Note 14
Fluid Dynamics
Sections Covered in the Text: Chapter 15, except 15.6
To complete our study of fluids we now examine
fluids in motion. For the most part the study of fluids
in motion was put into an organized state by scientists
working a generation after Pascal and Torricelli. We
shall see here that much of the physics of fluids is
encapsulated in the statement of Bernoulli’s principle.
The study of fluid flow was driven by the demands
of the industrial revolution. It was vital for progress
and profit that the movement of water, steam and oil
through pipes, and the movement of aerofoils through
the air, were understood mathematically. Today, the
physics of fluids in motion is of special interest in the
various branches of the environmental and life
sciences.
Flow Rate
When a fluid occupies a pipe of cross sectional area A
and flows with average speed v, the rate of flow Q is
given by
Q = Av .
…[14-1]
The units of Q are m 3.s–1. These are units of volume.s–1,
so flow rate is the same as volume rate.
The paths of€particles in a fluid moving with laminar
flow are called streamlines. Streamlines never cross
one another (Figure 14-1).
Fluids in Motion
There are two types of fluid motion called laminar flow
and turbulent flow.
A fluid will execute laminar flow when it is moving at
low velocity. The particles of the fluid follow smooth
paths that do not cross, and the rate of fluid flow
remains constant in time. This is the easiest type of
flow to describe mathematically.
A fluid will execute turbulent flow when it is moving
above a certain critical velocity. Strings of vortices
form in the fluid, resulting in highly irregular motion.
A white water rapid is a good example. A system
moving irregularly is difficult to describe mathematically, so we shall not be concerned with turbulent
flow here.
An Ideal Fluid
Since a fluid is in general a complicated medium to
describe mathematically, even in laminar flow, we
shall assume for simplicity that the fluid is ideal. By
ideal we mean
1 The fluid is moving in laminar flow; viscous forces
between adjacent layers are negligible.
2 The flow is steady, that is, the flow rate does not
change with time.
3 The fluid has a uniform density and is thus incompressible.
4 The flow is irrotational, that is, the angular momentum about any point is zero; in common parlance
the fluid does not “swirl”.
Figure 14-1. Particles in a fluid moving with laminar flow
follow streamlines that do not cross.
The Equation of Continuity
A fluid moving in laminar flow in a flow tube (that
may be a pipe) can be shown to satisfy a simple relationship. Consider an ideal fluid flowing through a
tube of variable cross-section (Figure 14-2). In an
elapsed time ∆t, a volume A 1v1∆t of fluid crosses area
A1, and a volume A 2v2∆t crosses area A 2. Since the
fluid is incompressible and the streamlines do not
cross the volume of fluid crossing A1 must equal the
volume of fluid crossing A2, so A 1v1∆t = A2v2∆t, from
which it follows that A 1v1 = A2v2 or Q 1 = Q2. This
means that the flow rate
Q = Av = const .
…[14-2]
An important consequence of eq[14-2] is that if the
cross sectional area of the flow tube is reduced at
some point,
€ then the flow speed increases. Eq[14-2] is
14-1
Note 14
known as the equation of continuity. 1
Example Problem 14-1
Speed of Blood Flow in Capillaries
The radius of the aorta is about 1.0 cm and the blood
flowing through it has a speed of about 30.0 cm.s–1.
Calculate the average speed of the blood in the
capillaries using the fact that although each capillary
has a diameter of about 8.0 x 10–4 cm, there are
literally billions of them so that their total cross
section is about 2000 cm 2.
Solution:
From the equation of continuity the speed of blood in
the capillaries is
v1 A1 0.30(m.s−1 ) × 3.14 × (0.010) 2 (m 2 )
v2 =
=
A2
2.0 ×10−1 (m 2 )
= 5.0 x 10–4 m.s–1
€
or about 0.5 mm.s–1. This is a very low speed.
Bernoulli’s Equation
Figure 14-2. Illustration of the equation of continuity. The
flow tube of a fluid is shown at two positions 1 and 2. At no
time does fluid enter or leave the flow tube.
The equation of continuity can be applied to explain
the various rates of blood flow in the body. Blood
flows from the heart into the aorta from which it
passes into the major arteries; these branch into the
small arteries (arterioles), which in turn branch into
myriads of tiny capillaries. The blood then returns to
the heart via the veins. Blood flow is fast in the aorta,
but quite slow in capillaries. When you cut a finger
(capillary) the blood oozes, or flows very slowly. We
can show this by means of a numerical example.
1
The equation of continuity can be thought of as a statement of
the conservation of fluid. As the fluid flows through the pipe the
volume of fluid remains constant; it neither increases nor decreases.
14-2
Daniel Bernoulli (1700-1782), a Swiss mathematician
and scientist, lived a generation after Pascal and
Torricelli. The equation he derived is a more general
statement of the laws and principles of fluids we have
examined thus far.
Bernoulli allowed for the flow tube to undergo a
possible change in height (Figure 14-3). Consider
points 1 and 2. Let point 1 be at a height y1 and let v1,
A1 and p1 be the speed of the fluid, cross sectional area
of the tube and pressure of the fluid at that point.
Similarly let v2, A2 and p 2 be the same variables at
point 2. The actual system is the volume of fluid in the
flow tube.
In an elapsed time ∆t the amount of fluid crossing A 1
is ∆V1 = A1v1∆t and the amount of fluid crossing A2 is
∆V2 = A2v2∆t. But from the equation of continuity, A1v1
= A 2v2. So the volume of fluid crossing either area is
the same; let us simply write it as
ΔV = AvΔt .
Fluid is moved in the flow tube as the result of the
work done on the fluid by the surrounding fluid (the
environment).
€ The net work W done on the fluid in
the elapsed time ∆t is
Note 14
ρ
p +  v 2 + ρgy = const .
2
…[14-6]
Many of the “principles” and “laws” we have seen
can be shown to be special cases of Bernoulli’s equation.
€ We shall consider a number of them.
Many homes and buildings in the colder climates are
heated by the circulation of hot water in pipes. Even if
they are not explicitly aware of it, architects must
ensure that their designs conform to Bernoulli’s
principle in order to avoid system failure. Let us
consider an example.
Figure 14-3. Illustration of Bernoulli’s equation.
W = (F2 − F1 )Δx = ( F2 − F1 )
Example Problem 14-2
Application of Bernoulli’s Principle
ΔV
= ( p2 − p1)ΔV .
A
This work goes into achieving two things:
1 changing the kinetic energy of the fluid between
the two points by the amount:
€
 ρΔV  2 2
ΔK = 
(v − v ) .
 2  2 1
…[14-3]
2 changing the gravitational potential energy of the
fluid between the two points by the amount mg∆h
or €
ΔU = ρΔVg( y 2 − y1 ) .
…[14-4]
W = ( p2 − p1 )ΔV = ΔK + ΔU ,
Since
( p1 − p2 )ΔV
€
2
v1 A1
−1 π (0.020m)
v2 =
= 0.50(m.s )
A2
π (0.013m) 2
= 1.2 m.s–1.
eq[14-5]:
1
p2 = p1 + ρg( y1 − y 2 ) + ρ (v12 − v 22 ) .
2
Substituting p1 = 3.0 x 105 Pa, ρwater = 1000 kg.m–3, g =
 ρΔV  2
2
9.8 m.s –2, y1 = 0, y2 = 5.0 m, v1 = 0.50 m.s–1, v2 = 1.2 m.s–1
=
(v − v ) + ρΔVg( y 2 − y1 ) .
 2  2 1
€ we get
p2 = 2.5 x 105 Pa.
Dividing through by ∆V we obtain the general form of
Bernoulli’s equation:
€
ρ
 ρ
p1 +  v12 + ρgy1 = p2 +  v 22 + ρgy 2 .
2
2
…[14-5]
We can put this equation into the simpler form:
€
Solution:
Let the basement be level 1 and the second floor be
level 2. We can obtain the flow speed by applying the
equation of continuity, eq[14-2]:
€ To find the pressure p2 we use Bernoulli’s equation,
€ by substituting eqs[14-3] and [14-4]:
we have,
€
Water circulates throughout a house in a hot water
heating system. If the water is pumped at a speed of
0.50 m.s –1 through a pipe of diameter 4.0 cm in the
basement under a pressure of 3.0 atm, what will be
the flow speed and pressure in a pipe of diameter 2.6
cm on the second floor 5.0 m above?
Thus p 2 < p 1. Obviously, the pipe on the second floor
of the house must be able to withstand less pressure
than the pipe in the basement.
Eqs[14-5] and [14-6] have the look of conservation of
energy expressions including work, because that, in
effect, is what they are.
14-3
Note 14
1 Torricelli’s Theorem
Consider the container filled with fluid in Figure 14-4.
A distance h below the surface of the fluid a small
hole allows fluid to escape. What is the velocity of the
outflowing fluid?
A glass container of height 1.0 m is full of water. A
small hole appears on the side of the container at the
bottom (as shown in Figure 14-4). What is the speed of
the water flowing out the hole?
Solution:
This is a straightforward application of Torricelli’s
theorem and eq[14-7]. The speed of the water is
h
Figure 14-4. A container of fluid with a small hole a distance
h below the surface allowing fluid to escape.
Applying Bernoulli’s equation we have
Example Problem 14-3
Applying Torricelli’s Theorem
€
ρ
p +  v 2 + ρgy = const .
2
v = 2gh = 2 × 9.80(m.s−1 ) ×1.0(m) = 4.43 m.s–1
The value measured would be less than this if the
fluid (for example, shampoo) had a significant
viscosity.
2 A Fluid at Rest
Assuming a container of normal laboratory size, the
atmospheric pressure p is essentially the same at its
top€and bottom, so we can cancel p and write to a
good approximation
For a fluid at rest (static fluid) the speed v in eq[14-6]
is zero. Bernoulli’s equation then reduces to
ρ 2
 v + ρgy = const .
2
This is just Pascal’s law; see eq[13-3] in Note 13. The
point to be made here is that Bernoulli’s equation is a
more general
statement of the physics of fluids than is
€
Pascal’s law.
But if the container is not vanishingly small then the
fluid velocity is essentially zero at the top (at y = 0). So
the velocity
v of the outflowing fluid (at y = –h) is
€
given by
ρ 2
 v + ρg(−h) = 0 ,
2
from which it follows that
€
v = 2gh .
…[14-7]
This is what is known as Torricelli’s theorem. Note that
the velocity is independent of the fluid density. Note
too that our
€treatment here neglects the effect of fluid
viscosity (that would otherwise reduce the speed). 2
2
Eq[14-5] is the same expression as obtained for the final velocity
of an object dropped from rest at a height h (Note 09).
14-4
p + ρgy = const .
3 Pressure in a Flowing Fluid
We studied hydrostatic pressure in Note 13. We can
show that in a moving fluid the pressure is dependent on velocity.
A fluid flowing along a horizontal level experiences
a constant gravitational potential (y = constant). Bernoulli’s equation therefore becomes
ρ
p +  v 2 = const .
2
…[14-8]
This means that as the speed v of the fluid increases
the pressure p must fall. This result forms the
principle
€ of operation of many practical devices. One
is the Venturi tube, which we consider next.
Note 14
The Venturi Tube
The Venturi tube (Figure 14-5) is a device that is used
to measure the flow rate Q of a fluid. It consists of two
sections of different cross sectional areas A1 and A 2
that are known with good precision. The difference in
pressure in the fluid in the two sections (p1 – p2) is
measured with a built-in manometer.
Example Problem 14-4
Using a Venturi Tube
A Venturi tube is used to measure the flow of water. It
has a main diameter of 3.0 cm tapering down to a
throat diameter of 1.0 cm. The pressure difference p1 –
p2 is measured to be 18 mm Hg. Calculate the velocity
v1 of the fluid input and the flow rate Q.
Solution:
The pressure difference in Pa is, using the conversion
expression eq[13-4]:
p1 – p2 = 18 mm Hg ≡ 1.333 x 105 (Pa.m–1)x18x10 –3 (m)
= 24.0 x 102 Pa.
Therefore from eq[14-9] the speed of the input fluid is
v1 = A2
2( p1 − p2 )
ρ ( A12 − A22 )
= 24.6 cm.s –1.
Multiplying
by A1 we obtain the flow rate Q:
€
Q = π (1.5cm) 2 × 24.6(cm.s−1 ) = 174 cm3.s–1
Figure 14-5. The Venturi tube.
From Bernoulli’s equation with y2 = y1 we have
€
ρ
p1 − p2 =  (v 22 − v12 ) .
2
Substituting the equation of continuity to eliminate v2
we can rearrange and solve for v1. The flow rate Q is
therefore
€ given by
Q = A1v1 = A1 A2
2( p1 − p2 )
. …[14-9]
ρ ( A12 − A22 )
Since the density of the fluid ρ is also known, as are
the areas A1 and A 2, Q can be calculated once (p1 – p 2)
is
€ measured with the manometer.
These results, too, would be quite in error if the fluid
had a non-negligible viscosity. The viscosity of water
has a negligible effect in this example.
The Aerofoil
Bernoulli’s equation helps to explain why an airplane
is equipped with wings to help it stay in the air. An
airplane wing is an example of an aerofoil (Figure 146). In a moving stream of air, the air travels more
quickly over the top surface of an aerofoil than over the
bottom surface. According to Bernoulli’s principle the
pressure on the top surface is less than the pressure on
the bottom surface, contributing to a net upward force
called aerodynamic lift. When an airplane is flying at
constant altitude and speed, the upward aerodynamic
lift balances the downward gravitational force and
prevents the plane from falling. This is sometimes
called the Bernoulli effect.
A glider moving on an air track is a kind of aerofoil.
On the top surface of a glider the pressure is atmospheric pressure. In the space between the bottom
14-5
Note 14
surface and the surface of the air track the pressure of
the air is higher than atmospheric. The difference in
pressure accounts for the aerodynamic lift that keeps
the glider from contacting the air track surface and
grinding to a halt.
€
wing. From eq[14-5] we can write
ρ  2
ρ  2
pabove +  air v above
= pbelow +  air v below
 2 
 2 
…[14-10]
where “above” and “below” refer to the airplane
wing. The difference in pressure accounts for the aerodynamic lift, that is, the difference in pressure is equal
to the resultant force per unit area on the wing. The
magnitude of force must just equal the weight of the
airplane. Thus
pbelow − pabove
F 2.0 ×10 6 (kg) × 9.80(m.s−2 )
= =
A
1200(m 2 )
= 16,300 Pa.
€
Rearranging eq[14-10] we can write
p
− pabove  2
2
v above
=  below
 + v below .
ρ /2


Figure 14-6. Streamlines around an aerofoil.
Example Problem 14-5
Aerodynamic Lift
An airplane has a mass of 2.0 x 106 kg and the air
flows past the lower surface of the wings at 100 m.s–1.
If the wings have a surface area of 1200 m2, how fast
must the air flow over the upper surface of the wing if
the plane is to stay in the air? Consider only the
Bernoulli effect.
Solution:
The wing is a surface moving horizontally through
the air. The effect is the same as if the wing were stationary and the air were flowing horizontally over the
14-6
Taking the density of air as the value at the Earth’s
surface, i.e., 1.29 kg.m–3 we have
€
2
v above
=
16300
+ 10 4 .
(1.29 /2)
Evaluating and taking the square root we have finally
€
vabove = 190 m.s–1.
Thus the air must flow over the upper surface of the
wing nearly twice as fast as past the lower surface.
Note 14
To Be Mastered
•
Definitions: laminar flow, turbulent flow, equation of continuity
•
Physics of: Bernoulli’s Equation
•
•
Physics of: the Venturi tube
Physics of: the aerofoil
ρ
 ρ
p1 +  v12 + ρgy1 = p2 +  v 22 + ρgy 2
2
2
€
Typical Quiz/Test/Exam Questions
1.
State Benoulli’s law relating pressure p, velocity v and density ρ in a moving fluid.
2.
Sketch a Venturi tube, labelling the important features.
3.
Explain briefly what a Venturi tube is used for, and how it is used.
4.
5.
6.
7.
8.
14-7