derivation of flow equations

ENV5056 Numerical Modeling ofFlow and Contaminant Transport in Rivers

Derivation of Flow Derivation of Flow

Equations

Asst. Prof. Dr. Orhan GÜNDÜZ

Navier-Stokes Equations

General 3-D equationsof incompressible fluid flow

General 1-D equations

2

Saint-VenantEquations

Further simplifications possiblebased on dimensionality, time dependency and uniformity of flow

General 1-D equationsof channel flow(can be simplified from N-S eqnsor derived separately)

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.

3

In a three-dimensional cartesian coordinate system, the conservation of mass equation coupled with the Navier-Stokes 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.

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

4

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.

Dot product of two vectors:

Velocity vector:

v i j ku wυ= + +� � �

�

Some fundamental concepts

i

j

k

Unitvectors

5

[ ]

[ ]1 2

1 2

1 1 2 2

1

, ,...,

, ,...,

a b ...

n

n

n

i i n n

i

a a a a

b b b b

a b a b a b a b=

=

=

⋅ = = + + +∑�

�

Del operator:

i j kx y z

∂ ∂ ∂∇ = + +

∂ ∂ ∂

� � �

Some fundamental concepts

Gradient of a scalar field, f:

i j kf f f

fx y z

∂ ∂ ∂∇ = + +

∂ ∂ ∂

� � �

6

Divergence of a vector field, v:

v i j k

v v

u w

u wdiv

x y z

υ

υ

= + +

∂ ∂ ∂= ∇ ⋅ = + +

∂ ∂ ∂

� � �

�

�

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)

Navier-Stokes Equations of Fluid Flow

( ) 0vt

ρρ

∂+ ∇ ⋅ =

∂

�

�

Continuity Equation

7

ft

vv vρ σ

∂ + ⋅∇ = ∇ ⋅ +

∂

�

�

�

� �

Momentum Equation

where ρ is fluid density, is flow velocity vector, is stress tensoris body forces and is del operator.

v�

σ�

f�

∇


xx xy xz

yx yy yz

σ τ τ

σ τ σ τ

=

8

zx zy zzτ τ σ

where σ are the normal stresses and τ are shear stresses acting on the fluid


( ) ( ) ( )0

u w

t x y z

ρ ρ ρυ ρ∂ ∂ ∂ ∂+ + + =

∂ ∂ ∂ ∂

Conservation of mass equation:

9

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.

yxxx zxx

u u u uu g

t x y z x y z

τσ τρ υ ω ρ

∂ ∂ ∂∂ ∂ ∂ ∂+ + + = + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂

xy yy zyu g

τ σ τυ υ υ υρ υ ω ρ

∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = + + +


10

xy yy zy

yu gt x y z x y z

τ σ τυ υ υ υρ υ ω ρ

∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂

yzxz zzz

w w w wu g

t x y z x y z

ττ σρ υ ω ρ

∂ ∂∂ ∂ ∂ ∂ ∂+ + + = + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂

gx, gy and gz are gravitational acceleration along x, y and z axes. The terms σand 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.


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:

11

~u

yτ

∂

∂For Newtonian fluids:

Three assumptions needed to define stress terms:

1. The stress tensor is a linear function of the strain rates.2. The fluid is isotropic.

equations in terms of the velocities:


22 v

3xx

uP

xσ µ µ

∂= − + − ∇

∂

�

xy yx

u

y x

υτ τ µ

∂ ∂= = +

∂ ∂

12

22 v

3yy P

y

υσ µ µ

∂= − + − ∇

∂

�

22 v

3zz

wP

zσ µ µ

∂= − + − ∇

∂

�

yz zy

w

z y

υτ τ µ

∂ ∂= = +

∂ ∂

zx xz

w u

x zτ τ µ

∂ ∂ = = +

∂ ∂

2 2 2

2 2 2

1

3x

u u u u P u u u u wu w g

t x y z x x y z x x y z

υρ υ ρ µ µ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = − + + + + + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂


Substituted and simplified:

13

2 2 2

2 2 2

1

3y

P u wu w g

t x y z y x y z y x y z

υ υ υ υ υ υ υ υρ υ ρ µ µ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = − + + + + + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

2 2 2

2 2 2

1

3z

w w w w P w w w u wu w g

t x y z z x y z z x y z

υρ υ ρ µ µ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = − + + + + + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

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

Incompressible Newtonian Fluid

14

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:

0u w

x y z

υ∂ ∂ ∂+ + =

∂ ∂ ∂

2 2 2

2 2 2x

u u u u p u u uu v w g

t x y z x x y zρ ρ µ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = − + + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂


(Incompressible Flow of Newtonian Fluids)

15

2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

x

y

z

t x y z x x y z

u v v v p v v vu v w g

t x y z y x y z

u w w w p w w wu v w g

t x y z z x y z

ρ ρ µ

ρ ρ µ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = − + + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = − + + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂


(Incompressible Flow of Newtonian Fluids)

0v∇ ⋅ =

∂ �

�

�

� � �

Continuity Equation

16

2gt

vv v vpρ ρ µ

∂ + ⋅∇ = −∇ + + ∇

∂

�

�

� � �

Momentum Equation

UnsteadyAcceleration

ConvectiveAcceleration

PressureGradient

ViscosityOtherbodyforces

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

17

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.


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

18

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.

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).


19

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.

Assumptions made in derivation of


1. The flow is one-dimensional. The water depth and flowvelocity vary only in the direction of flow. Therefore, the flowvelocity is constant and the water surface is horizontal across

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velocity is constant and the water surface is horizontal acrossany section perpendicular to the direction of flow.

2. The flow is assumed to vary gradually along the channel sothat the hydrostatic pressure distribution prevails and verticalaccelerations can be neglected.

3. The channel bottom slope is small.

4. The channel bed is stable such that there is no changein bed elevations in time.

5. The Manning and Chezy equations, which are used in

Assumptions made in derivation of


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5. The Manning and Chezy equations, which are used inthe definition of channel resistance factor in steady,uniform flow conditions, are also used to describe theresistance to flow in unsteady, non-uniform flowapplications.

6. The fluid is incompressible and of constant densitythroughout the flow.

22

Cross-sectional view (X-Z plane)

Plan view (X-Y plane)

Cross-sectional view (Y-Z plane)

23

Cross-sectional view (Y-Z plane)

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

( )DB ∂ �

�

Reynolds Transport Theorem

24

( )sys

cv cs

DBbd b V n dA

Dt tρ ρ

∂= ∀ + ⋅

∂ ∫ ∫�

�

where ρ and are density and volume of fluid. is velocity vector. 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.

∀ V��

n�

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:

25

the left hand-side of equation becomes zero:

( )0cv cs

d V n dAt

ρ ρ∂

= ∀ + ⋅∂ ∫ ∫

�

�

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.


= −∑ ∑Rate of change of mass Mass inflow Mass out flow

This general equation of continuity can be given for the

26

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.

( )( ) ( )

Q AdxQ qdx Q dx

x t

ρρ ρ

∂ ∂+ − + =

∂ ∂


Applying the assumption of constant density and rearranging produces the conservation form of the

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rearranging produces the conservation form of the continuity equation, which is valid for any irregular cross section

0A Q

qt x

∂ ∂+ − =

∂ ∂

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

Momentum Equation

28

( )cv cs

F Vd V V n dAt

ρ ρ∂

= ∀ + ⋅∂

∑ ∫ ∫� � � �

�

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

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)

Momentum Equation

29

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)

g p f e wF F F F F F= + + + +∑� � � � � �

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

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The corresponding weight of the fluid can be expressed as

W d g gAdxρ ρ= ∀ =

where g is the gravitational acceleration. The component of weight in the direction of flow becomes the gravity force

singF gAdxρ 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:

Momentum Equation

sin( ) tan( )θ θ≈

which is also equal to the slope of the channel bed, S . Therefore, for a small

31

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

g oF gAdxSρ=

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:

FFFF +−=

32

pbprplp FFFF +−=

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

Momentum Equation

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

( )pldF g y w bdwρ= −

33

( )0

y

plF g y w bdwρ= −∫

Using the Taylor's series expansion of the hydrostatic force on the leftend, Fpl, and ignoring the higher order terms, one might obtain thehydrostatic force on the right end of the control volume as:

pl

pr pl

FF F dx

x

∂ = +

∂

Momentum Equation

The differential of Fpl with respect to x in the above equation is computed using the Leibnitz rule for differentiation

( )( )( ) ( )

0

00

y

plg y w bF dy d

dw g y y b g y bx x dx dx

ρρ ρ

∂ −∂= + − + −

∂ ∂∫

34

0x x dx dx∂ ∂

where the second and third terms on the right hand-side of theequation 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

( )0 0

y y

plF y bg bdw g y w dw

x x xρ ρ

∂ ∂ ∂= + −

∂ ∂ ∂∫ ∫

Momentum Equation

The first integral in above equation can be simplified as

0 0

y yy y y

g bdw g bdw g Ax x x

ρ ρ ρ∂ ∂ ∂

= =∂ ∂ ∂∫ ∫

35

∫=

y

bdwA

0

since

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

( )0

y

pb

bF g y w dw dx

xρ

∂= −

∂ ∫

Momentum Equation

pl pl

p pl pl pb pb

F FF F F dx F dx F

x x

∂ ∂ = − + + = − +

∂ ∂

Substituting:

36

0 0

( ) ( )

y y

p

y b bF g A g y w dw dx g y w dw dx

x x xρ ρ ρ ∂ ∂ ∂

= − + − + − ∂ ∂ ∂ ∫ ∫

The resultant pressure force acting on the control volume can be written as:

p

yF g Adx

xρ

∂= −

∂

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

37

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:

ffo SPAggRS )/(ρρτ ==

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

Momentum Equation

342

22

/R

VnSf

µ=

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.

38

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

dxgASF ff ρ−=

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

Momentum Equation

39

is the eddy loss slope representing the loss of energy due to an abrupt contraction or expansion

x

AQ

g

KS ec

e∂

∂=

2

2

)/(

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:

dxgASF ee ρ−=

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

40

τw, and written as τwBdx. The wind shear stress is defined as the product of a wind shear factor, Wf and fluid density

fw Wρτ −=

2

rrff

VVCW =

where Cf is a shear stress coefficient and Vr is the velocity of fluid relative to the boundary, which can be written as

Momentum Equation

ωcoswr VA

QV −=

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

41

stress given in above equation, the wind shear force acting on the control volume takes the final form given below

BdxWF fw ρ−=

Finally, the sum of the five forces define the total force on the left-hand side of the momentum equation

BdxWdxgASdxgASdxx

ygAdxgASF fefo ρρρρρ −−−

∂

∂−=∑

Momentum Equation

( )cv cs

F Vd V V n dAt

ρ ρ∂

= ∀ + ⋅∂

∑ ∫ ∫� � � �

�

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

42

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).

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:

( ) ( )x

inlet

V V n dA VQ qdxρ ρ β βυ⋅ = − +∫∫� �

�

43

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.

∫∫= dAAV

2

2

1υβ

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

Momentum Equation

The momentum outflow from the control volume is also a function of massoutflow from the control volume, which can be defined as the Taylor seriesexpansion of mass inflow. Hence, the momentum outflow from the controlvolume is computed as:

( )VQβ∂ ∫∫� �

�

44

( )( )

outlet

VQV V n dA VQ dx

x

βρ ρ β

∂ ⋅ = +

∂ ∫∫� �

�

Thus, net momentum outflow across the control surface

( )( ) ( )

( ) x x

cs

VQ VQV V n dA VQ qdx VQ dx q dx

x x

β βρ ρ β βυ ρ β ρ βυ

∂ ∂ ⋅ = − + + + = − −

∂ ∂ ∫� �

�

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

QV d dxρ ρ

∂ ∂∀ =∫

�

45

cv

QV d dx

t tρ ρ

∂ ∂∀ =

∂ ∂∫�

When all terms are combined and substituted back into the momentum equation:

dxt

Qdx

x

VQqdxBWdxgASdxgASdx

x

ygAdxgAS xfefo

∂

∂+

∂

∂−−=−−−

∂

∂− ρ

ββυρρρρρρ

)(

( )cv cs

F Vd V V n dAt

ρ ρ∂

= ∀ + ⋅∂

∑ ∫ ∫� � � �

�

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

02

=+−

++−

∂

∂+

∂

∂+

∂

∂BWqSSS

x

ygA

x

AQ

t

Qfxefo υβ

β )/(

46

∂∂∂ xxt

The water depth term in this equation can be replaced by the water surface elevation (stage), h, using the equality

zyh +=

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

x

z

x

y

x

h

∂

∂+

∂

∂=

∂

∂

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

oSx

y

x

h−

∂

∂=

∂

∂

The momentum equation can now be expressed in terms of h by

47

The momentum equation can now be expressed in terms of h by

02

=+−

++

∂

∂+

∂

∂+

∂

∂BWqSS

x

hgA

x

AQ

t

Qfxef υβ

β )/(

Saint-Venant Equations

The continuity and momentum equations are always addressed together and form the conservation form of the Saint-Venant equations

0=−∂

∂+

∂

∂q

x

Q

t

A

48

02

=+−

++

∂

∂+

∂

∂+

∂

∂BWqSS

x

hgA

x

AQ

t

Qfxef υβ

β )/(

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

Saint-Venant Equations – Steady State

Under steady state assumption, Saint-Venant equations simplify to:

0dQ

qdx

− =

49

dx

2( / )

0f e x f

d Q A dhgA S S q W B

dx dx

ββ υ

+ + + − + =

derivation of flow equations

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