Heat Transfer Essay Example

Heat Transfer Essay The University of Sydney Aerospace, Mechanical and Mechatronic Engineering MECH3260 Thermal Engineering Heat Transfer Quiz 3 2007 Time: 40 minutes Answer ONE question only. Question 1 Consider the cylindrical receiver in a solar thermal power plant shown below. The receiver is made of opaque material and has a diameter D = 8m and length L = 14m. At a particular time, the heliostats direct a concentrated solar flux of q†S = 80 kW/m2 onto the receiver. The ambient air has a temperature T( = 300K and there is no wind. Under these conditions, the surface of the receiver has a temperature of Ts = 700K, an emissivity of ( = 0. and a solar absorptivity (S = 0. 8. i) Calculate the heat loss from the vertical surface of the receiver due to convection. Assume effects of curvature are negligible. ii) Calculate the heat loss from the vertical surface of the receiver due to radiative emission. Neglect irradiation from the surroundings. iii) Determine the collector efficiency. iv) If a wind started to blow, what would happen to the surface temperature of the receiver? Would you expect the collector efficiency to increase, decrease or remain the same? Explain your answers. (Note – assume all other conditions remain the same. ) [pic] Question 2 The roof of a refrigerated truck compartment is of composite construction, consisting of a layer of foamed urethane insulation (t2 = 40mm, ki = 0. 03W/mK) sandwiched between aluminium alloy panels (t1 = 6mm, kp = 200W/mK). The length and width of the roof are L = 12m and W = 4m respectively. The temperature of the refrigerated space within the truck is –15? C. The solar absorptivity and the emissivity of the outer surface are (S = 0. 3 and ( = 0. 7. Consider conditions for which the truck is moving at a speed of V = 90km/h, the air temperature is T( = 30? C and the solar irradiation is GS = 900W/m2. We will write a custom essay sample on Heat Transfer specifically for you for only $16.38 $13.9/page Order now We will write a custom essay sample on Heat Transfer specifically for you FOR ONLY $16.38 $13.9/page Hire Writer We will write a custom essay sample on Heat Transfer specifically for you FOR ONLY $16.38 $13.9/page Hire Writer Assume turbulent flow over the entire roof. Take the average convection coefficient on the inner surface of the roof to be 0. 5W/m2K. i) Derive an equation for the average temperature of the outer surface Ts,o in the form and hence show that Ts,o ( 302K = 29? C. Assume a film temperature of 300K and neglect irradiation from surroundings. (No marks if your equation is not in the above form) ii) Determine the corresponding heat load imposed on the refrigeration system. iii) The average convection coefficient assumed for the inner surface of the roof is relatively low. Explain why this is so. [pic] [pic] [pic] Heat Transfer Essay Example Heat Transfer Essay Consider the cylindrical receiver in a solar thermal power plant shown below. The receiver is made of opaque material and has a diameter D = 8m and length L = 14m. At a particular time, the heliostats direct a concentrated solar flux of q†S = 80 kW/m2 onto the receiver. The ambient air has a temperature T( = 300K and there is no wind. Under these conditions, the surface of the receiver has a temperature of Ts = 700K, an emissivity of ( = 0. and a solar absorptivity (S = 0. 8. i) Calculate the heat loss from the vertical surface of the receiver due to convection. Assume effects of curvature are negligible. ii) Calculate the heat loss from the vertical surface of the receiver due to radiative emission. Neglect irradiation from the surroundings. iii) Determine the collector efficiency. iv) If a wind started to blow, what would happen to the surface temperature of the receiver? Would you expect the collector efficiency to increase, decrease or remain the same? Explain your answers. (Note – assume all other conditions remain the same. ) [pic] Question 2 The roof of a refrigerated truck compartment is of composite construction, consisting of a layer of foamed urethane insulation (t2 = 40mm, ki = 0. 03W/mK) sandwiched between aluminium alloy panels (t1 = 6mm, kp = 200W/mK). The length and width of the roof are L = 12m and W = 4m respectively. The temperature of the refrigerated space within the truck is –15? C. The solar absorptivity and the emissivity of the outer surface are (S = 0. 3 and ( = 0. 7. Consider conditions for which the truck is moving at a speed of V = 90km/h, the air temperature is T( = 30? C and the solar irradiation is GS = 900W/m2. We will write a custom essay sample on Heat Transfer specifically for you for only $16.38 $13.9/page Order now We will write a custom essay sample on Heat Transfer specifically for you FOR ONLY $16.38 $13.9/page Hire Writer We will write a custom essay sample on Heat Transfer specifically for you FOR ONLY $16.38 $13.9/page Hire Writer Assume turbulent flow over the entire roof. Take the average convection coefficient on the inner surface of the roof to be 0. 5W/m2K. i) Derive an equation for the average temperature of the outer surface Ts,o in the form and hence show that Ts,o ( 302K = 29? C. Assume a film temperature of 300K and neglect irradiation from surroundings. (No marks if your equation is not in the above form) ii) Determine the corresponding heat load imposed on the refrigeration system. iii) The average convection coefficient assumed for the inner surface of the roof is relatively low.

Dead Poets Society John Keating Essays

Dead Poets Society John Keating Essays Dead Poets Society John Keating Paper Dead Poets Society John Keating Paper I deeply remember the English teacher He is really a great teacher. His new teaching method brought vitality Mr.. Keating. To the old school. I still remember that he said, Even though it may seem sill or wrong, you may try. I wont forget he made the students tear some pages of their books on which he thought what the author said was wrong. He told the students that, Now when you read, dont Just consider what the author thinks, consider what you think. Boys, you must strive to find your own voice. I wont forget that he stood on the desk ,Why do I stand here? To feel taller than you? I stand on my desk to remind myself that we must constantly force ourselves to look at things In a different way, The world looks different from up here. If you dont believe It, stand up here and try It. All of you, take turns. I wont forget at the end of the movie the students who love him so much that they dont want him to leave stood on their desks, called him, Oh. Captain, my captain' The death of Nell make me feel sorrow. HIS father loved IM, but not in the right way. If he could listen and understand the thought of Neil, maybe the tragedy wont happen. It is Mr.. Keating who teaches the students to follow their dreams and to be creative and individual. We really need teachers like Mr.. Keating. I know from him that it is important to stand ones ground no matter how hard it may be. l went to the woods because I wanted to live deliberately I wanted to live deep and suck out all the marrow of life To put to rout all that was not life And not when I came to die, discover that I had not lived This story takes place in 1959 at Walton School, a prep school for boys. John Keating, a former student at the school, begins working in the school as an English teacher. With his unique teaching methods he inspires his students to be themselves, and act like individuals. He teaches them to think for themselves and to *Seize the day* With the boys having a new way of viewing life, they follow there thoughts, and form a new Dead Poets Society. Mr.. Keating classes drew a lot of attention from the administration. Eventually Neil commits suicide, and the school investigates into why this happened. The boys were then intimidated into saying that Mr.. Keating was a bad influence to Neil, and he was forced to leave the school. MR. passion, 95, Keating, fatty , NORM Eating Dead Poets Society -John Keating By Nancy on the desk do I stand here? To feel taller than you? I stand on my desk to remind myself that we must constantly force ourselves to look at things in a different way. The world looks different from up here. If you dont believe it, stand up here and ray it. All of you, take turns. I wont forget at the end of the movie the students who love him so much that they dont want him to leave stood on their desks, called him, Oh, captain, my captain! The death of Neil make me feel sorrow. His father loved maybe the tragedy wont happen. It is Mr.. Keating who teaches the students to follow wanted to live deep and suck out all the marrow of life! Like individuals. He teaches them to think for themselves and to Baize the day. MR. Keating?+Effie, passion, DEED IDLER Eating

Bsiness statistic assignment Example | Topics and Well Written Essays - 1250 words

Bsiness statistic - Assignment Example For example, when the factors that are contributing most to the overhead costs are known, it becomes easy to control such costs. This report will involve analysis of different variables that are believed to affect overhead costs, including number of setups, machines hours and labour hours. Some of the methods of analysis that will be used include descriptive statistics, regression analysis, and use of histograms and graphs. Section A - characteristics of all the variables Calculating measures of location (averages) and dispersion (variation) N Mean Std. Deviation Variance Overhead costs 24 172.79 17.939 321.824 Direct Labour hours 24 1134.67 64.629 4176.928 Machine hours 24 1402.58 213.135 45426.514 No. of setups 24 201.42 13.990 195.732 Valid N (listwise) 24 Table 2: descriptive statistics (averages and variation) Using all 24 months’ data to construct a frequency distribution and histogram for the machine hours Machine Hours Frequency 1000 0 1100 5 1100 0 1200 1 1200 0 1300 2 1300 0 1400 4 1500 3 1600 6 1700 2 1800 1 Table 2: Frequency Distribution Table Figure 1: Histogram for the machine hours Estimating the median and other descriptive statistics for the machine hours Column1 Mean 1402.583 Standard Error 43.50599 Median 1428 Mode 1300 Standard Deviation 213.135 Sample Variance 45426.51 Kurtosis -0.97697 Skewness -0.22011 Range 710 Minimum 1060 Maximum 1770 Sum 33662 Count 24 Table 3: Median and Other Descriptive Statistics for machine hrs A brief Report Determination of descriptive statistics is very essential in decision-making regarding cost. Among the four variables, the highest level of dispersion was reported in machine hours, as shown by the standard deviation and variance, implying that use of machines is likely to significantly affect variations in overhead cost. Number of set ups have the least dispersion, possibly meaning that this variable may not significantly cause cost variation (Table 2). Table 2 shows the grouped frequency histogram for the machine hours, where it is evident that most of the hours lie between 1001 to 1100 and 1501 to1600. The shape of the histogram exhibits a bimodal distribution or rather it has two peaks. This is a problem that results from stratification and it means that the data could have originated from two different sources. It is, therefore, important to identify these sources and analyse the data separately if appropriate. Figure 3 shows the descriptive statistics for the machine hours. The minimum and the maximum values are very useful in quickly identifying the range between which hours fall, which in this case is 1060(min) and 1770(max). The standard error gives the extent to which the distribution is exposed to sample error, which in this case amounts to 43.5 (Mann, 1995). Section B Graphs and analysis of the linear relationships in the variables related to the overhead costs. Figure 2: OH costs vs. direct labour hours Figure 2: Overheads costs are directly related to the direct l abour hours, which means that an increase in direct labour hours leads to an increase in overhead costs. The equation y=2.5352x +

Resistance Essay Example | Topics and Well Written Essays - 500 words

Resistance - Essay Example Here, the connection between gender and violence can be established if gender relations is understood as a discourse â€Å"signifying relationships of power† (Jacobs, Jacobson and Marchbank, 2000, p.3). In both cases of violence and gender disparity, thus an exercise of power is involved. On the contrary, resistance is â€Å"a reflection of the potential for subversion and contestation† within the premises â€Å"of established order† (Jacobs, Jacobson and Marchbank, 2000, p.3). This is to suggest that resistance is more concerned with opposing the undemocratic exercise of power, which is often part of the establishment, rather than the simple or violent exercise of power. War has been the greatest exercise of power in human history. As the ethics of war has changed from protecting civilians to deliberately targeting them, the most vulnerable groups, which have lost their â€Å"personal security† are women, children and the ethnic minorities (Jacobs, Jacobs on and Marchbank, 2000, p.5). There is a continuation of this situation in non-war conditions as well in the form of domestic violence, as far as women are concerned. On the other hand, it has also been argued, â€Å"women have been central in democratizing processes† (Jacobs, Jacobson and Marchbank, 2000, p.9).

The Importance Of Biodiversity And Effects Environmental Sciences Essay

The Importance Of Biodiversity And Effects Environmental Sciences Essay The importance of biodiversity has lack know by peoples. Human is actually relying on the biodiversity to survive because biodiversity provide useful and necessary resources such as food and raw material of medicine. Besides, the biodiversity is important role that working on ecosystem. It is kind of transportation of ecosystem service. The ecosystem provides services such as protection of soil thus the biodiversity loss is related to the reduction of ecosystem services. One of the main causes of biodiversity is deforestation. Deforestation rate increase day by day, the species lose their habitat and lack of food to survive. Human hew the forests to get the resources and also develop the human activities such as agriculture and tourism. More than fifty percent of Asian land area use to develop the agriculture. Human being should find out the solution to improve the biodiversity loss and reduction of ecosystem services. Reforestation improves the rate of deforestation but do not have effects on biodiversity loss. Therefore, human being should understand the importance of biodiversity and the consequences of biodiversity loss to increase the awareness and consideration of human to biodiversity issues. The importance of biodiversity and effects of deforestation on biodiversity loss Biodiversity means the species of life on our mother earth and the ecological processes that support the diversity of ecosystems. The goal of this research paper is to understand the importance of biodiversity and the effects of deforestation on biodiversity. The biodiversity and ecosystem issues compare to other environmental issues, peoples have less consideration to biodiversity issues. As the point above mentioned, the biodiversity is related to the condition of global ecosystem. The species of organisms on earth are facing the threat of habitat loss and extinction. If human do not take any action to improve this situation, we would lose many precious species. There are different areas of species of life in the world such as mammals, reptiles, birds, amphibians, fish and plants. Each of the different species of life has few or more species already extinct. For examples, red colobus monkey, dodo, and phantom shiner. There have no scientists or study can tell us how many species of life are in the world because our mother earth is a complex world and the relevant organizations discover new species to science ceaselessly until now. Therefore, we do not know there is how many species we are losing exactly. There are some experts forecast that the speed of biodiversity loss nowadays is between thousand and ten thousand higher than the natural extinction rate. Human may not understand well the influence of biodiversity to the entire human race and global ecosystems. Human obtain many resources from biodiversity, such as food, timber, raw materials of medicine and souvenir. In addition of those basic provisions, the biodiversity play the important roles of maintaining the ecosystem of the world. It provides the important services, such as cleaning water resources, protection of soil, reduction of global diseases and dedication of maintain the stability of climate. Biodiversity also contributes social benefits to us like tourism and entertainment development. However, human pay attention to what and how many benefits they can get from the biodiversity and ecosystem, the value of biodiversity and ecosystem would neglected by human. Nowadays, science has better perceptive to the biodiversity effects and ecosystem services. There are several study of biodiversity effects show that the biodiversity loss is relevant with the reduction of ecosystem services (e.g., Balvanera et al.2006, Cardinale et al.2007, Thompson et al.2009 as cited in Thompson, Okabe, Tylianakis, Kumar, Brockerhoff, Schellhorn, Parrotta Nasi, 2011). Thompson and colleagues (2009) found that seventy six percent of twenty one studies states that there is direct relationship between the richness of species and the production of ecosystem services. Based on all these studies, the biodiversity play the role of transportation of ecosystem services. Therefore, the damage of biodiversity may influence the cycling of species, and the transportation of ecosystem services. When the ecosystem services reduce, the environmental issues such as climate change will be more serious. From the above points of view, we can understand how important the biodiversity are and human could not survive without biodiversity and ecosystem service. One of the main reasons that cause biodiversity loss is deforestation. The animals lose their habitat and the biodiversity of forests are damage when deforestation happens. The area which has the highest rate of deforestation is Southeast Asia. According to a study of land use change in Asia, fifty percent of the Asian land is under the development of agriculture or the development of human activities ( Zhao, Peng, Jiang, Tian, Lei Zhou, 2006) In South Asia, seventy three percent of the land area is used as the development of agriculture. Therefore, the animals have no habitats and enough food to survive. For example, Singapore has ninety five percent of deforestation rate over the past one hundred and eighty three thus there has more than twenty eight percent of species has eliminated. Another country that have high rate of biodiversity loss caused by deforestation is Thailand (Zhao et al., 2006). Brooks, Mittermeier, Fonseca, Rylands, Konstant, Flick, Pilgrim, Oldfield, Magin Taylor (2001) stated that Nearly half the worlds vascular plant species and one-third of terrestrial vertebrates are endemic to twenty five of hotspots of biodiversity, each of which has at least one thousand and five hundred endemic plant species. None of these hotspots have more than one-third of their pristine habitat remaining. Historically, they covered twelve percent of the lands surface, but today their intact habitat covers only one point four of the land. Based on this article, we can know that the endemic of hotspots are facing the problem of extinction because of the habitat loss. When the habitat areas decrease, the species have highest possibility of extinction problems. Discussion As we know that, the biodiversity loss is more and more serious nowadays. Human should find out the solution to reduce the biodiversity loss. To reduce the extinction of biodiversity, we should decrease the rate of deforestation of the world. One of the solutions of deforestation is reforestation. Although reforestation does improve the rate of deforestation, it does not fully renovate the species richness and ecosystem of a forest. There is a study about the value of plantation forests; it finds that the species richness of reforestation is not as abundant as the aboriginal forest (Gardner, Hernandez, Barlow Peres, 2008). In my opinion, the best way to decrease the rate of deforestation is improve the consideration and awareness of human being. Government and the relevant organization should encourage peoples to concern about this topic and let children have the concepts of biodiversity and deforestation. Peoples should know well about the seriousness and consequences of biodiversity loss and the reduction of ecosystem thus they will appreciate and conserve the nature. Besides, Government should be stricter in land management to prevent peoples or some companies over extract the forests to acquire self-benefits. In addition, we can use science to improve the biodiversity loss and reduction of ecosystem services. For example, Sustainable forest management (SFM) is ecosystem management that work on the conservation of environment, such as biodiversity issues. They try to use scientific method to find out the most effective way and translating it into policy to conserve and renovate the biodiversity loss and reduction of ecosystem services (Thompson et al., 2011). Finally, there are also simple solutions such as do not buy the souvenirs that made of animals, less use of air-conditioner and buy timber from the legal company. Although those simple ways look not as effective as scientific method, peoples should make contribution to environmental issues. Conclusion The biodiversity loss has facing the highest rate over the past years. The government and organization of environment should not be the only who responsible to the conservation of environment. Human should have more knowledge about the importance of biodiversity and ecosystem service thus the awareness of human to biodiversity will increase. Human being should be more intelligent to maintain and protect our mother earth. If the rate of biodiversity loss and reduction of ecosystem continue increase, none of us can endure the consequences. Therefore, human being must be more consideration to this issue. Not only the government, should parents play a good role of model to their children to make the better environment for our next generation.

Maturity :: essays research papers

Maturity   Ã‚  Ã‚  Ã‚  Ã‚  Maturity is not something that happens overnight. Maturity comes through trial and error; it’s a process which has to be perfected. Life throws curveballs and there are always multiple ways to deal with the situations. Being mature is making the right choices and knowing right from wrong.   Ã‚  Ã‚  Ã‚  Ã‚  When you turn eighteen you are expected to just be mature. You’re allowed to buy a house, a car, get a credit card, it seems like you can do whatever you want. Maturity basically gets thrown at you, so hopefully you are prepared for it. High school responsibilities, like getting a job, budgeting your time and learning to drive teach you skills that you can use in your adult life, if you choose to apply them. Maturity is about choice. You can be given every skill that you need to survive and if you are mature enough you will use them.   Ã‚  Ã‚  Ã‚  Ã‚  For me, driving a car, going to Europe, graduating, and having a job have all contributed to my maturity. I have to make life and death choices every day while driving, and my choices affect other’s lives- and that bring a lot of maturity. I went to Europe basically by myself, with no parent there to hold my hand. I had to make good choices, be safe and be mature while I was half way across the world from my parents. I am graduating this year as well. After this summer I am going to college, being a real â€Å"adult† and I will be able to make my own choices. Maturity will help me keep my head on straight while I am being my own boss. For the past two years I have held the same job. This job has taught me responsibility, how to deal with money and people and basic life skills. If I were not mature I would not be able to hold a job for that long.

Hydraulic Machines

30 CHAPTER 5 TURBINES 5. 1 Introduction Hydraulic turbines are the machines which use the energy of water and convert it to mechanical energy. The mechanical energy developed by a turbine is used in running an electric generator which is directly coupled to the shaft of the turbine. The electric generator thus develops electric power, which is known as hydro-electric power. 5. 2 Elements of Hydraulic Power Plants Fig. 5. 1 General Layout of a Hydraulic Power Plant 31 Fig. 5. 1 shows a general layout of hydraulic power plant, in which an artificial storage reservoir formed by constructing a dam has been shown. 5. 3General Classification of Turbines Turbines are hydraulic machines that convert energy into rotating mechanical energy which in turn generators to produce electrical energy. Originally developed from the water wheels, hydraulic turbines are the prime mortars of importance in modern water power development. According to their hydraulic action, turbines are broadly divided int o two classes. (1) Impulse Turbine: Impulse turbines are more efficient for high heads. At the inlet to the turbine runner, pressure head can be completely converted into kinetic head in the form of a jet of water issuing from one or more nozzles.The free jet will be at atmospheric pressure before as well as after striking the vanes. The turbines are regulated by nozzles which may be a simple straight flow type or a deflector type. The impulse turbines are commonly represented by Pelton Wheels. Turgo turbine is also an impulse turbine but with different buckets, when compared with pelton. Turgo and cross flow turbines are relatively new developments in this class. The main advantages of these turbines are: †¢ They can be easily adopted to power variation with almost constant efficiency. †¢The penstock overpressure and the runner overspeed control are easier. †¢ The turbine enables an easier maintenance. †¢ Due to the jet the manufacturer of these turbines impose a better solid particle control, conducting, consequently, to a lower abrasion effect. (2) Reaction Turbine: A turbine can be made to rotate under the action of water flowing under pressure through the runner. In such turbines the penstocks, the inlet passage to the runner, passage between the runner vanes, all form a continuous passage for the flow under a pressure which continuously decreases from inlet to outlet.The turbine runner directly converts both kinetic energy as well as the pressure energy into mechanical energy. Reaction turbines are represented in modern practice by two principal types: the Francis turbine where the flow is directed radial to the runner axis and the Propeller type 32 where the flow is axial to the runner axis. Propeller turbines may be fixed blade or adjustable blade types. Kaplan turbine has adjustable blades. The main advantages of these turbines are: †¢ It needs lesser installation space. †¢ It provides a greater net head and a better prot ection against downstream high flood levels. It can have greater runner speed. †¢ It can attain greater efficiencies for high power values. In order to distinguish different turbines, the hydraulically salient features like pressure, head, flow direction and magnitude, speed and power etc. The general classification of hydraulic turbines is illustrated in Fig. 5. 2. Hydraulic Turbines Impulse Turbines Pelton Turgo Reaction Turbines Cross-Flow Fixed -Blade Propeller Kaplan Francis Tubular Deriaz Bulb Fig. 5. 2 General Classification of Turbines 5. 4 Number of Units It is normally cost effective to have a minimum number of units at a given small ydropower installation. Multiple units may, however, be necessary from the operational point of view so that even one unit breaks down or is in the routine maintenance, the power generation can be achieved to a certain extent. The efficiency curves of turbines show that the 33 efficiency of power generation from hydraulic turbines conside rably decreases at low flow ratios or power ratios. In multiple units, it is possible to maintain the higher efficiency even in low flows and the low loads by running a certain number of the units at a time depending upon the available discharge and the load demand.Multiple units thus, make the most effective use of water where the flow as well as the load variations are significant. 5. 5 Limits of Use of Turbine Types The selection of best turbines for any particular small hydropower site depends on the site characteristics, the dominant beings the head and the available flow. There are some limits on the range of these parameters in the selection of turbines. Each turbine type is best suited to a certain range of pressure head and the flow rate. For instance, Pelton wheels operate with low flows discharged under great pressures where as Propeller turbines are effective in high flows under low heads.Francis turbines fall in the medium category covering a wide range of different hea ds and discharges. The common practice of SHP systems is to develop standard unit sizes of equipment that will operate over a range of heads and flows. Either charts or nomographs are used to select appropriate units for site specific application. One such chart showing the head-flow range of normal SHP schemes applicable to each type of turbine is given in Fig5. 3. The graph also indicates the approximate power generation for each combination of the head and the discharge applicable to SHP schemes. Fig. . 3 Head-Flow Ranges for Different Turbines 34 5. 6 Pelton Wheel Pelton wheel is well suited for operating under high heads. A pelton turbine has one or more nozzles discharging jets of water which strike a series of buckets mounted on the periphery of a circular disc. The runner consists of a circular disc with a number of buckets evenly spaced round its periphery. The buckets have a shape of a double semi-ellipsoidal cups. The pelton bucket is designed to deflect the jet back thro ugh 165 ° which is the maximum angle possible without the return jet interfering with the next bucket.Fig. 5. 4 Pelton bucket General arrangement of a pelton wheel is shown in the Fig. 5. 5. For SHP schemes, Pelton wheels are easier to fabricate and are relatively cheaper. The turbines are in general, not subjected to the cavitation effect. The turbines have access to working parts so that the maintenance or repairs can be effected in a shorter time. Fig. 5. 5 General Arrangement of a Pelton Wheel 35 Traditionally, micro hydro pelton wheels were always single jet because of the complexity and the cost of flow control governing of more than one jet.Advantages of multi-jet: -Higher rotational speed -Smaller runner -Less chance of blockage Disadvantages of multi-jet: -Possibility of jet interference on incorrectly designed systems -Complexity of manifolds 5. 7 Francis Turbine Francis turbine is a mixed flow type, in which water enters the runner radially at its outer periphery and le aves axially at its center. Fig. 5. 6 illustrates the Francis turbine. The runner blades are profiled in a complex manner and the casing is scrolled to distribute water around the entire perimeter of the runner.The water from the penstock enters a scroll case which completely surrounds the runner. The purpose of the scroll case is to provide an even distribution of water around the circumference of the turbine runner, maintaining an approximately constant velocity for the water so distributed. The function of guide vane is to regulate the quantity of water supplied to the runner and to direct water on to the runner at an angle appropriate design. A draft tube is a pipe or passage of gradually increasing cross sectional area which connects the runner exit to the tail race. Fig. 5. 6 Francis Turbine 36 . 8 Kaplan Turbine It is an axial flow turbine which is suitable for relatively low heads. From Fig. 5. 7, it will be seen that the main components of Kaplan turbine such as scroll casi ng, guide vanes, and the draft tube are similar to those of a Francis turbine. Fig. 5. 7 5. 9 Kaplan Turbine Specific Speed The specific speed of any turbine is the speed in r. p. m of a turbine geometrically similar to the actual turbine but of such a size that under corresponding conditions it will develop 1 metric horsepower when working under unit head. Ns = NP H5/ 4 †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦(5. 1) where Ns = specific speedP = power in HP 5. 10 Characteristic Curves The turbines are generally designed to work at particular values of H,Q,P,N and ? o which are known as the designed conditions. It is essential to determine exact behaviour of the turbines under the varying conditions by carrying out tests either on the actual turbines or on their small scale models. The results of these tests are usually graphically represented and the resulting curves are known as characteristic curves. 37 -constant head characteristic curves -constant speed characteristic curves -constan t efficiency characteristic curvesIn order to obtain constant head characteristics curves the tests are performed on the turbine by maintaining a constant head and a constant gate opening and the speed is varied by changing the load on the turbine. A series of values of N are thus obtained and corresponding to each value of N, discharge Q and the output power P are measured. A series of such tests are performed by varying the gate opening, the head being maintained constant at the previous value. From the data of the tests the values of Qu, Pu, nu and ? o are computed for each gate opening. Then with Nu as abscissa the values ofQu, Pu and ? o for each gate opening are plotted. The curves thus obtained for pelton wheel and the reaction turbines for four different gate openings are shown in Fig. 5. 8. Fig. 5. 8 Constant head characteristics for Pelton wheel and reaction turbines 38 5. 11 Cavitation in turbines When the pressure in any part of the turbine reaches the vapour pressure of the flowing water, it boils and small bubbles of vapour form in large numbers. These bubbles are carried along by the flow, and on reaching the high pressure zones these bubbles suddenly collapse as the vapour condenses to liquid again.The alternate formation and collapse of vapour bubbles may cause severe damage to the surface which ultimately fails to fatigue and the surface becomes badly scored and pitted. This phenomenon is known as cavitation. In order to determine whether cavitation will occur in any portion of the turbine, D. Thomas has developed a dimensionless parameter called Thomas'cavitation factor ? which is expressed as ?= Ha ? Hv ? Hs H †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦(5. 2) where Ha = atmospheric pressure head Hv = vapour pressure head Hs = suction pressure head For Francis turbines: Critical cavitation factor ? c = 0. 625 (Ns/444)2 †¦Ã¢â‚¬ ¦.. (5. 3) For Propeller turbines: ? c = 0. 28 + [ Example 5. 1 1 Ns 3 ( )] 7. 5 444 †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.. (5. 4) Est imate the maximum height of straight conical draft tube of a 18000 h. p. Francis turbine running at 150 r. p. m under a net head of 27 m. The turbine is installed at a station where the effective atmospheric pressure is 10. 6 m of water. The draft tube must sink at least 0. 77 m below the tail race. Ns = NP H5/ 4 = 327 39 ?c = 0. 625 (Ns/444)2 = 0. 339 Cavitation factor ? = Ha -Hv -Hs H Ha -Hv = 10. 6 m, H =27 m 0. 339 = 10. 6 -Hs 27 Hs = 1. 45 m Max length of the draft tube = 1. 45 + 0. 7 = 2. 22 m 5. 12 Governing of Turbines All the modern turbines are directly coupled to the electric generators. The generators are always required to run at constant speed irrespective of the variations in the load. This constant speed N of the generator is given by the expression N= 60 f p †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.. (5. 5) where f = frequency (usually 50) p = numbers of pairs of poles 5. 13 Water Hammer A gate or valve at the end of the penstock pipe controls the discharge to the turbine. As s oon as this governor regulated gate opening is altered, the pipe flow has to be adjusted to the new magnitude of flow.In doing so, there are rapid pressure oscillations in the pipe, often accompanied by a hammering like sound. Hence this phenomenon is called as water hammer. 5. 14 Jet Speed The velocity of flow of the jet depends upon the total net head H at the base of the nozzle and is given by the nozzle equation: v = C v ? 2gH where the discharge coefficient velocity of the nozzle is taken as 0. 95. †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦(5. 6) 40 5. 15 Bucket speed V= ? DN 60 †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦(5. 7) The bucket speed should be half of the jet speed. In practice, losses in the turbine cause the maximum efficiency to occur at slightly less than a half, typically 0. 46. V =0. 6 v 5. 16 Design of Pelton Wheel Runner diameter: Runner diameter can be found out from the rpm equation. D= 38 ? H N †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦. (5. 8) where N = runner speed(rpm) H = net head Nozzle diameter: The nozzle diameter is given by the nozzle equation: d = 0. 54 ? Q 0. 5 1 ? 0. 25 H n jet †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦. (5. 9) Jet ratio: Jet ratio D/d is a size parameter for the turbine. It has a value in a range of 10 to 24. For the high efficiency Pelton wheel design, the ratio of the runner diameter to the nozzle should be more than 9. Number of buckets: The number of buckets required for the efficient operation of the Pelton turbine is calculated as:N buc = 0. 5 ? D + 15 d †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.. (5. 10) In practice, the selection and the detail design of the turbine units are carried out by the manufactures based on the model performances. 41 Example 5. 2 Powerhouse is equipped with a vertical shaft pelton turbine. The generator is provided with 6 pairs of poles. Design discharge is 1. 4 m3/s and net head is 425 m. The turbine will provide 6500 hp. Take coefficient of nozzle as 0. 95. Determine (a) the specific speed (b) velocity of jet (c) jet diameter (d) pitch circle diameter of the wheel (e) number of buckets (a) N= 60 f p = 60 x 50/6 = 500 rpmNs = = NP H5/ 4 500 6500 4255 / 4 = 20. 9 Use single jet pelton turbine (b) velocity of jet v = C v ? 2gH = 0. 95 ? 2Ãâ€"9. 81Ãâ€"425 = 86. 75 m/s (c) jet diameter d = 0. 54 ? = 0. 54 ? Q 0. 5 1 ? 0. 25 H n jet 1. 4 0. 5 425 0. 25 ? = 0. 14 m =14 cm 1 1 42 (d) pitch circle diameter D= 38 ? H N = 1. 57 m (e) Number of buckets N buc = 0. 5 ? N buc = 0. 5 ? D + 15 d 1. 57 + 15 0. 14 = 20. 6 = 21 5. 17 Work done of Pelton Wheel In turbines, the water flows on to the runner, which itself is rotating with a certain speed. The water flows over the runner and leaves the runner at its outlet point.We can speak of absolute velocity of water before it flows in the runner, the relative velocity of water w. r. t the runner and again the absolute velocity of water after it has left the runner. In order to ascertain the relationship between these velocities, the velocity vector diagram prove to be very useful. Fig. 5. 9 shows the velocity triangles at the tips of the bucket of a pelton wheel. At the outlet tip velocity triangles are different depending upon the magnitude of u corresponding to which it is slow, medium or fast runner. Inlet velocity diagram VVri Vai =Vwi 43 Outlet velocity diagram V Vwo ? ? Vfo Vro Vao Fig. 5. 9 Velocity triangles V = bucket velocity Vai = absolute velocity of jet at inlet tip Vai = Cv 2 gH1 Vao = absolute velocity of jet at outlet tip Vri = relative velocity of jet at inlet = Vai-V Vro = relative velocity of jet at outlet = k. Vri Vwi = velocity of whirl at inlet =Vai Vwo = velocity of whirl at outlet = V-VroCos ? Vfo = velocity of flow at outlet Mass/sec m =? Q =? a Vai=? ?/4 d2 Vai †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦(5. 11) Workdone on the bucket/sec (power developed by turbine) P = m (Vwi -Vwo) V †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦(5. 12)Maximum hydraulic efficiency ? h max = 1 (1 + kCos? ) 2 †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦(5. 13) The hydraulic efficiency is maximum when the bucket speed is equal to half of the velocity of jet. 44 Example 5. 3 The head available at entrance to the nozzle supplying a pelton wheel is 300 m and the coefficient of velocity for the nozzle is 0. 98. The wheel diameter is 1. 8 m and the nozzle diameter is 125 mm. The buckets deflect the jet through 165 °. Assuming the relative velocity of the jet is reduced by 16%, calculate the theoretical speed in rev per min for the maximum hydraulic efficiency.What is the hydraulic efficiency when running at this speed, and what is the power developed? Deflection angle =165 ° = (180- ? ) ? = 15 ° k = 0. 84 hf hn H'=Vai2/2g H For max hydraulic efficiency V/Vai =0. 5 Vai = Cvv2gH1 = 75 m/s V =Vai/2 = 37. 5 m/s V =? DN/60 N =60V/? D = 398 rpm ?hmax = 1/2 (1+ k Cos ? ) = 90. 55 % mass/sec = m = ? Q = ? ?/4 d2 Vai = 920 kg/sec H1 45 Inlet diagram: V Vri From velocity diagram Vwi = Vai =75 m/s Vri = Vai- V =37. 5 m/s Vwi = Vai Outlet diagram: Vwo = V -Vro Cos ? V = V – k Vri Cos ? Vwo Vro Power = m V ( Vwi -Vwo) = 234600 Watts Vao =7 m/s 46 CHAPTER 6CENTRIFUGAL PUMP 6. 1 Introduction Centrifugal pumps are classified as rotodynamic type of pumps in which dynamic pressure is developed which enables the lifting of liquids from a lower to a higher level. The basic principle on which a centrifugal works is that when a certain mass of liquid is made to rotate by an external force, it is thrown away from the central axis of rotation and a centrifugal head is impressed which enable it to rise to a higher level. Now, if more liquid is constantly made available at the centre of rotation, a continuous supply of liquid at a higher level may be ensured.Since in these pumps the lifting of the liquid is due to centrifugal action, these pumps are called ‘centrifugal pumps'. 6. 2 Advantages of centrifugal pumps over reciprocating pumps The main advantage of a centrifugal pump is that its discharging capacity is very much greater than a reciprocating pump which can handle relatively small quantity of liquid only. A centrifugal pump can be operated at very high speeds without any danger of separation and cavitation . The maintenance cost of a centrifugal pump is low and only periodical check up is sufficient .But for a reciprocating pump the maintenance cost is high because the parts such as valves etc. , may need frequent replacement. 6. 3 Component Parts of a Centrifugal Pump The main component parts of a centrifugal pump are: -impeller -casing -suction pipe -delivery pipe 47 Fig. 6. 1 Component part of a centrifugal pump 6. 4 Workdone by the Impeller The expression of the workdone by the impeller of a centrifugal pump on the liquid flowing through it may be derived in the same way as for a turbine. The liquid enters the impeller at its centre and leaves at its periphery. Fig. 6. shows a portion of the impeller of a centrifugal pump with one vane and the velocity triangles at the inlet and outlet tips of the vane. V is absolute velocity of liquid, u is tangential velocity of th e impeller, Vr is relative velocity of liquid, Vf is velocity of flow of liquid, and Vw is velocity of whirl of the liquid at the entrance to the impeller. Similarly V1,u1,Vr1,Vf1 and Vw1 represent their counterparts at the exit point of the impeller. 48 Fig. 6. 2 Velocity triangles for an impeller vane ? = the impeller vane angle at the entrance ? = the impeller vane angle at the outlet = the angle between the directions of the absolute velocity of entering liquid and the peripheral velocity of the impeller at the entrance ? = the angle between the absolute velocity of leaving liquid and the peripheral velocity of the impeller at the exit point Work done per second by the impeller on the liquid may be written as Work done = W ( Vw1 u1 – Vw u) ————-(6. 1) g where W kg of liquid per second passes through the impeller. Since the liquid enters the impeller radially ? = 90 and hence Vw = 0. Thus equation (6. 1) becomes Work done = W (Vw1u1) —â⠂¬â€Ã¢â‚¬â€Ã¢â‚¬â€-(6. ) g 6. 5 Head of a Pump The head of a centrifugal pump may be expressed in the following two ways: (a) Static head (b) Manometric head (or total head or gross head) 49 Fig. 6. 3 Head on a centrifugal pump (a) Static Head Static head is the vertical distance between the liquid surfaces in the pump and the tank to which the liquid is delivered by the pump. Static head (or lift) Hs = hs + hd where hs = static suction lift hd = static delivery lift †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.. (6. 3) 50 (b) Manometric Head Manometric head is the total head that must be produced by the pump to satisfy external requirements.If there are no energy losses in the impeller and the casing of the pump, then the manometric head Hm will be equal to the energy given to the liquid by the V ? 1u1 g impeller, i. e Hm = . But if losses occur in the pump then V ? 1u1 ? losses of head in the pump g †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.. (6. 4) Applying Bernoulli’s equation between the points, O at the liqui d surface in the pump and 1 in the suction pipe just at the inlet to the pump (i. e. , at the centre line of the pump), the following expression is obtained if the liquid surface in the sump is taken as datum. 0= p s Vs2 + + hs + h f s ? 2g ps Vs2 ? [ + hs + h fs ] 2g ? †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦. (6. 5) where ps is the pressure at point l ; Vs is the velocity of flow in the suction pipe ; hs is the suction lift and hfs is the head loss in the suction pipe which includes the head loss due to friction and the other minor losses. It may however be pointed out that if the pump is situated below the level of the liquid surface in the sump, hs will be negative. Equation (6. 5) indicates that at the inlet to the pump there is always a suction or vacuum pressure developed which will be recorded by the vacuum gauge provided at this point as shown in Fig. . 3. The head expressed by equation (6. 5) is called the suction head of the pump. Also, applying Bernoulli’s equation between points 1 and 2, which is just at the outlet of the impeller and is assumed to be at the same level as point 1, then since the impeller imparts a head equal to (Vw1u1/g) to the liquid the following expression is obtained: p s V s2 V? 1u1 p 2 V12 + + = + + hLi ? 2g g ? 2g †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.. (6. 6) where p2 is the pressure and V1 is the absolute velocity of the liquid leaving the impeller and hLi is the loss of head in the impeller. 51 6. Specific Speed of Centrifugal Pumps In order to compare the performance of different pumps, it is necessary to have some term which will be common to all centrifugal pumps. The term used for this purpose is the specific speed. The specific speed of a centrifugal pump is the speed at which the specific pump must run to deliver unit quantity against unit head, the efficiency being the same as the actual pump. Ns = NQ H 3/ 4 †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦(6. 7) where Ns= specific speed N = rotational speed(rpm) H = total head 6. 7 Performance of Pumps- Charact eristic CurvesA pump is usually designed for one speed, flow rate and head in actual practice, the operation may be at some other condition of head on flow rate, and for the changed conditions, the behaviour of the pump may be quite different. Therefore, in order to predict the behaviour and performance of a pump under varying conditions, tests are performed and the results of the tests are plotted. The curves thus obtained are known as the characteristic curves of the pump. The following three types of characteristic curves are usually prepared for the centrifugal pumps : a) Main and operating characteristics. (b) Constant efficiency or Muschel curves . (c) Constant head and constant discharge curves. Main and Operating Characteristics In order to obtain the main characteristic curves of a pump it is operated at different speeds. For each speed the rate of flow Q is varied by means of a delivery valve and for the different values of Q the corresponding values of manometric head Hm, shaft H. P. , P , and overall efficiency ? are measured or calculated. The same operation is repeated for different speeds of the pump. Then Q v/s Hm ; Q v/s P and Q v/s ? urves for different speeds are plotted, so that three sets of curves, as shown in Fig. 6. 4 are obtained, which represent the 52 main characteristics of a pump. The main characteristics are useful in indicating the performance of a pump at different speeds. During operation a pump is normally required to run at a constant speed, which is its designed speed, (same as the speed of the driving motor). As such that particular set of main characteristics which corresponds to the designed speed is mostly used in the operations of a pump and is, therefore, known as the operating characteristics.A typical set of such characteristics of a pump is shown in Fig. 6. 5 Fig. 6. 4 Main characteristics of a centrifugal pump Fig. 6. 5 Operating characteristic curves of a centrifugal pump 53 6. 8 Parallel or Series Operation of Pu mps Pumps in series Centrifugal pumps generate a relatively low head delivering a fairly high rate of discharge. Normally a pump with a single impeller can be used to deliver the required discharge against a maximum head of about 100 m. But if the liquid is required to be delivered against a still larger head then it can be done by using two or more pumps in series. Fig. 6. 6 Three stage centrifugal pumpIf the required head is more than that can be provided by one pump, the pumps are connected in series. The same discharge passes through both pumps but the head developed by one pump add the other. The total head developed is obtained by adding together the value of the head of each pump corresponding to the relevant discharge. Pumps in Parallel The multi-stage pumps or the pumps in series as described earlier are employed for delivering a relatively small quantity of liquid against very high heads. However, when a large quantity of liquid is required to be pumped against a relativel y small head, then it may 4 not be possible for a single pump to deliver the required discharge. In such cases two or more pumps are used which are so arranged that each of these pumps working separately lift the liquid from a common sump and deliver it to a common collecting pipe through which it is carried to the required height Fig. 6. 7. Since in this case each of the pumps deliver the liquid against the same head, the arrangement is known as pumps in parallel. If Q1, Q2, Q3†¦.. , Qn are the discharging capacities of n pumps arranged in parallel then the total discharge delivered by these pumps will beQt = (Q1+Q2+Q3+†¦Ã¢â‚¬ ¦+Qn) †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦. (6. 8) If the discharging capacity of all the n pumps is same, equal to Q , then the total discharge delivered by these pumps will be Qt = nQ Fig. 6. 7 Two centrifugal pumps arranged in parallel 55 A centrifugal pump, having four stages in parallel, delivers 11m3/min of Example 6. 1 liquid against a head of 24. 7m , the diameter of the impeller being 225mm and the speed 1700 rpm. A pump is to be made up with a number of identical stages in series, of similar constriction to those in the first pump, to run at 1250 rpm, and to deliver 14. 5 m3/min, against a head of 248m.Find the number of stages required for the second pump. 1st Pump 2nd Pump Q = 11m3/min Q = 14. 5m3/min H = 24. 7 m H = 248 m N = 1700 rpm N =1250 rpm D = 225 mm Specific speed N s = NQ H 3/ 4 Q for one pump = 11/4 = 2. 75 m3/min Ns = 1700 2. 75 24. 7 3 / 4 = 254 For 2nd pump, with identical stages in series i. e multi-stage pump, if each stage is similar to those of each stage is similar to those of the first pump. The specific speed of each stage Ns = 254 Ns = 254 = NQ H 3/ 4 1250 14. 5 H 3/ 4 H = 49. 64 m Total head required = 248 m No of stages required = 248/49. 64 = 5 stages 56 CHAPTER 7DIMENSIONAL ANALYSIS, HYDRAULIC SIMILITUDE AND MODEL INVESTIGATION 7. 1 Dimensional Analysis Dimensional analysis is a mathematical method of obtaining the equations, changing units, determining a convenient arrangement of variable of a physical relation. In an equation expressing a physical relationship between quantities, absolute numerical and dimensional equality must exit. In general, all such physical relationships can be reduced to the fundamental quantities of mass M, length L and time T. It is a based on the assumption that the phenomenon can be expressed by a dimensionally homogeneous equation, with certain variable.The dimensional analysis is widely used in research work for developing design criteria and also for conducting model tests. 7. 2 Dimensions and Units All physical quantities are measured by comparison. This comparison is always made with respect to some arbitrarily fixed value for each independent quantity, called dimension(e. g. , length, mass, time, etc. ). Since there is no direct relationship between these dimensions, they are called fundamental dimensions. Some other quantities such as area , volume, velocity, force etc. can not be expressed in terms of fundamental dimensions and thus may be alled derived dimensions. There are two systems for fundamental dimensions namely FLT (i. e force, length, time) and MLT (i. e. , mass, length, time). One common system employed in dimensional analysis is the M,L,T system. Table is a listing of some of the quantities used in fluid flow, together with their symbols and dimensions. 57 Quantity Symbol Dimensional Form Length l L Time t T Mass m M Velocity v L T-1 Acceleration a L T-2 Force F M L T-2 Pressure P M L-1 T-2 Discharge Q L3 T-1 Power P M L2 T-3 W,E M L2 T-2 Density ? M L-3 Dynamic viscosity  µ M L-1 T-1 Kinematic viscosity ? L2 T-1 Surface tension M T-2 Work,energy 7. 3 Methods of Dimensional Analysis The methods of dimensional analysis are: -Buckingham's ? theorem -Ralyeigh's method Buckingham's ? Theorem If there are ‘n' variables in a dimensionally homogeneous equation, and if these variables contain ‘m' fu ndamental dimensions such as (M,L,T) , they may be grouped into (nm) non-dimensional independent ? terms. Mathematically, if a variable x1 depends upon independent variables x2, x3,x4, †¦. ,xn, the functional equation may be written as x1 = f (x2, x3, x4, †¦. , xn) The equation may be written in its general form as 58 f1 ( x1, x2, x3,†¦Ã¢â‚¬ ¦. xn) = C In this equation there are ‘n' variables. If there are ‘m' fundimental dimensions, the according to ? theorem f2 ( ? 1,? 2,? 3,†¦.. , ? n-m ) = C1 e. g Q = f (d,H, µ,? ,g) f1 (Q, d, H,  µ, ? , g) = C n = 6; m = 3; (n-m) = 3 f2 ( ? 1, ? 2, ? 3) = C1 Procedure 1. First of all, write the functional relationship with the given data. 2. Then write the equation in its general form. 3. Choose ‘m' repeating variables and write separate expressions for each term. Every ? term will contain the repeating variables and one of the remaining variables. The repeating variables are written in exponential form . 4.With the help of the principle of dimensional homogeneity, find out the values of the exponents by obtaining simultaneous equations. 5. Substitute the value of these exponents in the ? term. 6. After the ? terms are obtained, write the functional relation in the required form. e. g Q = f ( d, H,  µ, ? , g) f1 (Q, d, H,  µ, ? , g) =C n = 6 , m = 3 , (n-m) = 3 f2 (? 1,? 2,? 3)=C1 Choose ? , g, d as repeating variable, ?1 = ? a1 gb1 dc1 Q ? 2 = ? a2 gb2 d c2 H ? 3 = ? a3 gb3 dc3  µ 59 Selection of Repeating Variable 1. The variables should be such that none of them is dimensionless. 2. No two variables should have the same dimensions. . Independent variables should be as far as possible, be selected as repeating variable. ? > fluid property ? > flow characteristics l > geometric characteristics Example 7. 1 A V-notch weir is a vertical plate with a notch angle ? cut into the top of it and placed across an open channel. The liquid in the channel is backed up and forced to flow through the notch. The discharge Q is some function of the elevation H of upstream liquid surface above the bottom of the notch. In addition it depends upon gravity and upon the velocity of approach Vo to the weir. Determine the form of discharge equation: ?V ? Q = gH 5/2 f ? ,? ? ? gH ? ? ? Q = f (H, g , Vo , ? ) f1 ( Q, H ,g ,Vo, ? ) = C Choose g and H as repeating variables n = 5; n-m =3 ; m=2 ?1 = Ha1 gb1 Q = (L)a1 (LT-2)b1 L3 T-1 ?2 =Ha2 gb2 Vo = (L)a2 (LT-2)b2 LT-1 ?3 = ? (M)o (L)o (T)o = (L)a1 (LT-2)b1 L3 T-1 a1+b1+3 =0 ? a1 = -5/2 -2b1-1 =0 ? b1 = -1/2 ?1 = H-5/2 g-1/2 Q = Q vg H 5/2 (M)o (L)o (T)o = (L)a2 (LT-2)b2 L T-1 a2 = -1/2 60 b2 = -1/2 ?2 =H-1/2 g-1/2 Vo= Vo vgH Q f2 ( gH 5/ 2 Q gH 5/ 2 , Vo ,? gH ) = C1 Vo ,? gH ) Vo ,? gH ) =f( Q = gH 5/2 f ( Example 7. 2 Q = VD 2 f [ Prove that the discharge over a spillway is given by the relation gD H ,] V D where V= velocity of flowD = depth of throat H = Head of water g = Acceleration due to gravity Q= f (V,D,H,G) f1 (Q,V,D,H ,G) = C Choose V and D as repeating variables n = 5, m =2, n-m =3 ?1= Va1 Db1 Q = (LT-1)a1 (L)b1 (L3T-1) ?2 = Va2 Db2 H = (LT-1)a2 (L)b2 (L) ?3 = Va3 Db3 g = (LT-1)a3 (L)b3 (LT-2) M0L0T 0 = (LT-1)a1 (L)b1 (L3T-1) 0 = -a1-1 ; a1 = -1 0 = a1+b1+3 ; b1= -2 ?1 = V -1 D -2 Q 61 ?1 = Q VD2 M0L0T 0 =(LT-1)a2 (L)b2 (L) 0 = -a2 a2+b2+1 =0 ; b2 = -1 ?2 = V0 D-1 H = H/D M0L0T 0 = (LT-1)a3 (L)b3 (LT-2) 0 = -a3-2 ; a3 = -2 0 = a3+b3+1 ; b3 =1 ?3 = V-2 D g gD V = f2 ( gD H , )=0 VD 2 D V Q VD 2 Q , = f( gD H ,) V D Q = VD 2 f ( 7. 4 D H ,) V D HYDRAULIC MODELS Hydraulic models, in general, may be either true models or distorted models. True models have all the significant characteristics of the prototype reproduced to scale (geometrically similar) and satisfy design restrictions (kinematic and dynamic similitude). Model-prototype comparisons have clearly shown that the correspondence of behaviour is often well beyond expected limitations, as has been attested by the successful operation of many s tructures designed from model tests. 7. 5 Hydraulic Similitude To know the complete working and behaviour of the prototype, from its model, there hould be a complete similarity between the prototype and its scale model. This similarity is 62 known as hydraulic similitude. From the subject point of view, the following three types of hydraulic similitude are important. (1) Geometric similitude (2) Kinematic similitude (3) Dynamic similitude GEOMETRIC SIMILITUDE The model and the prototype are identical in shape, but differ only in size. (The ratios of all the corresponding linear dimensions are equal) . Let L = Length of the prototype B = Breadth of the prototype D = Depth of the prototype l,b,d = corresponding values of the model L Linear ratio Lr =Area ratio Ar = ( Volume ratio Vr = ( l = BD = bd L B D )2 = ( )2 = ( )2 l b d L B D ) 3 = ( ) 3 = ( )3 l b d KINEMATIC SIMILITUDE The model and the prototype have identical motions. ( The ratios of the velocities at corresponding points a re equal) Let V1 = velocity of liquid in prototype at point 1 V 2 = velocity of liquid in prototype at point 2 v1,v2 = corresponding values of the model Velocity ratio Vr = V1 V2 = = †¦.. v1 v 2 DYNAMIC SIMILITUDE The model and prototype have identical forces. (The ratios of the corresponding forces acting at corresponding points are equal). F Force ratio Fr = 1 = f1 F2 †¦.. f2 63 7. 6 CLASSIFICATION OF MODELS (1) Undistorted model (2) Distorted model Undistorted model A model which is geometrically similar to the prototype is known as undistorted model. Distorted model Model does not have complete geometric similarity with the prototype, is known as distorted model. 7. 7 Comparison of an Undistorted Model and the Prototype If the model is to be overall similar to the prototype, then all the three similarities (i. e, geometric, kinematic, dynamic ) should exist. But this is not possible in actual practice, as it is difficult to exist two types of similarities simultaneousl y.In general, and undistorted model of a prototype is made keeping in view the geometric similarity only and the remaining similarities are then compared for the scale ratio. 7. 8 Velocity of Water in Prototype for the Given Velocity of an Undistorted Model Consider an undistorted model geometrically similar to a proposed prototype like a weir, dam, spillway etc. Let h = head of water over the model v = velocity of water at a point in the model H,V = corresponding values for the prototype 1/s = scale ratio of the model to the prototype Velocity of water in the model v = Cv v2ghVelocity of water on the corresponding point in the prototype V = Cv v2gH C v 2 gh v = = V C v 2 gH h 1 = H s V = v vs Example 7. 3 The velocity at a point on a spillway model of a dam is 1. 3m/sec for a prototype of model ratio 1:10. What is the velocity at the corresponding point in the prototype? 64 Velocity in the model v = 1. 3 m/s Model ratio, 1/s = 1/10 s = 10 Velocity in the prototype V = v vs V =1. 3 x v10 = 4. 11 m/sec 65 References: 1. Hydraulics and Fluid Mechanics P. N. Modi and S. M. Seth 2. Hydraulics, Fluid Mechanics and Hydraulic Machines R. S. Khurmi 3. Fluid Mechanics Victor L. Streeter