# 1 web notes: lect6.ppt.

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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 (1) Surface of liquid (2) Just outside hole v 2 = ? m.s -1 y1y1 y2y2 Draw flow tubes v 1 ~ 0 m.s -1 p 1 = p atm p 2 = p atm h = (y 1 - y 2 ) What is the speed with which liquid flows from a hole at the bottom of a tank?

4 Assume liquid behaves as an ideal fluid and that Bernoulli's equation can be applied p 1 + ½  v 1 2 +  g y 1 = p 2 + ½  v 2 2 +  g y 2 A small hole is at level (2) and the water level at (1) drops slowly  v 1 = 0 p 1 = p atm p 2 = p atm  g y 1 = ½  v 2 2 +  g y 2 v 2 2 = 2 g (y 1 – y 2 ) = 2 g h h = (y 1 - y 2 ) v 2 =  (2 g h) Torricelli formula (1608 – 1647) This is the same velocity as a particle falling freely through a height h

5 (1) (2) FF mm h v 1 = ? What is the speed of flow in section 1 of the system?

6 Assume liquid behaves as an ideal fluid and that Bernoulli's equation can be applied for the flow along a streamline p 1 + ½  v 1 2 +  g y 1 = p 2 + ½  v 2 2 +  g y 2 y 1 = y 2 p 1 – p 2 = ½  F (v 2 2 - v 1 2 ) p 1 - p 2 =  m g h A 1 v 1 = A 2 v 2  v 2 = v 1 (A 1 / A 2 )  m g h = ½  F { v 1 2 (A 1 / A 2 ) 2 - v 1 2 } = ½  F v 1 2 {(A 1 / A 2 ) 2 - 1}

7 How does a siphon work? Q: What do we know? Continuous flow Pressure in top section > 0 otherwise there will be a vacuum p C  0 Focus on falling water not rising water p atm - p C  0 p atm   g y C yCyC p C

8 C B A D yAyA yByB yCyC Assume that the liquid behaves as an ideal fluid, the equation of continuity and Bernoulli's equation can be used. y D = 0 p A = p atm = p D

9 p C + ½  v C 2 +  g y C = p D + ½  v D 2 +  g y D From equation of continuity v C = v D p C = p D +  g (y D - y C ) = p atm +  g (y D - y C ) The pressure at point C can not be negative p C  0 and y D = 0 p C = p atm -  g y C  0 y C  p atm / (  g) For a water siphon p atm ~ 10 5 Pa g ~ 10 m.s -1  ~ 10 3 kg.m -3 y C  10 5 / {(10)(10 3 )} m y C  10 m Consider points C and D and apply Bernoulli's principle.

10 p A + ½  v A 2 +  g y A = p D + ½  v D 2 +  g y D v D 2 = 2 (p A – p D ) /  + v A 2 + 2 g (y A - y D ) p A – p D = 0 y D = 0 assume v A 2 << v D 2 v D =  (2 g y A ) 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

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 Lift F L drag F D Resultant F R Motion of object through fluid Fluid moving around stationary object FORCES ACTING ON OBJECT MOVING THROUGH FLUID Forward thrust by engine

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

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

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

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

17 Professional golf drive Initial speed v 0 ~ 70 m.s -1 Angle ~ 6° Spin  ~ 3500 rpm Range ~ 100 m (no Magnus effect) Range ~ 300 m (Magnus effect)

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21 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 m 3.s -1. The blood has a viscosity of 2.084  10 -3 Pa.s and a density of 1.06  10 3 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. (iii) 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).

22 Solution radius R = 4.00  10 -3 m volume flow rate Q = 1.00  10 -6 m 3.s -1 viscosity of blood  = 2.084  10 -3 Pa.s density of blood  = 1.060  10 -3 kg.m -3 (i) v = ? m.s -1 (ii)  p = ? Pa (iii) R e = ?

23 (i) Equation of continuity: Q = A v A =  R 2 =  (4.00  10 -3 ) 2 = 5.03  10 -5 m 2 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  R 4 / (8  L) L = 0.100 m  P = 8  L Q / (  R 4 )  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 R e =  v L /  where L = 2 R (diameter of artery) R e = (1.060  10 3 )(1.99  10 -2 )(2)(4.00  10 -3 ) / (2.084  10 -3 ) R e = 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) R e very small  laminar flow (R e < 2000)

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