1.  Newton’s Second Law of Motion: Control-Volume Approach 
Chapter 5
 Newton’s Second Law of Motion: Control-Volume Approach 
Dr Sharif F Zaman
Assistant Professor
Chemical and Materials Eng. Dept.
King Abdul Aziz University
Jeddah, Saudi Arabia

2. INTEGRAL RELATION FOR LINEAR MOMENTUM
Newton’s second law of motion may be stated as follows:
The time rate of change of momentum of a system is equal to the net force acting on the system and takes place in the direction of the net force.
Important part is this law describes both
Magnitude
Direction
Of the system.
Question : What is a system?

4. Momentum at time ‘t+Δt’
From Newton’s 2nd law “ --sum of the forces acting on the control volume
Momentum at time ‘t+Δt’
Momentum at time ‘t’
Subtracting above two equations and divide by ‘Δt’
Taking the limit Δt  0
expresses the net rate of momentum efflux across the control surface during the time interval Δt.
MOMENTUM OUT- MOMENTUM IN
Rate of change of momentum in the control volume

5. Body Force ?
Surface force ?
The total force acting on the control volume consists both of
Surface forces due to interactions between the control-volume fluid, and its surroundings through direct contact, and
Body forces resulting from the location of the control volume in a force field.
The gravitational field and its resultant force are the most common examples of this latter type ( i.e. Body Force)

6. Dot product

7. V.n = - ve {momentum entering the system}
V.n = + ve {momentum leaving the system}
The rate of accumulation of linear momentum within the control volume may be expressed as
This extremely important relation is often referred to in fluid mechanics as the
momentum theorem.

8. For the fluid in the control volume, the forces to be employed in these equations are those acting on the fluid.
When applying any or all of the above equations, it must be remembered that each term has a sign with respect to the positively defined x, y, and z directions.
The determination of the sign of the surface integral should be considered with special care, as both the velocity component (vx) and the scalar product (v: n) have signs.

10. The first step is the definition of the control volume.
One choice for the control volume, of the several available, is all fluid in the pipe at a given time. The control volume chosen in this manner is designated in Figure 5.4, showing the external forces imposed upon it.

12. Over all momentum balance in x and y direction

13. +Y +X
Control surface integration =>

15. Recall that we were to evaluate the force exerted on the pipe rather than that on the fluid. The force sought is the reaction to B and has components equal in magnitude and opposite in sense to Bx and By. The components of the reaction force, R, exerted on the pipe are -

16. Our control-volume boundary will be selected as the interior of the tank and scoop.
As the train is moving with a uniform velocity, there are two possible choices of coordinate systems.
A coordinate system either
=> fixed in space or
=> moving with the velocity of the train, v0.

19. Example 3

20. Vj = velocity of jet = 12 m/s
Aj = cross sectional area of the jet = m2.