Roosevelt And Roosevelt s President Essay - 1495 Words

Throughout the beginning of the 1900’s, Roosevelt became immensely favored and adored by the majority of United States citizens. After Theodore Roosevelt served his terms of presidency from 1901-1909, he declared that he would not accept a renomination for another term. With being in control of the Republican Party and also becoming quite favored, Roosevelt was able to name who his successor would be. With having being so popular and trustworthy, Roosevelt commanded the Republican Party to stay loyal to his ways by nominating and supporting Roosevelt’s secretary of war, William Howard Taft. When Roosevelt left office, he stated, â€Å"I have the profound satisfaction of knowing that he [Taft] will do all in his power to further every one of the great causes for which I have fought and that he will persevere in every one of the great governmental policies in which I most firmly believe† (6). However, when election time came around, many Americans were not impressed by the actions that Taft had taken and the way he carried himself. Many Americans were against Taft and believed that he was not competent enough to take on the job. When Taft ran for his second term in 1912, his opposers, betrayal of the Republican Party and his seemingly lack of political knowledge and training cost him the election. Mid-way through his first term, Taft had changed his motives and wanted to remove so called â€Å"disloyal† members from the Republican party. Around this time, Taft was also in aShow MoreRelatedPresident Roosevelt s President Of The United States Essay1267 Words   |  6 PagesWhen President McKinley was assassinated Vice President Theodore Roosevelt became President of the United States at the age of 43. He is the youngest president in our nation’s history. 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Compared to Roosevelt, Taft was not as energetic and ardent; his policies caused much controversy from the Republican Party and Roosevelt himself (Brinkley 606). In addition to Taft’s passivity towards Congress

Feasibility of the National Health Insurance in South Africa

In the current society; South Africa’s health care has been left with a major gap, in quality and accessibility, between the wealthy and the impoverished, in the country. With the wealthier minority having access to the private sector of medicine, a sector with a higher and better level of health care, and the poverty-stricken majority having limited access to only the public sector, the sector funded by the government. Therefore the current South African government, the ANC, has proposed a policy called the NHI; that looks to lessen the gap between the public and private sectors of medicine, by: improving the accessibility to health services for all South Africans, irrespective of whether they are employed or not, to pool funds so that equity and social solidarity will be achieved through the creation of a single fund, to obtain services on behalf on the entire population and to strengthen the under-resourced and strained public sector so as to improve health systems perfor mance. (The South African Medical Journal, 2013) This essay will be discussing what the NHI is and the feasibility of the NHI, with reference to the current level of South African health care, South African culture, socio-economic background, medicalisation, Due to the apartheid regime of South Africa prior to 1994; the South African health care system has been left fragmented with the level of health care provided to the individual being based on their racial backgrounds, with the white minorityShow MoreRelatedThe Feasibility of the National Health Insurance in South Africa514 Words   |  2 PagesHealth care is a basic human right that every person is entitled to receive. Health care programs exist throughout the world, with the intention of providing quality health care to all members of society. 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Fluid Flow Viscosity Shear - Free Sample

Questions: VISCOSITY :Shear1. (a) A yield stress material is found to be best described by the Bingham-Plastic model. Explain the terms which are required to represent the material in this model and suggest an alternative model for materials which display the following characteristics: shear-thinning, shear-thickening. Illustrate your flow models using diagrams of shear stress against shear rate and apparent viscosity against shear rate. What are the limitations of the model when applied to shear-thinning fluids and can it be modified to account for a yield stress ? Your answer should refer to the following terms and define each of them: shear stress ; shear rate ; apparent viscosity ; zero-shear viscosity; consistency. Suggest a flow model for (i) shear-thinning, (ii) shear-thickening fluids and (iii) Newtonian fluids and explain your answer using appropriate diagrams of shear stress against shear rate and apparent viscosity against shear rate. 1. (b) A pressure drop of 105 N/m2 is developed when a fluid flows through a straight circular pipe 20 m in length and 25 mm in diameter. The fluid is Newtonian with a shear viscosity of 1 Pa s. A chemical is added to the fluid and its flow properties are changed. A graph of shear stress against shear rate and apparent viscosity against shear rate shows (i) that the changed fluid is shear-thinning, with a shear-thinning power-law index = 1/3 ; and (ii) that its apparent viscosity is the same as that of the original fluid at a shear rate of 1000 s-1. Calculate the volumetric flowrate for the shear-thinning fluid at the original value of the pressure drop.1. (c) A power-law fluid has a density of 1075 kg/m3. It is pumped at a rate of 2500 kg/hour through a pipe of internal diameter 25 mm.The flow is laminar and the power law constants are K2 = 3 Pa.sn and n = 0.5. Estimate the pressure drop over a 10 m straight length of pipe and the centre-line velocity for these conditions.2. (a) You are provided with information from a viscometer for 3 different fluids. The information consists of the results of measurements of shear stress for different values of shear rate. Using appropriate diagrams, explain how you would use this information to characterise the fluids in terms of their non-Newtonian flow characteristics according to (i) the Bingh am-Plastic; (ii) the shear-thinning fluid and (iii) the shear-thickening fluid models, respectively. Suggest a suitable model for the shear-thinning fluid and discuss any limitations it may have.2.(b) A material flows in a pipe of 0.15m diameter at a velocity of 0.5 ms-1. The relation between shear stress and shear rate is = 2.5 0.2 (SI Units)As a result of temperature change the Power Law index becomes 0.25 but the apparent viscosity is unchanged when the shear rate is 100 s-1. Calculate the percentage increase in the mean velocity of the material at the same pressure drop and at the new temperature.(b) A horizontal pipe of circular cross-section and 600 mm diameter carries water under a head of 30 m with a velocity of flow of 3 m s-1. If the pipe turns through a 75 degree bend, calculate the magnitude and direction of the resultant force on the bend.NEWTONS SECOND LAW:Bernoulli1. (a) Explain how the Bernoulli Energy Equation can be obtained from considerations of the forces act ing on a streamtube of fluid. Ensure that your answer explains the significance of the terms in the Equation and its limitations. What principle could be used to create a differential head flowmeter based on this Equation ?1. (b) Water flows through a pipe of inside diameter 200 mm at a rate of 100 m3h-1. The flow abruptly enters a section which reduces the pipe diameter to 150 mm, for which the head loss is equivalent to 0.2 velocity heads based on the smaller pipe. If the gauge pressure is 80 kNm-2 upstream of the reducing section, find the force needed to hold the section in position.2. (a) You are required to create a differential head flowmeter based on a convergence in a section of pipe. Starting with an expression for the forces acting on an ideal fluid, show how you would estimate the volumetric flowrate based on measurements of differential pressure. How would you modify your answer to account for the fact that a real fluid will have viscosity and how would you ensure that energy degradation is minimised? 2.(b) A manometer uses a manometric fluid of density 1075 kg/m3 to measure the pressure drop across an orifice plate with a throat diameter of 75 mm. The orifice plate is placed inside a vertical pipe with a diameter of 225 mm and oil with a density of 860 kg/m3 is flowing upwards inside the pipe. The deflection of the manometer fluid is 0.5 m and the discharge coefficient of the orifice is 0.659. What is the flowrate of the oil? 3. (a) Beginning with Newtons Second Law calculate the force required to stabilise a 90o horizontal pipe bend against movement due to hydrodynamic reaction forces. State any assumptions you would make and explain how you would calculate the direction of the force. 3. (b) A jet of water of 22.5 cm diameter impinges normally on a flat plate moving at 0.6 m s-1 in the same direction as the jet. If the discharge is 0.14 m3 s-1 find the force and the work done per second on the plate.4. (a) Explain how you would create a differen tial head flowmeter based on convergence at an orifice plate placed in a section of pipe. Starting with an expression for the forces acting on an ideal fluid, show how you would estimate the volumetric flowrate based on measurements of differential pressure. 4. (b) A horizontal venturi meter with a discharge coefficient of 0.96 is to be used to measure the flowrate of water up to 0.025m3s-1 in a pipe of internal diameter 100 mm. The meter is connected to a differential manometer containing mercury (Specific Gravity,SG = 13.6). If the maximum allowable difference in mercury levels is 80 cm, what is the diameter of the throat ?5. (a) Using Newtons Second Law as a starting point, explain how you would create a flowmeter based on a converging section of pipe for a real (non-ideal) fluid. Your answer must explain how the degradation of energy is minimised and how you would estimate the volumetric flowrate based on measurements of differential pressure.5. (b) Obtain an expression for the force exerted by a jet of liquid which leaves a nozzle and strikes a stationary flat plate normally with a velocity v. How would this expression be modified if the plate were to be moving in the same direction as the jet with a velocity u ? Explain any assumptions which you make.5. (c) Two pressure gauges are located at tapping points 50 cm apart on a vertical Venturi tube which has an inlet diameter of 150 mm, a throat diameter of 70 mm and a discharge coefficient of 0.98. If a liquid of density 1,000 kgm-3 flows upward through the Venturi tube at a rate of 0.075 m3s-1 what is the difference in reading of the two pressure gauges ?5. (d) Droplets of oil (density = 960 kg/m3) are dispersed as an emulsion in a solution with a density and shear viscosity the same as water. Calculate how long an 80 m spherical droplet will take to rise from the bottom of a tank to the surface 1.4 m above in still liquid. Neglect acceleration effects and state any assumptions you make.RUSHTON TURBINE1. ( a) Explain what factors influence the amount of power input required for fluid agitation and mixing in a standard Rushton turbine. Use this mixer configuration to explain why scale-up under conditions of same torque per unit volume is equivalent to performing scale-up at constant tip speed in the fully turbulent region of mixing. Explain why the same mixing time criterion can be prohibitively expensive. 1. (b). You are required to agitate water with a standard Rushton turbine. The tank diameter is 2 m and you are required to work to a tip speed design criterion of 3 m s-1. Assuming a Power Number of 6, calculate:(i) the power required per unit volume of fluid(ii) the speed that the impeller should be driven at in a geometrically similar 4 m diameter tank on the basis of scale-up at equal power per unit volume State any assumptions you make 2. (a) Describe the configuration of a standard Rushton turbine and use this mixer configuration to explain (i) why the same torque per unit volume scale-up criterion is equivalent to performing scale-up at constant tip speed in the fully turbulent region of mixing; and (ii) why the conditions under which scale-up based on the same mixing time criterion could prove very expensive. (b). A horizontal pipe with an inside diameter of 200 mm has a 180o U-bend and carries a fluid of density 900 kgm-3 at a rate of 150 m3h-1. Find the force exerted by the liquid on the bend if the gauge pressures upstream and downstream of the bend are 100 kPa and 80 kPa, respectively.3. (a) What are the main parameters influencing power input for fluid agitation and mixing. Explain why the same mixing time scale-up criterion may be prohibitively expensive. Refer to the fully turbulent region of mixing and the standard Rushton-Turbine configuration in your answer. 3(b). Tests on a small scale tank 0.3 m in diameter show that a blending process between two miscible liquids (both aqueous with properties the same as water) is complete after 1 minute when using an impeller speed of 250 revolutions per minute.It is decided to scale up the process to a tank of 2.5 m diameter using the criterion of constant tip speed.(i) What speed should be chosen for the larger impeller ? (ii) What power will be required ? (iii) What will be the blend time in the large tank ? State any assumptions you make. Assume a standard Rushton-turbine configuration, and a Power Num ber of 6.LAMINAR/TURBULENT FLOW: Reynolds Number1. (a) Explain the meaning of the terms friction factor, Reynolds Number and relative roughness and how they are used in the construction of the Moody plot. Verify the relationship between friction factor and Reynolds Number given in the Moody plot for the laminar flow region. 1. (b) A foam consists of droplets of oil (density 960 kg/m3) which are dispersed as an emulsion in a solution whose density and viscosity is the same as water. Calculate how long an 80 micron diameter ( 1 m = 1 x 10-6 m ) spherical drop will take to rise from the bottom of a tank to the surface 1.4 m above in a still liquid (neglect acceleration effects).2. (a) The flow of a viscous fluid past a spherical particle is characterised by the Reynolds Number, Re. Explain the form of the curve obtained when Re is plotted against the Drag Coefficient. Ensure that your answer includes:(i) an explanation of the terms involved in Re and the Drag Coefficient, (ii) an expla nation of the term 'separation', (iii) an account of the changing drag force, F, on the particle. 2. (b) A water softener consists of a vertical cylindrical pipe 0.5 m long and 50 mm internal diameter packed with an ion-exchange resin consisting of spherical particles whose diameter is 1 mm. The bed porosity is 0.33. The column runs full of water, under a head of 0.2 m, but water trickles out slowly from the bottom of the column which is supported by a perforated plate. Calculate the flowrate and verify any assumptions which you make. Answers: 1. a) Bingham plastic can be defined as such kind of viscoplastic materials which acts as the rigid body at minimum stresses and acts as the viscous flow at maximum stresses. It is widely used in drilling engineering for model of mathematical flow of the mud. The Bingham plastic model has been showcased below; The alternative model is Carreau model which displays the shear-thinning and shear thickening characteristics of the model. The Carreau model is such kind of Newtonian generalized fluid where the viscosity of the material always depends on rate of shear of the material. The equation of the Carreau model is given by, , where the material coefficients are n, and is the viscosity at zero rate, is the viscosity at the shear rate of infinite, relaxation time is and n is the power index. The above diagram showcases the graph plot between the shear rate and the dynamic viscosity. The equation, = K( )n1 is given then reduced form of the Carreau model where k is consistency factor. The above plot depicts the relationship between the shear rate and the viscosity. The plot showcases the decreases of the viscosity of the fluid of shear thinning materials with the increasing of the shear rate. Limitations The limitation of the model is that it cannot be fitted for any kind of polymer solutions in minimum and maximum shear range. The ranges between the lower and higher shear ranges showcase the zero shear viscosity and the infinite values of shear viscosity. This kind of model is best suited for the medium level of the polymer industry. i) Shear thinning model The above diagram showcases the shear thinning model which describes the plot between the viscosity and the shear rate. ii) Shear thickening model Herchel-Bulkley fluid model showcases the shear thickening condition of the viscosity. The plot showcases the relationship between the shear rate and the viscosity in Herchel-Bulkley fluid model. iii) Newtonian fluids The example of Newtonian fluids is the Bingham model which is described below in the plot. b) The volumetric flow arte equation is given by, (8u/D= 4Q/R^3) = 8*1/3 /25 =4Q/ 3.14*(0.02)^3= 6.70*10^-7 m^3/s. c) The flow characteristics is denoted by, 8u/D = 8 / 25 = 0.32 s^-1 (8u/D)n-1 = 0.32 (0.5-1)= 1.76s ap = K(8u/D)^-0.5 = 3*0.32^-0.5=1.020 Pa sAnd Re = uD/ ap = 25*1/1.020= 24.50So, the pressure drop is given by, PF = 4f(L/di) u2 /2 = 2*0.0075(25)*0.01/25 = 1.5 Pa. 2. a) The above diagram depicts the plot between shear rate and the shear stress of the three kinds of fluids such as shear thickening, shear thinning and the Bingham plastic. Shear thickening plot showcases that the rate of the shear increases with the increasing of the dilatants fluid. The shear thinning fluid can be called as the Pseudo plastic fluid materials. The Bingham plastic graph plot does not pass through origin point of the graph. Limitations The limitation of shear thickening fluid that it cannot pass through any kinds of solutions. b)The calculation is given by, = /P = 100/2.5 = 40*100 = 4000 m/sc) FX = -P1A1 - P2A2 cos(b) - d Q [V1 + V2 cos(b)] and Fy = P2 A2 sin(b) + d V2 Q sin(b) so, F can be denoted as F= (Fx2 + Fy2)1/2 , FX = 706.5*sin(75) = 682.42 Newton's Second Law 1. a) The Bernoulli Energy Equation is stated that the motion of fluid is increase by decreasing its pressure or its potential energy. The mathematical expression of Bernoullis theorem is like:PV+ m/2v2 = Constant, This is a correct statement to proceed with steam fluid for different cross sectional areas. Now the volume V at any point of this steam flow is the sum of its kinetic energy and potential energy. The equation looks like that V= mv2/2 + PVNow the energy of a fluid is moving from one point to another point with same motion then its energy equation is like P1V+mv12 = P2V +mv22 By using m=v, Bernoulli energy equation can obtain, that is: P1-p2 = /2(v22 v11) This equation derives the Bernoulli energy equation with the help of Newtons second law. This theorem is used in fluid mechanics for steady streamline regions of flow. It can control the motion of fluid by decreasing the pressure of air or potential energy of fluid. The Bernoulli theorem is limited for heat transformation of a particle including its mechanical transformation. This theorem will reduce the thermal energy. With the help of nonlinear relationship in between flow and pressure, an accuracy of flow measurement can change the pressure of flow meter. This common relationship in between pressure and fluid flow can control the flow meter pressure. This is a successful approach of Bernoullis theorem to create the main principle of flow meter. b)The velocity for larger and smaller pipes isV1= 4Q/3.14d12 = 4*100/3600/3.14*0.22 = 0.884 ms -1 AndV2= 1.67 ms -1 This is the head loss based on the velocity of smaller pipeHL= 0.2 * 1.57^2/2g= 0.025The pressure in 200 mm diameter including 80kNm-2 including 150 mm diameter pipe is found by applying Bernoullis equationP2= P1 + /2(V12 +V22) gHL = 80 * 10^3 +500*(0.884^2-1.57^2) 1000*g*0.025 = 78.913 Nm-2 The upstream and downstream pressure are, P1a1 = 2513 NAnd P2a2 = 1394 NThe force in the X direction isFX = 1000* 100/3600*(1.57-0.884) 2513 + 1394 = -1100 NSo, 1.1 kNm-2 is the opposite direction to hold the reducing sector in position. 2. a) Figure: section of a pipe Here static head = P/yDynamic head = (P/y + v^2/2g)Procedures Venturis Nossles OrificesIt depends on Flowrate Fluid properties Element geometry Figure: flow meter based on pipe Bernoullis equation of energy conservation P1 + 1/21U12 =P2 + 1/22U22 = constant =P0 P0 = total pressure at medium Total sum of dynamic and static pressure is same throughout the whole pipe p = P1 P2 =/2(U22 U12) Volumetric flow rate is defined as Q A1 = pipes area A2= pipes flow area P1 = upstream pressure P2 = restriction pressure = density C= correction factor Viscosity of fluid is basically the measurement of the resistance to steady deformation by tensile stress or shear stress. Real fluid has also effected by tensile stress and shear stress. So it is ensured that fluid has viscosity. From the Bernoullis equation it is ensured that the degradation of energy is minimized. b) Density of manometric fluid = 1075 kg/m3 Diameter = 75 mm The diameter of orifice plate = 225 mm Density of orifice plate = 860 kg/ m3 Deflection of manometer fluid = 0.5 m Discharge coefficient of the orifice = 0.659 Flow rate of the oil = Q = VA = V * D2/4 [A = D2/4] Where Q = flow rate V= Viscosity A = Area D= Diameter of the pipe Flow rate of oil = 0.5** 2252/4 0.659* * 752/4 = /4(0.5 * 2252 0.659 * 752) = 16969.019 mm = 16.969 m3 s-1 The flow rate of oil is 16.97 m3 s-1 3. a) The pressure required to stabilize the pipe against movement is done by applying Bernoulli in between V1 and V2 points is as follows P1 + V12 = p2 + V12 So the pressure forces Fpx1 = p1A1 = 1200 N Fpy2 = p2A2 Assume the inlet pressure of V1 is 4m/s and the outlet pressure V2 is 16 m/s So the direction will go through from inlet pressure to outlet pressure that is from V1 to V2 So, v = (4^2 + 16^2) = 16.49 m/s And Fm = m2v= 659.7 N Now the pressure for resolve section Fmy = 659.7 sin 75.96 = 640 N Fmx = 659.7 cos 75.96 = 160 N So, the total force is acting at X direction is, 1200+160= 1360 N The force acting at y direction = 0 + 640= 640 N Here no initial force is acting towards Y direction so the direction of the actual force is acting towards X Axis. b) The resulting force of the jet is F= m v1 {2(1-cos)} F = 45 N 4 a) A differential head flow meter based on convergence at an orifice plate placed in a section of pipe Here P = Pressure D = Density Q = Flow rate Flow meter is basically a process which is used to determine the volume of the gas which is passing through the nozzle with the per unit time. The creation of differential head flow meter is depended on three procedures. These procedures are basically Venturis procedure, Nozzles procedure and Orifices procedure. The orifice procedure is dependent on the Flow rate of the fluid, various properties of the fluid and the geometry of element. Volumetric flow rate Figure: Orifice meter pipe The volumetric flow rate is basically defined by the below equation: Q= flow rate C= correction factor = density A1 = pipes area A2= flow area of the pipe P1 = upstream pressure P2 = restriction pressure b) Discharge coefficient of venture meter = 0.96 Flow rate of water = 0.025 m3 s-1 Internal diameter = 100 mm Specific gravity = 13.6 Maximum difference = 80 cm [in mercury] The equation of flow rate Q = u2A2 = Cd {2R (pm p) g/p}* A1A2/ (A12 A22) Where A= Area P = density G= gravity A2 = 13.6 * 2* 80/0.025 * 1002 = 33 meter The diameter of the throat is 33 meter. 5. a) The most common flow meter is obtained from Newtons second law. In numerical form the common formula is written as: m1 = m2 = A1V1/v2 or A2V2/v1 If the V is constant then the equation is simplified to a new form A1V1= A2V2 Using Bernoullis theorem the flow meter of a pipe for a real fluid is represented as V1 A1= V2A2 or V1= A2V2 /A1 b) Suppose a Nozzle has a n inlet area is 0.005 m2 and the discharge diameter is combined with its atmosphere. The gauge bar including inlet of the nozzle is 3 Bar. Since the area of vertical place is coming from Fv =0 by using gauge pressures, so the final force is Zero at exit point. Fpx2 = 0, so FH = 1500 0= 1500 N to the right angle. c) The Bernoullis equation has been stated below, p1/+v12/2g=p2/g+v22/2g+z2 Continuity equation has been mentioned below, A1v1=a2v2 After the rearrangement equation of the differential pressure has bee stated below, P1-p2=g(v1^2/2g((a1/a2)^2-1)+z2-z1) Equation of the flow rate is Q=Cdav1 Differences between two pressure gauges= P1-P2=500x (.075/(.98x3, 141x.15^2/4) ^2x ((.15/.07)^4-1)+2x9.8x.5) =290,232.18 NRA^-2 =2902 KNm-^2 Rushton Turbine 1. a) The factors are the Reynolds number, power number, flow meter and the blend time of the dimensionless. Scale up under the conditions of same torque per unit volume is equivalent to performing scale up at the constant speed in the turbulent region of mixing because, torque per volume can be termed as the mixing intensity in fluid velocities and it related to the effective motion of the mixer. b) i)The equation is given by, Where P0 is the dimensionless power number, N is the speed of the rotational, is the fluid density. So, P= 6*3^3*2 = 324 Pa ii) so, N^3 = P/p0 D^5 = 324/(6*4^5)= 0.03 m/s where, Where P0 is the dimensionless power number, N is the speed of the rotational, is the fluid density. 2. a) The factors are the Reynolds number, power number, flow meter and the blend time of the dimensionless. Scale up under the conditions of same torque per unit volume is equivalent to performing scale up at the constant speed in the turbulent region of mixing because, torque per volume can be termed as the mixing intensity in fluid velocities and it related to the effective motion of the mixer. b) The equation is given by, FX = -P1A1 - P2A2 cos(b) - d Q [V1 + V2 cos(b)] and Fy = P2 A2 sin(b) + d V2 Q sin(b)FX = -100*200- 80*200cos (180)- 150(900)= -139000 Fy = 0 so, F = 139000 pa 3. a) The main parameters are which influencing the power input for the agitation and mixing of the fluids are mixing time and the circulation time. The same mixing time scale-up criterion may be prohibitively expensive because of larger tank mixture. b) i)The required speed can be calculated by, N^3 = P/p0 D^5 = 100/(6*0.3^5) = 5.29*10^-5 ii) The required power can be calculated by, = 6*250 30.3^5 = 227812.5 Pa iii) So, the required blend time can be calculated by, 227812.5 * 5.29*10^-5*227812.5^-(1.69/5.29) = 0.234 Laminar/Turbulent Flow: Reynoldss Number 1. a) The head loss or various pressure losses due to the loss of the friction can be called as the friction factor. Friction factor can also be referred to as the Darcy friction factor or the Moody friction factor. Renolds number can be defined as such kind of dimensionless quantities which can help to predict the similar kinds of flow patterns in various fluid situations. The ratio between the roughness of the duct and the diameter of the hydraulic can be referred to as the relative roughness. Which can be expressed as r = k / dh The above diagram depicts the moody plot between the friction factor and the Reynolds number. b) The length of the oil droplet can be calculated by, mg= d Where m is the mass of the oil droplets, d is diameter of the droplet and is length of the droplet. So, volume can be calculated by, 4/3 r3 = 4/3*3.41*(80*10^-4) = 0.036 m^3 So, length can be calculated by, 0.036/ (3.41* 80*10^-4) = 1.31 2. a) The above plot showcases the relationship between the Reynolds number and the drag coefficient. The separation can be termed as the relationship between the removal of the impurities and the components of the substances. b) Flow rate can be calculated by, * (pipe diameter) 2* velocity = *3.14*(50)2* 0.33 = 647.625 References Bourne, M. (2002).Food texture and viscosity. San Diego: Academic Press. Bowdler, D. and Leventhall, H. (2011).Wind turbine noise. Brentwood, Essex: Multi-Science Pub. Boyce, M. (2006).Gas turbine engineering handbook. Boston: Gulf Professional Pub. Chapman, S. and Cowling, T. (1970).The mathematical theory of non-uniform gases. [Cambridge, Eng.]: Cambridge University Press. Cohen, H., Rogers, G. and Saravanamuttoo, H. (1987).Gas turbine theory. Burnt Mill, Harlow, Essex, England: Longman Scientific Technical. Du Bois, W. (1996).The souls of Black folk. Charlottesville, Va.: University of Virginia Library. Falcone, M. and Makridakis, C. (2001).Numerical methods for viscosity solutions and applications. Singapore: World Scientific. Fleming, W. and Soner, H. (2006).Controlled Markov processes and viscosity solutions. New York: Springer. Fowler, H. and Gowers, E. (1965).A dictionary of modern English usage. New York: Oxford Univ. Press. Giampaolo, T. (2006).Gas turbine handbook. Lilburn, GA: Fairmount Press. Glasstone, S., Laidler, K. and Eyring, H. (1941).The theory of rate processes. New York: McGraw-Hill Book Company. Hudson, W. (n.d.).Green mansions. Champaign, Ill.: Project Gutenberg. King, P. (1981).The turbine. [Dunedin, N.Z.: P. King. Lefebvre, A. (1983).Gas turbine combustion. Washington: Hemisphere Pub. Corp. Maugham, W. (1936).Of human bondage. Garden City, N.Y.: Doubleday. Maugham, W. (1995).The moon and sixpence. Champaign, Ill.: Project Gutenberg. Neilson, R. (1902).The steam turbine. London: Longmans, Green, and Co. Oates, G. (1997).Aerothermodynamics of gas turbine and rocket propulsion. Reston, VA: American Institute of Aeronautics and Astronautics. Philbrick, W. (2009).The mostly true adventures of Homer P. Figg. New York: Blue Sky Press. Tuchman, B. (1962).The guns of August. New York: Macmillan. Tuchman, B. (1978).A distant mirror. New York: Knopf. TURBINE., (1904).Die Turbine. Zeitschrift fur modernen Schnellbetrieb, fur Dampf-, Gas-, Wind- Wasserturbinen. (Organ der Turbinentechnischen Gesellschaft.) Jahrg. 1. Hft. 1-Jahrg. 9. Hft. 24. Okt. 1904-Sept. 1913. Berlin. Walsh, P. and Fletcher, P. (1998).Gas turbine performance. Oxford: Blackwell Science. Yeats, W. (1956).The collected poems of W.B. Yeats. New York: Macmillan.

Robert Frost Essays (782 words) - Robert Frost,

Robert Frost Robert Lee Frost was born in San Francisco, California, on March 26, 1874 and was the son of William Prescott Frost and Isabelle Moodie Frost. After his father died in 1885, the family returned to Lawrence, Massachusetts, which was the home of Frost's grandparents. There he grew up through his high school years. After less than a year at Dartmouth College, he left to work in textile mill and to marry Elinor White, a high school classmate. When his academic experience at Harvard disappointed him, Frost returned to Lawrence and had a variety of jobs. Finally, he became a chicken farmer in Derry, New Hampshire, on property that he bought from his grandfather. In 1912, Frost took his family to England, hoping that the residence there would help advance his poetic career. A British publisher accepted his first two volumes of verse, A Boy's Will (1913) and North of Boston (1914). Both were published in the United States in 1915, the year the Frost family returned him and settled on a farm in Franconia, New Hampshire. He then became a summer farmer and poet-teacher, just like he was in Derry. Except for brief periods at the University of Michigan and Harvard, he spent his academic years 1916-1963 mainly at Amherst College. Meanwhile, as he was finishing the poem collection New Hampshire (1923), he decided that most of his living should be done in Vermont, where he helped create and sustain the Writers' Conference at Middlebury College's Bread Loaf School of English. Frost's eventual poetic success was counter-pointed by much personal grief and loss. Several of the Frost children were stillborn or died in infancy - they are remembered in the poem Home Burial. Frost's son committed suicide and his daughter became insane. After his wife's death in 1938, the poet lived either alone or with friends. He died in Boston on January 29, 1963. Frost kept his religious faith mostly to himself or confided it only to close friends (Smith). When it entered his poetry at all, it was usually in a very guarded fashion. Earlier poems such as Sitting by a Bush in Broad Daylight and Not All There imply religious attitudes, and later ones - A Masque of Mercy, Accidentally on Purpose, and Kitty Hawk - are explicitly religious. The ?dark? poems - Spring Pools, A Leaf Trader, Design and The Draft Horse - expressing tragic moods rather than hard-won convictions, and the poems of endurance, like Stopping by Woods on a Snowy Evening, seem more deeply felt and more perfectly executed. And it seems Frost knew instinctively that they would have more appeal in a naturalistic age. Robert Frost, an established American poet, lived to become his country's unofficial poet laureate. He won the Pulitzer Prize for poetry four times and was awarded the Bollingen Prize posthumously. The U.S. Senate honored him on his 75th and 85th birthdays, and he had a prominent part in the inauguration ceremony for President John F. Kennedy in 1961, speaking the poem The Gift Outright, which he had written for the occasion. The poem, Stopping by Woods on a Snowy Evening, is about a man, or the author, that was going through his hectic life and than all the sudden, one evening, he actually stopped to look at his surroundings. He realized how beautiful his life and this world was and that sometimes there's too much going on to enjoy this. This poem is a metaphor for life. So many people are involved in so many things that they can never enjoy what's happening right now in their lives. The author sits for a minute, studies his surroundings for once and then realizes that there is too much to do to just sit there. He finishes the poem by saying, ?The woods are lovely, dark and deep, But I have promises to keep, And miles to go before I sleep, And miles to go before I sleep.? This is the part where he comes back to his senses and realizes that he can't just sit there, that he must return to the real world and finish what has to be done in life before he can actually stop. The reason I picked this poem is because I can completely relate to it. Sometimes, in