Transcript lect5

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
FLUID FLOW
STREAMLINE – LAMINAR FLOW
TURBULENT FLOW
REYNOLDS NUMBER
Streamlines for
fluid passing an
obstacle
streamlines
v
Velocity profile for the laminar
flow of a non viscous liquid
Velocity of particle
- tangent to streamline
Turbulent flow indicted by the swirls – eddies and vortices
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 Re.
Re =  v L / 

v
L
density of fluid
average flow velocity over the cross section of the
pipe
dimension characterising a cross section
Re is a dimensionless number
Re =  v L / 
[Re]  [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
Re < ~ 2000
laminar flow
~ 2000 < Re < ~ 3000 unstable laminar / turbulent
Re > ~ 2000
turbulent
Re =  v L / 
Sydney Harbour Ferry
Re =  v L / 
 = 103 kg.m-3
Re = (103)(5)(10) / (10-3)
v = 5 m.s-1
Re = 5x107
L = 10 m
 = 10-3 Pa.s
Spermatozoa swimming
Re =  v L / 
Spermatozoa swimming
Re =  v L / 
 = 103 kg.m-3
v = 10-5 m.s-1
Re = (103)(10-5)(10x10-6) / (10-3)
L = 10 mm
Re = 10-4
 = 10-3 Pa.s
Household plumbing pipes
Typical household pipes are about 30 mm in diameter and water flows
at about 10 m.s-1
Re ~ (10)(3010-3)(103) / (10-3) ~ 3105
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)
Re ~ (0.2)(1010-3)(103) / 10-3) ~ 2103
The method of swimming is quite different for
Fish (Re ~ 10 000) and sperm (Re ~ 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.
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?
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.
EQUATION OF CONTINUITY (conservation of mass)
A1
A2


v2
v1
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.
EQUATION OF CONTINUITY (conservation of mass)
m1 = m2
 V1 =  V2
 A1 v1 t =  A2 v2 t
A1 v1 = A2 v2
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
Applications
Rivers
Circulatory systems
Respiratory systems
Air conditioning systems
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.
Setup
radius of aorta RA = 10 mm = 1010-3 m
radius of capillaries RC = 410-3 mm = 410-6 m
speed of blood thru. aorta vA = 300 mm.s-1 = 0.300 m.s-1
speed of blood thru. capillaries RC = 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
where AA and AC are cross sectional areas of aorta & capillaries respectively.
aorta
capillaries
Action
AC = AA (vA / vC) =  RA2 (vA / vC)
AC = (1010-3)2(0.300 / 510-4) m2 = 0.20 m2
If N is the number of capillaries then
AC = N  RC2
N = AC / ( RC2) = 0.2 / { (410-6)2}
N = 4109