South valley University
Faculty of Engineering
Mechanical Power Engineering Dep.
Fluid Mechanics 2
Course Code MPEG222
Second Semester
Fall 2019/2020
By
Dr. Eng./Ahmed Abdelhady
Mobile: 01118501269
Email: a_abdelhady80@eng.sve.edu.eg
Lecture No. 3
Momentum Analysis of Flow Systems
Objectives of this Chapter
Define the various kinds of forces and
moments acting on a control volume
Use control volume analysis to determine the
forces associated with fluid flow
Use control volume analysis to determine the
moments caused by fluid flow and the torque
transmitted
Momentum Alysis of Flow Systems
Most engineering problems, including those associated
with fluid flow, can be analyzed using one of three basic
approaches:
Differential Approaches.
Experimental Approaches.
Finite Control Volume Approach
Applications of the Momentum Equation
➢ Force due to the flow of fluid a round a pipe bend.
➢ Force on a nozzle at the outlet of a pipe.
➢ Impact of a jet on a plane surface.
➢ Force due to flow round a curved vane.
Newton’s Laws and Conservation of Momentum
Newton’s first law states that: a body at rest remains
at rest, and a body in motion remains in motion at the
same velocity in a straight path when the net force
acting on it is zero.
Newton’s second law states that: the acceleration of
a body is proportional to the net force acting on it and
is inversely proportional to its mass.
Newton’s third law states that: when a body exerts a
force on a second body, the second body exerts an
equal and opposite force on the first.
For a rigid body of mass m, Newton’s second law is
expressed as:
Where: is the net force acting on the body and is
the acceleration of the body under the influence of .
The product of the mass and the velocity of a body
is called the linear momentum or just the
momentum of the body.
The momentum of a rigid body of mass m moving with
a velocity is:
Momentum = m*
The rate of change of the momentum of a body is equal to:
Fx = max =d(mvx)/dt
Newton’s second law for rotating rigid bodies is:
is the net moment or torque applied on the body, I is
the moment of inertia of the body about the axis of
rotation, and is the angular acceleration.
The rate of change of angular momentum:
For a rigid body rotating about a fixed x-axis, the
angular momentum equation can be written in scalar
form as:
Types of Control Volume
Moving Control Volume:
Example, if the airplane is moving at 500 km/h to the left, and
the velocity of the exhaust gases is 800 km/h to the right relative
to the ground, the velocity of the exhaust gases relative to the
nozzle exit is:
Fixed Control Volume:
Deforming Control Volume:
Forces Acting on A control Volume
Body Forces:
that act throughout the entire body of the control
volume (such as , fluid and metal weight, gravity,
electric, and magnetic forces)
Surface Forces:
that act on the control surface (such as pressure and
viscous forces and reaction forces at points of contact).
The Body Force
The Surface Force
The Linear Momentum Equation
Newton’s second law can be stated as:
the sum of all external forces acting on a system is equal
to the time rate of change of linear momentum of the
system.
For Steady Flow
During steady flow, the amount of momentum within
the control volume remains constant
Momentum-Flux Correction Factor, β
For uniform Flow; β =1.0
Steps in Analysis
➢ Draw a control volume.
➢ Chose the co-ordinate axis system.
➢ Calculate the total force.
➢ Calculate the pressure force.
➢ Calculate the body force.
➢ Calculate the resultant force.
Example 1
Consider a pipe bend with a constant cross section lying in
the horizontal plane and turning through an angle of θ .
Solution
Step in Analysis:
1. Control Volume:
2. Co-ordinate axis system:
Solution
3. Calculate the total force:
In the x-direction:
In the y-direction:
Solution
4. Calculate the pressure force:
5. Calculate the body force
Solution
6. Calculate the resultant force:
And the resultant force
on the fluid is given by:
The direction of application is:
The force on the bend is the same magnitude but in the
opposite direction:
Example 2
Impact of a Jet on a Plane (Fixed Plane):
We will first consider a jet hitting a flat plate (a plane) at an angle
of 90°, as shown in the figure below. We want to find the reaction
force of the plate i.e. the force the plate will have to apply to stay
in the same position
Solution
Step in Analysis:
1 & 2. Control volume and Co-
ordinate axis are shown in the
figure below.
3. Calculate the total force:
As the system is symmetrical the
forces in the y direction cancel i.e.
Solution
4. Calculate the pressure force:
The pressure force is zero as the pressure at both the inlet
and the outlets to the control volume are atmospheric.
5. Calculate the body force:
As the control volume is small we can ignore the body
force due to the weight of gravity.
6. Calculate the resultant force:
Exerted on the fluid.
The force on the plane is the same magnitude but in
the opposite direction
Example 3
Force on a curved vane:
This case is similar to that of a pipe, but the analysis is
simpler because the pressures are equal - atmospheric ,
and both the cross-section and velocities (in the
direction of flow) remain constant. The jet, vane and
co-ordinate direction are arranged as in the figure
below.
Example 4
A reducing elbow is used to deflect water flow at a rate of 30
kg/s in a horizontal pipe upward by an angle θ = 45° from the
flow direction while accelerating it. The elbow discharges
water into the atmosphere. The cross sectional area of the
elbow is 150 cm2 at the inlet and 25 cm2 at the exit. The
elevation difference between the centers of the exit and the
inlet is 40 cm. The mass of the elbow and the water in it is 50
kg. Determine the anchoring force needed to hold the elbow in
place. Take the momentum-flux correction factor to be 1.03.
Solution
The weight of the elbow and the water in it is:
Solution
Note that: the magnitude of the anchoring force is 1.18 kN,
and its line of action makes –39.7° from +x direction. Negative
value for FRx indicates the assumed direction is wrong.
