fluid mechanics 2010-2011 finite control volume

7/29/2019 Fluid Mechanics 2010-2011 Finite Control Volume

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FINITE CONTROL VOLUME ANALYSIS


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Conservation of Mass The Continuity Equation

Finite Control Volume Analysis

Derivation of the Continuity Equation

Fixed, Nondeforming Control Volume

Moving, Nondeforming Control Volume


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Finite Control Volume Analysis

Many practical problem in fluid mechanic require analysis of

behavior of contents of a finite region in space (control

volume)

The base of this analysis method are some fundamental

principles of physics, namely conservation of mass, Newtons

second law of motion and the law of thermodynamic.

Resultant techniques are powerful and applicable to a wide

variety of fluid mechanical circumstances that require

engineering judgment.

The finite control volume formulas are easy to interpret

physically and not difficult to use.


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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 themoments caused by fluid flow and the torque

transmitted.

Objective


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What is a System ??

A system is a collection of matter of fixed identity (alwaysthe same atom or fluid particle), which may move, flowand interact with its surroundings.


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What is a Control Volume ??

A control volume is a volume in space through whichfluid may flow.


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A system is defined as a collection of unchanging

contents, so the conservation of mass principle for asystem is simply stated as;

time rate of change of the system mass = 0or

)1.5....(....................0Dt

Dmsys

Derivation of the Continuity Equation


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Figure 5.1: System and control volume at 3 different instances of

time.(a) system and control at time t-t.(b) system and control volume at t, coincident condition.(c) system and control volume at t+t.


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Figure 5.1b shows a system and a fixed, non-deforming controlvolume that are coincident at an instant of time.

(considering conservation of mass B:mass, b = 1)

Thus, from Reynolds Transport Theorem:

)2.5.......(..........111222 VAVAtm

DtDm cvsys

Time rate of change

of the mass of thecoincident system

=

time rate of change of

the mass of the contentsof the coincident controlvolume

+

net rate of flow of mass

through the controlsurface


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)3.5.......(

to;modifiedbecan5.2Eq.

outlet,andinletmultiplehasvolumecontroltheIf

,volumecontrolinmassofamountThe

mBecause

cv

inininoutoutout

sys

cv

cv

VAVAdtDt

Dm

dm

d


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)4.5(..........0

t

mcv

When the flow is steady;

0

cvd

t

So that, for steady flow;


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Eq. 5.5 is restricted to fixed, non-deforming control volumes having

uniform properties across the inlets and outlets, with the velocitynormal to the inlet and outlet areas.

)5.5.....(0

inininoutout

cv

out VAVAdt

The control volume expression for conservative of mass,

commonly called the continuity equation, is obtained bycombining Eq.5.1 and 5.3


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For a steady, uniform flow with one entrance and one exit

If the density is constant in CV, the derivative /= 0even if the flow is unstable.

The continuity equation then reduce to

=

=


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Mass flow rate, through a section of a volume having area A is definedas;

QAVm

: fluid densityV: component of f luid velocity normal to area A

Q = VA : volume f lowrate

Note: symbols used to denote mass,m (slugs or kg), and mass flowrate,m (slugs/s or kg/s)


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A

VdA

VA

If the velocity is considered uniformly distributed (1-D flow) overthe section area, then;

VV

Often the fluid velocity across a section area A is not uniform.

Average value of the component of velocity normal to sectionarea involved:


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The cross-sectional area of the test section

of a large water tunnel is 100 ft2. For a test

velocity of 50 ft/s, what volume flow rate

capacity in minute is needed?

Exercise 1


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Example 1: conservation of mass: steady , incompresible flow

Seawater flows steadily through a simple conical-shapenozzle at the end of a fire horse. If the nozzle exit velocitymust be at least 20m/s, determine the minimum pumping

capacity required in m

3

/s.

Fixed, Nondeforming Control Volume


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Solution example 1:

21

111222

mthat withSo

0

becomes(1)Eqoutflows,oneandinflowoneisThere

)1.......(0

givetovolume

controlthisofcontentsthetoappliedis5.5Equation

mmAV

VAVA

VAVApdtinininoutoutout

cv


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12

12

1122

(4)and(3)EqFrom

)4......(..........Therefore

ible.incompressconsideredmaybe,examplein thisasspeed,lowatflowLiquid

)3.........(obtainWe

,mBecause

QQ

QQ

Q


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thenD),-(1uniformconsideredis(2),section,plane

exitnozzleat theondistributivelocitythe,simplicityforIf,exit.

nozzleat theflowratevolumethetoequaliscapacitypumpingThe

2221 AVQQ

./0251.0/1000

40

4)/20(

4

32

22 smmmm

mmsmDV


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The example problem illustrate that when the flow issteady, the time rate of change of the mass of the

contents of the control volume is zero and the net

amount of mass flow rate, through the control surface

is therefore also zero.

)9.5(..........0

zeroalsoissurfacecontrolethrough thQ,flowrate,volumeofamountnettheible,incompressalsoisflowsteadytheIf

)8.5.........(0

inout

inout

QQ

mm


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)11.5..(..........

flow,ibleincompressforand

)10.5.......(

(2),and(1)sectionsatvolumecontrolthroughflowing

fluidspecificaofstreamoneonlyinvolvingflowsteadyFor

2211

222111

VAVAQ

VAVAm


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Exercise 2

Air flows steadily between two cross sections in a

long, straight section of 0.25 m inside diameter pipe.

The static temperature and pressure at eachsection are indicated in Fig 2. If the average air

velocity at section (2) is 320 m/s, determine the

average air velocity at section (1).


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Solution exercise 2

For steady flow between section (1) and (2),

Or

Thus

2 = 1

1 =2

1

2

1

2. Eq.1

111 = 222


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Combining Eq. 1 and 2, and observing that A1 = A2

1 =2

1

1

22

= (127 kPa)/(690 kPa) x (300K/252K) x (320 m/s)= 70.1 m/s

Assuming that under the conditions of this problem, air behaves as an ideal

gas we use the ideal gas equation of state;

2

1=

2

1

1

2eq2


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When a moving control volume is used, the velocity relativeto the moving control volume (relative velocity) is animportant flow field variable.

The relative velocity, W, is the f luid velocity seen by anobserver moving with the control volume.

The control volume velocity, Vcv, is the velocity of thecontrol volume as seen from a fixed coordinate system.

Moving, Nondeforming Control Volume


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The absolute velocity, V is the f luid velocity seen by astationary observer in a fixed coordinate system.

The velocities are related to each other by the vectorequation;

V = W + VCV

V VCV

W


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The cv expression for conservation of mass (the continuityequation) for a moving, non-deforming control volume isthe same as that for stationary control volume, providedthe absolute velocity is replaced by the relative velocity.

)13.5(..........0 inininoutout

cv

out WAWAt

This equation is restricted to cv having uniform propertiesacross the inlets and outlets, with the velocity normal tothe inlet and outlet areas.


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Example 2:

Water enters a rotating lawn sprinkler through its base at thesteady rate of 1000 ml/s as sketched in Figure 5.4. If the exit

area of each of the two nozzles is 30mm2, determine theaverage speed of water leaving the nozzle, relative to thenozzle if(a) the rotary sprinkler head is stationary, (b) thesprinkler head rotates at 600 rpm and(c) the sprinkler head

accelerates from 0 to 600 rpm.


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Solution example 2:

From the equation, for steady flow:

)1.....(0

inininoutout

cv

out WAWAt

Because there is only one inflow [at the base ofrotating arm, section (1)] and two outflows [the twonozzles at the tips of the arm, section (2) and (3), eachhave the same area and fluid velocity], Eq. 1 becomes;


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smmmliterml

mmmlitermsmlW

or

A

Q

WWAAWAQ

WAWAWA

WAWAWA

/7.16)30)(2)(/1000(

)/10)(/001.0)(/1000(

2W

thatfollowsitand,With

0

becomes2Eq.,withflowibleincompressforHence,

)2........(..........0

2

2263

2

2

2

323211

113322

321

111333222


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The value of W2is independent of the speed of rotation of

the sprinkler head and represents the average velocity ofthe water exiting from each nozzle with respect to thenozzle for cases (a), (b) and (c).

The velocity of water discharging from each nozzle, whenviewed from a stationary reference (i.e., V2), will vary asthe rotation speed of the sprinkler head varies since fromEq. 5.12,

V2 = W2 U

fluid mechanics 2010-2011 finite control volume

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