Elementary Fluid Dynamics:
The Bernoulli Equation
CEE 331
March 14, 2016
Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Bernoulli
Along a Streamline
p   a   gkˆ
p
dz
   as   g
s
ds
z
Separate acceleration due to gravity. Coordinate
system may be in any orientation!
k is vertical, s is in direction of flow, n is normal.
Component of g in s direction
Note: No shear forces!
Therefore flow must be
frictionless.
Steady state (no change in
p wrt time)
k
i
x
n̂
ŝj
y
Bernoulli
Along a Streamline
p
dz
   as  
s
ds
dV V ds V
as 

 V
dt
s dt s
Can we eliminate the partial derivative?
Chain rule
Write acceleration as derivative wrt s
0 (n is constant along streamline, dn=0)
p
p
dp  ds  dn
 dp ds  p s and dV ds  V s
s
n
dp
dV
dz

 V

ds
ds
ds
Integrate F=ma Along a
Streamline
dp
dV
dz

 V

ds
ds
ds
Eliminate ds
dp  VdV   dz  0
Now let’s integrate…


dp

  VdV  g  dz  0
dp 1 2
 V  gz  C p
 2
1
p  V 2   z  C p '
2
But density is a function
of ________.
pressure
If density is constant…
Bernoulli Equation
 Assumptions needed for Bernoulli Equation
Frictionless
Steady
Constant density (incompressible)
Along a streamline
 Eliminate the constant in the Bernoulli equation?
_______________________________________
Apply
at two points along a streamline.
 Bernoulli equation does not include
Mechanical energy to thermal energy
 ___________________________
Heat transfer, Shaft Work
 ___________________________
Bernoulli Equation
The Bernoulli Equation is a
statement of the conservation
of ____________________
Mechanical Energy
p
1 2
 gz  V  C p

2
p.e.
k.e.
V2
z
 C p"

2g
p
p

 Pressure head
z  Elevation head
V2
 Velocity head
2g
Hydraulic Grade Line
p
 z  Piezometric head

V 2 Energy Grade Line
z


2 g Total head
p
Bernoulli Equation: Simple Case
(V = 0)
Reservoir (V = 0)
z
Pressure datum
Put one point on the surface,
one point anywhere else
V2
z
 C p"

2g
p
p1

 z1 
p2

p2
z1 - z2 =
g
 z2
Elevation datum
We didn’t cross any streamlines
so this analysis is okay!
Same as we found using statics
1
2
Hydraulic and
Energy
Grade
Mechanical energy
Lines (neglecting losses for now)
V2
z   z  2 g  C p"
p
Mechanical Energy The 2 cm diameter jet is
Conserved
5 m lower than the
V2
2g
p

z
surface of the reservoir.
V 2 What is the flow rate
2 g (Q)?
Teams
z
Elevation datum
Atmospheric pressure
Pressure datum? __________________
Bernoulli Equation: Simple Case
(p = 0 or constant)
What is an example of a fluid experiencing
a change in elevation, but remaining at a
Free jet
constant pressure? ________
V12 p2
V22
 z1 

 z2 

2g 
2g
p1
V12
V22
z1 
 z2 
2g
2g
V2 = 2 g ( z1 - z2 ) + V12
Bernoulli Equation Application:
Stagnation Tube
 What happens when the
water starts flowing in the
channel?
 Does the orientation of the
tube matter? _______
Yes!
 How high does the water
rise in the stagnation tube?
 How do we choose the
points on the streamline?
Stagnation point
V2
z
 Cp"

2g
p
Bernoulli Equation Application:
Stagnation Tube
1a-2a V = f(Dp)
_______________
Same streamline
1b-2a V = f(Dp)
V2
z
 Cp"

2g
1b
p
_______________
Crosses || streamlines
1a-2b V = f(z2)
_______________
Doesn’t cross
_____________
streamlines
V12 p2
V22
 z1 

 z2 

2g 
2g
p1
1a
2b z
2a
In all cases we don’t know p1
V12
 z2
2g
V1  2 gz2
Stagnation Tube
Great for measuring __________________
EGL (defined for a point)
Q   V  dA
How could you measure Q?
Could you use a stagnation tube in a
pipeline?
What problem might you encounter?
How could you modify the stagnation tube to
solve the problem?
Pitot Tubes
Used to measure air speed on airplanes
Can connect a differential pressure
transducer to directly measure V2/2g
Can be used to measure the flow of water
in pipelines Point measurement!
Pitot Tube
Stagnation pressure tap
Static pressure tap
V12 p2
V22
 z1 

 z2 

2g 
2g
p1
2
V
1
V1 = 0
z1 = z2
V
2

 p1  p2 
Connect two ports to differential pressure transducer.
Make sure Pitot tube is completely filled with the fluid
that is being measured.
Solve for velocity as function of pressure difference
Relaxed Assumptions for
Bernoulli Equation
Frictionless (velocity not influenced by viscosity)
Small energy loss (accelerating flow, short distances)
Steady
Or gradually varying
Constant density (incompressible)
Small changes in density
Along a streamline
Don’t cross streamlines
Bernoulli
Normal to the Streamlines
p   a   gkˆ
p
dz
   an   g
n
dn
Separate acceleration due to
gravity. Coordinate system
may be in any orientation!
Component of g in n direction
k
ŝ
n̂
Bernoulli
Normal to the Streamlines
p
dz
   an   g
n
dn
an 
dp 
2
V
R
R is local radius of curvature
n is toward the center of the radius of curvature
0 (s is constant normal to streamline)
p
p
ds  dn
s
n
dp
V2
dz
 
 g
dn
R
dn
 dp dn  p n
Integrate F=ma Normal to the
Streamlines
dp
V2
dz
 
 g
dn
R
dn
2
dp
V
   dn  gdz  C


n


R


p V 2
  dn  gz  Cn
  R
V

p    dn   gz  Cn"
 R
2
Multiply by dn
Integrate
(If density is constant)
Pressure Change Across
Streamlines
1 V 2
  dn  z  Cn '
 g R
p
V

p    dn   gz  Cn"
 R
2
If you cross streamlines that
are straight and parallel,
p + r gz = C and the
then ___________
hydrostatic
pressure is ____________.
p  C
2
1
p

 C12
2
rdr   gz  Cn"
n
V (r )  C1r
dn  dr
r 2   gz  Cn"
As r decreases p ______________
decreases
r
End of pipeline?
What must be happening when a horizontal
pipe discharges to the atmosphere?
V

p    dn   gz  Cn"
 R
2
Try applying statics… (assume straight streamlines)
Streamlines must be curved!
Nozzle Flow Rate: Find Q
Crossing streamlines
z2 , z3  0
z
1
z1  h
D1=30 cm
90 cm
Along streamline
p1 , p3  0
h
Q
Coordinate system
3
2
h=105 cm
gage pressure
Pressure datum____________
D2=10 cm
Solution to Nozzle Flow
1 V
 
 dn  z  Cn '
 g R
p
p1

 z1 
p2

 z2
h=105 cm
h
Now along the streamline
V2
z
 Cp"

2g
p
h
V32
V22 p3
 z2 
  z3 

2g 
2g
p2
z
1
2
p2

D1=30 cm
3
Q
2
D2=10 cm
V22 V32
h

2g 2g
Two unknowns…
_______________
Mass conservation
Q  V2 A2  V3 A3
4Q
V2 
 d 22
Solution to Nozzle Flow (continued)
2
8Q
8Q
h 2 4 
g d 2 g 2 d34
8 1 1 2
h
 4 Q
2  4
g  d3 d 2 
hg 2
Q
1 1
8 4  4 
 d3 d 2 
z
1
2
h=105 cm
D1=30 cm
3
Q
2
D2=10 cm
Incorrect technique…
z
1
1 V 2
  dn  z  Cn '
 g R
p
h=105 cm
3
Q
p
2
V
z
 Cp"

2g
D1=30 cm
2
D2=10 cm
p3
V32
1 V2
  dn  z1   z3 
 g R

2g
p1
V32
h
2g
Cp"  Cn '
The constants of integration
are not equal!
Bernoulli Equation Applications
Stagnation tube
Pitot tube
Free Jets
Orifice
Venturi
Sluice gate
Sharp-crested weir
Applicable to contracting
streamlines
(accelerating
flow).
Ping Pong Ball
Why does the ping pong
ball try to return to the
center of the jet?
What forces are acting on
the ball when it is not
centered on the jet?
1 V 2
  dn  z  Cn '
 g R
p
V2
z
 Cp"

2g
How does the ball choose the
distance above the source of
the jet?
p
Teams
n
r
Summary
By integrating F=ma along a streamline we
found…
mechanical
That energy can be converted between pressure,
elevation, and velocity
That we can understand many simple flows by
applying the Bernoulli equation
However, the Bernoulli equation can not be
applied to flows where viscosity is large, where
mechanical energy is converted into thermal
energy, or where there is shaft work.
Jet Problem
How could you choose your elevation
datum to help simplify the problem?
How can you pick 2 locations where you
know enough of the parameters to solve for
the velocity?
You have one equation (so one unknown!)
Jet Solution
The
2
cm
diameter
jet
is
5
m
lower
than
the
surface
of
the
z
reservoir. What is the flow rate (Q)?
z
p

V12
p2 V22

 z1 

 z2
 2g
 2g
V2
2g
Elevation datum
z2 = - 5 m
Are the 2 points on the
same streamline?
V2  2 g   z2 
p1
Q  V2
 d 22
4

 d 22
4
2 g   z2 
What is the radius of curvature at the
end of the pipe?
2
p V

 R

dn  gz  Cn
p V2
 z  gz  Cn
 R
h=105 cm
D1=30 cm
3
Q
Uniform velocity
Assume R>>D2
dn  dz

V2 
 z  g    Cn
 
R
p
V2
g
R
z
1
p V 2
  dn  gz  Cn
  R
V2
R
g
p1  p2  0
2
D2=10 cm
1
2
R
Example: Venturi
Example: Venturi
How would you find the flow (Q) given the pressure drop
between point 1 and 2 and the diameters of the two sections?
You may assume the head loss is negligible. Draw the EGL and
the HGL over the contracting section of the Venturi.
How many unknowns?
What equations will you use?
Dh
1
2
Example Venturi
V12
p2
V22
 z1 

 z2 
1
2g
2
2g
p1
V22 V12





2g 2g
p1
V1 A1  V2 A2
p2
4

d  
p1 p2 V 


1  2  
d  


2g 
 1 

2
2
V2 
2 g ( p1  p2 )

 1  d 2 d1 
Q  Cv A2
4
 1  d 2 d1 4
V1
d12
 V2

d 22
4
4
V1d12  V2 d 22
V1  V2

2 g ( p1  p2 )

Q  VA
d 22
d12