Download Derivation of Flow Equations

ENV5056 Numerical Modeling of
Flow and Contaminant Transport in Rivers
Derivation of Flow
Equations
Asst. Prof. Dr. Orhan GÜNDÜZ
General 3-D equations
of incompressible fluid flow
General 1-D equations
of channel flow
(can be simplified from N-S eqns
or derived separately)
Navier-Stokes
Equations
Saint-Venant
Equations
Further simplifications possible
based on dimensionality, time dependency and uniformity of flow
2
Three-dimensional Hydrodynamic
Equations of Flow
The three-dimensional hydrodynamic equations of fluid
flow are the basic differential equations describing the
flow of a Newtonian fluid.
In a three-dimensional cartesian coordinate system, the
conservation of mass equation coupled with the NavierStokes equations of motion in x, y and z dimensions form
the general hydrodynamic equations.
They define a wide range of flow phenomena from
unsteady, compressible flows to steady, incompressible
flows.
3
Three-dimensional Hydrodynamic
Equations of Flow
Developed by Claude-Louis Navier and George Gabriel Stokes,
these equations arise from applying Newton's second law to
fluid motion, together with the assumption that the fluid
stress is the sum of a diffusing viscous term (proportional to
the gradient of velocity), plus a pressure term.
Although Navier-Stokes equations only refer to the equations
of motion (conservation of momentum), it is commonly
accepted to include the equation of conservation of mass.
These four equations all together fully describe the
fundamental characteristics of fluid motion.
4
Some fundamental concepts
j
Velocity vector:
v = u i + υ j + wk
i
Unit
vectors
k
Dot product of two vectors:
a = [ a1 , a2 ,..., an ]
b = [b1 , b2 ,..., bn ]
n
a ⋅ b = ∑ ai bi = a1b1 + a2b2 + ... + an bn
i =1
Del operator:
∂ ∂ ∂ ∇=
i+
j+ k
∂x
∂y
∂z
5
Some fundamental concepts
Gradient of a scalar field, f:
∂f ∂f ∂f ∇f =
i+
j+ k
∂x
∂y
∂z
Divergence of a vector field, v:
v = u i + υ j + wk
∂u ∂υ ∂w
div v = ∇ ⋅ v =
+
+
∂x ∂y ∂z
In Euclidean space, the dot product of two unit vectors is simply the
cosine of the angle between them. (Ex: i.i=1x1xcos0=1,
i.j=1x1xcos90=0, i.k=1x1xcos90=0)
6
Navier-Stokes Equations of Fluid Flow
∂ρ
+ ∇ ⋅ ( ρ v) = 0
∂t
 ∂v 
ρ  + v ⋅∇v  = ∇ ⋅ σ + f
 ∂t
Continuity Equation
Momentum Equation

σ
v
where
ρ
is
fluid
density,
is
flow
velocity
vector,
is stress tensor
f is body forces and ∇ is del operator.
7
Navier-Stokes Equations of Fluid Flow
σ xx τ xy τ xz 


σ = τ yx σ yy τ yz 
 τ zx τ zy σ zz 


where σ are the normal stresses and τ are shear stresses acting on the fluid
8
Navier-Stokes Equations of Fluid Flow
Conservation of mass equation:
∂ρ ∂ ( ρ u ) ∂ ( ρυ ) ∂ ( ρ w)
+
+
+
=0
∂t
∂x
∂y
∂z
where ρ is the density of the fluid, t is time and u, υ and w are the components of
the velocity vector in x, y and z coordinates.
9
Navier-Stokes Equations of Fluid Flow
 ∂u
∂u
∂u
∂u  ∂σ xx ∂τ yx ∂τ zx
+
+
+ ρ gx
ρ  + u +υ + ω  =
∂x
∂y
∂z 
∂x
∂y
∂z
 ∂t
 ∂υ
∂υ
∂υ
∂υ  ∂τ xy ∂σ yy ∂τ zy
ρ +u
+υ
+ω
+
+
+ ρgy
=
∂x
∂y
∂z  ∂x
∂y
∂z
 ∂t
 ∂w
∂w
∂w
∂w  ∂τ xz ∂τ yz ∂σ zz
ρ +u
+υ
+ω
+
+
+ ρ gz
=
∂x
∂y
∂z  ∂x
∂y
∂z
 ∂t
gx, gy and gz are gravitational acceleration along x, y and z axes. The terms σ
and τ are normal and shear stresses acting on the fluid, respectively. The
first subscript in the notation indicates the direction of the normal to the
plane on which the stress acts, and the second subscript indicates the
direction of the stress.
10
Navier-Stokes Equations of Fluid Flow
The general equations of motion include both velocities
and stresses as unknown variables. Using the relationships
derived for a compressible Newtonian fluid, one can
express the normal and shear stress components in these
equations in terms of the velocities:
For Newtonian fluids:
∂u
τ~
∂y
Three assumptions needed to define stress terms:
1. The stress tensor is a linear function of the strain rates.
2. The fluid is isotropic.
11
Navier-Stokes Equations of Fluid Flow
σ xx
∂u 2
= − P + 2µ
− µ∇ v
∂x 3
σ yy
∂υ 2
= − P + 2µ
− µ∇ v
∂y 3
∂w 2
σ zz = − P + 2 µ
− µ∇ v
∂z 3
τ xy = τ yx
τ yz
 ∂u ∂υ 
= µ +

∂
y
∂
x


 ∂υ ∂w 
= τ zy = µ 
+

∂
z
∂
y


 ∂w
∂u 
τ zx = τ xz = µ 
+ 
 ∂x ∂z 
12
Navier-Stokes Equations of Fluid Flow
Substituted and simplified:
 ∂ 2u ∂ 2u ∂ 2u  1  ∂  ∂u ∂υ ∂w  
 ∂u
∂u
∂u
∂u 
∂P
ρ  + u +υ + w  = − + ρ gx + µ  2 + 2 + 2  + µ   +
+

∂
t
∂
x
∂
y
∂
z
∂
x
∂
x
∂
y
∂
z
3
∂
x
∂
x
∂
y
∂z  




 
 ∂ 2υ ∂ 2υ ∂ 2υ  1  ∂  ∂u ∂υ ∂w  
 ∂υ
∂υ
∂υ
∂υ 
∂P
ρ +u
+υ
+w
+ ρgy + µ  2 + 2 + 2  + µ   +
+
=−

∂
t
∂
x
∂
y
∂
z
∂
y
∂
x
∂
y
∂
z
∂
y
∂
x
∂
y
∂
z
3





 
 ∂ 2 w ∂ 2 w ∂ 2 w  1  ∂  ∂u ∂υ ∂w  
 ∂w
∂w
∂w
∂w 
∂P
ρ +u
+υ
+w =−
+ ρ gz + µ  2 + 2 + 2  + µ   +
+

∂
t
∂
x
∂
y
∂
z
∂
z
∂
x
∂
y
∂
z
3
∂
z
∂
x
∂
y
∂
z





 
13
Incompressible Newtonian Fluid
Most fluids of practical importance such as water are
incompressible, which means that their densities are
constant for a wide range of flows. This is a reasonable
assumption except for certain extreme situations such
as the cases where the fluid is under profound
pressures. Since the density of fluid is constant, the
continuity equation for incompressible flows can be
simplified as:
∂u ∂υ ∂w
+
+
=0
∂x ∂y ∂z
14
Navier-Stokes Equations of Fluid Flow
(Incompressible Flow of Newtonian Fluids)
 ∂ 2u ∂ 2u ∂ 2 u 
 ∂u
∂u
∂u
∂u 
∂p
ρ  + u + v + w  = − + ρ gx + µ  2 + 2 + 2 
∂x
∂y
∂z 
∂x
∂y
∂z 
 ∂t
 ∂x
 ∂ 2v ∂ 2v ∂ 2v 
 ∂u
∂v
∂v
∂v 
∂p
ρ  + u + v + w  = − + ρgy + µ  2 + 2 + 2 
∂x
∂y
∂z 
∂y
 ∂t
 ∂x ∂y ∂z 
 ∂2w ∂2w ∂2w 
 ∂u
∂w
∂w
∂w 
∂p
ρ +u
+v
+ w  = − + ρ gz + µ  2 + 2 + 2 
∂x
∂y
∂z 
∂z
∂y
∂z 
 ∂t
 ∂x
15
Navier-Stokes Equations of Fluid Flow
(Incompressible Flow of Newtonian Fluids)
∇⋅v = 0
Continuity Equation
 ∂v 
2
ρ  + v ⋅∇v  = −∇p + ρ g + µ∇ v
 ∂t
Unsteady
Acceleration

Convective
Pressure Other
Acceleration Gradient body
forces
Momentum Equation
Viscosity
16
Saint-Venant Equations of Channel Flow
The flow of water through stream channels is a distributed
process since the flow rate, velocity and depth vary spatially
throughout the channel.
Estimates of flow rate or water level at certain locations in
the channel system may be obtained using a set of
equations that define the conservation of mass and
momentum along this channel.
This type of a model is based on partial differential
equations that allow the flow rate and water level to be
computed as a function of space and time.
17
Saint-Venant Equations of Channel Flow
From a theoretical standpoint, the true flow process in the
river system varies in all three spatial coordinate directions
(longitudinal, lateral and transverse) as well as time.
If a three-dimensional system is used, the resulting equations
(modified Navier-Stokes equations for channel flow) would
be very complex, and would require considerable amount of
field data, which is also spatially variable.
In field applications, most of this data could only be
described approximately, thus rendering the three
dimensional solutions susceptible to data errors.
18
Saint-Venant Equations of Channel Flow
However, for most practical purposes, the spatial variations in
lateral and transverse directions can be neglected and the flow in a
river system can be approximated as a one-dimensional process
along the longitudinal direction (i.e., in the direction of flow).
The Saint Venant equations that were derived in the early 1870s
by Barre de Saint-Venant, may be obtained through the application
of control volume theory to a differential element of a river reach.
The Navier-Stokes equations can be simplified for one-dimensional
flow. However, it is more intuitive if these equations are derived
from an elemental volume of fluid along a channel.
19
Assumptions made in derivation of
Saint-Venant Equations of Channel Flow
1. The flow is one-dimensional. The water depth and flow
velocity vary only in the direction of flow. Therefore, the flow
velocity is constant and the water surface is horizontal across
any section perpendicular to the direction of flow.
2. The flow is assumed to vary gradually along the channel so
that the hydrostatic pressure distribution prevails and vertical
accelerations can be neglected.
3. The channel bottom slope is small.
20
Assumptions made in derivation of
Saint-Venant Equations of Channel Flow
4. The channel bed is stable such that there is no change
in bed elevations in time.
5. The Manning and Chezy equations, which are used in
the definition of channel resistance factor in steady,
uniform flow conditions, are also used to describe the
resistance to flow in unsteady, non-uniform flow
applications.
6. The fluid is incompressible and of constant density
throughout the flow.
21
Cross-sectional view (X-Z plane)
22
Plan view (X-Y plane)
Cross-sectional view (Y-Z plane)
23
Reynolds Transport Theorem
The equations of mass and momentum conservation can
be derived starting from the Reynolds Transport
Theorem . If Bsys and b are defined as extensive and
intensive parameters of the system, respectively, the
Reynolds Transport Theorem for a fixed non-deforming
control volume can be stated as
DBsys
∂
= ∫ ρ bd ∀ + ∫ ρ b V ⋅ n dA
Dt
∂t cv
cs
(
)
where ρ and ∀ are density and volume of fluid. V is
velocity vector. n is the unit vector normal to the
control surface and t is time. In this equation, the
intensive parameter is defined as extensive property per
unit mass. Therefore, if mass is the extensive property,
the intensive property becomes unity.
24
Continuity Equation (Conservation of Mass)
The continuity equation can be obtained by
simplifying the Reynolds Transport Theorem with the
extensive property being the mass of the system.
Since the total mass in a system is always constant,
the left hand-side of equation becomes zero:
∂
0 = ∫ ρ d ∀ + ∫ ρ V ⋅ n dA
∂t cv
cs
(
)
This implies that the time rate of change of mass in the
control volume is equal to the difference in cumulative
mass inflow and outflow from the control surfaces of
the control volume.
25
Continuity Equation (Conservation of Mass)
Rat e of change of m ass = ∑ Mass inflow − ∑ Mass out flow
This general equation of continuity can be given for the
particular case of an open channel with an irregular
geometry. As seen in figure, the inflow to the control
volume is the sum of the flow Q entering the control
volume at the upstream end of the channel and the
lateral inflow q entering the control volume as a
distributed flow along the side of the channel. In this
case, the lateral inflow has the dimensions of flow per
unit length of channel.
26
Continuity Equation (Conservation of Mass)
∂Q
∂ ( ρ Adx)
ρ (Q + qdx) − ρ (Q +
dx) =
∂x
∂t
Applying the assumption of constant density and
rearranging produces the conservation form of the
continuity equation, which is valid for any irregular
cross section
∂A ∂Q
+
−q =0
∂t ∂x
27
Momentum Equation
It is stated in the Newton's second law for a system that the
time rate of change of the linear momentum of the system is
equal to the sum of the external forces acting on the system.
Using this law in the Reynolds Transport Theorem yields
∂
∑ F = ∂t ∫ ρVd ∀ + ∫ ρV V ⋅ n dA
cv
cs
(
)
In this case, the intensive property is the velocity of the
fluid. This equation reveals that the sum of the forces
applied on the system is equal to the time rate of change
of momentum stored within the control volume plus the
net flow of momentum across the control surfaces
28
Momentum Equation
For an open channel flow, there are five different forces
acting on the control volume:
1. the gravity force along the channel due to weight of
water (Fg)
2. the pressure force (Fp)
3. the friction force along the bottom and sides of the
channel (Ff)
4. the contraction/expansion force due to abrupt changes
in channel cross section (Fe)
5. the wind shear force (Fw)
∑ F = Fg + Fp + Ff + Fe + Fw
29
Momentum Equation
Gravity Force: The gravity force acting on the control volume shown in
figure is a function of the volume of the fluid, which may be given as
d ∀ = Adx
The corresponding weight of the fluid can be expressed as
W = ρ d ∀g = ρ gAdx
where g is the gravitational acceleration. The component of weight in the
direction of flow becomes the gravity force
Fg = ρ gAdx sin θ
where θ is the angle of inclination of the channel bed. With the
assumption of a small angle of inclination of the channel bed, the sine of
the angle can be approximated as the tangent of the angle:
30
Momentum Equation
sin(θ ) ≈ tan(θ )
which is also equal to the slope of the channel bed, So. Therefore, for a small
angle of inclination of the channel bed, gravity force acting on the control
volume can be written as
Fg = ρ gAdxSo
31
Momentum Equation
Pressure Force: The pressure force is the resultant of the hydrostatic force on
the left side of the control volume (Fpl), the hydrostatic force on the right side
of the control volume (Frl) and the pressure force exerted by the banks on the
control volume (Fpb) as can be seen from Figure 1.b:
Fp = Fpl − Fpr + Fpb
If an element of fluid of thickness dw at
an elevation of w from the bottom of
the channel is immersed at depth y-w,
the incremental hydrostatic pressure on
this element is computed as ρg(y-w).
The corresponding incremental
hydrostatic force is then calculated as
32
Momentum Equation
dFpl = ρ g ( y − w ) bdw
where b is the width of the element across the channel. Integrating this
force over the cross section gives the total hydrostatic force on the left
end of the control volume
y
Fpl = ∫ ρ g ( y − w ) bdw
0
Using the Taylor's series expansion of the hydrostatic force on the left
end, Fpl, and ignoring the higher order terms, one might obtain the
hydrostatic force on the right end of the control volume as:
∂Fpl 

Fpr =  Fpl +
dx 
∂x


33
Momentum Equation
The differential of Fpl with respect to x in the above equation is computed
using the Leibnitz rule for differentiation
∂Fpl
∂x
y
=∫
0
∂ ( ρ g ( y − w) b )
∂x
dw + ρ g ( y − y ) b
dy
d0
+ ρ g ( y − 0) b
dx
dx
where the second and third terms on the right hand-side of the
equation are evaluated as zero. The partial derivative term can be
expanded using the multiplication rule of differentiation and the
differential of Fpl with respect to x takes the following form
∂Fpl
y
y
∂y
∂b
= ∫ ρ g bdw + ∫ ρ g ( y − w ) dw
∂x
∂x
∂x
0
0
34
Momentum Equation
The first integral in above equation can be simplified as
y
y
∂y
∂y
∂y
=
=
ρ
g
bdw
ρ
g
bdw
ρ
g
A
∫0 ∂x
∫
∂x 0
∂x
y
since
∫
A = bdw
0
The pressure force exerted by the banks on the control volume is
related to the rate of change of the width of the channel through
the element dx and is given as
y
∂b 
Fpb =  ∫ ρ g ( y − w ) dw dx
∂x 
0
35
Momentum Equation
Substituting:
∂Fpl 
∂Fpl

Fp = Fpl −  Fpl +
dx  + Fpb = −
dx + Fpb
∂x
∂x


y

y
∂y
∂b 
∂b 
Fp = −  ρ g
A + ∫ ρ g ( y − w) dw  dx +  ∫ ρ g ( y − w) dw  dx
∂
x
∂x 
∂x 
0

0
The resultant pressure force acting on the control volume can be written as:
Fp = − ρ g
∂y
Adx
∂x
36
Momentum Equation
Friction Force: The friction forces
created by shear stress along the
bottom and sides of the control
volume can be defined in terms of
the bed shear stress τo and can be
given as -τoPdx, where P is the
wetted perimeter. In accordance
with the assumptions stated
previously, the bed shear stress
can be defined in terms of the
steady uniform flow formula:
τ o = ρgRS f = ρg ( A / P) S f
where R is the hydraulic radius defined by the ratio of the flow area and the
wetted perimeter and Sf is the friction slope, which is derived from the
Manning's equation and given as
37
Momentum Equation
Sf =
n 2V 2
µ 2R 4 / 3
n is the Manning's roughness coefficient, V is the flow velocity and µ is a
constant, which is equal to 1.49 in British units and 1.0 in SI units.
Based on the definition of bed shear stress given in above equation, the
friction force acting on the control volume takes the final form given below
F f = − ρgAS f dx
38
Momentum Equation
Contraction/Expansion Force: Abrupt contraction or expansion of the
channel causes energy loss through turbulence. These losses can be
considered similar to the losses in a pipe system. The magnitude of these
losses is a function of the change in velocity head, V2/2g=(Q/A)2/2g, through
the length of the channel. The forces associated with these eddy losses can
be defined similar to friction force except the term Sf is replaced by Se, which
is the eddy loss slope representing the loss of energy due to an abrupt
contraction or expansion
K ec ∂(Q / A) 2
Se =
2g
∂x
where Kec is the non-dimensional expansion or contraction coefficient that
is defined as negative for channel expansion and positive for channel
contraction. Therefore, contraction/expansion force term becomes:
Fe = − ρgAS e dx
39
Momentum Equation
Wind Shear Force: The wind
shear force is caused by the
frictional resistance of wind
against the free surface of the
water. It can be defined as a
function of wind shear stress,
τw, and written as τwBdx. The
wind shear stress is defined as
the product of a wind shear
factor, Wf and fluid density
τ w = − ρW f
Wf =
C f Vr Vr
2
where Cf is a shear stress coefficient and Vr is the velocity of fluid relative to
the boundary, which can be written as
40
Momentum Equation
Vr =
Q
− Vw cos ω
A
Vw is the wind velocity and ω is the angle that wind direction makes with the
direction of average fluid velocity, (Q/A). Based on the definition of wind shear
stress given in above equation, the wind shear force acting on the control
volume takes the final form given below
Fw = − ρW f Bdx
Finally, the sum of the five forces define the total force on the left-hand side of
the momentum equation
∑
F = ρgAS o dx − ρgA
∂y
dx − ρgAS f dx − ρgAS e dx − ρW f Bdx
∂x
41
Momentum Equation
∂
∑ F = ∂t ∫ ρVd ∀ + ∫ ρV V ⋅ n dA
cv
cs
(
)
The two momentum terms on the right-hand side represent the rate of
change of storage of momentum in the control volume and the net outflow
of momentum across the control surface, respectively
The net momentum outflow is the sum of momentum outflow minus the
momentum inflow to the control volume. The mass inflow rate to the
control volume is the sum of both stream inflow and the lateral inflow and
is defined as -ρ(Q + qdx).
42
Momentum Equation
The momentum inflow to the control volume is computed by multiplying the
two mass inflow rates by their respective velocities and a momentum
correction factor, β:
∫∫ V ρ (V ⋅ n )dA = − ρ (βVQ + βυ x qdx)
inlet
where υx is the average velocity of lateral inflow in the direction of main
channel flow. The momentum coefficient, β, accounts for the non-uniform
distribution of velocity at a specific channel cross section.
β =
1
V
2
υ
∫∫
A
2
dA
where υ is the velocity of the fluid in a small elemental area dA in the
channel cross section. Generally, the value of the momentum coefficient
ranges from 1.01 for straight prismatic channels to 1.33 for river valleys with
floodplains
43
Momentum Equation
The momentum outflow from the control volume is also a function of mass
outflow from the control volume, which can be defined as the Taylor series
expansion of mass inflow. Hence, the momentum outflow from the control
volume is computed as:
∂ ( β VQ ) 

V
ρ
(
V
⋅
n
)
dA
=
ρ
β
VQ
+
dx 

∫∫
∂
x


outlet
Thus, net momentum outflow across the control surface
∂ ( β VQ ) 
∂ ( β VQ ) 


ρ
ρ
β
βυ
ρ
β
ρ
βυ
V
(
V
⋅
n
)
dA
=
−
VQ
+
qdx
+
VQ
+
dx
=
−
q
−
(
)
x


 x
 dx
∫cs
∂x
∂x 



44
Momentum Equation
The time rate of change of momentum stored in the control volume is written
as a function of the volume of the elemental channel length dx. The
momentum associated with this elemental volume can be written as ρVAdx, or
ρQdx and the time rate of change of momentum is given as
∂ ∂Q
V
ρ
d
∀
=
ρ
dx
∂t cv∫
∂t
When all terms are combined and substituted back into the momentum
equation:
∂
∑ F = ∂t ∫ ρVd ∀ + ∫ ρV V ⋅ n dA
cv
cs
(
ρgAS o dx − ρgA
)
∂y
∂( βVQ ) 
∂Q

dx − ρgAS f dx − ρgAS e dx − W f Bρdx = − ρ  βυ x q −
dx
dx + ρ
∂x
∂
x
∂
t


45
Momentum Equation
This equation is simplified and rearranged to the following form if all terms are
divided by ρdx and V is replaced by Q/A
∂Q ∂( βQ 2 / A)
 ∂y

+
+ gA
− S o + S f + S e  − βqυ x + W f B = 0
∂t
∂x
 ∂x

The water depth term in this equation can be replaced by the water surface
elevation (stage), h, using the equality
h= y+z
where z is the channel bottom above a datum such as mean sea level as
seen in figure. The derivative of this with respect to x is written as
∂h ∂y ∂z
+
=
∂x ∂x ∂x
46
Momentum Equation
However, the term ∂z/∂x is equal to the negative of the slope of the channel,
so the equation can also be written as
∂h ∂y
=
− So
∂x ∂x
The momentum equation can now be expressed in terms of h by
∂Q ∂( βQ 2 / A)
 ∂h

+
+ gA
+ Sf + Se  − βqυ x + Wf B = 0
∂t
∂x
 ∂x

47
Saint-Venant Equations
The continuity and momentum equations are always addressed together
and form the conservation form of the Saint-Venant equations
∂A ∂Q
−q = 0
+
∂t
∂x
∂Q ∂( βQ 2 / A)
 ∂h

+
+ gA
+ Sf + Se  − βqυ x + Wf B = 0
∂t
∂x
 ∂x

These equations are the governing equations of one-dimensional
unsteady flow in open channels and were originally developed by the
French scientist Barre de Saint-Venant in 1872
48
Saint-Venant Equations – Steady State
Under steady state assumption, Saint-Venant equations simplify to:
dQ
−q =0
dx
d ( β Q 2 / A)
 dh

+ gA  + S f + Se  − β qυ x + W f B = 0
dx
 dx

49