SO513 Lesson 3: Navier-Stokes equation applications
Du
 p   gkˆ   2u ,
Dt
has exact, integral solutions associated with it. However, because of the non-linear advection
term  u   u , and the second-order Laplacian operator in the friction term, exact solutions are
not easy to come by. Let’s make some simplifications to the equation and see if we can apply
the resulting form in any real-world settings.
The incompressible form of the Navier-Stokes equation of motion, 
In some real-world settings, friction is not all that important, at least in some parts of a flow. For
example, examine horizontal flow around a vertical telephone pole. Looking down from above,
streamlines for this flow might look something like:
Around the pole, there is a thin boundary layer where friction is important. Also, in the lee of the
pole, there is boundary layer separation where slow flow near the pole erupts into the main flow,
leading to downstream vorticity (eddies) and turbulence. In these areas, friction is important, but
not elsewhere.
In areas where friction is unimportant, the Navier-Stokes equations become:
u
1
  u    u   p  g
t

This form of the equation is called Euler’s equation of motion, and is valid for inviscid flows
(that can be either compressible or incompressible).
q2
Using the following vector identity,  u    u      u , where q is speed of the fluid
2
2
ˆ
( q  u  u , q  u  u ), and g   gk  gz , the Euler equation becomes
 q2
 1
u
    gz   p    u  0 .
t
 2
 
If density is constant, then
1

p  
p

. If the flow is steady,
u
 0 . Then the equation
t
becomes
 q2 p

    gz     u  0
 2 

 q2 p

Dot this equation with a streamline, dr , to get dr      gz   dr    u   0 . dr is
 2 

parallel to u (by definition of a streamline), so dr will be perpendicular to   u . We are left
 q2 p

with dr      gz   0 , which is valid along a streamline for an inviscid, constant-density,
 2 

q2 p
steady fluid. The derivation yields
  gz  C , where C is a constant, which is Bernoulli’s
2 
Equation for a steady, inviscid, constant density flow.
Applications of Bernoulli’s Equation
1. Flow through a restriction (like flow through a channel or a canyon)
q2 p
  gz  C . Let the velocity u  Uiˆ at the
2 
point A far upstream of the constriction. This means the speed at A is U ( qA  U ). Bernoulli’s
Bernoulli’s equation tell us along a streamline,
U 2 pA
q 2 p

 gz A  C . At the constriction B, B  B  gzB  C . Since A and
2

2

B are along the same streamline, the constants in the two Bernoulli equations are equal. Thus,
U 2 pA
q2 p

 gz A  B  B  gzB . Simplifying the equation (zA = zB), we get
2

2

equation becomes
p A  pB 

q
2
2
B
 U 2  . Since qB  U (from mass conservation), we see that p A  pB .
Bernoulli’s equation has thus told us that the reason the low accelerated from A to B was
because of a change in pressure.
2. Horizontal flow around a vertical cylinder (like a telephone pole, or a thunderstorm updraft,
which sometimes behaves like a cylinder!)
Here upstream flow with speed U approaches the cylinder. The center streamline deflects
significantly around the obstacle, while the far-away streamlines experience slight deflection.
Point “S” is a stagnation point where velocity goes to zero. At points A and B, because
streamlines get closer together, expect velocity to increase. Assume constant density and
inviscid, steady flow. Bernoulli equation upstream is
U 2 pup

 gz  C
2

At the stagnation point S,
pS

 gz  C . The constant C is the same constant because we’re on
U2
U 2 pup
p
, or that the

 gz  S  gz . We get that pS  pup  
2
2


pressure at the stagnation point is higher than it is upstream. At the top (and bottom, by
the same streamline. Thus,
symmetry) of the cylinder,
qB2 pB

 gz  C . Still the same C (the same streamline), so
2

 2 2
qB2 pB
U 2 pup

 gz 

 gz . This gives that pB  pup  U  qB  , which says that pB  pup
2
2

2

2
2
because U  qB  0 .
3. Wind storm over a Walmart.
q2 p
Bernoulli equation along a streamline is
  gz  C . Let’s work with perturbation
2 
pressure, the departure of pressure from a mean, background state. p  p  p0  z  .
p0
   g , or p0    gz . Bernoulli’s equation
z
q2 p
q2 p  p0
q2 p     gz 
q 2 p
becomes
  gz  
 gz  
 gz    C . Comparing
2 
2

2

2 
velocity upstream with velocity over the roof of Walmart, we see that the velocity over the roof
is much faster (again, because of closer streamlines and mass conservation). This requires a
large, negative perturbation pressure, which for a sudden wind storm, will not allow for the
Background pressure is in hydrostatic balance,
pressure to equilibrate from within the store (which was in hydrostatic balance) and above the
roof (which is much smaller). The resulting sudden pressure difference causes the roof to lift off,
which immediately equalizes the pressure difference, after which the roof will collapse back into
the store.
4. Bernoulli ventilation of prairie dog burrows.
Prairie dog burrows have at least two openings, one of which will be elevated. Faster wind
speed above the elevated burrow creates a negative perturbation pressure there, when compared
to the other opening(s). Thus, pressure gradient drives a slow ventilation flow through the
burrow. (Friction within the burrow is probably important, so do not apply Bernoulli’s equation
down there).
5. Bernoulli ventilation of underground cities.
Similar to prairie dogs, the Hittites built vast underground cities (some up to 20 stories deep!) in
the mountains of Turkey. They built ventilation shafts in the mountain side and connected the
cities to lateral shafts with outlets at lower elevations. The Taliban had similar structures at Tora
Bora, and the Byzantines and Romans also built underground cities.