Transcript Elementary Fluid Dynamics: The Bernoulli Equation

Elementary Fluid Dynamics:
The Bernoulli Equation
CVEN 311 Fluid Dynamics
Streamlines
Steady State
Bernoulli
Along a Streamline
z
- Ñp = r a + g kˆ
-
(eqn 2.2)
¶p
dz
= r as + g
¶s
ds
Separate acceleration due to
gravity. Coordinate system
may be in any orientation!
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
nˆ
sˆj
y
Bernoulli
Along a Streamline
¶p
dz
= r as + g
¶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)
¶p
¶p
dp = ds + dn
and dV ds = ¶V ¶s
\
dp
ds
=
¶
p
¶
s
¶s
¶n
dp
dV
dz
= rV
+g
ds
ds
ds
V
2
d
V
(
)
1
dV
=
ds 2 ds
Integrate F=ma Along a
Streamline
dp 1
= r
ds 2
d (V 2 )
ds
dz
+g
ds
1
dp + r d (V 2 ) + g dz = 0
2
dp 1
2
+
d
V
(
) + g ò dz = 0
ò r 2ò
Eliminate ds
Now let’s integrate…
But density is a function
of ________.
pressure
dp 1 2
ò r + 2 V + gz = C
If density is constant…
1
p + r V 2 +g z = C
2
Along a streamline
Bernoulli Equation

Assumptions needed for Bernoulli Equation
Inviscid (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
r
2
p.e. k.e.
p
V2
+z+
=C
g
2g
p
= Pressure head
g
z = Elevation head
V2
= Velocity head
2g
p
+ z = Piezometric head
g
p
V2
+z+
= Total head
g
2g
Hydraulic and Energy Grade
Lines (neglecting losses for now)
z
p

z
The 2 cm diameter jet is
5 m lower than the
surface of the reservoir.
V 2 What is the flow rate
2 g (Q)?
What about the free jet?
Elevation datum
Atmospheric pressure
Pressure datum? __________________
Bernoulli Equation: Simple Case
(V = 0)
h
 Reservoir
(V = 0)
Pressure datum
one point on the surface,
one point anywhere else
1
 Put
p
V2
+z+
=C
g
2g
p1
p2
+ z1 = + z2
g
g
p2
z1 - z2 =
g
Elevation datum
We didn’t cross any streamlines
so this analysis is okay!
Same as we found using statics
2
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? ________
p1
V12 p2
V22
+ z1 +
= + z2 +
g
2g g
2g
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
p
V2
+z+
=C
g
2g
Bernoulli Equation Application:
Stagnation Tube
 1a-2a
 _______________
Same streamline
p
V2
+z+
=C
g
2g
1b
x
 1b-2a
 _______________
Crosses streamlines
1a
2a
 1a-2b
 _______________
Doesn’t cross
_____________
streamlines
p1
V12 p2
V22
+ z1 +
= + z2 +
g
2g g
2g
1. We can obtain V1 if p1
and (z2-z1) are known
2. z2 is the total energy!
2b z
Stagnation Tube
 Great
for measuring __________________
EGL (defined for a point)
Q = òV ×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?
Bernoulli
Normal to the Streamlines
- Ñp = r a + g kˆ
-
¶p
dz
= r an + g
¶n
dn
Separate acceleration due to
gravity. Coordinate system
may be in any orientation!
Component of g in n direction
k
nˆ
sˆ
Bernoulli
Normal to the Streamlines
¶p
dz
= r an + g
¶n
dn
2
V
an =
R
centrifugal force. R is local radius of curvature
n is toward the center of the radius of curvature
0 (s is constant along streamline)
¶p
¶p
dp = ds + dn
¶s
¶n
dp
V2
dz
=r
+g
dn
R
dn
\ dp dn = ¶p ¶n and dV dn = ¶V ¶n
Integrate F=ma Normal to the
Streamlines
dp
V2
dz
=r
+g
dn
R
dn
2
ó dp ó V
+ ô dn + ògdz = C
ô
õ r õ R
p ó V2
+ ô dn + gz = C
r õ R
2
V
ó
p + r ô dn + g z = C
õ R
Multiply by dn
Integrate
If density is constant…
Normal to streamline
Pressure Change Across
Streamlines
2
V
ó
p + r ô dn + g z = C
õ R
If you cross streamlines that
are straight and parallel,
p + g z = C and the
then ___________
hydrostatic
pressure is ____________.
2
1
p- r C
ò rdr + g z = C
n
V (r ) = C1r
dn = - dr
r C12 2
pr +g z = C
2
As r decreases p ______________
decreases
r
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
p1
V12 p2
V22
+ z1 +
= + z2 +
g
2g g
2g
2
V
1
V1 = 0
z1 = z2
2
V=
( p1 - p2 )
r
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
Viscous energy loss must be small
 Steady
Or gradually varying
 Constant
density (incompressible)
Small changes in density
 Along
a streamline
Don’t cross streamlines
Bernoulli Equation Applications
 Stagnation
 Pitot
tube
tube
 Free Jets
 Orifice
 Venturi
 Sluice gate
 Sharp-crested weir
Applicable to contracting
streamlines (accelerating
flow).
Teams
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?
Teams
n
How does the ball choose the
distance above the source of
the jet?
r
Summary
 By
integrating F=ma along a streamline we
found…
 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 or
where mechanical energy is converted into
thermal energy.
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 actually
find 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)?
Elevation datum
z
p
g
V2
2g
What about the free jet?
p1 V12
p2 V22
+
+ z1 = +
+ z2
g 2g
g 2g
z2 = - 5 m
Example: Venturi
Example: Venturi
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.
h
1
2
Example Venturi
p1
V12
p2
V22
+ z1 +
=
+ z2 +
g1
2g g2
2g
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