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Concept of Drag

Concept of Drag. Viscous Drag and 1D motion. Concept of Drag. Drag is the retarding force exerted on a moving body in a fluid medium It does not attempt to turn the object, simply to slow it down

ainsley-leblanc
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Concept of Drag

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  1. Concept of Drag Viscous Drag and 1D motion

  2. Concept of Drag • Drag is the retarding force exerted on a moving body in a fluid medium • It does not attempt to turn the object, simply to slow it down • It is a function of the speed of the body, the size (and shape) of the body, and the fluid through which it is moving

  3. Drag Force Due to Air • The drag force due to wind (air) acting on an object can be found by: FD = ½ ρ CDV2A where: FD = drag force (N) CD = drag coefficient (no units) V = velocity of object (m/s) A = projected area (m2) ρ = density of air (kg/m3) {1.2 kg/m3}

  4. Drag Coefficient: CD • The drag coefficient is a function of the shape of the object and the Reynolds Number (Re) • Re = (velocity * length-scale) / (kinematic viscosity) • For a spherical shape the drag coefficient ranges from 0.1 to 300, as shown on the next slide.

  5. Drag Coefficient for Spheres

  6. Dropping a Ping Pong Ball • If you dropped a ping pong ball down the stairwell in this building (height 50 feet), and the stairwell had a vacuum in it, how long would it take for the ping pong ball to hit the floor? • If you left the air in the stairwell would it take longer, shorter, or the same time to hit the bottom?

  7. FD=f(v) FG=mg Looking at the ball in detail • Drawing a Free Body Diagram (FBD) of the ball is shown to the right • Since all the drag force is doing is slowing the ball down, it is directly vertical and upwards

  8. Numerical Analysis • If you have two data points (time, position), then you can approximate the velocity of the body. • Given the points (2 s, -15m) and (2.1 s, -17m), what is the approximate velocity at 2.1 seconds? • If the next data point is (2.2 s, -19.05m), what is the velocity at 2.2 seconds?

  9. Solution

  10. Now Find Acceleration • Given the velocities at 2.1s and 2.2s, what is the acceleration at 2.2s? • Data points are (time, velocity): • (2.1s, -20 m/s) • (2.2s, -20.5 m/s)

  11. Acceleration Solution

  12. Continuing the process • The ultimate goal of this numerical analysis is to find the drag force on the body • Now that we have the acceleration, we can find the total force acting on the body (F=ma), the force of gravity (Fg=mg), and Drag Force (FD)

  13. Implementing this in Matlab (Section 8.3) • Suppose you have two arrays defined, one with the elevation of the ping pong ball and the other with the time intervals • You can use numerical methods to determine the drag coefficient by estimating the acceleration of the ball at each time step

  14. Central Differencing • To find the average velocity over a time interval, you divide the change in position by the change in time. • This is called forward differencing. • A better estimate is central differencing, where you estimate the velocity at a point by considering the points on either side of it.

  15. Sample Data • Type the sample data into Matlab and plot Height versus time • Use central differencing to find the velocity at the points 0.1 through 0.9

  16. The plotted data

  17. Central Differencing for Velocity • To find the velocity at time 0.1 seconds, use the formula: >> t1=[0:.1:1]; >> h1=[0 -.05 -.2 -.44 -.76 -1.17 -1.65 -2.19 -2.78 -3.41 -4.08]; >> v1=(h1(3:10)-h1(1:8))./(t1(3:10)-t1(1:8)) v1 = -1.0000 -1.9500 -2.8000 -3.6500 -4.4500 -5.1000 -5.6500 -6.1000

  18. Central Differencing for Acceleration • To find the acceleration at time 0.2 seconds, use the formula: • Notice you must be careful with the indices >> a1=(v1(3:8)-v1(1:6))./(t1(4:9)-t1(2:7)) a1 = -9.0000 -8.5000 -8.2500 -7.2500 -6.0000 -5.0000

  19. Finding the Drag Force • With the net acceleration known at each time step, the acceleration due to drag can be found be subtracting the acceleration due to gravity. • Then the drag force, at each time step, can be found from Newton’s second law (for a mass of 0.0025 kg). >> a2=a1-(-9.8) a2 = 0.8000 1.3000 1.5500 2.5500 3.8000 4.8000 >> f2=a2*.0025 f2 = 0.0020 0.0032 0.0039 0.0064 0.0095 0.0120

  20. Finding the Drag Coefficient • If you know the velocity at each time step and the drag force at each time step, you can plot F versus v2 and fit a straight line to the data. • Care must be taken to make sure you line up the correct force (a 1x6 array) with the correct velocity (a 1x8 array). Force 1 matches velocity 2, etc.

  21. Using Matlab >> v2=v1(2:7) v2 = -1.9500 -2.8000 -3.6500 -4.4500 -5.1000 -5.6500 >> v3=v2.^2 v3 = 3.8025 7.8400 13.3225 19.8025 26.0100 31.9225 >> plot(v3,f2,v3,f2,'o')

  22. Using Least Squares • Cast the data points into array form using the base equation and then solve using matrix math. • Note that currently V2 and FD are row vectors and need to be transposed. >> k=v3'\f2' k = 3.5928e-004 For this problem, \ is the least squares operator!

  23. Final Form of Drag Equation • The final form of the drag equation can be written, which will allow the drag force to be calculated at any velocity. • where: • FD is in Newtons • V is in meters per second

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