1.
web notes: lect5.ppt

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. EQUATION OF CONTINUITY (conservation of mass) A 1 A 2
A simple model
EQUATION OF CONTINUITY
(conservation of mass) A 1 A 2 DV Dt
What changes ?
volume flow rate?
Speed? Q= DV Dt Q=

4. EQUATION OF CONTINUITY (conservation of mass)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.

5. A1 v1 = A2 v2 =Q=V/  t =constant
EQUATION OF CONTINUITY (conservation of mass)
m1 = m2
 V1 =  V2
 A1 v1 t =  A2 v2 t
A1 v1 = A2 v2 =Q=V/  t =constant
Applications
Rivers
Circulatory systems
Respiratory systems
Air conditioning systems

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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.
How would you implement the problem solving scheme
Identify Setup Execute Evaluate ?

7. AC = AA (vA / vC) =  RA2 (vA / vC) AC = = 0.20 m2
Identify / Setup
aorta RA = 1010-3 m AA = cross sectional area of aorta = ?
capillaries RC = 410-6 m AC = cross sectional area of capillaries = ?
N = number of capillaries = ?
speed thru. aorta vA = m.s-1
speed thru. capillaries vC = 510-4 m.s-1
Assume steady flow of an ideal fluid and apply the equation of continuity
Q = A v = constant  AA vA = AC vC
Execute
AC = AA (vA / vC) =  RA2 (vA / vC)
AC = = m2
If N is the number of capillaries then
AC = N  RC2
N = AC / ( RC2) = 0.2 / { (410-6)2}
N = 4109
aorta
capillaries

8. How does a perfume spray work? What is the venturi effect?
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. Daniel Bernoulli (1700 – 1782)

10. A1 A1 A2
Fluid flow

11. Relationship between speed & density of streamlines?v2 v1 v1
Streamlines closer
Relationship between speed & density of streamlines?
How about pressure?

12. A1 A1 A2 v2 v1 v1
Low speed
Low KE
High pressure
high speed
high KE
low pressure
Streamlines closer
Low speed
Low KE
High pressure

13. p large
p large
p small
v small
v large
v small

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. VENTURI EFFECT ?

16. Flow tubes Eq Continuity high pressure (patm) VENTURI EFFECT
low pressure
velocity increased
pressure decreased
high
pressure
(patm)

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. force
high speed
low pressure
force

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

20. energy densities Bernoulli’s Principle Conservation of energy
Work W = F d = p A d = p V W / V = p
KE K = ½ m v2 = ½  V v K / V = ½  v2
PE U = m g h =  V g h U / V =  g h
Along a streamline
p + ½  v2 +  g h = constant

21. Bernoulli’s PrincipleY X t i m e 2
What has changed
between 1 and 2? r t i m e 1

22. A better model: Bernoulli’s

23. Mass element m moves from (1) to (2)
Derivation of Bernoulli's equation
Mass element m moves from (1) to (2)
m =  A1 x1 =  A2 x2 =  V where V = A1 x1 = A2 x2
Equation of continuity A V = constant
A1 v1 = A2 v2 A1 > A2  v1 < v2
Since v1 < v2 the mass element has been accelerated by the net force
F1 – F2 = p1 A1 – p2 A2
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.

24. K = ½ m v22 - ½ m v12 = ½  V v22 - ½  V v12
U = m g y2 – m g y1 =  V g y2 =  V g y1
Wnet = F1 x1 – F2 x2 = p1 A1 x1 – p2 A2 x2
Wnet = p1 V – p2 V = K + U
p1 V – p2 V =
½  V v22 - 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)

25. for any point along a flow tube or streamline
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 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)