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Unformatted text preview: FUNDAMENTALS OF
FLUID MECHANICS
Chapter 6 Flow Analysis
Using Differential Methods
JyhCherng Shieh
Department of BioIndustrial Mechatronics Engineering
National Taiwan University
12/24/2007
1 MAIN TOPICS
Fluid Element Kinematics
Fluid
Conservation of Mass
Conservation
Conservation of Linear Momentum
Conservation
Inviscid Flow
Inviscid
Some Basic, Plane Potential Flow
Some 2 Motion of a Fluid Element
Fluid Translation: The element moves from one point to another.
Fluid
Fluid Rotation: The element rotates about any or all of the x,y,z axes.
Fluid
Fluid Deformation:
Fluid
Angular Deformation:The element’s angles between the sides
Angular
change.
Linear Deformation:The element’s sides stretch or contract.
Linear 3 velocity
Fluid Translation velocity and acceleration The velocity of a fluid particle can be expressed
The
r
r
r
r
v
V = V ( x , y, z , t ) = u i + v j + wk Velocity field The total acceleration of the particle is given by
The r
r
r
r
r
r DV ∂V ∂V dx ∂V dy ∂V dz
a=
=
+
+
+
Dt
∂t ∂x dt ∂y dt ∂z dt
dx
dy
dz
= u,
= v, = w
dt
dt
dt
r
r
r
r
r
r DV ∂V
∂V
∂V
∂V
+v
+w
⇒a=
=
+u
Dt r ∂t
∂x
∂y
∂z Acceleration field r DV
a=
is called the material , or substantial derivative.
Dt 4 Physical Significance
r
r
r
r
r
r DV
∂V
∂V
∂V ∂V
a=
+v
+w
+
=u
Dt
∂x
∂y
∂z
∂t
Total
Acceleration
Of a particle Local
Acceleration
Convective
Acceleration r
r
r ∂V
v
r DV
a=
= (V ⋅ ∇)V +
Dt
∂t 5 Scalar Component
∂u
∂u
∂u
∂u
+u
+v
+w
∂z
∂t
∂x
∂y
∂v
∂v
∂v
∂v
ay =
+u
+v
+w
∂t
∂x
∂y
∂z
∂w
∂w
∂w
∂w
az =
+u
+v
+w
∂t
∂x
∂y
∂z ax = Rectangular
coordinates system ∂Vr
∂Vr Vθ ∂Vr V 2 θ
∂V
ar =
+ Vr
+
−
+ Vz r
∂t
∂r
r ∂θ
r
∂z
∂Vθ
∂Vθ Vθ ∂Vθ Vr Vθ
∂Vθ
aθ =
+ Vr
+
+
+ Vz
∂t
∂r
r ∂θ
r
∂z
∂Vz Vθ ∂Vz
∂Vz
∂Vz
az =
+ Vr
+
+ Vz
∂r
∂t
r ∂θ
∂z Cylindrical
coordinates system
6 Linear Translation
All points in the element have
All
the same velocity (which is only
true if there are o velocity
gradients), then the element will
simply translate from one
position to another. 7 1/2
Linear Deformation 1/2 The shape of the fluid element, described by the angles at
The
its vertices, remains unchanged, since all right angles
continue to be right angles.
A change in the x dimension requires a nonzero value of
change ∂u / ∂x
A ……………… y ∂ v / ∂ y
………………
A ……………… z ∂ w / ∂ z
……………… 8 2/2
Linear Deformation 2/2 The change in length of the sides may produce change in
The
volume of the element.
The change in ⎛ ∂u ⎞
δV = ⎜ δx ⎟( δyδz )( δt )
⎝ ∂x ⎠ The rate at which the δV is changing per
unit volume due to gradient ∂u/ ∂x
1 d (δV ) ∂u
=
δV dt
∂x If ∂v/ ∂y and ∂w/ ∂z are involved Volumetric dilatation rate r
1 d (δV ) ∂u ∂v ∂w
=
+
+
= ∇⋅V
δV dt
∂x ∂y ∂z 9 1/4
Angular Rotation 1/4
δα
ωOA = lim
The angular velocity of line OA
δt → 0 δ t
⎛ ∂v ⎞
⎜ ⎟δxδt
∂v
For small angles tan δα = δα = ⎝ ∂x ⎠
=
δt
&
δx
∂x ωOA ∂v
=
CCW
∂x ωOB ∂u
=
∂y CW
“” for CCW 10 2/4
Angular Rotation 2/4 The rotation of the element about the zaxis is defined as the
average of the angular velocities ωOA and ωOB of the two
mutually perpendicular lines OA and OB.
1
1 ⎛ ∂v ∂u ⎞
ωz = (ωOA + ωOB ) = ⎜
⎟
⎜ ∂x − ∂y ⎟
2
2⎝
⎠
1 ⎛ ∂w ∂v ⎞
ωx = ⎜
⎜ ∂y − ∂z ⎟
⎟
2⎝
⎠
In vector form 1 ⎛ ∂u ∂w ⎞
ωy = ⎜ − ⎟
2 ⎝ ∂z ∂x ⎠ r
v
v
r
ω = ωx i + ω y j + ωz k
11 3/4
Angular Rotation 3/4
1 ⎛ ∂u ∂w ⎞
1 ⎛ ∂w ∂v ⎞
1 ⎛ ∂v ∂u ⎞
⎟
ωy = ⎜
−
ωx = ⎜
−
⎟
ωz = ⎜ − ⎟
⎜ ∂y ∂z ⎟
2 ⎝ ∂z ∂x ⎠
2⎝
2 ⎜ ∂x ∂y ⎟
⎠
⎝
⎠
r 1 ⎡ ⎛ ∂w ∂v ⎞ r ⎛ ∂u ∂w ⎞ r ⎛ ∂v ∂u ⎞ r ⎤
ω = ⎢⎜
⎜ ∂y − ∂z ⎟ i + ⎜ ∂z − ∂x ⎟ j + ⎜ ∂x − ∂y ⎟ k ⎥
⎜
⎟
⎟
2 ⎢⎝
⎝
⎠
⎝
⎠
⎠
⎣
⎦
r1
r 1 ⎡ ∂w ∂v ⎤ r 1 ⎡ ∂u ∂w ⎤ r 1 ⎡ ∂v ∂u ⎤ r
r1
ω = curlV = ∇ × V = ⎢
− ⎥i + ⎢ −
⎥ j + 2 ⎢ ∂x − ∂y ⎥ k
2
2
2 ⎣ ∂y ∂z ⎦
2 ⎣ ∂z ∂x ⎦
⎣
⎦ Defining vorticity r
r
ζ = 2ω = ∇ × V r
Defining irrotation ∇ × V = 0
12 4/4
Angular Rotation 4/4
r
i
r1
r 1∂
r1
ω = curlV = ∇ × V =
2
2
2 ∂x
u r
j
∂
∂y
v r
k
∂
∂z
w 1 ⎡ ∂w ∂v ⎤ r 1 ⎡ ∂u ∂w ⎤ r 1 ⎡ ∂v ∂u ⎤ r
=⎢
− ⎥i + ⎢ −
⎥ j + 2 ⎢ ∂x − ∂y ⎥ k
2 ⎣ ∂y ∂z ⎦
2 ⎣ ∂z ∂x ⎦
⎣
⎦ 13 Vorticity
Defining Vorticity ζ which is a measurement of the rotation of a
Defining
fluid element as it moves in the flow field: r
r
r
r
ζ = 2 ω = curl V = ∇ × V r
r 1 ⎡ ⎛ ∂w ∂v ⎞ r ⎛ ∂u ∂w ⎞ r ⎛ ∂v ∂u ⎞ r ⎤ 1
ω = ⎢⎜
⎜
⎟
⎜ ∂y − ∂z ⎟ i + ⎜ ∂z − ∂x ⎟ j + ⎜ ∂x − ∂y ⎟ k ⎥ = 2 ∇ × V
⎟
2 ⎢⎝
⎝
⎠
⎝
⎠⎥
⎠
⎣
⎦ In cylindrical coordinates system:
In
r r ⎛ 1 ∂ Vz ∂ Vθ ⎞ r ⎛ ∂ Vr ∂ Vz ⎞ r ⎛ 1 ∂ rVθ 1 ∂ Vr ⎞
−
−
−
∇ × V = er ⎜
⎟
⎟ + eθ ⎜
⎟ + ez ⎜
∂r ⎠
∂z ⎠
r ∂θ ⎠
⎝ ∂z
⎝ r ∂r
⎝ r ∂θ 14 1/2
Angular Deformation 1/2
Angular deformation of a particle is given by the sum of the two
Angular
angular deformation
δγ = δα + δβ
⎛
∂u ⎞
∂u
δξ = ⎜ u +
δy ⎟ δt − uδt =
δ yδ t
⎜
⎟
∂y ⎠
∂y
⎝ δα = δη / δ x δβ = δξ / δ y ∂v ⎞
∂v
⎛
δxδt
δη = ⎜ v +
δx ⎟ δt − vδt =
∂x
∂x ⎠
⎝
ξ（Xi）η（Eta） Rate of shearing strain or the rate of angular deformation
⎛ ∂v δx
∂u δy ⎞
δt +
δt ⎟
⎜
∂x δx
∂y δy ⎟
δα + δβ
⎝
⎠ = ... = ⎛ ∂ v + ∂ u ⎞
&
γ = lim
= lim
⎜
⎜ ∂x ∂y ⎟
⎟
δt → 0
δt → 0
δt
δt
⎝
⎠ 15 2/2
Angular Deformation 2/2
The rate of angular deformation in xy plane
⎛ ∂v ∂u ⎞
⎜
⎜ ∂x + ∂y ⎟
⎟
⎝
⎠ The rate of angular deformation in yz plane
⎛ ∂w ∂v ⎞
⎜
⎜ ∂y + ∂z ⎟
⎟
⎝
⎠ The rate of angular deformation in zx plane
⎛ ∂w ∂u ⎞
+
⎜
⎟
∂x ∂z ⎠
⎝ 16 Example 6.1 Vorticity
For a certain twodimensional flow field th evelocity is given by
For
r
r
r
2
2
V = 4 xy i + 2 ( x − y ) j Is this flow irrotational? 17 Example 6.1 Solution
u = 4 xy v = x2 − y2 1 ⎛ ∂w ∂v ⎞
⎜
⎜ ∂y − ∂z ⎟ = 0
⎟
2⎝
⎠
1 ⎛ ∂u ∂w ⎞
ωy = ⎜
−
⎟=0
2 ⎝ ∂z ∂x ⎠ w=0 ωx = This flow is irrotational 1 ⎛ ∂v ∂u ⎞
ωz = ⎜ − ⎟ = 0
2 ⎜ ∂x ∂y ⎟
⎝
⎠ 18 1/5
Conservation of Mass 1/5 With field representation, the property fields are defined
With
by continuous functions of the space coordinates and time.
To derive the differential equation for conservation of
To
mass in rectangular and in cylindrical coordinate system.
The derivation is carried out by applying conservation of
The
mass to a differential control volume.
With the control volume representation of the conservation of mass vr
∂
∫CV d V + ∫CS V ⋅ n dA = 0
∂t
The differential form of continuity equation???
19 2/5
Conservation of Mass 2/5
The CV chosen is an infinitesimal cube with sides of length δx, δ y,
The
and δ z. ∂
∂ρ
∫ CV ρdV = ∂t δxδyδz
∂t
∂ (ρu ) δx
ρu  ⎛ dx ⎞ = ρu +
x +⎜ ⎟
∂x 2
⎝2⎠
Net rate of mass
Outflow in xdirection ρu  = ρu −
⎛ δx ⎞ x −⎜ ⎟
⎝2⎠ ∂ (ρu ) δx
∂x 2 20 3/5
Conservation of Mass 3/5
Net rate of mass
Outflow in xdirection ∂ (ρu ) δx ⎤
∂ (ρu ) δx ⎤
∂ (ρu )
⎡
⎡
= ⎢ρu +
⎥δyδz − ⎢ρu − ∂x 2 ⎥δyδz = ∂x δxδyδz
∂x 2 ⎦
⎣
⎦
⎣ Net rate of mass
Outflow in ydirection
Net rate of mass
Outflow in zdirection ∂ (ρv )
δxδyδz
= ⋅⋅⋅⋅⋅ =
∂y
∂ (ρw )
= ⋅⋅⋅⋅⋅ =
δxδyδz
∂z 21 4/5
Conservation of Mass 4/5
Net rate of mass
Outflow ⎡ ∂ (ρu ) ∂ (ρv ) ∂ (ρw ) ⎤
⎢ ∂x + ∂y + ∂z ⎥δxδyδz
⎣
⎦ The differential equation for conservation of mass
r
∂ρ ∂ (ρu ) ∂ (ρv ) ∂ (ρw ) ∂ρ
+
+
+
=
+ ∇ ⋅ ρV = 0
Continuity equation
∂t
∂x
∂y
∂z
∂t 22 5/5
Conservation of Mass 5/5 Incompressible fluid
Incompressible r
∂u ∂v ∂w
+
+
= ∇⋅V = 0
∂x ∂y ∂z
Steady flow
Steady
r
∂ (ρu ) ∂ (ρv ) ∂ (ρw )
+
+
= ∇ ⋅ ρV = 0
∂x
∂y
∂z 23 Example 6.2 Continuity Equation
The velocity components for a certain incompressible, steady flow
The
field are
u = x 2 + y2 + z2
v = xy + yz + z
w=? Determine the form of the z component, w, required to satisfy the
continuity equation. 24 Example 6.2 Solution
The continuity equation ∂u ∂v ∂w
+
=0
+
∂x ∂y ∂z ∂u
= 2z
∂x
∂v
=x+z
∂y
∂w
= − 2 x − ( x + z ) = − 3x − z
∂z
z2
⇒ w = − 3xz −
+ f (x, y)
2
25 Conservation of Mass
1/3
Cylindrical Coordinate System 1/3 The CV chosen is an infinitesimal cube with sides of
The
length dr, rdθ, and dz.
dr
dz
The net rate of mass flux out through the control surface
The
∂ ρ Vr ∂ ρ Vθ
∂ ρ Vz ⎤
⎡
⎢ρ Vr + r ∂ r + ∂ θ + r ∂ z ⎥ δ rδθδ z
⎣
⎦ The rate of change of mass
The
inside the control volume ∂ρ
rd θ drdz
∂t
26 Conservation of Mass
2/3
Cylindrical Coordinate System 2/3 The continuity equation
∂ ρ 1 ∂ ( rρ Vr ) 1 ∂ (ρ Vθ ) ∂ (ρ Vz )
+
+
+
=0
∂t r
∂r
r ∂θ
∂z By “Del” operator
r ∂ r1∂ r∂
∇ = er
+ eθ
+k
∂r
r ∂θ
∂z The continuity equation becomes
r
∂ρ
+ ∇ ⋅ ρV = 0
∂t
27 Conservation of Mass
Cylindrical Coordinate System 3/3 Incompressible fluid
Incompressible
r
1 ∂ ( rVr ) 1 ∂ ( Vθ ) ∂ ( Vz )
+
+
= ∇⋅V = 0
r ∂r
r ∂θ
∂z
Steady flow
Steady r
1 ∂ ( rρ Vr ) 1 ∂ (ρ Vθ ) ∂ (ρ Vz )
+
+
= ∇ ⋅ ρV = 0
r
∂r
r ∂θ
∂z 28 1/6
Stream Function 1/6
Streamlines ? Lines tangent to the instantaneous velocity vectors at
Streamlines
every point.
Stream function Ψ(x,y) [Psi] ? Used to represent the velocity
Stream
component u(x,y,t) and v(x,y,t) of a twodimensional
incompressible flow.
Define a function Ψ(x,y), called the stream function, which relates
Define
the velocities shown by the figure in the margin as ∂ψ
u=
∂y ∂ψ
v=−
∂x
29 2/6
Stream Function 2/6
The stream function Ψ(x,y) satisfies the twodimensional form of
The
the incompressible continuity equation
∂u ∂v
∂ 2ψ
∂ 2ψ
+
=0⇒
−
=0
∂x ∂y
∂ x ∂ y ∂ y∂ x Ψ(x,y) ? Still unknown for a particular problem, but at least we
have simplify the analysis by having to determine only one
unknown, Ψ(x,y) , rather than the two function u(x,y) and v(x,y). 30 3/6
Stream Function 3/6
Another advantage of using stream function is related to the fact that
Another
line along which Ψ(x,y) =constant are streamlines.
How to prove ? From the definition of the streamline that the slope
How
at any point along a streamline is given by
dy ⎞
v
=
⎟
dx ⎠ streamline
u Velocity and velocity component along a streamline
31 4/6
Stream Function 4/6 The change of Ψ(x,y) as we move from one point (x,y)
The
to a nearly point (x+dx,y+dy) is given by
∂ψ
∂ψ
dψ =
dx +
dy = − vdx + udy
∂x
∂y
>> d ψ = 0
>> − vdx + udy = 0
Along a line of constant Ψ dy ⎞
v
=
⎟
dx ⎠ streamline
u
This is the definition for a streamline. Thus, if we know the function Ψ(x,y) we can
plot lines of constant Ψto provide the family of streamlines that are helpful in
visualizing the pattern of flow. There are an infinite number of streamlines that make up
a particular flow field, since for each constant value assigned to Ψa streamline can be
drawn. 32 5/6
Stream Function 5/6
The actual numerical value associated with a particular streamline is
The
not of particular significance, but the change in the value of Ψ is
related to the volume rate of flow.
For a unit depth, the flow rate across AB is
For
y2
y 2 ∂ψ
ψ2
q = ∫ udy = ∫
dy = ∫ d ψ = ψ 2 − ψ 1
y1
y1 ∂ y
ψ1
For a unit depth, the flow rate across BC is
For
ψ
x
x ∂ψ
q = ∫ vdx = − ∫
dx = − ∫ d ψ = ψ 2 ψ 1
2 x1 2 x1 1 ∂x ψ2 33 6/6
Stream Function 6/6
Thus the volume flow rate between any two streamlines can be
Thus
written as the difference between the constant values of Ψ defining
two streamlines.
The velocity will be relatively high wherever the streamlines are
The
close together, and relatively low wherever the streamlines are far
apart. 34 Stream Function
Cylindrical Coordinate System
Cylindrical For a twodimensional, incompressible flow in the rθ
For
plane, conservation of mass can be written as: ∂ ( rv r ) ∂v θ
+
=0
∂r
∂θ
The velocity components can be related to the stream
The
function, Ψ(r,θ) through the equation 1 ∂ψ
vr =
r ∂θ and ∂ψ
vθ = −
∂r
35 Example 6.3 Stream Function
The velocity component in a steady, incompressible, two
The
dimensional flow field are u = 2y v = 4x Determine the corresponding stream function and show on a sketch
several streamlines. Indicate the direction of glow along the
streamlines. 36 Example 6.3 Solution
From the definition of the stream function ∂ψ
u=
= 2y
∂y ∂ψ
v=−
= 4x
∂x ψ = y 2 + f1 (x) ψ = −2 x + y
2 2 ψ = − 2 x 2 + f 2 (y) Ψ=0 ψ = −2 x + y + C
2 2 For simplicity, we set C=0
2 2 y
x
−
=1
ψ ψ/2 Ψ≠0
37 Conservation of Linear Momentum
Applying Newton’s second law to control volume
Applying r
r DP
r
r
r
F=
Psystem = ∫
Vdm = ∫
Vρd V
M ( system )
V ( system )
Dt SYS ( ) r
r
r
r
r
r D V δm
⎛ ∂V
∂V
∂V
∂V ⎞
δF =
= δm ⎜
⎟
⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟
Dt
⎠
⎝
r
r
DV
= δm
= δm a
Dt For a infinitesimal system of mass dm, what’s the The
infinitesimal
The
differential form of linear momentum equation? 38 1/2
Forces Acting on Element 1/2
The forces acting on a fluid element may be classified as body
forces and surface forces; surface forces include normal forces and
normal
tangential (shear) forces. r
r
r
δF = δFS + δFB
r
r
r
= δFsx i + δFsy j + δFsz k
r
r
r
+ δFbx i + δFby j + δFbz k Surface forces acting on a fluid
element can be described in terms
of normal and shearing stresses. δF1
δFn
δF2
τ 1 = lim
τ 2 = lim
σ n = lim
δt → 0 δA
δt → 0 δA
δt → 0 δA 39 2/2
Forces Acting on Element 2/2
⎛ ∂σ xx ∂τ yx ∂τzx ⎞
δFsx = ⎜
⎟
⎜ ∂x + ∂y + ∂z ⎟δxδyδz
⎠
⎝
⎛ ∂τ xy ∂σ yy ∂τ zy ⎞
δFsy = ⎜
⎜ ∂x + ∂y + ∂z ⎟δxδyδz
⎟
⎝
⎠
⎛ ∂τ xz ∂τ yz ∂σzz ⎞
δFsz = ⎜
⎟
⎜ ∂x + ∂y + ∂z ⎟δxδyδz
⎠
⎝
δFbx = ρg x δxδyδz
δFby = ρg y δxδyδz Equation of Motion δFbz = ρg z δxδyδz
40 Equation of Motion
δFx = δma x δFy = δma y δFz = δma z ⎛ ∂u
∂σ xx ∂τ yx ∂τ zx
∂u
∂u
∂u ⎞
+
+
= ρ⎜ + u
+v
+w ⎟
ρg x +
⎜ ∂t
∂x
∂y
∂z
∂x
∂y
∂z ⎟
⎝
⎠
∂τ xy ∂σ yy ∂τzy
⎛ ∂v
∂v
∂v
∂v ⎞
ρg y +
+
+
= ρ⎜ + u
+v
+w ⎟
⎜ ∂t
∂x
∂y
∂z
∂x
∂y
∂z ⎟
⎝
⎠
⎛ ∂w
∂τ xz ∂τ yz ∂σzz
∂w
∂w
∂w ⎞
ρg z +
+
+
= ρ⎜
⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟
⎟
∂x
∂y
∂z
⎝
⎠ These are the differential equations of motion for any
fluid satisfying the continuum assumption.
How to solve u,v,w ?
41 Double Subscript Notation for Stresses τ xy The direction of the stress The direction of the
normal to the plane
on which the stress
acts
42 Inviscid Flow
Shearing stresses develop in a moving fluid because of the viscosity
Shearing
of the fluid.
For some common fluid, such as air and water, the viscosity is small,
For
and therefore it seems reasonable to assume that under some
circumstances we may be able to simply neglect the effect of
viscosity.
Flow fields in which the shearing stresses are assumed to be
Flow
negligible are said to be inviscid, nonviscous, or frictionless.
Define the pressure, p, as the negative of the normal stress − p = σ xx = σ yy = σzz
43 Euler’s Equation of Motion
Under frictionless condition, the equations of motion are
reduced to Euler’s Equation:
⎛ ∂u
∂p
∂u
∂u
∂u ⎞
ρg x −
= ρ⎜
⎜ t + u ∂x + v ∂y + w ∂z ⎟
⎟
∂x
⎝
⎠
⎛ ∂v
∂p
∂v
∂v
∂v ⎞
ρg y −
= ρ⎜
⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟
⎟
∂y
⎝
⎠
ρg z − ⎛ ∂w
∂w
∂w ⎞
∂p
∂w
= ρ⎜
+u
+v
+w
⎟
⎜ ∂t
∂x
∂y
∂z ⎟
∂z
⎝
⎠ r
r
r⎞
⎛ ∂V
r
ρg − ∇p = ρ ⎜
+ ( V ⋅ ∇) V ⎟
⎜ ∂t
⎟
⎝
⎠ 44 1/3
Bernoulli Equation 1/3 Euler’s equation for steady flow along a streamline is
Euler r
r
r
ρg − ∇ p = ρ ( V ⋅ ∇ ) V Selecting the coordinate system with the zaxis vertical so that
the acceleration of gravity vector can be expressed as r
g = − g∇z ( ) ( ) ( r
r 1 rr
r
r
V⋅∇ V = ∇ V⋅V − V × ∇× V
2 − ρg∇ z − ∇ p = ρ 2 ) Vector identity …. rr
r
r
∇(V ⋅ V ) − ρ (V × ∇ × V )
45 2/3
Bernoulli Equation 2/3 ( r
r
∇p 1
2
+ ∇(V ) + g∇z = V × ∇ × V
ρ2 ) r
r
r
V × ∇ × V perpendicular to V ( ) r
r
⋅ d s ∇p r 1
r
r
vr
2
⋅ d s + ∇(V )⋅ d s + g∇z ⋅ d s = V × ∇ × V ⋅ d s
ρ
2 [( With )] s
r
r
r
d s = dx i + dy j + dzk r ∂p
∂p
∂p
∇p ⋅ d s =
dx + dy + dz = dp
∂x
∂y
∂z 46 3/3
Bernoulli Equation 3/3
r
r
∇p r 1
2
⋅ d s + ∇(V ) ⋅ d s + g∇z ⋅ d s = 0
ρ
2 dp 1
+ d (V 2 ) + gdz = 0
ρ2 dp V 2
+
+ gz = cons tan t
Integrating … ∫
ρ
2
For steady inviscid, incompressible fluid ( commonly called ideal
fluids) along a streamline
along p V2
+
+ gz = cons tan t
ρ2 Bernoulli equation
47 1/4
Irrotational Flow 1/4 Irrotation ? The irrotational condition is r
∇ ×V = 0
In rectangular coordinates system
∂v ∂u ∂w ∂v ∂u ∂w
−
=
−
=
−
=0
∂x ∂y ∂y ∂z ∂z ∂x In cylindrical coordinates system
1 ∂ Vz ∂ Vθ ∂ Vr ∂ Vz 1 ∂ rVθ 1 ∂ Vr
=
−
=
=0
−
−
r ∂θ
r ∂r
r ∂θ
∂z
∂z
∂r
48 2/4
Irrotational Flow 2/4 A general flow field would not be irrotational flow.
general
A special uniform flow field is an example of an irrotation
special
flow 49 3/4
Irrotational Flow 3/4
A general flow field
A solid body is placed in a uniform stream of fluid. Far away from
solid
the body remain uniform, and in this far region the flow is
irrotational.
The flow around the body remains irrotational except very near the
The
boundary.
Near the boundary the
Near
velocity changes rapidly from
zero at the boundary (noslip
condition) to some relatively
large value in a short distance
from the boundary.
Chapter 9
50 4/4
Irrotational Flow 4/4
A general flow field
Flow from a large reservoir enters a pipe through a streamlined
Flow
entrance where the velocity distribution is essentially uniform. Thus,
at entrance the flow is irrotational. (b)
In the central core of the pipe the flow remains irrotational for some
In
distance.
The boundary layer will develop along the wall and grow in
The
thickness until it fills the pipe.
Viscous forces are dominant Chapter 8
51 1/3
Bernoulli Equation for Irrotational Flow 1/3 The Bernoulli equation for steady, incompressible, and inviscid
The
flow is p V2
+
+ gz = cons tan t
ρ2 The equation can be applied between any two points on the same
The
streamline. In general, the value of the constant will vary from
streamline
streamline to streamline.
streamline
Under additional irrotational condition, the Bernoulli equation ?
Under
the
Starting with Euler’s equation in vector form r 1 rr
r
r
r
r
1
( V ⋅ ∇ ) V = − ∇ p − gk = ∇ V ⋅ V − V × ∇ × V
ρ
2 ( ) ( ) ZERO Regardless of the direction of ds 52 2/3
Bernoulli Equation for Irrotational Flow 2/3 r
∇ ×V = 0 With irrotaional condition
irrotaional r 1 rr
r
r
r
r
1
( V ⋅ ∇) V = − ∇p − gk = ∇ V ⋅ V − V × ∇ × V
ρ
2
r
1
1
1 rr
2
∇ V ⋅ V = ∇ V = − ∇ p − gk
2
ρ
2 ( ( ) () ) ( ) r
⋅ dr rv
1
1
v
v
2
∇ (V ) ⋅ d r = − ∇ p ⋅ d r − gk ⋅ d r
2
ρ
1
dp
dp 1
2
>> d (V ) = −
− gdz >>
+ d (V 2 ) + gdz = 0
2
ρ
ρ2
53 3/3
Bernoulli Equation for Irrotational Flow 3/3 Integrating for incompressible flow
2 dp V
∫ ρ + 2 + gz = con tan t p V2
+
+ gz = cons tan t
ρ2 This equation is valid between any two points in a steady,
incompressible, inviscid, and irrotational flow.
2 2 p1 V1
p 2 V2
+
+ z1 =
+
+ z2
2g
γ 2g
γ 54 Velocity Potential Φ(x,y,z,t) 1/4
The stream function for twodimensional incompressible
The
flow is Ψ(x,y) which satisfies the continuous condtions
For an irrotational flow, the velocity components can be
For
expressed in terms of a scalar function ψ(x,y,z,t) as ∂φ
u=
∂x ∂φ
v=
∂y ∂φ
w=
∂z where ψ(x,y,z,t) is called the velocity potential.
55 Velocity Potential Φ(x,y,z,t) 2/4
In vector form
In r
V = ∇φ
For an incompressible flow
For
r
∇⋅V = 0 called a potential flow For incompressible, irrotational flow
r
r
∂ 2φ ∂ 2φ ∂ 2φ
V = ∇φ ⇒ ∇ ⋅ V = ∇ 2φ = 2 + 2 + 2 = 0
∂z
∂y
∂x
Laplace’s equation
Laplacian operator 56 Velocity Potential Φ(x,y,z,t) 3/4
Inviscid, incompressible, irrotational fields are governed
Inviscid
by Laplace’s equation.
This type flow is commonly called a potential flow.
This
To complete the mathematical formulation of a given
To
problem, boundary conditions have to be specified. These
are usually velocities specified on the boundaries of the
flow field of interest. 57 Velocity Potential Φ(x,y,z,t) 4/4
In cylindrical coordinate, r, θ, and z
In
1 ∂φ
∂φ
∂φ
vz =
vr =
vθ =
r ∂θ
∂z
∂r 1 ∂ ⎛ ∂φ ⎞ 1 ∂ 2 φ ∂ 2 φ
+ 2 =0
⎜r
⎟+ 2
2
r ∂r ⎝ ∂r ⎠ r ∂θ
∂z 58 Potential Flow Theory
Velocity potential ψ (x,y,z,t) exists only for irrotational
flow.
Irrotationality may be a valid assumption for those regions
of a flow in which viscous forces are negligible.
In an irrotational flow, the velocity field may be defined
by the potential function, ψ(x,y,z,t), the theory is often
referred to as potential flow theory. 59 1/2
Laplace’s Equation 1/2 For twodimensional, incompressible flow
u= ∂ψ
∂y v=− ∂ψ
∂x (1) For twodimensional, irrotational flow
u= ∂φ
∂x v= ∂φ
∂y (2) ∂u ∂v
∂ 2ψ ∂ 2ψ
(1) + irrotational condition … ∂ y − ∂ x = 0 ⇒ ∂ x 2 + ∂ y 2 = 0
(2) + continuity equation … ∂u ∂v
∂ 2φ ∂ 2φ
+
=0⇒
+ 2 =0
2
∂x ∂y
∂x
∂y
60 2/2
Laplace’s Equation 2/2 For a twodimensional incompressible flow, we can define
a stream function Ψ; if the flow is also irrotational, Ψ
will satisfy Laplace’s equation.
For an irrotational flow, we can define a velocity potential
Φ; if the flow is also incompressible, Φ will satisfy
Laplace’s equation.
Any function ψ or Ψ that satisfies Laplace’s equation
represents a possible twodimensional, incompressible,
irrotational flow field.
61 Φ and Ψ 1/2
For Ψ=constant, dΨ =0 and
For ∂ψ
∂ψ
dψ =
dx +
dy = 0
∂x
∂y The slope of a streamline – a line of constant Ψ
The
dy ⎞
∂ψ / ∂x v
=
⎟ =−
dx ⎠ ψ
∂ψ / ∂y u Along a line of constant ψ, dψ =0 and d φ =
Along ∂φ
∂φ
dx +
dy = 0
∂x
∂y The slope of a potential line – a line of constant ψ
The
dy ⎞
∂φ / ∂x
u
=−
⎟ =−
dx ⎠ φ
∂φ / ∂y
v Line of constant Ψ and constant ψ are orthogonal. 62 Φ and Ψ 2/2
The lines of constant ψ (called
The
equipotential lines) are orthogonal to lines
of constant Ψ (streamlines) at all points
where they intersect..
For any potential flow a “flow net” can be
For
drawn that consists of a family of
streamlines and equipotential lines.
Velocities can be estimated from the flow
Velocities
net, since the velocity is inversely
proportional to the streamline spacing
Flow net for 90° bend
63 Example 6.4 Velocity Potential and
1/2
Inviscid Flow Pressure 1/2
The twodimensional flow of a nonviscous, incompressible fluid in
The
the vicinity of the 90° corner of Figure E6.4a is described by the
stream function
ψ = 2 r 2 sin 2 θ Where Ψ has units of m2/s when r is in meters. (a) Determine, if
possible, the corresponding velocity potential. (b) If the pressure at
point (1) on the wall is 30 kPa, what is the pressure at point (2)?
Assume the fluid density is 103 kg/m3 and the xy plane is
horizontal – that is, there is no difference in elevation between
points (1) and (2). 64 Example 6.4 Velocity Potential and
2/2
Inviscid Flow Pressure 2/2 65 Example 6.4 Solution1/2
1 ∂ψ
= 4r cos 2θ
r ∂θ
∂ψ
vθ = −
= −4 r sin 2θ
∂r vr = vr = ∂φ
= 4 r cos 2 θ ⇒ φ = 2 r 2 cos 2 θ + f1 ( θ )
∂r vθ = 1 ∂φ
= − 4 r sin 2 θ ⇒ φ = 2 r 2 cos 2 θ + f 2 ( r )
r ∂θ φ = 2 r 2 cos 2 θ + C
Let C=0 2 φ = 2 r cos 2 θ 66 Example 6.4 Solution2/2
Bernoulli equation between points (1) and (2) with no
elevation change p1 V12 p 2 V2 2
ρ
+
=
+
⇒ p 2 = p1 + ( V12 − V2 2 )
γ
2g
γ
2g
2 V 2 = v r 2 + v θ2
V12 = ... = 16m 2 / s 2 p 2 = ... = 36kPa V2 2 = ... = 4 m 2 / s 2 φ = 2 r 2 cos 2 θ = 4 r 2 cos θ sin θ = 4 xy
67 Some Basic, Plane Potential Flow
Since Laplace’s equation is linear, various solutions can be added to
Since
obtain other solution – that is , if ψ1(x,y,z) and ψ2(x,y,z) are two
solutions to Laplace’s equation, then ψ= ψ1 + ψ2 is also solution.
The practical implication of this result is that if we have certain
The
basic solution we can combine them to obtain more complicated and
interesting solutions.
Several basic velocity potentials, which describe some relatively
Several
simple flows, will be determined.
These basic velocity potential will be combined to represent
These
complicate flows. 68 1/2
Uniform Flow 1/2 A uniform flow is a simplest plane flow for which the
uniform
streamlines are all straight and parallel, and the magnitude
of the velocity is constant.
u=U and v=0
∂φ
= U,
∂x ∂ψ
= U,
∂y ∂φ
=0
∂y ∂ψ
=0
∂x ⇒ φ = Ux + C ⇒ ψ = Uy 69 2/2
Uniform Flow 2/2 For a uniform flow of constant velocity V, inclined to an
For
angle α to the xaxis.
ψ = ( V cos α ) y − ( V sin α ) x
φ = − ( V sin α ) y − ( V cos α ) x 70 1/2
Source and Sink 1/2 For a source flow ( from origin radially) with volume flow
For
rate per unit depth m (m=2π r vr )
(m=2 m
vr =
2 πr m
θ
vθ= 0 ⇒ ψ =
2π m
φ=
ln r
2π 1 ∂ψ
∂ψ
vr =
and
vθ = −
r ∂θ
∂r
∂φ
1 ∂φ
vr =
and
vθ =
∂r
r ∂θ
71 2/2
Source and Sink 2/2 For a sink flow (toward origin radially) with volume flow
For
rate per unit depth m
m
vr = −
2πr m
vθ = 0 ⇒ ψ = − θ
2π m
φ = − ln r
2π 1 ∂ψ
∂ψ
vr =
and
vθ = −
r ∂θ
∂r
∂φ
1 ∂φ
vr =
and
vθ =
∂r
r ∂θ 72 Example 6.5 Potential Flow  Sink
A nonviscous, incompressible fluid flows between wedgeshaped
nonviscous
walls into a small opening as shown in Figure E6.5. The velocity
potential (in ft2/s), which approximately describes this flow is φ = − 2 ln r
Determine the volume rate
of flow (per unit length) into
the opening. 73 Example 6.5 Solution
The components of velocity ∂φ
2
vr =
=−
∂r
r 1 ∂φ
=0
vθ =
r ∂θ The flowrate per unit width
π/6 π
q = v r Rdθ = ... = − = −1.05ft 2 / s
3
0 ∫ 74 Vortex
A vortex represents a flow in which the streamlines are
vortex
concentric circles.
Vortex motion can be either rotational or irrotational.
Vortex
For an irrotational vortex (ccw, center at origin) with
For
vortex strength K
At r=0, the velocity
At r=0, the velocity
becomes infinite. singularity K
vr= 0
vθ =
r
⇒ ψ = − K ln r
φ = Kθ 75 1/2
Free Vortex 1/2 Free (Irrotational) vortex (a) is that rotation refers to the
Free
orientation of a fluid element and not the path followed by
the element.
A pair of small sticks were
pair
placed in the flow field at
location A, the sticks would
rotate as they as they move to
location B. 76 2/2
Free Vortex 2/2 One of the sticks, the one that is aligned the streamline,
One
would follow a circular path and rotate in a
counterclockwise direction
The other rotates in a clockwise
The
direction due to the nature of the
flow field – that is, the part of the
stick nearest the origin moves
faster than the opposite end.
The average velocity of the two
The
sticks is zero.
77 Forced Vortex
If the flow were rotating as a
If
rigid body, such that vθ=K1r
where K1 is a constant.
Force vortex is rotational and
Force
cannot be described with a
velocity potential.
Force vortex is commonly
Force
called a rotational vortex. 78 Combined Vortex
A combine vortex is one with a forced vortex as a central
combine
core and a velocity distribution corresponding to that of a
free vortex outside the core. K
vθ =
r
v θ = rω r > r0
r ≤ r0 79 1/3
Circulation 1/3 Circulation Γ is defined as the line integral of the
Circulation
tangential velocity component about any closed curve
fixed in the flow:
rr
r
Γ = V ⋅ d s = 2 ω Z dA = ( ∇ × V ) Z dA ∫ ∫ c A ∫ A r
ds where the
is an element vector tangent to the curve
and having length ds of the element of arc. It’s positive
ds
corresponds to a c.c.w. direction of integration around the
curve. 80 2/3
Circulation 2/3 For irrotational flow , Γ =0
For rr
Γ = ∫ V ⋅ ds =
c ∫ A r
r
( ∇ × V ) dA = ∫ ∇ φ ⋅ d s = ∫ d φ = 0
c c For irrotational flow , Γ =0 The circulation around any path that does not
The circulation around any path that does not
include the singular point at the origin will be
include the singular point at the origin will be
zero.
zero. 81 3/3
Circulation 3/3 For free vortex
For K
vθ =
r K
( rdθ) = 2πK
Γ=∫
0
r
K = Γ/2 π
2π Γ
ψ = − ln r
2π Γ
φ=
θ
2π The circulation around any path that encloses
The circulation around any path that encloses
singularities will be nozero.
singularities will be nozero.
82 Example 6.6 Potential Flow – Free Vortex
A liquid drains from a large tank through a small opening as
liquid
illustrated in Figure E6.6. A vortex forms whose velocity
distribution away from the tank opening can be approximated as that
of a free vortex having a velocity potential Γ
φ=
θ
2π
Determine an expression
relating the surface shape
to the strength of the
vortex as specified by the
circulation Γ.
83 Example 6.6 Solution
Since the free vortex represents an irrotational flow field, the
Bernoulli equation p1 V12
p 2 V2 2
+
+ z1 =
+
+ z2
γ
γ
2g
2g
2 2 V1
V2
=
+ zs
2g
2g
1 ∂φ
Γ
vθ =
=
r ∂θ 2πr At free surface p 1 = p 2 = 0 Far from the origin at
point (1), V1=vθ=0 Γ2
zs = − 2 2
2π r g
84 1/2
Doublet 1/2 For a doublet ( produced mathematically by allowing a
For
source and a sink of numerically equal strength to merge)
with a strength m
The combined stream function for the pair is
m
ψ = − (θ1 − θ2 )
2π
tan θ1 − tan θ2
⎛ 2πψ ⎞
tan ⎜ −
⎟ = tan (θ1 − θ2 ) =
1 + tan θ1 tan θ 2
⎝ m⎠ tan θ1 = r sin θ
(r cos θ − a ) and tan θ2 = r sin θ
(r cos θ + a ) m
⎛ 2πψ ⎞ 2ar sin θ
⎛ 2ar sin θ ⎞
⇒ tan ⎜ −
⇒ ψ = − tan −1 ⎜ 2
⎟= 2
⎟
m ⎠ (r − a 2 )
2π
r − a2 ⎠
⎝
⎝ 85 2/2
Doublet 2/2
a →0 m 2ar sin θ
mar sin θ
ψ=−
=− 2 2
2
2
2π r − a
π(r − a ) The socalled doublet is formed by letting a→0，m→∞ r
1
→
2
2
r −a
r K sin θ
ψ=−
r
ma
K=
π K cos θ
φ=
r
86 Streamlines for a Doublet
Plots of lines of constant Ψ reveal that the streamlines for
Plots
a doublet are circles through the origin tangent to the x
axis. 87 Summary 88 Superposition of Elementary Plane
1/2
Flows 1/2
Potential flows are governed by Laplace’s equation, which
is a linear partial differential equation.
Various basic velocity potentials and stream function, ψ
and Ψ, can be combined to form new potentials and
stream functions. ψ 3 = ψ1+ ψ 2 φ 3 = φ1 + φ 2 89 Superposition of Elementary Plane
2/2
Flows 2/2
Any streamline in an inviscid flow field can be considered
as a solid boundary, since the conditions along a solid
boundary and as streamline are the same – that is, there is
no flow through the boundary or the streamline.
We can combine some of the basic velocity potentials or
stream functions to yield a streamline that corresponds to a
particular body shape of interest, that combination can be
used to describe in detail the flow around that body.
Methods of superposition
90 1/4
HalfBody : Uniform Stream + Source 1/4
ψ = ψ uniform− flow + ψ source = Ur sin θ + m
θ
2π m
φ = φuniform− flow + φsource = Ur cos+ ln r
2π 1 ∂ψ
m
vr =
= U cos θ +
r ∂θ
2πr
∂ψ
vθ = −
= − U sin θ
∂r The stagnation point occurs at x=b
m
m
vr = U +
=0⇒b=
− 2πb
2πU The combination of a uniform flow and a
source can be used to describe the flow
around a streamlined body placed in a
uniform stream. 91 2/4
HalfBody : Uniform Stream + Source 2/4 The value of the stream function at the stagnation point can be
obtained by evaluating Ψ at r=b θ=π
m
m
ψ stagnation =
= πbU
2
2
The equation of the streamline passing through the stagnation
point is πbU = Ur sin θ + bUθ
b( π − θ) The streamline can be replaced by a solid boundary.
r=
The body is open at the downstream end, and thus is
sin θ
called a HALFBODY. The combination of a uniform called a HALFBODY. The combination of a uniform
flow and a source can be used to describe the flow
around a streamlined body placed in a uniform stream.
92 3/4
HalfBody : Uniform Stream + Source 3/4 b ( π − θ)
r=
⇒ y = b ( π − θ)
sin θ
As θ 0 or θ= π the halfwidth approaches ±bπ. The width
or
of the halfbody asymptotically approach 2πb.
The velocity components at any point m
1 ∂ψ
vr =
= U cos θ +
r ∂θ
2 πr ∂ψ
vθ = −
= − U sin θ
∂r
b= m
2 πU 2 Um cos θ ⎛ m ⎞
b
b2 ⎞
2
2
2
2
2⎛
V = v r + vθ = U +
+⎜
⎟ = U ⎜1 + 2 cos θ + 2 ⎟
⎜
r
r⎟
πr
⎝ 2πr ⎠
⎝
⎠ 93 4/4
HalfBody : Uniform Stream + Source 4/4 With the velocity known, the pressure at any point can be
determined from the Bernoulli equation
1
12
2
p 0 + ρU = p + ρV
2
2
Far from the body Where elevation change have been neglected. 94 Example 6.7 Potential Flow – HalfBody
The shape of a hill arising from a plain can be approximated with
The
the top section of a halfbody as is illustrated in Figure E6.7a. The
height of the hill approaches 200 ft as shown. (a) When a 40 mi/hr
wind blows toward the hill, what is the magnitude of the air velocity
at a point on the hill directly above the origin [point (2)]? (b) What
is the elevation of point (2) above the plain and what is the
difference in pressure between point (1) on the plain far from the hill
and point (2)? Assume an air density of 0.00238 slugs/ft3? 95 Example 6.7 Solution1/2
The velocity is
⎛
b
b2 ⎞
V = U 1 + 2 cos θ + 2 ⎟
⎜
r
r⎟
⎝
⎠
2 2⎜ At point (2), θ=π/2 b ( π − θ) π b
r=
=
sin θ
2
⎛
b2 ⎞
⎟ = U 2 ⎛1 + 4 ⎞ = ... = 47.4mi / hr
V = U 1+
⎟
⎜
2
⎜ ( πb / 2) 2 ⎟
⎝ π⎠
⎝
⎠
2 2⎜ πb 200ft
=
= 100ft
The elevation at (2) above the plain is y 2 =
2
2
96 Example 6.7 Solution2/2
From the Bernoulli equation
p1 V12
p 2 V2 2
+
+ z1 =
+
+ z2
γ
2g
γ
2g
ρ2
p1 − p 2 = ( V2 − V12 ) + γ ( y 2 − y1 ) = ... = 9.31lb / ft 2 = 0.0647 psi
2
⎛ 5280ft / mi ⎞
V1 = ( 40mi / hr )⎜
⎟ = 58.7ft / s
3600s / hr ⎠
⎝
⎛ 5280ft / mi ⎞
V2 = ( 47.4 mi / hr )⎜
⎟ = 69.5ft / s
3600s / hr ⎠
⎝ 97 1/3
Rankine Oval: Uniform Stream + Doublet 1/3 m
ψ = Ur sin θ − (θ1 − θ2 )
2π
m
φ = Ur cos θ − (ln r1 − ln r2 )
2π ψ = Ur sin θ − m
⎛ 2ar sin θ ⎞
tan −1 ⎜ 2
⎟
2
2π
⎝ r −a ⎠ ⎞
m
2ay
−1 ⎛
⎟
⎜
tan
ψ = Uy −
⎜ x 2 + y2 − a 2 ⎟
2π
⎠
⎝ The corresponding streamlines for this flow field are obtained by
setting Ψ=constant. It is discovered that the streamline forms a
closed body of length 2l and width 2h. Rankine ovals 98 2/3
Rankine Oval: Uniform Stream + Doublet 2/3 The stagnation points occur at the upstream and downstream ends of
The
the body. These points can be located by determining where along
the x axis the velocity is zero.
The stagnation points correspond to the points where the uniform
The
velocity, the source velocity, and the sink velocity all combine to
give a zero velocity.
The locations of the stagnation points depend on the value of a, m,
The
and U.
Dimensionless
The body halflength 2l
1/ 2 ⎛ ma
2⎞
l=⎜
+a ⎟
⎠
⎝ πU 1/ 2 l ⎛m
⎞
⇒ =⎜
+ 1⎟
a ⎝ πUa
⎠ 99 3/3
Rankine Oval: Uniform Stream + Doublet 3/3 The body halfwidth, h, can be obtained by determining the value of
The
y where the y axis intersects the Ψ=0 streamline. Thus, with Ψ=0,
x=0, and y=h.
The body halfwidth 2h
2
⎤ ⎡ ⎛ πUa ⎞ h ⎤
h2 − a2
2πUh
h 1 ⎡⎛ h ⎞
h=
tan
⇒ = ⎢⎜ ⎟ − 1⎥ tan ⎢2⎜
⎟⎥
2a
m
a 2 ⎢⎝ a ⎠
⎥ ⎣ ⎝ m ⎠a⎦
⎣
⎦ Dimensionless 100 1/4
Flow around a Circular Cylinder 1/4
When the distance between the sourcesink pair approaches zero, the
When
shape of the rankine oval becomes more blunt and in fact
approaches a circular shape.
⎛ a2 ⎞
K sin θ
ψ = Ur ⎜1 − 2 ⎟ sin θ
ψ = Ur sin θ −
⎜ r⎟
⎝
⎠
r
K cos θ
⎛ a2 ⎞
φ = Ur cos θ +
φ = Ur ⎜1 + 2 ⎟ cos θ
⎜ r⎟
r
⎝
⎠
In order for the stream
Ψ=constant for r=a
function to represent flow
around a circular cylinder Ψ=0 for r=a K=Ua2 101 2/4
Flow around a Circular Cylinder 2/4
The velocity components On the surface of the cylinder (r=a) ⎛ a2 ⎞
∂φ 1 ∂ψ
= U⎜1 − 2 ⎟ cos θ
vr =
=
⎜ r⎟
∂r r ∂θ
⎝
⎠ vr = 0 ⎛ a2 ⎞
∂ψ
1 ∂φ
= − U⎜1 + 2 ⎟ sin θ
=−
vθ =
⎜ r⎟
∂r
r ∂θ
⎝
⎠ v θ = −2 U sin θ The pressure distribution on the cylinder surface can be obtained
from the Bernoulli equation 12
1
2
p 0 + ρU = p s + ρv θs 2
2
2
Far from the body 1
p s = p 0 + ρU 2 (1 − 4 sin 2 θ)
2
v θ = −2 U sin θ 102 3/4
Flow around a Circular Cylinder 3/4
On the upstream part of the
cylinder, there is approximate
agreement between the potential
flow and the experimental results.
Because of the viscous boundary
layer that develops on the cylinder,
the main flow separates from the
surface of the cylinder, leading to
the large difference between the
theoretical, frictionless solution
and the experimental results on the
downstream side of the cylinder.
103 4/4
Flow around a Circular Cylinder 4/4 2π ∫
= − p sin θadθ
∫ Fx = − ps cos θadθ
Fy 0
2π 0 s 104 Flow around a Circular Cylinder + Free
1/4
Vortex 1/4
Adding a free vortex to the stream function or velocity
Adding
potential for the flow around a cylinder.
⎛ a2 ⎞
Γ
ψ = Ur ⎜1 − 2 ⎟ sin θ − ln r
⎜ r⎟
2π
⎝
⎠ ⎛ a2 ⎞
Γ
θ
φ = Ur ⎜1 + 2 ⎟ cos θ +
⎜ r⎟
2π
⎝
⎠ The circle r=a will still be a streamline, since the
streamlines for the added free vortex are all circular.
The tangential velocity
on the surface of the
cylinder ∂ψ
vθ = −
∂r r =a Γ
= −2 U sin θ +
2πa
105 Flow around a Circular Cylinder + Free
2/4
Vortex 2/4
This type of flow could be approximately created by
This
placing a rotating cylinder in a uniform stream.
Because of the presence of viscosity in any real fluid, the
Because
fluid in contacting with the rotating cylinder would rotate
with the same velocity as the cylinder, and the resulting
flow field would resemble that developed by the
combination of a uniform flow past a cylinder and a free
vortex. 106 Flow around a Circular Cylinder + Free
3/4
Vortex 3/4
A variety of streamline patterns
variety
can be developed, depending
on the vortex strength Г.
The location of stagnation
The
points on a circular cylinder (a)
without circulation; (b, c, d)
with circulation. sin θstag
vθ = 0 Γ
=
4πUa
θ = θstag
107 Flow around a Circular Cylinder + Free
4/4
Vortex 4/4
For the cylinder with circulation, the surface pressure, ps, is
For
obtained from the Bernoulli equation
12
1⎛
Γ⎞
p 0 + ρU = ps + ρ⎜ − 2U sin θ +
⎟
2
2⎝
2πa ⎠ 2 Γ2 ⎞
1 2⎛
2Γ sin θ
2
− 2 2 2⎟
ps = p 0 + ρU ⎜1 − 4 sin θ +
⎜
πaU
2
4π a U ⎟
⎝
⎠
2π Drag
Lift Fx = − ∫ ps cos θadθ = 0
0 2π Fy = − ∫ ps sin θadθ = −ρUT
0 108 1/2
Example 6.8 Potential Flow – Cylinder 1/2 When a circular cylinder is placed in a uniform stream, a stagnation
When
point is created on the cylinder as is shown in Figure E6.8a. If a
small hole is located at this point, the stagnation pressure, pstag, can
be measured and used to determine the approach velocity, U. (a)
Show how pstag and U are related. (b) If the cylinder is misaligned
by an angle α (Figure E6.8b), but the measured pressure still
interpreted as the stagnation pressure, determine an expression for
the ratio of the true velocity, U, to the predicted velocity, U’. Plot
this ratio as a function of α for the range 20 °≦α≦20°.
°≦α≦ 109 2/2
Example 6.8 Potential Flow – Cylinder 2/2 110 Example 6.7 Solution1/2
The Bernoulli equation between a point on the stagnation streamline
upstream from the cylinder and the stagnation point
p 0 U 2 pstag
+
=
γ 2g
γ The difference between the pressure at the
stagnation point and the upstream pressure ⎡2
⎤
⇒ U = ⎢ ( pstag − p 0 )⎥
⎣ρ
⎦ 1/ 2 If the cylinder is misaligned by an angle, α, the pressure actually
measured, pa, will be different from the stagnation pressure.
⎡2
⎤
U ' = ⎢ ( p a − p 0 )⎥
⎣ρ
⎦ 1/ 2 1/2 ⎛ pstag − p 0 ⎞
U(true)
⇒
=⎜
⎜ p −p ⎟
⎟
U' (predicted ) ⎝ a
0⎠ 111 Example 6.7 Solution2/2
On the surface of the cylinder (r=a) v θ = −2U sin θ
The Bernoulli equation between a point upstream if the cylinder and
the point on the cylinder where r=a, θ=α.
12
1
p 0 + ρU = p a + ρ( −2U sin α ) 2
2
2 1
p a − p 0 = ρU 2 (1 − 4 sin 2 α)
2
1
p stag − p 0 = ρU 2
2
U
= (1 − 4 sin 2 α)1 / 2
U' 112 Viscous Flow
To incorporate viscous effects into the differential analysis
of fluid motion
General equation of motion
⎛ ∂u
∂σ xx ∂τ yx ∂τ zx
∂u
∂u
∂u ⎞
+
+
= ρ⎜
ρg x +
⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟
⎟
∂x
∂y
∂z
⎝
⎠
∂τ xy ∂σ yy ∂τ zy
⎛ ∂v
∂v
∂v
∂v ⎞
ρg y +
+
+
= ρ⎜
+u
+v
+w ⎟
⎜ ∂t
∂x
∂y
∂z
∂x
∂y
∂z ⎟
⎠
⎝
⎛ ∂w
∂τ xz ∂τ yz ∂σ zz
∂w
∂w
∂w ⎞
ρg z +
+
+
= ρ⎜
⎟
⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟
∂x
∂y
∂z
⎠
⎝ StressDeformation Relationship
Stress
113 1/2
StressDeformation Relationship 1/2 The stresses must be
The
expressed in terms of the
velocity and pressure
field.
Cartesian coordinates r
∂u
2
σ xx = − p − µ ∇ ⋅ V + 2µ
∂x
3
r
∂v
2
σ yy = − p − µ ∇ ⋅ V + 2µ
∂y
3
r
∂w
2
σ zz = − p − µ ∇ ⋅ V + 2µ
∂z
3
⎛ ∂v ∂u ⎞
τ xy = τ yx = µ ⎜
⎜ ∂x + ∂y ⎟
⎟
⎝
⎠
τ xz = τ zx ⎛ ∂w ∂u ⎞
= µ⎜
+
⎟
⎝ ∂x ∂z ⎠ τ yz = τ zy ⎛ ∂w ∂v ⎞
= µ⎜
⎜ ∂y + ∂z ⎟
⎟
⎝
⎠ 114 2/2
StressDeformation Relationship 2/2
∂v r
σ rr = − p + 2µ
∂r
⎛ 1 ∂v θ v r ⎞
+
σ θθ = − p + 2µ ⎜
⎟
r⎠
⎝ r ∂θ
∂v z
σ zz = − p + 2µ
∂z
⎛ ∂ ⎛ v ⎞ 1 ∂v r
τ rθ = τ θ r = µ ⎜ r ⎜ θ ⎟ +
⎜ ∂r r
⎝
⎠ r ∂θ
⎝
⎛ ∂v θ 1 ∂v z ⎞
+
τ θz = τ zθ = µ ⎜
⎟
r ∂θ ⎠
⎝ ∂z
⎛ ∂v r ∂v z ⎞
τ rz = τ zr = µ ⎜
+
⎟
∂r ⎠
⎝ ∂z ⎞
⎟
⎟
⎠ Cylindrical polar coordinates
Introduced into the differential
equation of motion….
115 1/5
The NavierStokes Equations 1/5 These obtained equations of motion are called the NavierThese
Stokes Equations.
Cartesian coordinates
r ⎞ ⎤ ∂ ⎡ ⎛ ∂u ∂v ⎞ ⎤
Du
∂p
∂ ⎡ ⎛ ∂u 2
− ∇ ⋅ V ⎟⎥ +
ρ
= ρg x −
+
⎢µ ⎜
⎢µ ⎜ 2
⎜ ∂y + ∂x ⎟ ⎥ +
⎟
Dt
∂x ∂x ⎣ ⎝ ∂x 3
⎠ ⎦ ∂y ⎢ ⎝
⎠⎥
⎣
⎦
r ⎞⎤
∂p
∂ ⎡ ⎛ ∂u ∂v ⎞ ⎤ ∂ ⎡ ⎛ ∂v 2
Dv
ρ
= ρg y −
+
⎢µ ⎜
⎜ ∂ y + ∂ x ⎟ ⎥ + ∂ y ⎢µ ⎜ 2 ∂ y − 3 ∇ ⋅ V ⎟ ⎥ +
⎟
⎜
⎟
∂y ∂x ⎢ ⎝
Dt
⎢⎝
⎠⎥
⎠⎥
⎦
⎣
⎦
⎣ ∂
∂z ⎡ ⎛ ∂w ∂u ⎞ ⎤
+
⎟⎥
⎢µ ⎜
⎣ ⎝ ∂x ∂z ⎠ ⎦ ⎡ ⎛ ∂v ∂w ⎞ ⎤
⎢µ ⎜
⎜ ∂z + ∂y ⎟ ⎥
⎟
⎢⎝
⎠⎥
⎦
⎣
r ⎞⎤
∂ ⎡ ⎛ ∂w
∂u ⎞ ⎤ ∂ ⎡ ⎛ ∂v ∂w ⎞ ⎤ ∂ ⎡ ⎛ ∂w 2
∂p
Dw
⎟⎥ +
ρ
= ρg z −
+
+
+
− ∇ ⋅ V ⎟⎥
⎟⎥ +
⎢µ ⎜
⎢µ ⎜
⎢µ ⎜ 2
∂ z ∂ x ⎣ ⎝ ∂ x ∂ zx ⎠ ⎦ ∂ y ⎢ ⎜ ∂ z ∂ y ⎟ ⎥ ∂ z ⎣ ⎝ ∂ z
Dt
3
⎠⎦
⎠⎦
⎣⎝
∂
∂z 116 2/5
The NavierStokes Equations 2/5
Cylindrical polar coordinates
2
⎛ ∂v r
∂v r v θ ∂v r v θ
∂v ⎞
ρ⎜
+ vr
+
−
+ vz r ⎟
⎜ ∂t
r ∂θ
r
∂r
∂z ⎟
⎝
⎠ ⎡ 1 ∂ ⎛ ∂v r ⎞ v r
1 ∂2v r
2 ∂v θ ∂ 2 v r ⎤
∂p
=−
+ ρg r + µ ⎢
−2
+
⎜r
⎟− 2 + 2
⎥
2
r ∂r ⎝ ∂r ⎠ r
∂r
r ∂θ
r ∂θ
∂z 2 ⎥
⎢
⎦
⎣
v ∂v
vv
∂v
∂v ⎞
⎛ ∂v
ρ⎜ θ + v r θ + θ θ + r θ + v z θ ⎟
r ∂θ
r
∂r
∂z ⎠
⎝ ∂t
⎡ 1 ∂ ⎛ ∂v θ ⎞ v θ 1 ∂ 2 v θ 2 ∂v r ∂ 2 v θ ⎤
1 ∂p
+2
+
=−
+ ρg θ + µ ⎢
⎜r
⎟− 2 + 2
⎥
2
r ∂θ
r ∂r ⎝ ∂r ⎠ r
r ∂θ
r ∂θ
∂z 2 ⎥
⎢
⎣
⎦
v ∂v
∂v
∂v ⎞
⎛ ∂v
ρ⎜ z + v r z + + θ z + v z z ⎟
r ∂θ
∂r
∂z ⎠
⎝ ∂t
⎡ 1 ∂ ⎛ ∂v z
∂P
=−
+ ρg z + µ ⎢
⎜r
∂z
⎢ r ∂r ⎝ ∂r
⎣ 2
2
⎞ 1 ∂ vz ∂ vz ⎤
+
⎟+ 2
2
2⎥
∂z ⎥
⎠ r ∂θ
⎦ 117 3/5
The NavierStokes Equations 3/5 Under incompressible flow with constant viscosity
Under
conditions, the NavierStokes equations are reduced to:
⎛ ∂2u ∂2u ∂2u ⎞
⎛ ∂u
∂u
∂u
∂u ⎞
∂p
⎜
ρ⎜
+ 2⎟
⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ = − ∂x + ρ g x + µ ⎜ 2 +
⎟
2
∂z ⎟
∂y
⎝
⎠
⎝ ∂x
⎠
⎛ ∂2v ∂2v ∂2v ⎞
⎛ ∂v
∂v
∂v
∂v ⎞
∂p
⎜
⎟
+
ρ⎜
⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ = − ∂y + ρ g y + µ ⎜ 2 +
⎟
2
2⎟
∂z ⎠
∂y
⎝
⎠
⎝ ∂x
⎛ ∂2w ∂2w ∂2w ⎞
⎛ ∂w
∂w
∂w
∂w ⎞
∂p
⎜
⎟
+
+
ρ⎜
⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ = − ∂z + ρ g z + µ ⎜
⎟
2
2
2⎟
∂z ⎠
∂y
⎝
⎠
⎝ ∂x 118 4/5
The NavierStokes Equations 4/5 Under frictionless condition, the equations of motion are
Unde frictionless
reduced to Euler’s Equation:
⎛ ∂u
∂p
∂u
∂u
∂u ⎞
⎜
⎟=−
+ ρg x
ρ⎜
+u
+v
+w
⎟
∂x
∂x
∂y
∂z ⎠
⎝ ∂t
⎛ ∂v
∂v
∂v
∂v ⎞
∂p
ρ⎜
⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ = − ∂y + ρ g y
⎟
⎝
⎠
⎛ ∂w
∂w
∂w
∂w ⎞
∂p
ρ⎜
⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ = − ∂z + ρ g z
⎟
⎝
⎠ r
r
DV
ρ
= ρg − ∇p
Dt 119 5/5
The NavierStokes Equations 5/5 The NavierStokes equations apply to both laminar and
The
turbulent flow, but for turbulent flow each velocity
component fluctuates randomly with respect to time and
this added complication makes an analytical solution
intractable.
The exact solutions referred to are for laminar flows in
The
which the velocity is either independent of time (steady
flow) or dependent on time (unsteady flow) in a welldefined manner.
120 Some Simple Solutions for Viscous,
Incompressible Fluids
A principal difficulty in solving the NavierStokes
principal
equations is because of their nonlinearity arising from the
convective acceleration terms.
convective
There are no general analytical schemes for solving
There
nonlinear partial differential equations.
There are a few special cases for which the convective
There
acceleration vanishes. In these cases exact solution are
often possible. 121 Steady, Laminar Flow between Fixed
1/6
Parallel Plates 1/6
Consider flow between the two horizontal, infinite
Consider
parallel plate.
For this geometry the fluid particle move in the x
For
direction parallel to the pates, and there is no velocity
in the y or z direction – that is, v=0 and w=0. 122 Steady, Laminar Flow between Fixed
2/6
Parallel Plates 2/6
From the continuity equation that ∂u/∂x=0.
From
There would be no variation of u in the z direction for
There
infinite plates, and for steady flow so that u=u(y).
The NavierStokes equations reduce to
The ⎛ ∂ 2u ⎞
∂p
0=−
+ µ⎜ 2 ⎟
⎜ ∂y ⎟
∂x
⎝
⎠
∂p
∂p
− ρg
0=−
0=−
∂z
∂y
123 Steady, Laminar Flow between Fixed
3/6
Parallel Plates 3/6
∂p
− ρg
∂y
∂p
0=−
∂z
⎛ ∂2u ⎞
∂p
+ µ⎜ 2 ⎟
0=−
⎜ ∂y ⎟
∂x
⎝
⎠
0=− Integrating p = − ρ gy + f1 (x ) Integrating 1 ⎛ ∂p ⎞ 2
u=
⎜
⎟ y + c1 y + c 2
2µ ⎝ ∂ x ⎠ c1 ? c2 ?
124 Steady, Laminar Flow between Fixed
4/6
Parallel Plates 4/6
With the boundary conditions u=0 at y=h u=0 at y=h
With
1 ⎛ ∂p ⎞ 2
c 2 = 0, c1 = −
⎟h
⎜
2µ ⎝ ∂ x ⎠
Velocity distribution ( 1 ⎛ ∂p ⎞ 2
2
u=
⎜
⎟y −h
2µ ⎝ ∂ x ⎠ )
125 Steady, Laminar Flow between Fixed
5/6
Parallel Plates 5/6
Shear stress distribution
Shear
⎛ ∂p ⎞
=⎜
⎟y
⎝ ∂x ⎠ τ yx Volume flow rate
Volume
q= ∫ h udy = −h ∫ h −h 1 ⎛ ∂p ⎞ 2
2h 3
2
⎜
⎟ ( y − h ) dy = −
2µ ⎝ ∂ x ⎠
3µ ⎛ ∂p ⎞
⎜
⎟
∂x ⎠
⎝ p − p1
∆p
∂p
= cons tan t = 2
=−
∂x
l
l 2 h 3∆p
>> q =
3µ l 126 Steady, Laminar Flow between Fixed
6/6
Parallel Plates 6/6
Average velocity
Average
Vaverage q
h 2 ∆p
=
=
2h
3µ l Point of maximum velocity
Point
du
=0
dy at y=0 u = u max h 2 ⎛ ∂p ⎞ 3
=U=−
⎜
⎟ = Vaverage
2µ ⎝ ∂ x ⎠ 2
127 1/3
Couette Flow 1/3
Since only the boundary conditions have changed, there
Since
is no need to repeat the entire analysis of the “both
plates stationary” case.
1 ⎛ ∂p ⎞ 2
u=
⎟ y + c1 y + c 2
⎜
2µ ⎝ ∂ x ⎠ c1 ? c2 ? 128 2/3
Couette Flow 2/3
The boundary conditions for the moving plate case are
The
u=0 at y=0
u=U at y=b
U
1 ⎛ ∂p ⎞
⇒ c1 =
−
⎜
⎟b
b 2µ ⎝ ∂ x ⎠ Velocity distribution
b2
P=
2µ U ⎛ ∂p ⎞
⎜
⎟
⎝ ∂x ⎠ c2 = 0 Uy
1 ⎛ ∂p ⎞ 2
1 ⎛ ∂p ⎞
+
u=
⎜
⎟ by
⎜
⎟y −
b
2µ ⎝ ∂ x ⎠
2µ ⎝ ∂ x ⎠
u
y
b2
=−
U b 2µ U ⎛ ∂p ⎞⎛ y ⎞ ⎡ ⎛ b ⎞ ⎤
⎜
⎟ ⎜ ⎟ ⎢1 − ⎜ ⎟ ⎥
∂x ⎠⎝ b ⎠ ⎢ ⎜ y ⎟ ⎥
⎝
⎣ ⎝ ⎠⎦
129 3/3
Couette Flow 3/3
Simplest type of Couette flow U = ri ω b = ro − ri τ = µ ri ω /( ro − r ) y
∂p
=0⇒ u=U
∂x
b This flow can be approximated by
the flow between closely spaced
concentric cylinder is fixed and
the other cylinder rotates with a
constant angular velocity.
Flow in the narrow gap
of a journal bearing.
130 Example 6.9 Plane Couette Flow
A wide moving belt passes through a
wide
container of a viscous liquid. The belt
moves vertically upward with a constant
velocity, V0, as illustrated in Figure
E6.9a. Because of viscous forces the belt
picks up a film of fluid of thickness h.
Gravity tends to make the fluid drain
down the belt. Use the Navier Stokes
equations to determine an expression for
the average velocity of the fluid film as it
is dragged up the belt. Assume that the
flow is laminar, steady, and fully
developed.
131 Example 6.9 Solution1/2
Since the flow is assumed to be fully developed, the only velocity
component is in the y direction so that u=w=0.
From the continuity equation
∂v
= 0 , and for steady flow, so that v=v(x)
∂y
∂p
∂p
=0
=0
dv
γ
∂x
∂z
= x + c1
2 0 = −ρg + µ dv dx 2 Integrating dx τ xy µ ⎛ dv ⎞
= µ⎜
⎟
⎝ dx ⎠
132 Example 6.9 Solution2/2
τ xy = 0 at x = h
c1 = − γh
µ v = V0 at x = 0 Integrating γ 2 γh
v=
x−
x + c2
2µ
2µ γ 2 γh
v=
x−
x + V0
2µ
2µ 133 1/5
Steady, Laminar Flow in Circular Tubes 1/5 Consider the flow through a horizontal circular tube of
Consider
radius R.
∂v z
v r = 0,
vθ = 0 ⇒
= 0 v z = v z (r )
∂z 134 Steady, Laminar Flow in Circular Tubes 2/5
Navier – Stokes equation reduced to
∂p
0 = −ρg sin θ −
∂r
1 ∂p
0 = −ρg sin θ −
r ∂θ
⎛ 1 ∂ ⎛ ∂v z
∂p
0=−
+ µ⎜
⎜ r ∂r ⎜ r ∂r
∂z
⎝
⎝ p = −ρg (r sin θ) + f1 (z )
Integrating ⎞⎞
⎟⎟
⎟
⎠⎠ p = −ρgy + f1 (z ) Integrating 1 ⎛ ∂p ⎞ 2
vz =
⎜ ⎟ r + c1 ln r + c 2
4µ ⎝ ∂z ⎠ c1 ? c2 ?
135 Steady, Laminar Flow in Circular Tubes 3/5
At r=0, the velocity vz is finite. At r=R, the velocity vz
is zero.
1 ⎛ ∂p ⎞ 2
c1 = 0, c 2 = −
⎜ ⎟R
4µ ⎝ ∂z ⎠ Velocity distribution ( 1 ⎛ ∂p ⎞ 2
2
vz =
⎜ ⎟r −R
4µ ⎝ ∂z ⎠ )
136 4/5
Steady, Laminar Flow in Circular Tubes 4/5 The shear stress distribution
The
τ rx du
r ⎛ ∂p ⎞
=µ
=⎜
⎟
dr
2 ⎝ ∂x ⎠ Volume flow rate
Volume
π R 4 ⎛ ∂p ⎞
Q = u z 2 π rdr = ..... = −
⎜
⎟
8µ ⎝ ∂ z ⎠
0 ∫ R p 2 − p1
∂p
= cons tan t =
= − ∆p / l
l
∂z π R 4 ⎛ ∂p ⎞ π R 4 ⎛ ∆ p ⎞ π ∆ pD 4
>> Q = −
⎜
⎟=
⎜
⎟=
8µ ⎝ ∂ z ⎠
8µ ⎝ l ⎠ 128 µ l 137 5/5
Steady, Laminar Flow in Circular Tubes 5/5 Average velocity
Average
Vaverage Q
Q
R 2 ∆p
=
=
=
2
A πR
8µ l Point of maximum velocity
Point
dv z
=0
dr at r=0
2 v max vz
R ∆p
⎛r⎞
=−
= 2 Vaverage ⇒
=1− ⎜ ⎟
4µ l
v max
⎝R⎠ 2 138 1/2
Steady, Axial, Laminar Flow in an Annulus 1/2 For steady, laminar flow in circular tubes
1 ⎛ ∂p ⎞ 2
vz =
⎜ ⎟ r + c1 ln r + c 2
4µ ⎝ ∂z ⎠ c1 ? c2 ? Boundary conditions
vz = 0 , at r = ro
vz = 0 , at r = ri 139 2/2
Steady, Axial, Laminar Flow in an Annulus 2/2 The velocity distribution
2
ri2 − ro
1 ⎛ ∂p ⎞ ⎡ 2 2
r⎤
vz =
ln ⎥
⎜ ⎟ ⎢ r − ro +
4µ ⎝ ∂z ⎠ ⎢
ln( ro / ri ) ro ⎥
⎣
⎦ The volume rate of flow
2
π ⎛ ∂p ⎞ ⎡ 4 4 ( ro − ri2 ) 2 ⎤
Q = v z ( 2πr )dr = − ⎜ ⎟ ⎢ ro − ri −
⎥
8µ ⎝ ∂z ⎠ ⎢
ln( ro / ri ) ⎥
ri
⎣
⎦ ∫ ro The maximum velocity occurs at r=rm
∂v z
=0
∂r 1/ 2 ⎡
⎤
rm = ⎢
⎥
⎢ 2 ln( ro / ri ) ⎥
⎣
⎦
2
ro − ri2 140 ...
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This note was uploaded on 11/17/2011 for the course MECHANICAL 303 taught by Professor Shieh during the Spring '11 term at California Coast University.
 Spring '11
 Shieh
 Mechatronics

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