Download The equation of continuity is often expressed as

VIS U AL P HYS ICS
S ch ool of P h ysi cs
U n i v er si t y of S yd n ey Au st r a l i a
FLUID FLOW
IDEAL FLUID
EQUATION OF CONTINUITY
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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)
Consider an ideal fluid flowing through a pipe of varying cross
sectional area A. The volume V1 of fluid and mass m1 flowing
past (1) in a very small time interval t is
V1 = A1 v1 t
m1 = 1 A1 v1 t
Similarly the volume and mass of fluid flowing past (2) in time t
is
V2 = A2 v2 t
m2 = 2 A2 v2 t
When the flow is steady all the material which goes past (1) must
go past (2) in the same time (or else it will be continually piling up
somewhere) and since the fluid is incompressible its density does
not change
1 = 2 = 
Therefore we must have
m1 = m2
 A1 v1 t =  A2 v2 t
!
A1 v1 = A2 v2
If the fluid is approximately incompressible, i.e. if its density never
changes by very much, then the equation of continuity, as we
quoted it, is approximately true.
The quantity A v which measures the volume of the fluid that
flows past any point of the tube divided by time is called the
volume flow rate Q = dV/dt.
!
The equation of continuity is often expressed as
Q = A v = constant
 if A decreases then v increases
 if A increases then v decreases
A1
A2


v2
v1
In complicated patterns of streamline flow, the streamlines
effectively define flow tubes. So the equation of continuity says
that where streamlines crowd together the flow speed must
increase.
Applications
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In flowing rivers, when going from deep to shallow, the flow
speed increases (often becoming turbulent) "still water runs
deep".
A river flows slowly and languidly through a meadow where it
is broad, but speeds up to torrential speed when passing a
narrows.
In the circulatory system of the blood there is a branching
effect. When a fluid flows past a Y-junction made up of pipes
of the same diameter, the total cross-sectional area after the
branch is twice that before the branch, so the flow speed
must fall to half. Conversely, if it is important to keep the flow
speed up, the pipes after the branch must have half the
cross-sectional area of those before. (Note: blood will clot if
its speed falls too low.)
Blood flow – blood flows from the heart into the aorta then
into the major arteries (32). These branch into smaller
arteries (arterioles) that branch into a myriad of tiny
capillaries and then the blood returns to the heart via the
veins.
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Air conditioning systems must also be built with
consideration for the branch effect.
Also the tube structure of the respiratory system is
remarkably similar to that of the circulatory system.
Blood flowing through our body
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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.
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