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1 web notes: lect5.ppt.

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1 1 web notes: lect5.ppt

2 2 Assume an IDEAL FLUID Fluid motion is very complicated. However, by making some assumptions, we can develop a useful model 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.

3 3 EQUATION OF CONTINUITY (conservation of mass) A 1 A 2 A simple model V tV t V tV t Q=Q= Q=Q= What changes ? volume flow rate? Speed?

4 4 A 1 A 2 v 1 v 2 In complicated patterns of streamline flow, the streamlines effectively define flow tubes. where streamlines crowd together the flow speed increases. EQUATION OF CONTINUITY (conservation of mass)

5 5 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 =Q=  V/  t =constant EQUATION OF CONTINUITY (conservation of mass) Applications Rivers Circulatory systems Respiratory systems Air conditioning systems

6 6 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  mm but there are literally billions of them. The average speed of blood through the capillaries is ~ 5  m.s -1. Calculate the effective cross sectional area of the capillaries and the approximate number of capillaries. How would you implement the problem solving scheme Identify Setup Execute Evaluate ?

7 7 Identify / Setup aorta R A = 10  m A A = cross sectional area of aorta = ? capillaries R C = 4  m A C = cross sectional area of capillaries = ? N = number of capillaries = ? speed thru. aorta v A = m.s -1 speed thru. capillaries v C = 5  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 aorta capillaries Execute A C = A A (v A / v C ) =  R A 2 (v A / v C ) A C = = 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  ) 2 } N = 4  10 9

8 8 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?

9 9 Daniel Bernoulli (1700 – 1782)

10 10 A1A1 A2A2 A1A1 Fluid flow

11 11 A1A1 A2A2 v1v1 v2v2 A1A1 v1v1 Streamlines closer Relationship between speed & density of streamlines? How about pressure?

12 12 A1A1 A2A2 v1v1 v2v2 A1A1 v1v1 Low speed Low KE High pressure high speed high KE low pressure Streamlines closer Low speed Low KE High pressure

13 13 v small v large p large p small

14 14 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?

15 15 VENTURI EFFECT ?

16 16 velocity increased pressure decreased low pressure high pressure (p atm ) VENTURI EFFECT Flow tubes Eq Continuity

17 17 What happens when two ships or trucks pass alongside each other? Have you noticed this effect in driving across the Sydney Harbour Bridge?

18 18 high speed low pressure force

19 19 artery External forces are unchanged Flow speeds up at constriction Pressure is lower Internal force acting on artery wall is reduced Arteriosclerosis and vascular flutter Artery can collapse

20 20 Work W = F d = p A d = p V W / V = p KE K = ½ m v 2 = ½  V v 2 K / V = ½  v 2 PE U = m g h =  V g h U / V =  g h Bernoulli’s Principle Conservation of energy Along a streamline p + ½  v 2 +  g h = constant energy densities

21 21  X Y time 1 time 2 Bernoulli’s Principle What has changed between 1 and 2?

22 22 A better model: Bernoulli’s

23 23 Mass element m moves from (1) to (2) m =  A 1  x 1 =  A 2  x 2 =   V where  V = A 1  x 1 = A 2  x 2 Equation of continuity A V = constant A 1 v 1 = A 2 v 2 A 1 > A 2  v 1 < v 2 Since v 1 < v 2 the mass element has been accelerated by the net force F 1 – F 2 = p 1 A 1 – p 2 A 2 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. Derivation of Bernoulli's equation

24 24  K = ½ m v ½ m v 1 2 = ½   V v ½   V v 1 2  U = m g y 2 – m g y 1 =   V g y 2 =   V g y 1 W net = F 1  x 1 – F 2  x 2 = p 1 A 1  x 1 – p 2 A 2  x 2 W net = p 1  V – p 2  V =  K +  U p 1  V – p 2  V = ½   V v ½   V v   V g y 2 -   V g y 1 Rearranging p 1 + ½  v  g y 1 = p 2 + ½  v  g y 2 Applies only to an ideal fluid (zero viscosity)

25 25 Bernoulli’s Equation for any point along a flow tube or streamline p + ½  v 2 +  g y = constant Dimensions p [Pa] = [N.m -2 ] = [N.m.m -3 ] = [J.m -3 ] ½  v 2 [kg.m -3.m 2.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)


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