Post on 07-Apr-2020
NS applied to special cases
. ij
DVg
Dtρ ρ σ= + ∇
�
�
Class 10
• The fluid at rest: Hydrostatics
• Newtonian fluid
• Incompressible flow
• Inviscid flow: The Euler and Bernoulli equations
Special casesThe fluid at rest: Hydrostatics
If the fluid is at rest:
• Viscous stresses vanish:
0 ( )ij for i jσ = ≠
• Normal stresses equal to the hydrostatic pressure.
pσ σ σ= = = −xx yy zz pσ σ σ= = = −
p gρ∇ =�
Hence the NS equations become
In expended form: (we take the z coordinate as up)
0,p
x
∂=
∂0,
p
y
∂=
∂
pg
zρ
∂= −
∂
Newtonian fluid
Stokes’ postulates for Newtonian fluid:
• The fluid is continuous, and its stress tensor is at most a linear
function of the strain rate.
• The fluid is isotropic, i.e., its properties are independent of direction,
and therefore the deformation law is independent of the axes in
which it is expressed.
• When the strain rate is zero, the deformation law must reduce to the
hydrostatic pressure condition.
, :
1
0
ij ij ij
ij
ij
p
if i j
if i j
σ δ δ
δ
δ
= −
= =
= =
Kronecker delta function.
Newtonian fluid...
General deformation law for a Newtonian viscous fluid (Stokes (1845)):
.ji
ij ij ij
j i
uup V
x xσ δ µ δ λ
∂∂= − + + + ∇ ∂ ∂
� �
:λ Coefficient of bulk viscosity, it is independent of This
comes only in the compressible flow.
.µcomes only in the compressible flow.
Stokes’ hypothesis:2
03
λ µ+ =
. 0V∇ =�
Incompressible flow:
Navier-Stokes Equation for a Newtonian Fluid
: 2 .x
u u u u p ux u v w g V
t x y z x x x
u v w u
y y x z x z
ρ ρ µ λ
µ µ
∂ ∂ ∂ ∂ ∂ ∂ ∂ + + + = − + + ∇ + ∂ ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂ ∂ + + + ∂ ∂ ∂ ∂ ∂ ∂
�
: y
v v v v p v uy u v w g
t x y z y x x yρ ρ µ
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = − + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
2 .
yt x y z y x x y
v v wV
y y z z yµ λ µ
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂+ ∇ + +
∂ ∂ ∂ ∂ ∂
�
:
2 .
z
w w w w p w uz u v w g
t x y z z x x z
v w wV
y z y z z
ρ ρ µ
µ µ λ
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + + + = − + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂ + + + ∇
∂ ∂ ∂ ∂ ∂
�
For incompressible flow if the viscosity is constant, the NS equations reduce to:
2DVg p V
Dtρ ρ µ= − ∇ + ∇
�
�
�
But in general viscosity could be a function of temperature and pressure.
Inviscid flow ( )0µ =
The NS equations reduce to the Euler’s equation: DV
g pDt
ρ ρ= − ∇
�
�
Euler’s equation for unidirectional flow: ρ ρ∂ ∂ ∂
+ = − ∂ ∂ ∂
�w w pw g
t z z
Further for steady flow:
( )
w pw g
z zρ ρ
∂ ∂ = − −
∂ ∂
( )2
1
2
w pg
z zρ ρ
∂ ∂⇒ = − −
∂ ∂Integrating w.r.t. z (we take the z coordinate as up) along a streamline
between points 1 and 2, we obtain
2 2
1 1 1 2 2 2
1 1
2 2p w gz p w gzρ ρ ρ ρ+ + = + +
Bernoulli’s Equation for steady, inviscid and incompressible flow.
Bernoulli’s Equation...
21
2p w gz Constρ ρ+ + =
Pressure
Energy
Kinetic
Energy
Potential
Energy
Divide by we get ,gρ2
p w2
2
p wH z Const
g gρ= + + =
Total head = Pressure+ Kinetic + Static
head head head
Bernoulli’s Equation for Real Fluid
However, all the real fluids are viscous and hence offer resistance to flow. Thus
there are always some losses in the flows.
Assumptions in deriving the Bernoulli’s equation:
• Fluid is inviscid
• Incompressible
• Flow is steady
• Flow is irrotational
The Bernoulli’s equation for real fluids between points 1 and 2 is given by
2 2
1 1 2 2
1 2 ,2 2
L
p w p wz z h
g g g gρ ρ+ + = + + +
where hL is the loss of energy between points 1 and 2.
Exact solution of the NS equations
Objective: To find out the exact solution of the NS equations for some
laminar, Newtonian viscous flows.
Contents:
• Conditions at the solid boundary
• Laminar flow velocity profile• Laminar flow velocity profileI. Steady pressure-driven flow through two parallel plates (Channel
flow).
II. Steady wall-driven flow through two parallel plates (Couette flow)
III. Steady pressure-driven flow through circular pipe (Poiseuille flow).
Conditions at a solid boundary
Normal component of velocity: Obviously the normal component of velocity must be zero! This is called no-
penetration condition.
Tangential component of velocity: Most fluids ‘stick’ to boundary. The tangential component of velocity is same
as that of the boundary. If the boundary is at rest, this is zero. This is called
no-slip condition.no-slip condition.
Some complex fluids (suspensions, pastes, etc.) show apparent slip at the wall.
Shear stress at the wall From Newton’s law of viscosity:
0y
du
dyτ µ
=
=
y
u(y)
u=0
Pressure-driven flow through two parallel plates (Channel flow/Poiseuille flow)
2 2
2 2
2 2
2 2
u u u p u uu v
t x y x x y
v v v p v vu v g
t x y y x y
ρ µ
ρ µ ρ
∂ ∂ ∂ ∂ ∂ ∂+ + = − + + ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂ ∂+ + = − + + − ∂ ∂ ∂ ∂ ∂ ∂
Assumption: wallτAssumption:
• Steady
• Incompressible
• Neglecting the gravity
• Fully developed flow, i.e., v=0, u=u(y)
2
2,
0 ( ).
p u
x y
pp p x
y
µ ∂ ∂
= ∂ ∂
∂= ⇒ =
∂
x
y
y=hwallτ
Hence,2
2
dp d u
dx dyµ=
Integrating the above equation w.r.t y, we obtain
1(1)
du dpy c
dy dxµ= +
Integrating again, we obtain21
(2)2
dpu y cy d
dxµ= + +
2 dxµ
The constants are obtained using the no-slip boundary conditions at the walls
0, 0,
0, .
u at y
u at y h
= =
= =
1,
2
0
dpc h
dx
d
µ= −
=Substituting the values of these constants,
we get( )21
2
1
2
dpu y hy
dx
du dp hy
dy dx
µ
µ
= −
= −
Velocity parabolic for this case.
Ratio of maximum velocity to average velocity:
Velocity is maximum, when centerline of the channel.0 ,2
du hy
dy= ⇒ =
2
max
1
8
dpu h
dxµ= −
Volume flow rate, Q, is given by
31
12
hdp
Q udy hdxµ
= = −∫0
12 dxµ∫
Thus, average velocity,3
2
1
112
1 12avg
dph
Q dpdxu h
A h dx
µ
µ
−
= = = −×
max 3
2avg
u
u=Hence,
Drop of pressure head for a given length:
2
2 1 2 12
12
. . . ,
12( )
avg
avg
udp
dx h
Integrating w r t x we get
up p x x
h
µ
µ
µ
= −
− = − −
1 2 2 12
1 2
2
12( )
,
12
avg
f
avg
f
u Lp p where L x x
h
If h is the drop of pressure head then
u Lp ph
g gh
µ
µ
ρ ρ
⇒ − = = −
−= =
21
Shear stress distribution:
1( 2 )
2
1
2wall
du dph y
dy dx
dph
dx
τ µ
τ
= = − −
⇒ = −
See how we are getting this expression at both the walls??
2 2
2 2
2 2
2 2
u u u p u uu v
t x y x x y
v v v p v vu v g
t x y y x y
ρ µ
ρ µ ρ
∂ ∂ ∂ ∂ ∂ ∂+ + = − + + ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂ ∂+ + = − + + − ∂ ∂ ∂ ∂ ∂ ∂
Assumption: U
Wall-driven flow between two parallel plates (Couette flow)
Assumption: • Steady
• Incompressible
• Neglecting the gravity
• Fully developed flow, i.e., v=0, u=u(y)
• No pressure gradient2
20
u
yµ ∂
= ∂
x
y
y=h0
U
duc
dy=
u cy d= + 0
0, 0,
, .
u at y
u U at y h
= =
= =
Boundary conditions
0 , 0U
c dh
= =
0
0
0
Uu y
h
Udu
dy h
Udu
dy hτ µ µ
=
=
= =dy h
Pressure-driven flow through circular pipe (Hagen-Poiseuille flow)
Assumption: • Steady
• Incompressible
• Neglecting the gravity
• Fully developed flow, i.e., 0, 0, ( )r z zu u u u rφ= = =
1 1zd du dpr
r dr dr dzµ
=
2
2
. . . ,
1
2
. . . ,
1ln
4
z
z
rr dr dr dz
Integrating w r t r weobtain
du dp rr c
dr dz
Integrating w r t r weobtain
dpu r c r d
dz
µ
µ
µ
=
= +
= + +
0, 0,
0, .
z
z
u at r
duat r R
dr
= =
= =
Boundary conditions
2 21( )
4
1
2
1
2
z
z
z
dpu R r
dz
du dpr
dr dz
du dpr
dr dz
µ
µ
τ µ
= − −
=
= − = −
R
Volume flow rate,
0
2
R
zQ u rdrπ= ∫
max( )
2( )
z
z avg
u
u=
( )
( )
2
max
2
1
4
1
8
z
z avg
dpu R
dz
dpu R
dz
µ
µ
= −
= −
( )2
1 2
2
8
32
z avg
avg
f
udp
dz R
u Lp ph
g gD
µ
µ
ρ ρ
= −
−= =
Hagen-Poiseuille Formula
Navier - Stokes equation:
We consider an incompressible , isothermal Newtonian flow (density ρ =const, viscosity μ =const), with a velocity field ))()()(( x,y,z, w x,y,z, vx,y,zuV =
r
Incompressible continuity equation:
0=∂∂
+∂∂
+∂∂
zw
yv
xu eq1.
Navier - Stokes equation:
vector form: VgPDt
VD rrr
2∇++−∇= μρρ
x component:
)( 2
2
2
2
2
2
zu
yu
xug
xP
zuw
yuv
xuu
tu
x ∂∂
+∂∂
+∂∂
++∂∂
−=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂∂ μρρ eq2.
y component:
)( 2
2
2
2
2
2
zv
yv
xvg
yP
zvw
yvv
xvu
tv
y ∂∂
+∂∂
+∂∂
++∂∂
−=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂∂ μρρ eq3.
z component:
)( 2
2
2
2
2
2
zw
yw
xwg
zP
zww
ywv
xwu
tw
z ∂∂
+∂∂
+∂∂
++∂∂
−=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂∂ μρρ eq4.
Cylindrical coordinates ),,( zr θ : We consider an incompressible , isothermal Newtonian flow (density ρ =const, viscosity μ =const), with a velocity field .uuuV zr ),,( θ=
r
Incompressible continuity equation:
0)(1)(1
=∂∂
+∂
∂+
∂∂
zuu
rrru
rzr
θθ eq a)
r-component:
⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
−∂∂
+−⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
++∂∂
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+−∂∂
+∂∂
+∂∂
2
2
22
2
22
2
211zuu
ru
rru
rur
rrg
rP
zuu
ruu
ru
ruu
tu
rrrrr
rz
rrr
r
θθμρ
θρ
θ
θθ
eq b)
θ -component:
⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
+∂∂
+−⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
++∂∂
−=
⎟⎠⎞
⎜⎝⎛
∂∂
++∂∂
+∂∂
+∂∂
2
2
22
2
22
2111zuu
ru
rru
ru
rrr
gPr
zu
uruuu
ru
ru
ut
u
r
zr
r
θθθθθ
θθθθθθ
θθμρ
θ
θρ
eq c)
z-component:
⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
++∂∂
−=
⎟⎠⎞
⎜⎝⎛
∂∂
+∂∂
+∂∂
+∂∂
2
2
2
2
2
11zuu
rrur
rrg
zP
zuuu
ru
ruu
tu
zzzz
zz
zzr
z
θμρ
θρ θ
eq d)