Download - 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
$"
-
7/25/2019 Ch15 Differential Momentum Balance
2/20
Differential Chee 223 15.2
Reminder: Definition of stress
Tensile causes elongation Compressive causes shrinkage
%% % %& &
Stress= force per unit area (=F!"
Normal stress acts perpendicular to the surface (F=normal force).
%
% &
Shear stress acts tangentially to the surface (F=tangential force).
-
7/25/2019 Ch15 Differential Momentum Balance
3/20
Differential Chee 223 15.3
Forces acting on a differential element (#$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.
'
''
-
7/25/2019 Ch15 Differential Momentum Balance
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#& #""#" =
-
7/25/2019 Ch15 Differential Momentum Balance
5/20
Differential Chee 223 15.5
#"#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#"
$$y
$$
#
$$t
$ +
+
+
=
+
+
+
$component of momentum e%uation:
(15.)a!
(15.)*!
(15.)c!
-
7/25/2019 Ch15 Differential Momentum Balance
6/20
Differential Chee 223 15.
Stress ) Deformation relationship
!n general the stresses are linearly related to the rates of deformation"
(shear stress) = (#iscosity)x(rate of shear strain)
!n $artesian coordinates% for the &' case"
"
$2+
y
$2+
#
$2+
"""
yyy
###
+=
+=
+=
+
==
+==
+
==
"
$
#
$
y$
"
$
#
$
y
$
#"#""#
"y"yy"
y#y##y
(15.,!
y
$
#
y# =eminder" Neton*s la in one direction"
-
7/25/2019 Ch15 Differential Momentum Balance
7/20Differential Chee 223 15.)
*a+ier$Sto,es E%uations
Ta+ing into account the stress'deformation relationships (,-. /.0) and
ma+ing the folloing assumptions"
The fluid has constant density
The flo is laminar throughout
The fluid is Netonian
e obtain the Na#ier'Sto+es ,-uations"
-
7/25/2019 Ch15 Differential Momentum Balance
8/20Differential Chee 223 15.,
*a+ier$Sto,es E%uations
+
+
++
=
+
+
+
2
#
2
2
#
2
2
#
2
##
"#
y#
##
"
$
y
$
#
$g
#
+
"
$$
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
$
$component :
(15.a!
(15.*!
(15.c!
-
7/25/2019 Ch15 Differential Momentum Balance
9/20Differential Chee 223 15.
*a+ier$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
$
$component :
(15.1/a!
(15.1/*!
-
7/25/2019 Ch15 Differential Momentum Balance
10/20Differential Chee 223 15.1/
*a+ier$Sto,es E%uations
+
+
++
=
+
+
+
2
"
2
2
"
2
2
""
""
""r
"
"$$
r1
r$r
rr1g
"+
"
$$
$
r
$
r
$$t
$$component : (15.1/c!
-
7/25/2019 Ch15 Differential Momentum Balance
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.
-
7/25/2019 Ch15 Differential Momentum Balance
12/20Differential Chee 223 15.12
E&le1: 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
-
7/25/2019 Ch15 Differential Momentum Balance
13/20Differential Chee 223 15.13
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.
-
7/25/2019 Ch15 Differential Momentum Balance
14/20Differential Chee 223 15.1
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$2R
/"=
-
7/25/2019 Ch15 Differential Momentum Balance
15/20Differential Chee 223 15.15
E&le 2: -ressure dri+en (-oiseuille" flo/
0et/een parallel plates
The figure belo shos a fluid of #iscosity that flos in the x direction
beteen to rectangular plates% hose idth is #ery large in the zdirection hen compared to their separation in the y direction. Such a
situation could occur in a die hen a polymer is being extruded at the exit
into a sheet% hich is subse-uently cooled and solidified. 3e ill
determine the relationship beteen the flo rate and the pressure drop
beteen the inlet and exit% together ith se#eral other -uantities of
interest.
2h
-
7/25/2019 Ch15 Differential Momentum Balance
16/20Differential Chee 223 15.1
E&le 2: -ressure dri+en flo/ 0et/een parallel plates
No sol#e the folloing problem"
6 highly #iscous fluid ha#ing a #iscosity of
-
7/25/2019 Ch15 Differential Momentum Balance
17/20Differential Chee 223 15.1)
Summar' of some useful results
Steady
pressure dri#en%laminar flo
beteen fixed
parallel plates
2h
( )22# hy#
'
2
1$
= here
!''(
'
#
' 21=
=
='
2h$2
ma#4# 3'h$2
ae = aema#4# $
23$ =
3
'h2
0
3
=
0
$elocity 'rofile6
$olumetric flo rate6
-
7/25/2019 Ch15 Differential Momentum Balance
18/20Differential Chee 223 15.1,
Summar' of some useful results Steady% laminar% rag ($ouette) flo beteen parallel plates"
$
*y
x
0
*
y$$#=
*$2
1
0
=
$elocity rofile6
$olumetric flo rate6
-
7/25/2019 Ch15 Differential Momentum Balance
19/20Differential Chee 223 15.1
Summar' of some useful results
Steady% pressure
dri#en% laminar flo incircular tubes
( )22" Rr"
'
1$
= here
!''(
'
"
' 21=
=
(
'R
$
2
ma#4"
= ,'R
$
2
ae
= aema#$2$
=
2
ma#
"
R
r
1$
$
=
,
'R
=
$elocity 'rofile6
$olumetric flo rate6
R
"
r
'1 '2
-
7/25/2019 Ch15 Differential Momentum Balance
20/20
Differential Chee 223 15 2/
Summar' of some useful results
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
"
$"
$"