Fluids in Motion Supplement I

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Fluid Drag at Low Reynolds Number For very low Reynolds Number (Re < 0.1 – this regime is called creeping flow), the drag coefficient for a sphere has been shown to be: C d = 24 Re where Re is the Reynolds Number = Dvρ/η. Plugging this into the drag equation on the previous page, and using the cross-sectional area of a sphere ⎛1 2⎞ 2⎛1 2⎞ 24 Fd = A⎜ ρv ⎟Cd = πR ⎜ ρv ⎟ ⎝2 ⎠ ⎝2 ⎠ 2 / (note that D = 2R has been used here). There is a lot of cancellation and we get the final expression for the Stokes drag, or drag on a sphere at very low Reynolds Number: Fd = 6πη Rv This expression was calculated by Stokes in 1850 for the drag force on a sphere of radius R moving with speed v relative to a medium with viscosity η. This expression only applies to a sphere, but other shapes have drag forces that have similar forms at low Reynolds Number. Objects with densities ρ 1 that differ from the density, ρ 2 , of the surrounding medium will either sink (if ρ 1 more dense than ρ 2 ) or float (if ρ 1 less dense than ρ 2 ). For objects sedimenting in fluids three forces acting on the particle: – Gravity, F g = mg = ρ 1 Vg – Buoyant force, F b = m fluid displaced g = ρ 2 Vg – Hydrodynamic drag, F d ( Rvρ η) Sedimentation and Terminal Velocity The drag force increases with speed and so if a particle starts at rest, it will accelerate until the drag force equals the sum of the gravitational and buoyant forces. At this point the object will move with a constant velocity. Choosing down to be negative, the force balance is then: ∑ Fy = Fd + Fb − Fg = 0 F d F b Plugging in for F g and F b , we get: F + ρ gV − ρ gV = d 2 1 0 ρ 1 ρ 2 We can rearrange to get the fluid drag at terminal velocity is given by: Fd ( ρ ρ ) = − 1 2 gV F g
Sedimentation of a Sphere at high Re Suppose a spherical object with density ρ 1 is sinking (or floating) within a fluid. At high Reynolds Number, the drag force is given by the modified Bernoulli drag: 2⎛1 2⎞ Fd = πR ⎜ ρ fluidv ⎟ ⎝2 ⎠ Plugging this into the sedimentation force balance on the previous page and using the formula for the volume and area of a sphere: 2 1 2 4 3 πR ⎛ ⎜ ρ2v ⎞ ⎟( 0.44) = ( ρ1−ρ2) g ⎛ ⎜ πR ⎞ ⎟ ⎝2 ⎠ ⎝3 ⎠ There is a lot of cancellation here and one may then solve for the terminal velocity of the sedimenting sphere. v 8 Rg ⎛ ρ − ρ ⎞ g ⎝ ⎠ 1 2 = ⎜ ⎟ 3 C d ρ2 ( 0.44) Example S3 Sinking speed of a cannonball in water Suppose you have a 3 kg iron cannonball sinking in water. What is its terminal velocity? First, find the radius of the cannonball. Since the density of iron is 7860 kg/m 3 : So: Since this radius is fairly large and the sinking speed (which we know from experience) will be reasonably large, say 1 m/s, the Reynolds number = Dvρ/η ~ (2*0.045)(1)(1000)/(.001) ~ 90,000. One can thus use the high Re formula for sedimentation: v We finally get: R cannonball 8 Rg ⎛ ρ −ρ ⎞ 8 (0.0450)(9.80) ⎛7860 −1000 ⎞ ⎜ ⎟ ⎜ ⎟ 3 ⎝ ⎠ 3 (0.44) ⎝ 1000 ⎠ 1 2 = = Cd ρ2 m 4 V 3 cannonball π R cannonball ρ = = 3 = 3m 33.00 ( kg) 3 3 0.0450 m 3 4πρ = 4π 7860 kg/m = ( ) v = 4.28 m/s
Page 1 and 2: Fluids in Motion Supplement I Cutne
Page 3 and 4: Fluids in Motion Supplement I Some
Page 5: Drag on an Object Bernoulli’s Law

Fluids in Motion Supplement I

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