Momentum Equation PPT

Ch 5 Momentum and forces in fluid flow

Outline

Momentum equation

Applications of the 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 2nd law:

( )dt

VmdFr

r=Σ

∑Fr

Vmr

= sum of forces on fluid system

= linear momentum of system

Also, -------Impulse-momentum principle sVmddtF )(

rr=Σ

6.1 Development of the Momentum Principle

111 tuuA δρ=

222 tuuA δρ=

Momentum entering

Momentum leaving

Force = rate of change of momentum

ttuuAtuuAF

δδρδρ )( 111222 −

=

Q= A1 u1 = A2 u2,

)( 12 uuQF −= ρ

Fig 6.1 Momentum in a flowing fluid

Momentum equation for two-and three-dimensional flow

The force in the x-direction

Fig 6.2 Two dimensional

flow in a streamtube


The force in the y-direction

Fig 6.2 Two dimensional flow in a streamtube


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

)( 12 uuQF rrr−= ρ

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

)( 1122 uuQF rrrββρ −=

β—— momentum correction factor

Momentum correction factor

velocityMeantimeunitperMasstimeunitpermomentumTrue ××= β

AVdAu 22 /∫=β

The momentum equation

)( 1122 VVQFrrr

ββρ −=

the resultant force

22yx FFF +=

the angle which this force acts at is given by

the total force, FT, is given by the sum of these forces

PBRT FFFFrrrr

++=

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 controlvolume;

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.4. Force due to flow round a curved vane.

Force due to Jet Striking Surface

Force by Flow Round a Pipe-bend

Example 1

Find the horizontal thrust of the water on each meter of width of the sluice gate shown in the Fig., given y1=2.2 m, y2=0.4 m, and y3=0.5 m. Neglect friction. ( 6.4.2)

Solution

Solution

The flow rate

Momentum equation

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

Momentum

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.

Solution:

Example• Given: Figure• Find: Horizontal force required to

hold plate in position• Solution:

kNQVFVVQF

9.43.12*4.0*999)( 12

===−=−

ρρ

smpV

gVp

gVzp

gVzp

AB

BA

BB

BAA

A

/3.12999/75000*22

2

222

22

===

=

++=++

ρ

γ

γγ

T=15 oC

Q=0.4 m3/s

pA=75 kPa

F

B

iVrr

V=

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 gateSolution: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 momentum and continuity equations

Forces on Transitions

Example

Example: Energy Equation(energy loss)

datum

2 m4 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?

p o u t Lh z h= + L p o u th h z= -

2.4 mcs1

cs2

2 2

2 2in in o u t o u t

in in P o u t o u t T Lp V p Vz h z h h

g ga a

g g+ + + = + + + +

What is hL?

Why can’t I draw the cs at the end of the pipe?

Example: Energy Equation(pressure at pump outlet)

datum

2 m4 m

50 L/shP = 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 ____ ____.

cs1

cs2

2 2



g ga a

g g+ + + = + + + +


How do we get the velocity in the pipe?

How do we get the frictional losses?

What about α?

Kinetic Energy Correction Term: α

α is a function of the velocity distribution in the pipe.For a uniform velocity distribution ____For laminar flow ______For turbulent flow _____________

Often neglected in calculations because it is so close to 1


datum

2 m4 m

50 L/shP = 10 m

V = 1.6 m/sα = 1.05hL = 1.44 m

⎥⎦

⎤⎢⎣

⎡−−−= m) (1.44

)m/s (9.812m/s) (1.6

(1.05)m) (2.4m) (10)N/m (9810 2

23

2p

2.4 m

2

2o u t o u t

P o u t o u t Lp Vh z h

ga

g= + + +

Example: Energy Equation(Hydraulic Grade Line - HGL)

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).Plot the pressure as piezometric head (height water would rise to in a piezometer)How?

Example: Energy Equation(Energy Grade Line - EGL)

datum

2 m4 m

50 L/s

2.4 m

2

2p V

ga

g+

2 2



g ga a

g g+ + + = + + + +

HP = 10 m

p = 59 kPa

What is the pressure at the pump intake?

Entrance loss

Exit lossLoss due to shear

EGL (or TEL) and HGL

gV

zp2

EGL2

αγ

++= zp+=

γHGL

What is the difference between EGL defined by Bernoulli and EGL defined here?

The energy grade line may never be horizontal or slope upward (in direction of flow) unless energy is added (______)The decrease in total energy represents the head loss or energy dissipation per unit weightEGL and HGL are ____________and lie at the free surface for water at rest (reservoir)Whenever the HGL falls below the point in the system for which it is plotted the local

EGL (or TEL) and HGL

z

Example HGL and EGL

z = 0

pump

energy grade line

hydraulic grade line

velocity head

pressure head

elevation

datum

2gV 2

α

γp

2 2



g ga a

g g+ + + = + + + +

See you next time.

Momentum Equation PPT

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