African Methodist Episcopal and the African Methodist Episcopal Zion churches claimed southern membership in the hundreds of thousands, far outstripping that of any other organizations. They were quickly joined in 1870 by a new southern-based denomination, the Colored (Christian) Methodist Episcopal Church. The church was founded by indigenous southern black leaders. Finally, in 1894 black Baptists formed the National Baptist Convention which was an organization that is the largest black religious organization still today in the United States(8). The blossoming of the Harlem Renaissance was during 1918 through 1937. Harlem is located just north of Central Park. Harlem formally was a white residential district but by the early 1920’s, it was the becoming of a virtually black city. Harlem was a catalyst for artistic experimentation and a nightly popular nightlife destination. This was also an economic opportunity in New York(8). The Harlem Renaissance was a phase of a larger new Negro movement that emerged in the early 20th century. The movement raised issues affecting the lives of African Americans through various forms of literature, art, and drama. Its influence spread throughout the nation and beyond that included writers and philosophers. Between the 1920’s and 1930’s, about 750,000 African Americans left the south and migrated to the north to take advantage of this movement. The Harlem Renaissance appealed to a mixture audience. The literature appealed to the African American middle class and to whites. Many critics point out that the Harlem Renaissance could not escape its history and culture. Its attempt to create a new one separates from the foundational elements of White, European culture. Social foundations of this movement included the great migration of African Americans from rural to urban areas(8). The Harlem Renaissance was exposure to the African American Art and culture. It is also unusual among literary and artistic movements for its close relationship to civil rights. The Harlem renaissance set the stage for the civil rights movement of the 1950’s and the 60’s. This was very much black culture exposure. The African American artists intended to express themselves freely, no matter what the black public or white public thought. Since the 1980s, New Orleans has been the area for a new school of jazz players, among them trumpeter Wynton Marsalis and his brothers, saxophonist Branford Marsalis and trombonist Delfeayo Marsalis. New Orleans has brought widespread attention to jazz and a new appreciation of the city and its jazz tradition(7). In the 1890’s it was the beginning of the Mississippi Delta Blues. The blues is the generation of American Music. By the 1900’s New Orleans Jazz was introduced. Jazz was first originally accepted in France before anywhere else. Jazz was played by whites and blacks. Both races played jazz together. Jazz was for the middle class African American people(7). In conclusion, during the years of 1870 and 1920, African Americans still encountered a lot of continued discrimination against blacks in American and the separation of race. Through the Jim Crow laws and the segregation, Blacks kept it together through religion, and even music. When looking back at our history, African Americans had a great deal to do with the shaping of America today. Bibliography 1) â€Å"Jim Crow Law (United States [1877-1954]). † Encyclopedia Britannica Online. Encyclopedia Britannica, n. d. Web. 09 May 2013. 2) â€Å"Primary Documents in American History. † 15th Amendment to the Constitution: Primary Documents of American History (Virtual Programs & Services, Library of Congress). N. p. , n. d. Web. 09 May 2013. 3) Fifteenth Amendment. † History. com. A&E Television Networks, n. d. Web. 09 May 2013. 4) â€Å"Compromise of 1877. † History. com. A&E Television Networks, n. d. Web. 09 May 2013. 5) â€Å"Compromise of 1877. † Information about The. N. p. , n. d. Web. 09 May 2013. 6) â€Å"The KKK and Racial Problems. † The KKK and Racial Problems. N. p. , n. d. Web. 09 May 2013. 7) â€Å"A History of Jazz Music. † A History of Jazz Music. N. p. , n. d. Web. 09 May 2013. 8) Notes.

Solution Manual for Fluid Mech Cengel Book

Chapter 6 Momentum Analysis of Flow Systems Chapter 6 MOMENTUM ANALYSIS OF FLOW SYSTEMS Newton’s Laws and Conservation of Momentum 6-1C Newton’s first law states that â€Å"a body at rest remains at rest, and a body in motion remains in motion at the same velocity in a straight path when the net force acting on it is zero. † Therefore, a body tends to preserve its state or inertia. Newton’s second law states that â€Å"the acceleration of a body is proportional to the net force acting on it and is inversely proportional to its mass. Newton’s third law states â€Å"when a body exerts a force on a second body, the second body exerts an equal and opposite force on the first. † r 6-2C Since momentum ( mV ) is the product of a vector (velocity) and a scalar (mass), momentum must be a vector that points in the same direction as the velocity vector. 6-3C The conservation of momentum principle is expressed as â€Å"the momentum of a system remains constant when the net force acting on it is zero, and thus the momentum of such systems is conserved†.The momentum of a body remains constant if the net force acting on it is zero. 6-4C Newton’s second law of motion, also called the angular momentum equation, is expressed as â€Å"the rate of change of the angular momentum of a body is equal to the net torque acting it. † For a non-rigid body with zero net torque, the angular momentum remains constant, but the angular velocity changes in accordance with I? = constant where I is the moment of inertia of the body. 6-5C No.Two rigid bodies having the same mass and angular speed will have different angular momentums unless they also have the same moment of inertia I. Linear Momentum Equation 6-6C The relationship between the time rates of change of an extensive property for a system and for a control volume is expressed by the Reynolds transport theorem, which provides the link between the r system and control volume concepts. The linear momentum equation is obtained by setting b = V and thus r B = mV in the Reynolds transport theorem. -7C The forces acting on the control volume consist of body forces that act throughout the entire body of the control volume (such as gravity, electric, and magnetic forces) and surface forces that act on the control surface (such as the pressure forces and reaction forces at points of contact). The net force acting on a control volume is the sum of all body and surface forces. Fluid weight is a body force, and pressure is a surface force (acting per unit area). -8C All of these surface forces arise as the control volume is isolated from its surroundings for analysis, and the effect of any detached object is accounted for by a force at that location. We can minimize the number of surface forces exposed by choosing the control volume such that the forces that we are not interested in remain internal, and thus they do not complicate the analysis. A well-chosen cont rol volume exposes only the forces that are to be determined (such as reaction forces) and a minimum number of other forces. 6-9C The momentum-flux correction factor ? nables us to express the momentum flux in terms of the r r r r & ? V (V ? n )dAc = ? mV avg . The value of ? is unity for uniform mass flow rate and mean flow velocity as ? Ac flow, such as a jet flow, nearly unity for turbulent flow (between 1. 01 and 1. 04), but about 1. 3 for laminar flow. So it should be considered in laminar flow. 6-1 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.Chapter 6 Momentum Analysis of Flow Systems 6-10C The momentum equation for steady one-dimensional flow for the case 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 control volume, and first term on the right hand side is the incoming momentum flux and the second term is the outgoing momentum flux by mass. 6-11C In the application of the momentum equation, we can disregard the atmospheric pressure and work with gage pressures only since the atmospheric pressure acts in all directions, and its effect cancels out in every direction. -12C The fireman who holds the hose backwards so that the water makes a U-turn before being discharged will experience a greater reaction force since the numerical values of momentum fluxes across the nozzle are added in this case instead of being subtracted. 6-13C No, V is not the upper limit to the rocket’s ultimate velocity. Without friction the rocket velocity will continue to increase as more gas outlets the nozzle. 6-14C A helicopter hovers because the strong downdraft of air, caused by the overhead propeller blades, manifests a momentum in the air stream.This momentum must be countered by the helicopter lift force. 6-15C As the air density decreases, it requires more energy for a helicopter to hover, because more air must be forced into the downdraft by the helicopter blades to provide the same lift force. Therefore, it takes more power for a helicopter to hover on the top of a high mountain than it does at sea level. 6-16C In winter the air is generally colder, and thus denser. Therefore, less air must be driven by the blades to provide the same helicopter lift, requiring less power. 6-2 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc.Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-17C The force required to hold the plate against the horizontal water stream will increase by a factor of 4 when the velocity is doubled since & F = mV = ( ? AV )V = ? AV 2 and thus the force is proportional to the square of the velocity. 6-18C The accele ration will not be constant since the force is not constant. The impulse force exerted by & water on the plate is F = mV = ( ? AV )V = ?AV 2 , where V is the relative velocity between the water and the plate, which is moving. The plate acceleration will be a = F/m. But as the plate begins to move, V decreases, so the acceleration must also decrease. 6-19C The maximum velocity possible for the plate is the velocity of the water jet. As long as the plate is moving slower than the jet, the water will exert a force on the plate, which will cause it to accelerate, until terminal jet velocity is reached. 6-20 It is to be shown that the force exerted by a liquid jet of velocity V on a stationary nozzle is & proportional to V2, or alternatively, to m 2 . Assumptions 1 The flow is steady and incompressible. 2 The nozzle is given to be stationary. 3 The nozzle involves a 90 ° turn and thus the incoming and outgoing flow streams are normal to each other. 4 The water is discharged to the atmo sphere, and thus the gage pressure at the outlet is zero. Analysis We take the nozzle as the control volume, and the flow direction at the outlet as the x axis. Note that the nozzle makes a 90 ° turn, and thus it does not contribute to any pressure force or momentum flux & term at the inlet in the x direction. Noting that m = ?AV where A is the nozzle outlet area and V is the average nozzle outlet velocity, the momentum equation for steady one-dimensional flow in the x direction reduces to r r r & & & & F= ? mV ? ? mV > FRx = ? m out V out = ? mV ? ? out ? in where FRx is the reaction force on the nozzle due to liquid jet at the nozzle outlet. Then, & m = ? AV & > FRx = ? mV = AVV = AV 2 & & or FRx = ? mV = ? m & & m m2 =? ?A ? A Therefore, the force exerted by a liquid jet of velocity V on this & stationary nozzle is proportional to V2, or alternatively, to m 2 . Liquid Nozzle V FR 6-3 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc.Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-21 A water jet of velocity V impinges on a plate moving toward the water jet with velocity ? V. The force required to move the plate towards the jet is to be determined in terms of F acting on the stationary plate. Assumptions 1 The flow is steady and incompressible. 2 The plate is vertical and the jet is normal to plate. 3 The pressure on both sides of the plate is atmospheric pressure (and thus its effect cancels out). Fiction during motion is negligible. 5 There is no acceleration of the plate. 6 The water splashes off the sides of the plate in a plane normal to the jet. 6 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Analysis We take the plate as the control volume. The relative velocity between the plate and the jet is V when the plate is st ationary, and 1. 5V when the plate is moving with a velocity ? V towards the plate. Then the momentum equation for steady one-dimensional flow in the horizontal direction reduces to r r r & & & & F= ? mV ? ? mV > ? FR = ? mi Vi > FR = miVi ? out ? in Stationary 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 required to hold the plate stationary against the oncoming water jet becomes 2. 25 times when the jet velocity becomes 1. 5 times. Discussion Note that when the plate is stationary, V is also the jet 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 off its sides) per unit time also increases by 50%. 1/2V VWaterjet 6-4 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-22 A 90 ° elbow deflects water upwards and discharges it to the atmosphere at a specified rate. The gage pressure at the inlet of the elbow and the anchoring force needed to hold the elbow in place are to be determined. v Assumptions 1 The flow is steady, frictionless, incompressible, and irrotational (so that the Bernoulli equation 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 momentum-flux 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 (a) 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-outlet steady flow 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) (1000 kg/m 3 )[? (0. 1 m) 2 / 4] Noting that V1 = V2 and P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed 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 = (1000 kg/m3 )(9. 81 m/s 2 )(0. 35 m)? 1 ? 1000 kg ? /s2 ? = 3. 434 kN/m = 3. 434 kPa ? ? r r r & & (b) The momentum equation for steady one-dimensional flow 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 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 y axes become & & FRx + P1,gage A1 = 0 ? ?m(+V1 ) = ? ?mV & & FRz = ? m(+V 2 ) = ? mV z x FRz 2 35 cm Solving for FRx and FRz, and substituting the given 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 ? ? FRy 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 Discussion Note that the magnitude of the anchoring force is 136 N, and its line of action makes 143 ° from the positive x direction. Also, a negative value for FRx indicates the assumed direction is wrong, and should be reversed. 6-5 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc.Limited distribution permitted only to teachers and educators for course preparatio n. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-23 An 180 ° elbow forces the flow to make a U-turn and discharges it to the atmosphere at a specified rate. The gage pressure at the inlet of the elbow and the anchoring force needed to hold the elbow in place are to be determined. v Assumptions 1 The flow is steady, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli equation 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 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 (a) 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 b eing 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 = 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) (1000 kg/m 3 )[? (0. 1 m) 2 / 4] Noting that V1 = V2 and P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed 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 = (1000 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 flow 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 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 = ? m(? V 2 ) ? ? m(+V1 ) = ? 2 ? mV FRz = 0 Solving for FRx and substituting the given values, & FRx = ? 2 ? mV ? P1, gage A1 ? 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 negative x direction. Discussion Note that a negative value for FRx indicates the assumed direction is wrong, and should be reversed. FRx 1 6-6 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.Chapter 6 Momentum Analysis of Flow Systems 6-24E A horizontal water j et strikes a vertical stationary plate normally at a specified velocity. For a given anchoring force needed to hold the plate in place, the flow rate of water is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water 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 splattered water is the atmospheric pressure which is disregarded since it acts on the entire control surface. The vertical forces and momentum fluxes are not considered since they have no effect on the horizontal reaction force. 5 Jet flow is nearly 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 We take the plate as the control volume such that it contains the entire plate and cuts through the water jet and the support bar normally, and the direction of flow as the positive direction of x axis. The momentum equation for steady one-dimensional flow in the x (flow) direction reduces in this case o r r r & & & & F= ? mV ? ? mV > ? FRx = ? mV1 > FR = mV1 ? ? out ? in We note that the reaction force acts in the opposite direction to flow, and we should not forget the negative & sign for forces and velocities in the negative x-direction. Solving for m and substituting 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 under stated assumptions 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 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparati on. 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 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. 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 coordina te 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-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. 0150 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 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 ? ? ? ? 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 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 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 assumed direction is wrong. 6-8 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 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 densi ty 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 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-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 ? ? The 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)[(12cos110 ° – 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, and 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 accelerated by a nozzle strikes the back surface of a cart moving horizontally at a constant velocity. The braking force and the power wasted by the brakes are to be determined. . Assumptions 1 The flow is steady and incompressible. 2 The water splatters off the sides of the plate in all direction s 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 splattered water is the atmospheric pressure which is disregarded since it acts on all surfaces. 4 Fiction during motion is negligible. 5 There is no acceleration of the cart. 7 The motions of the water jet and the cart are horizontal. 6 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Analysis We take the cart 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 forget the negative sign for forces and velocities in the negative x-direction. Substituting the given values, ? 1N & Fbrake = ? mV r = ? (25 kg/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 distance and the distance traveled by the cart per unit time is the cart velocity, the power 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 power wasted is equivalent to the maximum power 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 Fl ow 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 fail is to be determined. Analysis The braking force was determined 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 splitter, such that half of the flow is diverted up ward at 45 °, and the other half is directed down.The force required to hold the splitter in place is to be determined. vEES 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 before and after the split is the atmospheric pressure which is disregarded since it acts on all surfaces. 3 The gravitational effects are disregarded. 4 Jet flow is nearly 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 force 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)? ? = ? 1135 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 1135 lbf must be applied to the splitter in the opposite direction to flow to hold it in place. No holding force is necessary in the vertical direction.This can also be concluded from the symmetry. Discussion In reality, the gravitational effects will cau se the upper stream to slow down and the lower stream to speed up after the split. But for short 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 180 ° in increments of 10 ° is to be investigated. g=32. 2 â€Å"ft/s2† rho=62. 4 â€Å"lbm/ft3† V_dot=100 â€Å"ft3/s† V=20 â€Å"ft/s† m_dot=rho*V_dot F_R=-m_dot*V*(cos(theta)-1)/g â€Å"lbf† ?,  ° 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 8000 7000 6000 5000 & m , lbm/s 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 624 0 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 80 100 120 140 160 180 ?,  ° 6-13 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 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 a tmospheric 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 The 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, a n 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 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 6-32 A 90 ° reducer elbow deflects water downwards into a smaller diameter pipe. The resultant 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 gravitational 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-outlet 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 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 6-33 A wind turbine with a given span diameter and efficiency is subjected to steady winds. 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 independent of wind speed. 3 The frictional effects are negligible, and thus none of the incoming kinetic energy is converted to thermal energy. Wind 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 potential of the wind is its kinetic 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 converted 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 choose the control volume around the wind turbine such that the wind is norm al 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-direction ? ? 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 tower where the wind turbine is installed, and the bending moment it generates at the bottom of the tower 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 c ourse 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 curved plate, which deflects the water back to 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 Friction between the plate and the surface it is on is negligible (or the friction force can be included in the required force to hold the plate). 4 There is no splashing 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 w ater 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). 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. 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 applied on the plate in the negative x direction to hold it in place. Discussion Note that a nega tive 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 Waterjet 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 effec ts 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 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 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 assu me 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 while trying to extinguish a fire.The average water outlet velocity and the resistance force required 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 taken 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 (thi s way the pressure force and the momentum flux at the inlet are in the vertical direction, with no contribution to the force balance in the horizontal direction), 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). The average 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 applied 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 t he 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 done by a single person. This demonstrates why several 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 o f 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 plate. 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 along 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 limiting 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 Comp anies, 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. The 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 â€Å"kg/m3† D=0. 05 â€Å"m† V_jet=30 â€Å"m/s† Ac=pi*D^2/4 V_r=V_jet-V_plate m_dot=rho*Ac*V_jet F_R=m_dot*V_r â€Å"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 ar e using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-39E A fan moves air at sea level at a specified rate. The force required to hold the fan and the minimum power input required for the fan are to be determined. v Assumptions 1 The flow of air is steady and incompressible. 2 Standard atmospheric conditions exist so that the pressure at sea level is 1 atm. Air leaves the fan at a uniform velocity at atmospheric pressure. 4 Air approaches the fan through a large area at atmospheric pressure with negligible velocity. 5 The frictional effects are negligible, and thus the entire mechanical power input is converted to kinetic energy of air (no conversion 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 hyperbolic cylinder bounded by streamlines on the sides with air entering through the large cross-section (section 1) and the fan located at the narrow cross-section at the end (section 2), and let its centerline be the x axis. The density, mass 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 frict ion 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 pump, 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 input required for the fan. Discussion The actual power input 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 loaded. The volumetric air flow rate and the required power input during unloaded hover, and the rpm and the required power input during loaded 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 conversion to thermal energy through frictional effects). 5 The change in air pressure with elevation 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 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 bottom (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 span 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 unloaded mode become V 2,unloaded = m unloaded g = ? A (10,000 kg)(9. 81 m/s 2 ) (1. 18 kg/m 3 )(176. 7 m 2 ) = 21. 7 m/s Sea 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 repeat 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 overhead blade rotational velocity, the rpm 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 rpm) = 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.  © 200 6 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 mountain where the air density considerably lower than that at sea level. The blade rotational velocity to hover at the higher altitude 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 locatio n 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 momentum-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 mountain. 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 ? mountain ?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 sluice gate by raising or lowering a vertical plate. A relation for the force acting on a sluice gate of width w for steady and uniform flow is to be developed. Assumptions 1 The flow 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 free 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 free 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 relations 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 inner 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