Impact of a Jet Issues - Lab Report Example

Lab Report: Impact of a Jet Presented by: Names (Id) Names (Id) Names (Id) Introduction Water turbines are a widely used for hydro-power generation. Hydro power accounts for over 25% of total power generation (Heshmati, 2012). The power generation is based on very simple principle which lies in the impact of a jet whereby the force of water impacting on a cap results in the torque that turns the turbine. This is the most important. Research on the factors affecting the force of impact is therefore an important venture since this enhances maximum power production. According to Burke( n.d.), the shape of the cap has a great impact on the power produced. An experimental investigation into how the effects of a cap affects the power is a very important venture. The shape of the cap simply determines the deflection of the jet on collision with the cap. This deflection is what determines the force impact of the jet on the turbine hence the torque produced (Kothandaraman & Rudramoorthy, 2006). One such a turbine where water hits on the buckets is the Pelton† wheel, one or more water jets are directed tangentially on to vanes or buckets that are fastened to the rim of the turbine disc as shown below. An experimental design is set to determine the impact of the jet on the vanes and how the Pelton turbine produces maximum results. Figure 1: Pelton hydro-power turbine where one or more water jets hit the buckets tangentially. Aims of the experiment Understand how deflection of the jet generates a force on buckets. Determine the most effective vane/jet deflection (flat, conical and hemispherical vanes). To compare the practical/real results and the ideal/theoretical/expected results for the forces produced on the vanes. Apparatus and procedure The apparatus were set as shown as shown in the figure below. Figure 2: The “impact of a jet H8” apparatus setting with the flat plate on. Safety and precaution taken The machines are heavy as heavy as 225 kg therefore caution was taken when moving them Hands were made dry to avoid electrical shocks Apparatus were never used within 2.4 metres of a main electrical outlet to avoid electrical problems Challenges Some of the activities that required very close attention to a successful experiment included Turning the adjusting knob above the spring align the grooves on the tally with the top plate. Starting and stopping the timer and adding weights when beam moves to horizontal Recording the scale readings corresponding to the groove on the jockey weight Procedure A. Flat plate (90o) A) The apparatus was set as shown in figure 2 with the flat plate B) The lever was leveled to the balance position (zero position) with the jockey weight C) Water was admitted through the bench supply valve and the rate of flow increased to the maximum D) Observed force on the lever scale was recorded E) The mass of the flow rate was established by collecting the water over a timed flow F) Time of the water flow rate was recorded to the table as well G) The procedure was repeated for varying lengths of the jockey y and recorded as shown B. Conical plate (120o) a) The apparatus were set as shown in figure 2 using a conical plate b) The procedure in A above was repeated C. Hemispherical cup (180) a) The apparatus were set as shown in figure 2 using a hemispherical plate b) The procedure in A above was repeated Results and Calculations Diameter of nozzle, D = 10.0 mm. Cross sectional area of nozzle, A =πD2 /4 = 78.5 mm2 Implying A= 7.85×10−5m2 Height of vane above nozzle tip s = 35 mm = 0.035 m Distance from centre of vane to pivot of lever L= 150 mm= 0.15m Mass of jockey weight, M = 0.600 kg Weight of jockey weight, W = Mg = 0.600×9.81 = 5.89 N When the jockey weight is moved a distance y mm from its zero position, the force F on the vane which is required to restore balance is given by: F×150 = W×y Inserting the value of W, namely 5.89 N, gives589×y F=.or F = 0.03924 y N/150 The mass flow rate m in the jet is found by timing the collection of a known mass of   water. The velocity u1 of the jet as it leaves the nozzle is found from the volumetric flow rate Q and the cross sectional area A of the nozzle. The velocity u0 with which the jet strikes the vane is slightly less than u1because of the deceleration due to gravity. This effect may be calculated from the expression Inserting the value s = 0.035 m leads to the result Recorded values of quantity collected, measured time, and jockey displacement y are presented in tables together with the ensuing calculations below. Mass flow rate is computed by the total mass divided by the time taken as shown below in an illustration. M= weight/time Eg M=5/28.6 kg/s This was used for volumetric computations Volumetric flow rate Q is mass flow rate divided by the density, which is 1000 kg/m3 for water. Q=m/1000 m3/s Table 2: Results for the flat plate, conical and cup vane Volume (L) Time (s) Y (m) W (kg) m (kg/s u (m/s) uo (m/s) muo (N) Force (N) cup 5 28.6 0.141 0.6 0.174825175 #REF! #REF! #REF! 5.535096 cup 5 39.32 0.107 0.6 0.12716175 #REF! #REF! #REF! 4.200392 cup 5 45.46 0.093 0.6 0.109986802 #REF! #REF! #REF! 3.650808 cup 15 196.81 0.05 0.6 0.076215639 #REF! #REF! #REF! 1.9628 Flat 15 65.72 0.087 0.6 0.228241023 #REF! #REF! #REF! 3.415272 Flat 15 75.85 0.034 0.6 0.197758734 #REF! #REF! #REF! 1.334704 Flat 5 165.97 0.018 0.6 0.030125926 #REF! #REF! #REF! 0.706608 Flat 5 190.45 0.015 0.6 0.02625361 #REF! #REF! #REF! 0.58884 Conical 15 68 0.114 0.6 0.220588235 #REF! #REF! #REF! 4.475184 Conical 15 88.56 0.092 0.6 0.169376694 #REF! #REF! #REF! 3.611552 Conical 15 103.74 0.069 0.6 0.14459225 #REF! #REF! #REF! 2.708664 Conical 15 111.44 0.035 0.6 0.134601579 #REF! #REF! #REF! 1.37396 A comparative plot for the flat plate (90o), conical plate (120o) and the hemispherical cup (180o) is as shown below Figure1: The experimental result for jet impact at different angles. The figure shows a highest jet impact at 180o and lowest impact at 90o. Taking into account that the flat plate represents 90 degrees of exit, conical plate 120 and the cup 180 degrees of momentum, the comparison between the experimental results and the ideal results was as shown in the tables below. A comparison between experimental and ideal jet impact for 90, 120 and 180 degrees Figure2: Comparison between experimental and ideal result for jet impact at 90o Figure 3: Comparison between experimental and ideal result for jet impact at 120o Figure 4: Comparison between experimental and ideal result for jet impact at 180o Discussion From the comparative results, it can be seen that highest impact for a given rate of delivery of momentum by the change occurs at 180o. This can be explained mathematically using the approaches below Thus taking to the beta approach and taking into account that beta is the cosine of the angle of deflection, then from the equation above for F, it can be shown/seen that F is maximum for a maximum –ve β1. This is at 180 degrees where β1= -1 hence the force being at the maximum. The force reduces as β1 becomes more positive. From this, therefore it is expected that at 60o, the force will be less and even less at that. What this means is that, a theoretically mathematical expectation would be that the jet impact force tends to zero as the angle of deflection approaches zero. This is theoretically true since zero deflection is attainable when the plate is absolutely parallel and in fact no contact. Comparison between experimental and theoretical results In order to compare the ideal and the experimental results shown above, the nth value was computed using the formula below From this, the ideal force is computed and plotted as shown in the results above. From the comparison between ideal/theoretical results and experimental results, it can be seen that ideal results show a higher impact compared to experimental. This error can be explained by as described below. The error seen in 90o and 120o is the same. This implies that the sources of error are the same. The same sources apply to 180o including some other. The overall experimental sources of error include those as shown below: An ideal situation assumes a frictionless environment. However, a real environment has friction which slows down motion. Thus, losses will always occur due to friction. Ideal computations assume standard temperature and pressure which is not true for a real experimental environment. This results to differences between the experimental and ideal density of water. Small density discrepancies case significant discrepancies in results. Although all controls should be put in place to avoid basic human error, it is practically within reasonable sense that slight variations in readings may arise. However small, the human error may cause the error. Apart from the errors mentioned above, in 180o, the water flowing out of the vane and water flowing into the vanes do have a contact (touching effect). This results into the large error enough to cause change in slope, but not just displacement of the slope like the one seen where chances for “touching are reduced.” Modern Applications The impact of a jet, since it produces force, has very valued real life applications. One of such applications is in water powered turbines as shown in the figure below Figure 5: An illustration of water powered turbine From the figure, it can be seen that designers tend to take the hemispherical design because it maximizes impact. By the turning (i.e impact), the power is generated and given by the equation below. The turbine efficiency is found by dividing the measured force by ideal force resulting from the jet impact. This power is the cleanest energy. Conclusion The shape of the cap determines the angle of deflection of the jet. The angle of deflection determines the force, hence impact of the jet. This force is then applied in relevant fields according to need. From the experiment, it is explained, including mathematically, that hemispherical shapes produce the highest impact of the jet. This is due to the resultant deflection through 180o. This explains why the cups of a water driven turbine are hemispherical: to maximize the impact of a jet. References Burke, L. Hydroelectricity and power electronics. Heshmati, A. (2012). Economic fundamentals of power plant performance. Abingdon, Oxon: Routledge. Kothandaraman, C. & Rudramoorthy, R. (2006). Fluid mechanics and machinery. New Delhi: New Age International (P) Ltd., Publishers. Read More

Conical plate (120o) a) The apparatus were set as shown in figure 2 using a conical plate b) The procedure in A above was repeated C. Hemispherical cup (180) a) The apparatus were set as shown in figure 2 using a hemispherical plate b) The procedure in A above was repeated Results and Calculations Diameter of nozzle, D = 10.0 mm. Cross sectional area of nozzle, A =πD2 /4 = 78.5 mm2 Implying A= 7.85×10−5m2 Height of vane above nozzle tip s = 35 mm = 0.035 m Distance from centre of vane to pivot of lever L= 150 mm= 0.

15m Mass of jockey weight, M = 0.600 kg Weight of jockey weight, W = Mg = 0.600×9.81 = 5.89 N When the jockey weight is moved a distance y mm from its zero position, the force F on the vane which is required to restore balance is given by: F×150 = W×y Inserting the value of W, namely 5.89 N, gives589×y F=.or F = 0.03924 y N/150 The mass flow rate m in the jet is found by timing the collection of a known mass of   water. The velocity u1 of the jet as it leaves the nozzle is found from the volumetric flow rate Q and the cross sectional area A of the nozzle.

The velocity u0 with which the jet strikes the vane is slightly less than u1because of the deceleration due to gravity. This effect may be calculated from the expression Inserting the value s = 0.035 m leads to the result Recorded values of quantity collected, measured time, and jockey displacement y are presented in tables together with the ensuing calculations below. Mass flow rate is computed by the total mass divided by the time taken as shown below in an illustration. M= weight/time Eg M=5/28.

6 kg/s This was used for volumetric computations Volumetric flow rate Q is mass flow rate divided by the density, which is 1000 kg/m3 for water. Q=m/1000 m3/s Table 2: Results for the flat plate, conical and cup vane Volume (L) Time (s) Y (m) W (kg) m (kg/s u (m/s) uo (m/s) muo (N) Force (N) cup 5 28.6 0.141 0.6 0.174825175 #REF! #REF! #REF! 5.535096 cup 5 39.32 0.107 0.6 0.12716175 #REF! #REF! #REF! 4.200392 cup 5 45.46 0.093 0.6 0.109986802 #REF! #REF! #REF! 3.650808 cup 15 196.81 0.05 0.6 0.

076215639 #REF! #REF! #REF! 1.9628 Flat 15 65.72 0.087 0.6 0.228241023 #REF! #REF! #REF! 3.415272 Flat 15 75.85 0.034 0.6 0.197758734 #REF! #REF! #REF! 1.334704 Flat 5 165.97 0.018 0.6 0.030125926 #REF! #REF! #REF! 0.706608 Flat 5 190.45 0.015 0.6 0.02625361 #REF! #REF! #REF! 0.58884 Conical 15 68 0.114 0.6 0.220588235 #REF! #REF! #REF! 4.475184 Conical 15 88.56 0.092 0.6 0.169376694 #REF! #REF! #REF! 3.611552 Conical 15 103.74 0.069 0.6 0.14459225 #REF! #REF! #REF! 2.708664 Conical 15 111.44 0.035 0.6 0.

134601579 #REF! #REF! #REF! 1.37396 A comparative plot for the flat plate (90o), conical plate (120o) and the hemispherical cup (180o) is as shown below Figure1: The experimental result for jet impact at different angles. The figure shows a highest jet impact at 180o and lowest impact at 90o. Taking into account that the flat plate represents 90 degrees of exit, conical plate 120 and the cup 180 degrees of momentum, the comparison between the experimental results and the ideal results was as shown in the tables below.

A comparison between experimental and ideal jet impact for 90, 120 and 180 degrees Figure2: Comparison between experimental and ideal result for jet impact at 90o Figure 3: Comparison between experimental and ideal result for jet impact at 120o Figure 4: Comparison between experimental and ideal result for jet impact at 180o Discussion From the comparative results, it can be seen that highest impact for a given rate of delivery of momentum by the change occurs at 180o. This can be explained mathematically using the approaches below Thus taking to the beta approach and taking into account that beta is the cosine of the angle of deflection, then from the equation above for F, it can be shown/seen that F is maximum for a maximum –ve β1.

This is at 180 degrees where β1= -1 hence the force being at the maximum. The force reduces as β1 becomes more positive.

Read More