8. Jet and Film - iypt solutions

Levitation Mingyu Kang Team Korea Phenomenon • Investigate the effect  Oscillation • Produce the maximum angle of tilt Force Analysis 𝐹𝐷 Bernoulli 𝐹𝑃 𝑦 𝜃 𝑥 𝑚𝑔 Coanda 𝑢 𝜃 Flow Chart Airstream Velocity Profile Momentum flow conservation Gaussian curve distribution Pitot tube experiment 𝒚Oscillation 𝒙Oscillation Drag force Bernoulli effect Gravity Oscillation period 𝒚-Equilibrium Point Optimization (Maximum angle of tilt) Force equilibrium Coanda effect (Boundary layer thickness) Oscillation period 𝒙-Equilibrium Point Airstream velocity Mass of ball Free Airstream Velocity Profile Momentum flow conservation of free jet 𝑑𝑝 𝐹= 𝑑𝑡 Free jet: 𝑢 𝐴 𝐹 𝐹0 𝑈 𝐴0 𝐹𝑒𝑥𝑡 = 𝐹 − 𝐹0 = 0 𝑑𝑝 𝑑𝑡 ∴ Momentum flow is a CONSERVED quantity 𝑑𝑝 𝑑𝑉 𝐹= =𝜌 𝑢 = 𝜌𝐴𝑢2 𝑑𝑡 𝑑𝑡 𝐹0 = 𝜌𝐴0 𝑈 2 Velocity at the Center Assume that the velocity is even at the narrow area of the center. 𝑢0 By momentum flow conservation 𝑟2 𝑦 𝜌𝑄0 𝑈 = 𝜌𝑄𝑢0 𝜌 𝜋𝑟1 2 𝑈 2 = 𝜌(𝜋𝑟2 2 )𝑢0 2 𝜑 𝑈 𝑟1 𝑟1 = 𝑟2 − 𝑦 tan 𝜑 𝑈 tan 𝜑 ∴ 𝑢0 = 𝑈 − 𝑦 𝑟2 Experiment: Velocity at the Center 𝑦 Pitot tube anemoemeter Result 20 18 16 14 12 airstream velocity 10 (m/s) 8 𝑢0 = 𝑈 − 𝐾𝑦 6 4 2 0 0 3 6 9 Height from the opening (cm) 12 15 Result 30 25 20 airstream velocity 15 (m/s) 10 5 0 0 3 6 9 Height from the opening (cm) 12 15 Velocity Profile: Gaussian Curve Assume Gaussian curve profile: 𝑢0 𝑢(𝑥) 𝑥 𝑢 𝑥 = 𝑥2 − 𝑢0 𝑒 2𝑐 2 By momentum flow conservation 𝜌𝑈 2 ∙ 𝜋𝑟0 2 = 𝜌 ∞ = 𝜌𝑢0 2 𝑥2 − 2 𝑒 𝑐 𝑢 𝑥 2 𝑑𝐴 ∙ 2𝜋𝑥𝑑𝑥 0 𝑈𝑟 = 𝜌𝑢0 2 ∙ 𝜋𝑐 2 𝑟0 𝑈 ∴𝑐= 𝑢0 0 𝑢 𝑥, 𝑦 = (𝑈 1 𝐾 2 2 − (1− 𝑦) 𝑥 2 𝑈 2𝑟 0 − 𝐾𝑦)𝑒 Experiment: Velocity Profile 𝑥 Result 20 18 16 14 12 Airstream velocity 10 (m/s) 8 6 4 2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 Radial distance from the center (cm) 4.5 5 Error Airstream velocity at the opening is weaker at the edge! 𝑈 Velocity at the Opening 20 15 Airstream velocity at the 10 opening (m/s) 5 0 0 0.5 1 Radial distance from the center (cm) 1.5 Experiment Experimental Setting 𝑈 𝜃 JAVA Image Processing Controlling the Mass of the Ball Clay 𝑚 Drag Force and 𝒚- Oscillation Drag Force 𝑏 𝑢 𝑥 = (𝑈 𝐹𝐷 = 𝑈 𝑟 0 1 𝐾 2 2 − (1− 𝑦) 𝑥 2 𝑈 2𝑟 − 𝐾𝑦)𝑒 0 𝑑𝐹𝐷 = 1 𝜌𝐶𝐷 𝑢2 𝑑𝐴 2 1 𝐾 2 2 𝑏 1 − (1− 𝑦) 𝑥 2 2 𝑈 𝑟 = 𝜌𝐶𝐷 (𝑈 − 𝐾𝑦) 𝑒 0 ∙ 2𝜋𝑥𝑑𝑥 2 0 𝑏2 𝐾 2 1 − 2 1−𝑈 𝑦 2 2 = 𝜋𝜌𝐶𝐷 𝑈 𝑟0 (1 − 𝑒 𝑟0 ) 2 𝒚- Oscillation Model 𝑚𝑦 = 𝐹𝐷 − 𝑚𝑔 2 𝜋𝜌𝐶𝐷 𝑈 𝑟0 𝑦= 2𝑚 2 1− 𝑏2 𝐾 2 − 2 1− 𝑦 𝑈 𝑒 𝑟0 −𝑔 𝑈 = 16m/s, 𝐾 = 38.04s −1 , 𝑦 0 = 0.18m, 𝑦 ′ 0 = 0 Theory: 𝒚- Oscillation 𝑇 ≈ 0.83s Experiment: 𝒚- Oscillation 𝑇 ≈ 0.93s ± 0.066s Theory: y-Equilibrium Position 𝑚𝑦 = 𝐹𝐷 − 𝑚𝑔 = 0 1 𝜋𝜌𝐶𝐷 𝑈 2 𝑟0 2 1 2 2 𝑏2 𝐾 − 2 1− 𝑦𝑒𝑞 𝑈 − 𝑒 𝑟0 − 𝑚𝑔 = 0 Experiment: 𝒚-Equilibrium Position 0.3 Cause of Error 1. Gaussian curve assumption 2. Compression effect 0.25 0.2 yEquilibrium 0.15 position (m) 0.1 Theory Experiment 0.05 0 0 4 8 12 16 20 Airstream velocity at the opening (m/s) 24 Compression of the Airstream When 𝑦-displacement is low…… 𝐴𝑒𝑓𝑓 𝑢𝑒𝑓𝑓 Experiment Setting: Compression of the Airstream 𝑦 Force sensor Experiment Result: Compression of the Airstream 1.2 Compression! 1 0.8 Drag Force 0.6 (N) 0.4 0.2 0 0 1 2 Distance between the air pump and the ball (cm) 3 Pressure Force and 𝒙- oscillation Bernoulli Effect 𝑏 𝑃𝑏−𝑥 𝑢𝑏−𝑥 𝑦 𝑈 𝑥𝑈 1 2 1 2 𝑃 + 𝜌𝑈 = 𝑃 + 𝜌𝑢 𝑏+𝑥 𝑃𝑏+𝑥 0 2 2 𝑏+𝑥 𝑢𝑏+𝑥 𝑃 + 1 𝜌𝑈 2 = 𝑃 + 1 𝜌𝑢 2 0 𝑏−𝑥 2 2 𝑏−𝑥 𝐹𝑏𝑒𝑟𝑛𝑜𝑢𝑙𝑙𝑖 = 𝜋𝑏 2 (𝑃𝑏−𝑥 − 𝑃𝑏+𝑥 ) 𝜌 2 = −𝜋𝑏 (𝑢𝑏−𝑥 2 − 𝑢𝑏+𝑥 2 ) 2 Bernoulli Effect 2 𝐹𝑏𝑒𝑟𝑛𝑜𝑢𝑙𝑙𝑖 𝜋𝑏 =− 𝜌 𝑈 − 𝐾𝑦 2 𝐾 2 𝑏−𝑥 2 2 − 1−𝑈 𝑦 𝑟0 (𝑒 𝐾 2 𝑏+𝑥 2 − 1− 𝑦 𝑈 𝑟0 −𝑒 ) Coanda Effect 𝑑𝑃 𝑑𝐹 = −𝐴𝑑𝑃 = − 𝐴𝑑𝑥 𝑑𝑥 𝑑𝐹 = 𝑎𝑑𝑚 = 𝑎𝜌𝐴𝑑𝑥 𝑃 𝑑𝑃 = −𝜌𝑎 𝑑𝑥 𝑃 − 𝑑𝑃 𝑑𝑥 𝑢2 𝑎=− 𝑏 ∴ 𝑑𝑃 𝑑𝑥 = 𝑢2 𝜌 𝑏 How thick is 𝒅𝒙? 𝑢 Viscous force 𝛿 𝑦 No-slip Condition Boundary Layer Thickness 𝛿 ≈ 0.382𝑦/𝑅𝑒𝑦 0.2 𝜌𝑢 𝜇 𝑅𝑒𝑦 = 𝑦 ( : kinematic viscosity ) 𝜇 𝜌 Coanda Effect 𝑈 = 16m/s: 𝛿 ≈ 1.2mm 𝑢𝑏−𝑟 𝛿 𝑢𝑏+𝑟 𝛿 𝑑𝑃 𝑢2 =𝜌 𝑑𝑥 𝑏 𝑥 𝑢 = 𝑢𝑏+𝑥 𝛿 𝜌𝑢𝑏+𝑥 2 0 2 ∆𝑃𝑟𝑖𝑔ℎ𝑡 = 𝑥 𝑑𝑥 2 𝑏𝛿 𝛿 𝜌𝛿 = − 𝑢𝑏+𝑥 2 3𝑏 ∆𝑃𝑙𝑒𝑓𝑡 = 𝜌𝛿 − 𝑢𝑏−𝑥 2 3𝑏 Coanda Effect 𝑈 = 16m/s: 𝛿 ≈ 1.2mm 𝐹𝑐𝑜𝑎𝑛𝑑𝑎 = 𝜋𝑏 2 (∆𝑃𝑙𝑒𝑓𝑡 − ∆𝑃𝑟𝑖𝑔ℎ𝑡 ) 2 = −𝜋𝑏 ∙ 𝜌𝛿 (𝑢𝑏−𝑥 2 3𝑏 − 𝑢𝑏+𝑥 2 ) 2𝛿 𝐹𝑏𝑒𝑟𝑛𝑜𝑢𝑙𝑙𝑖 3𝑏 = ≈ 0.04𝐹𝑏𝑒𝑟𝑛𝑜𝑢𝑙𝑙𝑖 Bernoulli effect is the major one! 𝒙-Oscillation Model 𝑥 2 𝜋𝑏 = −1.04 𝜌 𝑈 − 𝐾𝑦 2𝑚 − 𝐾 2 𝑏−𝑥 2 2 − 1−𝑈𝑦 𝑟0 (𝑒 𝐾 2 𝑏+𝑥 2 − 1− 𝑦 𝑈 𝑟0 𝑒 ) 𝑈 = 16m/s, 𝐾 = 38.04s −1 , 𝑥 0 = 0.001m, 𝑥 ′ 0 = 0 Theory: 𝒙-Oscillation 𝑇 ≈ 0.16s Experiment: 𝒙-Oscillation 𝑇 ≈ 0.18s ± 0.0056s Experiment: 𝒙-Equilibrium Position 12 10 𝐹𝑃 = 𝑚𝑔 sin 𝜃 8 x -Equilibrium 6 position (mm) 4 2 0 0 5 10 15 20 25 Angle of tilt (º) 30 35 40 Motion of the Ball Theory Experiment Optimization Theory: Optimization 𝑦 𝐹𝐷 𝐹𝐷 = 𝑚𝑔 cos 𝜃 𝐹𝑃 = 𝑚𝑔 sin 𝜃 𝐹𝑃 = tan 𝜃 𝐹𝐷 𝐹𝑃 𝑥 𝜃 𝑚𝑔 𝜃 𝑦, 𝑥 that achieves 𝐹𝑃 max( ) 𝐹𝐷 ⇒ OPTIMIZATION! Airstream velocity optimization 60 50 40 Maximum angle of tilt 30 (º) 20 10 0 0 4 8 12 16 Airstream velocity at the opening (m/s) 20 Mass of the ball Optimization 60 50 Maximum 40 angle of tilt 30 (º) 20 10 0 0 1 2 3 4 Mass of the ball (g) 5 6 Experiment: Optimization Maximum Angle of Tilt 𝜽𝒎𝒂𝒙 = 𝟓𝟎° 𝐹𝑃 max = tan 50° 𝐹𝐷 𝟓𝟎° Experiment: Optimization 𝑈 = 7~8m/s 𝑚 = 2.8g 𝑦𝑒𝑞 = 8.4cm 𝑥𝑒𝑞 = 1.3cm 𝑥𝑒𝑞 𝑦𝑒𝑞 𝟓𝟎° 𝐹𝑃 max = tan 50° 𝐹𝐷 Conclusion Velocity Profile Gaussian Curve & Momentum flow Conservation 𝒚- Oscillation Drag Force 𝐹𝐷 𝒙- Oscillation Bernoulli + Coanda Pressure Force 𝐹𝑃 Optimization max( 𝐹𝑃 𝐹𝐷 ), 50° Thank You! Spin of the ball 𝝎 = 0.74rev/s=4.65rad/s 𝒗 = 𝒃𝝎 = 𝟎. 𝟎𝟗𝟑m/s ≪ 8.5m/s (airstream velocity at equilibrium point) Irregular direction Small rotation speed NEGLIGENT! Turbulence and Bernoulli Effect 𝑣 𝑃 1 2 𝑃 + 𝜌𝑣 = constant 2 Throughout the field! Assumption: 𝜵 × 𝒗 = 𝟎 (Irrotational) But in my model… 𝑏 𝑃𝑏+𝑥 𝑃𝑏−𝑥 𝑢𝑏−𝑥 𝑢𝑏+𝑥 𝑦 𝑈 𝑥𝑈 1 2 1 𝑃0 + 𝜌𝑈 = 𝑃𝑏+𝑥 + 𝜌𝑢𝑏+𝑥 2 2 2 1 2 1 𝑃0 + 𝜌𝑈 = 𝑃𝑏−𝑥 + 𝜌𝑢𝑏−𝑥 2 2 2 Bernoulli equation applied in a SINGLE streamline Although cause of error, still VALID in turbulent situations! Ignoring gravity term in Bernoulli equation 𝜌𝑔ℎ 1 2 𝜌𝑣 2 8.5m/s at 8cm height 2 𝑔ℎ = 0.784 𝑚 2 𝑠 1 2 2 𝑚 𝑣 = 75.9 2 𝑠 2 Reynolds number 𝜌𝑈𝐿 𝑅𝑒 = 𝜇 𝜇 𝜌 =(kinematic viscosity of air at 27℃)=1.57 × 10−5 m2/s 𝑈 = 8.5m/s 𝐿 = 0.04m 𝑅𝑒 = 2.7 × 10 𝐶𝐷 ≈ 0.5 4 Conservative Drag Force Under assumptions that… 1. Velocity of the ball can be ignored compared to velocity of the airstream 2. Drag force is a decreasing function of height We can apply ‘drag force field’ just like gravitational field Starting point of the ball does not matter! Equilibrium position / stable condition would not significantly matter 𝐸𝑚𝑒𝑐 How measure maximum angle of tilt? • Ten times of trial • When stably levitated for more than ten seconds, PASS! Period and Amplitude: 𝒚-Oscillation 𝑦 0 = 0.17m 𝑦 0 = 0.18m 𝑦 0 = 0.19m 𝑦 0 = 0.20m Period and Amplitude: 𝒙-Oscillation 𝑥 0 = 0.0005m 𝑥 0 = 0.001m 𝑥 0 = 0.002m 𝑥 0 = 0.003m Drag force change due to motion of the ball 𝑢𝑏𝑎𝑙𝑙,𝑦 = (𝑦0 − 𝑦𝑒𝑞 )(2𝜋 𝑇 ) ≈ 0.10m/s 𝑦 𝑢𝑏𝑎𝑙𝑙,𝑥 = 𝑥0 (2𝜋 𝑇 ) ≈ 0.04m/s 𝑥 𝑈 = 16m/s Gaussian Curve Assumption Assume Gaussian curve profile: 𝑢0 𝑢(𝑥) 𝑥 𝑢 𝑥 = 𝑥2 − 2 𝑢0 𝑒 2𝑐 2 𝑑𝑢(𝑥) 𝑢0 − 𝑥 2 = − 2 𝑥𝑒 2𝑐 𝑑𝑥 𝑐 1 = − 2 𝑥𝑢(𝑥) 𝑐 𝑈𝑟 0 𝑑𝑢(𝑥) ∝ −𝑥𝑢(𝑥) 𝑑𝑥 Experimental proof: Coanda < Bernoulli Liquid Nitrogen Outer layer: No interaction with the ball surface JAVA code public void obtainFilteredImage() { for (int w=1; w<=width-1; w++) { for (int h=1; h<=height-1; h++) { green[w][h]>20) } } } } if (red[w][h]>50 && red[w][h]-blue[w][h]>20 && red[w][h]{ copyImg.setRGB(w,h,RED.getRGB()); else { copyImg.setRGB(w,h,WHITE.getRGB()) JAVA code public double[] center() { int xsum, ysum, count = 0;;; double[] c = new double[2]; for (int w=1; w<=width-1; w++){ for (int h=1; h<=height-1; h++){ if(new Color(copyImg.getRGB(w,h)).equals(new Color(255,0,0)) ){ xsum += w; ysum += h; count += 1; } } } c[0] = xsum/count; c[1] = ysum/count; return c; } JAVA code public static double rotateX(double x, double y) { double xReal = x-referX; double yReal = referY-y; } return xReal*Math.cos(deg)+yReal*Math.sin(deg); public static double rotateY(double x, double y) { double xReal = x-referX; double yReal = referY-y; } return yReal*Math.cos(deg)-xReal*Math.sin(deg);

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