Philosophy - Kants Universal Law Formation Of The Categorical Imperati

Philosophy - Kants Universal Law Formation of the Categorical Imperative Kantian philosophy outlines the Universal Law Formation of the Categorical Imperative as a method for determining morality of actions. This formula is a two part test. First, one creates a maxim and considers whether the maxim could be a universal law for all rational beings. Second, one determines whether rational beings would will it to be a universal law. Once it is clear that the maxim passes both prongs of the test, there are no exceptions. As a paramedic faced with a distraught widow who asks whether her late husband suffered in his accidental death, you must decide which maxim to create and based on the test which action to perform. The maxim "when answering a widow's inquiry as to the nature and duration of her late husbands death, one should always tell the truth regarding the nature of her late husband's death" (M1) passes both parts of the Universal Law Formation of the Categorical Imperative. Consequently, according to Kant, M1 is a moral action. The initial stage of the Universal Law Formation of the Categorical Imperative requires that a maxim be universally applicable to all rational beings. M1 succeeds in passing the first stage. We can easily imagine a world in which paramedics always answer widows truthfully when queried. Therefore, this maxim is logical and everyone can abide by it without causing a logical impossibility. The next logical step is to apply the second stage of the test. The second requirement is that a rational being would will this maxim to become a universal law. In testing this part, you must decide whether in every case, a rational being would believe that the morally correct action is to tell the truth. First, it is clear that the widow expects to know the truth. A lie would only serve to spare her feelings if she believed it to be the truth. Therefore, even people who would consider lying to her, must concede that the correct and expected action is to tell the truth. By asking she has already decided, good or bad, that she must know the truth. What if telling the truth brings the widow to the point where she commits suicide, however? Is telling her the truth then a moral action although its consequence is this terrible response? If telling the widow the truth drives her to commit suicide, it seems like no rational being would will the maxim to become a universal law. The suicide is, however, a consequence of your initial action. The suicide has no bearing, at least for the Categorical Imperative, on whether telling the truth is moral or not. Likewise it is impossible to judge whether upon hearing the news, the widow would commit suicide. Granted it is a possibility, but there are a multitude of alternative choices that she could make and it is impossible to predict each one. To decide whether rational being would will a maxim to become a law, the maxim itself must be examined rationally and not its consequences. Accordingly, the maxim passes the second test. Conversely, some people might argue that in telling the widow a lie, you spare her years of torment and suffering. These supporters of "white lies" feel the maxim should read, "When facing a distraught widow, you should lie in regards to the death of her late husband in order to spare her feelings." Applying the first part of the Universal Law Formation of the Categorical Imperative, it appears that this maxim is a moral act. Certainly, a universal law that prevents the feelings of people who are already in pain from being hurt further seems like an excellent universal law. Unfortunately for this line of objection, the only reason a lie works is because the person being lied to believes it to be the truth. In a situation where every widow is lied to in order to spare her feelings, then they never get the truth. This leads to a logical contradiction because no one will believe a lie if they know it a lie and the maxim fails. Perhaps the die-hard liar can regroup and test a narrower maxim. If it is narrow

H.J. Heinz Case Essays

H.J. Heinz Case Essays H.J. Heinz Case Paper H.J. Heinz Case Paper With its headquarters in Pittsburgh, Pennsylvania, Heinz has offices located on six of the seven continents. Heinz focuses on marketing their products with an emphasis on health, wellness, and sustainability. Along with manufacturing high quality food products adapted to unique consumer traits in various regions of the world, Heinz also contributes generously to charitable efforts In parts of the world that need help most. Recently, Heinz received an award for ranking number one In customer satisfaction for the eleventh year In a row by the American Customer Satisfaction Index (Heinz, n. D. ). The Company, as it calls itself, has high standards for the food it manufactures and for its corporate operations resulting in a ole-model of a company raising the bar in global business operations. History and Global Footprint Heinz founder Henry John (H. J) Heinz created a foundation for company success by Instilling values of team bulling, collaboration, Innovation, v ision, results-based operations, and integrity in his employees. By following his mission statement, To do a common thing uncommonly well brings success, Heinz has been successful at introducing new products to the market for over 140 years (Heinz, n. . ). The H. J. Heinz Company was founded in 1869 in Sharpeners, Pennsylvania. It was incorporated on July 17, 1900, and grew to become a business partnership in 1905. Currently, Heinz headquarters are In Pittsburgh, Pennsylvania. 3 Heinz Is striving to develop globally while positively Impacting the world. Through Its than 30 countries with natural hybrid tomato seeds (H. J. Heinz Annual Report and ask, 2010). This enables farmers to produce higher yields of quality tomatoes without genetic modification. It has been particularly successful in China, and the Company recently began a program partnering with the United States Agency for International Disaster in the economically afflicted Upper Nile region of Egypt. Heinz also partners with Lucy Lieu as spokeswoman for the Heinz Encountering program to provide packages of powdered vitamins and minerals that can be mixed into normal meals for infants and children living in developing nations. The Heinz Microscope program has helped three million children fight the threat of iron deficiency, anemia, and vitamin and mineral malnutrition in 15 countries (H. J. Heinz Annual Report and ask, 2010). The program was recently introduced into North America and Africa as it continues to grow, with a goal of reaching one million additional children by the end of this year. Heinz is committed to impacting the world through the Hindered and Heinz Encountering programs. Industry and Products As previously mentioned, Heinz is a global leader in the global market of food products. With sixty percent of its sales generated outside the United States, Heinz is an example of a company that has prospered from globalization (Heinz ASK Annual Report). : The products that Heinz manufactures and markets fall under the three categories of condiments, frozen foods, and infant foods. Product availability varies around the world, as do the ingredients. For example, popular Heinz products strutted in the United States include Smart Ones and T. G. I. Fridays frozen entrees, Pickles, Classics pasta sauce, and ketchup. Ketchup is also available in Poland, but it is made with different ingredients, is packaged differently, and is called Puddings (Heinz H. J. HEINZ NC: INDUSTRY ANALYSIS 4 ASK Annual Report). Heinz manufactures and markets products that meet the standards of individual countries, and the needs and wants of consumers in these countries. Whines innovative marketing techniques assist in generating product success in the global food industry. In February 2010, Heinz announced the latest addition to the Heinz Ketchup product line. The new Dip Squeeze ketchup, which enables consumers to squeeze ketchup onto food by removing the tip of the small, ketchup bottle-shaped packet, or peel back the wrapper to dip their food in the sauce. In recent efforts to promote the new Dip Squeeze product, Heinz purchased a used truck, installed a custom kitchen with double-stacked convention ovens, food warmers, sinks, and a freezer. The truck was branded with the projects slogan, Heinz Ketchup Road Trip, and the campaign was pitched on social media sites including Twitter and Faceable. The road trip began in mid-November in Pittsburgh, spent Thanksgiving in New York, and will make its way to Philadelphia before a final stop down south in Dallas. Visitors to the Heinz Ketchup Road Trip truck receive a free serving of Ore-Ida crinkle cut fries or Ore-Ida sweet potato fries (both Heinz public relations and communications for Heinz North America, explains the on-the-go campaign was very fitting for the Dip Squeeze Ketchup packets. She says, Since it was really made for eating on the go, we wanted to create an environment where people could experience it on the go (Vega, 2010). Geographic Coverage and International Sales Totals The Heinz Company sales extend across the globe and generated about $10. Billion internationally in the last year (Heinz ASK Annual Report). Heinz divides its markets into the five regions of North America, U. S. Foddering, Europe, Asia/Pacific and Rest of World. The Asia/Pacific region includes Australia, New Zealand, India, Japan, China, South Korea, 5 Indonesia, and Singapore. The Rest of World region includes Africa, Latin America, and the Middle East. The $10. 5 billion earned in the last year was th e sum of North At-enemas $3,192, 219,000, U. S Food services $1 Rupees Sat/pacific and Rest of worlds revenues (H. J. Heinz Annual Report and ask, 2010). These totals indicate that North America is Whines strongest market, closely followed by the European region. The Rest of World region includes many developing nations thus generated significantly lower revenue. A future agenda is the expansion of Heinz products in developing nations through the creation of useful, economical products and targeted marketing. Emerging markets that Heinz has operations in include China, India, Latin America, and Russia. Heinz is also exploring new markets in the Philippines, Turkey, Vietnam, ND Brazil. On November 30, 2010, Heinz announced that it would participate in the 2010 Credit Issue Holiday Conference on December 7, 2010. During the conference, the Company will present an overview of Whines strategies, results, and businesses around the world represented by Margaret Noel, Senior Vice President, Investor Relations (The Wall Street Journal Market Watch) (H. J. Heinz Company to Participate in the 2010 Credit Issue Holiday Conference, 2010). Global Production, Distribution Facilities Supply Chain The Company owns a total of 67 factories and leases eight across the globe (H. J. Heinz Annual Report and ask, 2010). In North America, 20 factories are owned and four are leased, in Europe 21 are owned and none are leased, in the Asia-Pacific region 20 are owned and two are leased, and in the Rest of World six factories are owned and two are leased. Whines intricate supply chain enables its products to command vast global coverage and accessibility. The products are sold through Whines own sales organizations, through 6 and independent grocery accounts, convenience stores, bakeries, pharmacies, mass merchants, club stores, foddering distributors and institutions including hotels, saturates, hospitals, healthcare facilities, and certain government agencies (H. J. Heinz Annual Report and ask, 2010). The Company has set a goal of delivering more than $1 billion in cost savings over the next five years through global supply chain initiatives designed to achieve economies of scale and reduce costs by leveraging people, process, and technology (H. J. Heinz Annual Report and ask, 2010). Multinational Market Regions and Market Groups Throughout the varying entities of the Heinz Corporation at least one thing remains constant, high expectations pertaining to corporate and social responsibility. Their pride in responsibility, along with their strict outline of rules and business standards, allows for swift and easy adaptations to local and regional group regulations (such as: NONFAT, the EX., CAFTAN, SEAN, etc). Although countries and regions have differing health standards and regulations for food, Heinz avoids complications with the countries respective economic unions or organizations due to the Companys nature of business. If a firm in the food industry wishes to compete on a global scale, the foremost outside force it will contend with and adapt to is culture. One way in which a country expresses its culture is through food. Since cultures change from region to region, one can see that it would be virtually impossible for a firm competing on a global scale to have one product which transcends all markets and appeals to everyones preferences. Therefore, Heinz does not have one product that it must mold to different rules and regulations in order to sell it to different markets. Heinz competes on a global scale through the acquisition of companies that have already established demand for products in their market. These acquisitions enable Heinz to bypass the political red 7 ape which most companies introducing products into a foreign marketplace must overcome. Therefore Heinz does not have to adapt, but maintain, the business practices of the firms they acquire. Although Heinz may not have direct involvement with economic unions or other similar organizations, 22 of their international locations are certified by the International Organization for Standardization (SO). The ISO is, a network of the national standards institutes of 163 countries, one member per country, with a Central Secretariat in Geneva, Switzerland, that coordinates the system (FAQ, n. D. ). Specifically Heinz adheres to ISO 14001 which provides firms with a framework in order to manage effectively their business in hopes of assuring employees and stakeholders that they are working for, or investing in, an environmentally conscious company. Their involvement with the ISO illustrates Whines emphasis on the responsibility to stakeholders and consumers. Cultural Issues In the United States, Heinz is an iconic brand known for the distinct glass ketchup bottles that are a mainstay in restaurants around the country. The simple condiment has become a staple to pair with many different kinds of food and is seen as much of necessity as salt and pepper to keep readily available. Heinz products as a whole have adapted and grown over the years to become well integrated with society needs. In terms of global development, Heinz has succeeded in finding new ways of innovation in product offerings to cater to specific demographics. In order to accommodate to the tastes of the specific area, Heinz acquires an existing food company that has already proven successful in the market or shows potential for future growth. Recently, Heinz acquired Foodstuff, a leading 8 Chinese manufacturer of premium branded soy sauces and fermented bean curd Heinz completes acquisition of Foodstuff, 2010). Foodstuff has pre-existing brands that have been well established in an economy that is projecting an annual growth of seven to eight percent in the coming years. North America In the United States, Heinz is the best-selling brand of ketchup in the country (H. J. Heinz Company, Inc. , 2010). There are many heritage brands associated with common condiments used by Americans. These include items such as vinegar, cocktail sauce, chili sauce, pickles, relishes, mustard, and Heinz 578 Sauce. In the past year Heinz has introduced 200 new products in the past year, including Smart Ones breakfast sandwiches and Ore-Ida microwaveable mashed potatoes. Heinz also developed a variety of frozen products from successful preexisting brands, such as T. G. I. Fridays by creating foods with similar tastes that customers can enjoy at home. Since the U. S. Food market is heavily saturated, Heinz is focusing on developing new items from existing ideas to promote growth in the industry. Heinz has been a part of Canadian culture for the past 100 years specializing in condiments, pasta sauces, canned beans, pasta, refrigerated dressings, and infant foods. Their largest presence in this country is in infant food markets; they have enjoyed the number one position in this market for 70 years (H. J. Heinz Annual Report and ask, 2010). The Company offers a wide range of products, from Nature Infant Formula to toddler snacks. More recently, Heinz has begun to deliver these items in traditional and organic varieties to accommodate parental health concerns. Europe In the United Kingdom, Heinz is inextricably linked with British history and culture to the extent that it is believed to be a U. K. Company. Items such as Heinz Baked Beans have become a century old tradition, along with other products such as HP sauce, Lea Perrine Sauce, and Heinz Salad Cream. Heinz Salad Cream was the first brand developed exclusively for the U. K. Market and was developed in their London kitchens in 1925 (H. J. Heinz Inc. 2010). Heinz frequently uses the strategy of taking successful products and exporting them into markets where there is a vacancy. This is especially true in the European market. For example, Puddings is one of Pollards largest producers of ketchup and it is now offered in the United Kingdom. Also, Italys Plasmas baby food brand has become the gold standard for infiltration products. It has dominated the marketplace due to the all-natural recipes as well as expanded lines of Baby Specials that parents seek. The Baby Specials line promotes easier digestion and contains hypoallergenic nutrients. Heinz also positioned this line in Russia and has experienced increased adaptation in this market as well (H. J. Heinz Inc. 2010). Transferring successful products into different markets is a common Heinz business technique. Asia Pacific Heinz has learned from experience that it is not beneficial to sell one product globally and expect high assimilation into different cultures. China and Indonesia are two areas that are growing rapidly and their culture is developing significantly. The Asian Heinz label reported a 44% increase in sales last year and is expected to continue to grow (Bayle 2008). In this region, Heinz sells beverage syrup and chili sauce and has acquired the successful soy sauce brand BBC. Since its creation in 1988, BBC has become the second largest soy sauce brand in this region behind the Japanese brand, Kinsman. Heinz has also done a major overhaul in the packaging, flavoring, marketing, and distribution of these products to generate more than $200 million in 10 sales in the past year. To promote further growth and make the products more ideally accepted in these cultures, Heinz has inspired chefs to create innovative recipes using BBC (Bayle 2008). Heinz has been successful in the Asia Pacific region because they have adapted to the unique tastes of the culture and have used innovative marketing techniques such as inspiring procrastinating recipes. Political and Legal Environment The Companys performance is impacted by political and economic conditions in the nations in which it operates. The factors and conditions include changes in applicable laws and regulations, including changes in food and drug laws, accounting tankards and critical accounting estimates, environmental laws, and taxation requirements. Other factors include import and export restrictions, nationalization, hyperinflation environments, terrorist acts and political unrest. Venezuela, a country included in the Rest of World market, is a country of concern for Heinz. This environment of Venezuela creates risk for market penetration and business operations. Other political and legal factors listed by The Heinz Company that can occur and adversely affect financial results are disruption of supply chain, factors affecting the cost of production, transportation, and distribution such as increased energy costs; increased pension, labor, and people-related expenses; food safety issues, environmental, legal, tax and other regulations. Foreign currency exchange rate exposure, the failure to successfully integrate acquisitions and Joint ventures into existing operations and the failure to gain applicable regulatory approval for such transactions or divestitures can also adversely affect the Company 11 (H. J. Heinz Annual Report and ask, 2010). These factors are most frequently encountered when attempting to enter an emerging market. On November 19, 2010, the H. J. Heinz Company reported higher second quarter operating income, net income, and earnings per share resulting from strong sales from growth in Emerging Markets and global ketchup (Courier, A. Styles, T, 2010). Emerging Markets were responsible for a 10. 2 percent sales growth driven by infant nutrition products, ketchup, and nutritional beverages as well as increased pricing, primarily in Latin America (Courier, A. , Styles, T, 2010). For the quarter, Emerging Markets were responsible for 15 percent of total sales (Courier, A. , Styles, T, 2010). Globally , sales of ketchup grew 3. Percent, led by Russia where Heinz is the number one brand (The Wall Street Journal Market Watch). However, currency movements adversely affected sales in Europe and Venezuela. In an online article from Bloomberg Business (Devourer, C, 2010), it was reported that sales from the European market decreased 5. 2 percent year-over-year to $798 million primarily resulting from unfavorable currency effects. The report also stated that the Rest of World market saw sales decrease 19 percent to $120 million due to the devaluation of the Venezuelan currency late in the third quarter of Fiscal 2010 (Proactive Investors, 2010). Emerging Markets If a firm wishes to expand and maximize its growth potential it must analyze what markets they are currently in and which ones they have yet to enter. They should then determine which markets show the most promise for profitability and identify these as their emerging markets. According to the Business Dictionary, emerging markets are, new market structures arising from digitization, deregulation, globalization, and open-standards, that are shifting the balance of economic power from the sellers to the buyers. For Heinz, emerging markets are crucial because they generated about of their 2009 total sales (H. J. Heinz Annual Report and ask, 2010). Their primary emerging markets include India, Indonesia, Latin America, Poland, and China; sales in these markets grew by 8. % in the fiscal 2009 year and continue to show expansion and growth (H. J. Heinz Annual Report and ask, 2010). According to the Heinz website one of the four keys in its business strategy is acceleration and growth in emerging markets (H. J. Heinz Annual Report and ask, 2010). They seek to do this by investing for double-digit sales and profit growth, leveraging infr astructure and expanding distribution, winning the global customers, applying modern trade tactics, growing through health and wellness (as well as taste), and by leveraging themselves through innovation (H. J. Heinz Annual Report and ask, 2010). On a macro level Heinz generally expands its company throughout the world through major mergers, acquisitions, and Joint ventures. Heinz seeks companies with established products in the region they wish to expand to. Then seek to compose a deal where they can merge or acquire the company; thus attaining a low-risk market entry. Understanding Whines emerging market strategy is exemplified in the establishment of business in India. India has become one of the leading emerging markets for numerous industries due to its large population (1 , people in July 2010), and advancements in technology, transportation, education, and government (CIA, n. . ). Since the culture and cuisine are exponentially different than Whines home country, the Company has had to invest in products which will meet the preferences of the Indian consumer. In India, Heinz offers a variety of food products such as Indian chutneys, energy drinks, instant (or ready-to-eat and ready-took) meals, and most importantly nut ritional beverage mixes that enhance child growth and 13 placement. The Indian consumer is health conscious thus one of the main sources of revenue for Heinz is its nutritional beverages. In order to manage its Indian market Heinz must have locations for offices, factories, and distribution in India. Heinz currently has a head office in Iambi, in addition to four branch offices, two factories in Assistant and Aligarhs, and 119 distributors dispersed throughout India (Map, n. D). This complex system of management is necessary in order to be successful in a foreign market. This allows Heinz to decrease communication time between distributors and management, which could prove to be essential in problem solving. The dispersed web of Heinz factories and distributors decreases costs because it reduces shipping expenses. Having in country sites allows them to distribute their product with minimal shipping and transportation costs; and it also greatly shortens the distribution time. India is only one example of the value of the emerging market to Heinz. Adaptation and implementation is necessary for the Companys global success. Heinz has similar other important markets such as Poland, China, Latin America, and Indonesia to name a few. Growth in sales in China can be accredited to major acquisitions of Foodstuff, Long Fond, and their involvement in nutrition and baby foods. Poland hosts Whines Puddings brand and is the countrys main food processor for prepared meals and also is its largest producer of ketchup. There are many examples as to how Heinz stays competitive in emerging markets. They continue to seek growth in these markets through acquisition, mergers, adaptation, research and development, local supply, distribution, and management systems, and through the implementation of high standards which must be adhered to by all segments of their large corporation. The future in these emerging markets looks promising for Heinz as they expect to maintain their #1 or #2 market 14 share in over 50 countries through an expected contribution of 20% of sales (from emerging markets) by 2013 (H. J. Heinz Annual Report and ask, 2010). Foreign Currency Marketplace Currency exchange rates have a high impact on Whines present and future focus. In emerging markets, acquisitions and foreign exchange translation rates had a favorable impact on the 30% sales growth in 2010 (H. J. Heinz Annual Report and ask, 2010). For Heinz, assets and liabilities are translated at the exchange rate in effect at ACH year-end. Income statements are translated at the average rate of exchange prevailing during the year and translation adjustments are included within shareholders equity (H. J. Heinz Annual Report and ask, 2010). To date, it appears that foreign currency exchange rate has not completely hindered Heinz from pursuing potential markets where they feel that their company would be well positioned. Due to the fact that H. J. Heinz Inc. Is located in the United States, there are no currency exchange issues. During the PAYOFF, sales in Canada increased by 1. 3% due to favorable Canadian exchange rates. This is an improvement over the past year where sales were down as result of the dollar weakening against the Canadian currency. This favorable exchange rate in 2010 also resulted in an increase in gross profit from PAYOFF (H. J. Heinz Annual Report and ask, 2010). During the fiscal year of 2010, transaction-related currency cross-rates in the U. K. Hindered the Companys gross profit margin (H. J. Heinz Annual Report and ask, 2010). For the continent as a whole, gross profit declined to $1. 25 billion largely due to unfavorable foreign exchange translation rates and increased commodity costs (H. J. Heinz Annual Report and ask, 2010). This also led to a large decrease in sales due to the unfavorable rate. The areas that were affected the most by this slide were the frozen food plants in Europe resulting from increased commodity costs and higher manufacturing costs (H. J. Heinz Annual Report and ask, 2010). Asia Pacific In this region, foreign exchange translation rates were very favorable, which led to an $83 million gross profit increase for Heinz (H. J. Heinz Annual Report and ask, 2010). Improvements in productivity also motivated this large increase. Operating income of he firm increased as a result of good exchange rates and an increase in marketing investments. Venezuela In Venezuela, Heinz has a currency control board that is responsible for the translation of foreign currencies. Concerning imports, Heinz has established good relations to obtain U. S. Dollars for the official exchange rate for items such as ingredients, packaging, manufacturing equipment, and other necessary inputs. There is an unregulated parallel market for exchanging EVE to U. S. Dollars, but this Company does not enter into such transactions. The official exchange rate has been axed for several years at 2. 5 Veto 1 U. S. Dollar, despite a large increase in inflation. Heinz has recorded a $62 million currency translation loss due to the currency devaluation (H. J. Heinz Annual Report and ask, 2010). However, even with these regulatory institutions in place to combat issues that may arise, the country of Venezuela is still a large liability. The government o f Venezuela recently expropriated Owens-Illinois, the worlds largest producer of glass containers. Hugo Chapel claimed to the media that [Owens-Illinois] was exploiting workers and damaging the environment, which is why the government is taking over the local unit. In the media release, it was cited that Chapel even momentarily forgot the name of the company he was taking over. 16 This demonstrates that Venezuelan dictator has ruthless motives that affect his decisions. Heinz must continue to monitor these developments in order to prevent incurring future damage. Global Market Entry Strategies Whines primary method of market entry is through acquisition of existing companies in a particular market. The Companys recent acquisition trends have involved Beams Big Emerging Markets) and the BRICK (Brazil, Russia, India, and China) countries.

Herbert Hoover Biography and Presidency

Herbert Hoover Biography and Presidency Herbert Hoover (1874-1964) served as Americas thirty-first president. Before turning to politics, he served as a mining engineer in China. He and his wife Lou were able to escape the country when the Boxer Rebellion broke out. During World War I, he was quite effective organizing Americas war relief efforts. He was then named as the Secretary of the Commerce for two presidents: Warren G. Harding and Calvin Coolidge. When he ran for the presidency in 1928, he handily won with 444 electoral votes.   Here is a quick list of fast facts for Herbert Hoover. For more in depth information, you can also read the Herbert Hoover Biography Birth August 10, 1874 Death October 20, 1964 Term of Office March 4, 1929-March 3, 1933 Number of Terms Elected 1 Term First Lady Lou Henry Chart of the First Ladies Herbert Hoover Quote Every time the government is forced to act, we lose something in self-reliance, character, and initiative.Additional Herbert Hoover Quotes Major Events While in Office Stock Market Crash (1929) Start of the Great Depression (1929) Hawley-Smoot Tariff (1930)Bonus Army March (1932)Lame Duck Amendment Ratified (1933) The stock market crashed on Black Thursday, October 24, 1929, only seven months after Hoover had taken office. Five days later, on October 29th, Black Tuesday happened devastating stock prices even further. This was the beginning of the Great Depression that would effect countries around the world. Unemployment levels in the United States hit twenty-five percent.   When the Hawley-Smoot Tariff was passed in 1930, Hoovers goal was to protect the American farming industry. However, the real effect of this tariff was that foreign countries countered with high tariffs of their own.   In 1932, a Bonus March happened in Washington. Veterans had previously been awarded insurance under President Calvin Coolidge that was to be paid out after twenty years. However, because of the economic devastation of the Great Depression, over 15,000 veterans went to Washington D.C. to demand immediate payouts of their bonus insurance. They were virtually ignored by Congress. The Marchers ended up living in shantytowns around the US Capitol. To deal with this situation, Hoover sent in the military under  General Douglas MacArthur to get the veterans to move. The military used tanks and tear gas to get the veterans to leave.   Hoover lost reelection by a wide margin as he was blamed for much of the fallout and dire situations for many Americans during the Great Depression.   States Entering Union While in Office None Related Herbert Hoover Resources: These additional resources on Herbert Hoover can provide you with further information about the president and his times. Causes of the Great DepressionWhat actually caused the Great Depression? Here is a list of the top five most commonly agreed upon causes of the Great Depression. Chart of Presidents and Vice PresidentsThis informative chart gives quick reference information on the presidents, vice-presidents, their terms of office, and their political parties. Other Presidential Fast Facts Calvin CoolidgeFranklin D. RooseveltList of American Presidents

Economics Assignment Essay Example | Topics and Well Written Essays - 1750 words

Economics Assignment - Essay Example In the final stage, the paper would explain several further businesses and operational strategies that the company can adopt in future. With the help of these theories, the company would be able to improve its business in future. The learnt from the essay would help to understand the importance of economics in the current state of business affairs. Contents Contents 3 1. Introduction 4 1.1 Report Brief 4 1.2 Company Background 4 2. Literature Review 5 2.1 Demand Theory 5 2.2 Market Structure Theory 6 2.2.1 Differentiation and Revenue Maximizing Theory 7 2.3 Long Run Theory 7 3. Case Study 8 3.1 P&G and Theory of Demand 8 3.2 P&G and Market Structure Theory 10 3.3. P&G and Long Run Theory 12 4. Conclusion and Recommendations 13 Reference List 15 1. Introduction 1.1 Report Brief This essay will analyze the business of the famous American consumer goods company named Procter & Gamble (P&G). The researcher would analyze the business performance of the company on the basis of economic the ories by scrutinizing the company’s annual report. In the later stage, the researcher would explain the business and operations strategies undertaken by the company through the analysis of its corporate activities. 1.2 Company Background P&G, the famous consumer goods company in America has its headquarters in Cincinnati, Ohio, United States. The products which are produced and sold by the company are primarily categorized in terms of three segments. These are cleanup agents, pet foods and individual care products. The company enjoys a high brand value in the market, with annual sales of about $83.68 billion (2012). The organization also enjoys a wide social presence in the market. It operates in almost all the countries in the world and as recorded in 2012, its operating income was $13.29 billion (P&G, 2012). The following context of the paper would explain the performance of the company by analyzing its annual report on the basis of economic theories. 2. Literature Review 2 .1 Demand Theory Demand for a commodity or a service in the market is the want of an individual backed by proper purchasing power. If a demand is created by a single household in an economy, then it becomes an individual demand. The horizontal integration of all the demand curves is termed as the market demand curve. According to the law or Theory of demand, considering the other factors affecting demand to be constant (Ceteris Paribus), the quantity demanded for a product is inversely related to its prices and vice versa (Mullerat, 2011). Thus, a market demand curve is always negatively sloped in nature with the assumption of Ceteris Paribus. Figure 1 Demand Curve Price Demand Curve (Market or Individual) Quantity (Source: Authors Creation) A shift in the demand occurs when other factors affecting demand, apart from price, changes. These other factors include tastes and preferences of the consumers, price of substitutes and price of complements along with income of the consumers. F igure 2 Shifts in Demand Curve Price Initial Demand Curve New Demand Curve Quantity Demanded (Source: Authors Creation) The above diagram shows a shift in the demand due to changes in any of the demand determinants. 2.2 Market Structure Theory The market structure theory explains the exact type of market in which organizations can operate. The market structural differentiation is mainly categorized on the basis of the strength of the seller and buyer in a

Containerizations in Maritime Shipping Essay Example | Topics and Well Written Essays - 250 words

Containerizations in Maritime Shipping - Essay Example It is worth mentioning that this particular system assists such organizations in attaining superior competitive position through providing them with numerous advantages. In this regard, one of the primary advantages of containerization in maritime shipping is to protect goods from different hazardous conditions while delivering them to other locations. In this regard, the system is most preferable towards conserving safety of the goods that to be transported in destinations located throughout different parts of the world. Most significantly, the containerization aspect also ensures to save a major portion of transportation costs as the system tends to seal intact and load huge quantity of finished goods or logistics from diverse locations (YouTube, n.d.). With regard to the roles and functions possess by an effective containerization system, in most often cases, the process seems to be parallel with air freight industry. This can be justified with reference to the fact that the system of containerization is often viewed to be a freight transport process, which involves the usage of effective and standardized techniques of shipping containers through diverse sorts of mediums including, shipyard, truck and airlines. Notteboom, T., & Rodrigue, J. P. (2009). The future of containerization: perspectives from maritime and inland freight distribution. Retrieved from http://people.hofstra.edu/jean-paul_rodrigue/downloads/future_containerization_tn_jpr_draft%20final.pdf YouTube. (n.d.). Pport of long beach: life of ocean container - learn international shipping & export doc. Retrieved from

Football game Essay Example for Free

Football game Essay The crowd sat on the edge, waiting impatiently for the next pass, the next tackle. This intense game had both teams fighting for control of the ball, struggling to score the next touchdown. The crowd was alive with concentration, which was then immediately interrupted by deafening cheers as soon as their team scored. As a fan, football from my perspective has always been a phenomenal sport that creates an intense, fun, and lively social atmosphere. It gives a chance for friends and family to bond over cheering their team on, while jokingly trash talking others in a casual manner. At a particular Eagles vs. Cowboys game I attended last year, this experience is exactly what I received. The lights, the screams, and the smell of delicious food all served to create a fun event. The highlight of the night, however, was when quarterback Michael Vick got hit extremely hard in the back field. It should have been a late hit but no whistles were blown. The crowd was immediately silenced by the sudden knockout. No one had expected such a hard hit. We fans craned our neck desperately trying to get a better view of what was going on. I remember being completely in awe of the violent nature of football that was revealed to me that night. I was surprised to find that such a dangerous game was even being encouraged to play! But there was one thing that was undeniable for sure— that hit was one moment I would surely never forget. From that game on, I always continued to follow the events and highlights of football. The league put out a lot of new rules for the NFL, to ensure the safety of its players. One of these rules was that a defenseless player cannot be tackled. So for example, if a receiver goes up for the ball, he cannot be tackled in the air. Another rule is the helmet to helmet tackle. A defender is not allowed to make a hit on a player, making contact from his helmet to the defenders. This can result in a serious concussion, and for long time purposes, the players can develop brain damage. To also make the game safer, kickoffs are now issued on the 30 yard line opposed to the traditional 20 yard line. This makes the return possibility very low, because the kick usually ends up deep in the end zone. This new rule change enforced on the kickoffs also accounts of less injuries, because it is one less play being executed. Special rules for quarterbacks have also been placed. Quarterbacks are looked at as the captains of their team. They know the offense’s strengths and weaknesses the best so they can make the right calls in the right situation. These players cannot be injured whatsoever, because of their impact on the game. That is why new rules, such as late tackles, are being enforced on defenders that try to scare the quarterbacks. When the new rules first came out, I supported and respected the fact that they were concerned about the player’s safety. The scientific discoveries about head trauma and mental illness made sense to me, and protecting the players from it seemed like a great idea. Soon the rules were being strictly enforced in the pros. When my friend and I went to the Cowboys and Eagles game, one of the first things that happened was a foul called for hitting helmets with a player. I approved of the call, it seemed fair. But out of the corner of my eye, I saw my friend shaking his head and say under his breath, â€Å"Man, football just isn’t the same anymore. † I shrugged it off— it wasn’t my fault he wasn’t enjoying the game. But later that night, his words tossed around in my head. Something did seem different about that game. But what exactly did he mean by not being the same? This question led me to reminisce back to earlier that game, when Vick first got late hit in the back field. The one thing I was certain of was that the beauty of the sport was created by the suspense of the game.

Three Strikes And Youre Out Law Essay -- essays research papers

Three Strikes You're Out Law We have all heard of the newest anti-crime law, the "Three strikes and you’re out" law. It wasn’t easy getting this law from the bill stage in Sacramento to the law stage, because it is not a criminal friendly law. Meaning that this law’s purpose is to bring pain, suffering, and intimidation to criminals. Our state government was basically ran by the Assembly Speaker Willie Brown, now mayor of San Francisco. Brown had the power to choose who sat on what committee in the house, and using this he could terminate any bill he did not agree with. And with this attitude it took a lot of patients and perseverance by the people trying to pass this bill. But how did the bill become a bill? I will answer this question with help of the Kimber Reynolds story. Monday, June 29, 1992 in Fresno, California a young woman was brutally murdered outside The Daily Planet, a restaurant patronized by the local young people. The girl was visiting home for the summer after being in the Los Angeles area attending school. Her and a friend were getting into their car when two guys on a motorcycle rode up next to Kimber Reynolds blocking her in, taking her purse, and beating her into submission. The story made the 11 o’clock news only minutes after her father had gone to bed. When police ran a background check on the two suspected men, Joeseph Micheal Davis and Douglas Walker, both men had recently been released on parole with multiple offenses on their records. Unfortunately Davis was never brought in because when police were attempting to arrest him he began firing, wounding unsuspecting police officers and ultimately being killed. Douglas Walker was convicted of accessory to murder. Mike Reynolds, Kimber’s father, went on the radio on a local radio show called the Ray Appleton Show, KMJ 580. There he would discuss his outrage about how he was sick of repeat offenders being locked up only to be released after a fraction of the sentence was completed. He swore to the people listening that he was going to do something about the problem, even if it takes him forever. Listening to that show was Fresno Assemblyman Bill Jones (R). He was interested in the issue and arranged a meeting with Mike. They discussed ideas about how they could solve this problem. With that in mind Mike used some connections and g... ...victions. There were repeated warnings about the cost to implement the new law, but few have addressed the other side of the equation and the savings to the state, in lives and in dollars. Had our 1993 crime rate continued unaffected over these past few years, nearly 815,000 additional crimes would have been committed in California, including 217,000+ violent crimes. We would have suffered more than 4,000 homicide victims; 6000+ women would have been victims of rape. Also the savings in dollars is between $5.8 billion and $15.5 billion since the enactment of the "Three Strikes" law. There has been swift and dramatic impact on crime since the enactment of the "Three Strikes" law. The crime rate has dropped more than 30%. But there are other factors that play a part in this reduction like crime prevention, and community policing. However there has been a significant drop in the crime rate. Also the predictions about cost, over populating and others have not come true. With all of the opposition out there trying to tear this law down I believe that California can not afford to do without this law because it is saving our state money and lives.

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