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Lecture 5
Equation of motions for a liquid. Inviscid and Viscous flowNavier-Stockes equation
Differential analysis of Fluid Flow
• The aim: to produce differential equation describing the motion of fluid in detail
Description of forces
Forces
Body forces – distributed through the element, e.g. Gravity
b mδ δ=F g
Surface forces – result of interaction with the surrounding elements: e.g. Stress
Linear forces: Surface tension
Normal stress
Shearing stresses
Stress acting on a fluidic element
• normal stress0
lim nn A
FAδ
δσδ→
=
• shearing stresses
1 21 20 0
lim limA A
F FA Aδ δ
δ δτ τδ δ→ →
= =
Stresses: double subscript notation• normal stress: xxσ
• shearing stress: xy xzτ τ
normal to the plane
direction of stress
sign convention: positive stress is directed in positive axis directions if surface normal is pointing in the positive direction
Stress tensor
11 12 13
21 22 23
31 32 33
σ τ ττ τ σ τ
τ τ σ
⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
• To define stress at a point we need to define “stress vector”for all 3 perpendicular planes passing through the point
zyxzyx
F zxyxxxx δδδττσδ )(
∂∂
+∂∂
+∂∂
=
Force on a fluid element• To find force in each direction we need to sum all forces
(normal and shearing) acting in the same direction
)()(zuw
yuv
xuu
tu
zyxg zxyxxx
x ∂∂+
∂∂+
∂∂+
∂∂=
∂∂
+∂∂
+∂∂
+ ρττσρ
aF mδδ =
Acceleration (“material derivative”)
)()(zvw
yvv
xvu
tv
zxyg zyxyyy
y ∂∂+
∂∂+
∂∂+
∂∂=
∂∂
+∂∂
+∂∂
+ ρττσ
ρ
)()(zww
ywv
xwu
tw
xyzg xzyzzz
z ∂∂+
∂∂+
∂∂+
∂∂=
∂∂
+∂∂
+∂∂+ ρττσρ
Differential equation of motion
,x x
y y
z z
F maF ma m x y z
F ma
δ δδ δ δ δ δ δ
δ δ
=
= =
=
Inviscid flow
• no shearing stress in inviscid flow, so xx yy zzp σ σ σ− = = =
• equation of motion is reduced to Euler equations
( )
( )
( )
xxx
yyy
zzz
u u u ug u v wx t x y z
v v v vg u v wy t x y z
w w w wg u v wx t x y z
σρ ρ
σρ ρ
σρ ρ
∂ ∂ ∂ ∂ ∂− = + + +
∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂
− = + + +∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂
− = + + +∂ ∂ ∂ ∂ ∂
(pt
ρ ρ ∂⎡ ⎤− = + ⋅⎢ ⎥∂⎣ ⎦Vg V V∇ ∇)
Bernoulli equation• let’s write Euler equation for a steady flow along a streamline
(pρ ρ− = ⋅g V V∇ ∇)
g z= − ∇g 12( ) ( ) ( )⋅ = ⋅ − × ×V V V V V V∇ ∇ ∇
( ) ( )2
z p ρρ ρ− − = ⋅ − × ×g V V V V∇ ∇ ∇ ∇
• now we multiply it by ds along the streamline
( ) ( )2
z d p d d dρρ ρ− ⋅ − ⋅ = ⋅ ⋅ − × × ⋅g s s V V s V V s∇ ∇ ∇ ∇
221 ( ) 0
2 2dp p Vd V gdz or gz constρ ρ+ + = + + =
Irrotational Flow• Analysis of inviscide flow can be further simplified if we
assume if the flow is irrotational:1 02
; ;
zv ux y
v u w v u wx y y z z x
ω⎛ ⎞∂ ∂
= − =⎜ ⎟∂ ∂⎝ ⎠∂ ∂ ∂ ∂ ∂ ∂
= = =∂ ∂ ∂ ∂ ∂ ∂
• Example: uniform flow in x-direction:
Bernoulli equation for irrotational flow
• Thus, Bernoulli equation can be applied between any two points in the flow field
( ) ( )2
z p ρρ ρ− − = ⋅ − × ×g V V V V∇ ∇ ∇ ∇
always =0, not only along a stream line
2
2p V gz constρ+ + =
Velocity potential• equations for irrotational flow will be satisfied automatically if
we introduce a scalar function called velocity potential such that:
u v wx y zφ φ φ∂ ∂ ∂
= = =∂ ∂ ∂
φ=V ∇
• As for incompressible flow conservation of mass leads to:20, 0φ⋅ = =∇ V ∇
Laplace equation2 2 2
2 2 2 0x y zφ φ φ∂ ∂ ∂+ + =
∂ ∂ ∂
• This type of flow is called potential flow
• In cylindrical coordinates:
2 2
2 2 2
1 1 1 0r z rr r z r r r r zφ φ φ φ φ φφ θ
θ θ∂ ∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞= + + + + =⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠
∇
Some basic potential flows
• As Laplace equation is a linear one, the solutions can be added to each other producing another solution;
• stream lines (y=const) and equipotential lines (f=const) are mutually perpendicular
along streanline
along const
dy vdx u
dy ud dx dy udx vdyx y dx vφ
φ φφ=
=
∂ ∂= + = + ⇒ = −∂ ∂
Both f and y satisfy Laplace’s equation
u vy x y y x x
ψ ψ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞= ⇒ = −⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠
Uniform flow• constant velocity, all stream lines are straight and
parallel0
0
Ux y
Ux
Uy x
Uy
φ φ
φψ ψ
ψ
∂ ∂= =
∂ ∂=
∂ ∂= =
∂ ∂=
( cos sin )( cos sin )
U x yU y x
φ α αψ α α= += −
Source and Sink
• Let’s consider fluid flowing radially outward from a line through the origin perpendicular to x-y planefrom mass conservation:
(2 ) rr v mπ =
1 02m
r r rφ φ
π θ∂ ∂
= =∂ ∂
1 02m
r r rψ ψθ π
∂ ∂= =
∂ ∂
ln2m rφπ
=
2mψ θπ
=
singularity point
Vortex• now we consider situation when the
stream lines are concentric circles i.e. we interchange potential and stream functions:
lnK
K rφ θψ== −
01
rvKv
r r rθφ ψθ
=∂ ∂
= = − =∂ ∂
• circulation
0C C C
ds ds dφ φΓ = ⋅ = ∇ ⋅ = =∫ ∫ ∫V
• in case of vortex the circulation is zero along any contour except ones enclosing origin 2
0
( ) 2
ln2 2
K rd Kr
r
π
θ π
φ θ ψπ π
Γ = =
Γ Γ= = −
∫
for potential flow
Shape of a free vortex
2φ θ
πΓ
=
2
2p V gz constρ+ + =
2 21 2
2 2V V z
g g= +
at the free surface p=0:
2
2 28z
r gπΓ
= −
Doublet• let’s consider the equal strength, source-sink pair:
1 2( )2mψ θ θπ
= − −
12 2
2 sintan ( )2m ar
r aθψ
π−= −
−
if the source and sink are close to each other:
sin
cos
Kr
Kr
θψ
θφ
= −
=
K – strength of a doublet
Summary
Superposition of basic flows• basic potential flows can be combined to form new
potentials and stream functions. This technique is called the method of superposition
• superposition of source and uniform flow
sin cos ln2 2m mUr Ur rψ θ θ φ θπ π
= + = +
Superposition of basic flows
• Streamlines created by injecting dye in steadily flowing water show a uniform flow. Source flow is created by injecting water through a small hole. It is observed that for this combination the streamline passing through the stagnation point could be replaced by a solid boundary which resembles a streamlined body in a uniform flow. The body is open at the downstream end and is thus called a halfbody.
Rankine Ovals
• a closed body can be modeled as a combination of a uniform flow and source and a sink of equal strength
1 2 1 2sin ( ) cos (ln ln )2 2m mUr Ur r rψ θ θ θ φ θπ π
= − − = − −
Flow around circular cylinder
• when the distance between source and sink approaches 0, shape of Rankine oval approaches a circular shape
sinsin
coscos
KUrr
KUrr
θψ θ
θφ θ
= −
= +
Potential flows• Flow fields for which an incompressible
fluid is assumed to be frictionless and the motion to be irrotational are commonly referred to as potential flows.
• Paradoxically, potential flows can be simulated by a slowly moving, viscous flow between closely spaced parallel plates. For such a system, dye injected upstream reveals an approximate potential flow pattern around a streamlined airfoil shape. Similarly, the potential flow pattern around a bluff body is shown. Even at the rear of the bluff body the streamlines closely follow the body shape. Generally, however, the flow would separate at the rear of the body, an important phenomenon not accounted for with potential theory.
Viscous Flow
dzdwp
dydvp
dxdup
zz
yy
xx
µσ
µσ
µσ
2
2
2
+−=
+−=
+−=
)(
)(
)(
dxdw
dzdu
dydw
dzdv
dxdv
dydu
xzzx
zyyz
yxxy
+==
+==
+==
µττ
µττ
µττ
for viscous flow normal stresses are not necessary the same in all directions
Navier-Stokes Equations
0
)()(
)()(
)()(
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
=∂∂+
∂∂+
∂∂
∂∂
+∂∂
+∂∂
++∂∂
−=∂∂
+∂∂
+∂∂
+∂∂
∂∂
+∂∂
+∂∂
++∂∂
−=∂∂
+∂∂
+∂∂
+∂∂
∂∂
+∂∂
+∂∂
++∂∂
−=∂∂
+∂∂
+∂∂
+∂∂
zw
yv
xu
zw
yw
xwg
zp
zww
ywv
xwu
tw
zv
yv
xvg
yp
zvw
yvv
xvu
tv
zu
yu
xug
xp
zuw
yuv
xuu
tu
z
y
x
µρρ
µρρ
µρρ
• 4 equations for 4 unknowns (u,v,w,p)• Analytical solution are known for only few cases
Steady Laminar Flow between parallel plates
zp
gyp
yu
xp
xuwv
∂∂
−=
−∂∂
−=
∂∂
+∂∂
−=
=∂∂
⇒==
0
0
)(0
00;0
2
2
ρ
µ
0
)()(
)()(
)()(
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
=∂∂
+∂∂
+∂∂
∂∂
+∂∂
+∂∂
++∂∂
−=∂∂
+∂∂
+∂∂
+∂∂
∂∂
+∂∂
+∂∂
++∂∂
−=∂∂
+∂∂
+∂∂
+∂∂
∂∂
+∂∂
+∂∂
++∂∂
−=∂∂
+∂∂
+∂∂
+∂∂
zw
yv
xu
zw
yw
xwg
zp
zww
ywv
xwu
tw
zv
yv
xvg
yp
zvw
yvv
xvu
tv
zu
yu
xug
xp
zuw
yuv
xuu
tu
z
y
x
µρρ
µρρ
µρρ
zp
gyp
yu
xp
xuwv
∂∂
−=
−∂∂
−=
∂∂
+∂∂
−=
=∂∂
⇒==
0
0
)(0
00;0
2
2
ρ
µ
)(211
00;0
212
1
xfgyp
CyCyxpuCy
xp
dydu
xuwv
+−=
++∂∂
=⇒+∂∂
=
=∂∂
⇒==
ρµµ
Boundary condition (no slip) 0)()( =−= huhu
)(21 22 hy
xpu −∂∂
=µ
Velocity profile
Flow rate )(32)(
21 3
22
xphdyhy
xpudyq
h
h
h
h ∂∂−=−
∂∂== ∫∫
−
−− µµ
What is maximum velocity (umax) and average velocity?
No-slip boundary condition
• Boundary conditions are needed to solve the differential equations governing fluid motion. One condition is that any viscous fluid sticks to any solid surface that it touches.
• Clearly a very viscous fluid sticks to a solid surface as illustrated by pulling a knife out of a jar of honey. The honey can be removed from the jar because it sticks to the knife. This no-slip boundary condition is equally valid for small viscosity fluids. Water flowing past the same knife also sticks to it. This is shown by the fact that the dye on the knife surface remains there as the water flows past the knife.
Couette flow
)(211
00;0
212
1
xfgyp
CyCyxpuCy
xp
dydu
xuwv
+−=
++∂∂
=⇒+∂∂
=
=∂∂
⇒==
ρµµ
Boundary condition (no slip) Ubuu == )(;0)0(
Please find velocity profile and flow rate
Boundary condition (no slip) Ubuu == )(;0)0(
)(21 2 byy
xp
byUu −
∂∂
+=µ
Velocity profile )1)((2
2
by
by
xp
Ub
by
Uu
−∂∂
−=µ
Dimensionless parameter P
Hagen-Poiseuille flow
4
2
;8
8
R pQl
R pvl
πµ
µ
∆=
∆=
Poiseuille law:
Laminar flow• The velocity distribution is
parabolic for steady, laminar flow in circular tubes. A filament of dye is placed across a circular tube containing a very viscous liquid which is initially at rest. With the opening of a valve at the bottom of the tube the liquid starts to flow, and the parabolic velocity distribution is revealed. Although the flow is actually unsteady, it is quasi-steady since it is only slowly changing. Thus, at any instant in time the velocity distribution corresponds to the characteristic steady-flow parabolic distribution.
Problems
• 6.8 An incompressible viscous fluid is placed between two large parallel plates. The bottom plate is fixed and the top moves with the velocity U. Determine: – volumetric dilation rate; – rotation vector; – vorticity; – rate of angular deformation.
yu Ub
=
• 6.74 Oil SAE30 at 15.6C steadily flows between fixed horizontal parallel plates. The pressure drop per unit length is 20kPa/m and the distance between the plates is 4mm, the flow is laminar.Determine the volume rate of flow per unit width; magnitude and direction of the shearing stress on the bottom plate; velocity along the centerline of the channel