Fluids in motion

1. Fluids in motion • Goals • Understand the implications of continuity for Newtonian fluids • Distinguish pressure and force for fluids in motion • Employ Bernoulli’s equation http://boojum.as.arizona.edu/~jill/NS102_2006/Lectures/Lecture12/turbulent.html

2. Case 1 M dA P1 A1 A10 Case 2 M dB P2 A2 A10 Pascal’s Principle: Example • Now consider the set up shown on right. • Mass M is placed on right piston, A10 > A1 = 2A1 • How do dA and dB compare? • Equilibrium when pressures at P(left & right) are equal and P1= P2 F1 / A1= F2 / A2 r (A1dA) g/ A1= r(A2d2) g/ A2 dA = dB

3. Fluids in Motion • Real flow vs. ideal flow • non-steady / steady state • compressible / incompressible • rotational / irrotational • viscous / frictionless

4. Types of Fluid Flow • Laminar flow • Each particle of the fluid follows a smooth path • The paths of the different particles never cross each other • The path taken by the particles is called a streamline • Turbulent flow • An irregular flow characterized by small whirlpool like regions • Turbulent flow occurs when the particles go above some critical speed

5. Types of Fluid Flow • Laminar flow • Each particle of the fluid follows a smooth path • The paths of the different particles never cross each other • The path taken by the particles is called a streamline • Turbulent flow • An irregular flow characterized by small whirlpool like regions • Turbulent flow occurs when the particles go above some critical speed

6. Continuity • The mass or volume per unit time of an ideal fluid moving past point 1 equals that moving past point 2 • Flow obeys continuityor mass conservation Volume flow rate (m3/s) Q = A·v is constant along tube. A1v1 = A2v2 Mass flow rate is just r Q (kg/s)

7. Example problem • The figure shows a water stream in steady flow from a faucet. At the faucet the diameter of the stream is 1.00 cm. The stream fills a 1000 cm3 container in 20 s. Find the velocity of the stream 10.0 cm below the opening of the faucet. Q = A1v1 = A2v2 Q = DV / Dt =1000 x 10-6 / 20 m3/s v1 = Q / A1= 5 x 10-5 / 0.25p x 10-4 m/s v1 = Q / A1= 0.64 m/s K2 = K1+Dmgh= ½ Dmv12 + Dmgh v2 = (v12 + gh)½ = (0.642 + 9.8 x 0.1)½ = 1.2 m/s

8. A 2 A 1 v 1 v 2 Ideal Fluid Model (frictionless, incompressible) • Streamlines do not meet or cross • Velocity vector is tangent to streamline • Volume of fluid follows a “tube of flow” bounded by streamlines • Streamline density is proportional to velocity

9. (A) 2 v1 (B) 4 v1 (C) 1/2 v1 (D) 1/4 v1 Exercise Continuity • A housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2” diameter at some point in your house. v1 v1/2 • Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the 1” diameter pipe, how fast is the water going in the 1/2” pipe?

10. v1 v1/2 Exercise Continuity (A) 2 v1 (B) 4 v1 (C) 1/2 v1 (D) 1/4 v1 • For equal volumes in equal times then ½ the diameter implies ¼ the area so the water has to flow four times as fast. • But if the water is moving 4 times as fast then it has 16times as much kinetic energy. • Something must be doing work on the water

11. v1 v1 v1/2 v1/2 F1 F2 Exercise Continuity • Experimentally we observe a pressure drop at the neck and this can be recast as work (i.e., energy transfer) P DV = (F/A) (A Dx) = F Dx F1 and F2 maintain the pressure in the tube as the water flows

12. P1 P2 Conservation of Energy for an Ideal Fluid W = (P1– P2 ) DV = DK W = ½ Dm v22 – ½ Dm v12 = ½ (r DV) v22 – ½ (r DV)v12 (P1– P2 ) = ½ r v22 – ½ r v12 P1+ ½ r v12= P2+ ½ r v22= const. and with height variations (potential energy): Bernoulli Equation P1+ ½ r v12 + r g y1 = constant Smaller diameter implies a pressure drop

13. P0 = 1 atm Torcelli’s Law (Bernoulli in action) • The flow velocity v = (gh)½ where h is the depth from the top surface P + r g h + ½ r v2 = const AB P0 + r g h + 0 = P0 + 0 + ½ r v2 2g h = v2 d = ½ g t2 t = (2d/g)½ x = vt = (2gh)½(2d/g)½ = (4dh)½ d d A B d

14. Applications of Fluid Dynamics • Streamline flow around a moving airplane wing • Lift is the upward force on the wing from the air • Dragis the resistance • The lift depends on the speed of the airplane, the area of the wing, its curvature, and the angle between the wing and the horizontal • But Bernoulli’s Principle is not applicable (open system) and air is very compressible higher velocity lower pressure lower velocity higher pressure Note: density of flow lines reflects velocity, not density. We are assuming an incompressible fluid.

15. Fluids: A tricky problem • A beaker contains a layer of oil (green) with density ρ2 floating on H2O (blue), which has density ρ3. A cube wood of density ρ1and side length L is lowered, so as not to disturb the layers of liquid, until it floats peacefully between the layers, as shown in the figure. • What is the distance d between the top of the wood cube (after it has come to rest) and the interface between oil and water? • Hint: The magnitude of the buoyant force (directed upward) must exactly equal the magnitude of the gravitational force (directed downward). The buoyant force depends on d. The total buoyant force has two contributions, one from each of the two different fluids. Split this force into its two pieces and add the two buoyant forces to find the total force

16. Fluids: A tricky problem • A beaker contains a layer of oil (green) with density ρ2 floating on H2O (blue), which has density ρ3. A cube wood of density ρ1 and side length L is lowered, so as not to disturb the layers of liquid, until it floats peacefully between the layers, as shown in the figure. • What is the distance d between the top of the wood cube (after it has come to rest) and the interface between oil and water? • Soln: Fb = W1 = ρ1 V1 g = ρ1 L3 g = Fb2 + Fb3 = ρ2 d L2 g + ρ3 (L-d) L2 g ρ1 L = ρ2 d + ρ3 (L-d) (ρ1 - ρ3 ) L = (ρ2 - ρ3 ) d

17. For Thursday • Read Chapter 15.1 to 15.3