C h a p t e r

FLUID DYNAMICS

Learning Objectives
At the end of this chapter the students will be able to:

1. Understand that viscous forces in a fluid cause a retarding force on an object moving
through it.
2. Use Stokes’ law to derive an expression for terminal velocity of a spherical body
falling through a viscous fluid under laminar conditions.
3. Understand the terms steady (laminar, streamline) flow, incompressible flow, non
viscous flow as applied to the motion of an ideal fluid.
4. Appreciate that at a sufficiently high, velocity, the flow of viscous fluid undergoes a
transition from laminar to turbulence conditions.
5. Appreciate the equation of continuity Av = Constant for the flow of an ideal and
incompressible fluid.
6. Appreciate that the equation of continuity is a form of the principle of conservation of
mass.
7. Understand that the pressure difference can arise from different rates of flow of a
fluid (Bernoulli effect).
8. Derive Bernoulli’s equation in form P+ Vzpv2*pgh = constant.
9. Explain how Bernoulli effect is applied in the filter pump, atomizers, in the
flow of air over an aerofoil, Venturimeter and in blood physics.
10. Give qualitative explanations for the swing of a spinning ball.

he study of fluids in motion is relatively complicated, but analysis can be simplified by

making a few assumptions. The analysis is further simplified by the use of two important
conservation principles; the conservation of mass and the conservation of energy. The law of
conservation of mass gives us the equation of continuity while the law of conservation of
energy is the basis of Bernoulli’s equation. The equation of continuity and the Bernoulli’s
equation along with their applications in aeroplane and blood circulation are discussed in this
chapter.

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 6.1 VISCOUS DRAG AND STOKES’ LAW
The frictional effect between different layers of a flowing
fluid is described in terms of viscosity of the fluid. Viscosity
measures, how much force is required to slide one layer of
the liquid over another layer. Substances that do not flow
easily, such as thick tar and honey etc; have large
coefficients of viscosity, usually denoted by greek letter V|\
Substances which flow easily, like water, have ,small
coefficients of^ viscosity. Since liquids and gases have non
zero viscosity" a force is required if an object is to be
moved through them. Even the small viscosity of the air
causes a large retarding force on a car as it travels at high
speed. If you stick out your hand out of the window of a
fast moving car, you can easily recognize that considerable
force has to be exerted on your hand to move it through the
For Your Information air. These are typical examples of the following fact,
Viscosities of Liquids and Gases
at 30°C An object moving through a fluid experiences a
Viscosity retarding force called a drag force. The drag force
Material 10° (Nsm'2)
increases as the speed of the object increases.
Air 0:019
Acetone 0.295
Methanol 0.510
0.564
Even in the simplest cases the exact value of the drag
Benzene
Water 0.801 force is difficult to calculate. However, the case of a sphere
Ethanol 1.000 moving through a fluid is of great importance.
Plasma 1.6
Glycerin 6.29 The drag force F on a sphere of radius r moving slowly with
speed v through a fluid of viscosity r\ is given by Stokes’ law
as under.
F = 6 7tr| r v 61
( . )

At high speeds the force is no longer simply proportional to

speed.

6.2 TERMINAL VELOCITY

Consider a water droplet such as that of fog falling
vertically, the air drag on the water droplet increases with
speed. The droplet accelerates rapidly under the over
powering force of gravity which pulls the droplet downward.
However, the upward drag force on it increases as the
speed of the droplet increases. The net force on the
droplet is

128
 Net force = Weight - Drag force (6 .2 )

As the speed of the droplet continues to increase, the drag

force eventually approaches the weight in the magnitude.
Finally, when the magnitude of the drag force becomes
equal to the weight, the net force acting on the droplet is
zero. Then the droplet will fall with constant speed called
terminal velocity.
To find the terminal velocity vt in this case, we use Stokes
Law for the drag force. Equating it to the weight of the
drop, we have
mg = 6 nx\rvt

or Vl=jn9_ (6.3 ) Can You Do That?

6nr\r

The mass of the droplet is pV,

where volume

Substituting this value in the above equation, we get

vt = 2gr2p (6.4)
9n

Example 6.1: A tiny water droplet of radius 0.010 cm A table tennis ball can be made
suspended in the stream of air
descends through air from a high building. Calculate its coming from the nozzle of o hair
terminal velocity. Given that n for air = 19 x 10‘6kg m'1 s‘1 dryer.
and density of water P = 1000 kgm*3.

Solution:
r= 1.0 x 10'4m , P= 1000kgm*3 , n = 19 x io ^ k g m*1 s'1

Putting the above values in Eq. 6.4

2 x9.8ms"2x (1x10 4m fx 1000 kgm"3

9 x 1 9 x 1 0 *6kgm*1s*1

We get Terminal velocity = 1.1 m s'

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 6.3 FLUID FLOW
Moving fluids are of great importance. To learn about the
behaviour of the fluid in motion, we consider their flow
through the pipes. When a fluid is in motion, its flow can be
(a) Streamlines (laminar |p w }„ etthfeflstreamline or turbulent.

The flow is said to be streamline or laminar, if

Plate
every particle that passes a particular point,
moves along exactly the same path, as followed
by particles which passed that points earlier.
(b) Turbulent flow
Fig, 6.1
In this case each particle of the fluid moves along a smooth
path called a streamline as shown in Fig. 6.1 (a). The
different streamlines can not cross each other. This
condition is called steady flow condition. The direction of the
streamlines is the same as the direction of the velocity of the
fluid at that point. Above a certain velocity of the fluid flow,
the motion of the fluid becomes unsteady and irregular.
Under this condition the velocity of the fluid changes
abruptly as shown in Fig.6.1 (b). In this case the exact path
For Your Information of the particles of the fluid can not be predicted.

The irregular or unsteady flow of

the fluid is called turbulent flow.

We can understand many features of the fluid in motion by

Formula One racing cars have a
considering the behaviour of a fluid which satisfies the
streamlined design. following conditions.
The fluid is non-viscous i.e, there is no internal
frictional force between adjacent layers of fluid.
The fluid is incompressible, i.e., its density is constant.
The fluid motion is steady.
Dolphins have streamlined bodies
to assist their movement in water.

6.4 EQUATION OF CONTINUITY

Consider a fluid flowing through a pipe of non-uniform size.
The particles in the fluid move along the streamlines in a
steady state flow as shown in Fig. 6.2.
In a small time At, the fluid at the lower end of the tube
moves a distance AXi, with a velocity v*. If A J s the area of
cross section of this end, then the mass of the fluid
contained in the shaded region is:
A m 1 = P1A 1 AX1 = piA-tVt x A t
Where Pt is the density of the fluid. Similarly the fluid that
moves with velocity v2 through the upper end of the pipe
(area of cross section A2) in the same time At has a mass

A m 2 = P 2A 2v2 x A t

If the fluid is incompressible and the flow is steady, the

mass of the fluid is conserved. That is, the mass that flows
into the bottom of the pipe through A^ in a time At must be
equal to mass of the liquid that flows out through A 2 in the
same time. Therefore,
Am 1 = A m 2
or PiA 1v 1 = P2A 2v 2
This equation is called the equation of continuity. Since
density is constant for the steady flow of incompressible
fluid, the equation of continuity becomes
A 1v1 = A2v2 (6.5)

The product o f cross sectional area of the pipe

and the fluid speed at any point along the pipe
is a constant. This constant equals the volume
flow per second o f the fluid or simply flow rate.

Example: 6.2: A water hose with an internal diameter of

20 mm at the outlet discharges 30 kg of water in 60 s.
Calculate the water speed at th.e outlet. Assume the density
of water is 1000 kgm'3and its flow is steady.
As the water falls; its speed
increases and so its cross sectional
Solution: -— - area decreases as mandated by the
continuity equation.

Mass flow per second = = 0.5 kgs'1

60s
Cross sectional area A = n r 2

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The mass of water discharging per second through area A is.
mass
pAv
second

m ass/
v= second
or
PA
0.5 kgs'1
1000 kgm'3 x 3.14 x (10 xi O'3m)2
= 1.6 ms'1

6.5 BERNOULLI’S EQUATION

As the fluid moves through a pipe of varying cross section
and height, the pressure will change along the pipe.
Bernoulli’s equation is the fundamental equation in fluid
dynamics that relates pressure to fluid speed and height.
In deriving Bernoulli’s equation, we assume that the fluid is
incompressible, non viscous and flows in a steady state
manner. Let us consider the flow of the fluid through the
pipe in time t, as shown in Fig. 6.3.

Ax,

The force on the upper end of the fluid is P?A? where P? the
pressure and A 1 is the area of cross section at the upper
end. The work done on the fluid, by the fluid behind it, in
moving it through a distance Ax?, will be
W1= F? Ax? = P?A? Ax?

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Similarly the work done on the fluid at the lower end is
W2 - - F 2 A x 2 = - P 2A 2 Ax 2

Where P 2 is the pressure, A 2 is the area of cross section of

lower end and Ax2 is the distance moved by the fluid in the
same time interval t. The work W 2 is taken to be -ive as
this work is done against the fluid force.
The net work done = W = W 1 + W 2 '
or \A/ — P tAj Ax? — P 2A 2 ax2 ........ (6.6)
If v 1 and v 2 are the velocities at the upper and lower ends
respectively, then A stream of air passing over a tube
dipped in a liquid will cause the liquid
W = Vff - P 2A 2 v2t to rise in the tube as shown. This
effect is used in perfume bottles and
From equation of continuity (equation 6.5) paint sprayers.

A1V1 = A 2v 2
a v#-\/ /Volume of fluid ^
Hence, A m x t - A 2v2 x t - V \^uncjer consideration/
So, we have
W = ( P 1 - P 2) V (6.7)

if m is the mass and p is the density then V = ^

So equation 6.7 becomes

W = ( P 1 - P 2) j ............... (6 .8 )

Part of this work is utilized by the fluid in changing its K.E.

and a part is used in changing its gravitational P.E.

Change in K.E. = &(K.E.) = ~m v22 - .... (6.9)

Change in P.E. = A(PE.) = mgh2-m g h 1 . (6.10)

Where h1 and h2 are the heightsofthe upper and lower
ends respectively. A chimney works best when it is tall
and exposed to air currents, which
Applying, the law of conservation of energy to this.volume reduces the pressure at the top and
force the upward flow of smoke.
of the fluid, we get

133
 (Pi - P2) ™ = \ m v 2 ~ \ mv i + m 9 h 2 ~ m 'Q hi. (6 -1 1 )

rearranging the equation (6.11)

p i Jr\ Pvi + P9hi = p 2+-^Pv2 + P9h2

This is Bernoulli’s equation and is often expressed as:

1 2
P + -p v '+ p g h = constant (6 .12)

6.6 APPLICATIONS OF BERNOULLI’S

EQUATION

Torricelli’s Theorem
A simple application of Bernoulli’s equation is shown in
Fig.6.4. Fig. 6.4. Suppose a large tank of fluid has two small
orifices A and B on it, as shown in the figure. Let us find \
the speed with which the water flows from the orifice A.
Since the orifices are so small, the efflux speeds v2 and v3
For your information
will be much larger than the speed v-i of the top surface of
Water water. We can therefore, take as approximately zero.
Hence, Bernoulli’s equation can be written as:

Pi + p g/?! = P2 + ~ p v2 + P fl^ 2

But P^ = P2= atmospheric pressure

Therefore, the above equation becomes

v2 = p g ( h , - h 2) (6.13)

This is Torricelli’s theorem which states that;

The speed of efflux is equal to the velocity

A filter pump has a constriction in the
centre, so that a jet of water from the gained by the fluid in falling through the
tap flows faster here. This cause a distance (/»i - h 2) under the action o f gravity.
drop in pressure near it and air,
therefore, flows in from the side
tube. The air and water together are
expelled through the lower part of Nptice that the speed of the efflux of liquid is the same as
the pump.
the speed of a ball that falls through a height (h1- h2). The

134
top level of the tank has moved down a little and the P.E.
has been transferred into K.E. of the efflux of fluid. If the
orifice had been pointed upward as at B shown in Fig.6.4,
this K.E. would allow the liquid to rise to the level of
water tank. In practice, viscous-energy losses would alter
the result to some extent.

R e la tio n b e tw e e n S peed and P re s s u re o f the

F lu id
A result of the Bernoulli’s equation is that the pressure will
be low where the speed of the fluid is high. Suppose that
water flows through a pipe system as shown in Fig. 6.5.
Clearly, the water will flow faster at B than it does at A or C.
Assuming the flow speed at A to be 0.20 ms'1 and at B to be
2.0 ms'1, we compare the pressure at B with that at A.
Applying Bernoulli’s equation and noting that the average
P.E. is the same at both places, We have,

PA+±pvZ =P f l + i p v I (6.14)
.

Substituting vA = 0.20 ms'1 , vB = 2.0 ms'1

And P = 1000 kgm'3
We get PA-P B= 1980 Nm'2
This shows that the pressure in the narrow pipe where
streamlines are closer together is much smaller than in the
wider pipe. Thus,

Where the speed is high, the pressure will be low.

The lift on an aeroplane is due to this effect. The flow of air

around an aeroplane wing is illustrated in Fig. 6.6. The wing is
designed to deflect the air so that streamlines are closer
together above the wing than below it. We have seen in
Fig.6.6 that where the streamlines are forced closer together,__
, the speed is faster. Thus, air is travelling faster on the upper^
side of the wing than on the lower. The pressure will be lower'
at the top of the wing, and the wing will be forced upward.

Similarly, when a tennis ball is hit by a racket in such a way

that it spins as well as moves forward, the velocity of the

135
 air on one side of the ball increases (Fig. 6.7) due to spin
and air speed in the same direction as at B and hence, the
pressure decreases. This gives an extra curvature to the
ball known as swing which deceives an opponent player.

Venturi Relation
If one of the pipes has a much smaller diameter than the
other, as shown in Fig. 6.8, we write Bernoulli’s equation in
a more convenient form. It is assumed that the pipes are
horizontal so that pgA? terms become equal and can,
therefore, be dropped. Then
Fig. 6.7.

P i - p 2 = ± p v i - ± p * 2 = i P(v f - * j (615)

As the cross-sectional area A2 is small as compared to the

erea Au then from equation of continuity v1= (A /A *) v2, will
be small as compared to v2. Thus for flow from a large pipe
to a small pipe we can neglect v* on the right hand side of
equation 6.15. Hence,
P i- P 2= i p v i (6.16)
Fig. 6.8.
This is known as Venturi relation, which is used in Venturi-
meter, a device used to measure speed of liquid flow.

Interesting Information Example 6.3: Water flows down hill through a closed
vertical funnel. The flow speed at the top is 12.0 cms'1. The
flow speed at the bottom is twice the speed at the top. If
Atmospheric
the funnel is 40.0 cm long and the pressure at the top is
pressure 1.013 x 105 Nm'2, what is the pressure at the bottom?

Solution: Using Bernoulli’s equation

Pi + Pgh7+ ^ pVi = P 2+ pgh2 + pv2

The carburetor of a car engine uses

Or P2- Pi + pgh + -i p (vi - v \ )
a Venturi duct to feed the correct mix
of air and petrol to the cylinders. Air
is drawn through the duct and along
. where /7 = - f?2= the length of the funnel
a pipe to the cylinders. A tiny inlet at P2 = (1.013 x 105Nm'2) + (1000 kgrn3 x9.8 m s2 x0.4m)
the side of duct is fed with petrol.
The air through the duct moves very
fast, creating low pressure in the + [ i (lOOOkgm-3) x {(0.12 m s1)2- (0.24 m s1)2}]
duct, which draws petrol vapour into
the air stream.
= 1.05 x105 N m'2

136
 A stethoscope detects the instant at which the external
pressure becomes equal to the systolic pressure. At this
point the first surges of blood flow through the narrow
stricture produces a high flow speed. As a result the flow is
initially turbulent.
As the pressure drops, the external pressure eventually
equals the diastolic pressure. From this point, the vessel
no longer collapse during any portion of the flow cycle. The
flow switches from turbulent to laminar, and the gurgle in
the stethoscope disappears. This is the signal to record
diastolic pressure.

An object moving through a fluid experiences a retarding force known as drag force.
It increases as the speed of object increases.
A sphere of radius r moving with speed v through a fluid of viscosity rj experiences a
viscous drag force F given by Stokes’ law F = 6 n r\rv.
The maximum and constant velocity of an object falling vertically downward is called
terminal velocity.
An ideal fluid is incompressible and has no viscosity. Both air and water at low
speeds approximate to ideal fluid behaviour.
In laminar flow, layers of fluid slide smoothly past each other.
In turbulentTtewTbere is great disorder and a constantly changing flow pattern.
Conservation of mass in an incompressible fluid is expressed by the equation of
•>
continuity- A1v1= A2v2. = constant >» —
Applying the principles of conservation of mechanical energy to the steady flow of an
ideal fluid leads to Bernoulli’s equation.

P+i pv2 + pgh = constant

The effect of the decrease in pressure with the increase in speed of the fluid in a
horizontal pipe is known as Venturi effect.

138
 B lo o d F lo w
Blood is an incompressible fluid having a density nearly
equal to that of water. A high concentration (-50%) of red
blood cells increases its viscosity from three to five times
that of water. Blood vessels are not rigid. They stretch like
a rubber hose. Under normal circumstances the volume
of the blood is sufficient to keep the vessels inflated at all
times, even in the relaxed state between heart beats. This
means there is tension in the walls of the blood vessels
and consequently the pressure of blood inside is greater
than the external atmospheric pressure. Fig. 6.9 shows
the variation in blood pressure as the heart beats. The
pressure varies from a high (systolic pressure) of 120 torr
(1 torr = 133.3 Nm'2) to a low diastolic pressure) of about
75-80 torr between beats in normal, healthy person. The
numbers tend to increase with age, corresponding to the
decrease in the flexibility of the vessel walls.
The unit torr or mm of Hg is opted instead of SI unit of
pressure because of its extensive use in medical equipments.
An instrument called a sphygmomanometer measures
blood pressure dynamically (Fig. 6.10).

An inflatable bag is wound around the arm of a patient and

external pressure on the arm is increased by inflating the bag.
The effect is to squeeze the arm and compress the blood
vessels inside. When the external pressure applied becomes
larger than the systolic pressure, the vessels collapse, cutting
off the flow of the blood. Opening the release valve on the bag
graduallydecreasesthe fextefjnal gres-

137
 e jH r o
6.1 Explain what do you understand by the term viscosity?
6.2 What is meant by drag force? What are the factors upon which drag force acting
upon a small sphere of radius r, moving down through a liquid, depend?
6.3 Why fog droplets appear to be suspended in air?
6.4 Explain the difference between laminar flow and turbulent flow.
State Bernoulli’s relation for a liquid in motion and describe some of its applications.
A person is standing near a fast moving train. Is there any danger that he will fall
towards it?
Identify the correct answer. What do you infer from Bernoulli’s theorem?
Where the speed of the fluid is high the pressure will be low.
Where the speed of the fluid is high the pressure is also high.
This theorem is valid only for turbulent flow of the liquid.
6.8 Two row boats moving parallel in the same direction are pulled towards each other.
Explain.
6.9. Explain, how the swing is produced in a fast moving cricket ball.
6.10 Explain the working of a carburetor of a motorcar using by Bernoulli’s principle.
6.1 For which position will the maximum blood pressure in the body have the smallest
value, (a) Standing up right (b) Sitting (c) Lying horizontally (d) Standing on one’s
head?
6.12 In an orbiting space station, would the blood pressure in major arteries in the leg
ever be greater than the blood pressure in major arteries in the neck?

NUMERICAL PROBLEMS

6.1 Certain globular protein particle has a density of 1246 kg m'3. It falls through pure
water (r|=8.0 x 10'4Nrris) with a terminal speed of 3.0 cm h'1. Find the radius of
the particle.
(Ans: 1.6 x 10'6m)

6.2 Water flows through a hose, whose internal diameter is 1cm at a speed of 1ms‘1.
What should be the diameter of the nozzle if the water is to emerge at 21ms'1?

(Ans: 0.2 cm)

139
 6.3 The pipe near the lower end of a large water storage tank develops a small leak and
a stream of water shoots from it. The top of water in the tank is 15m above the point
of leak.
a) With what speed does the water rush from the hole?
b) If the hole has an area of 0.060 cm2, how much water flows out in one second?
(Ans: (a) 17 m s'1, (b) 102 cm3)

6.4 Water is flowing smoothly through a closed pipe system. At one point the speed of
water is 3.0 ms'! while at another point 3.0 m higher, the speed is 4.0 ms1. If the
pressure is 80 kPa at the lower point, what is pressure at the upper point?
(Ans: 47 kPa)
" 6.5 An airplane wing is designed so that when the speed of the air across the top of the
wing is 450 ms'1, the speed of air below the wing is 410 ms"1. What is the pressure
difference between the top and bottom of the wings? (Density of air = 1.29kgm'3)
(Ans: 22 kPa)
The radius of the aorta is about 1.0 cm and the blood flowing through it has a speed
of about 30 cms'1. Calculate the average speed of the blood in the capillaries using
the fact that although each capillary has a diameter of about 8 x 10"4 cm, there are
literally millions of them so that their total cross section is about 2000cm2.
(Ans: 5 x 104ms1)
6.7 How large must a heating duct be if air moving 3.0 ms'1 along it can replenish the air in
a room of 300 m3 volume every 15 min? Assume the air’s density remains constant.
(Ans: Radius = 19 cm)
6 8 An airplane design calls for a “lift” due to the net force of the moving air on the wing of
about 1000 Nm of wing area. Assume that air flows past the wing of an aircraft with
streamline flow. If the speed of flow past the lower wing surface is 160ms1, what is
the required speed over the upper surface to give a “lift” of 1000Nm'2? The density of
air is 1.29 kgm'3 and assume maximum thickness of wing to be one metre.
(Ans: 165 ms'1)
6.9 What gauge pressure is required in the city mains for a stream from a fire hose
connected to the mains to reach a vertical height of 15.0 m?
(Ans: 1.47x10s Pa)

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