Time management Research Paper Example | Topics and Well Written Essays - 2000 words

Time management - Research Paper Example 8 Conclusion†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.. 9 References†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦...11 Time management Introduction Time management entails exercising prudent control on the amount of time spent in performing several activities in order to improve productivity, efficiency and increase effectiveness (Pausch, 2008). Time management includes several tools, techniques and skills that enable individuals and organizations to accomplish specific activities within the time limit. The main aspects of time management include setting and adhering to priorities and minimizing time spent on non-core activities (Carroll, 2012). Time management is about utilizing the available time effectively through effective scheduling of time, setting clear goal s, prioritizing of activities, delegation of the tasks and reviewing the time spent on different tasks. According to Pausch (2008), the management must organize the workspace or environment in order to eliminate destructions and ensure maximum concentration. Eisenhower time management method classifies tasks in to either urgent or important tasks while action priority matrix determines the priority of tasks depending on the efforts and perceived benefits to the organization. According to Pareto principle (80:20 rule), 80 percent of the output is generated by 20 percent of the efforts. The Pareto analysis is useful in prioritizing the tasks since 80 percent of the results can be attributed to 20 percent of tasks completed. This paper will discuss benefits of time management, various skills in time management and methods of time management. The paper will also discuss ways of avoiding procrastination. Carroll (2012) asserts that effective planning entails preparing work lists of activ ities that must be completed during a certain time. These include the ‘to do lists’ that prioritize activities depending on their importance and urgency to the organization. The pending tasks must take accomplished first and new tasks must not start until the previous or already started tasks are complete. Accordingly, the individual or management must set goals and objectives that aim at ensuring efficiency of work. The goals and objectives must be specific, attainable and measurable (Croft, 1996). The goals must have clear deadlines that will guide the efforts committed towards accomplishment of the specific tasks. The management must motivate employees towards attainment of the set goals within the pre-determined deadlines. Accordingly, the management must delegate authority and responsibilities in order to ensure employees take accountability of their tasks in completing the tasks within the deadlines (Pausch, 2008). The employees must spend the appropriate and allo cated time on the right activities depending on the priorities in order to meet any strict deadlines and ensure efficient workflow in the organization (Croft, 1996). Benefits of time management Time management involves making the possible best use of the available time while doing the right things at the right time. One of the benefits of time management is that it makes individuals more disciplined and punctual. The individuals are capable of preparing ‘

Scientific Writing Essay Example | Topics and Well Written Essays - 500 words

Scientific Writing - Essay Example Innate immunity comprises mainly of the skin and mucosal surfaces. The main function of this immune is to prevent entry of specific non-self antigens. In many cases, compromising of innate immunity occurs and consequently pathogens gain entry into the body. The remedy then is the acquired immunity, which comprise the cellular immunity, (lymphocytes, macrophages, natural killer cells, and antigen-presenting cell) and humoral immunity (mediated by macromolecules). It works by recognizing disease causing pathogens and triggering mechanisms that destroy them. (Weinstein 40) Many diseases that kill people in this century originate from stress as opposed to improper diet, and pathogen resistance to drugs. More investment on in making people less stressed than in pharmaceutical firms it to be considered by any nation that want healthy and effective citizens. An immune system in its right condition is able to keep pathogens at bay and maintain the body health. Stress remarkably interferes the immune system by altering its functioning. It suppresses the immune system function and over time causes tearing down of this important system. Stress is the feeling of inability to deal with a life-threatening situation. It leads to emotional, psychological, and physical problems. The physical problems include heart disease, high blood pressure, chest diseases, and heart rate abnormalities (Harrington 26). The adrenal glands release adrenaline when someone is under stressful conditions. It responds well to short term stress through fight or flight response. Adrenaline stimulates the heart rate, contracting blood vessels’ dilating air passages thus increasing a subsequent flow of blood to the muscles and the amount of oxygen intake in the lungs (Weinstein 43). These responses affect the digestive system by inhibiting the digestion. If stress is prolonged, the production of

Therapeutic Recreation Models

Therapeutic Recreation Models Therapeutic Recreation seeks to promote the capacity and ability of groups and individuals to make self determined and responsible choices, in light of their needs to grow, to explore new perspectives and possibilities, and to realise their full potential. Within this assignment I am going to critically compare and evaluate the use of the following models in the Therapeutic Recreation Service: The Leisure Ability Model and the Health Promotion/ Health Protection Model. In doing so I will firstly describe the two models in detail and then critically compare and evaluate them both and their use in the therapeutic recreation service. The Leisure Ability Model: Every human being needs, wants, and deserves leisure. Leisure presents opportunities to experience mastery, learn new skills, meet new people, deepen existing relationships, and develop a clearer sense of self. Leisure provides the context in which people can learn, interact, express individualism, and self-actualize (Kelly, 1990). A large number of individuals are constrained from full and satisfying leisure experiences. It then follows that many individuals with disabilities and/or illnesses may experience more frequent, severe, or lasting barriers compared with their non-disabled counterparts, simply due to the presence of disability and/or illness. The Leisure Ability Models underlying basis stems from the concepts of: (a) learned helplessness vs. mastery or self-determination; (b) intrinsic motivation, internal locus of control, and causal attribution; (c) choice; and (d) flow. Learned Helplessness: Learned helplessness is the perception by an individual that events happening in his or her life are beyond his or her personal control, and therefore, the individual stops trying to effect changes or outcomes with his or her life (Seligman, 1975). They will eventually stop wanting to participate in activity or participate in any other way. They will learn that the rules are outside of their control and someone else is in charge of setting the rules. Their ability to take a risk will be diminished and they will learn to be helpless. Learned helplessness may present a psychological barrier to full leisure participation and it may, conversely, be unlearned with the provision of well-designed services. Intrinsic Motivation, Internal Locus of Control, and Causal Attribution: All individuals are intrinsically motivated toward behaviour in which they can experience competence and self-determination. As such, individuals seek experiences of incongruity or challenges in which they can master the situation, reduce the incongruity, and show competence. This process is continual and through skill acquisition and mastery, produces feelings of satisfaction, competence, and control. An internal locus of control implies that the individual has the orientation that he or she is responsible for the behaviour and outcomes he or she produces (Deci, 1975). Typically individuals with an internal locus of control take responsibility for their decisions and the consequences of their decisions, while an individual with an external locus of control will place responsibility, credit, and blame on other individuals. An internal locus of control is important for the individual to feel self-directed or responsible, be motivated to continue to seek challenges, and develop a sense of self-competence. http://dw.com.com/redir?tag=rbxira.2.a.10destUrl=http://www.cnet.com/b.gif Attribution implies that an individual believes that he or she can affect a particular outcome (Deci, 1975; Seligman, 1975). An important aspect of the sense of accomplishment, competence, and control is the individuals interpretation of personal contribution to the outcome. Without a sense of personal causation, the likelihood of the individual developing learned helplessness increases greatly. Choice: The Leisure Ability Model also relies heavily on the concept of choice, choice implies that the individual has sufficient skills, knowledge, and attitudes to be able to have options from which to choose, and the skills and desires to make appropriate choices. Lee and Mobily (1988) stated that therapeutic recreation services should build skills and provide participants with options for participation. Flow: When skill level is high and activity challenge is low, the individual is quite likely to be bored. When the skill level is low and the activity challenge is high, the individual is most likely to be anxious. When the skill level and activity challenge are identical or nearly identical, the individual is most able to achieve a state of concentration and energy expenditure that Csikszentmihalyi (1990) has labeled flow. Treatment Services During treatment services, the client generally has less control over the intent of the programs and is dependent on the professional judgment and guidance provided by the specialist. The client experiences less freedom of choice during treatment services than any other category of therapeutic recreation service. The role of the specialist providing treatment services is that of therapist. Within treatment services, the client has minimal control and the therapist has maximum control. The specialist typically designates the clients level and type of involvement, with considerably little input from the client. In order to successfully produce client outcomes, the specialist must be able to assess accurately the clients functional deficits; create, design, and implement specific interventions to improve these deficits; and evaluate the client outcomes achieved from treatment programs. http://dw.com.com/redir?tag=rbxira.2.a.10destUrl=http://www.cnet.com/b.gifThe ultimate outcome of treatment services is to eliminate, significantly improve, or teach the client to adapt to existing functional limitations that hamper efforts to engage fully in leisure pursuits. Often these functional deficits are to the degree that the client has difficulty learning, developing his or her full potential, interacting with others, or being independent. The aim of treatment services is to reduce these barriers so further learning and involvement by the client can take place. Leisure Education: Leisure education services focus on the client acquiring leisure-related attitudes, knowledge, and skills. Participating successfully in leisure requires a diverse range of skills and abilities, and many clients of therapeutic recreation services do not possess these, have not been able to use them in their leisure time, or need to re-learn them incorporating the effects of their illness and/ or disability. Leisure education services are provided to meet a wide range of client needs related to engaging in a variety of leisure activities and experiences. (Howe, 1989, p. 207). The overall outcome sought through leisure education services is a client who has enough knowledge and skills that an informed and independent choice can be made for his or her future leisure participation. Leisure education means increased freedom of choice, increased locus of control, increased intrinsic motivation, and increased independence for the client. Recreation Participation: http://dw.com.com/redir?tag=rbxira.2.a.10destUrl=http://www.cnet.com/b.gif Recreation participation programs are structured activities that allow the client to practice newly acquired skills, and/or experience enjoyment and self-expression. These programs are provided to allow the client greater freedom of choice within an organized delivery system and may, in fact, be part of the individuals leisure lifestyle. The clients role in recreation participation programs includes greater decision making and increased self-regulated behaviour. The client has increased freedom of choice and his or her motivation is largely intrinsic. In these programs, the specialist is generally no longer teaching or in charge per se. The client becomes largely responsible for his or her own experience and outcome, with the specialist moving to an organizer and/or supervisor role. As Stumbo and Peterson (1998) noted, recreation participation allows the client an opportunity to practice new skills, experience enjoyment, and achieve self-expression. From a clinical perspective, recreation participation does much more. For instance, recreation opportunities provide clients with respite from other, more arduous, therapy services. Leisure education programs may focus on: (a) self-awareness in relation to clients new status; (b) learning social skills such as assertiveness, coping, and friendship making; (c) re-learning or adapting pre-morbid leisure skills; and (d) locating leisure resources appropriate to new interests and that are accessible. Recreation participation programs may involve practicing a variety of new leisure and social skills in a safe, structured environment. In designing and implementing these programs, the specialist builds on opportunities for the individual to exercise control, mastery, intrinsic motivation, and choice. The ultimate outcome would be for each client to be able to adapt to and cope with individual disability to the extent that he or she will experience a satisfying and independent leisure lifestyle, and be able to master skills to achieve flow. Health Promotion/ Health Protection Model: The Health Protection/Health Promotion Model (Austin, 1996, 1997) stipulates that the purpose of therapeutic recreation is to assist persons to recover following threats to health, by helping them to restore themselves or regain stability. (health protection), and secondly, optimising their potentials in order that they may enjoy as high a quality of health as possible (health promotion). Within this model (Austin, 1997, p. 144) states that â€Å"the mission of therapeutic recreation is to use activity, recreation, and leisure to help people to deal with problems that serve as barriers to health and to assist them to grow toward their highest levels of health and wellness The health promotion, health protection model is broken up into four broad concepts which are the humanistic perspective, high level wellness, stabilisation and actualisation and health. Humanistic Perspective: Those who embrace the humanistic perspective believe that each of us has the responsibility for his or her own health and the capacity for making self-directed and wise choices regarding our health. Since individuals are responsible for their own health, it is critical to empower individuals to become involved in decision-making to the fullest extent possible (Austin, 1997). High-Level Wellness: High-level wellness deals with helping persons to achieve as high a level of wellness as they are capable of achieving (Austin, 1997). Therapeutic Recreation professionals have concern for the full range of the illness-wellness continuum (Austin, 1997). http://dw.com.com/redir?tag=rbxira.2.a.10destUrl=http://www.cnet.com/b.gif Stabilization and Actualization Tendencies: The stabilizing tendency is concerned with maintaining the steady state of the individual. It is an adaptation mechanism that helps us keep stress in a manageable range. It protects us from biophysical and psychosocial harm. The stabilizing tendency is the motivational force behind health protection that focuses on efforts to move away from or avoid negatively valence states of illness and injury (Pender, 1996, p. 34). The actualization tendency drives us toward health promotion that focuses on efforts to approach or move toward a positively valence state of high-level health and well-being (Pender, 1996, p. 34). Health: King (1971) and Pender (1996) health encompasses both coping adaptively and growing and becoming. Healthy people can cope with lifes stressors. Those who enjoy optimal health have the opportunity to pursue the highest levels of personal growth and development. Under the Health Protection/Health Promotion Model, therapists* recognize that to help clients strive toward health promotion is the ultimate goal of therapeutic recreation. Further, therapists prize the right of each individual to pursue his or her highest state of well-being, or optimal health. TR practice is therefore based on a philosophy that encourages clients to attempt to achieve maximum health, rather than just recover from illness (Austin, 1997). The Component of Prescriptive Activities: When clients initially encounter illnesses or disorders, often they become self-absorbed. They have a tendency to withdraw from their usual life activities and to experience a loss of control over their lives (Flynn, 1980). Research (e.g., Langer Rodin, 1976; Seligman Maier, 1967) has shown that feelings of lack of control may bring about a sense of helplessness that can ultimately produce severe depression. At times such as this clients are encountering a significant threat to their health and are not prepared to enjoy and benefit from recreation or leisure. For these individuals, activity is a necessary prerequisite to health restoration. Activity is a means for them to begin to gain control over their situation and to overcome feelings of helplessness and depression that regularly accompany loss of control. At this point on the continuum, Therapeutic Recreation professionals provide direction and structure for prescribed activities. Once engaged in activity, clients can begin to perceive themselves as being able to successfully interact with their environments, to start to experience feelings of success and mastery, and to take steps toward regaining a sense of control. Clients come to realise that they are not passive victims but can take action to restore their health. They are then ready to partake in the recreation component of treatment. The Recreation Component: Recreation is activities that take place during leisure time (Kraus. 1971). Client need to take part in intrinsically motivated recreation experiences that produce a sense of mastery and accomplishment within a supportive and nonthreatening atmosphere. Clients have fun as they learn new skills, new behaviors, new ways to interact with others, new philosophies and values, and new cognition about themselves. In short, they learn that they can be successful in their interactions with the world. Through recreation they are able to re-create themselves, thus combating threats to health and restoring stability. http://dw.com.com/redir?tag=rbxira.2.a.10destUrl=http://www.cnet.com/b.gif The Leisure Component: Whereas recreation allows people to restore themselves, leisure is growth promoting. Leisure is a means to self-actualisation because it allows people to have self-determined opportunities to expand themselves by successfully using their abilities to meet challenges. Feelings of accomplishment, confidence and pleasure result from such growth producing experiences. Thus leisure assumes an important role in assisting people to reach their potentials (Iso-Ahola, 1989). Core elements in leisure seem to be that it is freely chosen and intrinsically motivated. The Recreation and Leisure Components: Although recreation and leisure differ in that recreation is an adaptive device that allows us to restore ourselves and leisure is a phenomenon that allows growth, they share commonalities. Both recreation and leisure are free from constraint. Both involve intrinsic motivation and both provide an opportunity for people to experience a tremendous amount of control in their lives. Both permit us to suspend everyday rules and conventions in order to be ourselves and let our hair down. Both allow us to be human with all of our imperfections and frailties. It is the task of the therapeutic recreation professional to maintain an open, supportive, and nonthreatening atmosphere that encourages these positive attributes of recreation and leisure and which help to bring about therapeutic benefit (Austin, 1996). http://dw.com.com/redir?tag=rbxira.2.a.10destUrl=http://www.cnet.com/b.gif According to Bandura (1986), bolstered efficacy expectations allow clients to have confidence in themselves and in their abilities to succeed in the face of frustration. Thus, clients feel more and more able to be in control of their lives and to meet adversity as they move along the continuum toward higher levels of health. It is the role of the TR professional to help each client assume increasing levels of independence as he or she moves along the illness-wellness continuum. Of course, the client with the greatest dependence on the therapist will be the individual who is in the poorest health. At this point the stabilizing tendency is paramount while the client attempts to ward off the threat to health and to return to his or her usual stable state. At this time the therapist engages the client in prescriptive activities or recreation experiences in order to assist the client with health protection. During prescriptive activities the clients control is the smallest and the therapi sts is the largest. During recreation there is more of a mutual participation by the client and therapist. With the help of the therapist, the client learns to select, and participate in, recreation experiences that promote health improvement. Approximately midway across the continuum, the stabilising tendency reduces and the actualising tendency begins to arise. Leisure begins to emerge as the paramount paradigm. As the actualisation tendency increases, the client becomes less and less dependent on the therapist and more and more responsible for self-determination. The role of the therapist continues to diminish until the client is able to function without the helper. At this point the client can function relatively independently of the TR professional and there is no need for TR service delivery (Austin, 1997). Comparison of the use of the Leisure Ability Model to the Health Promotion/ Health Protection Model in Therapeutic Recreation Services: The role of the therapeutic recreation specialist, in order to reverse the consequences of learned helplessness, is to assist the individual in: (a) increasing the sense of personal causation and internal control, (b) increasing intrinsic motivation, (c) increasing the sense of personal choice and alternatives, and (d) achieving the state of optimal experience or flow. In theory, then, therapeutic recreation is provided to affect the total leisure behaviour (leisure lifestyle) of individuals with disabilities and/or illnesses through decreasing learned helplessness, and increasing personal control, intrinsic motivation, and personal choice. This outcome is accomplished through the specific provision of treatment, leisure education, and recreation participation services which teach specific skills, knowledges, and abilities, and take into consideration the matching of client skill and activity challenge. Another strength is the Models flexibility. One level of flexibility is with the three components of service. Each component of service is selected and programmed based on client need. That is, some clients will need treatment and leisure education services, without recreation participation. Other clients will need only leisure education and recreation participation services. Clearly, services are selected based on client need. In addition, programs conceptualized within each service component are selected based on client need. flexibility allow the specialist to custom design programs to fit the needs of every and any client group served by therapeutic recreation. The ultimate goal of leisure lifestyle remains the same for every client, but since it is based on the individual, how the lifestyle will be implemented by the individual and what it contains may differ. As such, the content of the Leisure Ability Model is not specific to any one population or client group, nor is it confined to any specific service or delivery setting. Some authors, including Kinney and Shank (1989), have reported this as a strength of the Model. According to the model, intervention may occur in a wide range of settings and addresses individuals with physical, mental, social, or emotional limitations (Peterson Gunn, p. 4). The intervention model is conceptually divided into three phases along a continuum of client functioning and restrictiveness. The three phases of therapeutic recreation intervention are arranged in a sequence, from greater therapist control to lesser therapist control, and from lesser client independence to greater client independence. This arrangement is purposeful and is meant to convey that the ultimate aim of the appropriate leisure lifestyle is that it be engaged in independently and freely. Summary The Health Protection/Health Promotion Model contains three major components (i.e., prescribed activities, recreation, and leisure) that range along an illness-wellness continuum. According to their needs, clients may enter anywhere along the continuum. The model emphasizes the active role of the client who becomes less and less reliant on the TR professional as he or she moves toward higher levels of health. Initially, direction and structure are provided through prescriptive activities to help activate the client. During recreation, the client and therapist join together in a mutual effort to restore normal functioning. During leisure, the client assumes primary responsibility for his or her own health and well-being. Evaluation of both models and there use in therapeutic recreation services: The overall intended outcome of therapeutic recreation services, as defined by the Leisure Ability Model, is a satisfying, independent, and freely chosen leisure lifestyle. In order to facilitate these perceptions, therapeutic recreation specialists must be able to design, implement, and evaluate a variety of activities that increase the persons individual competence and sense of control. In relation to leisure behaviour, Peterson (1989) felt that this includes improving functional abilities, improving leisure-related attitudes, skills, knowledge, and abilities, and voluntarily engaging in self-directed leisure behaviour. Thus, the three service areas of treatment, leisure education, and recreation participation are designed to teach specific skills to improve personal competence and a sense of accomplishment. Csikszentmihalyi (1990) summed up the importance of these perceptions: In the long run optimal experiences add up to a sense of mastery-or perhaps better, a sense of participation in determining the content of life-that comes as close to what is usually meant by happiness as anything else we can conceivably imagine (p. 4). The therapeutic recreation specialist must be able to adequately assess clients skill level (through client assessment) and activity requirements (through activity analysis) in order for the two to approximate one another. Given Decis (1975) theory of intrinsic motivation which includes the concept of incongruity, therapeutic recreation specialists may provide activities slightly above the skill level of clients in order to increase the sense of mastery. When this match between the activity requirements and client skill levels occurs, clients are most able to learn and experience a higher quality leisure. To facilitate this, therapeutic recreation specialists become responsible for comprehending and incorporating the: (a) theoretical bases (including but not limited to internal locus of control, intrinsic motivation, personal causation, freedom of choice, and flow); (b) typical client characteristics, including needs and deficits; (c) aspects of quality therapeutic recreation program delivery process (e.g., client assessment, activity analysis, outcome evaluation, etc.); and (d) therapeutic recreation content (treatment, leisure education, and recreation participation). These areas of understanding are important for the therapeutic recreation specialist to be able to design a series of coherent, organized programs that meet client needs and move the client further toward an independent and satisfactory leisure lifestyle. Again, the success of that lifestyle is dependent on the client gaining a sense of control and choice over leisure options, and having an orientation toward intrinsic motivation, an internal locus of control, and a personal sense of causality. The Leisure Ability Model provides specific content that can be addressed with clients in order to facilitate their development, maintenance, and expression of a successful leisure lifestyle. Each aspect of this content applies to the future success, independence, and well-being of clients in regard to their leisure. http://dw.com.com/redir?tag=rbxira.2.a.10destUrl=http://www.cnet.com/b.gif The client has reduced major functional limitations that prohibit or significantly limit leisure involvement (or at least has learned ways to overcome these barriers); understands and values the importance of leisure in the totality of life experiences; has adequate social skills for involvement with others; is able to choose between several leisure activity options on a daily basis, and make decisions for leisure participation; is able to locate and use leisure resources as necessary; and has increased perceptions of choice, motivation, freedom, responsibility, causality, and independence with regard to his or her leisure. These outcomes are targeted through the identification of client needs, the provision of programs to meet those needs, and the evaluation of outcomes during and after program delivery. A therapeutic recreation specialist designs, implements, and evaluates services aimed at these outcomes Austin (1989) objected to the Leisure Ability Model on the basis that is supporting a leisure behaviour orientation, instead of the therapy orientation. A number of authors have objected to the Leisure Ability Model, having observed that its all-encompassing approach is too broad and lacks the focus needed to direct a profession (Austin, p. 147). Austin advocated an alignment of therapeutic recreation with allied health and medical science disciplines, rather than leisure and recreation professionals The Model in Practice The Health Protection/Health Promotion Model may be applied in any setting (i.e., clinical or community) in which the goal of therapeutic recreation is holistic health and well-being. Thus, anyone who wishes to improve his or her level of health can become a TR client. TR professionals view all clients as having abilities and intact strengths, as well as possessing intrinsic worth and the potential for change. Through purposeful intervention using the TR process (i.e., assessment, planning, implementation, evaluation), therapeutic outcomes emphasize enhanced client functioning. Typical therapeutic outcomes include increasing personal awareness, improving social skills, enhancing leisure abilities, decreasing stress, improving physical functioning, and developing feelings of positive self-regard, self-efficacy and perceived control (Austin, 1996).

Peer Pressures of High School :: Peer Pressure Essays

Glaring down at the reddish glow coming from the tip of the cigarette, I found out that I was in a peer pressure situation. Peer Pressure can be a huge problem for some young adults. It can sometimes be positive, but most of the time it ¡Ã‚ ¯s negative and destructive. Smoking is just one of the peer pressures someone can go through. Alcohol and staying out late can also be huge peer pressures in high school. I know this because I have experienced them for myself. Drinking, smoking and staying out late were constant peer pressures throughout my high school career. Looking down at the cigarette and being encouraged by my friend to take a hit off of it, I knew that smoking was not something I wanted to do at that time in my life. Although smoking wasn ¡Ã‚ ¯t a huge peer pressure for me, it can be for others. Some of my friends did give into the pressure and are now addicted to cigarettes, and wish they hadn ¡Ã‚ ¯t give in to that peer pressure in high school. I would have say th at during high school, smoking was the most persistent peer pressure. It was at every party and gathering. Although it was there all the time sometimes alcohol would  ¡Ã‚ °rear its ugly head ¡Ã‚ ± at some of the parties. Drinking was probably the most dangerous peer pressure. It was extremely illegal for an underage adult to be caught drinking during this time. I never experienced this peer pressure during high school because I didn ¡Ã‚ ¯t hang around those types of people during that time. They were the types of people who didn ¡Ã‚ ¯t think it was a  ¡Ã‚ °party ¡Ã‚ ± unless there was alcohol involved. I have seen drinking totally deteriorate people, because it got the best of them. Some of my friends totally changed after they started drinking. At first it was just a social thing to do at parties, but then lead on to drinking during their  ¡Ã‚ °spare time ¡Ã‚ ±. It affected their grades and their overall behavior. I do think this was the most dangerous peer pressure in high school, but there was always the pressure to stay out late. Staying out late was a peer pressure I gave into on several occasions during high school. I know it affected my grades many times, and also made me late for school more than once. In high school you could always spot the students who stayed out late.

Solution Manual for Fluid Mech Cengel Book

Chapter 6 Momentum Analysis of Flow Systems Chapter 6 MOMENTUM ANALYSIS OF FLOW SYSTEMS Newton’s Laws and Conservation of Momentum 6-1C Newton’s first law states that â€Å"a body at rest remains at rest, and a body in motion remains in motion at the same velocity in a straight path when the net force acting on it is zero. † Therefore, a body tends to preserve its state or inertia. Newton’s second law states that â€Å"the acceleration of a body is proportional to the net force acting on it and is inversely proportional to its mass. Newton’s third law states â€Å"when a body exerts a force on a second body, the second body exerts an equal and opposite force on the first. † r 6-2C Since momentum ( mV ) is the product of a vector (velocity) and a scalar (mass), momentum must be a vector that points in the same direction as the velocity vector. 6-3C The conservation of momentum principle is expressed as â€Å"the momentum of a system remains constant when the net force acting on it is zero, and thus the momentum of such systems is conserved†.The momentum of a body remains constant if the net force acting on it is zero. 6-4C Newton’s second law of motion, also called the angular momentum equation, is expressed as â€Å"the rate of change of the angular momentum of a body is equal to the net torque acting it. † For a non-rigid body with zero net torque, the angular momentum remains constant, but the angular velocity changes in accordance with I? = constant where I is the moment of inertia of the body. 6-5C No.Two rigid bodies having the same mass and angular speed will have different angular momentums unless they also have the same moment of inertia I. Linear Momentum Equation 6-6C The relationship between the time rates of change of an extensive property for a system and for a control volume is expressed by the Reynolds transport theorem, which provides the link between the r system and control volume concepts. The linear momentum equation is obtained by setting b = V and thus r B = mV in the Reynolds transport theorem. -7C The forces acting on the control volume consist of body forces that act throughout the entire body of the control volume (such as gravity, electric, and magnetic forces) and surface forces that act on the control surface (such as the pressure forces and reaction forces at points of contact). The net force acting on a control volume is the sum of all body and surface forces. Fluid weight is a body force, and pressure is a surface force (acting per unit area). -8C All of these surface forces arise as the control volume is isolated from its surroundings for analysis, and the effect of any detached object is accounted for by a force at that location. We can minimize the number of surface forces exposed by choosing the control volume such that the forces that we are not interested in remain internal, and thus they do not complicate the analysis. A well-chosen cont rol volume exposes only the forces that are to be determined (such as reaction forces) and a minimum number of other forces. 6-9C The momentum-flux correction factor ? nables us to express the momentum flux in terms of the r r r r & ? V (V ? n )dAc = ? mV avg . The value of ? is unity for uniform mass flow rate and mean flow velocity as ? Ac flow, such as a jet flow, nearly unity for turbulent flow (between 1. 01 and 1. 04), but about 1. 3 for laminar flow. So it should be considered in laminar flow. 6-1 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.Chapter 6 Momentum Analysis of Flow Systems 6-10C The momentum equation for steady one-dimensional flow for the case of no external forces is r r r & & F= ? mV ? ? mV ? ? out ? in where the left hand side is the net force acting on the control volume, and first term on the right hand side is the incoming momentum flux and the second term is the outgoing momentum flux by mass. 6-11C In the application of the momentum equation, we can disregard the atmospheric pressure and work with gage pressures only since the atmospheric pressure acts in all directions, and its effect cancels out in every direction. -12C The fireman who holds the hose backwards so that the water makes a U-turn before being discharged will experience a greater reaction force since the numerical values of momentum fluxes across the nozzle are added in this case instead of being subtracted. 6-13C No, V is not the upper limit to the rocket’s ultimate velocity. Without friction the rocket velocity will continue to increase as more gas outlets the nozzle. 6-14C A helicopter hovers because the strong downdraft of air, caused by the overhead propeller blades, manifests a momentum in the air stream.This momentum must be countered by the helicopter lift force. 6-15C As the air density decreases, it requires more energy for a helicopter to hover, because more air must be forced into the downdraft by the helicopter blades to provide the same lift force. Therefore, it takes more power for a helicopter to hover on the top of a high mountain than it does at sea level. 6-16C In winter the air is generally colder, and thus denser. Therefore, less air must be driven by the blades to provide the same helicopter lift, requiring less power. 6-2 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc.Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-17C The force required to hold the plate against the horizontal water stream will increase by a factor of 4 when the velocity is doubled since & F = mV = ( ? AV )V = ? AV 2 and thus the force is proportional to the square of the velocity. 6-18C The accele ration will not be constant since the force is not constant. The impulse force exerted by & water on the plate is F = mV = ( ? AV )V = ?AV 2 , where V is the relative velocity between the water and the plate, which is moving. The plate acceleration will be a = F/m. But as the plate begins to move, V decreases, so the acceleration must also decrease. 6-19C The maximum velocity possible for the plate is the velocity of the water jet. As long as the plate is moving slower than the jet, the water will exert a force on the plate, which will cause it to accelerate, until terminal jet velocity is reached. 6-20 It is to be shown that the force exerted by a liquid jet of velocity V on a stationary nozzle is & proportional to V2, or alternatively, to m 2 . Assumptions 1 The flow is steady and incompressible. 2 The nozzle is given to be stationary. 3 The nozzle involves a 90 ° turn and thus the incoming and outgoing flow streams are normal to each other. 4 The water is discharged to the atmo sphere, and thus the gage pressure at the outlet is zero. Analysis We take the nozzle as the control volume, and the flow direction at the outlet as the x axis. Note that the nozzle makes a 90 ° turn, and thus it does not contribute to any pressure force or momentum flux & term at the inlet in the x direction. Noting that m = ?AV where A is the nozzle outlet area and V is the average nozzle outlet velocity, the momentum equation for steady one-dimensional flow in the x direction reduces to r r r & & & & F= ? mV ? ? mV > FRx = ? m out V out = ? mV ? ? out ? in where FRx is the reaction force on the nozzle due to liquid jet at the nozzle outlet. Then, & m = ? AV & > FRx = ? mV = AVV = AV 2 & & or FRx = ? mV = ? m & & m m2 =? ?A ? A Therefore, the force exerted by a liquid jet of velocity V on this & stationary nozzle is proportional to V2, or alternatively, to m 2 . Liquid Nozzle V FR 6-3 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc.Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-21 A water jet of velocity V impinges on a plate moving toward the water jet with velocity ? V. The force required to move the plate towards the jet is to be determined in terms of F acting on the stationary plate. Assumptions 1 The flow is steady and incompressible. 2 The plate is vertical and the jet is normal to plate. 3 The pressure on both sides of the plate is atmospheric pressure (and thus its effect cancels out). Fiction during motion is negligible. 5 There is no acceleration of the plate. 6 The water splashes off the sides of the plate in a plane normal to the jet. 6 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Analysis We take the plate as the control volume. The relative velocity between the plate and the jet is V when the plate is st ationary, and 1. 5V when the plate is moving with a velocity ? V towards the plate. Then the momentum equation for steady one-dimensional flow in the horizontal direction reduces to r r r & & & & F= ? mV ? ? mV > ? FR = ? mi Vi > FR = miVi ? out ? in Stationary plate: ( Vi = V and Moving plate: ( Vi = 1. 5V and & mi = ? AVi = ? AV ) > FR = ? AV 2 = F & mi = ? AVi = ? A(1. 5V ) ) > FR = ? A(1. 5V ) 2 = 2. 25 ? AV 2 = 2. 25 F Therefore, the force required to hold the plate stationary against the oncoming water jet becomes 2. 25 times when the jet velocity becomes 1. 5 times. Discussion Note that when the plate is stationary, V is also the jet velocity. But if the plate moves toward the stream with velocity ? V, then the relative velocity is 1. 5V, and the amount of mass striking the plate (and falling off its sides) per unit time also increases by 50%. 1/2V VWaterjet 6-4 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-22 A 90 ° elbow deflects water upwards and discharges it to the atmosphere at a specified rate. The gage pressure at the inlet of the elbow and the anchoring force needed to hold the elbow in place are to be determined. v Assumptions 1 The flow is steady, frictionless, incompressible, and irrotational (so that the Bernoulli equation is applicable). The weight of the elbow and the water in it is negligible. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 4 The momentum-flux correction factor for each inlet and outlet is given to be ? = 1. 03. Properties We take the density of water to be 1000 kg/m3. Analysis (a) We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction) and the vertical coordinate by z.The continuity equation for this one-inlet one-outlet steady flow system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = ? AV , the mean inlet and outlet velocities of water are & & 25 kg/s m m = = = 3. 18 m/s 2 ? A ? (? D / 4) (1000 kg/m 3 )[? (0. 1 m) 2 / 4] Noting that V1 = V2 and P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as V1 = V 2 = V = P V12 P V2 1 + + z1 = 2 + 2 + z2 > P ? P2 = ? g ( z2 ? z1 ) > P , gage = ? g ( z2 ? z1 ) 1 1 ? g 2 g ? g 2 g Substituting, ? ? 1 kN 2 ? P , gage = (1000 kg/m3 )(9. 81 m/s 2 )(0. 35 m)? 1 ? 1000 kg ? /s2 ? = 3. 434 kN/m = 3. 434 kPa ? ? r r r & & (b) The momentum equation for steady one-dimensional flow is F= ? mV ? ? mV . We let the x- ? ? out ? in and z- components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive directions. We also use gage pressures to avoid dealing with the atmospheric pressure which acts on all surfaces. Then the momentum equations along the x and y axes become & & FRx + P1,gage A1 = 0 ? ?m(+V1 ) = ? ?mV & & FRz = ? m(+V 2 ) = ? mV z x FRz 2 35 cm Solving for FRx and FRz, and substituting the given values, & FRx = ? ?mV ? P1, gage A1 ? N = ? 1. 03(25 kg/s)(3. 18 m/s)? ? 1 kg ? m/s 2 ? = ? 109 N ? ? ? (3434 N/m 2 )[? (0. 1 m) 2 / 4] ? ? ? ? = 81. 9 N ? ? FRy FRx = tan -1 Water 25 kg/s FRx 1 ? 1N & FRy = ? mV = 1. 03(25 kg/s)(3. 18 m/s)? ? 1 kg ? m/s 2 ? and 2 2 FR = FRx + FRy = (? 109) 2 + 81. 9 2 = 136 N, ? = tan -1 81. 9 = ? 37 ° = 143 ° ? 109 Discussion Note that the magnitude of the anchoring force is 136 N, and its line of action makes 143 ° from the positive x direction. Also, a negative value for FRx indicates the assumed direction is wrong, and should be reversed. 6-5 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc.Limited distribution permitted only to teachers and educators for course preparatio n. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-23 An 180 ° elbow forces the flow to make a U-turn and discharges it to the atmosphere at a specified rate. The gage pressure at the inlet of the elbow and the anchoring force needed to hold the elbow in place are to be determined. v Assumptions 1 The flow is steady, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli equation is applicable). The weight of the elbow and the water in it is negligible. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 4 The momentumflux correction factor for each inlet and outlet is given to be ? = 1. 03. Properties We take the density of water to be 1000 kg/m3. Analysis (a) We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as b eing the positive direction) and the vertical coordinate by z.The continuity equation for this one-inlet one-outlet steady flow system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = ? AV , the mean inlet and outlet velocities of water are & & 25 kg/s m m = = = 3. 18 m/s 2 ? A ? (? D / 4) (1000 kg/m 3 )[? (0. 1 m) 2 / 4] Noting that V1 = V2 and P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as V1 = V 2 = V = P V12 P V2 1 + + z1 = 2 + 2 + z2 > P ? P2 = ? g ( z2 ? z1 ) > P , gage = ? g ( z2 ? z1 ) 1 1 ? g 2 g ? g 2 g Substituting, ? ? 1 kN 2 ? P , gage = (1000 kg/m3 )(9. 81 m/s2 )(0. 70 m)? 1 ? 1000 kg ? m/s2 ? 6. 867 kN/m = 6. 867 kPa ? ? r r r & & (b) The momentum equation for steady one-dimensional flow is F= ? mV ? ? mV . We let the x- ? ? out ? in and z- components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive directions. We also use gage pressures to avoid dealing with the atmospheric pressure which acts on all surfaces. Then the momentum equations along the x and z axes become & & & FRx + P1,gage A1 = ? m(? V 2 ) ? ? m(+V1 ) = ? 2 ? mV FRz = 0 Solving for FRx and substituting the given values, & FRx = ? 2 ? mV ? P1, gage A1 ? 1N = ? 2 ? 1. 03(25 kg/s)(3. 18 m/s)? 1 kg ? m/s 2 ? = ? 218 N ? ? ? (6867 N/m 2 )[? (0. 1 m) 2 / 4] ? ? 2 z x FRz Water 25 kg/s 35 cm and FR = FRx = – 218 N since the y-component of the anchoring force is zero. Therefore, the anchoring force has a magnitude of 218 N and it acts in the negative x direction. Discussion Note that a negative value for FRx indicates the assumed direction is wrong, and should be reversed. FRx 1 6-6 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.Chapter 6 Momentum Analysis of Flow Systems 6-24E A horizontal water j et strikes a vertical stationary plate normally at a specified velocity. For a given anchoring force needed to hold the plate in place, the flow rate of water is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water splatters off the sides of the plate in a plane normal to the jet. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure which is disregarded since it acts on the entire control surface. The vertical forces and momentum fluxes are not considered since they have no effect on the horizontal reaction force. 5 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Properties We take the density of water to be 62. 4 lbm/ft3. Analysis We take the plate as the control volume such that it contains the entire plate and cuts through the water jet and the support bar normally, and the direction of flow as the positive direction of x axis. The momentum equation for steady one-dimensional flow in the x (flow) direction reduces in this case o r r r & & & & F= ? mV ? ? mV > ? FRx = ? mV1 > FR = mV1 ? ? out ? in We note that the reaction force acts in the opposite direction to flow, and we should not forget the negative & sign for forces and velocities in the negative x-direction. Solving for m and substituting the given values, & m= FRx 350 lbf = V1 30 ft/s ? 32. 2 lbm ? ft/s 2 ? ? 1 lbf ? ? ? = 376 lbm/s ? ? Then the volume flow rate becomes V& = & m ? = 376 lbm/s 62. 4 lbm/ft 3 = 6. 02 ft 3 /s Therefore, the volume flow rate of water under stated assumptions must be 6. 02 ft3/s.Discussion In reality, some water will be scattered back, and this will add to the reaction force of water. The flow rate in that case will be less. m 1 FRx = 350 lbf Waterjet 6-7 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparati on. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-25 A reducing elbow deflects water upwards and discharges it to the atmosphere at a specified rate.The anchoring force needed to hold the elbow in place is to be determined. v Assumptions 1 The flow is steady, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The weight of the elbow and the water in it is considered. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 4 The momentumflux correction factor for each inlet and outlet is given to be ? = 1. 03. Properties We take the density of water to be 1000 kg/m3. Analysis The weight of the elbow and the water in it is W = mg = (50 kg)(9. 1 m/s 2 ) = 490. 5 N = 0. 4905 kN We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordina te by x (with the direction of flow as being the positive direction) and the vertical coordinate by z. The continuity equation for this one-inlet one-outlet steady flow system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = ? AV , the inlet and outlet velocities of water are & 30 kg/s m V1 = = = 2. 0 m/s ? A1 (1000 kg/m 3 )(0. 0150 m 2 ) & 30 kg/s m V2 = = = 12 m/s ? A2 (1000 kg/m 3 )(0. 025 m 2 ) Taking the center of the inlet cross section as the reference level (z1 = 0) and noting that P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as ? V 2 ? V12 ? ? V22 ? V12 ? P V12 P V2 1 ? ? ? + + z1 = 2 + 2 + z2 > P ? P2 = ? g ? 2 1 1 ? 2 g + z2 ? z1 ? > P , gage = ? g ? 2 g + z2 ? ?g 2 g ? g 2 g ? ? ? ? Substituting, ? (12 m/s) 2 ? (2 m/s) 2 ? 1 kN ? = 73. 9 kN/m 2 = 73. 9 kPa P , gage = (1000 kg/m3 )(9. 81 m/s 2 )? + 0. 4 1 2 ? 1000 kg ? m/s 2 ? 2(9. 81 m/s ) ? ? The momentum equation for steady one-dimensional flow is & & ? F = ? mV ? ? ? mV . We let the x- and out in r r r z- components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive directions. We also use gage pressures to avoid dealing with the atmospheric pressure which acts on all surfaces. Then the momentum equations along the x and z axes become & & & FRx + P1,gage A1 = ? mV 2 cos ? ? ? mV1 and FRz ? W = ? mV 2 sin ? 2 25 cm2 Solving for FRx and FRz, and substituting the given values, & FRx = ? m(V 2 cos ? ? V1 ) ? P1, gage A1 ? 1 kN = 1. 03(30 kg/s)[(12cos45 ° – 2) m/s]? ? 1000 kg ? m/s 2 ? ? (73. 9 kN/m 2 )(0. 0150 m 2 ) = ? 0. 908 kN ? ? ? Water 30 kg/s 45 ° FRz FRx 150 m2 W 1 ? ? 1 kN ? & FRz = ? mV 2 sin ? + W = 1. 03(30 kg/s)(12sin45 ° m/s)? ? 1000 kg ? m/s 2 ? + 0. 4905 kN = 0. 753 kN ? ? 0. 753 2 2 2 2 -1 FRz FR = FRx + FRz = (? 0. 908) + (0. 753) = 1. 18 kN, ? = tan = tan -1 = ? 39. 7 ° FRx ? 0. 908 Discussion Note that the magnitude of the anchoring force is 1. 18 kN, and its line of action makes –39. 7 ° from +x direction. Negative value for FRx indicates the assumed direction is wrong. 6-8 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-26 A reducing elbow deflects water upwards and discharges it to the atmosphere at a specified rate. The anchoring force needed to hold the elbow in place is to be determined. v Assumptions 1 The flow is steady, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The weight of the elbow and the water in it is considered. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. The momentumflux correction factor for each inlet and outlet is given to be ? = 1. 03. Properties We take the densi ty of water to be 1000 kg/m3. Analysis The weight of the elbow and the water in it is W = mg = (50 kg)(9. 81 m/s 2 ) = 490. 5 N = 0. 4905 kN We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction) and the vertical coordinate by z. The continuity equation for this one-inlet one-outlet steady flow system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = ?AV , the inlet and outlet velocities of water are & 30 kg/s m = = 2. 0 m/s V1 = ? A1 (1000 kg/m 3 )(0. 0150 m 2 ) & 30 kg/s m V2 = = = 12 m/s ? A2 (1000 kg/m 3 )(0. 0025 m 2 ) Taking the center of the inlet cross section as the reference level (z1 = 0) and noting that P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as ? V 2 ? V12 ? ? V22 ? V12 ? P V12 P V2 1 ? ? ? + + z1 = 2 + 2 + z2 > P ? P2 = ? g ? 2 1 1 ? 2 g + z2 ? z1 ? > P , gage = ? g ? 2 g + z2 ? ?g 2 g ? g 2 g ? ? ? ? or, P , gage = (1000 kg/m3 )(9. 81 m/s2 )? 1 ? ? ? (12 m/s)2 ? (2 m/s)2 2(9. 81 m/s ) ? 1 kN ? = 73. 9 kN/m 2 = 73. 9 kPa + 0. 4 1000 kg ? m/s 2 ? ? The momentum equation for steady one-dimensional flow is & & ? F = ? ?mV ? ? ? mV . We let the xout in r r r and y- components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive directions. We also use gage pressures to avoid dealing with the atmospheric pressure which acts on all surfaces. Then the momentum equations along the x and z axes become & & FRx + P1,gage A1 = ? mV 2 cos ? ? ? mV1 and & FRy ? W = ? mV 2 sin ? Solving for FRx and FRz, and substituting the given values, & FRx = ? m(V 2 cos ? V1 ) ? P1, gage A1 ? 1 kN = 1. 03(30 kg/s)[(12cos110 ° – 2) m/s]? ? 1000 kg ? m/s 2 ? FRz ? ? ? (73. 9 kN/m 2 )(0. 0150 m 2 ) = ? 1. 297 kN ? ? ? ? 1 kN ? + 0. 4905 kN = 0. 8389 kN & = ? mV 2 sin ? + W = 1. 03(30 kg/s)(12sin110 ° m/s)? 2 ? ? 1000 kg ? m/s ? ? 2 25 cm2 110 ° 2 2 FR = FRx + FRz = (? 1. 297) 2 + 0. 8389 2 = 1. 54 kN and FRz 0. 8389 = tan -1 = ? 32. 9 ° FRx ? 1. 297 Discussion Note that the magnitude of the anchoring force is 1. 54 kN, and its line of action makes –32. 9 ° from +x direction. Negative value for FRx indicates assumed direction is wrong, and should be reversed. ? = tan -1 FRz FRx Water 1 30 kg/s 50 m2 W 6-9 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-27 Water accelerated by a nozzle strikes the back surface of a cart moving horizontally at a constant velocity. The braking force and the power wasted by the brakes are to be determined. . Assumptions 1 The flow is steady and incompressible. 2 The water splatters off the sides of the plate in all direction s in the plane of the back surface. The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure which is disregarded since it acts on all surfaces. 4 Fiction during motion is negligible. 5 There is no acceleration of the cart. 7 The motions of the water jet and the cart are horizontal. 6 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Analysis We take the cart as the control volume, and the direction of flow as the positive direction of x axis. The relative velocity between the cart and the jet is V r = V jet ?Vcart = 15 ? 10 = 10 m/s 15 m/s 5 m/s Therefore, we can assume the cart to be stationary and the jet to move Waterjet with a velocity of 10 m/s. The momentum equation for steady onedimensional flow in the x (flow) direction reduces in this case to r r r & & & & F= ? mV ? ? mV > FRx = ? mi Vi > Fbrake = ? mV r FRx ? ? out ? in We note that the brake force acts in the opposite direction to flow, and we should not forget the negative sign for forces and velocities in the negative x-direction. Substituting the given values, ? 1N & Fbrake = ? mV r = ? (25 kg/s)(+10 m/s)? ? 1 kg ? m/s 2 ? ? ? = ? 250 N ? ?The negative sign indicates that the braking force acts in the opposite direction to motion, as expected. Noting that work is force times distance and the distance traveled by the cart per unit time is the cart velocity, the power wasted by the brakes is 1 kW ? ? & W = FbrakeV cart = (250 N)(5 m/s)? ? = 1. 25 kW ? 1000 N ? m/s ? Discussion Note that the power wasted is equivalent to the maximum power that can be generated as the cart velocity is maintained constant. 6-10 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Fl ow Systems 6-28 Water accelerated by a nozzle strikes the back surface of a cart moving horizontally. The acceleration of the cart if the brakes fail is to be determined. Analysis The braking force was determined in previous problem to be 250 N. When the brakes fail, this force will propel the cart forward, and the accelerating will be a= F 250 N ? 1 kg ? m/s 2 ? = m cart 300 kg ? 1N ? ? ? = 0. 833 m/s 2 ? ? Discussion This is the acceleration at the moment the brakes fail.The acceleration will decrease as the relative velocity between the water jet and the cart (and thus the force) decreases. 5 m/s 15 m/s 300 kg Waterjet FRx 6-11 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-29E A water jet hits a stationary splitter, such that half of the flow is diverted up ward at 45 °, and the other half is directed down.The force required to hold the splitter in place is to be determined. vEES Assumptions 1 The flow is steady and incompressible. 2 The water jet is exposed to the atmosphere, and thus the pressure of the water jet before and after the split is the atmospheric pressure which is disregarded since it acts on all surfaces. 3 The gravitational effects are disregarded. 4 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Properties We take the density of water to be 62. 4 lbm/ft3. Analysis The mass flow rate of water jet is & & m = ? V = (62. lbm/ft 3 )(100 ft 3 /s) = 6240 lbm/s We take the splitting section of water jet, including the splitter as the control volume, and designate the entrance by 1 and the outlet of either arm by 2 (both arms have the same velocity and mass flow rate). We also designate the horizontal coordinate by x with the direction of flow as being the positive direction and the vertical coordinate by z. r r r & & The momentum equation for steady one-dimensional flow is F= ? mV ? ? mV . We let ? ? out ? in the x- and y- components of the anchoring force of the splitter be FRx and FRz, and assume them to be in the & & positive directions.Noting that V2 = V1 = V and m 2 = 1 m , the momentum equations along the x and z 2 axes become & & & FRx = 2( 1 m)V 2 cos ? ? mV1 = mV (cos ? ? 1) 2 & & FRz = 1 m(+V 2 sin ? ) + 1 m(? V 2 sin ? ) ? 0 = 0 2 2 Substituting the given values, 1 lbf ? ? FRx = (6240 lbm/s)(20 ft/s)(cos45 ° – 1)? ? = ? 1135 lbf 32. 2 lbm ? ft/s 2 ? ? FRz = 0 The negative value for FRx indicates the assumed direction is wrong, and should be reversed. Therefore, a force of 1135 lbf must be applied to the splitter in the opposite direction to flow to hold it in place. No holding force is necessary in the vertical direction.This can also be concluded from the symmetry. Discussion In reality, the gravitational effects will cau se the upper stream to slow down and the lower stream to speed up after the split. But for short distances, these effects are indeed negligible. 20 ft/s 100 ft/s FRz 45 ° 45 ° FRx 6-12 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-30E Problem 6-29E is reconsidered.The effect of splitter angle on the force exerted on the splitter as the half splitter angle varies from 0 to 180 ° in increments of 10 ° is to be investigated. g=32. 2 â€Å"ft/s2† rho=62. 4 â€Å"lbm/ft3† V_dot=100 â€Å"ft3/s† V=20 â€Å"ft/s† m_dot=rho*V_dot F_R=-m_dot*V*(cos(theta)-1)/g â€Å"lbf† ?,  ° 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 8000 7000 6000 5000 & m , lbm/s 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 624 0 6240 6240 6240 6240 6240 6240 6240 6240 FR, lbf 0 59 234 519 907 1384 1938 2550 3203 3876 4549 5201 5814 6367 6845 7232 7518 7693 7752 FR, lbf 000 3000 2000 1000 0 0 20 40 60 80 100 120 140 160 180 ?,  ° 6-13 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-31 A horizontal water jet impinges normally upon a vertical plate which is held on a frictionless track and is initially stationary. The initial acceleration of the plate, the time it takes to reach a certain velocity, and the velocity at a given time are to be determined.Assumptions 1 The flow is steady and incompressible. 2 The water always splatters in the plane of the retreating plate. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the a tmospheric pressure which is disregarded since it acts on all surfaces. 4 The tract is nearly frictionless, and thus fiction during motion is negligible. 5 The motions of the water jet and the cart are horizontal. 6 The velocity of the jet relative to the plate remains constant, Vr = Vjet = V. 7 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is egligible, ? ? 1. Properties We take the density of water to be 1000 kg/m3. Analysis (a) We take the vertical plate on the frictionless track as the control volume, and the direction of flow as the positive direction of x axis. The mass flow rate of water in the jet is & m = ? VA = (1000 kg/m 3 )(18 m/s)[? (0. 05 m) 2 / 4] = 35. 34 kg/s The momentum equation for steady one-dimensional flow in the x (flow) direction reduces in this case to r r r & & & & F= ? mV ? ? mV > FRx = ? mi Vi > FRx = ? mV ? ? out ? in where FRx is the reaction force required to hold the plate in place.When the plate is released, a n equal and opposite impulse force acts on the plate, which is determined to ? 1N & Fplate = ? FRx = mV = (35. 34 kg/s)(18 m/s)? ? 1 kg ? m/s 2 ? ? ? = 636 N ? ? Then the initial acceleration of the plate becomes a= Fplate m plate = 636 N ? 1 kg ? m/s 2 ? 1000 kg ? 1 N ? ? ? = 0. 636 m/s 2 ? ? 18 m/s 1000 kg Waterjet Frictionless track This acceleration will remain constant during motion since the force acting on the plate remains constant. (b) Noting that a = dV/dt = ? V/? t since the acceleration a is constant, the time it takes for the plate to reach a velocity of 9 m/s is ? t = ? V plate a = (9 ? ) m/s 0. 636 m/s 2 FRx = 14. 2 s (c) Noting that a = dV/dt and thus dV = adt and that the acceleration a is constant, the plate velocity in 20 s becomes V plate = V0, plate + a? t = 0 + (0. 636 m/s 2 )(20 s) = 12. 7 m/s Discussion The assumption that the relative velocity between the water jet and the plate remains constant is valid only for the initial moments of motion when the plate velocity is low unless the water jet is moving with the plate at the same velocity as the plate. 6-14 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-32 A 90 ° reducer elbow deflects water downwards into a smaller diameter pipe. The resultant force exerted on the reducer by water is to be determined. Assumptions 1 The flow is steady, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The weight of the elbow and the water in it is disregarded since the gravitational effects are negligible. 3 The momentum-flux correction factor for each inlet and outlet is given to be ? 1. 04. Properties We take the density of water to be 1000 kg/m3. Analysis We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction) and the vertical coordinate by z. The continuity equation for this one-inlet one-outlet steady flow system is & & & & m1 = m 2 = m = 353. 4 kg/s. Noting that m = ? AV , the mass flow rate of water and its outlet velocity are 2 & m = ? V1 A1 = ? V1 (? D1 / 4) = (1000 kg/m 3 )(5 m/s)[? (0. 3 m) 2 / 4] = 353. 4 kg/s & & 353. kg/s m m = = = 20 m/s 2 ? A2 D 2 / 4 (1000 kg/m 3 )[? (0. 15 m) 2 / 4] The Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as V2 = P V12 P V2 1 + + z1 = 2 + 2 + z2 ? g 2 g ? g 2 g > ? V 2 ? V22 ? ? P2 = P + ? g ? 1 1 ? 2 g + z1 ? z2 ? ? ? Substituting, the gage pressure at the outlet becomes ? (5 m/s)2 ? (20 m/s)2 1 kPa ? 1 kN ? P2 = (300 kPa) + (1000 kg/m 3 )(9. 81 m/s 2 )? + 0. 5 = 117. 4 kPa 2 ? 1000 kg ? m/s 2 1 kN/m 2 ? 2(9. 81 m/s ) ? ? The momentum equation for steady one-dimensional flow is & & ? F = ? ?mV ? ? ? mV . We let the xout in r r and z- components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive directions. Then the momentum equations along the x and z axes become & FRx + P1,gage A1 = 0 ? ? mV1 & FRz ? P2,gage A2 = ? m(? V 2 ) ? 0 Note that we should not forget the negative sign for forces and velocities in the negative x or z direction. Solving for FRx and FRz, and substituting the given values, ? 1 kN & FRx = ? ?mV1 ? P1, gage A1 = ? 1. 04(353. 4 kg/s)(5 m/s)? ? 1000 kg ? m/s 2 ? ? ? (0. 3 m) 2 ? ? (300 kN/m 2 ) = ? 23. 0 kN ? 4 ? ? ? (0. 15 m) 2 ? + (117. 4 kN/m 2 ) = ? 5. 28 kN ? ? FRz ? 1 kN & FRz = ? ? mV 2 + P2, gage A1 = ? 1. 04(353. 4 kg/s)(20 m/s)? ? 1000 kg ? m/s 2 ? and 2 2 FR = FRx + FRz = (? 23. 0) 2 + (? 5. 28) 2 = 23. 6 kN FRx 30 cm Water 5 m/s ? = tan -1 FRz ? 5. 28 = tan -1 = 12. 9 ° FRx ? 23. 0 Discussion The magnitude of the anchoring force is 23. 6 kN, and its line of action makes 12. 9 ° from +x direction. Negative values for FRx and FRy indicate that the assumed directions are wrong, and should be reversed. 15 cm 6-15 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-33 A wind turbine with a given span diameter and efficiency is subjected to steady winds. The power generated and the horizontal force on the supporting mast of the turbine are to be determined. vEES Assumptions 1 The wind flow is steady and incompressible. 2 The efficiency of the turbine-generator is independent of wind speed. 3 The frictional effects are negligible, and thus none of the incoming kinetic energy is converted to thermal energy. Wind flow is uniform and thus the momentum-flux correction factor is nearly unity, ? ? 1. Properties The density of air is given to be 1. 25 kg/m3. Analysis (a) The power potential of the wind is its kinetic energy, & which is V2/2 per unit mass, and mV 2 / 2 for a given mass flow rate: ? 1 m/s ? V1 = (25 km/h)? ? = 6. 94 m/s ? 3. 6 km/h ? & m = ? 1V1 A1 = ? 1V1 Wind V1 1 2 D V2 ?D 2 4 2 = (1. 25 kg/m 3 )(6. 94 m/s) ? (90 m) 2 4 2 = 55,200 kg/s V (6. 94 m/s) & & & W max = mke1 = m 1 = (55,200 kg/s) 2 2 ? 1 kN ? ? 1000 kg ? m/s 2 ? 1 kW ? 1 kN ? m/s ? = 1330 kW ? ? FR Then the actual power produced becomes & Wact = ? wind turbineW max = (0. 32)(1330 kW) = 426 kW (b) The frictional effects are assumed to be negligible, and thus the portion of incoming kinetic energy not converted to electric power leaves the wind turbine as outgoing kinetic energy. Therefore, V2 V2 & & & & mke 2 = mke1 (1 ? ? wind turbine ) > m 2 = m 1 (1 ? ? wind turbine ) 2 2 or V 2 = V1 1 ? ? wind turbine = (6. 94 m/s) 1 – 0. 32 = 5. 72 m/s We choose the control volume around the wind turbine such that the wind is norm al to the control surface at the inlet and the outlet, and the entire control surface is at the atmospheric pressure.The momentum r r r & & equation for steady one-dimensional flow is F= ? mV ? ? mV . Writing it along the x-direction ? ? out ? in (without forgetting the negative sign for forces and velocities in the negative x-direction) and assuming the flow velocity through the turbine to be equal to the wind velocity give ? 1 kN & & & FR = mV 2 ? mV1 = m(V 2 ? V1 ) = (55,200 kg/s)(5. 72 – 6. 94 m/s)? ? 1000 kg ? m/s 2 ? ? ? = ? 67. 3 kN ? ? The negative sign indicates that the reaction force acts in the negative x direction, as expected.Discussion This force acts on top of the tower where the wind turbine is installed, and the bending moment it generates at the bottom of the tower is obtained by multiplying this force by the tower height. 6-16 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for c ourse preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-34E A horizontal water jet strikes a curved plate, which deflects the water back to its original direction.The force required to hold the plate against the water stream is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure, which is disregarded since it acts on all surfaces. 3 Friction between the plate and the surface it is on is negligible (or the friction force can be included in the required force to hold the plate). 4 There is no splashing of water or the deformation of the jet, and the reversed jet leaves horizontally at the same velocity and flow rate. Jet flow is nearly uniform and thus the momentum-flux correction factor is nearly unity, ? ? 1. Properties We take the density of w ater to be 62. 4 lbm/ft3. Analysis We take the plate together with the curved water jet as the control volume, and designate the jet inlet by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of incoming flow as being the positive direction). The continuity equation for this one-inlet one-outlet steady & & & flow system is m1 = m 2 = m where & m = ? VA = ? V [? D 2 / 4] = (62. 4 lbm/ft 3 )(140 ft/s)[? (3 / 12 ft) 2 / 4] = 428. lbm/s r r r & & The momentum equation for steady one-dimensional flow is F= ? mV ? ? mV . Letting the ? ? out ? in reaction force to hold the plate be FRx and assuming it to be in the positive direction, the momentum equation along the x axis becomes & & & FRx = m(? V 2 ) ? m(+V1 ) = ? 2mV Substituting, 1 lbf ? ? FRx = ? 2(428. 8 lbm/s)(140 ft/s)? ? = ? 3729 lbf 2 ? 32. 2 lbm ? ft/s ? Therefore, a force of 3729 lbm must be applied on the plate in the negative x direction to hold it in place. Discussion Note that a nega tive value for FRx indicates the assumed direction is wrong (as expected), and should be reversed.Also, there is no need for an analysis in the vertical direction since the fluid streams are horizontal. 2 140 ft/s Waterjet FRx 1 140 ft/s 3 in 6-17 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-35E A horizontal water jet strikes a bent plate, which deflects the water by 135 ° from its original direction. The force required to hold the plate against the water stream is to be determined.Assumptions 1 The flow is steady and incompressible. 2 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure, which is disregarded since it acts on all surfaces. 3 Frictional and gravitational effec ts are negligible. 4 There is no splattering of water or the deformation of the jet, and the reversed jet leaves horizontally at the same velocity and flow rate. 5 Jet flow is nearly uniform and thus the momentum-flux correction factor is nearly unity, ? ? 1. Properties We take the density of water to be 62. 4 lbm/ft3.Analysis We take the plate together with the curved water jet as the control volume, and designate the jet inlet by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of incoming flow as being the positive direction), and the vertical coordinate by z. The continuity equation for & & & this one-inlet one-outlet steady flow system is m1 = m 2 = m where & m = ? VA = ? V [? D 2 / 4] = (62. 4 lbm/ft 3 )(140 ft/s)[? (3 / 12 ft) 2 / 4] = 428. 8 lbm/s r r r & & The momentum equation for steady one-dimensional flow is F= ? mV ? ? mV . We let the x- ? ? out ? in nd z- components of the anchoring force of the plate be FRx and FRz, and assu me them to be in the positive directions. Then the momentum equations along the x and y axes become & & & FRx = m(? V 2 ) cos 45 ° ? m(+V1 ) = ? mV (1 + cos 45 °) & (+V 2 ) sin 45 ° = mV sin 45 ° & FRz = m Substituting the given values, 1 lbf ? ? FRx = ? 2(428. 8 lbm/s)(140 ft/s)(1 + cos45 °)? 2 ? ? 32. 2 lbm ? ft/s ? = ? 6365 lbf 1 lbf ? ? FRz = (428. 8 lbm/s)(140 ft/s)sin45 °? = 1318 lbf 2 ? ? 32. 2 lbm ? ft/s ? 2 140 ft/s Waterjet 135 ° FRz FRx 3 in 1 and 2 2 FR = FRx + FRz = (? 6365) 2 + 1318 2 = 6500 lbf , ? = tan -1 FRy FRx = tan -1 1318 = ? 1. 7 ° = 168. 3 ° ? 6365 Discussion Note that the magnitude of the anchoring force is 6500 lbf, and its line of action makes 168. 3 ° from the positive x direction. Also, a negative value for FRx indicates the assumed direction is wrong, and should be reversed. 6-18 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-36 Firemen are holding a nozzle at the end of a hose while trying to extinguish a fire.The average water outlet velocity and the resistance force required of the firemen to hold the nozzle are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water jet is exposed to the atmosphere, and thus the pressure of the water jet is the atmospheric pressure, which is disregarded since it acts on all surfaces. 3 Gravitational effects and vertical forces are disregarded since the horizontal resistance force is to be determined. 5 Jet flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Properties We take the density of water to be 1000 kg/m3.Analysis (a) We take the nozzle and the horizontal portion of the hose as the system such that water enters the control volume vertically and outlets horizontally (thi s way the pressure force and the momentum flux at the inlet are in the vertical direction, with no contribution to the force balance in the horizontal direction), and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction). The average outlet velocity and the mass flow rate of water are determined from V= V& A = V& ? D / 4 2 = 5 m 3 /min ? (0. 06 m) 2 / 4 1768 m/min = 29. 5 m/s & m = ? V& = (1000 kg/m 3 )(5 m 3 /min) = 5000 kg/min = 83. 3 kg/s (b) The momentum equation for steady one-dimensional flow is & & ? F = ? ?mV ? ? ? mV . We let out in r r r horizontal force applied by the firemen to the nozzle to hold it be FRx, and assume it to be in the positive x direction. Then the momentum equation along the x direction gives ? ? 1N ? = 2457 N & & FRx = mVe ? 0 = mV = (83. 3 kg/s)(29. 5 m/s)? ? 1kg ? m/s 2 ? ? ? Therefore, the firemen must be able to resist a force of 2457 N to hold t he nozzle in place. Discussion The force of 2457 N is equivalent to the weight of about 250 kg.That is, holding the nozzle requires the strength of holding a weight of 250 kg, which cannot be done by a single person. This demonstrates why several firemen are used to hold a hose with a high flow rate. FRz FRx 5 m3/min 6-19 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-37 A horizontal jet of water with a given velocity strikes a flat plate that is moving in the same direction at a specified velocity.The force that the water stream exerts against the plate is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water splatters in all directions in the plane of the plate. 3 The water jet is exposed to the atmosphere, and thus the pressure o f the water jet and the splattered water is the atmospheric pressure, which is disregarded since it acts on all surfaces. 4 The vertical forces and momentum fluxes are not considered since they have no effect on the horizontal force exerted on the plate. 5 The velocity of the plate, and the velocity of the water jet relative to the plate, are constant. Jet flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Properties We take the density of water to be 1000 kg/m3. Analysis We take the plate as the control volume, and the flow direction as the positive direction of x axis. The mass flow rate of water in the jet is & m = ? V jet A = ? V jet 10 m/s 30 m/s FRx 5 cm Waterjet ?D 4 2 = (1000 kg/m 3 )(30 m/s) ? (0. 05 m) 2 4 = 58. 9 kg/s The relative velocity between the plate and the jet is V r = V jet ? V plate = 30 ? 10 = 20 m/s Therefore, we can assume the plate to be stationary and the jet to move with a velocity of 20 m/s.The r r r & & F= ? mV ? ? mV . We let the horizontal momentum equation for steady one-dimensional flow is ? ? out ? in reaction force applied to the plate in the negative x direction to counteract the impulse of the water jet be FRx. Then the momentum equation along the x direction gives ? ? 1N ? & & ? FRx = 0 ? mVi > FRx = mV r = (58. 9 kg/s)(20 m/s)? ? 1kg ? m/s 2 ? = 1178 N ? ? Therefore, the water jet applies a force of 1178 N on the plate in the direction of motion, and an equal and opposite force must be applied on the plate if its velocity is to remain constant.Discussion Note that we used the relative velocity in the determination of the mass flow rate of water in the momentum analysis since water will enter the control volume at this rate. (In the limiting case of the plate and the water jet moving at the same velocity, the mass flow rate of water relative to the plate will be zero since no water will be able to strike the plate). 6-20 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Comp anies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.Chapter 6 Momentum Analysis of Flow Systems 6-38 Problem 6-37 is reconsidered. The effect of the plate velocity on the force exerted on the plate as the plate velocity varies from 0 to 30 m/s in increments of 3 m/s is to be investigated. rho=1000 â€Å"kg/m3† D=0. 05 â€Å"m† V_jet=30 â€Å"m/s† Ac=pi*D^2/4 V_r=V_jet-V_plate m_dot=rho*Ac*V_jet F_R=m_dot*V_r â€Å"N† Vplate, m/s 0 3 6 9 12 15 18 21 24 27 30 Vr, m/s 30 27 24 21 18 15 12 9 6 3 0 FR, N 1767 1590 1414 1237 1060 883. 6 706. 9 530. 1 353. 4 176. 7 0 1800 1600 1400 1200 1000 FR, N 800 600 400 200 0 0 5 10 15 20 25 30 Vplate, m/s 6-21PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you ar e using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-39E A fan moves air at sea level at a specified rate. The force required to hold the fan and the minimum power input required for the fan are to be determined. v Assumptions 1 The flow of air is steady and incompressible. 2 Standard atmospheric conditions exist so that the pressure at sea level is 1 atm. Air leaves the fan at a uniform velocity at atmospheric pressure. 4 Air approaches the fan through a large area at atmospheric pressure with negligible velocity. 5 The frictional effects are negligible, and thus the entire mechanical power input is converted to kinetic energy of air (no conversion to thermal energy through frictional effects). 6 Wind flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Properties The gas constant of air is R = 0. 3704 psi? ft3/lbm? R. The standard atmospheric pressure at sea level is 1 atm = 14. 7 psi.Analysis (a) We take the control volume to be a horizontal hyperbolic cylinder bounded by streamlines on the sides with air entering through the large cross-section (section 1) and the fan located at the narrow cross-section at the end (section 2), and let its centerline be the x axis. The density, mass flow rate, and discharge velocity of air are 14. 7 psi P ? = = = 0. 0749 lbm/ft 3 RT (0. 3704 psi ? ft 3 /lbm ? R)(530 R) & m = ? V& = (0. 0749 lbm/ft 3 )(2000 ft 3/min) = 149. 8 lbm/min = 2. 50 lbm/s V2 = V& A2 = V& 2 ? D 2 / 4 = 2000 ft 3 /min ? (2 ft) 2 / 4 = 636. 6 ft/min = 10. ft/s & & ? F = ? ?mV ? ? ? mV . Letting the out in The momentum equation for steady one-dimensional flow is r r r reaction force to hold the fan be FRx and assuming it to be in the positive x (i. e. , the flow) direction, the momentum equation along the x axis becomes 1 lbf ? ? & & FRx = m(V 2 ) ? 0 = mV = (2. 50 lbm/s)(10. 6 ft/s)? ? = 0. 82 lbf 2 ? 32. 2 lbm ? ft/s ? Therefore, a force of 0. 82 lbf must be applied (through frict ion at the base, for example) to prevent the fan from moving in the horizontal direction under the influence of this force. (b) Noting that P1 = P2 = Patm and V1 ? , the energy equation for the selected control volume reduces to ?P V2 ? ?P V2 ? & & & & & m? 1 + 1 + gz1 ? + W pump, u = m? 2 + 2 + gz 2 ? + W turbine + E mech,loss > ? ? ? ? 2 2 ? ? ? ? Substituting, V & & Wfan, u = m 2 2 2 V2 (10. 6 ft/s) 2 ? 1 lbf 1W ? & & Wfan,u = m 2 = (2. 50 lbm/s) ? ? = 5. 91 W 2 2 2 ? 32. 2 lbm ? ft/s 0. 73756 lbf ? ft/s ? Therefore, a useful mechanical power of 5. 91 W must be supplied to 2000 cfm air. This is the minimum required power input required for the fan. Discussion The actual power input to the fan will be larger than 5. 1 W because of the fan inefficiency in converting mechanical power to kinetic energy. Fan 1 2 24 in 6-22 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-40 A helicopter hovers at sea level while being loaded. The volumetric air flow rate and the required power input during unloaded hover, and the rpm and the required power input during loaded hover are to be determined. Assumptions 1 The flow of air is steady and incompressible. 2 Air leaves the blades at a uniform velocity at atmospheric pressure. 3 Air approaches the blades from the top through a large area at atmospheric pressure with negligible velocity. 4 The frictional effects are negligible, and thus the entire mechanical power input is converted to kinetic energy of air (no conversion to thermal energy through frictional effects). 5 The change in air pressure with elevation is negligible because of the low density of air. 6 There is no acceleration of the helicopter, and thus the lift generated is equal to the total weight. Air flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Properties The density of air is given to be 1. 18 kg/m3. Analysis (a) We take the control volume to be a vertical hyperbolic cylinder bounded by streamlines on the sides with air entering through the large cross-section (section 1) at the top and the fan located at the narrow cross-section at the bottom (section 2), and let its centerline be the z axis with upwards being the positive direction. r r r & & F= ? mV ? ? mV . Noting The momentum equation for steady one-dimensional flow is ? out ? in that the only force acting on the control volume is the total weight W and it acts in the negative z direction, the momentum equation along the z axis gives W & & ? W = m(? V 2 ) ? 0 > W = mV 2 = ( ? AV 2 )V 2 = ? AV 22 > V2 = ? A 1 where A is the blade span area, 15 m A = ? D / 4 = ? (15 m) / 4 = 176. 7 m 2 2 2 Then the discharge velocity, volume flow rate, and the mass flow rate of air in the unloaded mode become V 2,unloaded = m unloaded g = ? A (10,000 kg)(9. 81 m/s 2 ) (1. 18 kg/m 3 )(176. 7 m 2 ) = 21. 7 m/s Sea level 2 V&unloaded = AV 2,unloaded = (176. 7 m 2 )(21. m/s) = 3834 m 3 /s & munloaded = ? V&unloaded = (1. 18 kg/m 3 )(3834 m 3/s) = 4524 kg/s Load 15,000 kg Noting that P1 = P2 = Patm, V1 ? 0, the elevation effects are negligible, and the frictional effects are disregarded, the energy equation for the selected control volume reduces to ? P V2 ? ?P V2 ? V2 & & & & & & & m? 1 + 1 + gz1 ? + W pump, u = m? 2 + 2 + gz 2 ? + W turbine + E mech,loss > Wfan, u = m 2 ? ? ? ? 2 2 2 ? ? ? ? Substituting, ? V2 ? 1 kW ? (21. 7 m/s) 2 ? 1 kN & ? ? & = (4524 kg/s) W unloaded fan,u = ? m 2 ? ? = 1065 kW 2 ? 1 kN ? m/s ? 1000 kg ? m/s ? ? 2 ? 2 ? ? ? ? nloaded (b) We now repeat the calculations for the loaded helicopter, whose mass is 10,000+15,000 = 25,000 kg: V 2,loaded = m loaded g = ? A (25,000 kg)(9. 81 m/s 2 ) (1. 18 kg/m 3 )(176. 7 m 2 ) = 34. 3 m/s & mloaded = ? V&loaded = ? AV2, loaded = (1. 18 kg/m 3 )(176. 7 m 2 )(34. 3 m/s) = 7152 kg/s ? V2 ? (34. 3 m/s)2 & & = (7152 kg/s) Wloaded fan,u = ? m 2 ? ? 2 ? 2 ? ?loaded ? 1 kW ? 1 kN ? ? ? 1000 kg ? m/s 2 1 kN ? m/s ? = 4207 kW ? ? 6-23 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems Noting that the average flow velocity is proportional to the overhead blade rotational velocity, the rpm of the loaded helicopter blades becomes & V 2 = kn > V 2,loaded V 2, unloaded = & n loaded & n unloaded > & n loaded = V 2,loaded V 2, unloaded & n unloaded = 34. 3 (400 rpm) = 632 rpm 21. 7 Discussion The actual power input to the helicopter blades will be considerably larger than the calculated power input because of the fan inefficiency in converting mechanical power to kinetic energy. -24 PROPRIETARY MATERIAL.  © 200 6 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-41 A helicopter hovers on top of a high mountain where the air density considerably lower than that at sea level. The blade rotational velocity to hover at the higher altitude and the percent increase in the required power input to hover at high altitude relative to that at sea level are to be determined. Assumptions 1 The flow of air is steady and incompressible. 2 The air leaves the blades at a uniform velocity at atmospheric pressure. 3 Air approaches the blades from the top through a large area at atmospheric pressure with negligible velocity. 4 The frictional effects are negligible, and thus the entire mechanical power input is converted to kinetic energy of air. 5 The change in air pressure with elevation while hovering at a given locatio n is negligible because of the low density of air. 6 There is no acceleration of the helicopter, and thus the lift generated is equal to the total weight. Air flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Properties The density of air is given to be 1. 18 kg/m3 at sea level, and 0. 79 kg/m3 on top of the mountain. Analysis (a) We take the control volume to be a vertical hyperbolic cylinder bounded by streamlines on the sides with air entering through the large cross-section (section 1) at the top and the fan located at the narrow cross-section at the bottom (section 2), and let its centerline be the z axis with upwards being the positive direction. r r & & F= ? mV ? ? mV . Noting The momentum equation for steady one-dimensional flow is ? ? out ? in that the only force acting on the control volume is the total weight W and it acts in the negative z direction, the momentum equation along the z axis gives W & & ? W = m(? V 2 ) ? 0 > W = mV 2 = ( ? AV 2 )V 2 = ? AV 22 > V2 = ? A where A is the blade span area. Then for a given weight W, the ratio of discharge velocities becomes V 2,mountain V 2,sea = W / ? mountain A W / ? sea A = ? sea ? mountain = 1. 18 kg/m 3 0. 79 kg/m 3 = 1. 222Noting that the average flow velocity is proportional to the overhead blade rotational velocity, the rpm of the helicopter blades on top of the mountain becomes & n = kV 2 > & n mountain V 2, mountain = & n sea V 2,sea > & n mountain = V 2, mountain V 2,sea & nsea = 1. 222(400 rpm) = 489 rpm Noting that P1 = P2 = Patm, V1 ? 0, the elevation effect are negligible, and the frictional effects are disregarded, the energy equation for the selected control volume reduces to ? P V2 ? ?P V2 ? V2 & & & & & & & m? 1 + 1 + gz1 ? + W pump, u = m? 2 + 2 + gz 2 ? W turbine + E mech,loss > Wfan, u = m 2 ? ? ? ? 1 2 2 2 ? ? ? ? or V2 V2 V3 & & Wfan,u = m 2 = ? AV2 2 = ? A 2 = 2 2 2 1 2 ?A? ? ? W ? ? = ? ? ? A ? 3 1 2 ?A? ? ?W ? ? ? ? ? A ? 1 . 5 = W 1 . 5 2 ? A 15 m Then the ratio of the required power input on top of the mountain to that at sea level becomes & Wmountain fan,u 0. 5W 1. 5 / ? mountain A = & Wsea fan,u 0. 5W 1. 5 / ? sea A 2 ? mountain ?sea = 1. 18 kg/m3 = 1. 222 0. 79 kg/m3 Sea level Load 15,000 kg Therefore, the required power input will increase by 22. 2% on top of the mountain relative to the sea level.Discussion Note that both the rpm and the required power input to the helicopter are inversely proportional to the square root of air density. Therefore, more power is required at higher elevations for the helicopter to operate because air is less dense, and more air must be forced by the blades into the downdraft. 6-25 PROPRIETARY MATERIAL.  © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.Chapter 6 Momentum Analysis of Flow Systems 6-42 The flow rate in a channel is controlled by a sluice gate by raising or lowering a vertical plate. A relation for the force acting on a sluice gate of width w for steady and uniform flow is to be developed. Assumptions 1 The flow is steady, incompressible, frictionless, and uniform (and thus the Bernoulli equation is applicable. ) 2 Wall shear forces at surfaces are negligible. 3 The channel is exposed to the atmosphere, and thus the pressure at free surfaces is the atmospheric pressure. 4 The flow is horizontal. Water flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Analysis We take point 1 at the free surface of the upstream flow before the gate and point 2 at the free surface of the downstream flow after the gate. We also take the bottom surface of the channel as the reference level so that the elevations of points 1 and 2 are y1 and y2, respectively. The application of the Bernoulli equation between points 1 and 2 gives P1 V12 P V2 + + y1 = 2 + 2 + y 2 ? g 2 g ? g 2 g > V 22 ? V12 = 2 g( y1 ? y 2 ) (1)The flow is assumed to be incompressible and thus the density is constant. Then the conservation of mass relation for this single stream steady flow device can be expressed as V&1 = V&2 = V& > A1V1 = A2V 2 = V& > V1 = V& A1 = V& wy1 and V2 = V& A2 = V& wy 2 (2) Substituting into Eq. (1), ? V& ? ? wy ? 2 ? ? V& ? 2 g ( y1 ? y 2 ) & ? ? ? ? wy ? = 2 g ( y1 ? y 2 ) > V = w 1 / y 2 ? 1 / y 2 ? ? 1? 2 1 2 2 2 g ( y1 ? y 2 ) > V& = wy 2 2 2 1 ? y 2 / y1 (3) Substituting Eq. (3) into Eqs. (2) gives the following relations for velocities, V1 = y2 y1 2 g ( y1 ? y 2 ) 1? y2 / 2 y1 and V2 = 2 g ( y1 ? y 2 ) 2 2 1 ? y 2 / y1 (4) We choose the control volume as the water body surrounded by the vertical cross-sections of the upstream and downstream flows, free surfaces of water, the inner surface of the sluice gate, and the bottom surface of r r r & & F= ? mV ? ? mV . The the channel. The momentum equation for steady one-dimensional flow is ? ? out ? in force acting on the sluice gate FRx is horizontal

What does Eliza consider to be her real education

The play is an adaptation of the Greek myth of Pygmalion who fell in love with a statue as it was more real in the understanding of its own composition than the actual women he had observed and grown despondent to. It is a work that closely follows the relationship between society and linguistics, wherein the women is real, but has yet to have her manners sculptured. In particular, it highlights the role of convention and articulation in relation to identities, depicting this through the subject of Eliza.In this paper the author will be addressing the subject of the play and its central character, whilst examining the effects that learning the speech of, what was considered, correct English had on her. Main Body When first completing the text, it is clear that there is an irony in the play that brings forth the now famed social and political points to the surface. However, one may be forgiven for considering these points relevant in today’s society, though in a more fractured sense. This is because they relate to speech and language use in relation to social standing.// Although social standing in today’s liberal society is becoming an ever more redundant concept, using someone’s speech as an indication of someone‘s identity is still in evidence. This notion is apparent in the main plot line in which Eliza becomes entrapped to the perspective of a new language system. When adopting the role of the speaker, Eliza adopts a slowly differing identity that emerges with child like astonishment before she changes into what is essentially a different person. It does not continue to be a liberating and learning experience.Rather, the liberation of a woman hiding behind the veil of civility in a bid to expose it, perhaps showing the power of the human spirit over class in the process, is lost. That is to say, that on speaking the language through the conventions of class Eliza loses sight of the world through her former eyes and comes to view i t through her new language that cannot be escaped. Essentially, it is through this change in persona that the play delivers its moral warning and cutting implication in that the core of the human being cannot escape from the language that it uses to identify itself with.The language and convention used by those of high society is responsible for each of their perspectives and it is not the person or people‘s speaking the language. Essentially, if you are to change the person’s language, language use and perspective then they themselves will come to define themselves and their being according to the structural meaning inherent to the language that is used by that society. This is indicated throughout Eliza’s discussions and becomes the main rationale for all that she does.For example, in one part of the play she states that ‘’you know I can't go back to the gutter, as you call it, and that I have no real friends in the world but you and the Colonelâ⠂¬â„¢Ã¢â‚¬â„¢ (Shaw, 1998). This short extract shows the great division based upon the language being used and the fact that it is represented by a social reality, in this case being social standing. What is interesting about the use of language in relation to others is the way in which Eliza is accepted and rejected at different times during the play.For example, it first appears that Eliza is rejected from society as her language does not denote the correct social grouping, stock and/ or class. This is first justified as being because of her use of language, accent and the incorrect convention. However, it appears on later reading that the convention is of little consequence as she uses the same convention, but put to a different context. Rather, it is the response from others alone that make it something of note.At one point during the play she makes the assertion that speaking properly (meaning without a cockney accent) is simply learning to dance in a fashionable way, which acc entuates this point even further. Essentially, the assertion that she puts forward here relates to the realisation of the superficiality of language in its conventional format as both languages mean exactly the same thing from a pragmatic perspective.At this stage she is learning the meaning of language and the convention of getting from one thing to another via language use. She realises that the only difference is a superficial one as the functional meaning (cause and effect) is the same whichever language is spoken. Essentially, the only different in the language is the significance of the source of referents, which dictate a different context to convention.Therefore, her conclusion is that it is merely a state of fashion in which the dancer dances the same, but where one dancer adopts the fashionable style, the other is overlooked as being able to dance (Baudrillard, 1968). This conclusion relates to the elements of high society that come with the speakers of proper English and that are not afforded to those of a poorer language, such as cockney. Those that do not speak the language are simply those that do not speak of anything meaningful, when in reality there is simply a clash over the source of referential meaning.