cc501- ch3-jun2015

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CHAPTER 3 MOMENTUM EQUATIONS

CC501 – HYDRAULICS 2

3.1 Understand the fundamental principles of momentum equations in pipe flow analysis :

3.1.1 Describe Newton’s Second law of motion 3.1.2 Explain Newton’s Third Law of motion3.2 Understand the effect of forces caused by fluids in motion 3.2.1 Calculate the force exerted by a jet striking a static and moving flat plate. 3.2.2 Calculate the force due to the deflection of a jet by a static and moving

curved vane. 3.2.3 Determine the force exerted when a jet is deflected by a moving curved

vane. 3.2.4 determine the magnitude and the direction of the resultant forces

exerted on pipe bends and closed conduit.

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CHAPTER 3 : MOMENTUM EQUATIONS

Momentum Equations

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Momentum can be defined as "mass in motion."

All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion.

The amount of momentum that an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving.

Momentum depends upon the variables mass and velocity. In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object.

Momentum = mass x velocity

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Newton's FIRST laws of motion• First law: If an object experiences no net force, then its velocity is constant:

the object is either at rest (if its velocity is zero), or it moves in a straight line with constant speed (if its velocity is nonzero).

Newton's SECOND laws of motion• Second law: The acceleration a of a body is parallel and directly proportional

to the net force F acting on the body, is in the direction of the net force, and is inversely proportional to the mass m of the body, i.e., F = m a

Newton's THIRD laws of motion• Third law: When a first body exerts a force F1 on a second body, the second

body simultaneously exerts a force F2 = −F1 on the first body. This means that F1 and F2 are equal in magnitude and opposite in direction.

Newton's laws of motion are three physical laws that form the basis for classical mechanics. They describe the relationship between the forces acting on a body and its motion due to those forces.

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Impulse-momentum equation

When a force (Push or pull) is applied on the bodies it tries to change the state of rest or state of motion of those bodies. The amount of force applied is equal to the rate of change of momentum. Where momentum is the product of mass and velocity.

So :

Where product Ft is the impulse and it is equal to the change momentum.

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• Define the term impact of jet.

A fluid jet is a stream of fluid obtained from nozzle. When this jet strikes on flat or curved plate the momentum is changed and a hydrodynamic force is exerted.

So ‘Impact of Jet’ term refers to the study of the effect when a jet strikes on the plate or vane under the various conditions.

The force exerted by a jet striking a flat plate.

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The force exerted by a jet striking a static and moving flat plate.

The force exerted by a jet on a stationary flat plate held normal to the jet

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The force exerted by liquid jet on a moving plate which is vertical and moving along a jet.

Let the velocity of jet is v and velocity of plate is u

Fig. Fluid jet striking a moving fluid plate

Relative velocity = v – u Mass of water striking the plate (Relative ‘velocity of jet)

Force of striking the jet = Mass of water x Relative velocity

The expression for force exerted by a jet on stationary inclined flat plate.

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= Mass (Initial velocity - Final velocity)

(Values of velocities are taken in the direction of the Jets).

Example 1

A flat plate is struck normal by a jet of water 50mm in diameter with a velocity at 18m/s. Calculate

(a) The force on the plate when it is stationary

(b) The force on the plate when it moves in the same as the jet with a velocity of 6 m/s.

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• These components can be combined to give the resultant force:

• And the angle of this force:

The force due to the deflection of a jet by a curved vane• Consider the two dimensional system shown:• Since both momentum and force are vector quantities,

they can be resolving into components in the x and y directions

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A jet of water from a nozzle is deflected through an angle u = 60 from its original direction by a curved vane which enters tangentially without shock with mean velocity of 30 m/s and leaves with mean velocity of 25 m/s. If the discharge from the nozzle is 0.95 m /s.

Calculate the magnitude and direction of the resultant force on the vane if the vane is stationary.

Example 2

o

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A jet of water strikes a stationary curved vane without shock and is deflected 150° from its original direction. The discharge from the jet is

0.68 kg/s and the jet velocity is 24 m/s. Assume that there is no reduction of the relative velocity due to friction and determines the

magnitude and direction of reaction on vane.

NF

vvQF

NF

vvQF

Qm

y

y

x

x

16.8030sin2468.0

sinsin

45.302430cos2468.0

coscos

68.0

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12

o

yx NFFF

1545.30

16.8tan

52.3116.845.30

1

2222

Example 3

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Example

Figure below shows a jet of water with a velocity v1 striking a vane that is moving with a velocity v0. Determine the forces exerted by the vane on the water if v1 = 20 m/s and v0 = 8 m/s. The jet is 50 mm in diameter.


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The system with a moving vane can be converted into an equivalent stationary system as shown in Fig. 16.9(b) by defining an effective velocity ve and an effective volume flow rate Qe. We then have

where A1 is the area of the jet as it enters the vane. It is only the difference between the jet velocity and the vane velocity that is effective in creating a force on the vane.


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The force equations can be written in terms of ve and Qe. In the x direction,

In the y direction,

We know that

Then the reactions are calculated as :

1919Chapter 1 : Hydrostatic Forces

Pressure and Hydrostatic ForcesC4009 HYDRAULICS 1

Momentum Equation Applications :Forces on a Pipe Contraction or Nozzle

Flange connection


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FORCE EXERTED ONAPIPE BENDS AND CLOSED CONDUITSDetermine the magnitude and the direction of the resultantforces exerted on pipe bends and closed conduits.


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Linseed oil with a specific gravity of 0.93 enters the reducing bend shown in Fig. 16.7 with a velocity of 3 m/s and a pressure of 275 kPa. The bend is in a horizontal plane. Calculate the x and y forces required to hold the bend in place. Neglect energy losses in the bend.


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The fluid in the bend is shown as a free body in Fig. We must first develop the force equations for the x and y directions shown. The force equation for the x direction is

Algebraic signs must be carefully included according to the sign convention shown in Fig. Notice that all forces and velocity terms are the components in the x direction.


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In the y direction, the force equation is


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Example


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MOMENTUM AND FLUID FLOW

• From solid mechanics you will recognize F = ma• In fluid mechanics it is not clear what mass of moving fluid,

we

• In mechanics, the momentum of particle or object is defined as: Momentum = mv

• To determine the rate of change of momentum for a fluid we• will consider a stream tube (assuming steady non-uniform

flow).

CHAPTER 3 : MOMENTUM EQUATIONS 29

• The rate of change of momentum across the control volume:

• And according the Newton’s second law, this change of momentum per unit time will be caused by a force F, Thus:

• This is the resultant force acting on the fluid in the direction of motion.• By Newton’s third law,

“ the fluid will exert an equal and opposite reaction on its surroundings”

Daya Hentaman Jet

Bab 3 : Momentum Bendalir

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i. Plat Pegun Secara Normal


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ii. Plat pegun secara condong

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iii. Plat Bergerak Secara Normal


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Pop Kuiz

Nyatakan betul atau salah

1.Momentum bersamaan daya x halaju

2.Persamaan momentum bendalir diberikan oleh hukum momentum

yang kedua


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cc501- ch3-jun2015

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