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FLUID FLOW IDEAL FLUID BERNOULLI'S PRINCIPLE

FLUID FLOW IDEAL FLUID BERNOULLI'S PRINCIPLE. How can a plane fly? How does a perfume spray work? What is the venturi effect? Why does a cricket ball swing or a baseball curve?. web notes: lect6.ppt flow3.pdf. Daniel Bernoulli (1700 – 1782). A 1. A 1. A 2. v 2. v 1.

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FLUID FLOW IDEAL FLUID BERNOULLI'S PRINCIPLE

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  1. FLUID FLOW IDEAL FLUID BERNOULLI'S PRINCIPLE How can a plane fly? How does a perfume spray work? What is the venturi effect? Why does a cricket ball swing or a baseball curve? web notes: lect6.ppt flow3.pdf

  2. Daniel Bernoulli (1700 – 1782)

  3. A1 A1 A2 v2 v1 v1 Low speed Low KE High pressure high speed high KE low pressure Low speed Low KE High pressure

  4. p large p large p small v small v large v small

  5. In a serve storm how does a house loose its roof? Air flow is disturbed by the house. The "streamlines" crowd around the top of the roof faster flow above house reduced pressure above roof than inside the house room lifted off because of pressure difference. Why do rabbits not suffocate in the burrows? Air must circulate. The burrows must have two entrances. Air flows across the two holes is usually slightly different slight pressure difference forces flow of air through burrow. One hole is usually higher than the other and the a small mound is built around the holes to increase the pressure difference. Why do racing cars wear skirts?

  6. VENTURI EFFECT velocity increased pressure decreased low pressure high pressure (patm)

  7. force high speed low pressure force What happens when two ships or trucks pass alongside each other? Have you noticed this effect in driving across the Sydney Harbour Bridge?

  8. artery Flow speeds up at constriction Pressure is lower Internal force acting on artery wall is reduced External forces causes artery to collapse Arteriosclerosis and vascular flutter

  9. Bernoulli’s Equation • for any point along a flow tube or streamline • p + ½ v2 +  g y = constant • Dimensions • p [Pa] = [N.m-2] = [N.m.m-3] = [J.m-3] • ½ v2 [kg.m-3.m2.s-2] = [kg.m-1.s-2] = [N.m.m-3] = [J.m-3] • g h [kg.m-3 m.s-2. m] = [kg.m.s-2.m.m-3] = [N.m.m-3] = [J.m-3] • Each term has the dimensions of energy / volume or energy density. • ½ v 2 KE of bulk motion of fluid • g h GPE for location of fluid • p pressure energy density arising from internal forces within • moving fluid (similar to energy stored in a spring)

  10. Derivation of Bernoulli's equation Mass element m moves from (1) to (2) m = A1 x1 = A2 x2 = VwhereV = A1 x1 = A2 x2 Equation of continuity A V = constant A1v1 = A2v2A1 > A2 v1 < v2 Since v1 < v2 the mass element has been accelerated by the net force F1 – F2 = p1A1 – p2A2 Conservation of energy A pressurized fluid must contain energy by the virtue that work must be done to establish the pressure. A fluid that undergoes a pressure change undergoes an energy change.

  11. K = ½ m v22 - ½ m v12 = ½ Vv22 - ½ V v12 U = m g y2 – m g y1 = V g y2 = V g y1 Wnet = F1x1 – F2 x2 = p1A1 x1 – p2A2 x2 Wnet = p1V – p2V= K + U p1V – p2V = ½ Vv22 - ½ V v12 + V g y2 - V g y1 Rearranging p1 + ½ v12 + g y1 = p2 + ½ v22 + g y2 Applies only to an ideal fluid (zero viscosity)

  12. Ideal fluid Real fluid

  13. Flow of a liquid from a hole at the bottom of a tank (1) Point on surface of liquid y1 v2 = ? m.s-1 y2 (2) Point just outside hole

  14. 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

  15. How do you measure the speed of flow for a fluid? (1) (2) rF v1 = ? h rm

  16. 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}

  17. C yC A yA B yB D How does a siphon work? How fast does the liquid come out?

  18. Assume that the liquid behaves as an ideal fluid and that both the equation of continuity and Bernoulli's equation can be used. Heights: yD = 0 yByAyC Pressures: pA = patm = pD Consider a point A on the surface of the liquid in the container and the outlet point D. Apply Bernoulli's principle to these points Now consider the points C and D and apply Bernoulli's principle to these points From equation of continuity vC = vD The pressure at point C can not be negative

  19. 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 )

  20. pC + ½ vC2 + g yC = pD + ½ vD2 + g yD 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

  21. 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). • Semester 1, 2004 Exam question

  22. 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

  23. 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

  24. (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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