Ricardo Semler s Leadership Style - 820 Words

1. Describe Ricardo Semler’s leadership style. What do you think the advantages and drawbacks of his style might be? Ricardo Semler’s leadership style can be described as radical. This approach encourages followers to share their ideas and apply their creativity and ingenuity to reach the company’s goals. The employees have the opportunity to offer suggestions to improve the company. Also, every employee establishes their norms and their way of completing their job. They are free to choose their schedule, their level of salary, and who will be their boss. This model gives the employee a positive feedback since they are recognized for their good work and ideas. Some of the advantages the radical leadership style has are: †¢ The employees feel that they have greater responsibility and trust put on them, this can generate greater participation, commitment and a better job performance. †¢ Each employee is responsible for their own performance, creating self-commitment. †¢ There are no restrictions on resources or schedules that prevent them to carry out their work properly. †¢ Subordinates are free to choose their schedule and their level of salary. †¢ The employees have to assume the consequences of their actions. †¢ The type of communication is open, there are no defined controls or pressures. †¢ Each employee has the opportunity to vote in the election of the company’s leadership and new initiatives. The obstacles that can be found can also be some of the advantagesShow MoreRelatedRicardo Semler s Leadership Style813 Words   |  4 Pages1. Describe Ricardo Semler s leadership style. What do you think the advantages and drawbacks of his style might be? Ricardo Semler s leadership style can be described as radical. This approach encourages followers to share their ideas and apply their creativity and ingenuity to reach the company s goals. The employees have the opportunity to offer suggestions to improve the company. Also, every employee establishes their norms and their way of completing their job. They are free to choose theirRead MoreLeadership and Management2878 Words   |  12 Pagesincludes information about Recardo Semler and his organization Semco. It shows link between leadership and management. It also explains different types of leadership theories and management theories. It shows link between these theories. It also includes future leadership skills required in semco. And also shows method to plan the development of leadership skills. Answer: - 1 Ricardo Semler has situational leadership style. He is very close to Democratic leadership style. Philip (2003, p.63) describeRead MoreHow Do the Leader’s Management Styles Affect the Team’s Effective at Work?2315 Words   |  10 Pagesleader’s management styles affect the team’s effective at work? Dear Bill Yeung, We would like to inform that about our project topic is about the management styles affect the team’s effective at work. This project aims are to analyze different types of Leadership theories that identified behaviors that differentiated effective leaders from ineffective leaders. There have three types of Leadership theories. 1) Autocratic style 2) Democratic style 3) Laissez-faire style. First, what is â€Å"AutocraticRead MoreManagement Styles1872 Words   |  8 Pagesï » ¿Management Styles of Apple, Google, and SEMCO There are some businesses in the world that stand out more then others due to their success and ability to stay relevant in  their sectors. Three such companies, Apple, Google and Semco are like the all-star team  of corporations that command respect at the mention of their names. The reasons for their success are  illustrated in the business articles â€Å"Managing Without Managers,† â€Å"How Apple Got Everything Right by doing Everything Wrong,† and â€Å"WhereRead MoreRicardo Semler Semco: the Self-Managed Entreprise3880 Words   |  16 Pagesbased on 7 S model 12 5. Conclusions Recommendations 13 6. Bibliography 14 7. Appendix 15 SEMCO in brief SEMCO Group, originally founded in 1953 in Brazil by an Austrian engineer by the name of Antonio Curt Semler, went into business producing centrifuges for the vegetable oil industry. In the 60’s the company changed course to find success in producing components for the naval industry of Brazil, and over the next 20 years the company, still under control by Antonio Semler, enjoyedRead MoreManaging People4036 Words   |  17 Pages 4. EMPOWERMENT CONCEPT 11-13 5. GORES LTD – SIMILAR MANAGEMENT AS SEMCO 14-16 6. CONCLUSION 17 7. REFERENCES 18 INTRODUCTION âž ¢ Management is become very important in today s complex society, as more organizations are formed everyday. So management is very much essential in these organizations to plan organization direct and control operations to achieve the goal of organization. âž ¢ Terry and Rue management is definedRead MoreSemco Pumping Success Case Study4411 Words   |  18 PagesTheories and SEMCO† 8 2.1 â€Å"McGregor’s theory X and Theory Y† 8 2.2 Victor Vroom’s expectancy theory 11 2.3 Conclusion 12 3. Discuss the Concept of Empowerment and Its Relevance to SEMCO 13 3.1 The concept of empowerment 13 3.2 Empowerment s Relevance to SEMCO 15 3.3 Conclusion 17 4. Centaur Technologies INC Its Success/Failure 17 4.1 introduction 17 4.2 Reason which lead to the success of Centaur Company 19 4.3 Reason that might lead to the failure of the centaur company 20 Read MoreCompare and Contrast Semco to a ‘Classical Organisation’, I.E. One Which Has a Traditional Approach to Industrial Relations.2284 Words   |  10 Pagestechnology, tasks, and human components and accomplishes its purposes’ (Alajloni et al. 2010) â€Å"The purpose of work is not to make money. The purpose of work is to make the worker, whether working stiffs or top executives, feel good about life’ (Semler 2004) The present trend in human resource management and labour relations is to place more emphasis on employee involvement, productive employer -employee relations and mechanisms, and on practices that encourages them. The era of classical organisationRead MoreManaging People3937 Words   |  16 Pagestheir theories assumed that people could make rational and logical decisions while trying to maximize personal gains from their work situations. The classical school of management is based on scientific management which has its roots in Henri Fayol s work in France and the ideas of German sociologist Max Weber. Henri Fayol is known as the father of modern management. He developed 14 principles of management based on his management experiences. These principles provide modern-day managers with generalRead MoreThe Evolution of Management Thought2925 Words   |  12 Pagesand motion study evolved which would lead into the setting of standards and performance measurement. Unfortunately the majority of Taylor s excellent work is clouded by his misconception that workers restrict output from ‘the natural instinct and tendency of men to take it easy.’ He referred to it as ‘soldering’ Henri Fayol (1841-1925) writes about leadership and the ‘theory of management’ and Max Weber (1864-1920) develops organizational structures – Weber would be the originator of the term bureaucracy

The Effects Of Television On Children s Influence On Society

According to the statistics gathered by Nielson’s 20ll â€Å"State of the Trends in TV Viewing,† over 99% of Americans own at least one television and average a total of 34 hours 39 minutes of TV viewing per week (Citation). Outlets such as television shows and movies have the power to influence viewers both directly and indirectly in positive manners; being able to see someone like yourself has the ability to shape how you view yourself and the world. However, the problem lies in the fact that the majority of those seen on TV are heterosexual, white, men; this leaves little to no chances of positive representation and depth of those not included in that category. This commonly leads to stereotypes. Seeing oneself positively represented in terms of race, sexuality, and gender on television encourages a positive mental and physical body image, while the opposite has the ability to create a lower sense of both self-worth and self-esteem. In terms of television, â€Å"rep resentation refers to the authority granted by one group to another to monitor and uphold its interests.† (Citation). This being said, the group with the most power is in control of how said disadvantaged groups are viewed. As reported by a study conducted by UCLAs â€Å"2014 Hollywood Diversity Report: Making Sense of the Disconnect† of all the lead roles in broadcast shows only 5.1% belonged to minority actors despite minorities accounting for approximately 36.6% of the population. (Citation). The same can be seen withShow MoreRelatedViolence And Sex On Television898 Words   |  4 Pages Violence and Sex on Television: Effects on the Younger Audience In today’s society, the media is used greatly for communication, advertisement, information, and for numerous other reasons. The world has evolved by technological advances as well as by the type of content that is put out on the internet, radio, and especially on television. In particular, violence and sex are two of the most controversial content types that have been recently used loosely in the present as compared to theRead MoreThe Effects Of Television On America During The 1980s1289 Words   |  6 PagesImportant Effects of Television in America in the 1980s Television is one of the most popular ways to consume media. However, television wasn’t always the way it is today. Many changes took place in the television industry during the 1980s. Some of the important changes that took place in the 1980s were the new types of programming, the increase in the popularity of cable television, and new technology that was invented such as the remote control. Television had a huge effect on society through theRead MoreMass Media Is An Integral Part Of Everyday Life989 Words   |  4 PagesMass media is an integral part of everyday life. Society depends on the various forms of media for information, education, and personal entertainment. 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Children are spending more and more time in theRead MoreThe Effects Of Television Violence On Our Society1566 Words   |  7 Pagesthe media and does it affect our society. How we view television, has changed the world, no doubt in that. Turn your television set on and pick a channel at random; the odds are that half of the programs you come across will contain violent material. The statistics are overwhelming as I look on the internet, read articles, and look at the research. One of the things that most interests me in the violence on television, is the effects it has on children. Chi ldren learn by repetition while watchingRead MoreEssay Negative Effects of Television on Children1562 Words   |  7 PagesTelevision is a big part of today’s society. Everybody watches television, including the children. There is a potential problem with letting children watch television. Ask this question, would someone let their own child watch some of the programming that they watch, too? 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Part I Television plays a very large and influential role in spreading modern pop culture. It seems like there is no doubt that television are taking over and regulating many of our business, social value, and lifestyle. Its entertaining aspect led it to become a significant part of our lives. Since it has deeply rooted into our culture, we might believe that it always has been on our side. However, I personallyRead MoreThe Negative Effect of Television on People Essay1570 Words   |  7 Pages Of course, Television, one of the main sources of entertainment is pleasurable. Most adults and children find it very interesting to watch television programs. A good number of individuals in the society are unaware that their time, fervor and even their personal rights are sacrificed to watching television. Although television is enjoyable, accessible, cheap and attractive, most of the contents depicted by it such as violence and sexual intercourse have adverse influences on its viewers. ItRe ad MoreMedia Violence Essay1320 Words   |  6 Pagespublic issues society faces today. Television screens are loaded with the glamorization of weapon carrying. Violence constitute as amusing and trivialized. Needless portrayals of interpersonal violence spread across the television screens like wild fire. Televisions spew the disturbing events such as children being assaulted, husbands inflicting domestic abuse on their wives and children succumbing to abuse by their parents. Scenes of betrayal, anguish, infiltrate the television screen. Unfortunately

Solution Manual for Fluid Mech Cengel Book Free Essays

string(146) " of 4 when the velocity is doubled since F = mV = \( \? AV \)V = \? AV 2 and thus the force is proportional to the square of the velocity\." Chapter 6 Momentum Analysis of Flow Systems Chapter 6 MOMENTUM ANALYSIS OF FLOW SYSTEMS Newton’s Laws and Conservation of Momentum 6-1C Newton’s first law states that â€Å"a body at rest remains at rest, and a body in motion remains in motion at the same velocity in a straight path when the net force acting on it is zero. † Therefore, a body tends to preserve its state or inertia. Newton’s second law states that â€Å"the acceleration of a body is proportional to the net force acting on it and is inversely proportional to its mass. We will write a custom essay sample on Solution Manual for Fluid Mech Cengel Book or any similar topic only for you Order Now Newton’s third law states â€Å"when a body exerts a force on a second body, the second body exerts an equal and opposite force on the first. † r 6-2C Since momentum ( mV ) is the product of a vector (velocity) and a scalar (mass), momentum must be a vector that points in the same direction as the velocity vector. 6-3C The conservation of momentum principle is expressed as â€Å"the momentum of a system remains constant when the net force acting on it is zero, and thus the momentum of such systems is conserved†. The momentum of a body remains constant if the net force acting on it is zero. 6-4C Newton’s second law of motion, also called the angular momentum equation, is expressed as â€Å"the rate of change of the angular momentum of a body is equal to the net torque acting it. † For a non-rigid body with zero net torque, the angular momentum remains constant, but the angular velocity changes in accordance with I? = constant where I is the moment of inertia of the body. 6-5C No. Two rigid bodies having the same mass and angular speed will have different angular momentums unless they also have the same moment of inertia I. Linear Momentum Equation 6-6C The relationship between the time rates of change of an extensive property for a system and for a control volume is expressed by the Reynolds transport theorem, which provides the link between the r system and control volume concepts. The linear momentum equation is obtained by setting b = V and thus r B = mV in the Reynolds transport theorem. -7C The forces acting on the control volume consist of body forces that act throughout the entire body of the control volume (such as gravity, electric, and magnetic forces) and surface forces that act on the control surface (such as the pressure forces and reaction forces at points of contact). The net force acting on a control volume is the sum of all body and surface forces. Fluid weight is a body force, and pressure is a surface force (acting per unit area). -8C All o f these surface forces arise as the control volume is isolated from its surroundings for analysis, and the effect of any detached object is accounted for by a force at that location. We can minimize the number of surface forces exposed by choosing the control volume such that the forces that we are not interested in remain internal, and thus they do not complicate the analysis. A well-chosen control volume exposes only the forces that are to be determined (such as reaction forces) and a minimum number of other forces. 6-9C The momentum-flux correction factor ? nables us to express the momentum flux in terms of the r r r r ? V (V ? n )dAc = ? mV avg . The value of ? is unity for uniform mass flow rate and mean flow velocity as ? Ac flow, such as a jet flow, nearly unity for turbulent flow (between 1. 01 and 1. 04), but about 1. 3 for laminar flow. So it should be considered in laminar flow. 6-1 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution perm itted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-10C The momentum equation for steady one-dimensional flow for the case of no external forces is r r r F= ? mV ? ? mV ? ? out ? in where the left hand side is the net force acting on the control volume, and first term on the right hand side is the incoming momentum flux and the second term is the outgoing momentum flux by mass. 6-11C In the application of the momentum equation, we can disregard the atmospheric pressure and work with gage pressures only since the atmospheric pressure acts in all directions, and its effect cancels out in every direction. -12C The fireman who holds the hose backwards so that the water makes a U-turn before being discharged will experience a greater reaction force since the numerical values of momentum fluxes across the nozzle are added in this case instead of being subtracted. 6-13C No, V is not the upper limit to the rocket’s ultimate velocity. Without friction the rocket velocity will continue to i ncrease as more gas outlets the nozzle. 6-14C A helicopter hovers because the strong downdraft of air, caused by the overhead propeller blades, manifests a momentum in the air stream. This momentum must be countered by the helicopter lift force. 6-15C As the air density decreases, it requires more energy for a helicopter to hover, because more air must be forced into the downdraft by the helicopter blades to provide the same lift force. Therefore, it takes more power for a helicopter to hover on the top of a high mountain than it does at sea level. 6-16C In winter the air is generally colder, and thus denser. Therefore, less air must be driven by the blades to provide the same helicopter lift, requiring less power. 6-2 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-17C The force required to hold the plate against the horizontal water stream will increase by a factor of 4 when the velocity is doubled since F = mV = ( ? AV )V = ? AV 2 and thus the force is proportional to the square of the velocity. You read "Solution Manual for Fluid Mech Cengel Book" in category "Essay examples" 6-18C The acceleration will not be constant since the force is not constant. The impulse force exerted by water on the plate is F = mV = ( ? AV )V = ? AV 2 , where V is the relative velocity between the water and the plate, which is moving. The plate acceleration will be a = F/m. But as the plate begins to move, V decreases, so the acceleration must also decrease. 6-19C The maximum velocity possible for the plate is the velocity of the water jet. As long as the plate is moving slower than the jet, the water will exert a force on the plate, which will cause it to accelerate, until terminal jet velocity is reached. 6-20 It is to be shown that the force exerted by a liquid jet of velocity V on a stationary nozzle is proportional to V2, or alternatively, to m 2 . Assumptions 1 The flow is steady and incompressible. 2 The nozzle is given to be stationary. 3 The nozzle involves a 90 ° turn and thus the incoming and outgoing flow streams are normal to each other. 4 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. Analysis We take the nozzle as the control volume, and the flow direction at the outlet as the x axis. Note that the nozzle makes a 90 ° turn, and thus it does not contribute to any pressure force or momentum flux term at the inlet in the x direction. Noting that m = ? AV where A is the nozzle outlet area and V is the average nozzle outlet velocity, the momentum equation for steady one-dimensional flow in the x direction reduces to r r r F= ? mV ? ? mV FRx = ? m out V out = ? mV ? ? out ? in where FRx is the reaction force on the nozzle due to liquid jet at the nozzle outlet. Then, m = ? AV FRx = ? mV = AVV = AV 2 or FRx = ? mV = ? m m m2 =? ?A ? A Therefore, the force exerted by a liquid jet of velocity V on this stationary nozzle is proportional to V2, or alternatively, to m 2 . Liquid Nozzle V FR 6-3 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-21 A water jet of velocity V impinges on a plate moving toward the water jet with velocity ? V. The force required to move the plate towards the jet is to be determined in terms of F acting on the stationary plate. Assumptions 1 The flow is steady and incompressible. 2 The plate is vertical and the jet is normal to plate. 3 The pressure on both sides of the plate is atmospheric pressure (and thus its effect cancels out). Fiction during motion is negligible. 5 There is no acceleration of the plate. 6 The water splashes off the sides of the plate in a plane normal to the jet. 6 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Analysis We take the plate as the control volume. The relative velocity between the plate and the jet is V when the plate is 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 becomes 2. 25 times when the jet velocity becomes 1. 5 times. Discussion Note that when the plate is stationary, V is also the jet velocity. But if the plate moves toward the stream with velocity ? V, then the relative velocity is 1. 5V, and the amount of mass striking the plate (and falling off its sides) per unit time also increases by 50%. 1/2V V Waterjet 6-4 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-22 A 90 ° elbow deflects water upwards and discharges it to the atmosphere at a specified rate. The gage pressure at the inlet of the elbow and the anchoring force needed to hold the elbow in place are to be determined. v Assumptions 1 The flow is steady, frictionless, incompressible, and irrotational (so that the Bernoulli equation is applicable). The weight of the elbow and the water in it is negligible. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 4 The momentum-flux correction factor for each inlet and outlet is given to be ? = 1. 03. Properties We take the density of water to be 1000 kg/m3. Analysis (a) We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction) and the vertical coordinate by z. The continuity equation for this one-inlet one-outlet steady flow system is m1 = m 2 = m = 30 kg/s. Noting that m = ? AV , the mean inlet and outlet velocities of water are 25 kg/s m m = = = 3. 18 m/s 2 ? A ? (? D / 4) (1000 kg/m 3 )[? (0. 1 m) 2 / 4] Noting that V1 = V2 and P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as V1 = V 2 = V = P V12 P V2 1 + + z1 = 2 + 2 + z2 P ? P2 = ? g ( z2 ? z1 ) P , gage = ? g ( z2 ? z1 ) 1 1 ? g 2 g ? g 2 g Substituting, ? ? 1 kN 2 ? P , gage = (1000 kg/m3 )(9. 81 m/s 2 )(0. 35 m)? 1 ? 1000 kg ? /s2 ? = 3. 434 kN/m = 3. 434 kPa ? ? r r r (b) The momentum equation for steady one-dimensional flow is F= ? mV ? ? mV . We let the x- ? ? out ? in and z- components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive directions. We also use gage pressures to avoid dealing with the atmospheric pressure which acts on all surfaces. Then the momentum equations along the x and y axes become FRx + P1,gage A1 = 0 ? ?m(+V1 ) = ? ?mV FRz = ? m(+V 2 ) = ? mV z x FRz 2 35 cm Solving for FRx and FRz, and substituting the given values, FRx = ? ?mV ? P1, gage A1 ? N = ? 1. 03(25 kg/s)(3. 18 m/s)? ? 1 kg ? m/s 2 ? = ? 109 N ? ? ? (3434 N/m 2 )[? (0. 1 m) 2 / 4] ? ? ? ? = 81. 9 N ? ? FRy FRx = tan -1 Water 25 kg/s FRx 1 ? 1N FRy = ? mV = 1. 03(25 kg/s)(3. 18 m/s)? ? 1 kg ? m/s 2 ? and 2 2 FR = FRx + FRy = (? 109) 2 + 81. 9 2 = 136 N, ? = tan -1 81. 9 = ? 37 ° = 143 ° ? 109 Discussion Note that the magnitude of the anchoring force is 136 N, and its line of action makes 143 ° from the positive x direction. Also, a negative value for FRx indicates the assumed direction is wrong, and should be reversed. 6-5 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 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 zero. 4 The momentumflux correction factor for each inlet and outlet is given to be ? = 1. 03. Properties We take the density of water to be 1000 kg/m3. Analysis (a) We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as 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 eq uations along the x and z axes become FRx + P1,gage A1 = ? m(? V 2 ) ? ? m(+V1 ) = ? 2 ? mV FRz = 0 Solving for FRx and substituting the given values, FRx = ? 2 ? mV ? P1, gage A1 ? 1N = ? 2 ? 1. 03(25 kg/s)(3. 18 m/s)? 1 kg ? m/s 2 ? = ? 218 N ? ? ? (6867 N/m 2 )[? (0. 1 m) 2 / 4] ? ? 2 z x FRz Water 25 kg/s 35 cm and FR = FRx = – 218 N since the y-component of the anchoring force is zero. Therefore, the anchoring force has a magnitude of 218 N and it acts in the negative x direction. Discussion Note that a negative value for FRx indicates the assumed direction is wrong, and should be reversed. FRx 1 6-6 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-24E A horizontal water jet strikes a vertical stationary plate normally at a specified velocity. For a given anchoring force needed to hold the plate in place, the flow rate of water is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water splatters off the sides of the plate in a plane normal to the jet. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure which is disregarded since it acts on the entire control surface. The vertical forces and momentum fluxes are not considered since they have no effect on the horizontal reaction force. 5 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Properties We take the density of water to be 62. 4 lbm/ft3. Analysis We take the plate as the control volume such that it contains the entire plate and cuts through the water jet and the support bar normally, and the direction of flow as the positive direction of x axis. The momentum equation for steady one-dimensional flow in the x (flow) direction reduces in this case o r r r F= ? mV ? ? mV ? FRx = ? mV1 FR = mV1 ? ? out ? in We note that the reaction force acts in the opposite direction to flow, and we should not forget the negative sign for forces and velocities in the negative x-direction. Solving for m and substituting the given values, m= FRx 350 lbf = V1 30 ft/s ? 32. 2 lbm ? ft/s 2 ? ? 1 lbf ? ? ? = 376 lbm/s ? ? Then the volume flow rate becomes V = m ? = 376 lbm/s 62. 4 lbm/ft 3 = 6. 02 ft 3 /s Therefore, the volume flow rate of water under stated assumptions must be 6. 02 ft3/s. Discussion In reality, some water will be scattered back, and this will add to the reaction force of water. The flow rate in that case will be less. m 1 FRx = 350 lbf Waterjet 6-7 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 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 given to be ? = 1. 03. Properties We take the density of water to be 1000 kg/m3. Analysis The weight of the elbow and the water in it is W = mg = (50 kg)(9. 1 m/s 2 ) = 490. 5 N = 0. 4905 kN We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction) and the vertical coordinate by z. The continuity equation for this one-inlet one-outlet steady flow system is m1 = m 2 = m = 30 kg/s. Noting that m = ? AV , the inlet and outlet velocities of water are 30 kg/s m V1 = = = 2. 0 m/s ? A1 (1000 kg/m 3 )(0. 0150 m 2 ) 30 kg/s m V2 = = = 12 m/s ? A2 (1000 kg/m 3 )(0. 025 m 2 ) Taking the center of the inlet cross section as the reference level (z1 = 0) and noting that P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as ? V 2 ? V12 ? ? V22 ? V12 ? P V12 P V2 1 ? ? ? + + z1 = 2 + 2 + z2 P ? P2 = ? g ? 2 1 1 ? 2 g + z2 ? z1 ? P , gage = ? g ? 2 g + z2 ? ?g 2 g ? g 2 g ? ? ? ? Substituting, ? (12 m/s) 2 ? (2 m/s) 2 ? 1 kN ? = 73. 9 kN/m 2 = 73. 9 kPa P , gage = (1000 kg/m3 )(9. 81 m/s 2 )? + 0. 4 1 2 ? 1000 kg ? m/s 2 ? 2(9. 81 m/s ) ? ? The momentum equation for steady one-dimensional flow is ? F = ? mV ? ? ? mV . We let the x- and out in r r r z- components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive directions. We also use gage pressures to avoid dealing with the atmospheric pressure which acts on all surfaces. Then the momentum equations along the x and z axes become FRx + P1,gage A1 = ? mV 2 cos ? ? ? mV1 and FRz ? W = ? mV 2 sin ? 2 25 cm2 Solving for FRx and FRz, and substituting the given values, FRx = ? m(V 2 cos ? ? V1 ) ? P1, gage A1 ? 1 kN = 1. 03(30 kg/s)[(12cos45 ° – 2) m/s]? ? 1000 kg ? m/s 2 ? ? (73. 9 kN/m 2 )(0. 0150 m 2 ) = ? 0. 908 kN ? ? ? Water 30 kg/s 45 ° FRz FRx 150 m2 W 1 ? ? 1 kN ? FRz = ? mV 2 sin ? + W = 1. 03(30 kg/s)(12sin45 ° m/s)? ? 1000 kg ? m/s 2 ? + 0. 4905 kN = 0. 753 kN ? ? 0. 753 2 2 2 2 -1 FRz FR = FRx + FRz = (? 0. 908) + (0. 753) = 1. 18 kN, ? = tan = tan -1 = ? 39. 7 ° FRx ? 0. 908 Discussion Note that the magnitude of the anchoring force is 1. 18 kN, and its line of action makes –39. 7 ° from +x direction. Negative value for FRx indicates the assumed direction is wrong. 6-8 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-26 A reducing elbow deflects water upwards and discharges it to the atmosphere at a specified rate. The anchoring force needed to hold the elbow in place is to be determined. v Assumptions 1 The flow is steady, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The weight of the elbow and the water in it is considered. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. The momentumflux correction factor for each inlet and outlet is given to be ? = 1. 03. Properties We take the 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 inlet cross section as the reference level (z1 = 0) and noting that P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as ? V 2 ? V12 ? ? V22 ? V12 ? P V12 P V2 1 ? ? ? + + z1 = 2 + 2 + z2 P ? P2 = ? g ? 2 1 1 ? 2 g + z2 ? z1 ? P , gage = ? g ? 2 g + z2 ? ?g 2 g ? g 2 g ? ? ? ? or, P , gage = (1000 kg/m3 )(9. 81 m/s2 )? 1 ? ? ? (12 m/s)2 ? (2 m/s)2 2(9. 81 m/s ) ? 1 kN ? = 73. 9 kN/m 2 = 73. 9 kPa + 0. 4 1000 kg ? m/s 2 ? ? The momentum equation for steady one-dimensional flow is ? F = ? ?mV ? ? ? mV . We let the xout in r r r and y- components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive directions. We also use gage pressures to avoid dealing with the atmospheric pressure whic h acts on all surfaces. Then the momentum equations along the x and z axes become FRx + P1,gage A1 = ? mV 2 cos ? ? ? mV1 and FRy ? W = ? mV 2 sin ? Solving for FRx and FRz, and substituting the given values, FRx = ? m(V 2 cos ? V1 ) ? P1, gage A1 ? 1 kN = 1. 03(30 kg/s)[(12cos110 ° – 2) m/s]? ? 1000 kg ? m/s 2 ? FRz ? ? ? (73. 9 kN/m 2 )(0. 0150 m 2 ) = ? 1. 297 kN ? ? ? ? 1 kN ? + 0. 4905 kN = 0. 8389 kN = ? mV 2 sin ? + W = 1. 03(30 kg/s)(12sin110 ° m/s)? 2 ? ? 1000 kg ? m/s ? ? 2 25 cm2 110 ° 2 2 FR = FRx + FRz = (? 1. 297) 2 + 0. 8389 2 = 1. 54 kN and FRz 0. 8389 = tan -1 = ? 32. 9 ° FRx ? 1. 297 Discussion Note that the magnitude of the anchoring force is 1. 54 kN, and its line of action makes –32. 9 ° from +x direction. Negative value for FRx indicates assumed direction is wrong, and should be reversed. ? = tan -1 FRz FRx Water 1 30 kg/s 50 m2 W 6-9 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-27 Water accelerated by a nozzle strikes the back surface of a cart moving horizontally at a constant velocity. The braking force and the power wasted by the brakes are to be determined. . Assumptions 1 The flow is steady and incompressible. 2 The water splatters off the sides of the plate in all 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 cart. 7 The motions of the water jet and the cart are horizontal. 6 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Analysis We take the cart as the control vo lume, and the direction of flow as the positive direction of x axis. The relative velocity between the cart and the jet is V r = V jet ? Vcart = 15 ? 10 = 10 m/s 15 m/s 5 m/s Therefore, we can assume the cart to be stationary and the jet to move Waterjet with a velocity of 10 m/s. The momentum equation for steady onedimensional flow in the x (flow) direction reduces in this case to r r r F= ? mV ? ? mV FRx = ? mi Vi Fbrake = ? mV r FRx ? ? out ? in We note that the brake force acts in the opposite direction to flow, and we should not forget the negative sign for forces and velocities in the negative x-direction. Substituting the given values, ? 1N Fbrake = ? mV r = ? (25 kg/s)(+10 m/s)? ? 1 kg ? m/s 2 ? ? ? = ? 250 N ? ? The negative sign indicates that the braking force acts in the opposite direction to motion, as expected. Noting that work is force times distance and the distance traveled by the cart per unit time is the cart velocity, the power wasted by the brakes is 1 kW ? ? W = FbrakeV cart = (250 N)(5 m/s)? ? = 1. 25 kW ? 1000 N ? m/s ? Discussion Note that the power wasted is equivalent to the maximum power that can be generated as the cart velocity is maintained constant. 6-10 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of 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 fail is to be determined. Analysis The braking force was determined in previous problem to be 250 N. When the brakes fail, this force will propel the cart forward, and the accelerating will be a= F 250 N ? 1 kg ? m/s 2 ? = m cart 300 kg ? 1N ? ? ? = 0. 833 m/s 2 ? ? Discussion This is the acceleration at the moment the brakes fail. The acceleration will decrease as the relative velocity between the water jet and the cart (and thus the force) decreases. 5 m/s 15 m/s 300 kg Waterjet FRx 6-11 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-29E A water jet hits a stationary splitter, such that half of the flow is diverted 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 effects are disregarded. 4 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Properties We take the density of water to be 62. 4 lbm/ft3. Analysis The mass flow rate of water jet is m = ? V = (62. lbm/ft 3 )(100 ft 3 /s) = 6240 lbm/s We take the splitting section of water jet, including the splitter as the control volume, and designate the entrance by 1 and the outlet of either arm by 2 (both arms have the same velocity and mass flow rate). We also designate the horizontal coordinate by x with the direction of flow as being the positive direction and the vertical coordinate by z. r r r The momentum equation for steady one-dimensional flow is F= ? mV ? ? mV . We let ? ? out ? in the x- and y- components of the anchoring force of the splitter be FRx and FRz, and assume them to be in the positive directions. Noting that V2 = V1 = V and m 2 = 1 m , the momentum equations along the x and z 2 axes become FRx = 2( 1 m)V 2 cos ? ? mV1 = mV (cos ? ? 1) 2 FRz = 1 m(+V 2 sin ? ) + 1 m(? V 2 sin ? ) ? 0 = 0 2 2 Substituting the given values, 1 lbf ? ? FRx = (6240 lbm/s)(20 ft/s)(cos45 ° – 1)? ? = ? 1135 lbf 32. 2 lbm ? ft/s 2 ? ? FRz = 0 The negative value for FRx indicates the assumed direction is wrong, and should be reversed. Therefore, a force of 1135 lbf must be applied to the splitter in the opposite direction to flow to hold it in place. No holding force is necessary in the vertical direction. This can also be concluded from the symmetry. Discussion In reality, the gravitational effects will 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 â€Å"ft/s2† rho=62. 4 â€Å"lbm/ft3† V_dot=100 â€Å"ft3/s† V=20 â€Å"ft/s† m_dot=rho*V_dot F_R=-m_dot*V*(cos(theta)-1)/g â€Å"lbf† ?,  ° 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 8000 7000 6000 5000 m , lbm/s 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 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 horizon tal water jet impinges normally upon a vertical plate which is held on a frictionless track and is initially stationary. The initial acceleration of the plate, the time it takes to reach a certain velocity, and the velocity at a given time are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water always splatters in the plane of the retreating plate. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure which is disregarded since it acts on all surfaces. 4 The tract is nearly frictionless, and thus fiction during motion is negligible. 5 The motions of the water jet and the cart are horizontal. 6 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 Discussion The assumption that the relative velocity between the water jet and the plate remains constant is valid only for the initial moment s of motion when the plate velocity is low unless the water jet is moving with the plate at the same velocity as the plate. 6-14 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-32 A 90 ° reducer elbow deflects water downwards into a smaller diameter pipe. The resultant force exerted on the reducer by water is to be determined. Assumptions 1 The flow is steady, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The weight of the elbow and the water in it is disregarded since the gravitational effects are negligible. 3 The momentum-flux correction factor for each inlet and outlet is given to be ? 1. 04. Properties We take the density of water to be 1000 kg/m3. Analysis We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction) and the vertical coordinate by z. The continuity equation for this one-inlet one-outlet steady flow system is m1 = m 2 = m = 353. 4 kg/s. Noting that m = ? AV , the mass flow rate of water and its outlet velocity are 2 m = ? V1 A1 = ? V1 (? D1 / 4) = (1000 kg/m 3 )(5 m/s)[? (0. 3 m) 2 / 4] = 353. 4 kg/s 353. kg/s m m = = = 20 m/s 2 ? A2 D 2 / 4 (1000 kg/m 3 )[? (0. 15 m) 2 / 4] The Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as V2 = P V12 P V2 1 + + z1 = 2 + 2 + z2 ? g 2 g ? g 2 g ? V 2 ? V22 ? ? P2 = P + ? g ? 1 1 ? 2 g + z1 ? z2 ? ? ? Substituting, the gage pressure at the outlet becomes ? (5 m/s)2 ? (20 m/s)2 1 kPa ? 1 kN ? P2 = (300 kPa) + (1000 kg/m 3 )(9. 81 m/s 2 )? + 0. 5 = 117. 4 kPa 2 ? 1000 kg ? m/s 2 1 kN/m 2 ? 2(9. 81 m/s ) ? ? The momentum equation for steady one-dimensional flow is ? F = ? ?mV ? ? ? mV . We let the xout in r r and z- components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive directions. Then the momentum equations along the x and z axes become F Rx + P1,gage A1 = 0 ? ? mV1 FRz ? P2,gage A2 = ? m(? V 2 ) ? 0 Note that we should not forget the negative sign for forces and velocities in the negative x or z direction. Solving for FRx and FRz, and substituting the given values, ? 1 kN FRx = ? ?mV1 ? P1, gage A1 = ? 1. 04(353. 4 kg/s)(5 m/s)? ? 1000 kg ? m/s 2 ? ? ? (0. 3 m) 2 ? ? (300 kN/m 2 ) = ? 23. 0 kN ? 4 ? ? ? (0. 15 m) 2 ? + (117. 4 kN/m 2 ) = ? 5. 28 kN ? ? FRz ? 1 kN FRz = ? ? mV 2 + P2, gage A1 = ? 1. 04(353. 4 kg/s)(20 m/s)? ? 1000 kg ? m/s 2 ? and 2 2 FR = FRx + FRz = (? 23. 0) 2 + (? 5. 28) 2 = 23. 6 kN FRx 30 cm Water 5 m/s ? = tan -1 FRz ? 5. 28 = tan -1 = 12. 9 ° FRx ? 23. 0 Discussion The magnitude of the anchoring force is 23. 6 kN, and its line of action makes 12. 9 ° from +x direction. Negative values for FRx and FRy indicate that the assumed directions are wrong, and should be reversed. 15 cm 6-15 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to te achers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-33 A wind turbine with a given span diameter and efficiency is subjected to steady winds. The power generated and the horizontal force on the supporting mast of the turbine are to be determined. vEES Assumptions 1 The wind flow is steady and incompressible. 2 The efficiency of the turbine-generator is independent of wind speed. 3 The frictional effects are negligible, and thus none of the incoming kinetic energy is converted to thermal energy. Wind flow is uniform and thus the momentum-flux correction factor is nearly unity, ? ? 1. Properties The density of air is given to be 1. 25 kg/m3. Analysis (a) The power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and mV 2 / 2 for a given mass flow rate: ? 1 m/s ? V1 = (25 km/h)? ? = 6. 94 m/s ? 3. 6 km/h ? m = ? 1V1 A1 = ? 1V1 Wind V1 1 2 D V2 ?D 2 4 2 = (1. 25 kg/m 3 )(6. 94 m/s) ? (90 m) 2 4 2 = 55,200 kg/s V (6. 94 m/s) W max = mke1 = m 1 = (55,200 kg/s) 2 2 ? 1 kN ? ? 1000 kg ? m/s 2 ? 1 kW ? 1 kN ? m/s ? = 1330 kW ? ? FR Then the actual power produced becomes Wact = ? wind turbineW max = (0. 32)(1330 kW) = 426 kW (b) The frictional effects are assumed to be negligible, and thus the portion of incoming kinetic energy not converted to electric power leaves the wind turbine as outgoing kinetic energy. Therefore, V2 V2 mke 2 = mke1 (1 ? ? wind turbine ) m 2 = m 1 (1 ? ? wind turbine ) 2 2 or V 2 = V1 1 ? ? wind turbine = (6. 94 m/s) 1 – 0. 32 = 5. 72 m/s We choose the control volume around the wind turbine such that the wind is 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 assuming the flow velocity through the turbine to be equal to the wind velocity give ? 1 kN FR = mV 2 ? mV1 = m(V 2 ? V1 ) = (55,200 kg/s)(5. 72 – 6. 94 m/s)? ? 1000 kg ? m/s 2 ? ? ? = ? 67. 3 kN ? ? The negative sign indicates that the reaction force acts in the negative x direction, as expected. Discussion This force acts on top of the tower where the wind turbine is installed, and the bending moment it generates at the bottom of the tower is obtained by multiplying this force by the tower height. 6-16 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for 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 included in the required force to hold the plate). 4 There is no splashing of water or the deformation of the jet, and the reversed jet leaves horizontally at the same velocity and flow rate. Jet flow is nearly uniform and thus the momentum-flux correction factor is nearly unity, ? ? 1. Properties We take the density of 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 . Letting the ? ? out ? in reaction force to hold the plate be FRx and assuming it to be in the positive direction, the momentum equation along the x axis becomes FRx = m(? V 2 ) ? m(+V1 ) = ? 2mV Substituting, 1 lbf ? ? FRx = ? 2(428. 8 lbm/s)(140 ft/s)? ? = ? 3729 lbf 2 ? 32. 2 lbm ? ft/s ? Therefore, a force of 3729 lbm must be applied on the plate in the negative x direction to hold it in place. Discussion Note that a 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 splattered water is the atmospheric pressure, which is disregarded since it acts on all surfaces. 3 Frictional and gravitational effects are negligible. 4 There is no splattering of water or the deformation of the jet, and the reversed jet leaves horizontally at the same velocity and flow rate. 5 Jet flow is nearly uniform and thus the momentum-flux 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 distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-36 Firemen are holding a nozzle at the end of a hose while trying to extinguish a fire. The average water outlet velocity and the resistance force required of the firemen to hold the nozzle are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water jet is exposed to the atmosphere, and thus the pressure of the water jet is the atmospheric pressure, which is disregarded since it acts on all surfaces. 3 Gravitational effects and vertical forces are disregarded since the horizontal resistance force is to be determined. 5 Jet flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Properties We take the density of water to be 1000 kg/m3. Analysis (a) We take the nozzle and the horizontal portion of the hose as the system such that water enters the control volume vertically and outlets horizontally (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)? ? 1kg ? m/s 2 ? ? ? Therefore, the firemen must be able to resist a force of 2457 N to hold the nozzle in place. Discussion The force of 2457 N is equivalent to the weight of about 250 kg. That is, holding the nozzle requires the strength of holding a weight of 250 kg, which cannot be done by a single person. This demonstrates why several firemen are used to hold a hose with a high flow rate. FRz FRx 5 m3/min 6-19 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-37 A horizontal jet of water with a given velocity strikes a flat plate that is moving in the same direction at a specified velocity. The force that the water stream exerts against the plate is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water splatters in all directions in the plane of the plate. 3 The water jet is exposed to the atmosphere, and thus the pressure 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 Therefore, we can assume the plate to be stationary and the jet to move with a velocity of 20 m/s. The r r r F= ? mV ? ? mV . We let the horizontal momentum equation for steady one-dimensional flow is ? ? out ? in reaction force applied to the plate in the negative x direction to counteract the impulse of the water jet be FRx. Then the momentum equation along the x direction gives ? ? 1N ? ? FRx = 0 ? mVi FRx = mV r = (58. 9 kg/s)(20 m/s)? ? 1kg ? m/s 2 ? = 1178 N ? ? Therefore, the water jet applies a force of 1178 N on the plate in the direction of motion, and an equal and opposite force must be applied on the plate if its velocity is to remain constant. Discussion Note that we used the relative velocity in the determination of the mass flow rate of water in the momentum analysis since water will enter the control volume at this rate. (In the limiting case of the plate and the water jet moving at the same velocity, the mass flow rate of water relative to the plate will be zero since no water will be able to strike the plate). 6-20 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill 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 â€Å"kg/m3† D=0. 05 â€Å"m† V_jet=30 â€Å"m/s† Ac=pi*D^2/4 V_r=V_jet-V_plate m_dot=rho*Ac*V_jet F_R=m_dot*V_r â€Å"N† Vplate, m/s 0 3 6 9 12 15 18 21 24 27 30 Vr, m/s 30 27 24 21 18 15 12 9 6 3 0 FR, N 1767 1590 1414 1237 1060 883. 6 706. 9 530. 1 353. 4 176. 7 0 1800 1600 1400 1200 1000 FR, N 800 600 400 200 0 0 5 10 15 20 25 30 Vplate, m/s 6-21 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 6-39E A fan moves air at sea level at a specified rate. The force required to hold the fan and the minimum power input required for the fan are to be determined. v Assumptions 1 The flow of air is steady and incompressible. 2 Standard atmospheric conditions exist so that the pressure at sea level is 1 atm. Air leaves the fan at a uniform velocity at atmospheric pressure. 4 Air approaches the fan through a large area at atmospheric pressure with negligible velocity. 5 The frictional effects are negligible, and thus the entire mechanical power input is converted to kinetic energy of air (no conversion to thermal energy through frictional effects). 6 Wind flow is nearly uniform and thus the momentum-flux correction facto r can be taken to be unity, ? ? 1. Properties The gas constant of air is R = 0. 3704 psi? ft3/lbm? R. The standard atmospheric pressure at sea level is 1 atm = 14. 7 psi. Analysis (a) We take the control volume to be a horizontal hyperbolic cylinder bounded by streamlines on the sides with air entering through the large cross-section (section 1) and the fan located at the narrow cross-section at the end (section 2), and let its centerline be the x axis. The density, mass flow rate, and discharge velocity of air are 14. 7 psi P ? = = = 0. 0749 lbm/ft 3 RT (0. 3704 psi ? ft 3 /lbm ? R)(530 R) m = ? V = (0. 0749 lbm/ft 3 )(2000 ft 3/min) = 149. 8 lbm/min = 2. 50 lbm/s V2 = V A2 = V 2 ? D 2 / 4 = 2000 ft 3 /min ? (2 ft) 2 / 4 = 636. 6 ft/min = 10. ft/s ? F = ? ?mV ? ? ? mV . Letting the out in The momentum equation for steady one-dimensional flow is r r r reaction force to hold the fan be FRx and assuming it to be in the positive x (i. e. , the flow) direction, the momentum equation along the x axis becomes 1 lbf ? ? FRx = m(V 2 ) ? 0 = mV = (2. 50 lbm/s)(10. 6 ft/s)? ? = 0. 82 lbf 2 ? 32. 2 lbm ? ft/s ? Therefore, a force of 0. 82 lbf must be applie d (through friction at the base, for example) to prevent the fan from moving in the horizontal direction under the influence of this force. (b) Noting that P1 = P2 = Patm and V1 ? , the energy equation for the selected control volume reduces to ?P V2 ? ?P V2 ? m? 1 + 1 + gz1 ? + W 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 y ou are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-40 A helicopter hovers at sea level while being loaded. The volumetric air flow rate and the required power input during unloaded hover, and the rpm and the required power input during loaded hover are to be determined. Assumptions 1 The flow of air is steady and incompressible. 2 Air leaves the blades at a uniform velocity at atmospheric pressure. 3 Air approaches the blades from the top through a large area at atmospheric pressure with negligible velocity. 4 The frictional effects are negligible, and thus the entire mechanical power input is converted to kinetic energy of air (no conversion to thermal energy through frictional effects). 5 The change in air pressure with elevation is negligible because of the low density of air. 6 There is no acceleration of the helicopter, and thus the lift generated is equal to the total weight. Air flow is nearly uniform and thu s the momentum-flux correction factor can be taken to be unity, ? ? 1. Properties The density of air is given to be 1. 18 kg/m3. Analysis (a) We take the control volume to be a vertical hyperbolic cylinder bounded by streamlines on the sides with air entering through the large cross-section (section 1) at the top and the fan located at the narrow cross-section at the bottom (section 2), and let its centerline be the z axis with upwards being the positive direction. r r r F= ? mV ? ? mV . Noting The momentum equation for steady one-dimensional flow is ? out ? in that the only force acting on the control volume is the total weight W and it acts in the negative z direction, the momentum equation along the z axis gives W ? W = m(? V 2 ) ? 0 W = mV 2 = ( ? AV 2 )V 2 = ? AV 22 V2 = ? A 1 where A is the blade span area, 15 m A = ? D / 4 = ? (15 m) / 4 = 176. 7 m 2 2 2 Then the discharge velocity, volume flow rate, and the mass flow rate of air in the unloaded mode become V 2,unloaded = m unloaded g = ? A (10,000 kg)(9. 81 m/s 2 ) (1. 18 kg/m 3 )(176. 7 m 2 ) = 21. 7 m/s Sea level 2 Vunloaded = AV 2,unloaded = (176. 7 m 2 )(21. m/s) = 3834 m 3 /s munloaded = ? Vunloaded = (1. 18 kg/m 3 )(3834 m 3/s) = 4524 kg/s Load 15,000 kg Noting that P1 = P2 = Patm, V1 ? 0, the elevation effects are negligible, and the frictional effects are disregarded, the energy equation for the selected control volume reduces to ? P V2 ? ?P V2 ? V2 m? 1 + 1 + gz1 ? + W pump, u = m? 2 + 2 + gz 2 ? + W turbine + E mech,loss Wfan, u = m 2 ? ? ? ? 2 2 2 ? ? ? ? Substituting, ? V2 ? 1 kW ? (21. 7 m/s) 2 ? 1 kN ? ? = (4524 kg/s) W unloaded fan,u = ? m 2 ? ? = 1065 kW 2 ? 1 kN ? m/s ? 1000 kg ? m/s ? ? 2 ? 2 ? ? ? ? nloaded (b) We now repeat the calculations for the loaded helicopter, whose mass is 10,000+15,000 = 25,000 kg: V 2,loaded = m loaded g = ? A (25,000 kg)(9. 81 m/s 2 ) (1. 18 kg/m 3 )(176. 7 m 2 ) = 34. 3 m/s mloaded = ? Vloaded = ? AV2, loaded = (1. 18 kg/m 3 )(176. 7 m 2 )(34. 3 m/s) = 7152 kg/s ? V2 ? (34. 3 m/s)2 = (7152 kg/s) Wloaded fan,u = ? m 2 ? ? 2 ? 2 ? ?loaded ? 1 kW ? 1 kN ? ? ? 1000 kg ? m/s 2 1 kN ? m/s ? = 4207 kW ? ? 6-23 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems Noting that the average flow velocity is proportional to the overhead blade rotational velocity, the rpm of the loaded helicopter blades becomes V 2 = kn V 2,loaded V 2, unloaded = n loaded n unloaded n loaded = V 2,loaded V 2, unloaded n unloaded = 34. 3 (400 rpm) = 632 rpm 21. 7 Discussion The actual power input to the helicopter blades will be considerably larger than the calculated power input because of the fan inefficiency in converting mechanical power to kinetic energy. -24 PROPRIETARY MATERIAL.  © 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 lev el. 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. 222 Noting 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 Th erefore, the required power input will increase by 22. 2% on top of the mountain relative to the sea level. Discussion Note that both the rpm and the required power input to the helicopter are inversely proportional to the square root of air density. Therefore, more power is required at higher elevations for the helicopter to operate because air is less dense, and more air must be forced by the blades into the downdraft. 6-25 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-42 The flow rate in a channel is controlled by a sluice gate by raising or lowering a vertical plate. A relation for the force acting on a sluice gate of width w for steady and uniform flow is to be developed. Assumptions 1 The flow is steady, incompressible, frictionless, and uniform (and thus the Bernoulli equation is applicable. ) 2 Wall shear forces at surfaces are negligible. 3 The channel is exposed to the atmosphere, and thus the pressure at free surfaces is the atmospheric pressure. 4 The flow is horizontal. Water flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Analysis We take point 1 at the free surface of the upstream flow before the gate and point 2 at the free surface of the downstream flow after the gate. We also take the bottom surface of the channel as the reference level so that the elevations of points 1 and 2 are y1 and y2, respectively. The application of the Bernoul li equation between points 1 and 2 gives P1 V12 P V2 + + y1 = 2 + 2 + y 2 ? g 2 g ? g 2 g V 22 ? V12 = 2 g( y1 ? y 2 ) (1) The flow is assumed to be incompressible and thus the density is constant. Then the conservation of mass relation for this single stream steady flow device can be expressed as V1 = V2 = V A1V1 = A2V 2 = V V1 = V A1 = V wy1 and V2 = V A2 = V wy 2 (2) Substituting into Eq. (1), ? V ? ? wy ? 2 ? ? V ? 2 g ( y1 ? y 2 ) ? ? ? ? wy ? = 2 g ( y1 ? y 2 ) V = w 1 / y 2 ? 1 / y 2 ? ? 1? 2 1 2 2 2 g ( y1 ? y 2 ) V = wy 2 2 2 1 ? y 2 / y1 (3) Substituting Eq. (3) into Eqs. (2) gives the following relations for velocities, V1 = y2 y1 2 g ( y1 ? y 2 ) 1? y2 / 2 y1 and V2 = 2 g ( y1 ? y 2 ) 2 2 1 ? y 2 / y1 (4) We choose the control volume as the water body surrounded by the vertical cross-sections of the upstream and downstream flows, free surfaces of water, the inner surface of the sluice gate, and the bottom surface of r r r F= ? mV ? ? mV . The the channel. The momentum equation for steady one-dimensional flow is ? ? out ? in force acting on the sluice gate FRx is horizontal How to cite Solution Manual for Fluid Mech Cengel Book, Essay examples

Analysis Of Mr. Hsieh Fu Hua’s Leadership Traits Free Sample

Question: Disccuss about the Analysis of Mr. Hsieh Fu Huas Leadership Traits,Behaviour and Style. Answer: Introduction Leader plays an important role in every organisation. A true leader leads people with his or her charismatic personality and set examples before his or her followers so that all will follow unconditionally to leader in order to achieve the goals and objectives of the organisation. A true leader exhibits different traits and manages different situations successfully. Managing people in the organisation and avoiding risk on the way of achieving goal are the key responsibilities of the leader. This assignment makes an attempt to analyse Mr. Hsieh Fu Hua, the Chairman of UOB and the co-founder and advisor of PrimePartners Group of Singapore. His style of leadership in the change and critical situations are analysed in this assignment. Observation The leaders are the most important part of the organisation. They are the people to whom people follow. They are the people who lead the organisation with their innovative ideas and motivate the followers to work with the same norms that they set as the culture of the organisation. The leaders make the followers follow the things and focus on them. Trust is inspired by the leaders and they have a wide range of outlook. The true leaders believe in asking people what why and go on making the things original and different from others (chnangingminds.org, 2012). The current organisational culture is different from the traditional thinking of the organisational culture. The vision and mission of the organisation are only achieved when the leaders of the organisation formulate appropriate strategies. A true leader makes people understand and believe the vision of the organisation and need of establishing the culture of the organisation to achieve the goals of the organisation successfully(guides.wsj.com, 2015). Big Five Personality Traits The leader is judges with different characteristics he or she possesses. Self-confidence, responsibility, creative and hard working are some of the traits of the leaders. Big Five Personality Traits discusses about the five important behaviour of the leader. These traits are extraversion, openness, conscientiousness, neuroticism and agreeableness. Extraversion refers to the behaviour of the leader that shows that he or she favours socialization. High extraversion of the leader shows that leader has positive emotion. Openness reflects the attitude of the leader and his or her interest to remain attached to the things. Conscientiousness is a policy by which the leader makes coordination among the employees. Neuroticism is an attribute of the leader to deal the negative emotions of the employees of the members of the team. Agreeable refers to the interest of the leader to accept the suggestions of the followers. Leadership power The leaders have power to lead the people. These powers are the power of reward, coercive power, legitimate power, referent power expert power, informational power and connection power. Reward power shows how the leader has the ability to reward the followers for the compliance. This may be monetary and non monetary. Coercive power of the leader refers to the power of punishment and threatening to the employees when the outcome of the organisation does not come. Demotion, suspension, or dismiss from the services. Legitimate power refers to the power of the leader to influence the behaviour of the employees. Here the leader influences the people to manage and leads them to achieve the vision. Expert power refers to the ability of the leader to establish trust and respect among the followers. Informational power gives the leader knowledge about the followers of the organisation. Connection power helps the leader to gain influence by being acquainted with them Mr. Hsieh Fu Hua and his leadership styles, behaviour and traits Mr. Hsieh Fu Hua is the Chairman of UOB and the co-founder and advisor of PrimePartners Group of Singapore. Mr. Hua is also working as a Director of GIC also holds the position of Chairman of Tiger Airways Holdings.Hua also served as CEO of Singapore Exchange from 2003 to 2008. He served many non-profit organisations. National Gallery Singapore national Council of Social Service are also chaired by Hua. In this essay, the traits of leadership have been discussed with reference to Mr. Hua as an inspirational leader. His style of leadership in the change and critical situations are analysed in this assignment.(leadership.com.sg, 2015) Hsieh Fu Hua is famous for its charismatic leadership traits. He is a successful leader and his ability to manage the crisis is outstanding.Hua,in his corporate career, has contributed his experience and ability for the management of the crisis of the organisation. Hua is famous for its ability to manage the two most important crises. They are thePan-Electric crisis that happened in the year 2015 and another important crisis that he solved was the Asian Financial Crisis that occurred in thelate 90s. The ability to deal the Global Crisis was his most important contribution to the organisations he has served. Hua has tremendous ability to lead people. He has employed transformational leadership quality in his organisation. He believes that a leader should have ability to manage the risk and should have ability to solve the problems. Problem solving ability and interpersonal skills are two important factors that contribute the leader to become an effective leader. Hua has been very successful in managing the risks when the entire globe was facing financial crisis. He said that a leader has to be very constant and transparent in his or her principles. The leader should not change his or her core principles under no situation. The leader should remain constant and he or she should not make any change to the principles to favour anybody in the process of the decision making. The risk management and the principles of transparency of Hua have made him a successful leader. Traits of personality traits in Hua Hua possesses all the qualities of the big five personality traits of a leader. Hua is famous for his openness trait. He remains attached to people and becomes ready to face the challenges. Conscientiousness trait is also found in his leadership traits. He works to coordinate people. Neuroticism is another trait of the leader that helps him to deal all the negative situations. Hua is famous for his ability to face challenges in his tenure. He also got success for it. Leadership Power of Hua Hua is a strong leader. He believes in facing all situations bravely. He says that a leader should not bend. He has legitimate power to influence people to face the adverse situations. He has the power of expert to establish trust among his followers. The leaders have power to lead the people. These powers are the power of reward, coercive power, legitimate power, referent power expert power, informational power and connection power. Reward power shows how the leader has the ability to reward the followers for the compliance. This may be monetary and non monetary. Interview In the interview, I could find that Hua focused on the need of leaders decision to take the lead when the company passes through a very difficult position. As the leader, he managed to work effectively when the whole the world was passing through the Global Financial Crisis. During that period Hua was the chief executive of the Singapore Exchange (SGX). Hsieh Fu Hua has extensive ability and experiences in the financial sector. He has theability to handle stockbroking, banking that deals with investments and management of the funds. For his successful contribution to the banking and economic sectors, he has been appointed as the non-executive chairman of UOB, United Overseas bank. He has also been associated with thenon-profitorganisations. Hua advocates that the honesty and truthfulness make every leader stand on the front successfully and his ability to manage the crisis helps other followers of the organisation to follow the legacy the leaders leave his or her behind. Hsieh Fu Hua in his interview told that his leadership traits were developed in him when he read a book in his school career. The main character in the book, A man for All Seasons inspired him much who was an accomplished man. Hsieh Fu Hua told that constant is the supreme ability of a leader. The leader should be specifically constant in every complexity. He says that the leaders should have core values and based on that he or she should work. One should not compromise with the core values of the leaders. Hsieh Fu Hua, in his interview, told that there will be many situations arise in front of the leader when he or she has to be truth to the core value and true to himself or herself (Sandel, 2009).The leader in his interview expressed that integrity in the values and principles of life are very important. He says that it is very difficult to define integrity. He focuses on the importance of listening to the inner voice of oneself. There is always a greater need for taking decisions at different times. He says that one has to take aright decision at the right time. It is not that you are a leader and the leader has only ability to take the decisions. A follower has also required taking appropriate decision s in the functioning of the organisational goals (Low, 2009). The appropriate decision contributes generously to the achievement of the vision and mission of the company (Greyvenstein Cilliers, 2012). The leader says that the leader should listen to the followers and be reflective and should also accommodate in the situation. Hsieh Fu Hua says that every individual has its credibility and one should not divert from it. The leader should remain ready if he or she has to pay for his principles. But at the end, the leader will win and the truth and principles that he or she has designed honestly will bear good results. While answering the situations where he compromised with his core values and principles, Hsieh Fu Hua answered that the there are some situations that arise before the leaders when he finds himself or herself in the trouble. As a leader, one has to follow the culture of the organisation and the principles of the governance of the organisation cannot be violated (Balakrishnan, 2009). But in some instances, the leader has to think differently. In his interview, he discussed on his decisions of the huge investment. That decision was taken by him and he said that he owned it successfully. He said that the decision of making ahuge investment was not known to anybody. He could have kept it secret but he did not do that and kept everything transparent. This shows that as a leader the decisions taken by him or her should be owned. While answering to the life changing experience, Hua told that a leader has to be very sincere and transparent. He has faced many situations where he has confronted so many oppositions. He said that it is very easy for one to avoid the situations and favour the wrong to become popular among the co-workers. He said that this does not work at the end. The leader should be very clear to define the wrong and state the people when they do wrong. This has to be done at the time when the followers of the organisations doing wrong (Kannair, 2007). This has to be done very carefully by the leaders. Hua said that the most difficult situationsisto confront the oppositions. And the leader as a transformational leader needs to change the negativity among the followers not by just pointing to them what wrong they are doing rather by making them realise their mistakes (Bradberry, 2015). Thus he felt the confrontation to these situations brought life changing experience in him. While answering the question on his advice to the future leaders, he said that owning all the responsibility on his or her own head is quality important to take the followers to know about their dignity of labour. They should know that as the followers of the organisation they equal contribution. But the leader needs to remain in the front. Thus creates lot motivation in the mind of the followers (Poh, 2010). During any crisis, the leader has to communicate the message that he or she remains present with all. Hua also said that the leaders should apply his or her sense during the difficult situations. The leaders should be sensible and demonstrate control. When the leader becomes sensible and demonstrates control the followers follow him or her successfully. In that situation, the leaders should be with a risk management plan and this will help the leaders to manage the situations (Walton, 2008). In answering again to his advice on the future leaders, Hua said that the leader should not feel like a boss. Hua also commented on the need of managing the ego. He said that one should have an ego and he should apply this as per the need of the organisation. He is just merely the leader of the organisation. The leader should set the examples like a transformational leader. The leader should belikea true mentor (Wong, 2011). Like a democratic leader, the leader should also encourage the followers to offer their suggestions for the development of the organisation. The leader should motivate his followers well. The motivation of the leaders contributes to the pattern of the growth of the followers working style in the future days for the considerable contribution for the achievement of the organisational goals (Lim, 2011). Hua is an inspiration to me. His ability to manage people and risk, and to remain constant in every situation are the most effective traits that I have learnt from h im. Self-reflection and improvement The interview conducted on Hua offers extensive knowledge on the need of traits, behaviour and styles of leadership. The leader is a person who does something that is very new to all. The action initiated by the leader should be followed by the followers voluntary. In the interview I found that Hua is a transformational leader (Low, 2011). He has also possessed the traits of a democratic leader. In one instance he showed that sometimes he has to apply autocratic style of leadership. In the interview this has been found that the leader should have the ability to manage the risk and manage the people who are the important contributors for the achievement of the organisational goals. A have found that Hua works as a transformational leader and motivates people to work. He has tremendous positive ethos. He believes that trust and respect should there with the leader. The leader should be transparent and truthful. The inspiration of the leader helps the organisation to grow. The leader should guide and motivate people well so that followers will be persuaded well(Sanborn, 2016). The leader believes that in every moment the leader should practice the basic principles and traits of the leadership quality so that the best opportunity for the organisation will be made. Risk management The leader should have positive traits to handle the risks. The leader should be sensible and decisions should be taken by the leaders instantly to manage the risk. Thus it is very important for the leader should make a risk management plan. This will help the leader to take decisions and when it is required (Low, 2008). Decision making The leader should be very sincere in making decision. A leader should have some core principles and based on that he or she should work. Like a democratic leader, the leader of the organisation should involve the followers of the organisation to contribute with their rich experiences in the process of the decision making. Motivation The leader of the organisation inspires people to follow him or her. Thus the traits of the leader should motivate people to follow the leaders without any compulsion. By setting the example, the leader will motivate the followers to work on the way the organisation requires its followers to work. Communication skills A true leader is an effective communicator. The effective communication made by the leader will help him or her to communicate with people successfully. The emotional intelligence and the interpersonal skills are required to be developed by the leader to become a successful leader. As a leader I need to develop my emotional intelligence and interpersonal skills. This will help me to lead the follower of my organisation.(Nayar, 2013). Interpersonal skills Interpersonal skills are very important for a leader. The leader has to develop interpersonal skills. The leader should always make good communication with members of his or her organisation. This will create a better environment and all are committed to the achievement of the organisational goals (George-Godfrey, 2012). Developing my leadership quality Development of a leader is the enhancement of the ability of an individual to act and practice the principles of a leader in order to become successful in leading the people. The role of the leader includes alignment of the work, to set a significant objective and to be committed to the work and the group (Drucker, 2014). . The problem -solving ability and setting the right culture of the organisation are also very important for the leader. The leader needs to work for a favourable environment for the followers of the organisation so that the employees will be motivated and work for the achievement of the organisational goals. Need of a Two part- model This model helps the leader to become successful. Assessment, challenge support are three important requirements that will create an experience in the leader and that will help the leader become successful in the future(Velsor, et al., 2010). The challenges make a leader more experienced. In these difficult situations the leader should know how to handle the risks. A leader has to apply the presence of mind in a particular situation so that appropriate solution will be made for the risk. The leader should have strategic objectives to handle all these situations(Gray, 2014). The interview helped me to learn that I need to be a successful leader with the help of the leadership quality. I must have a transformational leadership style. I will use my democratic leadership style when I will take decision for the organisation. This will help to do the appropriate things on the appropriate time. I also require being constant and transparent in the process of leading my followers. The principles based on which I will lead the people will never bechanged.As a leader, I will motivate my followers to get the success. Conclusion Leaders are the most important people who contribute for the growth of the organisation. The leaders make themselves different from the other and work for the achievement of the organisational goals. The leaders with their vision and transformational leadership style encourage motivating people to work in the desired way. The true leaders need to have ability to manage the risks and problems. 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