ARO 3011 Fluid Dynamics/Low Speed
Aerodynamics
Lecture # 11
February 26-ish, 2026
Cauchy stress tensor (fluids or solids)
• Stresses are force/area. N/m2 or Pa.
• Viscous stresses denoted by .
• Index notation ij to indicate
direction.
• Nine stress components.
– xx, yy, zz are normal stresses.
– shear: zy is the stress in the ydirection on z-plane.
• Forces aligned with the coordinate
direction are positive.
 xx 
 yy  normal stresses
 zz 
 xy   yx 

 xz   zx  shear stresses
 yz   zy 
zy is the stress in the y-direction
on a z-plane
Pressure and viscous stress balance on the "cube"
( yx 
(p 
 yx 1
y 2
( zx 
y )xz
 zx 1
z )yz
z 2
p 1
x)yz
x 2
 ( yx 
 yx 1
y 2
(p 
 xx 1
 ( xx 
x)yz
x 2
( xx 
y )xz
p 1
x)yz
x 2
 xx 1
x)yz
x 2
z
y
x
 ( zx 
 zx 1
z )xy
z 2
Need to wrestle with
τii and P normal to
each face…
Net force in the x-direction is the sum of all the force components in that direction.
Viscous Stress balance on a “cube”: summing-up in "x"
x-direction balance of stresses
p 1 
p 1 


dx  dydz   p 
dx  dydz 
p

x
2

x
2




 1 
 1 


 yx  yx dy  dxdz   yx  yx dy  dxdz 
y 2 
y 2 


 1 
 1 


 zx  zx dz  dxdy   zx  zx dz  dxdy
z 2 
z 2 


Cancelling…
Surface forces in the xdirection due to respective
stress tensor components on
each face of the differential
control volume (6 faces, so
6 stresses)
 p  yx  zx 
 
dxdydz

y
z 
 x
Plugging back into F = ma:

 yx

Du
p 
   xx  zx 
Dt
x x
z
y
Navier equation: differential form of momentum eqn
x-direction
sum of forces
y-direction
sum of forces
z-direction
sum of forces
 p  yx  zx 
 Fx   Fx _ body  Fx _ surface  g x dxdydz   x  y  z dxdydz


  xy p  zy 

dxdydz
F

F

F


g
dxdydz



 y  y _ body  y _ surface
y
y
z 
 x
  xz  yz p 

F

F

F


g
dxdydz


 dxdydz
 z  z _ body  z _ surface z
y
z 
 x





F

F

F


g
   ij dxdydz
  body  surface
Body force (gravity)



 ( V )
   VV  g    ij
t


Divergence of the
stress tensor


DV

 g    ij
Dt
The Navier equations (plus continuity): 12 unknowns (9 components of stress 6 if by symmetry, 3 components of velocity, density via ideal gas law)… but
only 4 equations!
Stokes clarifies the stress tensor (Newtonian fluid)
•
(toothpaste)
•
(paint)
Reduction in the number of variables (12) by
relating stress tensor to strain-rate tensor.
For Newtonian fluid with constant properties
(Stokes’ relation)
 ui u j 
,i  j
 ij ~  



x

x
j
i


(Fancy “index” notation)
(quicksand)
u v
 xy   yx    (  )
y x
Newtonian fluid includes most common
fluids: air, other gases, water, gasoline
 xz   zx    (
And the on-diagonal terms of the stress tensor,
are just the negative of the pressure… pressure
is -1/3 of the trace of the stress tensor


  ij  p    V

Split pressure and viscous

 ij  V
u w
 )
z x
 yz   zy    (
v w
 )
z y
 Vi
Velocity gradient tensor V 
x j
Navier-Stokes equations (NSE) in Cartesian coordinates
Du
 2u  2u  2u p

  ( 2  2  2 )   g x
Dt
x
y
z
x
Dv
 2 v  2 v  2 v p

  ( 2  2  2 )   g y
Dt
x
y
z
y
Dw
 2 w  2 w  2 w p

  ( 2  2  2 )   g z
Dt
x
y
z
z



DV
2

  V  p  g
Dt
Inertial force/unit volume
Pressure force/unit volume
Viscous force/unit volume
Mass conservation (continuity)

 V  0
Incompressible NSE
written in vector form
Gravity force/unit volume





DV V
V
V
V

u
v
w
Dt
t  x
y
z
V  (u, v, w)
Summarizing the Navier-Stokes equations
1.
2.
3.
4.
5.
Newtonian fluid
Obeys Stokes’ hypothesis
Continuum
Isotropic viscosity (not a function of spatial coordinates)
If constant density --> velocity field is divergence-free
Navier-Stokes in conservative form
non-conservative form




 ( V )
   VV  p  g    V
t






DV

 p  g    V
Dt
  const



… in Cartesian component form:
Continuity
u v w
 
0
x y z
X-momentum
u
u
u
u
 2u  2u  2u p
 (  u  v  w )   ( 2  2  2 )   g x
t
x
y
z
x
y
z
x
Y-momentum
v
v
v
v
 2 v  2v  2 v p
 (  u  v  w )   ( 2  2  2 )   g y
t
x
y
z
x
y
z
y
Z-momentum
w
w
w
w
 2 w  2 w  2 w p
(  u
v
 w )   ( 2  2  2 )   g z
t
x
y
z
x
y
z
z
Navier-Stokes are highly non-linear because of the convective
acceleration terms: there is no general analytical solution
… but we do have 4 equations for 4 unknowns (u, v, w, p)
And in cylindrical coordinates
continuity
 1  (rvr ) 1  ( v )  ( v z )



)0
t r r
r 
z
r -momentum
   1  (rvr )  1  2 vr 2 v  2 vr 
vr
vr v vr v2
vr
p
(
 vr

  vz
)    g r    
 2
 2
 2
2
t
r
r  r
z
r

r
r

r
r


r


z 

 
θ -momentum
   1  (rv )  1  2v 2 vr  2 v 
v
v v v vr v
v
1 p
(
 vr


 vz
)
 g    
 2
 2 
 2
2
t
r
r 
r
z
r 
r  z 
 r  r r  r 
 1   v  v 1  2 v
1 p
2 vr  2 v 
...  
 g   
 2
 2 
r
 2  2
2
r 
r

r

r
r
r


r


z 



z -momentum
 1   vz  1  2 vz  2 v 
vz
vz v v z
v z
p
(
 vr

 vz
)    g z   
 2 
r
 2
2
t
r
r 
z
z
z 
 r r  r  r 
What about Euler?
Setting  = 0 in the Navier-Stokes equations, recover the Euler equations
u
u
u
u
p
u
 v  w )    g x
t
x
y
z
x
v
v
v
v
p
 (  u  v  w )    g y
t
x
y
z
y
w
w
w
w
p
(  u
v
 w )    g z
t
x
y
z
z
(
In vector form
Steady Euler




V

  (V  )V  p  g
t



 (V  )V  p  g
Gravitational
potential function
Valid for all inviscid
flows – incompressible
or compressible, even
supersonic

g   gz


 (V  )V  p  gz
An alternative path towards the Bernoulli eqn (1)






 1   

Vector identity
V   V   V V  V   V
2

1 2 
p
  V  V   V  
 gz
2


Case 1: going along a streamline



ds  dxiˆ  dyˆj  dzkˆ
 

2

  p 

 
1 
  gds  z  ds  V    V
ds   V  ds  
2
  





ds and V are parallel (along a streamline!), but V    V and ds are perpendicular
2
2
2
V
V
 2 V
2

ds   V 
dx 
dy 
dz  d V
x
y
z


z
z
z
ds  z  dx  dy  dz  dz
x
y
z

 


 
ds  V    V  0
An alternative path towards the Bernoulli eqn (2)

p
p
p
ds   p 
dx 
dy 
dz  dp
x
y
z
1
dp
2
Putting it all together:
dV 
 gdz  0
2

 
And finally assume incompressibility:
2
1
dp
V
p
2
d V   
 g  dz  0 
  gz  const

2

2 
Case 2: irrotational flow

 V  0 
1 2
p
V 
 gz
2

Still need incompressibility, but now, no longer need to be along a streamline!
V2 p
  gz  const
2 
Anywhere throughout an
irrotational velocity field!
Compare Euler and N-S: boundary conditions
• To solve a PDE (partial differential equation, we need to impose the correct
B.C. and initial conditions
• For Navier-Stokes eqns and Euler eqns, the corresponding BC are different
• NS are elliptical (if steady state) and mixed elliptical-hyperbolic (if unsteady),
and 2nd order (so, more BCs)
• Euler eqns are 1st order (fewer BC) and hyperbolic. So, different analytic or
computational methods to solve them. Euler solver “relaxes” wall boundary
condition from no-slip to slip in inviscid flow
No-slip B.C.
u=v=w=0
Navier-Stokes equation
or boundary layer
simplification of NS
Slip B.C.
w = 0, Vn = 0
Vn = wall-normal velocity
Inviscid flow (Euler) equation, =0
Euler vs. NS Wall BCs
• For inviscid flows, the velocity at the surface can be finite (nonzero --> “slip”)
• But because the flow cannot penetrate the surface, the velocity vector must be

tangent to the surface.
 
At the solid surface V  n  0

where n is outward normal
n

V
For a fluid in contact with a solid wall
in a viscous flow, the velocity of the
fluid must equal that of the wall
If solid wall is at rest, then for fluid
adjacent to the wall Vwall= 0.
If solid wall is moving at V = VP ,
then for fluid adjacent to the wall
Vwall= VP.
Summarizing…
•
•
•
•
Navier-Stokes are second order partial differential equations
Euler equations are first order PDEs
Both are non-linear
Neither is "easy" to solve
Navier-Stokes equations
Euler equation





V
2

  (V  )V  p  g   V
t




V

  (V  )V  p  g
t
=0
These are the extra terms
X-momentum
u
u
u
u
 2u  2u  2u p
 (  u  v  w )   ( 2  2  2 )   g x
t
x
y
z
x
y
z
x
Y-momentum
v
v
v
v
 2 v  2 v  2 v p
 (  u  v  w )   ( 2  2  2 )   g y
t
x
y
z
x
y
z
y
Z-momentum
w
w
w
w
 2 w  2 w  2 w p
(  u
v
 w )   ( 2  2  2 )   g z
t
x
y
z
x
y
z
z
u
u
u
u
p
u
 v  w )    g x
t
x
y
z
x
v
v
v
v
p
 (  u  v  w )    g y
t
x
y
z
y
w
w
w
w
p
(  u
v
 w )    g z
t
x
y
z
z
(
Applications of the Navier-Stokes Equations
• The equations are nonlinear partial differential equations
• No full analytical solution exists!
• Can be solved only for several simple flow conditions
• Numerical methods are the modern go-to tool: CFD (Computational Fluid
Dynamics) – with or without “modeling” of the turbulence
Example: fully developed Couette Flow
For the given geometry and BCs,
find the velocity and pressure
fields, and the shear force per unit
area acting on the bottom plate
Step 1: geometry, dimensions, fluid properties
Note: viscous flow --> can
not find a potential… but can
still find a stream function
Step 2: State assumptions and boundary conditions
Assume p/x=0
p/y=0
d 2u
 0  u ( y )  C1 y  C2
2
dy
Step 4: Simplify (continued)
Result: linear ODEs and algebraic relations for non-degenerated dependent variables
Step 5: Solve ODE and apply boundary conditions
y  0  u  0  C1 0  C2  C2  0
y  h  u  V  C1h  C1 
V
h
u( y)  V
Couette flow
y
h
For pressure, take p  p0 at z = 0, so integration constant is p0
p( z )  p0  gz
Step 6: Verify solution by substituting back into continuity
and momentum PDEs

u
 0,
x
v
 0,
y
y

Given the solution (u , v, w)  V , 0, 0 
 h


w
 0    V  0 Incompressible
z
  2u  2u  2u  p
 u
u
u
u 
   u  v  w     2  2  2    g x
x
y
z 
y
z  x
 t
 x


0  V
y
V

0  0  0    0  0  0  0   0
h
h

Momentum is satisfied
Step 7: Calculate the Stokes shear stress tensor at
the bottom plate (evaluate term by term)
Shear force per unit area acting on the wall
Note that w is equal and opposite to the
shear stress acting on the fluid yx
(Newton’s third law).
We have recovered    u , which we’ve been using ever
y
since Lecture 1!
Add nonzero pressure gradient to Couette Flow
Flow between two parallel plates: bottom one is fixed, and top is moving at velocity, U.
Constant nonzero dp/dx. Find the velocity profile between the plates. Assume laminar flow.
Velocity profiles
with different
values of dp/dx
( )
0
t
( )
• Fully developed flow
0
x
• Continuity
• Steady flow
• 2D flow --> w = 0
u v

0
x y
v
0
y
v0@ y0
thus v  0 everywhere
Nonzero pressure gradient Couette Flow: solution
1.
2.
3.
Apply BC
y = 0 --> u = 0
y = b --> u = U
1.
p
 g
y
2.
0
3.
0 = 0
= p0 - ρgy
If
dp
y
 0, back to Couette flow u  U ( )
dx
b
Waypoint
1. We have derived Euler, Navier and Navier-Stokes equations…
but rarely can we solve them.
1.
2.
A few examples of very simple viscous flows
More coming later, for boundary layers
2. Now switch gears and revisit Laplace's equation
1.
2.
Leverage linearity of Laplace, and superposition
"Elementary flows" and their superposition