Momentum Equation PPT

Outline

¾

Momentum equation

Objectives

•

After completing this chapter, you should be able to

• Identify 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.

6.1 Development of the Momentum Principle

Start with a modified-form of Newton’s 2

_law:

( )

dt

V

m

d

F

r

r

=

Σ

∑

V

m

r

= sum of forces on fluid system

= linear momentum of system

Also, ---Impulse-momentum principle

s

V

m

d

dt

F

r

=

(

r

)

Σ

6.1 Development of the Momentum Principle

Force = rate of change of momentum

t tu u A tu u A F δ δ ρ δ ρ ) ( _{2 2 2} − _{1 1 1} = Q= A₁ u₁ = A₂ u₂,

)

(

u

u

Q

F

=

ρ

−

Momentum equation for

two-and three-dimensional flow

The force in the x-direction

Fig 6.2 Two dimensional flow in a streamtube

Momentum equation for

two-and three-dimensional flow

The force in the y-direction

Momentum equation for

two-and three-dimensional flow

Total force exerted on the fluid Rate of change of momentum through the control volume

=

)

(

u

u

Q

F

r

=

ρ

r

−

r

For steady flow with one inlet and one outlet, the momentum equation is

)

(

u

u

Q

F

r

=

ρ

β

r

−

β

r

Momentum correction factor

velocity

Mean

time

unit

per

Mass

time

unit

per

momentum

True

=

β

×

×

A

V

dA

u

/

∫

=

β

The momentum equation

)

(

V

V

Q

the resultant force

F

F

F

=

+

the total force, F_T, is given by the sum of these forces P B R T

F

F

F

F

r

=

r

+

r

+

r

What are the forces

acting on the

fluid

in the control volume?

Step in Analysis with Momentum Equation

1. Draw a control volume: Based on the problem, selecting the stream between two gradually varied flow sections as the control volume;

2. Decide on co-ordinate axis system: Determining the

directions of co-ordinate axis, magnitudes and directions of components of all forces and velocities on each axis.

3. Plotting diagram for computation : Analyzing the forces on control volume and plotting the directions of all forces on the control volume.

4. Writing momentum equation and solving it: Substituting components of all forces and velocities on axes into

momentum equation and solving it. All the pressures are relative to the relative pressure.

Application of the Momentum Equation

1. Force due to the flow of fluid round a pipe bend.

2. Force on a nozzle at the outlet of a pipe.

3. Impact of a jet on a plane surface.

Example 1

Find the horizontal thrust of the water on each meter of

width of the sluice gate shown in the Fig., given y

=2.2

m, y

=0.4 m, and y

=0.5 m. Neglect friction. ( 6.4.2)

Solution

The flow rate

6.8

A reducing right-angled bend lies in a horizontal plane.

Water enters from the west with a velocity of 3 m/s and a

pressure of 30 kPa, and it leaves toward the north. The

diameter at the entrance is 500 mm and at the exit it is

400 mm. Neglecting any friction loss, find the magnitude

and direction of the resultant force on the bend.

Solution

Energy

Example

Water flows through a reducing 180°bend. The bend is shown in plan. Determine the magnitude of the force exerted on the bend in the x-direction. Assume energy losses to be negligible.

Example

• Given: Figure

• Find: Horizontal force required to

hold plate in position • Solution: kN QV F V V Q F 9 . 4 3 . 12 * 4 . 0 * 999 ) ( _{2 1} = = = − = −

ρ

ρ

ρ

γ

γ

γ

question

If the value of force calculated from

momentum equation is negative, what

does that mean? Does the magnitude of

the unknown force has anything or

nothing to do with that of the control

volume? How to select control volume in

the application?

directions are inverse；independency（when

there is no gravitation）；calculated

cross-section and solid wall

If the value of force calculated from momentum

equation is negative, what does that mean? Does the

magnitude of the unknown force has anything or

nothing to do with that of the control volume? How to

select control volume in the application?

Sluice Gate

Find: Force due to pressure on

face of gate

Solution:

Assume: v₁ and v₂ are uniform (so pressure is hydrostatic)

Application of the Energy, Momentum, and

Continuity Equations in Combination

In general, when solving fluid mechanics problems,

one should use all available equations in order to

derive as much information as possible about the

flow. For example, consistent with the approximation

of the energy equation we can also apply the

Forces on Transitions

Example: Energy Equation

(energy loss)

datum

2 m

4 m

An irrigation pump lifts 50 L/s of water from a reservoir and

discharges it into a farmer’s irrigation channel. The pump

supplies a total head of 10 m. How much mechanical energy

is lost?

h

=

z

+

h

h

=

h

-

z

2.4 m

cs

cs

a

a

g

g

What is h

?

Example: Energy Equation

(pressure at pump outlet)

datum

2 m

4 m

50 L/s

h

= 10 m

The total pipe length is 50 m and is 20 cm in diameter. The

pipe length to the pump is 12 m. What is the pressure in the

pipe at the pump outlet? You may assume (for now) that the

only losses are frictional losses in the pipeline.

2.4 m

We need _______ in the pipe, __, and ____ ____.

cs

cs

a

a

g

g

Example: Energy Equation

(pressure at pump outlet)

¾

How do we get the velocity in the pipe?

¾

How do we get the frictional losses?

Kinetic Energy Correction Term:

α

¾

α is a function of the velocity distribution in

the pipe.

¾

For a uniform velocity distribution ____

¾

For laminar flow ______

¾

For turbulent flow _____________

¾

Often neglected in calculations because it is so

close to 1

Example: Energy Equation

(pressure at pump outlet)

datum

2 m

4 m

50 L/s

h

= 10 m

V = 1.6 m/s

α = 1.05

h

= 1.44 m

⎥

⎦

⎤

⎢

⎣

⎡

−

−

−

=

(1.44

m)

)

m/s

(9.81

2

m/s)

(1.6

(1.05)

m)

(2.4

m)

(10

)

N/m

(9810

p

2.4 m

2

p

V

h

z

h

g

a

g

=

+

+

+

Example: Energy Equation

(Hydraulic Grade Line - HGL)

¾

We would like to know if there are any

places in the pipeline where the pressure is

too high (_________) or too low (water

might boil - cavitation).

¾

Plot the pressure as piezometric head

(height water would rise to in a piezometer)

Example: Energy Equation

(Energy Grade Line - EGL)

datum

2 m

4 m

50 L/s

2.4 m

2

p

V

g

a

g

+

a

a

g

g

H

= 10 m

p = 59 kPa

What is the pressure at the pump intake?

Entrance loss

Exit loss

Loss due to shear

EGL (or TEL) and HGL

g

V

z

p

2

EGL

α

γ

+

+

=

₌

_{p +}

_z

γ

HGL

What is the difference between EGL defined by Bernoulli

and EGL defined here?

¾

The energy grade line may never be

horizontal or slope upward (in direction of

flow) unless energy is added (______)

¾

The decrease in total energy represents the

head loss or energy dissipation per unit

weight

¾

EGL and HGL are ____________and lie at

the free surface for water at rest (reservoir)

¾

Whenever the HGL falls below the point in

the system for which it is plotted the local

z

Example HGL and EGL

z = 0

pump

energy grade line

hydraulic grade line

velocity head

pressure head

elevation

datum

2g

V

α

γ

p

a

a

g

g