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physicsyd.au/teach_res/jp/fluids09

http://www.physics.usyd.edu.au/teach_res/jp/fluids09. web notes: lect6.ppt. Ideal fluid. Real fluid. What is the speed with which liquid flows from a hole at the bottom of a tank?. v 1 ~ 0 m.s -1. (1) Surface of liquid. p 1 = p atm. Draw flow tubes. y 1. h = ( y 1 - y 2 ).

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physicsyd.au/teach_res/jp/fluids09

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  1. http://www.physics.usyd.edu.au/teach_res/jp/fluids09 web notes: lect6.ppt

  2. Ideal fluid Real fluid

  3. What is the speed with which liquid flows from a hole at the bottom of a tank? v1 ~ 0 m.s-1 (1) Surface of liquid p1 = patm Draw flow tubes y1 h = (y1 - y2) v2 = ? m.s-1 y2 (2) Just outside hole p2 = patm

  4. Assume liquid behaves as an ideal fluid and that Bernoulli's equation can be applied • p1 + ½ v12 + g y1 = p2 + ½ v22 + g y2 • A small hole is at level (2) and the water level at (1) drops slowly v1 = 0 • p1 = patmp2 = patm • g y1 = ½ v22 + g y2 • v22 = 2 g (y1 – y2) = 2 g h h = (y1 - y2) • v2 = (2 g h) Torricelli formula (1608 – 1647) • This is the same velocity as a particle falling freely through a height h

  5. What is the speed of flow in section 1 of the system? (1) (2) rF v1 = ? h rm

  6. Assume liquid behaves as an ideal fluid and that Bernoulli's equation can be applied for the flow along a streamline p1 + ½ v12 + g y1 = p2 + ½ v22 + g y2 y1 = y2 p1 – p2 = ½ F (v22 - v12) p1 - p2 = mg h A1v1 = A2v2 v2 = v1 (A1 / A2) mg h = ½ F { v12 (A1 / A2)2- v12 } = ½ Fv12 {(A1 / A2)2 - 1}

  7. pC yC Q: What do we know? Continuous flow Pressure in top section > 0 otherwise there will be a vacuum pC  0 Focus on falling water not rising water patm - pC  0 patm  gyC How does a siphon work?

  8. C yC A yA B yB Assume that the liquid behaves as an ideal fluid, the equation of continuity and Bernoulli's equation can be used. D yD = 0 pA = patm = pD

  9. Consider points C and D and apply Bernoulli's principle. pC + ½ vC2 + g yC = pD + ½ vD2 + g yD From equation of continuity vC = vD pC = pD + g (yD - yC) = patm + g (yD - yC) The pressure at point C can not be negative pC 0 and yD = 0 pC = patm - g yC 0 yC patm / (g) For a water siphon patm ~ 105 Pa g ~ 10 m.s-1  ~ 103 kg.m-3 yC 105 / {(10)(103)} m yC 10 m

  10. How fast does the liquid come out? Consider a point A on the surface of the liquid in the container and the outlet point D. Apply Bernoulli's principle pA + ½ vA2 + g yA = pD + ½ vD2 + g yD vD2 = 2 (pA – pD) / + vA2 + 2 g (yA - yD) pA – pD = 0 yD = 0 assume vA2 << vD2 vD = (2 gyA )

  11. FLUID FLOW MOTION OF OBJECTS IN FLUIDS How can a plane fly? Why does a cricket ball swing or a baseball curve? Why does a golf ball have dimples?

  12. FORCES ACTING ON OBJECT MOVING THROUGH FLUID Resultant FR Lift FL drag FD Forward thrust by engine Motion of object through fluid Fluid moving around stationary object

  13. Uniform motion of an object through an ideal fluid ( = 0) The pattern is symmetrical FR = 0 C B A D

  14. Drag force frictional drag (viscosity) pressure drag (eddies – lower pressure)

  15. motion of air Drag force due to pressure difference motion of object low pressure region rotational KE of eddies  heating effect  increase in internal energy  temperature increases NO CURVE Drag force is opposite to the direction of motion high pressure region

  16. motion of air Drag force due to pressure difference ball: curved flight motion of object flow speed (high) vair + v  reduced pressure v vair (vball) MAGNUS EFFECT flow speed (low) vair - v  increased pressure v Boundary layer – air sticks to ball (viscosity) – air dragged around with ball high pressure region low pressure region

  17. Professional golf drive Initial speed v0 ~ 70 m.s-1 Angle ~ 6° Spin  ~ 3500 rpm Range ~ 100 m (no Magnus effect) Range ~ 300 m (Magnus effect)

  18. Semester 1, 2004 Exam question • A large artery in a dog has an inner radius of 4.0010-3 m. Blood flows through the artery at the rate of 1.0010-6 m3.s-1. The blood has a viscosity of 2.08410-3 Pa.s and a density of 1.06103 kg.m-3. • Calculate: • (i) The average blood velocity in the artery. • (ii) The pressure drop in a 0.100 m segment of the artery. • The Reynolds number for the blood flow. • Briefly discuss each of the following: • (iv) The velocity profile across the artery (diagram may be helpful). • (v) The pressure drop along the segment of the artery. • (vi) The significance of the value of the Reynolds number calculated in • part (iii).

  19. Solution • radius R = 4.0010-3 m • volume flow rate Q = 1.0010-6 m3.s-1 • viscosity of blood  = 2.08410-3 Pa.s • density of blood  = 1.06010-3 kg.m-3 • v = ? m.s-1 • p = ? Pa • Re = ?

  20. Equation of continuity: Q = A v • A = R2 =  (4.0010-3)2 = 5.0310-5 m2 • v = Q / A = 1.0010-6 / 5.0310-5 m.s-1 = 1.9910-2 m.s-1 • (ii) Poiseuille’s Equation • Q = P R4 / (8 L) L = 0.100 m • P = 8 L Q / (R4) • P = (8)(2.08410-3)(0.1)(1.0010-6) / {()(4.0010-3)4} Pa • P = 2.07 Pa • (iii) Reynolds Number • Re = v L / where L = 2 R (diameter of artery) • Re= (1.060103)(1.9910-2)(2)(4.0010-3) / (2.08410-3) • Re = 81 • use diameter not length

  21. (iv) Parabolic velocity profile: velocity of blood zero at sides of artery (v) Viscosity  internal friction  energy dissipated as thermal energy  pressure drop along artery (vi) Re very small laminar flow (Re < 2000)

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