Friday, May 31, 2019

Essay --

Adham ElnashaiUnited States HistoryDuring the Civil War and the American renewing Americans didnt only have to deal with their republical policies, they too had to deal with their international relations with the rest of the countries in the World. The relations with other countries are also known as Foreign Policies. The overseas dimensions reflected onto the Civil War and Revolution Era. In 1861, 11 states seceded from the United States to form the Confederate States of America and over the socio-economic class of the following four years, America fought to bring the Confederate States back under control. During the Civil War the Confederacy frequently required international support for its cause, frequently calling upon foreign support on its cotton exports to obtain it. The Union, on the other hand, strove to prevent other nations from recognizing the Confederacy as a legitimate nation and from getting involved in the Civil War.In an attempt to starve the Confederate economy and to cut it off from its international supporters, the Union engaged in a blockade of Confederate ports a move that was of questionable legality in international law. Despite the Confederacys significant international commercial message bonds, the lack of definitive military victories for the South and the success of Union efforts to link the Confederacy with the institution of slavery finally prevented any of the European powers from officially recognizing or supporting the South. Figure 1 (Blockade Runners of the Civil War)One of Lincolns main strategies rested upon an effective blockade of the Souths 3,500 miles coastline, embracing a dozen main ports and nearly two hundred bays and maneuverable rivers. This was almost impossible for a nation with only so... ...ement of space to portraying the barbarities. The most renowned U.s. prohibitionist pioneer, Frances Willard of the WCTU, existed in England in 1896 she raised trusts for Armenian displaced people in Britain and Am erica. The United States, through resolutions, in the end chose to work by implication with different countries to secure the privileges of the Christian Armenians, as well as the wellbeing of American lives and property abroad. Protestant teachers gave by and large faultless reports of human rights ill-uses against Armenians. They and the American press, nonetheless, frequently stereotyped the roughshod and barbarian Turks, inferring that they were racially or religiously inclined to torture and homicide. In toons from the presidential fight, descriptions of Turks reflected and strengthened such biases. Figure 3 (Turks threatening to expel Armenians)

Thursday, May 30, 2019

The Computer :: Technology Internet Electronics Essays

The ComputerEven though J. David Bolter wrote Turings Man occidental Culture in the Computer Age in 1984, at least a century ago in computer years, his observations and concerns ab out(a) the electronic geezerhood are in no way obsolete. Bolter examines from a historical perspective how the computer will reshape our understanding of the human mind and our relationship with nature. By comparing the computer with defining technologies of previous ages, Bolter anticipates the some qualities and values of people in the digital age, his so-called Turings men. In doing so, he encourages those in the humanities to take an active role in shaping some of the perceptions of the emerging era.Humankind has created, used, and replaced innumerable technologies oer the past 10,000 years of written record. Turings Man concentrates on only four of these technologies the spindle and potters wheel of ancient Greece, the mechanical clock of Renaissance Western Europe, the steam engine of Indust rial Europe, and finally the computer of the electronic age. According to Bolter, each of these inventions defines or redefines mans role in relation to nature(13). Although Bolter considers the influences of these technologies on many facets of culture and society, his main efforts are devoted to understanding their implications for a cultures view of time, space, thought, and creation. The spindle and the potters wheel of Ancient Greece aim an intimate relationship between man and nature. The tools are seen more as an extension of the human hand than a barrier between craftsman and material. Bolter argues that the physique of the potter, fashioning his finely crafted, but still imperfect vessel on the rotating wheel, made a great impression on the thinkers of that time. The rotating nature of these tools, mimicking the great bankers bill paths followed by celestial bodies, led Aristotle to claim that circular motion . . . was natural, whereas motion in a straight line req uired further explanation(116). As a consequence, the Greeks adopted a cyclical view of time. The world did not progress forward in linear motion but repeated over and over. Rather than progressing, ideas and institutions would remain static or decay.The potter and his clay also served as a metaphor for divine creation. The world and its human inhabitants were fashioned out of imperfect materials by divine force. This material makes up the entirety of the world.

Wednesday, May 29, 2019

Are College Students Getting the Grades They Deserve? :: School Education Essays

Are College Students Getting the Grades They merit?Students at capital of Massachusetts University complain of variety deflationCollege students make for hard assumptive that they will get the put they deserve but this is not always the case. Caroline Boulanger, a sophomore business administration and management major at Boston University studies hard. In her freshmen economic science class, her final bulls eye was based on three exams. She received two As and an A- on these exams. However, at the end of the semester, she ended up with a final grade of a B-. She tried contacting her professor and he has still not gotten back to her, so she assumes her grade was deflated. Boulanger is not the alone someone who has had this problem as a result of the grading policy of Boston University. Ive heard that getting a 4.0 at this school is about as likely as loving the lottery or getting struck by lightening. It could be considered an act of god, say Haley Goucher, a freshman prem ed student at Boston University.If a student does receive a 4.0 at Boston University they are in the minority. In a survey of 100 Boston University students, only 23% received an A in any one of their classes and 0% of the students had a GPA of 4.0. Many of these students expressed that this sudden decline of grades made them fall asleep confidence in their work and themselves.People who did well in high school have trouble transitioning sometimes. They work hard and they still feel stupid, said Alex Corhan, a sophomore news media major at Boston University.There is no formal grading policy at Boston University but the university does hit several guidelines for professors to follow.Dont be a grade-inflater Grades should reflect the distribution of effort and success in the class If your distribution of grades is skewed toward the high end, it qualification imply that either an A does not require a high level of achievement in the course or you are demanding too little of your s tudents, the Boston University College of Arts and Sciences Information for cleverness Instructors states.Not only does Boston University strongly discourage grade inflation, but they evaluate their professors by how high they grade their students. This leads to problems for twain professors and students.The individual colleges monitor the grade distribution of courses and let instructors know if their grades seem to be considerably higher or lower than the norm.Are College Students Getting the Grades They Deserve? School Education EssaysAre College Students Getting the Grades They Deserve?Students at Boston University complain of grade deflationCollege students work hard assuming that they will get the grade they deserve but this is not always the case. Caroline Boulanger, a sophomore business administration and management major at Boston University studies hard. In her freshmen economics class, her final grade was based on three exams. She received two As and an A- on these e xams. However, at the end of the semester, she ended up with a final grade of a B-. She tried contacting her professor and he has still not gotten back to her, so she assumes her grade was deflated. Boulanger is not the only person who has had this problem as a result of the grading policy of Boston University. Ive heard that getting a 4.0 at this school is about as likely as winning the lottery or getting struck by lightening. It could be considered an act of god, said Haley Goucher, a freshman premed student at Boston University.If a student does receive a 4.0 at Boston University they are in the minority. In a survey of 100 Boston University students, only 23% received an A in any one of their classes and 0% of the students had a GPA of 4.0. Many of these students expressed that this sudden decline of grades made them lose confidence in their work and themselves.People who did well in high school have trouble transitioning sometimes. They work hard and they still feel stupid, sai d Alex Corhan, a sophomore journalism major at Boston University.There is no formal grading policy at Boston University but the university does give several guidelines for professors to follow.Dont be a grade-inflater Grades should reflect the distribution of effort and success in the class If your distribution of grades is skewed toward the high end, it might imply that either an A does not require a high level of achievement in the course or you are demanding too little of your students, the Boston University College of Arts and Sciences Information for Faculty Instructors states.Not only does Boston University strongly discourage grade inflation, but they evaluate their professors by how high they grade their students. This leads to problems for both professors and students.The individual colleges monitor the grade distribution of courses and let instructors know if their grades seem to be considerably higher or lower than the norm.

A Brain Drain Essay examples -- Biology Essays Research Papers

A Brain Drain Anthrax, arteriovenous malformation (cerebral), chronic inflammatory, polyneuropathy, dementia due to metabolous causes, encephalitis, epilepsy, febrile seizure (children), generalized tonic-clonic seizure, Guillain-Barre syndrome, hydrocephalus, inhalation anthrax, treatments involving leukemia and neuroblastoma, malignant diseases involving the head word or spinal cord, meningitis, multiple sclerosis, normal pressure hydrocephalus (NPH), pituitary tumor, polio, Reyes syndrome, subarachnoid brain hemorrhage, syphilis, tertiary, and so forth Seems like quite an extensive list, and yet the list goes on to include hundreds of dehabilitating diseases which can affect the central nervous system. One thing that all of these conditions wee in common is the potential to be discovered with the evidence from a procedure know as the lumbar puncture (spinal tap or cisternal puncture). over spring break, I had the wonderful opportunity to intern with a neurologist. During this t ime, I viewed the lumbar puncture. As painful and invasive as I had imagined it, it was not. I knowing that the lumbar puncture is actually a quite simple and absolutely amazing procedure, much less painful than an intravenous or drawing blood. at that place is an exorbidant amount of information which can be obtained from the cerebrospinal fluid. Lumbar punctures are an outpatient procedure, usually performed by a neurologist, pediatrician, or family doctor and lasting only a few minutes. The goal of the tap is to collect and measure the pressure of the cerebro spinal fluid (CSF) which surrounds, cusions, and protects the brain and spinal cord. The CSF is a dense water-like fluid secreted by the brain and flows through the skull and spine into the subara... ...e question and answer bulletin board. http//209.67.220.19/mayo/askphys/qa990215.htm3)Picture of Lumbar Puncture Tray.http//www.neocare.com/_borders/lpk.ht1.jpg4)Image of Lumbar Puncturehttp//night.medsch.ucla.edu/problem/w wid95/SPINLTP2.gif5)Lumbar Puncture Procedures, For med students. http//www.medstudents.com.br/proced/lumbpunc.htm6)Ask the Mayo Physician, More responses from the bulletin board.http//www.healthanswers.com/database/ami/converted/003515.html7)Healthanswers medical research library.http//www.healthanswers.com/database/ami/converted/003526.html8)Health central, More information from healthcentral, with image. http//www.healthcentral.com/mhc/img/img2930.cffm9) Medical review., An informative site with lumbar puncture information. http//cpmcnet.columbia.edu/news/review/archives/medrev_v1n1_0010.html

Tuesday, May 28, 2019

Introduction to Antiques and Breakables :: essays papers

Introduction to Antiques and BreakablesThis page is an introduction to antiques and breakables for the new lift upor, or just those curious about this field. Antiques and breakables as a hoby, or even a business, is very popular the world over. Every major city has antique shops, antique shows, and flea markets where antiques argon actively bought and sold.Many antiques also change hands finished the mails. Buyers and selers hook up by advertising and listings online, and in paper-based antiques publications. Its a lot of fun, and many who are in the antiques business started out as collectors. They enjoyed it so very much theyve decided to make it a full or part-time business. Those just starting out, however may have a few basic questions. What are antiques?The term antique was originally defined by U.S. Customs to be anything made made before 1830. This was later changed to anything over degree Celsius years old. This 100 years old criteria became a common standard to define s omething as an antique. Yet, most antiques shops and antiques shows today typically have a large percentage of store made in the 20th century.Today even high-end antiques dealers carry these items. They go by design, workmanship and rarity to define what is considered appropriate for an antique shop. Examples of items in this category are art pottery, art glass, jewelry from the turn of the century through the 1940s, and Tiffany silver produced in this century.What are Breakables?Breakables is a term that is sometimes used to grade more recent items from antiques. The term breakables also has some what of a connotation of popular culture. Things that were part of everyday life in a bygone date of reference now fondly remembered. There are thousands of categories, but some examples would include fountain pens, childrens lunch boxes, old movie memorabilia and comics books. It can also refer to moderate edition breakables. Those are things manufactured and marketed specifically as breakables as in collectors plates and Franklin mint items. Really though, breakables are anything that people collect. This could mean coins, antiques, rocks, Star Wars memorablia made in the 70s, or todays POGs that young kids (and who knows, probably adults too) collect. Who are the collectors?Today, throughout the world, there are millions of collectors that have collections in thousands of categories. There is collectors in all walks of life from multimillionaires that collect impressionist paintings to the average person, who might collect anything you can imagine.

Introduction to Antiques and Breakables :: essays papers

Introduction to Antiques and BreakablesThis page is an introduction to antiques and breakables for the new collector, or just those curious virtually this field. Antiques and breakables as a hoby, or even a business, is very popular the world over. Every major city has antique shops, antique shows, and flea markets where antiques ar actively bought and s grey-headed.Many antiques also change hands through the mails. Buyers and selers hook up by advertising and listings online, and in paper-based antiques publications. Its a lot of fun, and many who be in the antiques business started out as collectors. They enjoyed it so much theyve decided to make it a full or part-time business. Those just starting out, however whitethorn have a few basic questions. What are antiques?The term antique was originally defined by U.S. Customs to be anything made made originally 1830. This was later changed to anything over 100 years old. This 100 years old criteria became a common standard to defin e something as an antique. Yet, most antiques shops and antiques shows today typically have a large percentage of inventory made in the 20th century.Today even high-end antiques dealers carry these items. They go by design, workmanship and slenderness to define what is considered appropriate for an antique shop. Examples of items in this category are art pottery, art glass, jewelry from the turn of the century through the 1940s, and Tiffany silver produced in this century.What are Breakables?Breakables is a term that is sometimes used to distinguish more recent items from antiques. The term breakables also has some what of a connotation of popular culture. Things that were part of terrestrial life in a bygone era now fondly remembered. There are thousands of categories, but some examples would include fountain pens, childrens lunch boxes, old movie memorabilia and comics books. It can also refer to limited edition breakables. Those are things manufactured and marketed specifically as breakables as in collectors plates and Franklin mint items. Really though, breakables are anything that people collect. This could mean coins, antiques, rocks, Star Wars memorablia made in the 70s, or todays POGs that young kids (and who knows, probably adults too) collect. Who are the collectors?Today, throughout the world, there are millions of collectors that have collections in thousands of categories. There is collectors in all walks of life from multimillionaires that collect impressionist paintings to the average person, who might collect anything you can imagine.

Monday, May 27, 2019

Radio-the Movie

radiocommunication Topic 1 Psychological Disorders There are many different types of psychological disorders and even more people who fall infra the Diagnostic and Statistical Manual of Mental Disorders (Nevid, 2009, p. 524). In the photograph Radio, the main character is a young man name James Robert Kennedy. Also know as Radio, James is in the first scene of the movie shown pushing a shopping cart down a railroad track and is maunder under his breath and pass with an awkward limp. Immediately the viewer starts to lounge around the idea that there is something abnormal about him.When I looked at the DSM I notice that Radio fell under Axis II of mental retardation. Our book defines this as A generalized delay or impairment in the nurture of intellectual and adaptive skills or abilities (Nevid, 2009, p. 524). He has a very hard time socializing with others and comprehending simple information. At the beginning of the movie Radio is walking by the football field and the priva te instructor notices him. He walks over towards him and says, I just wanted to tell you Im sorry for what happened yesterday. Radio responds by nerve-racking to give back the football he found that was kicked over the fence.The day before, a few boys from the football team were caught locking him in the equipment shed and throwing footballs at it scaring him very badly. The coach opens the door and tries to help untie him but Radio doesnt seem to understand that the coach is only trying to help. You can tell here that his intellectual skills are severely impaired. Radio cannot care for himself and is very dependent on others in the movie for him to survive. One sample of this is on Christmas, he calls the coach and says What pants do I wear? because his mother is not home to dress him and he also ask the coach how the pants look over the phone.The coach takes Radio home one day after practice and meets his mother. He asks her what is wrong with him but the mother express that t he doctor didnt really diagnose him with anything, just that he was a little slower than others. She said he has a brother named Walter who is just fine. This movie is based off of the seventies so it is probable that then, we didnt have nearly as much information on mental retardations. I couldnt really get enough information to categorize him as anything more specific either nor am I educated enough on the melodic theme to make such an assumption.Radio clearly suffers from a psychological disorder and could not survive on his own. He couldnt adapt quickly enough to get the things he needs if he were not around someone he could depend on. He also lacks many intellectual skills and has many of the complications that I sympathize from the Google Health website. He cannot care for himself, interact with others appropriately, and is socially isolated.Work Cited Nevid, JS. (2009). Psychology concepts and applications. Boston, MA Houghton Mifflin Company. Google Health, www. google. co m/health

Sunday, May 26, 2019

Rivalry in the Oil and Gas Field Service Industry

Rivalry being present in any industry is obvious. Some industries have more than others and for different reasons. With all over 12,000 different companies in the Oil and Gas Field Services industry competition is high and is projected to only continue to increase. This is due to the crave of oil color and gas in the United States and the world. It is also because international firms are beginning to come in the United States to compete with US firms for business. In comparison, US firms operating(a) internationally generally make a significant amount more revenue from the business activity that goes on worldwide.Price competition is often the primary cistron in determining which contractor is awarded a contract, although quality of service, operational and safety performance, equipment suitability and availability, reputation, and technical expertise are also factors. (IBIS WORLD) Most of the contracts for service are awarded through competitive bidding. In order to compete f or profits it is imperative that companies in this industry look to create a competitive advantage and strait-laced business strategy. Large companies can offer a broad range of serve.Small firms can compete effectively by specializing in a especial(a) type of service or geographic area. (First Research) The Oil and Gas Field Service industries concentration is low. Even though their biggest companies do possess what seems to be great percentages of the market share they do not create a monopoly situation. The majority of the industry is small companies. About 78. 7% of the industry firms employ fewer than 20 people, and 95. 6% of firms employ fewer than 100 people. The overall level of strength for intensity of competitive rivalry in the Oil and Gas field services industry is high.The fact that it is hard to exit the industry creates higher rivalry. Due to the fact that oil and gas operations are highly energy and diligence intensive, fixed costs are high and market is hard to exit as leaving would require significant divestments of assets specific to the business. (Marketline) Many of these assets like equipment and machinery disparage causing the company to lose money. Also, fixed costs being high makes companies maintain their volume which escalates competition. The fact that the industry is growing and projected to continue this mode the more companies will enter the market making it more competitive.However, with increasing growth it also gives the companies already in existence a chance to improve income. The Mining Support industry has a low level of concentration, with the four largest firms accounting rougly 15. 7% of industry revenue. (Ibisworld) With the concentration being low the industry has many an(prenominal) companies that compete. In the Oil and Gas industry it is easy to swap products which creates low switching costs. When it comes to oil and gas many people choose whichever product or service is cheaper at that time so having repeat customers can be challenging.The overall products one company offers are not much different from others. The fact that this industry has low product differences creates higher competition. Higher competition is because there are not really any alternatives companies can use to attract customers. The deficiency of diversity also goes along with this idea. Some companies have technologic diversity over companies like Halliburton who uses Shale or have vertically integrated into other areas, but for the well-nigh part they are all acquiring oil and gas the same way.Main players activities are usually geographically and vertically integrated however most(prenominal) of them present similar business models. (Marketline) The oil and gas field service does have intermitten overcapacity creating more rivalry than normal at times. In this industry the demand fluctuates due to the market. At small periods of time companies supply will exceed demand. These firms will then compete m ore aggressively trying to get relinquish of the excess supply.

Saturday, May 25, 2019

History of Exxon Mobil Essay

Founded by John D. Rockefeller (1839-1937) in 1870, The Exxon Corporation developed from a nonher fossil oil company giant, which is regular crude oil Company. Reported monopoly of Standard oil over the oil industry in the early twentieth century proceeded to succession of criticism from politicians and even journalists. However, Exxon still remains the third largest company in the join States and reportedly to be the seventh largest in the world. (Fortune, April 28, 1997). It was Rockefeller anticipated a big potential of refining Pennsylvania crude oil.However internal combustion engines were not yet fully developed a substitute can be used which is kerosene to fuel lanterns. When Standard Oil was formed, it integrated all of the docks, pressure cars, warehouses, lumber resources, and other facilities it needed into its operations. Lucrative deals with railroads were made and that drove smaller refiners out of business. (Sampson, A. , 1975) Around 1878 when Rockefeller and pa rtner Henry Flagler (1830-1913) were in control of most of the provinces oil refining business.Because of its booming business in oil industry, Rockefellers was considered one of the five wealthiest men in the country. (Nevins, A. , 1953) In 1882 Rockefeller and his associates established the starting signal trust in the unite States, which consolidated all of Standard Oil Companys assets in the states under the New York Company, in which Rockefeller was the major shareholder. (Nevins, A. , 1953) Standard Oil began producing, refining and distributing oil in 1880s. Overseas trade had begun mostly in kerosene to Great Britain.The trust encountered challenges with the Sherman Antitrust Act of 1890, followed by an 1892 Ohio peremptory Court decision which forbade the trust to operate Standard of Ohio. The company then moved its base of its operations to New Jersey, which in 1899 became home to Standard Oil of New Jersey, or Jersey Standard. Jersey Standard later became Exxon Corpor ation (Wall, B. , 1988) In the 1920s, as the supply of crude oil began shifting its way from the United States and Latin America to the Middle East, Jersey Standard and other companies effectively used the same monopolistic practices that John D.Rockefeller had used Standard Oil utilize its rich resources in Iraq, Iran, and Saudi Arabia.This made oil prices stayed low and the United States and Europe became extremely dependent on oil fuels for industry and automobiles. (Nevins, A. , 1953) The Organization of oil Exporting Countries (OPEC) was formed to protect the interests of the producing countries and this led to Jersey Standard sought other sources of crude oil. The company discovered oil fields in Alaskas Prudhoe Bay and in the North Sea.Around the same time, in 1972, Standard Oil of New Jersey officially changed its name to Exxon Corporation. (Wall, B. , 1988) As the OPEC-induced oil dearth depleted much of Exxons reserves made them experience financial difficulties and a l ot of people suffered from this cause. In 1989 when a drunk Captain of the oil tanker Exxon Valdez ran aground in Alaskas Prince William Sound, doing immeasurable damage to the wildlife and to the companys public image. Eleven million gallons of oil spilled in the Alaskan cheer this made Exxon lost a share of the world oil market to its competitor, Royal Dutch/Shell in 1990.However this didnt stopped Exxons business when they agreed to a $15 billion development of three oil wells in Russia. (Clarke, 1997) Overall, Exxon started in the United States, which started as a regional marketer of kerosene that evolved to the biggest publicly traded petroleum and petrochemical project in the world. Their best known brand names are Exxon, Esso and Mobil. Such products that drive modern transportation, military force cities, lubricate industry and provide petrochemical building blocks that lead to thousands of consumer goods.

Friday, May 24, 2019

Solution Manual for Fluid Mech Cengel Book

Chapter 6 impulse abridgment of feed Systems Chapter 6 MOMENTUM ANALYSIS OF FLOW SYSTEMS Newtons Laws and saving of Momentum 6-1C Newtons first law states that a body at rest cadaver at rest, and a body in inquiry perseveres in motion at the equal hurrying in a straight path when the net press playacting on it is zero. in that locationfore, a body tends to preserve its state or inertia. Newtons entropy law states that the acceleration of a body is pro personaate to the net contract acting on it and is inversely proportionate to its lot. Newtons third law states when a body exerts a tweet on a second body, the second body exerts an refer and opposite constrict on the first. r 6-2C Since pulsation ( mV ) is the product of a vector ( stop number) and a scalar (mass), nerve impulse must(preno bital) be a vector that bucks in the same counsellor as the speeding vector. 6-3C The conservation of impulsion principle is expressed as the pulse of a system inhabit s cons false topazt when the net durability acting on it is zero, and hence the neural impulse of such systems is conserved.The urge of a body remains constant if the net describe acting on it is zero. 6-4C Newtons second law of motion, excessively c altogethered the angular momentum equivalence, is expressed as the arrange 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 focal ratio 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 go out have different angular momentums unless they besides have the same moment of inertia I. Linear Momentum Equation 6-6C The relationship amid the time puts of change of an extensive property for a system and for a image tawdriness is expressed by the Reynolds transport theorem, which provides the link among the r system and laterality good deal concepts. The linear momentum par is obtained by setting b = V and and then r B = mV in the Reynolds transport theorem. -7C The get outs acting on the control volume consist of body campaigns that act by means of forth the entire body of the control volume (such as gravity, electric, and magnetic casts) and surface forces that act on the control surface (such as the ram forces and chemical reception forces at points of contact). The net force acting on a control volume is the sum of all body and surface forces. Fluid system of weights is a body force, and tweet is a surface force (acting per unit of measurement bea). -8C All of these surface forces arise as the control volume is isolated from its surroundings for analysis, and the put up of any detached object is accounted for by a force at that location. We can minimize the number of surface forces subject by choosing the control volume such that the forces that we atomi c number 18 not interested in remain internal, and indeed they do not complicate the analysis. A well-chosen control volume exposes wholly the forces that be to be determined (such as reaction forces) and a minimum number of other forces. 6-9C The momentum-flux subject argona 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 geological period rate and mean mix velocity as ? Ac commingle, such as a reverse lightning meld, nearly unity for turbulent rate of flow (between 1. 01 and 1. 04), but about 1. 3 for laminal flow. So it should be con officered in laminar flow. 6-1 proprietorship MATERIAL. 2006 The McGraw-Hill Companies, Inc. trammel dispersal permitted except to t for each oneers and educators for furrow preparation. If you ar a scholarly person development this Manual, you are employ it without permission.Chapter 6 Momentum digest of hunt down Systems 6-10C Th e momentum equality for steady analogue 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 out deprivation momentum flux by mass. 6-11C In the application of the momentum equating, we can disregard the atmospheric squelch and work with boob pressures only since the atmospheric pressure acts in all perpetrations, and its belief cancels out in every worry. -12C The fireman who tallys the hose rearwards so that the peeing incurs a U-turn before being discharged entrust 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 rockets ultimate velocity. Without clang the rocket velocity will continue to increase as more gas out allows the no zzle. 6-14C A eggbeater hovers because the strong downdraft of circularise, caused by the overhead propeller blades, manifests a momentum in the advertize drift.This momentum must be countered by the helicopter lift force. 6-15C As the air density decreases, it requires more zip 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 throws more power for a helicopter to hover on the top of a high mountain than it does at ocean level. 6-16C In winter the air is generally colder, and then denser. Therefore, less air must be driven by the blades to provide the same helicopter lift, requiring less power. 6-2 branded MATERIAL. 2006 The McGraw-Hill Companies, Inc. contain scattering permitted only to teachers and educators for course preparation. If you are a student utilize this Manual, you are using it without permission. Chapter 6 Momentum abbreviation of fall Systems 6-17C The fo rce involve to hold the ordered series against the horizontal body of wet stream will increase by a factor of 4 when the velocity is doubled since & F = mV = ( ? AV )V = ? AV 2 and and so the force is relative to the square of the velocity. 6-18C The acceleration will not be constant since the force is not constant. The impulse force exerted by & water supply on the cuticle is F = mV = ( ? AV )V = ?AV 2 , where V is the congenator velocity between the water and the home base, which is lamentable. The headquarters acceleration will be a = F/m. just now as the plate begins to move, V decreases, so the acceleration must as well decrease. 6-19C The maximum velocity possible for the plate is the velocity of the water tarry. 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 eye socketed. 6-20 It is to be shown that the force exerted by a liquid jet of velocity V o n a unmoving 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 atmosphere, and thus the pot pressure at the way out is zero. abridgment We take the nozzle as the control volume, and the flow advocate at the out permit as the x axis. bank 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 statement. Noting that m = ?AV where A is the nozzle outlet area and V is the average nozzle outlet velocity, the momentum comparability for steady one-dimensional flow in the x program line 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 light 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 unsloped 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 o f 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 stationary, 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 vexs 2. 25 times when the jet velocity becomes 1. 5 times. Discussion line of descent 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 jostle 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 shopping mall 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 zer o. 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 grounds kg/m3. Analysis (a) We take the elbow as the control volume, and de reduceate the jinx by 1 and the outlet by 2. We also demo the horizontal coordinate by x (with the direction of flow as being the positive direction) and the upright piano coordinate by z.The persistence 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) ( molarity kg/m 3 )? (0. 1 m) 2 / 4 Noting that V1 = V2 and P2 = Patm, the Bernoulli equation for a streamline going through and through the center of the simplification 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 = ( pace kg/m3 )(9. 81 m/s 2 )(0. 35 m)? 1 ? honey oil 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 on the x and y axes become & & FRx + P1,gage A1 = 0 ? ?m(+V1 ) = ? ?mV & & FRz = ? m(+V 2 ) = ? mV z x FRz 2 35 cm Solving for FRx and FRz, and modify 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 magnitu de 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 preparation. 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 ze ro. 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 specialize 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/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 on 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 s hould 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 jet strikes a vertical stationary plate commonly 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 unresolved 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 thu s 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.Di scussion In reality, some water will be scattered hind end, 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 preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-25 A reducing elbow deflects water upwards and discharges it to the atmosphere at a specified rate.The anchoring force needed to 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 g iven to be ? = 1. 03. Properties We take the density of water to be 1000 kg/m3. Analysis The weight of the elbow and the water in it is W = mg = (50 kg)(9. 1 m/s 2 ) = 490. 5 N = 0. 4905 kN We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction) and the 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 lettuce ? ? ? 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 equati on is applicable). 2 The weight of the elbow and the water in it is considered. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. The momentumflux correction factor for each inlet and outlet is given to be ? = 1. 03. Properties We take the density of water to be 1000 kg/m3. Analysis The weight of the elbow and the water in it is W = mg = (50 kg)(9. 81 m/s 2 ) = 490. 5 N = 0. 4905 kN We take the elbow as the control volume, and designate the 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 in let 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 t he 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 baby carriage 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 directions in the plane of the back surface. The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the 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 pushcart. 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 station ary 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 P ROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-28 Water accelerated by a nozzle strikes the back surface of a cart moving horizontally. The acceleration of the cart if the brakes give way 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 bloodter, such that half of the flow is diverted upward 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 effect 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)? ? = ? 1 one hundred thirty-five 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 for ce of 1135 lbf must be utilise to the splitter in the opposite direction to flow to hold it in place. No retentivity force is necessary in the vertical direction.This can also be concluded from the symmetry. Discussion In reality, the gravitational effects will cause 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 ft/s2 rho=62. 4 lbm/ft3 V_dot=100 ft3/s V=20 ft/s m_dot=rho*V_dot F_R=-m_dot*V*(c os(theta)-1)/g 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 6240 6240 6240 6240 6240 6240 6240 6240 6240 FR, lbf 0 59 234 519 907 1384 1938 2550 3203 3876 4549 5201 5814 6367 6845 7232 7518 7693 7752 FR, lbf 000 3000 2000 1000 0 0 20 40 60 80 100 120 140 160 180 ?, 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 s platters in the plane of the retreating plate. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure which is disregarded since it acts on all surfaces. 4 The tract is nearly frictionless, and thus fiction during motion is negligible. 5 The motions of the water jet and the cart are horizontal. 6 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, an equal and opposite impulse force acts on the plate, which is determined to ? 1N & Fplate = ? FRx = mV = (35. 34 kg/s)(18 m/s)? ? 1 kg ? m/s 2 ? ? ? = 636 N ? ? Then the initial acceleration of the plate becomes a= Fplate m plate = 636 N ? 1 kg ? m/s 2 ? 1000 kg ? 1 N ? ? ? = 0. 636 m/s 2 ? ? 18 m/s 1000 kg Waterjet Frictionless track This acceleration will remain constant during motion since the force acting on the plate remains constant. (b) Noting that a = dV/dt = ? V/? t since the acceleration a is constant, the time it takes for the plate to reach a velocity of 9 m/s is ? t = ? V plate a = (9 ? ) m/s 0. 636 m/s 2 FRx = 14. 2 s (c) Noting that a = dV/dt and thus dV = adt and that the acceleration a is constant, the plate velocity in 20 s becomes V plate = V0, plate + a? t = 0 + (0. 636 m/s 2 )(20 s) = 12. 7 m/s Discussi on 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 out let 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 FR x 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 twist turbine with a given orthodontic braces diameter and efficiency is subjected to steady baksheeshs. The power generated and the horizontal force on the supporting mast of the turbine are to be determined. vEES Assumptions 1 The come up 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 turbi ne ) 2 2 or V 2 = V1 1 ? ? wind turbine = (6. 94 m/s) 1 0. 32 = 5. 72 m/s We ask the control volume around the wind turbine such that the wind is normal to the control surface at the inlet and the outlet, and the entire control surface is at the atmospheric pressure.The momentum r r r & & equation for steady one-dimensional flow is F= ? mV ? ? mV . Writing it along the x-direction ? ? out ? in (without forgetting the negative sign for forces and velocities in the negative x-direction) and presume 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 pillar where the wind turbine is installed, and the bending moment it generates at the render of the tower is obtained by multiplying this force by the tow er height. 6-16 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-34E A horizontal water jet strikes a 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 include 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 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). 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 . allow 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 negative value for FRx indicates the assumed direction is wrong (as expected), and should be reversed.Also, there is no need for an analysis in the vertical direction since the fluid streams are horizontal. 2 140 ft/s 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 splat tered water is the atmospheric pressure, which is disregarded since it acts on all surfaces. 3 Frictional and gravitational effects are negligible. 4 There is no splattering of water or the deformation of the jet, and the reversed jet leaves horizontally at the same velocity and flow rate. 5 Jet flow is nearly uniform and thus the momentum-flux 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 assume them to be in the positive directions. Then the momentum equations along the x and y axes become & & & FRx = m(? V 2 ) cos 45 ? m(+V1 ) = ? mV (1 + cos 45) & (+V 2 ) sin 45 = mV sin 45 & FRz = m Substituting the given values, 1 lbf ? ? FRx = ? 2(428. 8 lbm/s)(140 ft/s)(1 + cos45)? 2 ? ? 32. 2 lbm ? ft/s ? = ? 6365 lbf 1 lbf ? ? FRz = (428. 8 lbm/s)(140 ft/s)sin45? = 1318 lbf 2 ? ? 32. 2 lbm ? ft/s ? 2 140 ft/s Waterjet 135 FRz FRx 3 in 1 and 2 2 FR = FRx + FRz = (? 6365) 2 + 1318 2 = 6500 lbf , ? = tan -1 FRy FRx = tan -1 1318 = ? 1. 7 = 168. 3 ? 6365 Discussion Note that the magnitude of the anchoring force is 6500 lbf, and its line of action makes 168. 3 from the positive x direction. Also, a negative value for FRx indicates the assumed direction is wrong, and should be reversed. 6-18 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited dis tribution 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 plot of land essay 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 (this 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)? ? 1k g ? m/s 2 ? ? ? Therefore, the firemen must be able to resist a force of 2457 N to hold the nozzle in place. Discussion The force of 2457 N is equivalent to the weight of about 250 kg.That is, holding the nozzle requires the strength of holding a weight of 250 kg, which cannot be done by a angiotensin-converting enzyme 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 al l directions in the plane of the plate. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure, which is disregarded since it acts on all surfaces. 4 The vertical forces and momentum fluxes are not considered since they have no effect on the horizontal force exerted on the 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 The refore, 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 Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.Chapter 6 Momentum Analysis of Flow Systems 6-38 Problem 6-37 is reconsidered. 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 kg/m3 D=0. 05 m V_jet=30 m/s Ac=pi*D2/4 V_r=V_jet-V_plate m_dot=rho*Ac*V_jet F_R=m_dot*V_r N Vplate, m/s 0 3 6 9 12 15 18 21 24 27 30 Vr, m/s 30 27 24 21 18 15 12 9 6 3 0 FR, N 1767 1590 1414 1237 1060 883. 6 706. 9 530. 1 353. 4 176. 7 0 1800 1600 1400 1200 1000 FR, N 800 600 400 200 0 0 5 10 15 20 25 30 Vplate, m/s 6-21PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-39E A yellowish brown 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 specimen 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 mechanically skillful power input is converted to kinetic energy of air (no renewal 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 pr essure 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 friction at the base, for example) to prevent the fan from moving in the horizontal direction under the find 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 drop hover, and the rev 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 pith 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 r ate of air in the drop mode become V 2,drop = 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 reprize 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 & mlo aded = ? 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 energ y. -24 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-41 A helicopter hovers on top of a high 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 location is negligible because of the low density of air. 6 There is no acceleration of the helicopter, and thus the lift generated is equal to the total weight. Air flow is nearly uniform and thus the 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 System s 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 maven 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 o ne-dimensional flow is ? ? out ? in force acting on the sluice gate FRx is horizontal

Thursday, May 23, 2019

APPLICATION OF RISK MANAGMENGT TECHNIQUES Essay

In my opinion Windows Vista is a one or the nearly lack OS when comp bed to Windows 7. All desktops connect to an persistence standard switch via an Ethernet cable. While this can be a risk, it is not a major risk. The two large production facilities atomic number 18 connected to the headquarters via an external ISP. Even with the firew entirelys in place, there is no accountability if the inter-group communication they contract is in use by some(prenominal)one else. I would advise contacting the ISP and verifying if the radio link is shared with other users and take further action depending on their answer. The sales personnel connect via VPN software, but use their individual internet connection, usually let out of their home office. This can be very dangerous as they do not fall under the blanket of protection offered by the bigger offices and their terminals are at greater risk to be infected by a malicious user. The core idea of preventing risk is to safeguard the informa tion stored on the database server.The snipers and customers of the bon ton have private information stored there and the loss or leak of the data could be disaster to the confederation. I suggest the changes to be made to mitigate the risk of any unwanted personnel to gaining access to the cyberspace. There is not a lot of information given about the entirety of the network, so a great deal of this may not be necessary or already in place. I will use the mitigation risk technique for the Desktops/local LAN. Since the network is maintained via Active Directory, the bon ton should implement workgroups/user groups and control what workers have access to if a program, file, or other application is not part of a workers job, they have no reason to be able to access that file/application/etc. At the same time the workers should go through annual (if not bi-annual) information security training that understands how to protect their workstations, understand security policies and wh y they are in place.The company should also ensure that their switches, routers, and firewalls are endlessly up to date on the latestpatches. Another risk that the company has is the External ISP Line, since the company is relying on an outside source to provide network connection between the production facilities and their headquarters the best way to approach this risk is also with the mitigation technique. I understand the company is small and if they cant front the cost of their own line, they should be absolutely sure that no other users are gaining access to the line that is being provided for them. On top of that they should convert the technical environment by adding intrusion detection systems and ensuring all security features are always up to date. If possible I would suggest investing into a private line that they control to ensure security between the three sites, however outside of the initial investment there would also need to be maintenance costs. As long as the c ompany can ensure the line theyre currently using is secure, Id recommend gallop use as it is the less cost intensive. Another risk to look at is the Remote Users / Home Offices. This risk is critical as they are the most likely to be targeted for an attack. Just like the previous two risks, Id recommend a mitigation technique to lower this risk.The remote users sole(prenominal) use software to connect to the companys VPN, on their own ISP connection, in their home office. To start I would recommend a two-factor authentication to successfully log on to the VPN so even if the computer is stolen or infected, its hitherto relatively safe. At the same time since these are sales associates, I would recommend using a hard drive lock just like the previous reason, if the computer is stolen, the ability to glean information would be hampered. If the company can handle the expense they should look into purchasing a secure VPN from each sales associates ISP, this would back up ensure that there wouldnt be any outside eyes gleaning information from the sales associate connecting to the company. Using Active Directory, the sales associates terminal should be scanned to light upon sure all security implements are current and if not, they should be updated before being allowed to connect to the company network. This can help prevent malicious code being introduced to the company network. One amour that caught my attention is that there are three servers at Headquarters with very few uses. One thing that worries me is the possibility of no redundancy. If the Active Directory Server went down, no one would be able to access the network.Each server role should have redundancy to fill in if the primary server is to fail, this will helpensure the company is running efficiently, even during a server problem. This should be kept in mind as the company has sales representatives in all fifty states while the headquarters are in Indiana. So even in a standard eight hour day (9A M 5PM), there is still three hours of work to people on the west coast. If the servers were to go down, those sales reps would not be able to work effectively. On top of redundancy the company should look into some sort of backup. They have a lot of information and while its important to protect it, its also important to make sure its not lost. For a backup, Id recommend a transfer technique. There are many backups services available at an affordable price. To go with the backup I would recommend backing up the information at least once a week to ensure if work is lost, the company does not fall too far behind.

Wednesday, May 22, 2019

Shanghai Business Environment

Enterprises that want to invest in chinaware can stumble over an array of regulations that do not give them free extract of where they wish to locate. This situation has been changing, and Chinas membership in the World Trade Organization (WTO) should act as an different catalyst to make the investment climate freer in several industries. Enterprises can look forward to devising decisions on where to locate within China based on factors that they would function in the more-familiar open environment. To charm to businesses, cities need a good IT infrastructure, strong leadership, incentives, and livability (see Figure A).Figure A The size of Chinas market and Western enterprises desire to devil close to it means that Chinese cities do not generally compete for orthogonal investment with cities in other Asia/Pacific countries. Outside the manufacturing sector, near enterprises locate in China because they want to parcel out to China. Of all Chinas cities, Shanghai has gone fur thest toward the success factors for a global smart urban center. Its ambition to become a major financial center and pseud on the international stage by 2015 has fueled this drive. History has also given Shanghai many advantages.It is probably the most outward-looking of any Chinese urban center and has a strong political voice in Beijing. The latter has allowed Shanghai to lead the way in many initiatives because Chinas political leaders often use the city to test out sunrise(prenominal) ideas. As a result, many initiatives that started in Shanghai perplex now spread elsewhere in China. Shanghais characteristics Livability To most Western expatriates, Shanghai is perhaps the most livable of Chinese cities. The city continues to make strides to improve (e. g. , announcing new rules to allow foreigners to buy property for the first time).Measured against other big cities such as Singapore, Hong Kong, or Sydney in the Asia/Pacific region, Shanghai still scores low. From a globa l perspective, livability is one of Shanghais weakest areas. However, livability has a large inwrought component, and what appeals to Western tastes may not rank as important to the skilled Chinese workforce that an enterprise might what to attract. Incentives Shanghai has traditionally enjoyed a respectable chunk of foreign investment into China, in part because of its position as a testing ground for reform.In most cases, China has forced foreign investors to strict up in the city first. This advantage will diminish, and Shanghai will need to learn to play on a more-level playing field. The city has commit to spending, by 2005, 150 billion yuan (one-third of its total industrial investment) on expanding its high-tech sector. This investment targets software and integrated circuit manufacturing, and the city will bid some tax breaks for new operations and help for self-employed software designers. Keeping costs low is key in attracting new business.Shanghai will adopt to ba lance the inevitable rise in labor costs with suitable business incentives. Leadership Shanghais leaders know where they want to go during the next 10 or 15 years and what basic things they need to do to get there. However, they did not develop this vision in partnership with business or the community. Rather, as a command economy, decisions bring in been made by a select few behind closed doors. Thus, its leaders have greater ability to get things done quickly than leaders in democratic societies often have.Cities such as Shanghai can complete projects without long internal or earthly concern debate over infrastructure projects that might take years in the planning stages in other countries and involve a myriad of agencies. Shanghai has benefited tremendously (certainly compared to the rest of China) from the pedigree of its leaders. President Jiang Zemin and Premier Zhu Rongji are former leaders of Shanghai and have strongly supported the city. However, Shanghai knows that spen ding does not necessarily produce results.Ten years ago, it started pouring money into redeveloping the riverside Pudong district into an area of towering skyscrapers, designed to be Chinas version of Manhattan. Within a city of 16 million people, Pudong ended up as a ghost town. Slowly, occupation levels have risen, but largely because the central government has twisted the arm of foreign enterprises rather than use real incentives. Infrastructure Many of the citys hopes revolve around an ambitious project to link all of Shanghai to a demon fast data network, known as the Shanghai Infoport.Scheduled for completion in 2010, the project stands out not just for its scale but also for its attempt to play together many strands of existing infrastructure (telecommunications and cable television especially) into one cohesive network. If it succeeds, Shanghai will be among the few cities in the world to have achieved such a feat. Five main projects will rely on the Infoports high-speed infrastructure 1. Shanghai Information Interchange net profit A giant intranet for Shanghai with links to many kinds of information 2.Shanghai Society Security Network Designed to offer e-payment and checking and to facilitate the use of smart cards 3. Social electronic Data Interchange Network for Foreign Trade An export/import data exchange for foreign trade 4. Social Community Service Network Focused on the residential community 5. Gold Card and Commercial Value-Added Network Linking banks automated teller machines (ATMs) and payment systems With 3. 2 million users, Shanghai claims to have the largest cable TV network of any city in the world, and this local-access network has a central part in Infoport.Shanghais connections in national government compete an important role in keeping the project on track. Although cable TV networks in the rest of China were barred for a period from offer Internet or telecommunication services, Shanghai received a special dispensation to upgra de its cable TV networks to do just that. The municipal government claims that 1 million residents can now reach interactive services through their televisions. It wants the entire network to be interactive by 2004.In addition, Shanghai Telecom (part of China Telecom) is rolling out digital subscriber line services and installing the necessary in-building cabling to offer Ethernet wideband access. Mayor Xu Kuangdi talked recently of every home having broadband access by 2004. At the same time, Shanghai Telecom has worked on improving the quality of the core network to be ready for the deluge of new data traffic the Infoport will bring. By year-end 2000, it completed work on what it claims is the worlds largest local ATM network. Shanghai now uses a total of 320,000 kilometers (198,848 miles) of fibre-optic cable, with more than 4,000 optical nodes.Shanghai Telecom says it has deployed optical fiber in more than 90 percent of the citys residential areas. Shanghais challenges Buildin g an infrastructure represents only part of making the Infoport work. The real test is whether people will use it. Only in the last few months have the first real customers logged on, so its a shortsighted early to tell what the response will be. Cost may prove one prohibiting factor, and the local government may have to consider deep subsidies to advance more than just the wealthiest people to sign up. Content also remains a question.Experience from elsewhere in the world, especially Singapore, a world-class smart city where government has tried to link itself to all the people, shows that the bulk of a citys population generally does not have much interest in such projects. They may want video-on-demand, but filling out tax documents online doesnt really excite them. Bottom line Other Chinese cities, notably Beijing, have begun some of the initiatives down the stairs way in Shanghai and are rapidly improving their information infrastructure. Shanghai will likely continue to sta nd out as Chinas smartest city, accord to Gartners success factors.Shanghais early start and its ability to exploit openings created by the central governments policies will likely keep the city at the forefront of innovation for some time. Shanghais advanced, if incomplete, IT infrastructure makes it a good place for Western enterprises to locate central operations in China. In addition, Shanghai is a good place in which to experiment with business-to-business and business-to-consumer projects requiring advanced IT infrastructure in hopes of rolling them out to the rest of the country when the infrastructure permits.