Transcript Fluids - Northern Illinois University

Fluids
Eulerian View

 
r (r0 , t )
In a Lagrangian view each
body is described at each
point in space.
• Difficult for a fluid with many
particles.


 (r , t )
 
v (r , t )
In an Eulerian view the
points in space are
described.
• Bulk properties of density
and velocity
Compressibility

A change in pressure on a
fluid can cause deformation.

Compressibility measures
the relationship between
volume change and
pressure.
V
p
V
 p
V
B
Vp 1

V 
• Usually expressed as a bulk
modulus B

Ideal liquids are
incompressible.
Fluid Change



A change in a property like
pressure depends on the
view.
In a Lagrangian view the
total time derivative depends
on position and time.
An Eulerian view is just the
partial derivative with time.
• Points are fixed
dp p p dx p dy p dz




dt t x dt y dt z dt
dp p 

 v  p
dt t
d
 
 F v2k l  x  l 
dt t
2
dp
p

dt r const t
x
2
l  x2
2
Volume Change

Consider a fixed amount of
fluid in a volume dV.
• Cubic, Cartesian geometry
• Dimensions dx, dy, dz.

The change in dV is related
to the divergence.
• Incompressible fluids must
have no velocity divergence
v
d
dx  x dx
dt
x
v y
d
dy 
dy
dt
y
d
v
dz  z dz
dt
z
 vx v y vz 
d
dxdydz
dV  


dt
 x y z 

d
dV    v dV
dt
Continuity Equation

A mass element must remain
constant in time.
• Conservation of mass

Combine with divergence
relationship.

Write in terms of a point in
space.


   ( v )  0
t
dm  dV
d
d
dm  d V   0
dt
dt
d
ddV
dV  
dt
dt

d

dV    v dV  0
dt

d
   v  0
dt

 
 v      v  0
t
Stress

F

A

F
A
A stress measures the
surface force per unit area.
• A normal stress acts normal
to a surface.
• A shear stress acts parallel
to a surface.

A fluid at rest cannot support
a shear stress.
Force in Fluids

 
P(dS1 )

dS1
 
P(dS2 )
• Describe the stress P at any
point.
• Normal area vectors S form
a triangle.

dS 2




  dS1  dS2

P(dS1  dS2 )
Consider a small prism of
fluid in a continuous fluid.
The stress function is linear.
 
 
P(cdS )  cP(dS )


 
P(dS )  P(dS )

 
 
 
P(dS1 )  P(dS2 )  P(dS1  dS2 )
Stress Tensor

Represent the stress
function by a tensor.

 
P(dS )  P  dS
• Symmetric
• Specified by 6 components

If the only stress is pressure
the tensor is diagonal.

The total force is found by
integration.


 
P(dS )  P  dS  p1 dS


F   P  dS
S
Force Density

F    P  nˆ dS

S

F      PdV
The force on a closed
volume can be found
through Gauss’ law.
• Use outward unit vectors
V

f S    P

A force density due to stress
can be defined from the
tensor.
• Due to differences in stress
as a function of position
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