Chapter 5-Without Moment of Momentum

7/27/2019 Chapter 5-Without Moment of Momentum

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All rights reserved. 2003 A. Jaafar

Chapter 5: Finite Control

Volume AnalysisConservation of mass The continuity equation

Newtons 2nd Law The linear momentum and moment-of-

momentum equations

First law of thermodynamics The energy equation


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Conservation of Mass

Conservation of mass principle

As applied to a system

Time rate of change of the system mass = 0,

As applied to a coinciding system with a

CVTime rate of change of

the mass of the

coincident system

=

Time rate of change of

the mass of the contents

of the coincident controlvolume

+

Net rate of flow of

mass through the

control surfaces

0

Dt

DMsys

sys CSCV

dAVdt

VdDt

DnV


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(cont.)The control volume expression for

conservation of mass is commonly

known as the cont inui ty equation


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(cont.)

0

dAVdt CSCV nV

For a fixed, non-

deforming CV


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When the flow is steady

When the flow is steady and

incompressible


(cont.)

0inout mm

0inout QQ


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Forsteady flow involving only one

stream of a specific fluid flowing thru a

CV at sections (1) and (2)

Forsteady, incompressible flow

involving only one stream of a specificfluid flowing thru a CV at sections (1)

and (2)


(cont.)

222111 VAVAm

2211 VAVAQ


7/37All rights reserved. 2003 A. Jaafar

The control volume expression for

conservation of mass for a moving,

nondeforming control volume is thesame as that for a stationary control

volume, provided the absolute

velocity is replaced by the relative

velocity.


(cont.)

CVabsolute

CSCV

VWV

dAWVdt

where

0n



Example 1

The wind blows through a 7 ft X 10 ft garage door opening with a

speed of 5 ft/s as shown. Determine the average speed, V, of the

air through the two 3 ft X 4 ft openings in the windows.



Example 2

Storm sewer backup causes your basement to flood at the steady

rate of 1 in. of depth per hour. The basement floor area is 1500 ft2.

What capacity (gal/min) pump would you rent to

(a) keep the water accumulated in your basement at a constant

level until the storm sewer is blocked off?

(b) reduce the water accumulation in your basement at a rate of 3

in/hr even while the backup problem exist?



Newtons 2nd Law

Newtons 2nd Law of motion for a

systemTime rate of change of

the linear momentum ofthe system

= Sum of external forcesacting on the system

sysFVdDtD

sysV

CSCV

dAVdt

nVVV CVcoincidentofcontentsF



Newtons 2nd Law

Forces involved are the body force andsurface forces.

The only body force we are considering is

the one associated with the action ofgravity, i.e. the weight.

The surface forces are exerted on thecontents of the CV by the material just

outside the CV in contact with the materialjust inside the CV.



Newtons 2nd Law (cont.)

volumecontroltheofcontentsFnVVV dAVdt CSCV




Some notes on the Linear Momentum

Equation

It is directional; it can have components in

x,y,z coordinate directionsMomentum flow out of the control volume

involves a positive

If the CS is selected so that it is perpendicular

to the flow where fluid enters or leaves the CV,the surface force exerted at these locations by

fluid outside the control volume on fluid inside

will be due to pressure.

nV




When subsonic flow exits from a controlvolume into the atmosphere, atmosphericpressure prevails at the exit cross section.

Only external forces acting on the contents of

the control volume are considered in the linearmomentum equation

The force required to anchor an object willgenerally exist in response to the surfacepressure and/or shear forces acting on the CS,

to a change in linear momentum flow throughthe CV containing the object, and to theweight of the object and the fluid contained inthe CV.




Fluid flows can lead to a reaction

force in the following ways

Linear momentum flow variation indirection and/or magnitude

Fluid pressure forces

Fluid friction forces

Fluid weight



Example 3

Water enters the horizontal, circular cross section sudden

contraction nozzle at section (1) with a uniformly distributed

velocity of 25 ft/s and a pressure of 75 psi. The water exits from

the nozzle into the atmosphere at section (2) where the uniformly

distributed velocity is 100 ft/s. Determine the axial component of

the anchoring force required to hold the contraction in place.



Example 4

The thrust developed to propel the jet ski shown is a result of water

pumped through the vehicle and exiting as a high speed water jet.

For the conditions shown in the figure, what flowrate is needed to

produce a 300 lb thrust? Assume the inlet and outlet jets of water

are free jets.



More Examples

Water flows as two free jets from the tee attached to the pipe as

shown. The exit speed is 15 m/s. If viscous effects and gravity are

negligible, determine the x and y components of the force that the

pipe exerts on the tee.



More ExamplesA circular plate having a diameter of 300 mm is held perpendicular to

an axisymmetric horizontal jet of air having a velocity of 40 m/s

and a diameter of 80 mm as shown. A hole at the center of the

plate results in a discharge jet of air having a velocity of 40 m/s

and a diameter of 20 mm. Determine the horizontal component of

force required to hold the plate stationary.



More ExamplesA vertical circular cross section jet of air strikes a conical deflector

as shown. A vertical anchoring force of 0.1N is required to hold

the deflector in the place. Determine the mass (kg) of the

deflector. The magnitude of the velocity of the air remains

constant.



More ExamplesAir flows into the atmosphere from a nozzle and strikes a vertical

plate as shown. A horizontal force of 9 N is required to hold the

plate in place. Determine the reading on the pressure gage.

Assume the flow to be incompressible and frictionless.


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More ExamplesAir discharges from a 2-in.-diameter nozzle and strikes a curvedvane, which is in a vertical plane as shown. A stagnation tube

connected to a water U-tube manometer is located in the free air

jet. Determine the horizontal component of the force that the air jet

exerts on the vane. Neglect the weight of the air and all friction.


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First Law of

ThermodynamicsTime rate of increase of

the total stored energy

of the system=

Net time rate of energy

addition by heat

transfer into the system+

Net time rate of

energy addition

by work transfer

into the system

gzV

ue 2

2

innetQ

innetW

Represents the ways in

which energy isexchanged between the

CV contents and

surroundings because of a

temperature difference

Also known as power.+ when work is done

on the contents of the

CV by surroundings and

vice versa


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First Law of

Thermodynamics (cont.)

outin

QQQ innet

outin WWW innet


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Work transfer

In many instances, work is transferred by amoving shaft

Work can also occur at the CS when a forceassociated with the fluid normal stress acts

over a distance

Work transfer can also occur due to tangentialstress but is usually negligible

First Law of


shaftrotatingforntdisplacemeofrateXforce

shaftshaft

TW

W

CS dApW nVstressnormal


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First Law of


innetshaftinnet

2

2WQdAgz

VpudVe

t CSCV

nV

From normalstress From

moving

shaft


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If there is only one steady stream

entering and leaving the CV, and if all

the properties are assumed to be

uniformly distributed over the flow

cross-sectional areas involved

First Law of


innet

shaftinnet

22

2WQzzgVVppuum inoutinout

inout

inout

Valid for compressible and incompressible flow


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C i f th E


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Comparison of the Energy

Eqn with the Bernoulli

EquationFor steady, incompressible flow with zero

shaft power and negligible viscous forces

effects, the energy equation becomes theBernoulli Equation

For steady, incompressible flow with

friction,

inin

inoutout

out zV

pzV

p

22

22

0lossinnet quu inout

Loss of useful or available

energy that occurs in an

incompressible fluid flow

due to friction


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First Law of

Thermodynamics (cont.)Other related equations

Used for one dimensional steady-in-the-mean, incompressible flow withfriction and shaft work. Also knownas the Mechanical Energy Equat ionor the Extended Bernoul l i Equat ion

loss22 innetshaft

22

wgzVp

gz

Vpin

ininout

outout

minnetshaft

innetshaft

Ww


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First Law of

Thermodynamics (cont.)For non-uniform velocity distribution

(velocity profile at any cross section

where flow crosses the control surface is

not uniform),

where is the kinetic energy coefficient andis the average velocity.

loss22 innet

shaft

22

wgzVp

gzVp

inininin

outoutoutout

V


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Example 6

Air flows past an object in a pipe of 2-m diameter and exits as a free

jet as shown. The velocity and pressure upstream are uniform at

10 m/s and 50 N/m2, respectively. At the pipe exit, the velocity is

nonuniform as indicated. The shear stress along the pipe wall is

negligible.

(a) Determine the head loss associated with a particle as it flowsfrom the uniform velocity upstream of the object to the location in

the wake at the exit plane of the pipe.

(b) Determine the force that the air puts on the object.


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Example 7

What is the maximum possible power output of the hydroelectric

turbine shown in the figure.


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Example 8

Water is to be moved from one large reservoir to another at a higher

elevation. The loss in available energy associated with 2.5 ft3/s

being pumped from sections (1) to (2) is 61V2/2 where V is the

average velocity of water in the 8-in inside diameter piping

involved. Determine the amount of shaft power required.


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More Examples

Water flows steadily down the inclined pipe as shown. Determine thefollowing:

(a) The difference in pressure, p1-p2

(b) The loss between sections (1) and (2)

(c) The net axial force exerted by the pipe wall on the flowing water

between sections (1) and (2)


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More Examples

Water flows by gravity from one lake to another as shown at a steady

rate of 100 gpm. What is the loss in available energy associated

with this flow? If this same amount of loss is associated with

pumping the fluid from the lower lake to the higher one at the

same flowrate, estimate the amount of pumping power required.

Chapter 5-Without Moment of Momentum

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