Chapter 6
1
Chapter 6 Differential Analysis of Fluid Flow
Fluid Element Kinematics
Fluid element motion consists of translation, linear deformation, rotation, and angular deformation.
Types of motion and deformation for a fluid element.
Linear Motion and Deformation:
Translation of a fluid element
Linear deformation of a fluid element
1
Chapter 6
2
Change in δ∀ :
 ∂u 
δ x  (δ yδ z ) δ t
 ∂x 
δ∀ =
the rate at which the volume δ∀ is changing per unit volume due to the gradient ∂u/∂x is
∀)
 ( ∂u ∂x ) t  ∂u
1 d (dd
= lim
=


d t →0
dd
t
∀ dt

 ∂x
If velocity gradients ∂v/∂y and ∂w/∂z are also present, then
using a similar analysis it follows that, in the general case,
1 d (d∀ ) ∂u ∂v ∂w
=
+
+
= ∇⋅V
d∀ dt
∂x ∂y ∂z
This rate of change of the volume per unit volume is called
the volumetric dilatation rate.
Angular Motion and Deformation
For simplicity we will consider motion in the x–y plane,
but the results can be readily extended to the more general
case.
2
Chapter 6
3
Angular motion and deformation of a fluid element
The angular velocity of line OA, ωOA, is
δα
δ t →0 δ t
ωOA = lim
For small angles
tan δa
=
≈ δa
x ) δ xδ t
( ∂v ∂=
δx
∂v
δt
∂x
so that
 ( ∂v ∂x ) δ t  ∂v
ωOA lim
=
=


δ t →0
δ
t

 ∂x
Note that if ∂v/∂x is positive, ωOA will be counterclockwise.
Similarly, the angular velocity of the line OB is
δβ ∂u
=
ωOB lim
=
δ t →0 δ t
∂y
In this instance if ∂u/∂y is positive, ωOB will be clockwise.
3
Chapter 6
4
The rotation, ωz, of the element about the z axis is defined
as the average of the angular velocities ωOA and ωOB of the
two mutually perpendicular lines OA and OB. Thus, if
counterclockwise rotation is considered to be positive, it
follows that
1  ∂v ∂u 
ωz
=
 − 
2  ∂x ∂y 
Rotation of the field element about the other two coordinate
axes can be obtained in a similar manner:
1  ∂w ∂v 
wx
=
− 

2  ∂y ∂z 
wy
=
1  ∂u ∂w 
−


2  ∂z ∂x 
The three components, ωx,ωy, and ωz can be combined to
give the rotation vector, ω, in the form:
1
1
ω= ω x i + ω y j + ω z k=
curlV=
∇×V
2
2
since
i
j
k
1
1 ∂
∇×V =
2
2 ∂x
u
=
∂
∂y
v
∂
∂z
w
1  ∂w ∂v  1  ∂u ∂w  1  ∂v ∂u 
− i + 
−

 j +  − k
2  ∂y ∂z  2  ∂z ∂x 
2  ∂x ∂y 
4
Chapter 6
5
The vorticity, ζ, is defined as a vector that is twice the rotation vector; that is,
ς = 2ω = ∇ × ς
The use of the vorticity to describe the rotational characteristics of the fluid simply eliminates the (1/2) factor associated with the rotation vector. If ∇ × V = 0 , the flow is
called irrotational.
In addition to the rotation associated with the derivatives
∂u/∂y and ∂v/∂x, these derivatives can cause the fluid element to undergo an angular deformation, which results in a
change in shape of the element. The change in the original
right angle formed by the lines OA and OB is termed the
shearing strain, δγ,
δγ
= δα + δβ
The rate of change of δγ is called the rate of shearing strain
or the rate of angular deformation:
(𝜕𝜕𝜕𝜕 ⁄𝜕𝜕𝜕𝜕 )𝛿𝛿𝛿𝛿 + (𝜕𝜕𝜕𝜕⁄𝜕𝜕𝜕𝜕)𝛿𝛿𝛿𝛿
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
𝛿𝛿𝛿𝛿
𝛿𝛿𝛿𝛿
= lim
=�
�=
+
𝛿𝛿𝛿𝛿→0 𝛿𝛿𝛿𝛿
𝛿𝛿𝛿𝛿→0 𝛿𝛿𝛿𝛿
𝛿𝛿𝛿𝛿
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
𝛾𝛾̇𝑥𝑥𝑥𝑥 = lim
Similarly,
𝛾𝛾̇𝑥𝑥𝑥𝑥 =
𝛾𝛾̇𝑦𝑦𝑦𝑦 =
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
+
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
+
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
The rate of angular deformation is related to a corresponding shearing stress which causes the fluid element to
change in shape.
5
Chapter 6
6
The Continuity Equation in Differential Form
The governing equations can be expressed in both integral
and differential form. Integral form is useful for large-scale
control volume analysis, whereas the differential form is
useful for relatively small-scale point analysis.
Application of RTT to a fixed elemental control volume
yields the differential form of the governing equations. For
example for conservation of mass
∂ρ
dV
CV ∂t
∑ ρV ⋅ A = − ∫
CS
net outflow of mass
across CS
=
rate of decrease
of mass within CV
6
Chapter 6
7
Consider a cubical element oriented so that its sides are to
the (x,y,z) axes
∂

(ρu )dx  dydz
ρu +

inlet mass flux
ρudydz
∂x

outlet mass flux
Taylor series expansion
retaining only first order term
We assume that the element is infinitesimally small such
that we can assume that the flow is approximately one dimensional through each face.
The mass flux terms occur on all six faces, three inlets, and
three outlets. Consider the mass flux on the x faces
∂


x flux =
ρu
+
ρu
dx
dydz outflux − ρudydz influx
(
)


∂x

=
∂
(ρu )dxdydz
∂x
V
Similarly for the y and z faces
∂
y flux = (ρv)dxdydz
∂y
∂
z flux = (ρw )dxdydz
∂z
7
Chapter 6
8
The total net mass outflux must balance the rate of decrease
of mass within the CV which is
∂ρ
− dxdydz
∂t
Combining the above expressions yields the desired result
 ∂ρ ∂

∂
∂
(
u
)
(
v
)
(
w
)
ρ
+
ρ
+
ρ
+
 ∂t ∂x
 dxdydz = 0
y
z
∂
∂


dV
∂ρ ∂
∂
∂
+ (ρu ) + (ρv) + (ρw ) = 0
∂t ∂x
∂y
∂z
per unit V
differential form of continuity equations
∂ρ
+ ∇ ⋅ (ρV) = 0
∂t
ρ∇ ⋅ V + V ⋅ ∇ρ
Dρ
+ ρ∇ ⋅ V = 0
Dt
D ∂
= + V ⋅∇
Dt ∂t
Nonlinear 1st order PDE; ( unless ρ = constant, then linear)
Relates V to satisfy kinematic condition of mass conservation
Simplifications:
1. Steady flow: ∇ ⋅ (ρV) = 0
2. ρ = constant: ∇ ⋅ V = 0
8
Chapter 6
9
i.e.,
∂u ∂v ∂w
+ +
=0
∂x ∂y ∂z
3D
∂u ∂v
+
=0
∂x ∂y
2D
The continuity equation in Cylindrical Polar Coordinates
The velocity at some arbitrary point P can be expressed as
V = vr e r + vθ eθ + vz e z
The continuity equation:
vr ) 1 ∂ ( vθ ) ∂ ( vz )
∂r 1 ∂ ( r rrr
0
+
+
+
=
∂t r
∂r
∂z
r ∂θ
For steady, compressible flow
vr ) 1 ∂ ( vθ ) ∂ ( vz )
1 ∂ ( r rrr
0
+
+
=
∂r
∂z
r
r ∂θ
For incompressible fluids (for steady or unsteady flow)
1 ∂ ( rvr ) 1 ∂vθ ∂vz
0
+
+
=
r ∂r
r ∂θ ∂z
9
Chapter 6
10
The Stream Function
Steady, incompressible, plane, two-dimensional flow represents one of the simplest types of flow of practical importance. By plane, two-dimensional flow we mean that
there are only two velocity components, such as u and v,
when the flow is considered to be in the x–y plane. For this
flow the continuity equation reduces to
∂u ∂v
+
=0
∂x ∂y
We still have two variables, u and v, to deal with, but they
must be related in a special way as indicated. This equation
suggests that if we define a function ψ(x, y), called the
stream function, which relates the velocities as
∂y
∂y
u=
, v= −
∂y
∂x
then the continuity equation is identically satisfied:
∂  ∂y  ∂  ∂y  ∂ 2y
∂ 2y
−
= 0

+ −
=
∂x  ∂y  ∂y  ∂x  ∂x∂y ∂x∂y
Velocity and velocity components along a streamline
10
Chapter 6
11
Another particular advantage of using the stream function
is related to the fact that lines along which ψ is constant are
streamlines.The change in the value of ψ as we move from
one point (x, y) to a nearby point (x + dx, y + dy) along a
line of constant ψ is given by the relationship:
∂y
∂y
−vdx + udy =
dy = dx +
dy =
0
∂x
∂y
and, therefore, along a line of constant ψ
dy v
=
dx u
The flow between two streamlines
The actual numerical value associated with a particular
streamline is not of particular significance, but the change
in the value of ψ is related to the volume rate of flow. Let
dq represent the volume rate of flow (per unit width perpendicular to the x–y plane) passing between the two
streamlines.
∂y
∂y
dq = udy − vdx =
dx +
dy = dy
∂x
∂y
Thus, the volume rate of flow, q, between two streamlines
such as ψ1 and ψ2, can be determined by integrating to
yield:
11
Chapter 6
12
=
q
ψ2
∫ψ
= 2− 1
dψψψ
1
In cylindrical coordinates the continuity equation for incompressible, plane, two-dimensional flow reduces to
1 ∂ ( rvr ) 1 ∂vθ
0
+
=
r ∂r
r ∂θ
and the velocity components, vr and vθ, can be related to the
stream function, ψ(r, θ), through the equations
∂
1 ∂ψψ
, vθ = −
vr =
∂r
r ∂θ
Navier-Stokes Equations
Differential form of momentum equation can be derived by
applying control volume form to elemental control volume
The differential equation of linear momentum: elemental
fluid volume approach
12
Chapter 6
13
∑𝐹𝐹 =
𝜕𝜕
� 𝜌𝜌𝑉𝑉𝑑𝑑V + � 𝑉𝑉𝜌𝜌𝑉𝑉 ⋅ 𝑛𝑛�𝑑𝑑𝐴𝐴
𝜕𝜕𝜕𝜕 ����
��
�� ���������
𝐶𝐶𝐶𝐶 �
𝐶𝐶𝐶𝐶
(1)
𝜕𝜕
(2)
𝜕𝜕𝑉𝑉
𝜕𝜕𝜕𝜕
(1) = 𝜕𝜕𝜕𝜕 �𝜌𝜌𝑉𝑉�𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = � 𝜕𝜕𝜕𝜕 𝑉𝑉 + 𝜌𝜌 𝜕𝜕𝜕𝜕 � 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝜕𝜕
𝜕𝜕
𝜕𝜕
(2) = ���
�𝜌𝜌𝜌𝜌𝑉𝑉�
+
�𝜌𝜌𝜌𝜌𝑉𝑉�
+
�𝜌𝜌𝜌𝜌𝑉𝑉�
�� ��
��� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝜕𝜕𝜕𝜕���
𝜕𝜕𝜕𝜕���
𝜕𝜕𝜕𝜕 ���
�� ��
𝑥𝑥−face
𝜕𝜕𝑉𝑉
=�𝜌𝜌𝜌𝜌 𝜕𝜕𝜕𝜕 + 𝑉𝑉
𝑧𝑧−face
𝑦𝑦−face
𝜕𝜕𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
+ 𝜌𝜌𝜌𝜌
𝜕𝜕𝑉𝑉
𝜕𝜕𝜕𝜕
+ 𝑉𝑉
𝜕𝜕𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
+ 𝜌𝜌𝜌𝜌
𝜕𝜕𝑉𝑉
𝜕𝜕𝜕𝜕
+ 𝑉𝑉
𝜕𝜕𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
combining and making use of the continuity equation yields
𝜕𝜕𝜕𝜕
𝜕𝜕𝑉𝑉
∑𝐹𝐹 = �𝑉𝑉 ���+
∇ ⋅ �𝜌𝜌𝑉𝑉�
����
���� + 𝜌𝜌 �
𝜕𝜕𝜕𝜕
∴ ∑𝐹𝐹 = 𝜌𝜌
=0
𝐷𝐷𝑉𝑉
𝐷𝐷𝐷𝐷
𝜕𝜕𝜕𝜕
+ 𝑉𝑉 ⋅ ∇𝑉𝑉�� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 or
where ∑𝐹𝐹 = ∑𝐹𝐹body + ∑𝐹𝐹surface
∑f = ∑f𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 + ∑f𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
∑f = 𝜌𝜌
𝐷𝐷𝑉𝑉
𝐷𝐷𝐷𝐷
𝑉𝑉 ⋅ ∇ = 𝑢𝑢
𝐷𝐷
𝐷𝐷𝐷𝐷
=
𝜕𝜕
𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
+ 𝑣𝑣
𝜕𝜕
𝜕𝜕𝜕𝜕
+ 𝑉𝑉 ⋅ ∇
+ 𝑤𝑤
𝜕𝜕
𝜕𝜕𝜕𝜕
13
Chapter 6
14
Body forces are due to external fields such as gravity or
magnetics. Here we only consider a gravitational field; that
is,
∑ F body = d F gρav = ρgdxdydz
and g = −gk̂
for g↓ z↑
i.e., f body = −ρgk̂
Surface forces are due to the stresses that act on the sides of
the control surfaces
symmetric (σij = σji)
σij = - pδij + τij
2nd order tensor
normal pressure
viscous stress
=
-p+τxx
τyx
τzx
τxy
-p+τyy
τzy
δij = 1
δij = 0
i=j
i≠j
τxz
τyz
-p+τzz
As shown before for p alone it is not the stresses themselves that cause a net force but their gradients.
∂

∂
∂
dFx,surf =  (σ xx ) + (σ xy ) + (σ xz ) dxdydz
∂y
∂z
 ∂x

 ∂p ∂

∂
∂
= − + (τ xx ) + (τ xy ) + (τ xz ) dxdydz
∂y
∂z
 ∂x ∂x

14
Chapter 6
15
This can be put in a more compact form by defining vector
stress on x-face
τ x = τ xx î + τ xy ĵ + τ xz k̂
and noting that
 ∂p

dFx,surf = − + ∇ ⋅ τ x  dxdydz
 ∂x

∂p
per unit volume
fx,surf = − + ∇ ⋅ τ x
∂x
similarly for y and z
∂p
fy,surf = − + ∇ ⋅ τ y
∂y
τ y = τ yx î + τ yy ĵ + τ yz k̂
∂p
+ ∇ ⋅ τz
∂z
τ z = τ zx î + τ zy ĵ + τ zz k̂
fz,surf = −
finally if we define
τ ij = τ x î + τ y ĵ + τ z k̂
then
f surf = −∇p + ∇ ⋅ τ ij = ∇ ⋅ s ij
σ ij = − pδ ij + τ ij
15
Chapter 6
16
Putting together the above results
∑ f = f body + f surf = r
DV
Dt
f body = −ρgk̂
f surface = −∇p + ∇ ⋅ τ ij
=
a
DV ∂ V
=
+ V ⋅∇ V
Dt
∂t
ρ a = − ρ gkˆ − ∇p + ∇ ⋅τ ij
inertia
force
body
force surface
due to force due
gravity to p
surface force
due to viscous
shear and normal
stresses
16
Chapter 6
17
For Newtonian fluid the shear stress is proportional to the
rate of strain, which for incompressible flow can be written
𝜏𝜏𝑖𝑖𝑖𝑖 = 2𝜇𝜇𝜀𝜀𝑖𝑖𝑖𝑖 = 𝜇𝜇 �
𝜕𝜕𝑢𝑢𝑖𝑖
𝜕𝜕𝑥𝑥𝑗𝑗
+
𝜕𝜕𝑢𝑢𝑗𝑗
𝜕𝜕𝑥𝑥𝑖𝑖
�
where,
𝜇𝜇 = coefficient of viscosity
𝜀𝜀𝑖𝑖𝑖𝑖 = rate of strain tensor
𝜕𝜕𝜕𝜕
⎡
𝜕𝜕𝜕𝜕
⎢ 1 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
= ⎢ 2 �𝜕𝜕𝜕𝜕 + 𝜕𝜕𝜕𝜕�
⎢1 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
⎣2 � 𝜕𝜕𝜕𝜕 + 𝜕𝜕𝜕𝜕 �
1 𝜕𝜕𝜕𝜕
�
+
�
+
𝜕𝜕𝜕𝜕
2 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
�
𝜕𝜕𝜕𝜕
1 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
2 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
�
𝜌𝜌𝑎𝑎 = −𝜌𝜌𝜌𝜌𝑘𝑘� − ∇𝑝𝑝 + ∇ ⋅ �𝜏𝜏𝑖𝑖𝑖𝑖 �
1 𝜕𝜕𝜕𝜕
�
𝜕𝜕𝜕𝜕
�⎤
⎥
� + �⎥
2 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
⎥
𝜕𝜕𝜕𝜕
⎦
𝜕𝜕𝜕𝜕
2 𝜕𝜕𝜕𝜕
1 𝜕𝜕𝜕𝜕
+
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
Ex) 1-D flow
𝑑𝑑𝑑𝑑
𝜏𝜏 = 𝜇𝜇
𝑑𝑑𝑑𝑑
where,
∇ ⋅ �𝜏𝜏𝑖𝑖𝑖𝑖 � = 𝜇𝜇
2
𝜕𝜕𝑢𝑢𝑗𝑗
𝜕𝜕𝑢𝑢𝑖𝑖
𝜕𝜕 𝑢𝑢
𝜕𝜕 𝜕𝜕𝑢𝑢𝑗𝑗
⎞
+ � = 𝜇𝜇 ⎛ 2𝑖𝑖 +
�
𝜕𝜕𝑥𝑥𝑗𝑗 𝜕𝜕𝑥𝑥𝑗𝑗
𝜕𝜕𝑥𝑥
𝜕𝜕𝑥𝑥𝑖𝑖
𝜕𝜕𝑥𝑥
𝜕𝜕𝑥𝑥
𝑖𝑖 �𝑗𝑗
�𝑗𝑗
=0 ⎠
⎝ ∇2𝑉𝑉
𝜕𝜕
𝜌𝜌𝑎𝑎 = −𝜌𝜌𝜌𝜌𝑘𝑘� − ∇𝑝𝑝 + 𝜇𝜇∇2 𝑉𝑉
𝜌𝜌𝑎𝑎 = −∇(𝑝𝑝 + 𝛾𝛾𝛾𝛾) + 𝜇𝜇∇2 𝑉𝑉
∇ ⋅ 𝑉𝑉 = 0
Navier-Stokes Equation
Continuity Equation
17
Chapter 6
18
Four equations in four unknowns: V and p
Difficult to solve since 2nd order nonlinear PDE
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕2 𝑢𝑢
𝜕𝜕2 𝑢𝑢
𝜕𝜕2 𝑢𝑢
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕2 𝑣𝑣
𝜕𝜕2 𝑣𝑣
𝜕𝜕2 𝑣𝑣
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
x: 𝜌𝜌 � 𝜕𝜕𝜕𝜕 + 𝑢𝑢 𝜕𝜕𝜕𝜕 + 𝑣𝑣 𝜕𝜕𝜕𝜕 + 𝑤𝑤 𝜕𝜕𝜕𝜕 � = − 𝜕𝜕𝜕𝜕 + 𝜇𝜇 �𝜕𝜕𝑥𝑥2 + 𝜕𝜕𝑦𝑦2 + 𝜕𝜕𝑧𝑧 2 �
y: 𝜌𝜌 � 𝜕𝜕𝜕𝜕 + 𝑢𝑢 𝜕𝜕𝜕𝜕 + 𝑣𝑣 𝜕𝜕𝜕𝜕 + 𝑤𝑤 𝜕𝜕𝜕𝜕� = − 𝜕𝜕𝜕𝜕 + 𝜇𝜇 �𝜕𝜕𝑥𝑥2 + 𝜕𝜕𝑦𝑦2 + 𝜕𝜕𝑧𝑧 2 �
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕2 𝑤𝑤
𝜕𝜕2 𝑤𝑤
𝜕𝜕2 𝑤𝑤
z: 𝜌𝜌 � 𝜕𝜕𝜕𝜕 + 𝑢𝑢 𝜕𝜕𝜕𝜕 + 𝑣𝑣 𝜕𝜕𝜕𝜕 + 𝑤𝑤 𝜕𝜕𝜕𝜕 � = − 𝜕𝜕𝜕𝜕 − 𝜌𝜌𝜌𝜌 + 𝜇𝜇 � 𝜕𝜕𝑥𝑥2 + 𝜕𝜕𝑦𝑦2 + 𝜕𝜕𝑧𝑧 2 �
∂u ∂v ∂w
+ +
=0
∂x ∂y ∂z
Navier-Stokes equations can also be written in other coordinate systems such as cylindrical, spherical, etc.
There are about 80 exact solutions for simple geometries.
For practical geometries, the equations are reduced to algebraic form using finite differences and solved using computers.
18
Chapter 6
19
Ex) Exact solution for laminar incompressible steady flow
in a circular pipe
Use cylindrical coordinates with assumptions
𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕
𝜕𝜕𝜕𝜕
= 0 : Steady flow
= 0 : Fully-developed flow
𝑣𝑣𝑟𝑟 = 0 : Flow is laminar and parallel to the wall
𝑣𝑣𝜃𝜃 =
𝜕𝜕
𝜕𝜕𝜕𝜕
= 0 : Flow is axisymmetric with no swirl
Continuity equation:
1 𝜕𝜕(𝑟𝑟𝑣𝑣𝑟𝑟 )
𝑟𝑟
𝜕𝜕𝜕𝜕
+
1 𝜕𝜕𝑣𝑣𝜃𝜃
𝑟𝑟 𝜕𝜕𝜕𝜕
+
𝜕𝜕𝑣𝑣𝑧𝑧
𝜕𝜕𝜕𝜕
=0
Thus, (𝑣𝑣𝑟𝑟 , 𝑣𝑣𝜃𝜃 , 𝑣𝑣𝑧𝑧 ) satisfies the continuity equation
19
Chapter 6
20
Momentum equation:
𝜌𝜌 �
𝜕𝜕𝑣𝑣𝑟𝑟
𝜌𝜌 �
𝜕𝜕𝑣𝑣𝜃𝜃
𝜌𝜌 �
𝜕𝜕𝑣𝑣𝑧𝑧
or
𝜕𝜕𝜕𝜕
+ 𝑣𝑣𝑟𝑟
𝜕𝜕𝑣𝑣𝑟𝑟
+ 𝑣𝑣𝑟𝑟
𝜕𝜕𝑣𝑣𝜃𝜃
=−
𝜕𝜕𝜕𝜕
=−
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
1 𝜕𝜕𝜕𝜕
𝑟𝑟 𝜕𝜕𝜕𝜕
+ 𝑣𝑣𝑟𝑟
=−
𝜕𝜕𝜕𝜕
+
𝑣𝑣𝜃𝜃 𝜕𝜕𝑣𝑣𝑟𝑟
+
𝑣𝑣𝜃𝜃 𝜕𝜕𝑣𝑣𝜃𝜃
+
𝑣𝑣𝜃𝜃 𝜕𝜕𝑣𝑣𝑧𝑧
𝑟𝑟 𝜕𝜕𝜕𝜕
+ 𝜌𝜌𝑔𝑔𝑟𝑟 + 𝜇𝜇 �
𝜕𝜕𝜕𝜕
𝑟𝑟 𝜕𝜕𝜕𝜕
𝜕𝜕𝑣𝑣𝑧𝑧
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝑟𝑟 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝑟𝑟 𝜕𝜕𝜕𝜕
+
+ 𝜇𝜇 �
+ 𝑣𝑣𝑧𝑧
�𝑟𝑟
𝑣𝑣𝑟𝑟 𝑣𝑣𝜃𝜃
𝑟𝑟
1 𝜕𝜕
�𝑟𝑟
𝑟𝑟 𝜕𝜕𝜕𝜕
+ 𝑣𝑣𝑧𝑧
+ 𝜌𝜌𝑔𝑔𝑧𝑧 + 𝜇𝜇 �
0 = −𝜌𝜌𝜌𝜌 cos 𝜃𝜃 −
𝜕𝜕𝜕𝜕
𝑟𝑟
1 𝜕𝜕
+ 𝜌𝜌𝑔𝑔𝜃𝜃 + 𝜇𝜇 �
0 = −𝜌𝜌𝜌𝜌 sin 𝜃𝜃 −
0=−
−
𝑣𝑣𝜃𝜃2
1 𝜕𝜕
1 𝜕𝜕2 𝑣𝑣𝑟𝑟
𝑣𝑣𝑟𝑟
� − 𝑟𝑟 2 + 𝑟𝑟 2
𝜕𝜕𝑣𝑣𝜃𝜃
𝜕𝜕𝜕𝜕
�
𝜕𝜕𝑣𝑣𝜃𝜃
�
𝜕𝜕𝜕𝜕
𝑣𝑣𝜃𝜃
𝜕𝜕𝑣𝑣𝑧𝑧
1 𝜕𝜕2 𝑣𝑣𝑧𝑧
� + 𝑟𝑟 2
𝜕𝜕𝜕𝜕
−
1 𝜕𝜕2 𝑣𝑣𝜃𝜃
� − 𝑟𝑟 2 + 𝑟𝑟 2
𝜕𝜕𝜕𝜕
�
𝜕𝜕𝜃𝜃 2
𝜕𝜕𝜃𝜃 2
𝜕𝜕𝜃𝜃 2
+
2 𝜕𝜕𝑣𝑣𝜃𝜃
𝑟𝑟 2 𝜕𝜕𝜕𝜕
+
𝜕𝜕2 𝑣𝑣𝑧𝑧
𝜕𝜕𝑧𝑧 2
+
2 𝜕𝜕𝑣𝑣𝑟𝑟
𝑟𝑟 2 𝜕𝜕𝜕𝜕
𝜕𝜕2 𝑣𝑣𝑟𝑟
+
𝜕𝜕𝑧𝑧 2
�
𝜕𝜕2 𝑣𝑣𝜃𝜃
𝜕𝜕𝑧𝑧 2
�
�
(1)
𝑟𝑟 𝜕𝜕𝜕𝜕
1 𝜕𝜕
𝜕𝜕𝑣𝑣𝑧𝑧
where,
𝑔𝑔𝑟𝑟 = −𝑔𝑔 sin 𝜃𝜃
𝑔𝑔𝜃𝜃 = −𝑔𝑔 cos 𝜃𝜃
𝜕𝜕𝜕𝜕
+ 𝑣𝑣𝑧𝑧
�𝑟𝑟
𝑟𝑟 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
1 𝜕𝜕𝜕𝜕
�𝑟𝑟
𝜕𝜕𝜕𝜕
𝜕𝜕𝑣𝑣𝑧𝑧
𝜕𝜕𝜕𝜕
𝑟𝑟 𝜕𝜕𝜕𝜕
𝜕𝜕𝑣𝑣𝑟𝑟
𝜕𝜕𝑣𝑣𝑟𝑟
𝜕𝜕𝜕𝜕
(2)
��
(3)
Equations (1) and (2) can be integrated to give
𝑝𝑝 = −𝜌𝜌𝜌𝜌(𝑟𝑟 sin 𝜃𝜃 ) + 𝑓𝑓1 (𝑧𝑧) = −𝜌𝜌𝜌𝜌𝜌𝜌 + 𝑓𝑓1 (𝑧𝑧)
⇒ pressure 𝑝𝑝 is hydrostatic and 𝜕𝜕𝜕𝜕⁄𝜕𝜕𝜕𝜕 is not a function of 𝑟𝑟 or 𝜃𝜃
20
Chapter 6
21
Equation (3) can be written in the from
1 𝜕𝜕
𝜕𝜕𝑣𝑣𝑧𝑧
1 𝜕𝜕𝜕𝜕
�𝑟𝑟
�=
𝜇𝜇 𝜕𝜕𝜕𝜕
𝑟𝑟 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
and integrated (using the fact that 𝜕𝜕𝜕𝜕⁄𝜕𝜕𝜕𝜕 = constant) to
give
𝑟𝑟
𝜕𝜕𝑣𝑣𝑧𝑧
1 𝜕𝜕𝜕𝜕 2
=
� � 𝑟𝑟 + 𝐶𝐶1
𝜕𝜕𝜕𝜕
2𝜇𝜇 𝜕𝜕𝜕𝜕
Integrating again we obtain
B.C.
1 𝜕𝜕𝜕𝜕 2
𝑣𝑣𝑧𝑧 =
� � 𝑟𝑟 + 𝐶𝐶1 ln 𝑟𝑟 + 𝐶𝐶2
4𝜇𝜇 𝜕𝜕𝜕𝜕
𝑣𝑣𝑧𝑧 (𝑟𝑟 = 0) ≠ ∞ ⇒ 𝐶𝐶1 = 0
𝑣𝑣𝑧𝑧 (𝑟𝑟 = 𝑅𝑅) = 0 ⇒ 𝐶𝐶2 = −
∴ 𝑣𝑣𝑧𝑧 =
1
𝜕𝜕𝜕𝜕
� � 𝑅𝑅2
4𝜇𝜇 𝜕𝜕𝜕𝜕
1 𝜕𝜕𝜕𝜕
� � (𝑟𝑟 2 − 𝑅𝑅2 )
4𝜇𝜇 𝜕𝜕𝜕𝜕
⇒ at any cross section the velocity distribution is parabolic
21
Chapter 6
22
1) Flow rate 𝑄𝑄:
𝑅𝑅
𝑅𝑅
𝜋𝜋𝑅𝑅4 𝜕𝜕𝜕𝜕
� �
𝑄𝑄 = � 𝑣𝑣𝑧𝑧 𝑑𝑑𝑑𝑑 = 2𝜋𝜋 � 𝑣𝑣𝑧𝑧 𝑟𝑟𝑟𝑟𝑟𝑟 = −
8𝜇𝜇
𝜕𝜕𝜕𝜕
0
0
where, 𝑑𝑑𝑑𝑑 = (2𝜋𝜋𝜋𝜋)𝑑𝑑𝑑𝑑
If the pressure drops Δ𝑝𝑝 over a length ℓ:
2) Mean velocity 𝑉𝑉:
𝜋𝜋𝑅𝑅 4 Δ𝑝𝑝
𝑄𝑄 =
8𝜇𝜇ℓ
Δ𝑝𝑝
ℓ
=−
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝑄𝑄
1
𝜋𝜋𝑅𝑅4 Δ𝑝𝑝
𝑅𝑅 2 Δ𝑝𝑝
𝑉𝑉 = = � 2 � �
�=
𝐴𝐴
𝜋𝜋𝑅𝑅
8𝜇𝜇ℓ
8𝜇𝜇ℓ
3) Maximum velocity 𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚 :
𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚
⇒
𝑅𝑅2 𝜕𝜕𝜕𝜕
𝑅𝑅 2 Δ𝑝𝑝
= 𝑣𝑣𝑧𝑧 (𝑟𝑟 = 0) = − � � =
= 2𝑉𝑉
4𝜇𝜇 𝜕𝜕𝜕𝜕
4𝜇𝜇ℓ
𝑣𝑣𝑧𝑧
𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚
𝑟𝑟 2
=1−� �
𝑅𝑅
22
Chapter 6
23
4) Wall shear stress (𝜏𝜏𝑟𝑟𝑟𝑟 )𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 :
where
𝜕𝜕𝑣𝑣𝑟𝑟 𝜕𝜕𝑣𝑣𝑧𝑧
𝜕𝜕𝑣𝑣𝑧𝑧
𝜏𝜏𝑟𝑟𝑟𝑟 = 𝜇𝜇 �
+
� = 𝜇𝜇
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝑣𝑣𝑧𝑧
2𝑟𝑟
4𝑉𝑉𝑉𝑉
= 𝑣𝑣�
�−
�
=
−
𝑚𝑚𝑚𝑚𝑚𝑚
𝜕𝜕𝜕𝜕
𝑅𝑅 2
𝑅𝑅 2
=2𝑉𝑉
Thus, at the wall (i.e., 𝑟𝑟 = 𝑅𝑅),
(𝜏𝜏𝑟𝑟𝑟𝑟 )𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = −
and with 𝑄𝑄 = 𝜋𝜋𝑅𝑅 2 𝑉𝑉,
|(𝜏𝜏𝑟𝑟𝑟𝑟 )𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 | =
4𝜇𝜇𝜇𝜇
𝑅𝑅
4𝜇𝜇𝜇𝜇
𝜋𝜋𝑅𝑅 3
Note: Only valid for laminar flows. In general, the flow
remains laminar for Reynolds numbers, Re = 𝜌𝜌𝜌𝜌(2𝑅𝑅)⁄𝜇𝜇,
below 2100. Turbulent flow in tubes is considered in Chapter 8.
23
Chapter 6
24
Differential Analysis of Fluid Flow
We now discuss a couple of exact solutions to the NavierStokes equations. Although all known exact solutions
(about 80) are for highly simplified geometries and flow
conditions, they are very valuable as an aid to our understanding of the character of the NS equations and their solutions. Actually the examples to be discussed are for internal flow (Chapter 8) and open channel flow (Chapter
10), but they serve to underscore and display viscous flow.
Finally, the derivations to follow utilize differential analysis. See the text for derivations using CV analysis.
Couette Flow
boundary conditions
First, consider flow due to the relative motion of two parallel plates
Continuity
∂u
=0
∂x
Momentum
d 2u
0=µ 2
dy
u = u(y)
v=o
∂p ∂p
=
=0
∂x ∂y
or by CV continuity and momentum equations:
24
Chapter 6
25
ρu1∆y = ρu 2 ∆y
u1 = u2
∑ Fx =∑ uρV ⋅ d A = ρQ(u 2 − u1 ) = 0

dp 
dτ 

= p∆y −  p + ∆x ∆y − τ∆x +  τ + dy ∆x = 0
dx 

 dy 
dτ
=0
dy
d  du 
i.e.
µ  = 0
dy  dy 
d 2u
µ 2 =0
dy
from momentum equation
du
µ =C
dy
C
u = y+D
µ
u(0) = 0 ⇒ D = 0
U
u(t) = U ⇒ C = µ
t
U
u= y
t
du µU
τ=µ =
= constant
dy
τ
25
Chapter 6
26
Generalization for inclined flow with a constant pressure
gradient
Continutity
∂u
=0
∂x
Momentum
d 2u
∂
0 = − (p + γz ) + µ 2
∂x
dy
i.e.,
d 2u
dh
µ 2 =γ
dx
dy
u = u(y)
v=o
∂p
=0
∂y
h = p/γ +z = constant
plates horizontal
plates vertical
dz
=0
dx
dz
=-1
dx
which can be integrated twice to yield
du
dh
= γ y+A
dy
dx
dh y 2
µu = γ
+ Ay + B
dx 2
µ
26
Chapter 6
27
now apply boundary conditions to determine A and B
u(y = 0) = 0 ⇒ B = 0
u(y = t) = U
dh t 2
dh t
µU
µU = γ
+ At ⇒ A =
−γ
dx 2
t
dx 2
γ dh y 2 1  µU
dh t 
u ( y) =
+ 
−γ
µ dx 2 µ  t
dx 2 
γ dh
U
ty − y 2 + y
=−
2µ dx
t
(
)
This equation can be put in non-dimensional form:
u
γt 2 dh  y  y y
=−
1 −  +
U
2µU dx  t  t t
define: P = non-dimensional pressure gradient
γt 2 dh
p
h = +z
=−
2µU dx
γ
γz 2  1 dp dz 
Y = y/t
=−
+ 

2µU  γ dx dx 
u
⇒
= P ⋅ Y(1 − Y ) + Y
U
parabolic velocity profile
27
Chapter 6
28
u Py Py 2 y
=
− 2 +
U
t
t
t
t
q = ∫ udy
0
t
∫ U[ ]dy
q 0
u= =
t
t
tu t P
P
y
= ∫  y − 2 y 2 +  dy
U 0 t
t
t
=
Pt Pt t
− +
2 3 2
u P 1
t 2  dh  U
= + ⇒u=
− γ  +
U 6 2
12µ  dx  2
For laminar flow
ut
< 1000
ν
Recrit ∼ 1000
28
Chapter 6
29
The maximum velocity occurs at the value of y for which:
du
d u
P 2P
1
=0
0
y
=
=
−
+
 
dy
dy  U 
t t2
t
⇒y=
t
(P + 1) = t + t @ umax
2P
2 2P
∴ u max = u (y max ) =
note: if U = 0:
u
u max
=
for U = 0, y = t/2
UP U U
+ +
4
2 4P
P P 2
=
6 4 3
The shape of the velocity profile u(y) depends on P:
dh
1. If P > 0, i.e., < 0 the pressure decreases in the
dx
direction of flow (favorable pressure gradient) and the
velocity is positive over the entire width
γ
dh
d p
 dp
= γ  + z =
− γ sin θ
dx
dx  γ
 dx
a)
dp
<0
dx
b)
dp
< γ sin θ
dx
29
Chapter 6
30
1. If P < 0, i.e., dh dx > 0 the pressure increases in the direction of flow (adverse pressure gradient) and the velocity over a portion of the width can become negative
(backflow) near the stationary wall. In this case the
dragging action of the faster layers exerted on the fluid
particles near the stationary wall is insufficient to overcome the influence of the adverse pressure gradient.
dp
− γ sin θ > 0
dx
dp
> γ sin θ
dx
2. If P = 0, i.e.,
or
γ sin θ <
dp
dx
dh
= 0 the velocity profile is linear
dx
U
y
t
dp
Note: we derived
a)
= 0 and θ = 0
dx
this special case
dp
b)
= γ sin θ
dx
u
For U = 0 the form = PY (1 − Y ) + Y is not appropriate
U
u = UPY(1-Y)+UY
γt 2 dh
=−
Y(1 − Y ) + UY
2µ dx
γt 2 dh
Now let U = 0:
u=−
Y(1 − Y )
2µ dx
u=
30
Chapter 6
31
3. Shear stress distribution
Non-dimensional velocity distribution
u
u * = =P ⋅ Y (1 − Y ) + Y
U
u
is the non-dimensional velocity,
U
γ t 2 dh
P≡−
is the non-dimensional pressure
2 µU dx
where
u* ≡
Y≡
gradient
y
is the non-dimensional coordinate.
t
Shear stress
τ =µ
du
dy
In order to see the effect of pressure gradient on shear
stress using the non-dimensional velocity distribution, we
define the non-dimensional shear stress:
τ* =
Then
t
=
*
=
=
=
where
A≡
2µ
>0
ρUt
τ
1
ρU 2
2
Ud ( u U ) 2 µ du *
1
µ
=
1
td ( y t )
ρUt dY
2
ρU
2
2µ
( −2 PY + P + 1)
ρUt
2µ
( −2 PY + P + 1)
ρUt
A ( −2 PY + P + 1)
is a positive constant.
So the shear stress always varies linearly with Y across any
section.
31
Chapter 6
32
At the lower wall (Y = 0 ) :
*
τ=
A (1 + P )
lw
At the upper wall (Y = 1) :
*
τ=
A (1 − P )
uw
For favorable pressure gradient, the lower wall shear stress
is always positive:
1. For small favorable pressure gradient ( 0 < P < 1) :
*
τ lw* > 0 and τ uw
>0
2. For large favorable pressure gradient ( P > 1) :
*
τ lw* > 0 and τ uw < 0
τ
τ
( 0 < P < 1)
( P > 1)
For adverse pressure gradient, the upper wall shear stress is
always positive:
1. For small adverse pressure gradient ( −1 < P < 0 ) :
*
>0
τ lw* > 0 and τ uw
2. For large adverse pressure gradient ( P < −1) :
*
τ lw* < 0 and τ uw
>0
32
Chapter 6
33
τ
τ
( −1 < P < 0 )
( P < −1)
For U = 0 , i.e., channel flow, the above non-dimensional
form of velocity profile is not appropriate. Let’s use dimensional form:
γ t 2 dh
γ dh
u=
−
Y (1 − Y ) =
−
y (t − y )
2 µ dx
2 µ dx
Thus the fluid always flows in the direction of decreasing
piezometric pressure or piezometric head because
dh
γ
> 0, y > 0 and t − y > 0 . So if
is negative,
2µ
dx
dh
tive; if dx is positive, u is negative.
Shear stress:
du
γ dh  1 
t=
µ
=
−
t − y
dy
Since
 1 
t − y > 0 ,
 2 
u is posi-
2 dx 
2 
the sign of shear stress τ is always oppodh
site to the sign of piezometric pressure gradient dx , and the
magnitude of τ is always maximum at both walls and zero
at centerline of the channel.
33
Chapter 6
34
dh
< 0, τ > 0
dx
dh
>0, τ < 0
dx
For favorable pressure gradient,
For adverse pressure gradient,
τ
dh
<0
dx
τ
dh
>0
dx
Flow down an inclined plane
uniform flow ⇒ velocity and depth do not
change in x-direction
Continuity
du
=0
dx
34
Chapter 6
35
∂
d 2u
x-momentum 0 = − (p + γz ) + µ 2
∂x
dy
∂
y-momentum 0 = − (p + γz ) ⇒ hydrostatic pressure variation
∂y
dp
⇒
=0
dx
d2u
µ 2 = − γ sin θ
dy
du
γ
= − sin θy + c
dy
µ
γ
y2
u = − sin θ + Cy + D
µ
2
du
γ
γ
= 0 = − sin θd + c ⇒ c = + sin θd
dy y=d
µ
µ
u(0) = 0 ⇒ D = 0
γ
y2 γ
u = − sin θ + sin θ dy
µ
2 µ
=
γ
sin θ y(2d − y )
2µ
35
Chapter 6
36
u(y) =
g sin θ
y(2d − y )
2n
d
 2 y3 
γ
q = ∫ udy = sin qdy − 
2µ
3 0
0

d
=
V avγ
discharge per
unit width
1γ 3
d sin θ
3µ
θ 1γ 2
γd 2
= =
d siν θ =
siν θ
d 3µ
3ν
in terms of the slope So = tan θ ∼ sin θ
gd 2So
V=
3ν
Exp. show Recrit ∼ 500, i.e., for Re > 500 the flow will become turbulent
∂p
= − γ cos θ
∂y
Re crit =
Vd
∼ 500
ν
p = − γ cos θ y + C
p(d ) = p o = − γ cos θ d + C
36
Chapter 6
37
i.e.,
p = γ cos θ (d − y ) + p o
* p(d) > po
* if θ = 0
if θ = π/2
p = γ(d − y) + po
entire weight of fluid imposed
p = po
no pressure change through the fluid
37