Fluid in Motion
At any point in a body of a moving fluid, and at any moment of time, there is a velocity
associated with a small volume of the fluid as it goes past that point. The velocities at all
points in the fluid constitute a flow field, or simply a flow. In a steady flow, the flow
field does not change with time: the velocity at any point remains the same at all times.
Otherwise the flow is unsteady. Flows in which the velocity can fluctuate in time and
behave erratically are called turbulent flows. Flows tend to be steady if the fluid is
viscous and/or flowing slowly. Turbulent flows develop when the flow velocity is large
and the fluid has small viscosity (stickiness).
In a steady flow, if you follow the motion of a small piece of the fluid, it will trace out a
curve known as a stream line. The streamline can be made visible by introducing a small
amount of a dye into the fluid. At any point on a streamline, the tangent gives the
direction of the flow velocity at that point. A continuous bundle of streamlines forms a
stream tube. Fluid flows within the stream tube. A pipe is an example of a steam tube. A
steam tube can have varying cross-sections along its length.
Equation of continuity
We can derive a relation between the mass of the fluid flowing through a cross-section of
a steam tube in unit time, called the mass flow rate, and the flow velocity at that crosssection. Fix our attention on a cross-section area A in a stream tube, and calculate the
mass of fluid going through it in a time interval Δt.
A
1
If  is the flow velocity, this mass of liquid must come from upstream and at a distance
of no more than t from the cross-section. Those further away do not have enough time
to reach the cross-section.
t
Since the volume of this mass is V  t A , the volume going through the crosssection in unit time, called the volumetric flow rate, is given by
V
 A
t
The mass itself is m   t A , where  is the density of the fluid. Therefore the mass
flow rate is given by
m
 A
t
Example: If water flows out of a faucet at a velocity of 3 ft/s, and the cross-section of the
faucet is 0.6 in2, how long will it take to fill up a one gallon bucket? (1 gallon = 0.1337
ft3)
2
1
Solution: Flow area A  0.6    ft 2  4.2  10 3 ft 2
 12 
Volume flow rate = 3.0  4.2  10 3  1.26  10 2 ft 3 / s
0.1337
 10.6s
Time to fill 1 gallon =
1.26  10  2
Consider a stream tube that has varying cross-sections. The mass of fluid between the
point where the cross-section is A1 and another where the cross-section is A2 does not
change in time in steady flow. Therefore, the mass flowing into this region in one second
equals the mass flowing out, which leads to the equation of continuity:
11 A1   2 2 A2
where 1 ,  2 and 1 , 2 refer to the density and velocity at the two locations. If the fluid is
incompressible, (water is a good example) the density is the same everywhere, the
equation of continuity simplifies to
2
1 A1   2 A2
According to this, the wider the area, the slower is the flow. That’s why a river flows
more slowly as it empties into the ocean where the flow channel is wide, and rapids
develop where the channel becomes shallow.
Bernoulli Equation
Fluid pressure can vary along a stream tube if its cross-section is not uniform, because in
this case the velocity can change along the tube, and it takes a pressure difference to
cause the fluid to speed up or slow down.
Refer to the diagram below showing segment of a stream tube of an incompressible fluid
included between locations I and II where the cross-sections are A1 and A2 respectively,
and let the flow velocities and pressure be 1 , 2 and p1 , p2 . The body of fluid between I
and II are pushed by pressure forces p1 A1 and p2 A2 from the rest of the fluid in the
directions shown. After a lapse of time t , the segment has moved and now occupies
the space between the cross-sections in I’ and II’.
P2A2
P1A1
II’
II
I
 2 t
I’
1t
We now apply the work energy theorem to this body of fluid. The work done by the
pressure forces is
W  ( p1 A1 )(1t )  ( p2 A2 )(2 t )  ( p1  p2 )V
where V  1 A1t   2 A2 t
is the volume of fluid flowing through each cross-section in time t . The initial kinetic
energy of the segment can be written as
K i  K  from I to I   K  from I  to II 
while the final kinetic energy is
3
K f  K  from I  to II   K  from II to II 
The change in kinetic energy is
K  K f  K i  K  from II to II   K  from I to I  
1
V  22  1 V 12
2
2
Using the work energy theorem W  K , we obtain
p1 
1
1
12  p 2   22
2
2
which is a simple form of Bernoulli equation applied to incompressible flow with no
change in height.
According to Bernoulli theorem, if the flow increases along a stream tube, the pressure is
reduced. The pressure difference between upstream and downstream accelerates the fluid.
Example: Aneurism is the occurrence of an enlarged portion of the aorta. If the enlarged
portion has a cross-sectional area twice that of the normal cross-section, and if the blood
flow velocity is 0.5m/s in the normal portion, by how much is the pressure in the enlarged
portion exceed the normal portion, assuming the patient lies down horizontally.
Solution: Let 1 and 2 denote the normal and the abnormal portions of the aorta.
We have A2  2A1 and 1  0.5m / s
From the continuity equation for incompressible fluid,
A
1
 2  1 1   0.5  0.25m / s
A2
2
From Bernoulli equation,
p 2  p1 




1
1
 12   22   10 3  0.5 2  0.25 2  94 Pa
2
2
This can be converted to mm of mercury if we use the relation
1mm of mercury = 13.6×103×9.8×10-3=133Pa
The result is p 2  p1 
94
 0.7 mm
133
The excess pressure in the abnormal region adds to the risk posed by the already
weakened wall of the enlarged blood vessel.
4
A couple of applications of Bernoulli theorem follow.
Lift force on wings. When an aircraft moves forward, the shape of its wings causes
stream tubes over the wing to experience more contraction in cross-sectional area than
those below the wing. As a result, the increase of flow speed of air from upstream to
downstream is higher for stream tubes passing over the wing than those passing under.
The reduction in pressure from upstream to downstream is therefore also higher for
stream tubes over the wing. Since the upstream pressure is the same for the two kinds of
stream tubes, the pressure above the wing is less than the pressure below. The net
pressure force on the wing is therefore upward. This force is called a dynamic lift as it
depends on the air craft moving. The lift force is reduced if the speed of the aircraft
decreases.
The Magnus Effect For a ball travelling forward in air, the stream lines are symmetrical
placed on either side of the ball. If the ball is spinning, the air near its surface is dragged
along by the rotating surface, and creates stream lines as shown. The flow speed is higher
on one side than the other. This creates a pressure difference between the two sides. As a
result, the ball is deflected from its forward direction. This is the physics of a curved ball.
If the ball is replaced by a rotating cylinder, this can form the basis of propulsion, with
the cylinder behaving like a sail. The ship travels perpendicularly to wind direction.
5