ch15 differential momentum balance

7/25/2019 Ch15 Differential Momentum Balance

1/20

Differential Chee 223 15.1

Differential Momentum Balance

Rate of

momentumin

Rate of

momentumout

Rate of

accumulationof momentum- + =

Sum of forces

acting onsystem (15.!

"

y

#

!"y($!$m( 2

1#11# =

1. Estimation of net rate of momentum out of element

2. Estimation of forces acting on the element

!"y($!$m( 2

2#22# =

!y#($!$m( 2

5"55" =

!y#($!$m( 2

"" =

$#

$y

$"


2/20


Reminder: Definition of stress

Tensile causes elongation Compressive causes shrinkage

%% % %& &

Stress= force per unit area (=F!"

Normal stress acts perpendicular to the surface (F=normal force).

%

% &

Shear stress acts tangentially to the surface (F=tangential force).


3/20


Forces acting on a differential element (#$D"

x

z

yxx

yy

zz

zxzyxz

xy

yz

yx

The first subscript indicates the direction of the normal to the plane

on hich the stress acts. The second subscript indicates the direction of the stress.

'

''


4/20

Differential Chee 223 15.

Differential Momentum Balance

y

#

"

2. Estimation of forces acting on the element

"y2## "y& 1###1## =

"#3#y

y#5#"

"#(#y

y#& #""#" =


5/20


#"#y####

"#

y#

## g

"y#"

$$

y

$$

#

$$t

$ +

+

+

=

+

+

+

E%uations of Motion

&$component of momentum e%uation:

y

"yyy#yy

"

y

y

y

#

yg

"y#"

$$

y

$$

#

$$t

$+

+

+

=

+

+

+

'$component of momentum e%uation:

"""y"#""

""

y"

#" g

"y#"

$$y

$$

#

$$t

$ +

+

+

=

+

+

+

$component of momentum e%uation:

(15.)a!

(15.)*!

(15.)c!


6/20

Differential Chee 223 15.

Stress ) Deformation relationship

!n general the stresses are linearly related to the rates of deformation"

(shear stress) = (#iscosity)x(rate of shear strain)

!n $artesian coordinates% for the &' case"

"

$2+

y

$2+

#

$2+

"""

yyy

###

+=

+=

+=

+

==

+==

+

==

"

$

#

$

y$

"

$

#

$

y

$

#"#""#

"y"yy"

y#y##y

(15.,!

y

$

#

y# =eminder" Neton*s la in one direction"


7/20Differential Chee 223 15.)

*a+ier$Sto,es E%uations

Ta+ing into account the stress'deformation relationships (,-. /.0) and

ma+ing the folloing assumptions"

The fluid has constant density

The flo is laminar throughout

The fluid is Netonian

e obtain the Na#ier'Sto+es ,-uations"


8/20Differential Chee 223 15.,


+

+

++

=

+

+

+

2

#

2

2

#

2

2

#

2

##

"#

y#

##

"

$

y

$

#

$g

#

+

"

$$

y

$$

#

$$t

$

&$component :

+

+

++=

+

+

+

2

y

2

2

y

2

2

y

2

y

y

"

y

y

y

#

y

"

$

y

$

#

$g

y

+

"

$$

y

$$

#

$$t

$

'$component :

+

+

++

=

+

+

+

2

"

2

2

"

2

2

"

2

""

""

y"

#"

"

$

y

$

#

$g

"

+

"

$$y

$$

#

$$t

$

$component :

(15.a!

(15.*!

(15.c!


9/20Differential Chee 223 15.

*a+ier$Sto,es E%uations!n cylindrical (polar) coordinates"

++ ++=

+

+

+

2r

2

22r

2

22rrr

""

2

rrr

r

"$$

r2$

r1

r$

r$r

rr1g

r+

"

$$r

$$

r

$

r

$$t

$r$component :

+

+

+

++

=

++

+

+

2

2

r

22

2

22

"r

r

"

$$

r

2$

r

1

r

$

r

$rrr

1g

+

r

1

"

$

$r

$$$

r

$

r

$

$t

$

$component :

(15.1/a!

(15.1/*!


10/20Differential Chee 223 15.1/


+

+

++

=

+

+

+

2

"

2

2

"

2

2

""

""

""r

"

"$$

r1

r$r

rr1g

"+

"

$$

$

r

$

r

$$t

$$component : (15.1/c!


11/20Differential Chee 223 15.11

Solution -rocedure

. 1a+e reasonable simplifying assumptions (i.e. steady state%

incompressible flo% coordinate direction of flo)

2. 3rite don continuity and momentum (or Na#ier'Sto+es) e-uations

and simplify them according to the assumptions of Step .

&. !ntegrate the simplified e-uations.

4. !n#o+e boundary conditions in order to e#aluate integration constants

obtained in Step &.

5 No'slip condition

5 $ontinuity of #elocity

5 $ontinuity of shear stress

/. Sol#e for pressure and #elocity. eri#e shear stress distributions if

desired. 6pply numerical #alues.



E&ample1: Drag (ouette" flo/ 0et/een

t/o parallel plates

$onsider to flat parallel plates separated by a distance b as shon in

the figure. The top plate mo#es in the x'direction at a constant speed 7%hile the bottom plate remains stationary. The fluid beteen the plates is

assumed incompressible. 6s the top plate mo#es the fluid is dragged

along. This type of flo is often referred as Couette flow. !t has important

applications in lubrication applications (such as rotating 8ournal bearings)

and instruments for measurement of #iscosity.

9ro#e that the #elocity profile for this type of flo is linear. 3hat is the

#olumetric flo rate:

$

*y

x

0



Sample or,sheet

Step 1:State assumptions

' Steady'state (all deri#ati#es ith respect to time = ;)% incompressibleflo (= const.).' ecide on coordinate system% determine direction of flo% identify non'

zero #elocity components.

' !nspect for any other reasonable assumptions.

Step 2:3rite don continuity (chose from /.'/./) and Na#ier'Sto+es

e-uations (chose from /.< or /.;) for the appropriate coordinate

system and direction of flo.

Then simplify them% according to assumptions of Step .

Step #:!ntegrate the simplified Na#ier'Sto+es e-uation.



Sample or,sheet

Step : !dentify appropriate boundary conditions. se them to determine

the integration constants obtained abo#e.

Step 3:>btain #elocity profile.

Step 4(!f needed)" >btain #olumetric flo rate by integrating"

'For flo in channels (3=idth)"

' For flo through circular cross'sections"

Step 5 (!f needed)" >btain shear stress distributions% chosing theappropriate stress'deformation relationship% from e- (/.0) and simplifyingit.

=lateto

late*ottom# y$

0

rr$2R

/"=



E&ample 2: -ressure dri+en (-oiseuille" flo/

0et/een parallel plates

The figure belo shos a fluid of #iscosity that flos in the x direction

beteen to rectangular plates% hose idth is #ery large in the zdirection hen compared to their separation in the y direction. Such a

situation could occur in a die hen a polymer is being extruded at the exit

into a sheet% hich is subse-uently cooled and solidified. 3e ill

determine the relationship beteen the flo rate and the pressure drop

beteen the inlet and exit% together ith se#eral other -uantities of

interest.

2h



E&ample 2: -ressure dri+en flo/ 0et/een parallel plates

No sol#e the folloing problem"

6 highly #iscous fluid ha#ing a #iscosity of


17/20Differential Chee 223 15.1)

Summar' of some useful results

Steady

pressure dri#en%laminar flo

beteen fixed

parallel plates

2h

( )22# hy#

'

2

1$

= here

!''(

'

#

' 21=

=

='

2h$2

ma#4# 3'h$2

ae = aema#4# $

23$ =

3

'h2

0

3

=

0

$elocity 'rofile6

$olumetric flo rate6


18/20Differential Chee 223 15.1,

Summar' of some useful results Steady% laminar% rag ($ouette) flo beteen parallel plates"

$

*y

x

0

*

y$$#=

*$2

1

0

=

$elocity rofile6





Steady% pressure

dri#en% laminar flo incircular tubes

( )22" Rr"

'

1$

= here

!''(

'

"

' 21=

=

(

'R

$

2

ma#4"

= ,'R

$

2

ae

= aema#$2$

=

2

ma#

"

R

r

1$

$

=

,

'R

=

$elocity 'rofile6


R

"

r

'1 '2


20/20

Differential Chee 223 15 2/


Steady%

9ressuredri#en% 6xial%

Caminar flo

in an 6nnulus

'1 '2

ri

ro

+

= oio

2

o

2

i2

o

2

" r

r

ln!r7rln(

rr

rr"

'

1

$

=!r7rln(

!rr(rr

,

'

io

22

i

2

o

i

o

r

"

$"

$"

ch15 differential momentum balance

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