Impact of a water Jet laboratory - Lab Report Example

? Impact of a water Jet laboratory Impact of a water Jet laboratory In this experiment the key objectives is measuring theforce that is produced by a water jet when it strikes two types of vane, that is, a flat plate and a hemispherical cup, and comparing the results with the theory values obtained from the flux momentum in the jet. The experiment uses two deflectors, which can be categorized in two shapes. The two plates that are used in this experiment are hemispherical and flat shape plates. Force, flow rate, velocity, and the flux were calculated in this experiment. Introduction. Throughout the world, water turbines have been used in the generation of power. This happens when water that is under pressure strikes the turbine vane thus producing mechanical work. The force that is generated gives out rotational motion when the jet hits the vanes. A clear example of a water turbine is the Pelton wheel. This form of a turbine has more than one water jets which are normally tangentially directed towards vanes which are tightened on the turbine disc rim. The water creates an impact on the vanes producing a torque on the wheel. The torque makes the wheel to rotate thus developing power. The prediction of the pelton wheel’s output and determination of the optimum speed of rotation requires the understanding of the jet’s deflection to produce a force at the bucket and its relation to the momentum rate of the jet. This experiment explores the various forces that are exerted by a water jet on different plates. In this experiment, the measurement of the generated force when a water jet strikes a deflector was obtained. Theory. Whenever a horizontal water jet with a velocity v1 hits a freely moving plate, a force would be generated to the plate through the jet’s impact. This force, according to the theory of momentum is equal to the needed force in bringing back the plate in the initial position. This force should be same as the rate of momentum change of the flowing water towards that direction. In this regard, when F is a force of balancing needed to return the plate to the original position, it means that; F = ? O (v1-v2), where V2 includes the velocity of the jet in the direction that is horizontal after hitting the plate, V2 is certainly zero. This implies that F = ? O v1 F= , where is the jet’s cross-sectional area. The calculation of the force due to the jets impact on a plate requires the application of the momentum change principle. According to this principle, the force produced is equivalent to the rate of momentum change. This is given by the equation that described. Additionally, the volume of the rate of flow in the above equation could be calculated in the experiment by calculating the quantity of the volume in a specific time period. Thos equation would be given by; Q=v/t. V1 could be calculated in the experiment through identifying the nozzle velocity and making use of the equations of motion. V2, on the other hand, could be obtained by measuring the nozzle’s diameter and making use of the equation of motion. Therefore, Vnozzel= Q/A. V1 could be calculated using; V12 = Vnozzel2-2gs where by g represents acceleration due to gravity, S represents the distance in between the plates and the jet. From these calculations, the force for the hemispherical cup is F = 2 ? O v1. For the flat plate, F = ? O v1. Apparatus. In this experiment, the following apparatus were used; hydraulic bench, water jet apparatus, loading weight, stopwatch, weighing tank, and deflectors like the hemisphere, plate, and slope. Experiment procedures. In this experimental set up, the procedure was as follow. The flat plate was fitted on the apparatus. After fitting the cup, water was removed from the  cup through undoing the retaining screw and lifting it out. This is completed using the loose cover plate. The cover plate was fitted over the flit plate stem while holding it in position under the beam. Then, the retaining screw was screwed and tightened. It was then that the weigh-beam was set onto its datum position. This was achieved by first setting the jockey weight onto the beam in order to make sure the datum groove becomes zero on the scale. The adjusting nut was turned above the spring making sure the grooves on the tally come in line with the top plate. The bench pump was then switched on, and the bench supply valve opened for admission of water to the apparatus. The supply valve was fully opened alongside sliding the jockey weight on the beam making sure the tally goes back record the readings on the scale that corresponds to the groove that was on the jockey weight. The flow rate was measured through limiting the 15kg or 30kg collection of the water inside the bank weighing bench. The weights of 5kg, 10kg, lead to 15kg and 30 kg of water respectively. The weight’s mass was considered through filling the tank that was small with an equal quantity of water up to when the arm of the tank was at the position of balance. The weights were added, and the timer started. After this, the jockey weight was moved inside by about 15mm. The rate of flow was reduced until the beam was at a level. The beam was set to a position that was exact through having the jockey weight moved, and the scale reading recorded. The flow rate was measured. The step above was repeated up to when six reading were obtained on the flow range. In the final set, the jockey was set to approximately 10 from a zero position. At the flow rate, that was low the mass of water was reduced within the tank to 15 kg. The bench pump was switched off and the hemispherical cup fitted to the apparatus. The process above was repeated in checking the setting of the datum. The bench pump was switched, and the mass m of the weights of the jockey, nozzle diameter, vane distance out of the nozzle outlet was recorded in the table 1. Results. Table 1: Hemispherical data. Time (s) Distance (m) Mass (Kg) 22.22 150 12 22.97 140 12 24.56 130 12 24.66 120 12 25.10 110 12 26.22 100 12 Table 2: Flat plate data Time (s) Distance (m) Mass (Kg) 23.56 80 12 27.36 70 12 31.09 60 12 34.50 50 12 36.94 40 12 43.17 30 12 50.28 10 12 Data Analysis. Response to Question 1. Given that Force is given by the formula F = 4gy Newton, with the displacement of the Jockey measured in meters and g as mass of the jockey, the force for each condition is therefore: For the hemispherical cup ; F1 = 4 x 12 x 0.15 = 7.2N. F2 = 0.14 x 12 x 4 = 8.4N. F3= 0.13 x 12 x 4 = 7.8N. F4 = 0.12 x 12 x 4 = 7.2N. F5= 0.11x 12 x 4 = 6.6N. F6 = 0.1 x 12 x 4 = 6.0N. For the flat plate; F1 = 0.08 x 12 x 4 = 4.8N. F2 = 0.07 x 12 x 4 = 4.2N. F3= 0.06 x 12 x 4 = 3.6N. F4 = 0.05 x 12 x 4 = 3.0N. F3 = 0.04 x 12 x 4 = 2.4N. F2 = 0.03 x 12 x 3 = 1.8N. F1 = 0.02 x 12 x 2 = 1.2N. Response to Question 2. The mass flow rate is given by the equation Q = . For the Hemispherical cup : Q1 = 15/22.22 = 0.6751m3/s. Q2=15/22.97 = 0.6530 m3/s. Q3 = 15/24.56 = 0.6107 m3/s. Q4 = 15/24.66 = 0.6083 m3/s. Q5 = 15/25.10 = 0.5976 m3/s. Q6 = 15/26.22 = 0.5721 m3/s. For the Flat plate. Q1= 15/23.56 = 0.6367 m3/s. Q2 = 15/27.36 = 0.5482 m3/s. Q3 = 15/31.09 = 0.4825 m3/s. Q4 = 15/34.50 = 0.4348 m3/s. Q5 =15/36.94 = 0.4061 m3/s. Q6 = 15/43.17 = 0.3474 m3/s. Q7 = 15/50.28 = 0.2983 m3/s. Response to question 3. Velocity is given as the distance moved over the time used. Table 3: Velocity of water in the Hemispherical plate. Time (s) Distance (m) Velocity (m/s) 22.22 150 6.75 22.97 140 6.095 24.56 130 5.293 24.66 120 4.866 25.10 110 4.382 26.22 100 3.814 Table 4: Velocity of water in the Flat plate. Time (s) Distance (m) Velocity (m/s) 23.56 80 3.396 27.36 70 2.558 31.09 60 1.930 34.50 50 1.449 36.94 40 1.083 43.17 30 0.6949 50.28 10 0.1989 Response to question 4. The flux of water is given by F = ? O v1. From above v1 and Q for jet in different plates are known. ? for water = 103 kg/m3 is also given. This means that the flux of water would be as in the table 5. Table 5: Flux of water in the Hemispherical plate. Mass flow rate (Q) ? for water Velocity (m/s) Flux of water 0.6751. 103 6.75 469.4 0.6530 103 6.095 409.9 0.6107. 103 5.293 332.9 0.6083. 103 4.866 304.9 0.5976 103 4.382 269.7 0.5721 103 3.814 224.7 Table 6: Flux of water in the Flat plate. Mass flow rate (Q) ? for water Velocity (m/s) Flux of water 0.6367 103 3.396 222.7 0.5482 103 2.558 144.4 0.4825 103 1.930 95.92 0.4348. 103 1.449 64.89 0.4061 103 1.083 45.30 0.3474 103 0.6949 24.86 0.2983 103 0.1989 6.111 Table 9 Force F(hemispherical plate) Mass flow rate (Q) Flux of water in hemispherical plate F(flat plate) Mass flow rate (Q) Flux of water in flat plate 7.2. 0.6751. 469.4 4.8. 0.6367 222.7 8.4 0.6530 409.9 4.2 0.5482 144.4 7.8 0.6107. 332.9 3.6 0.4825 95.92 7.2 0.6083. 304.9 3.0 0.4348. 64.89 6.6 0.5976 269.7 2.4 0.4061 45.30 6.0 0.5721 224.7 1.8 0.3474 24.86 0.00 0.00 1.2 0.2983 6.111 Conclusion. In this experiment, the force calculated has a correlation with the measured force. The force has a relation that is directly proportion. In theory, the calculated force is expected to be similar to the force measured. Achieving this experimentally is not easy due to errors in the experiment and the experiment apparatus losses. This loses were utilized in the calculation of the flux for the two plates thus errors in the calculation. In the experiment, the flow rate for the plate that is hemispherical is the lowest. This means that the hemispherical plate require a long duration of time in order to increase the volumetric tank from 15 to 30 litres. From the plotted graph, it is observed that as the rate of flow increases, the force due to the jets impact on the two plates increase in both the measured and the predictive value of F. The display results do predict the equation of the momentum change in the force calculation. In addition to this, effect of the jet’s force on the plate that is flat is seen to be approximately half that of the hemispherical cup. This is normally predicted by the momentum equation in the calculation of force. The last point of the hemispherical plate is extremely significant for use in turbines than that on the flat plate. The water exiting the cup collides with water going into the hemispherical cup hence reducing the force. This implies that the hemispherical cup is normally made to be of an angle below one hundred and eighty degrees. References Douglas, J F., 1999. Gasiorek J M & Swaffield J A “Fluid Mechanics”, Pitman Ltd, London Massey B S., 2001. “Mechanics of Fluids”, Van Nostrand Reinhold Ltd, London   Lide, D., 2005. CRC A handbook of Chemistry and Physics. Boca Raton.CRC Press. Laurence, Harwood, C., 2000. Experimental organic chemistry: Principles and Practice (Illustrated ed.). Simon, E., & Jean, L., 2013. Campbell Essential Chemistry (5th Edition). Retrieved on 8th Oct 2012 from www.amazon.com. Singer, P., 1993. Life: Human. Practical ethics. Cambridge University Press. 1993. Syndrome. Web Reddy, U.M., et al, 2008. Prenatal Imaging: Ultrasonographies and the Magnetic Resonance Imaging. Obstet Gynecoll; vol 113(1): pp 144-158. Wiseman, F., et al, 2009. Human Molecular Genetics; vol 19: pp R76-R83. Webber, N,. 2008. “Fluid Mechanics for the Civil Engineers”, Chapman and Hall Ltd, London Read More