Fluid Report PresentationFluid Report Presentation

Onur ErkalOnur ErkalKorgun KoyunpınarKorgun Koyunpınar

Korhan TürkerKorhan TürkerHakan UzunerHakan Uzuner

ExperimentsExperiments

Our fluid dynamics experiment is made up Our fluid dynamics experiment is made up of two parts. of two parts.

1) Determination of the velocity distribution 1) Determination of the velocity distribution of the air jet at different cross-sections and of the air jet at different cross-sections and at different distances from the orifice to be at different distances from the orifice to be able to calculate the discharge, able to calculate the discharge, momentum flux and kinetic energy flux. momentum flux and kinetic energy flux.

2) Determination of the velocity distribution 2) Determination of the velocity distribution profiles of the air jet inside a pipe at profiles of the air jet inside a pipe at different cross-sections which have different cross-sections which have different distances from the pipe inlet. different distances from the pipe inlet.

In the first experiment we measured the cross-In the first experiment we measured the cross-sections 100mm, 300mm and 500mm away from sections 100mm, 300mm and 500mm away from the orifice. We used Bernoulli equations the orifice. We used Bernoulli equations

Where;Where;vv is the fluid flow speed at a point on a streamline, is the fluid flow speed at a point on a streamline, gg is the acceleration due to gravity, is the acceleration due to gravity, zz is the elevation of the point above a reference is the elevation of the point above a reference plane, with the positive plane, with the positive zz-direction in the direction -direction in the direction opposite to the gravitational acceleration, opposite to the gravitational acceleration, pp is the pressure at the point, is the pressure at the point,ρρ is the density of the fluid at all points in the fluid. is the density of the fluid at all points in the fluid.

And then we calculated theAnd then we calculated the Jet Discharge: Jet Discharge:

Jet Momentum:Jet Momentum:

Jet Kinetic Energy:Jet Kinetic Energy:

In the second experiment, we calculated the velocity In the second experiment, we calculated the velocity distributions and we tried to find out the location where fully distributions and we tried to find out the location where fully developed region starts. developed region starts. stagnationpressurestagnationpressure = = staticpressurestaticpressure + + dynamicpressuredynamicpressure

To calculate the velocity, we ignore the height differences and To calculate the velocity, we ignore the height differences and the velocity at the stagnation point is zero so the Bernoulli the velocity at the stagnation point is zero so the Bernoulli equation takes the form ofequation takes the form of

We took cross-sections 54 mm, 294 mm, 774 mm, 1574 mm We took cross-sections 54 mm, 294 mm, 774 mm, 1574 mm and 2534 mm away from the anti vortex vanes. At these cross-and 2534 mm away from the anti vortex vanes. At these cross-sections, we measured the pressures at 2.5 mm, 22.5 mm, 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 37.5 mm, 52.5 mm and 72.5 mm away from the tube bottom

For this part of the experiment, the critical For this part of the experiment, the critical part was to calculate the hpart was to calculate the h1 1 and hand h22 for for

different x values. Following is our recorded different x values. Following is our recorded data from our experiment.data from our experiment.

Values Values y y maxmax y y avgavg y y minmin z z maxmax z z avgavg z z minmin hh1 1

(mm)(mm)hh2 2

(mm)(mm)h h jet jet

(mm)(mm)x = x = 100100

285285 260260 235235 125125 9696 7373 9696 138138 7272

x = x = 300300

308308 260260 215215 146146 9696 4343 153153 163163 142142

x = x = 500500

330330 260260 190190 169169 9696 1818 162162 167167 158158

Using the above values and the Using the above values and the following formulas we calculated the following formulas we calculated the velocity of the flow at different velocity of the flow at different heights.heights.

ΔhΔh00 = h = h refref – h – h00

ΔPΔP00 = Δh.ρ.g = Δh.ρ.g

vv00 = √2.√ΔP / ρ = √2.√ΔP / ρ

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

z(mm) z(mm)  r(mm)r(mm)h(mm)h(mm) hhref ref (mm)(mm) ∆∆h (mm)h (mm) V (m/s)V (m/s)

9696 007272 168168 9696 35.1435.14

106106 1010 9696 168168 7272 30.4330.43

116116 2020 138138 168168 3030 19.6519.65

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

z(mm) z(mm) r(mm)r(mm)

h(mm)h(mm) hhref ref (mm)(mm) ∆∆h (mm)h (mm) V (m/s)V (m/s)

969600

142142 168168 2626 18.3018.30

109109 1313 153153 168168 1515 13.8913.89

122122 2626 163163 168168 55 8.028.02

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

z(mm) z(mm)  r(mm)r(mm)h(mm)h(mm) hhref ref (mm)(mm) ∆∆h (mm)h (mm) V (m/s)V (m/s)

9696 00158158 168168 1010 11.3411.34

120120 2424162162 168168 66 8.88.8

144144 4848167167 168168 11 3.583.58

Our next step was to calculate V / VOur next step was to calculate V / V00 and and R / R R / R 1/21/2 values. Following are our calculations and the table values. Following are our calculations and the table

we came up with.we came up with.For x = 100:For x = 100:r = 125 and r r = 125 and r 1/21/2 = 110. So, r / r = 110. So, r / r 1/21/2 = 125/110= = 125/110= 1.1361.136 v / vv / v00 = 19.65 / 35.14 = = 19.65 / 35.14 = 0.560.56  For x = 300:For x = 300:r = 146 and r r = 146 and r 1/21/2 = 121. So, r / r = 121. So, r / r 1/21/2 = 146/121 = = 146/121 = 1.2071.207v / vv / v00 = 8.02 / 18.30 = = 8.02 / 18.30 = 0,5770,577  For x = 500:For x = 500:r = 169 and r r = 169 and r 1/21/2 = 134. So, r / r = 134. So, r / r 1/21/2 = 169/134 = = 169/134 = 1.2611.261v / vv / v00 = 3.58 / 11.34 = = 3.58 / 11.34 = 0.3150.315

ValuesValues v / vv / v00 (-) (-) r / r r / r 1/2 1/2 (-) (-)

x = 100x = 100 0,560,56 1.1361.136

x = 300x = 300 0.4380.438 1.2071.207

x = 500 x = 500 0.3150.315 1.2611.261

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

Next step was to construct the Next step was to construct the Vj / V0 vs x / d graph. For Vj / V0 vs x / d graph. For this we did the following calculations.this we did the following calculations.

VVjj has to be calculated from the reference point h has to be calculated from the reference point hrefref. . Δh = 168 mm. Δh = 168 mm. ΔP = 168 x 10ΔP = 168 x 10-3 m.787 kg / m3-3 m.787 kg / m3.9.81 m / s.9.81 m / s22 = = 1297.03 N / m1297.03 N / m22

VVjj = √2.√ΔP / ρ = √2*1297.03 N / m = √2.√ΔP / ρ = √2*1297.03 N / m22 / 1.2 kg / m / 1.2 kg / m3 3 = = 46.49 46.49 m / sm / s

  x = 100: x = 100: vvjj / v / v00 = 46.49 / 35,14 = = 46.49 / 35,14 = 1,3231,323 and x / d = 100 / 30 = and x / d = 100 / 30 = 3,343,34  x = 300:x = 300:vvjj / v / v00 = 46.49 / 18.30 = = 46.49 / 18.30 = 2,542,54 and x / d = 300 / 30 = and x / d = 300 / 30 = 1010  x = 500:x = 500:vvjj / v / v00 = 46.49 / 11.34 = = 46.49 / 11.34 = 4.0994.099 and x / d = 500 / 30 = and x / d = 500 / 30 = 16,6716,67

ValuesValues vvjj / v / v0 0 (-)(-) x / d (-)x / d (-)

x = 100x = 100 1.3231.323 3,343,34

x = 300x = 300 2.5402.540 1010

x = 500x = 500 4.0994.099 16,6716,67

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

Jet dischargeJet dischargeDischarge through annulus:Discharge through annulus: δQ = 2 * (pi) * r * δr * v δQ = 2 * (pi) * r * δr * v

Total discharge:Total discharge: Q = 2 * (pi) * ∑ ( r * v * Q = 2 * (pi) * ∑ ( r * v * δr )δr )

for for x=100x=100 Q (total) = 0.044 mQ (total) = 0.044 m33/s/s

for for x=300x=300 Q (total) = 0.0241 mQ (total) = 0.0241 m33/s/s

for for x=500x=500 Q (total) = 0.0241 mQ (total) = 0.0241 m33/s/s

As expected volumetric flow rate decrases as As expected volumetric flow rate decrases as

horizontal distance x increases.horizontal distance x increases.

Jet MomentumJet Momentum Jet momentum:Jet momentum: δM = mass flow rate * velocity δM = mass flow rate * velocity

Mass flow rate: Mass flow rate: δQ * δQ * ρairρair

for for x=100x=100 M (total) = 1.283 mM (total) = 1.283 m kg/s2kg/s2 for for x=300x=300 M (total) = 0.309 M (total) = 0.309

mm kg/s2kg/s2 for for x=500x=500 M (total) = M (total) = 0.1878 m0.1878 m kg/s2kg/s2

The jet momentum decreases as the The jet momentum decreases as the horizontal distance x increases.horizontal distance x increases.

Jet Kinetic EnergyJet Kinetic Energy

Jet Kinetic Energy:Jet Kinetic Energy: δE = mass flow rate * 0.5 δE = mass flow rate * 0.5 velocity2velocity2

for for x=100x=100 E (total) = 16.35 JE (total) = 16.35 J

for for x=300x=300 E (total) = 1.778 JE (total) = 1.778 J

for for x=500x=500 E (total) = 0.702 JE (total) = 0.702 J

As expected the kinetic energy decreases as As expected the kinetic energy decreases as

the horizontal distance x increases.the horizontal distance x increases.

Conclusion (experiment 1)Conclusion (experiment 1) The velocity values at The velocity values at different horizontal distances at different horizontal distances at

different radiuses are calculated. As expected, the max different radiuses are calculated. As expected, the max velocity values are obtained at center (r=0). velocity values are obtained at center (r=0).

As the distance of x increases we see that the Q (volumetric As the distance of x increases we see that the Q (volumetric flow rate) decreases because overall velocity of the fluid flow rate) decreases because overall velocity of the fluid decreases.decreases. Same is true for jet momentum and jet kinetic Same is true for jet momentum and jet kinetic energy.energy.

The apparatus we’ve used is a proper device, which gave The apparatus we’ve used is a proper device, which gave us proper experiment results. Therefore, we didn’t have us proper experiment results. Therefore, we didn’t have any significant experimental error or unexpected any significant experimental error or unexpected experimental result. experimental result.

Experiment 2Experiment 2

Necessary formulas for the Necessary formulas for the calculations:calculations:

Δh = h ref – h Δh = h ref – h ΔP = Δh x ρ x g ΔP = Δh x ρ x g vv = √2 = √2 x x √Δ√ΔPP / ρ / ρ ReRe = = v x d v x d / ν/ ν

The first point x=54mmThe first point x=54mm

h ref = 170h ref = 170 Vaverage = 22.372Vaverage = 22.372 Re = 107557.7Re = 107557.7

The second point x=294mmThe second point x=294mm

h ref = 175h ref = 175 Vaverage = 24.164Vaverage = 24.164 Re = 116173.07Re = 116173.07

The third point x=774mmThe third point x=774mm

h ref = 180h ref = 180 Vaverage = 23.694Vaverage = 23.694 Re = 113913.46Re = 113913.46

The fourth point x=1574mmThe fourth point x=1574mm

h ref = 188h ref = 188 Vaverage = 25.462Vaverage = 25.462 Re = 122413.46Re = 122413.46

The fifth point x=2534mmThe fifth point x=2534mm

h ref = 194h ref = 194 Vaverage = 24.418Vaverage = 24.418 Re = 117394.23Re = 117394.23

Velocity ProfileVelocity Profile

0

10

20

30

40

50

60

70

80

0 5 10 15 20 25 30

Vecocity (m/s)

h (

mm

)

x=54

x=294

x=774

x=1574

x=2534

Turbulent FlowTurbulent Flow When we look at the Reynolds numbers we When we look at the Reynolds numbers we

calculated, we see clearly that this is a turbulent calculated, we see clearly that this is a turbulent flow, because all the Re values are much greater flow, because all the Re values are much greater than the critical number 2300.than the critical number 2300.

For x = 54 , Re = 107557.7For x = 54 , Re = 107557.7 For x = 294 , Re = 116173.07For x = 294 , Re = 116173.07 For x = 774 , Re = 113913.46For x = 774 , Re = 113913.46 For x = 1574, Re = 122413.46For x = 1574, Re = 122413.46 For x = 2534, Re = 117394.23For x = 2534, Re = 117394.23

So we can use the entrance length Le formula for So we can use the entrance length Le formula for the turbulent flow: Le = 4,4.Re average1/6.dthe turbulent flow: Le = 4,4.Re average1/6.d

Average Entrance LengthAverage Entrance Length

For x = 54 , Le = 2.2757 mFor x = 54 , Le = 2.2757 m For x = 294 , Le = 2.3051 mFor x = 294 , Le = 2.3051 m For x = 774 , Le = 2.2976 mFor x = 774 , Le = 2.2976 m For x = 1574, Le = 2.3253 mFor x = 1574, Le = 2.3253 m For x = 2534, Le = 2.3092 mFor x = 2534, Le = 2.3092 m

As a result our average entrance As a result our average entrance length value is: length value is: Le avg = 2.3025 mLe avg = 2.3025 m

Comparing Comparing To compare the experimental results with the To compare the experimental results with the

calculated results we use the formula:calculated results we use the formula: u = V centerline (1 – r / r0) 1/6 u = V centerline (1 – r / r0) 1/6

For r = 35 mm,For r = 35 mm, u = 27,55 m / s.(1 – 35 / 37,5) 1/6 = u = 27,55 m / s.(1 – 35 / 37,5) 1/6 = 17,543 m/s17,543 m/s. . Our experimental result was Our experimental result was 20.91 m/s20.91 m/s.. So; the So; the experimental error is 16.1%experimental error is 16.1%..

For r = 15 mm,For r = 15 mm, u = 27,32 m / s.(1 – 35 / 37,5) 1/6 = u = 27,32 m / s.(1 – 35 / 37,5) 1/6 = 25.3 m/s25.3 m/s. . Our experimental result was Our experimental result was 26.36 m/s26.36 m/s.. So; the So; the experimental error is 4.02%experimental error is 4.02%..