Euler’s Equation in
Fluid Mechanics
What is Fluid Mechanics?
• Fluid mechanics is the study of the macroscopic physical behavior of fluids.
• Fluids are specifically liquids and gases though some other materials and systems can
be described in a similar way.
• Problems involve calculating for various properties of the fluid as functions of space and
time, such as:
• Velocity
• Pressure
• Density
• Temperature
• Fluid dynamics is a branch of fluid mechanics which includes fields from:
• aerodynamics (the study of gases in motion)
• hydrodynamics (liquids in motion)
• These fields are used in calculating forces and moments on aircraft , the mass flow
of petroleum through pipelines, and prediction of weather patterns.
Euler’s Differential Equations of
Fluid Mechanics



( u )  0
t x
u
u
1 p
u

t
x
 x
 p
 p
(  )u (  )  0
t 
x 



ρ = density
u = velocity
p = pressure
γ = Cp / Cv
These equations are the very basis of fluid mechanics. This despite, their formulation
more than 250 years ago.
Here the Cp and the Cv represent the specific heats of the fluid at constant pressure
and constant volume, respectively.
For steady, isentropic, irrotational, inviscid flow, the transformed equation is linear
Rewriting the above equations in simplified
notation, we have:
t  ( u ) x  0
ut  uu x 
(
p

)t  u (
px
0
(2)
)x  0
(3)

p

(1)
From the third equation we can deduce that if we let
p  A
then it is automatically satisfied. Further, we can use this
relation to simplify the first two equations.

Differentiation and substitution into
equations one and two yields
t  u x  u x  0
ut  uu x  A
For simplicity, since
denote it as c.
A
 2
x  0
is a constant we will
The Hodograph Transformation
•A transformation of coordinates used in fluid
dynamics.
•In the physical plane, the independent variables are
the position coordinates x and t.
•In the hodograph plane, the independent variables are
the components of the velocity vector, ρ and u.
•Dependent variables (including position) are
determined from the velocity components.
Taking the first equation and applying the
Hodograph Transformation we get:
Start with the first equation rewritten:
 (  , x)
 (t , u )
 (t ,  )

u
0
 (t , x)
 (t , x)
 (t , x)
Applying the transformation:
 (  , x)
 (t , u )
 (t ,  )

u
0
(  , u )
(  , u )
(  , u )
Resulting in
xu  t   utu  0
Now, take the second equation and apply the
Hodograph Transformation:
Start with the second equation rewritten:
 (u , x)
 (t , u )
  2  (t ,  )
u
 c
0
 (t , x)
 (t , x)
 (t , x)
Applying the transformation:
 (u , x)
 (t , u )
  2  (t ,  )
u
 c
0
(  , u)
(  , u )
(  , u )
Resulting in
 x  ut   c
But by multiplying by -1:
 2
tu  0
x  ut  c
 2
tu  0
Therefore,
xu  t   utu
x   ut   c
 2
tu
And assuming that u and p are independent, we can let:
x u  xu
Taking partial derivatives and setting the two equal to each other we
then get
utu  t  t  t  utu  c
 2
tuu
Which after simplification produces
c
tuu  t  2t  0
 2
Streamline plot of potential flow around a cylinder
Physical Plane
Hodograph Plane
Hodograph Pros and Cons

Pros


An infinite area in the physical plane maps into a finite area in the
hodograph plane.
When the flow is steady, isentropic, irrotational, and inviscid, the
transformed equation is linear. This is the most important reason for
considering the hodograph transformation.

Cons

The transformation is not one-to-one; the fluid that flows above the
body, and the fluid that flows below the body usually transform to the
same area in the hodograph plane.
Questions