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1. EDUCATION HOLE PRESENTS
MECHANICS OF
FLUIDS
Module III: Dynamics of Fluid
Flow[Type the document
subtitle]

2. Euler’s equation of motion..................................................................................................... 1
Derivation.............................................................................................................................................................1
Bernoulli's Equation............................................................................................................... 4
Pressure/velocity variation ............................................................................................................4
Pilot Tube.............................................................................................................................. 5
Venturi meter........................................................................................................................ 8
Steady flow momentum equation.......................................................................................... 9
Newton's 2nd Law can be written .......................................................................................................................9
Forces exerted on a pipe bend ............................................................................................. 14
Resulting force due to Mass flow and Flow Velocity .........................................................................................14
Example - Resulting force on a bend due to mass flow and flow velocity.........................................................15
Resulting force due to Static Pressure ...............................................................................................................16
Example - Resulting force on a bend due to pressure .......................................................................................17
Euler’s equation of motion
This section is not a mandatory requirement. One can skip this section (if he/she does not like to
spend time on Euler's equation) and go directly to Steady Flow Energy Equation.
Using the Newton's second law of motion the relationship between the velocity and pressure
field for a flow of an in viscid fluid can be derived. The resulting equation, in its differential
form, is known as Euler’s Equation. The equation is first derived by the scientist Euler.
Derivation
let us consider an elementary parallelepiped of fluid element as a control mass system in a frame
of rectangular Cartesian coordinate axes as shown in Fig. 12.3. The external forces acting on a
fluid element are the body forces and the surface forces.

3. Fig 12.2 A Fluid Element appropriate to a Cartesian Coordinate System
used for the derivation of Euler's Equation
Let Xx, Xy, Xz be the components of body forces acting per unit mass of the fluid element along
the coordinate axes x, y and z respectively. The body forces arise due to external force fields like
gravity, electromagnetic field, etc., and therefore, the detailed description of Xx, Xy and Xz are
provided by the laws of physics describing the force fields. The surface forces for an in viscid
fluid will be the pressure forces acting on different surfaces as shown in Fig. 12.3. Therefore, the
net forces acting on the fluid element along x, y and z directions can be written as
Since each component of the force can be expressed as the rate of change of momentum in the
respective directions, we have

4. (12.5a)
(12.5b)
(12.5c)
s the mass of a control mass system does not change with time, is constant with time
and can be taken common. Therefore we can write Eqs (12.5a to 12.5c) as
(12.6a)
(12.6b)
(12.6c)
Expanding the material accelerations in Eqs (12.6a) to (12.6c) in terms of their respective
temporal and convective components, we get
(12.7a)
(12.7b)
(12.7c)
The Eqs (12.7a, 12.7b, 12.7c) are valid for both incompressible and compressible flow. By
putting u = v = w = 0, as a special case, one can obtain the equation of hydrostatics .
Equations (12.7a), (12.7b), (12.7c) can be put into a single vector form as
(12.7d)
(12.7e)
where the velocity vector and the body force vector per unit volume are defined as

5. Bernoulli's Equation
The Bernoulli equation states that,
Where
• points 1 and 2 lie on a streamline,
• the fluid has constant density,
• the flow is steady, and
• there is no friction.
Although these restrictions sound severe, the Bernoulli equation is very useful, partly because it
is very simple to use and partly because it can give great insight into the balance between
pressure, velocity and elevation
How useful is Bernoulli's equation? How restrictive are the assumptions governing its use? Here
we give some examples.
Pressure/velocity variation
Consider the steady, flow of a constant density fluid in a converging duct, without losses due to
friction (figure 14). The flow therefore satisfies all the restrictions governing the use of
Bernoulli's equation. Upstream and downstream of the contraction we make the one-dimensional
assumption that the velocity is constant over the inlet and outlet areas and parallel.
Figure 14. One-dimensional duct showing
control volume.
When streamlines are parallel the pressure is constant across them, except for hydrostatic head
differences (if the pressure was higher in the middle of the duct, for example, we would expect

6. the streamlines to diverge, and vice versa). If we ignore gravity, then the pressures over the inlet
and outlet areas are constant. Along a streamline on the centerline, the Bernoulli equation and the
one-dimensional continuity equation give, respectively,
These two observations provide an intuitive guide for analyzing fluid flows, even when the flow
is not one-dimensional. For example, when fluid passes over a solid body, the streamlines get
closer together, the flow velocity increases, and the pressure decreases. Airfoils are designed so
that the flow over the top surface is faster than over the bottom surface, and therefore the average
pressure over the top surface is less than the average pressure over the bottom surface, and a
resultant force due to this pressure difference is produced. This is the source of lift on an airfoil.
Lift is defined as the force acting on an airfoil due to its motion, in a direction normal to the
direction of motion. Likewise, drag on an airfoil is defined as the force acting on an airfoil due to
its motion, along the direction of motion.
An easy demonstration of the lift produced by an airstream requires a piece of notebook paper
and two books of about equal thickness. Place the books four to five inches apart, and cover the
gap with the paper. When you blow through the passage made by the books and the paper, what
do you see? Why?
Pilot Tube
A pilot tube is a pressure measurement instrument used to measure fluid flow velocity. The pilot
tube was invented by the French engineer Henri Pilot in the early 18th century[1]
and was
modified to its modern form in the mid-19th century by French scientist Henry Darcy.[2]
It
is widely used to determine the airspeed of an aircraft, water speed of a boat, and to measure
liquid, air and gas velocities in industrial applications. The pitot tube is used to measure the local
velocity at a given point in the flow stream and not the average velocity in the pipe or conduit.[3]
Theory of operation

7. The basic pitot tube consists of a tube pointing directly into the fluid flow. As this
tube contains fluid, a pressure can be measured; the moving fluid is brought to rest (stagnates) as
there is no outlet to allow flow to continue. This pressure is the stagnation pressure of the fluid,
also known as the total pressure or (particularly in aviation) the pitot pressure.
The measured stagnation pressure cannot itself be used to determine the fluid velocity (airspeed
in aviation). However, Bernoulli's equation states:
Stagnation pressure = static pressure + dynamic pressure
Which can also be written
Solving that for velocity we get:
NOTE: The above equation applies ONLY to fluids that can be treated as incompressible.
Liquids are treated as incompressible under almost all conditions. Gases under certain conditions
can be approximated as incompressible. See Compressibility.
Where:
• is fluid velocity;
• is stagnation or total pressure;
• is static pressure;
• and is fluid density.
The value for the pressure drop – or due to , the reading on the manometer:
Where:
• is the density of the fluid in the manometer
• is the manometer reading
The dynamic pressure, then, is the difference between the stagnation pressure and the static
pressure. The static pressure is generally measured using the static ports on the side of the
fuselage. The dynamic pressure is then determined using a diaphragm inside an enclosed
container. If the air on one side of the diaphragm is at the static pressure, and the other at the
stagnation pressure, then the deflection of the diaphragm is proportional to the dynamic pressure,
which can then be used to determine the indicated airspeed of the aircraft. The

8. diaphragm arrangement is typically contained within the airspeed indicator, which converts the
dynamic pressure to an airspeed reading by means of mechanical levers.
Instead of separate piltot and static ports, a pilot-static tube (also called a Prandtl tube) may be
employed, which has a second tube coaxial with the pilot tube with holes on the sides, outside
the direct airflow, to measure the static pressure.
Industry applications
Pitot tube from an F/A-18
In industry, the velocities being measured are often those flowing in ducts and tubing where
measurements by an anemometer would be difficult to obtain. In these kinds of measurements,
the most practical instrument to use is the pitot tube. The pitot tube can be inserted through a
small hole in the duct with the pitot connected to a U-tube water gauge or some other
differential pressure gauge for determining the velocity inside the ducted wind tunnel. One use of
this technique is to determine the volume of air that is being delivered to a conditioned space.
The fluid flow rate in a duct can then be estimated from:
Volume flow rate (cubic feet per minute) = duct area (square feet) × velocity (feet per minute)
Volume flow rate (cubic meters per second)
= duct area (square meters) × velocity (meters per second)
In aviation, airspeed is typically measured in knots.

9. Venturi meter
• In the upstream cone of the Venturi meter, velocity is increased, pressure is decreased
• Pressure drop in the upstream cone is utilized to measure the rate of flow through the
instrument
• Velocity is then decreased and pressure is largely recovered in the down stream cone
• Mostly used for liquids, water
Disadvantages of Venturi meter:
• Highly expensive
•Occupies considerable space (L/D ratio of appr. 50)
•Cannot be altered for measuring pressure beyond a maximum velocity.
Volumetric flow rate through a Venturi meter:

10. where Cv - Venturi coefficient
Sb - Cross sectional area of down stream
- Ratio of cs areas of upstream to that of down stream.
Pa-Pb - Pressure gradient across the Venturi meter
- Density of fluid
Steady flow momentum equation
We have all seen moving fluids exerting forces. The lift force on an aircraft is exerted by the
air moving over the wing. A jet of water from a hose exerts a force on whatever it hits. In
fluid mechanics the analysis of motion is performed in the same way as in solid mechanics -
by use of Newton's laws of motion. Account is also taken for the special properties of fluids
when in motion.
The momentum equation is a statement of Newton's Second Law and relates the sum of
the forces acting on an element of fluid to its acceleration or rate of change of momentum.
You will probably recognise the equation F = ma which is used in the analysis of solid
mechanics to relate applied force to acceleration. In fluid mechanics it is not clear what mass
of moving fluid we should use so we use a different form of the equation.
Newton's 2nd Law can be written
The Rate of change of momentum of a body is equal to the resultant force acting on the body,
and takes place in the direction of the force.
To determine the rate of change of momentum for a fluid we will consider a stream tube as
we did for the Bernoulli equation,
We start by assuming that we have steady flow which is non-uniform flowing in a stream
tube.

11. A stream tube in three and two-dimensions
In time a volume of the fluid moves from the inlet a distance , so the volume entering
the stream tube in the time is
this has mass,
and momentum
Similarly, at the exit, we can obtain an expression for the momentum leaving the steamtube:
We can now calculate the force exerted by the fluid using Newton's 2nd
Law. The force is
equal to the rate of change of momentum. So
We know from continuity that , and if we have a fluid of constant density,

12. i.e. , then we can write
For an alternative derivation of the same expression, as we know from conservation of mass
in a stream tube that
we can write
The rate at which momentum leaves face 1 is
The rate at which momentum enters face 2 is
Thus the rate at which momentum changes across the stream tube is
i.e.
This force is acting in the direction of the flow of the fluid.
This analysis assumed that the inlet and outlet velocities were in the same direction - i.e. a
one dimensional system. What happens when this is not the case?
Consider the two dimensional system in the figure below:

13. Two dimensional flow in a stream tube
At the inlet the velocity vector, , makes an angle, , with the x-axis, while at the
outlet make an angle . In this case we consider the forces by resolving in the directions
of the co-ordinate axes.
The force in the x-direction
And the force in the y-direction
We then find the resultant force by combining these vectorially:

14. And the angle which this force acts at is given by
For a three-dimensional (x, y, z) system we then have an extra force to calculate and resolve
in the z-direction. This is considered in exactly the same way.
In summary we can say:
Remember that we are working with vectors so F is in the direction of the
velocity. This force is made up of three components:
Force exerted on the fluid by any solid body touching the control
volume
Force exerted on the fluid body (e.g. gravity)
Force exerted on the fluid by fluid pressure outside the control volume

15. So we say that the total force, FT, is given by the sum of these forces:
The force exerted by the fluid on the solid body touching the control volume
is opposite to . So the reaction force, R, is given by
Forces exerted on a pipe bend
Resulting force due to Mass flow and Flow Velocity
The resulting force in x-direction due to mass flow and flow velocity can be expressed as:
Rx = m · v · (1 - cosβ) (1)
= ρ · A · v2
· (1 - cosβ) (1b)
= ρ · π · (d / 2)2
· v2
· (1 - cosβ) (1c)
where
Rx = resulting force in x-direction (N)
m = mass flow (kg/s)
v = flow velocity (m/s)

16. β = turning bend angle (degrees)
ρ = fluid density (kg/m3
)
d = internal pipe or bend diameter (m)
π = 3.14...
The resulting force in y-direction due to mass flow and flow velocity can be expressed as:
Ry = m · v · sinβ (2)
= ρ · A · v2
· sinβ (2b)
= ρ · π · (d / 2)2
· v2
· sinβ (2c)
Ry = resulting force in y direction (N)
The resulting force on the bend due to force in x- and y-direction can be expressed as:
R = (Rx
2
+ Ry
2
)1/2
(3)
where
R = resulting force on the bend (N)
Example - Resulting force on a bend due to mass flow and flow velocity
The resulting force on a 45o
bend with
• diameter 114 mm = 0.114 m
• water with density 1000 kg/m3
• flow velocity 20 m/s
can be calculated by as
Resulting force in x-direction:
Rx = 1000 (kg/m3
) · π · (0.114 (m) / 2)2
· 20 (m/s)2
· (1 - cos45)
= 1196 (N)
Resulting force in y-direction:

17. Ry = 1000 (kg/m3
) · π · (0.114 (m) / 2)2
· 20 (m/s)2
· sin45
= 2887 (N)
Resulting force on the bend
R = (1196 (N)2
+ 2887 (N)2
)1/2
= 3125 (N)
Note - if β is 90o
the resulting forces in x- and y-directions are the same.
Resulting force due to Static Pressure
The pressure and the end surfaces of the bend creates resulting forces in x- and y-directions.
The resulting force in x-direction can be expressed as
Rpx = p · A · (1- cos β) (4)
= p · π · (d / 2)2
(·1- cos β) (4b)
where
Rpx = resulting force due to pressure in x-direction (N)
p = gauge pressure inside pipe (Pa, N/m2
)
The resulting force in y-direction can be expressed as
Rpy = p · π · (d / 2)2
· sinβ (5)
where
Rpy = resulting force due to pressure in y-direction (N)
The resulting force on the bend due to force in x- and y-direction can be expressed as:
Rp = (Rpx
2
+ Rpy
2
)1/2
(6)
where
Rp = resulting force on the bend due to static pressure (N)

18. Example - Resulting force on a bend due to pressure
The resulting force on a 45o
bend with
• diameter 114 mm = 0.114 m
• pressure 100 kPa
can be calculated by as
Resulting force in x-direction:
Rx = 100 (kPa) · π · (0.114 (m) / 2)2
· (1 - cos45)
= 299 (N)
Resulting force in y-direction:
Ry = 100 (kPa) · π · (0.114 (m) / 2)2
· sin45
= 722 (N)
Resulting force on the bend
R = (299 (N)2
+ 722 (N)2
)1/2
= 781 (N)