Forms of The Momentum Equation
by
Eka Oktariyanto Nugroho
Basic Equation 5
Eka O. N.
5.1. MAIN DIFFERENTIAL FORMS OF THE MOMENTUM EQUATION
The momentum equation is obtained by equating the applied forces to the inertia force for a unit
volume of the fluid. The physical meaning and the mathematical expressions of these forces have
been developed in Chapters 4 and 5.
The different forms of the momentum equation corresponding to a number of cases encountered in
hydrodynamics are now presented.
5.1.1. Perfect Fluid: Equations of Euler
The first major approximation is to assume that the fluid is perfect. In this case the friction forces
are zero and the applied forces consist of gravity and pressure only. The momentum equation is
obtained directly from the expressions developed in Chapters 4 and 5, in the three-axis system OX,
O)', OZ, where OZ is assumed to be vertical (see Table 5-1). When the expressions du dt and
p * are expanded (see Section 4-4.1), the momentum equation takes the form along the OX axis
given by Equation 5-1.
Table 5. 1 The momentum equation.
Inertia forces per unit
volume
Pressure and gravity forces per unit
of volume of fluid
(see section 4-1.1)
(see Section 5-3.4)
du
dt
dv

dt


dw
dt
=

p*
x
=

p*
y
=

p*
z
Written in vector notation, these become

Recall
p*  p   gz
dV
 grad p*  0
dt
Inertia Forces
Applied Forces
Convective
inertia
Pressure
u
u
u 

 u
u
v
w

x
y
z 
x
 t
p
Local
inertia

Gravity
  gz 
(5-1)
Two similar equations may be written in the O Y and OZ directions. These are called the equations
of Euler. Such a system of equations associated with the continuity relationship
u v w
 
0
x y z
forms the basis of the largest part of the hydrodynamics dealing with a perfect incompressible fluid.
These equations are mathematically of the first order but are nonlinear (more specifically quadratic)
because of the convective inertia terms. This quadratic term is the cause of most mathematical
difficulties encountered in hvdrodvnamics.
It has been explained in Chapter I that it is possible to study hydrodynamic problems either in
Eulerian coordinates or in Lagrangian coordinates. It is recalled that the Lagrangian method
consists of following particles along their paths instead of dealing with particles at a given point.
This method is used, for example, in some studies of periodic gravity waves over a horizontal
bottom. If X, Y, Z are the volume or body forces, i.e., gravity, the Lagrangian equation along the OX
axis is written:
Forms of The Momentum Equation
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Basic Equation 5
Eka O. N.
1 p 
 2 x  x 
 2 y  y 
 2 z  z
X  2 
 Y  2 
Z  2 
 x0 
t  x0 
t  y0 
t  z0
p y0 and p z0 by permutation of x0 , y0 , z0 , which
are the coordinates of a particle at time t  t0 . These are called the equations of Lagrange.
Two similar equations give the value of
5.1.2. Viscous Fluid and the Navier-Stokes Equations
If the friction forces are introduced in the Eulerian equations, the Navier-Stokes equations are
obtained (see Section 4.4.1), as shown in Equation 5-2. The Navier-Stokes equations are the basis
of most problems in fluid mechanics dealing with liquid. They are second-order differential
equations because of the friction terms. and nonlinear because of the convective inertia terms.
These Navier-Stokes equations are written in a very concise manner with the aid of tensorial
notation. Although a knowledge of tensorial calculus is not required, it is given here as a guide to
further reading on the subject. Use is made of two subscripts, i and j , which indicate when an
operation is to be systematically repeated and which component of a vector quantity (such as V) is
being considered. When an index is repeated in a term, the considered quantity has to be summed
over the possible components. For example, the continuity equation
u v w
 
 0,
x y z
is
tensorial written: ui x j since the subscript i indicates that the quantity (here V) has to be
summed over the three components OX,OY,OZ. The three previous Navier-Stokes equations may
be writen simply as:
 ui

 t
uj
  p   gz 
ui 
 2ui


x j 
xi
x j x j
Here, the subscript i is called "free index" and indicates the component being considered; the
subscript j , called "dummy index," indicates repeated operations.
Inertia Forces
Local
inertia
Convective inertia
Applied Forces

Pressure
Friction
Gravity
u
u
u 
p
 u
u
v
w

x
y
z 
x
 t
  u  2u  2u 
 2  2  2 
y
z 
 x
 v
  2v  2v  2v 
 2  2  2 
y
z 
 x


u
v
v
v 
p
v
w 
x
y
z 
y
2
(5-2)
 t
 w
  p   gz 
 2w 2w 2w 
w
w
w 

u
v
w




 2 
 2 

x
y
z 
z
y 2
z 
 x
 t
These Navier-Stokes equations are often written in another way in order to emphasize the role of
the rotational component of motion. It is sufficient in this case to use the expression of the inertia
force demonstrated in Chapter 4, which yields (see Section 3.4.3) Equation 5-3.
Forms of The Momentum Equation
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Basic Equation 5
Eka O. N.
Applied Forces
Inertia Forces
Local
inertia

Pressure
Gravity
Convective inertia
Friction


  p   gz 
 V 
 u
 
 

2
w


v

  2u




x  2 
x

 t


Caused by variation
of kinetic energy
2
Caused by rotation
 v   V 2 

  p   gz 

 2  u  w    
  2 v


 t y  2 

y


 w   V 2 

  p   gz 


 2  v  u    
  2 w


 t z  2 
z



Inertia Forces
Local
inertia

Convective inertia
(5-3)
Applied Forces
Pressure
Gravity
Friction




V
 V
2

 grad 
   curlV   V   grad  p   gz    V

 t
 2 


Caused by variation
of kinetic energy
2
Caused by rotation
(5-4)
The three components of Equation 5-3 are more concisely written in the vector form of Equation 54, which may be transformed as
 V2

V
grad  
 p   gz    
   curlV   V+2 V
t
 2

In the case of a steady
 V t  0
irrorational flow
 curl V  0
of a perfect fluid
   0 , the
above equation gives at once:
 V2

grad  
 p   gz   0
2


Since the derivative of the sum in parentheses is zero in all directions, one obtains

V2
 p   gz  constant
2
which is the well-known Bernoulli equation.
5.1.3. The General Form of the Momentum Equation
It has been shown that the applied forces may be expressed independently of their physical nature
with the help of the tensor of rank two:
 xx  xy  xz
 xy  yy  yz
 xz  yz  zz
The main advantage of such a notation is that it is valid for any kind of fluid-perfect or real-and any
kind of motion-laminar or turbulent. It will be shown that if in the momentum equation the real
values u , v, w, and p are replaced by the average values u , v , w, and p in a turbulent flow, the
surface forces  and  include additional components due to the turbulent fluctuations.
Forms of The Momentum Equation
Page
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Basic Equation 5
Eka O. N.
Hence, the advantage of using the notations  and  exists in expressing general equations
which are independent of the nature of the flow. Equating the inertia forces to the applied forces
expressed in the manner shown in Chapter 5 yields Equation 6-5.
In practice, if OZ is vertical upward,
Inertia
Forces

X  0, Y  0, Z    g     z   gz 
Applied Forces
Volume
forces
Surface forces
  xx  xy  xz 
du
 X 



dt
y
z 
 x
  xy  yy  yz 
dv

Y 



dt
y
z 
 x


(5-5)
  xz  yz  zz 
dw
 Z 



dt
y
z 
 x
5.1.4. Synthesis of the Most Usual Approximations
Tables 5-2 and 5-3 recall the physical meaning of different terms and possible approximations
accepted in the studies of flow motions, which may be investigated in the following. Complex
disordered and random motions, even though also obeying the Navier-Stokes equation, cannot be
analyzed on a purely Newtonian deterministic approach. The motion is averaged and the friction
term
 2 V
is replaced by an empirical functional relationship proportional to
V in the case of
2
flow trough porous medium, or to V in the case of fully turbulent motion.
5.1.5. An Example of an Exact Solution of Navier-Stokes Equations: Flow on a Sloped Plane
It is to be expected that a general solution of the system of differential equations given by the
continuity and momentum principle does not exist. However, some exact solutions can be obtained
if the boundary conditions are simple. Examples where exact solutions may be obtained include
flow between parallel plates (i.e., the Couette flow, the Poiseuille flow), flow due to a rotating disk,
uniform unsteady flow over an infinite flat plate.
The very simple example of a two-dimensional steady uniform flow on an inclined plane of infinite
dimensions is given here as an example (Fig.5.1); the Navier-Stokes equation given in eq 5.1 may
be simplified in the following manner:
Since the motion is steady, u
t  0 and v t  0 . Since the motion is two-dimensional,
w  0 , and all derivatives with respect to z are zero. Since the motion is uniform and parallel to the
axis OX, v and all its derivatives are zero. All derivatives with respect to x are also zero. The
components of the gravitational force are X   gsin and Y    gcos . Since the flow is
uniform,
v  0 , and the continuity equation is reduced to u x  0 :
Forms of The Momentum Equation
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Basic Equation 5
Eka O. N.
Table 5. 2 Physical, mathematical characteristic and approximation of equations.
 V

 t
V 2
+grad 
 2
Physical meaning
Local inertia
Mathematical
characteristics
Approximation
First-order
linear term
= 0 In a
steady flow
Local inertia
Steady motion or
motion considered
as a succession of
steady motions
Unsteady motion


2
   curlV   V   grad  p   gz    V


Variation of
Rotational term
kinetic energy
with space
Convective inertia
Nonlinear (quadratic) term
Pressure
force
First-order
linear term
= 0 for
irrotational
motion; solution
given by a
harmonic
function
=0
=0
For slow motion
Gravity force
Friction force
Applied forces
Constant tem
Second-order
linear term
=0
in an ideal fluid
=0
in a gas
(with exceptions)
Table 5. 3 The momentum equation with some applications.
Convective Friction
Equations
inertia
Some
applications
Slow motion
Hydrostatics
Without friction
grad  p   gz   0
With friction
grad  p   gz   2 V=0
Irrotational
motion
Without friction
 V2

grad  
 p   gz   0
 2

Rotational
motion
With friction
 V2

grad  
 p   gz 
 2

    curlV   V+ 2 V
Slow motion
Without friction
With friction

V
 grad  p   gz   0
t

V
 grad  p   gz    2 V  0
t

 V2

V
 grad  
 p   gz   0
t
 2

Steady uniform flow
Flow in a porous medium
Nonuniform (convergent )
Steady flow at a constant
total energy. Calculation
of pressure in a twodimensional
flow net
General case of steady
motion ; laminar boundary
layer theories
Gravity wave (first-order
theory); water hammer
theory
Gravity wave damping
Irrotational
motion
Without friction
Most nonlinear wave
theories
Rotational
motion
Without friction
 V2

grad  
 p   gz 
 2

V

   curlV   V  0
t
Gravity wave theory of
Gerstner
With friction
General case
Tidal wave in an estuary
The Navier-Stokes equations are reduced to :
  2u 
 gsin    2   0
 y 
p
0     gcos
y
The second equation yields
p  pa   gycos
Forms of The Momentum Equation
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Basic Equation 5
Eka O. N.
pa is the atmospheric pressure. Hence the lines of equal pressure are parallel to the OX
axis. The boundary conditions are u  0 for y  d on the plane, and du dy  0 for y  0 at the
where
free surface. Taking into account these boundary conditions, the integration of
 2u
g

sin
2
y

gives successfully,
u
gsin

y
y


gsin 2

2
 =  and u  2  d  y 


which is the equation of a parabola. The discharge per unit of width is:
gsin
q   udy 
2
d
0
q
0
 d
2

 y 2 dy
d
gsin 3
d
3
The loss of energy per unit length may be given by the dissipation function  , which in this case is
  u y 
2
.
Figure 5. 1 Laminar flow on an inclined plane.
Hence the loss of energy per unit length is
 u 
  gsin  d 3

dy

dy

0
0  y 
3
d
2
d
2
This can also be obtained by determining the work done by friction forces Ff as follow;
2
d
 u 
u
F
du


du


0 f
0 y
0  y  dy
d
d
Forms of The Momentum Equation
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Basic Equation 5
Eka O. N.
5.2. DERIVATION MOMENTUM EQUATION ACCORDING TO TAYLOR’S SERIES
Conservation of Momentum for Fluids at a Point in Space and Time
We consider a fluid with hypothetical continuum properties averaged above the molecular level. We
assume that the fluid is incompressible.
Apply Newton’s Second Law:
 F  MA



Vector equation with 3 independent components
Acceleration has both time and spatially varying components.
Forces (Applied Forces) include:
o Gravitational forces
Gravitational Force is a body force which originates from the Earth’s gravitational field and
is proportional to the mass of the fluid.
dM   dV
o Pressure
Pressure Force is the surface force per unit area which acts normal to any surface (i.e. it
always pushes against any surface). It exists whether there is net fluid motion or not.
Pressure itself is a scalar and is direction independent.
o Viscous stresses
Viscous Forces are surface forces which result when gradients in motion exist. Essentially,
molecules resist one layer of fluid moving past another.
Must apply constitutive relationships in order to generate a sufficient number of equations (or
alternatively eliminate unknowns). For Newtonian Fluids:


Linear relationship between stress and velocity gradients
Valid for water, air, gasoline and most fluids of interest
Derivation of the Conservation of Momentum Equation at a Point in Space
Apply a fixed CV (Eulerian analysis)
w
v
u
C
z
G
H
D
z
p
F
B
E
A
y
x
p
p
dx
x
y
x
Figure 5. 2 Coordinate system for momentum equation.
Forms of The Momentum Equation
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Basic Equation 5
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Gravitational force is aligned with the z direction and for the defined CV
Fz  g   g  dV
Pressure force acts normal to each surface and is found by expressing the scalar value using
Taylor series about a point and multiplying by the surface area of each face
Net pressure force on faces ABCD and EFGH:


p
2 p
2
Fx  p  pdydz   p  dx  2  dx   ...  dydz
x
x


2
p
 p
Fx  p   dV  2 dxdV
x
x
Net pressure force on faces AEHD and BFGC:


p
2 p
2
Fy  p  pdxdz   p  dy  2  dy   ...  dxdz
y
y


2
p
 p
Fy  p   dV  2 dydV
y
y
Net pressure force on faces AEFB and DHGC:


p
2 p
2
Fz  p  pdxdy   p  dz  2  dz   ...  dxdy
z
z


p
2 p
Fz  p   dV  2 dzdV
z
z
Dynamic surface stresses are described by the stress tensor, as shown below
Figure 5. 3 Dynamic surface stresses system for momentum equation.
i = direction of plane
j = direction of stress
Forms of The Momentum Equation
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Basic Equation 5
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 ij can describe the stress on any surface with arbitrary orientation. It can be shown that the stress
tensor is symmetric; Dynamic stresses act only if the fluid at the molecularly averaged level
experiences relative motion
Sign convention for stresses: Stress is positive if it acts in the (positive/negative) coordinate
direction whose outward normal is in the (positive/negative) direction.
Net dynamic surface force in each direction is found by considering all surface stresses in each
direction and multiplying by the surface area of each face. Surface stresses on opposing faces can
be expressed in terms of each other by using Taylor series
Net dynamic stress force in the x direction:



 2 xx
2
Fx    xx dydz   xx  xx dx 
dx   ...  dydz
2 
x
x


2


 yx
  yx
2
  yx dxdz    yx 
dy 
dy   ...  dxdz

2


y
y


2



  zx
2
  zx dxdy   zx  zx dz 
dz   ...  dxdy
2 
z
z


Fx 
  2 xx
 2 yx
  xx  yx  yx 
 2 zx 



dy 
dz  dV  ...
 dV   2 dx 

y
y 
y 2
z 2
 x
 x

Net dynamic stress force in the y direction:
Fy 
Fy 


 xy
 2 xy
2
  xy dydz   xy 
dx 
dx

...


 dydz

x
x 2


2


 yy
  yy
2
  yy dxdz   yy 
dy 
dy   ...  dxdz

2


y
y


2


 zy
  zy
2
  zy dxdy   zy 
dz 
dz   ...  dxdy

2


z
z


  xy  yy  yy



y
y
 x
  2 xy
 2 yy
 2 zy 

dx 
dy 
dz  dV  ...
 dV  
2
2
2


x

y

z



Net dynamic stress force in the z direction:
Fz 
Fz 


 xz
 2 xz
2
  xz dydz   xz 
dx 
dx   ...  dydz
2 
x
x


2


 yz
  yz
2
  yz dxdz   yz 
dy 
dy

...


 dxdz

y
y 2


2




2
  zz dxdy   zz  zz dz  2zz  dz   ...  dxdy
z
z


 yz  zz
 
  xz 

y
y
 x
  2 xz
 2 yz

 2 zz 
dV

dx

dy

dz  dV  ...


 x 2

y 2
z 2



Forms of The Momentum Equation
Page
- 36
Basic Equation 5
Eka O. N.
Now applying the conservation of momentum law
  2 xx
 2 yx

 2 zx 
2 p
dV

dxdV

dx

dy

dz  dV
 2

2
2

x 2

x

y

z



2
2
2
2
   xy
  yy
  zy 
  xy  yy  zy 
Dv
p
 p
 dV
  dV  


dx 
dy 
dz  dV
 dV  2 dydV  
2
2
2

Dt
y
y
y 
y

x

y

z
 x


2
2
2
   xz
 yz  zz 
  yz
 
 2 zz 
Dw
p
 p
 dV
  dV  g  dV   xz 

dV

dzdV

dx

dy

dz  dV
 2

2
2

Dt
z
y
y 
z 2

x

y

z
 x


 dV
 yx  zx
 
Du
p
  dV   xx 

Dt
x
y
y
 x
Factoring out and letting the CV shrink to a point such that, dx0, dy0, dz0 and thus dV0,
leads us to the Conservation of Momentum Equation at a point in space and time

 yx  zx 
Du
p  
    xx 


Dt
x  x
y
y 

Dv
p   xy  yy  zy 
  



Dt
y  x
y
y 

 yz  zz 
 
Dw
p
    g    xz 


Dt
z
y
y 
 x
Expanding the acceleration terms leads to:
 u
u
u
u 
p
u v  w   
x
y
z 
x
 t
 yx  zx 
 
  xx 


y
y 
 x
 v
v
v
v 
p
u v  w   
x
y
z 
y
 t
  xy  yy  zy 




y
y 
 x


 yz  zz 
 
 w
w
w
w 
p
u
v
 w      g   xz 


x
y
z 
z
y
y 
 t
 x

Mass Local Acceleration Convective Acceleration + [ ] ´ Net Pressure+(Gravity)+Net Surface Force
=
To solve a fluid flow problem we currently have 11 unknown dependent variables:
 , u, v, w, p, xx , xy , xz , yy , yz , zz
However, we only have 5 equations:
D
=0
Dt

Conservation of Mass

Continuity

3 momentum conservation equations
u v w
 
0
x y z
We are 6 equations short of being able to start solving our system. Constitutive relationships must
now be applied to account for the physics removed by assuming a hypothetical continuum and
averaging molecular motion.
For a Newtonian incompressible isotropic fluid:
Forms of The Momentum Equation
Page
- 37
Basic Equation 5
Eka O. N.
 ui u j

 x j xi

 ij   



 u v 
xy   
    yx
 y x 
 v w 
 yz    
  zy
 z y 
 w x 
zx   

  zx
 x z 
where
u
x
v
 2
y
w
 2
z
 xx   2
 yy
zz
 = viscosity coefficient dependent on fluid type, temperature and pressure

Linear relationship for stress to rate of strain

No stress for flows with no relative motion (rigid body type motion)

No stress for flows with only pure rotation (rigid body rotation)
Substituting in for the stresses into the conservation of momentum statement and factoring out the
continuity equation leads to the Navier Stokes equations.


Conservation of Momentum - (Linear momentum)

Constitutive relationships - (for Newtonian fluids - linear stress/rate of strain)
Du
 
u 

 p  2 
Dt x 
x 

   u v      w u  


 
   
  X
y   y x   z   x z  
Du
1 p
 2u  2u  2u  2v  2w

2




Dt
 x
 x 2  y 2  z 2  yx  zx

 2u 2u 2u 
1 p
  u v w 
  2  2  2   


X


 x
x  x y z 
y
z 
 x
0
 u
u
u
u 
p
u v  w   
x
y
z 
x
 t
  2 u  2 u  2u 
 2  2  2 
z 
 x y
 v
v
v
v 
p
u v  w   
x
y
z 
y
 t
  2v  2v  2v 
 2  2  2 
 x y z 


 2w 2w 2w 
 w
w
w
w 
p
u
v
 w     g    2  2  2 
x
y
z 
z
y
z 
 t
 x

The Navier Stokes equations are valid for all incompressible isotropic Newtonian fluid flows. This
includes

Turbulent flows

Open channel and pipe flows that are normally depth or cross sectionally averaged
We can simplify the application of the Navier Stokes equations by assuming that density is almost
constant except when it involves the gravity term. When making the Boussinesq approximation, the
Navier Stokes equations are written as:
Forms of The Momentum Equation
Page
- 38
Basic Equation 5
Eka O. N.
u
u
u
u
1 p
u v  w  
t
x
y
z
0 x
   2 u  2 u  2u 
  2 2 2
0  x y z 
v
v
v
v
1 p
u v  w  
t
x
y
z
0 y
   2v  2v  2v 
  2 2 2
0  x y z 
w
w
w
w
1 p 
  2w 2w 2w 
u
v
w


g  2  2  2 
t
x
y
z
0 z 0
0  x
y
z 
The Boussinesq approximation essentially makes an approximation for the mass multiplying the
acceleration terms. Density varies little in natural water bodies. However in the gravity term,
gradients in density come into play and major circulation patterns can be driven by these terms. We
can not neglect small variations in density in these terms
In vector notation, the Navier Stokes equations with the Boussinesq approximation at a point are
expressed as:
DV
1
 ˆ  2
  p 
gk   V
Dt
0
0
0
where
V  uiˆ  vjˆ  wkˆ
Forms of The Momentum Equation
Page
- 39