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Fluid Report Presentation

Fluid Report Presentation. Onur Erkal Korgun Koyunpınar Korhan Türker Hakan Uzuner. Experiments. Our fluid dynamics experiment is made up of two parts.

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Fluid Report Presentation

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  1. Fluid Report Presentation Onur Erkal Korgun Koyunpınar Korhan Türker Hakan Uzuner

  2. Experiments Our fluid dynamics experiment is made up of two parts. 1) Determination of the velocity distribution of the air jet at different cross-sections and at different distances from the orifice to be able to calculate the discharge, momentum flux and kinetic energy flux. 2) Determination of the velocity distribution profiles of the air jet inside a pipe at different cross-sections which have different distances from the pipe inlet.

  3. In the first experiment we measured the cross-sections 100mm, 300mm and 500mm away from the orifice. We used Bernoulli equations Where; v is the fluid flow speed at a point on a streamline, g is the acceleration due to gravity, z is the elevation of the point above a reference plane, with the positive z-direction in the direction opposite to the gravitational acceleration, p is the pressure at the point, ρ is the density of the fluid at all points in the fluid.

  4. And then we calculated the • Jet Discharge: • Jet Momentum: • Jet Kinetic Energy:

  5. In the second experiment, we calculated the velocity distributions and we tried to find out the location where fully developed region starts. stagnationpressure = staticpressure + dynamicpressure To calculate the velocity, we ignore the height differences and the velocity at the stagnation point is zero so the Bernoulli equation takes the form of We took cross-sections 54 mm, 294 mm, 774 mm, 1574 mm and 2534 mm away from the anti vortex vanes. At these cross-sections, we measured the pressures at 2.5 mm, 22.5 mm, 37.5 mm, 52.5 mm and 72.5 mm away from the tube bottom

  6. Results For this part of the experiment, the critical part was to calculate the h1 and h2 for different x values. Following is our recorded data from our experiment.

  7. Acquired Data

  8. Calculations Using the above values and the following formulas we calculated the velocity of the flow at different heights. Δh0 = h ref – h0 ΔP0 = Δh.ρ.g v0 = √2.√ΔP / ρ

  9. Calculations After implemented the acquired data into the formulas listed above we came up with the following table for: x=100

  10. Calculations After implemented the acquired data into the formulas listed above we came up with the following table for: x=300

  11. Calculations After implemented the acquired data into the formulas listed above we came up with the following table for: x=500

  12. Calculations Our next step was to calculate V / V0 and R / R 1/2 values. Following are our calculations and the table we came up with. For x = 100: r = 125 and r 1/2 = 110. So, r / r 1/2 = 125/110= 1.136 v / v0 = 19.65 / 35.14 = 0.56 For x = 300: r = 146 and r 1/2 = 121. So, r / r 1/2 = 146/121 = 1.207 v / v0 = 8.02 / 18.30 = 0,577 For x = 500: r = 169 and r 1/2 = 134. So, r / r 1/2 = 169/134 = 1.261 v / v0 = 3.58 / 11.34 = 0.315

  13. Calculations

  14. Calculations With the help of the preceding table we drew the similarity curve of our flow:

  15. Calculations Next step was to construct the Vj / V0 vs x / d graph. For this we did the following calculations. Vj has to be calculated from the reference point href. Δh = 168 mm. ΔP = 168 x 10-3 m.787 kg / m3.9.81 m / s2 = 1297.03 N / m2 Vj = √2.√ΔP / ρ = √2*1297.03 N / m2 / 1.2 kg / m3 = 46.49 m / s x = 100: vj / v0 = 46.49 / 35,14 = 1,323 and x / d = 100 / 30 = 3,34 x = 300: vj / v0 = 46.49 / 18.30 = 2,54 and x / d = 300 / 30 = 10 x = 500: vj / v0 = 46.49 / 11.34 = 4.099 and x / d = 500 / 30 = 16,67

  16. Calculations

  17. Calculations From the preceding table we constructed Vj / V0 vs. x / d graph.

  18. Jet discharge Discharge through annulus: δQ = 2 * (pi) * r * δr * v Total discharge: Q = 2 * (pi) * ∑ ( r * v * δr ) for x=100 Q (total) = 0.044 m3/s for x=300 Q (total) = 0.0241 m3/s for x=500 Q (total) = 0.0241 m3/s As expected volumetric flow rate decrases as horizontal distance x increases.

  19. Jet Momentum • Jet momentum: δM = mass flow rate * velocity • Mass flow rate: δQ * ρair for x=100 M (total) = 1.283 mkg/s2for x=300 M (total) = 0.309 mkg/s2for x=500 M (total) = 0.1878 mkg/s2 The jet momentum decreases as the horizontal distance x increases.

  20. Jet Kinetic Energy Jet Kinetic Energy: δE = mass flow rate * 0.5 velocity2 for x=100  E (total) = 16.35 J for x=300 E (total) = 1.778 J for x=500 E (total) = 0.702 J As expected the kinetic energy decreases as the horizontal distance x increases.

  21. Conclusion (experiment 1) • The velocity values at different horizontal distances at different radiuses are calculated. As expected, the max velocity values are obtained at center (r=0). • As the distance of x increases we see that the Q (volumetric flow rate) decreases because overall velocity of the fluid decreases. Same is true for jet momentum and jet kinetic energy. • The apparatus we’ve used is a proper device, which gave us proper experiment results. Therefore, we didn’t have any significant experimental error or unexpected experimental result.

  22. Experiment 2 • Necessary formulas for the calculations: • Δh = h ref – h • ΔP = Δh x ρ x g • v = √2 x √ΔP / ρ • Re = v x d / ν

  23. The first point x=54mm • h ref = 170 • Vaverage = 22.372 • Re = 107557.7

  24. The second point x=294mm • h ref = 175 • Vaverage = 24.164 • Re = 116173.07

  25. The third point x=774mm • h ref = 180 • Vaverage = 23.694 • Re = 113913.46

  26. The fourth point x=1574mm • h ref = 188 • Vaverage = 25.462 • Re = 122413.46

  27. The fifth point x=2534mm • h ref = 194 • Vaverage = 24.418 • Re = 117394.23

  28. Velocity Profile

  29. Turbulent Flow • When we look at the Reynolds numbers we calculated, we see clearly that this is a turbulent flow, because all the Re values are much greater than the critical number 2300. • For x = 54 , Re = 107557.7 • For x = 294 , Re = 116173.07 • For x = 774 , Re = 113913.46 • For x = 1574, Re = 122413.46 • For x = 2534, Re = 117394.23 • So we can use the entrance length Le formula for the turbulent flow: Le = 4,4.Re average1/6.d

  30. Average Entrance Length • For x = 54 , Le = 2.2757 m • For x = 294 , Le = 2.3051 m • For x = 774 , Le = 2.2976 m • For x = 1574, Le = 2.3253 m • For x = 2534, Le = 2.3092 m • As a result our average entrance length value is: Le avg = 2.3025 m

  31. Comparing • To compare the experimental results with the calculated results we use the formula: • u = V centerline (1 – r / r0) 1/6 • For r = 35 mm, • u = 27,55 m / s.(1 – 35 / 37,5) 1/6 = 17,543 m/s. • Our experimental result was 20.91 m/s. • So; the experimental error is 16.1%. • For r = 15 mm, • u = 27,32 m / s.(1 – 35 / 37,5) 1/6 = 25.3 m/s. • Our experimental result was 26.36 m/s. • So; the experimental error is 4.02%.

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