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Investigation into the Effects of Fluid Flow through a Venturi - Essay Example

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This paper 'Investigation into the Effects of Fluid Flow through a Venturi' tells us that the Bernoulli principle is widely used in many applications such as keeping the rear wheels of race cars on the ground while they travel at extremely high speeds. Airplanes also use the Bernoulli equation for flight and balance in the sky…
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Investigation into the Effects of Fluid Flow through a Venturi
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Lecturer “Investigation into the effects of Fluid Flow through a Venturi” The Bernoulli principle is widely used in many applications such as keeping the rear wheels of race cars on the ground while they travel at extremely high speeds. Airplanes also use the Bernoulli equation for flight and balance in the sky. According to the law of energy conservation, energy is neither created nor destroyed. Energy can be said to include all forms of energy within the system. It may be in the form kinetic, heat, light and potential form. Concerning Bernoulli equation as a statement of energy conservation, the sum of pressure, kinetic and potential head of a fluid is constant if the condition of the fluid is non viscous, non compressible and of the liquid is of the same density. Increase in velocity of the fluid leads to decrease of the pressure exerted by the fluid. The equation defines is the relationship between the velocity of the fluid that is flowing in the pipe and the cross sectional area of this pipe. As the radius of the pipe decreases the velocity of the flowing fluid will increase and vice-versa. Fluid flow through a pipe will be used to study the Bernoulli Effect under specific requirements and constrictions. There will be calculations, results tables and plotted graphs that will be analyzed to arrive at the Bernoulli equation. The experiment requires a Bernoulli tube, Hydraulic bench, Stop watch and the venturi meter. The total energy=P/ ρg+v2/2g+z As P/ ρg is the static head = height of the fluid in piezometer tube Where: P: is the pressure at specific point ρ: is the density of that fluid And v2/2g is velocity head Where: V: velocity of the fluid specific point Z: the height above datum, but the tube is horizontal; that means the value of Z is zero g: is constant that equals 9.81 , it is the constant of gravitational acceleration Total energy is calculated by static head + velocity head From the equation; The velocity V=Q/A Where Q is the volumetric flow rate and A is the cross sectional area of the tube at specific point And Q = m/ ρ where m is the mass of flow rate which equals weight of the fluid in kilograms/time in seconds and ρ is the density of the fluid From the data: the weight of the fluid is 10 liters = 10 kilograms, (as the water density equal one) There were three patterns of fluid flow; slow flow, medium flow and full flow, each one of them has its time that will be stable for all reading at each station 1- for slow flow rate: the time obtained is 49.12 seconds 2- for medium flow rate: the recorded time is 32 seconds 3- for full flow rate: the recorded time is 29.05 seconds Therefore; To calculate the mass flow rate, the fluid weight which is 10 kilograms will be divided on the time of each pattern of flow in seconds. The resulting value will be divided on the on 1000 which is the density of the water to get the volumetric flow rate. The table below describes the recording measurements of piezometer at each station (each point has its cross sectional area) and the time recorded of each type of the flow patterns station (static head)Low flow (static head)Medium flow (static head)Full flow 1 0.231 m 0.176 m 0.133 m 2 0.229 m 0.173 m 0.129 m 3 0.216 m 0.141 m 0.087 m 4 0.190 m 0.082 m 0.011 m 5 0.193 m 0.088 m 0.022 m 6 0.203 m 0.116 m 0.058 m 7 0.207 m 0.129 m 0.078 m 8 0.218 m 0.145 m 0.092 m 9 0.222 m 0.153 m 0.103 m 10 0.225 m 0.159 m 0.110 m 11 0.229 m 0.166 m 0.117 m Time(seconds) 49.12 second 32 second 29.05 second Slow flow rate: The time calculated is 49.12 seconds, the weight of fluid is 10 kilograms, therefore mass rate= (10/49.2). To obtain the volumetric rate, the resulting value will be divided on 1000 which is the density of water, therefore volumetric rate = 0.000203583 X- axis Y-axis top Y-axis bottom Y-axis (top +bottom) Volumetric flow rate 0.000203583 = (10/49.12)/1000 m3/s station diameter mm area m sq Velocity velocity head (fluid velocity) m/s (pressure head) piezo reading (m) total head 1 25.4 0.000507 0.401544501 0.008218042 0.231 0.239218042 2 22.83 0.000409 0.497758098 0.01262809 0.229 0.24162809 3 18.48 0.000268 0.759638291 0.029411332 0.216 0.245411332 4 15.8 0.000196 1.038689091 0.054988534 0.19 0.244988534 5 16.4 0.000211 0.964848635 0.047448159 0.193 0.240448159 6 17.84 0.00025 0.814332248 0.033799032 0.203 0.236799032 7 19.24 0.000291 0.699598151 0.02494585 0.207 0.23194585 8 20.59 0.000333 0.611360546 0.019050037 0.218 0.237050037 9 21.83 0.000374 0.544339738 0.01510223 0.222 0.23710223 10 23.07 0.000418 0.487040818 0.012090151 0.225 0.237090151 11 25.4 0.000507 0.401544501 0.008218042 0.229 0.237218042 Medium flow rate: The time calculated is 32 seconds, the weight of fluid is 10 kilograms, therefore mass rate= (10/32). To obtain the volumetric rate, the resulting value will be divided on 1000 which is the density of water, therefore volumetric rate = 0.000313 X- axis Y-axis top Y-axis bottom Y-axis (top +bottom) Volumetric flow rate 0.000313 =(10/32)/1000 station diameter mm area m sq velocity velocity head (fluid velocity) m/s pressure head) piezo reading (m) total head 1 25.4 0.000507 0.616371 0.019364 0.176 0.195363556 2 22.83 0.000409 0.764059 0.029755 0.173 0.202754621 3 18.48 0.000268 1.166045 0.0693 0.141 0.210299716 4 15.8 0.000196 1.594388 0.129565 0.082 0.211565357 5 16.4 0.000211 1.481043 0.111799 0.088 0.199798539 6 17.84 0.00025 1.25 0.079638 0.116 0.195638124 7 19.24 0.000291 1.073883 0.058778 0.129 0.187778035 8 20.59 0.000333 0.938438 0.044886 0.145 0.189886172 9 21.83 0.000374 0.835561 0.035584 0.153 0.188584252 10 23.07 0.000418 0.747608 0.028487 0.159 0.187487116 11 25.4 0.000507 0.616371 0.019364 0.166 0.185363556 Graphic representation; Full flow rate: The time calculated is 29.05seconds, the weight of fluid is 10 kilograms, therefore mass rate= (10/29.05) To obtain the volumetric rate, the resulting value will be divided on 1000 which is the density of water, therefore volumetric rate = 0.000344 X- axis Y-axis top Y-axis bottom Y-axis (top +bottom) Volumetric flow rate 0.000344 =10/29.05)/1000 station diameter mm area m sq velocity velocity head (fluid velocity) m/s pressure head) piezo reading (m) total head 1 25.4 0.000507 0.678963 0.023496 0.133 0.156496 2 22.83 0.000409 0.841648 0.036105 0.129 0.165105 3 18.48 0.000268 1.284456 0.084089 0.087 0.171089 4 15.8 0.000196 1.756296 0.157216 0.011 0.168216 5 16.4 0.000211 1.631441 0.135658 0.022 0.157658 6 17.84 0.00025 1.376936 0.096634 0.058 0.154634 7 19.24 0.000291 1.182935 0.071322 0.078 0.149322 8 20.59 0.000333 1.033736 0.054465 0.092 0.146465 9 21.83 0.000374 0.920412 0.043178 0.103 0.146178 10 23.07 0.000418 0.823527 0.034567 0.11 0.144567 11 25.4 0.000507 0.678963 0.023496 0.117 0.140496 Graphical representation Analysis The ideal results that should be obtained is the constant values for the total head as Bernoulli equation stated, but there are few differences between the values, for example, the total head of slow flow rate is ranging from 0.23194585 and 0.245411332 , as the difference between the highest value and the lowest value is about 0.01346548. The difference is due to many aspects such as friction with the tube wall, viscosity of the fluid may be altered, turbulence of the flow There is a reverse relationship between the velocity of the fluid and the pressure exerted by the fluid, as by increasing the velocity of the fluid, there will be decrease of the pressure, and the decrease the of velocity is accompanied by an increase of the pressure, as the total head stays constant The relation between the cross sectional area and the pressure exerted by the fluid, is inversely proportional, as by increasing the cross sectional area, the pressure will decrease and the opposite is true. The relation between the pressure and the fluid flow rate is inversely proportional as the increase in the pressure exerted by the fluid, the flow will decrease and vice versa. Obtaining the results with difference in the total head values, points to presence of some important factors or restrictions that will influence the final values, these restrictions are: Lack of viscosity: as viscosity of the fluid will affect the flow rate due to friction with the tube wall, therefore the velocity will decrease and flow rate will be interrupted, this will lead to decreased or lost head. Non compressible or (low velocity): as fluid compression will actual affect the flow rate as the distance that fluid passed will be affected, hence the velocity head will be affected Lack of additional heat: as heat is energy, therefore, as any change in heat will affect the total energy calculation because the calculated energy is a sum of static head and velocity head The height is neglected: as a modified form of Bernoulli equation stats that: P1 + 1/2 ρ V12 + ρ gh1 = P1 + 1/2 ρ V22 + ρ gh2 + HL As the energy head at the entrance is equal to the energy head at the exit, when neglecting the value of height, the results will be accurate. Therefore the final form of the equation will be P1 + 1/2 ρ V12 = P1 + 1/2 ρ V22 + HL As HL is the total loss in the head flow The viscosity of the fluid it has been proved will affect the flow rate due to certain factors like friction with the wall of the tube. A drop in the velocity leads to a decrease of the head. References Tom Benson. (2010). Bernoulli equation. Available: http://wright.nasa.gov/airplane/bern.html. Last accessed 07 December 2011. Dr Andrew Sleigh & Dr Cath Noakes. (2009). CIVE1400: Fluid Mechanics. Available: http://www.efm.leeds.ac.uk/CIVE/CIVE1400/Section3/bernoulli-apps.htm. Last accessed 07 December 2011. Watts and Ferrer. (2000). Bernoulli Equation. Available: http://hyperphysics.phy- astr.gsu.edu/hbase/pber.html. Last accessed 08 December 2011. Allstar Network. (2008). Bernoullis Equation. Available: http://www.allstar.fiu.edu/aerojava/bernoulli.htm. Last accessed 9 December 2011. Engineeringtoolbox. (2010). Pitot Tubes. Available: http://www.engineeringtoolbox.com/pitot- tubes-d_612.html. Last accessed 8 December 2011. Read More
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