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1 http://www.physics.usyd.edu.au/teach_res/jp/fluids/wfluids.htm http://www.physics.usyd.edu.au/teach_res/jp/fluids/ web notes: lect5.ppt flow1.pdf flow2.pdf

2 FLUID FLOW STREAMLINE – LAMINAR FLOW TURBULENT FLOW REYNOLDS NUMBER

3 Velocity of particle -tangent to streamline streamlines Streamlines for fluid passing an obstacle v Velocity profile for the laminar flow of a non viscous liquid

4 Turbulent flow indicted by the swirls – eddies and vortices

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7 REYNOLDS NUMBER A British scientist Osborne Reynolds (1842 – 1912) established that the nature of the flow depends upon a dimensionless quantity, which is now called the Reynolds number R e. R e =  v L /   density of fluid v average flow velocity over the cross section of the pipe L dimension characterising a cross section

8 R e is a dimensionless number R e =  v L /  [R e ]  [kg.m -3 ] [m.s -1 ][m] [Pa.s] -1  [kg] [m -1 ][s -1 ][kg.m.s -2.m -2.s] -1 = [1] For a fluid flowing through a pipe – as a rule of thumb R e < ~ 2000 laminar flow ~ 2000 < R e < ~ 3000 unstable laminar / turbulent R e > ~ 2000 turbulent

9 R e =  v L /  Sydney Harbour Ferry

10 R e =  v L /   = 10 3 kg.m -3 v = 5 m.s -1 L = 10 m  = 10 -3 Pa.s R e = (10 3 )(5)(10) / (10 -3 ) R e = 5x10 7

11 R e =  v L /  Spermatozoa swimming

12 R e =  v L /  Spermatozoa swimming  = 10 3 kg.m -3 v = 10 -5 m.s -1 L = 10  m  = 10 -3 Pa.s R e = (10 3 )(10 -5 )(10x10 -6 ) / (10 -3 ) R e = 10 -4

13 Household plumbing pipes Typical household pipes are about 30 mm in diameter and water flows at about 10 m.s -1 R e ~ (10)(30  10 -3 )(10 3 ) / (10 -3 ) ~ 3  10 5 The circulatory system Speed of blood v ~ 0.2 m.s -1 Diameter of the largest blood vessel, the aorta L ~ 10 mm Viscosity of blood probably  ~ 10 -3 Pa.s (assume same as water) R e ~ (0.2)(10  10 -3 )(10 3 ) / 10 -3 ) ~ 2  10 3

14 The method of swimming is quite different for Fish (R e ~ 10 000) and sperm (R e ~ 0.0001) Modes of boat propulsion which work in thin liquids (water) will not work in thick liquids (glycerine) It is possible to stir glycerine up, and then unstir it completely. You cannot do this with water.

15 FLUID FLOW IDEAL FLUID EQUATION OF CONTINUITY How can the blood deliver oxygen to body so successfully? How do we model fluids flowing in streamlined motion?

16 IDEAL FLUID Fluid motion is usually very complicated. However, by making a set of assumptions about the fluid, one can still develop useful models of fluid behaviour. An ideal fluid is Incompressible – the density is constant Irrotational – the flow is smooth, no turbulence Nonviscous – fluid has no internal friction (  = 0 ) Steady flow – the velocity of the fluid at each point is constant in time.

17 EQUATION OF CONTINUITY (conservation of mass) In complicated patterns of streamline flow, the streamlines effectively define flow tubes. So the equation of continuity – where streamlines crowd together the flow speed must increase.

18 m 1 = m 2  V 1 =  V 2  A 1 v 1  t =  A 2 v 2  t A 1 v 1 = A 2 v 2 EQUATION OF CONTINUITY (conservation of mass) A v measures the volume of the fluid that flows past any point of the tube divided by the time interval volume flow rate Q = dV / dt Q = A v = constant if A decreases then v increases if A increases then v decreases

19 Applications Rivers Circulatory systems Respiratory systems Air conditioning systems

20 Blood flowing through our body The radius of the aorta is ~ 10 mm and the blood flowing through it has a speed ~ 300 mm.s -1. A capillary has a radius ~ 4  10 -3 mm but there are literally billions of them. The average speed of blood through the capillaries is ~ 5  10 -4 m.s -1. Calculate the effective cross sectional area of the capillaries and the approximate number of capillaries.

21 Setup radius of aorta R A = 10 mm = 10  10 -3 m radius of capillaries R C = 4  10 -3 mm = 4  10 -6 m speed of blood thru. aorta v A = 300 mm.s -1 = 0.300 m.s -1 speed of blood thru. capillaries R C = 5  10 -4 m.s -1 Assume steady flow of an ideal fluid and apply the equation of continuity Q = A v = constant  A A v A = A C v C where A A and A C are cross sectional areas of aorta & capillaries respectively. aorta capillaries

22 Action A C = A A (v A / v C ) =  R A 2 (v A / v C ) A C =  (10  10 -3 ) 2 (0.300 / 5  10 -4 ) m 2 = 0.20 m 2 If N is the number of capillaries then A C = N  R C 2 N = A C / (  R C 2 ) = 0.2 / {  (4  10 -6 ) 2 } N = 4  10 9


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