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Elementary Fluid Dynamics:The Bernoulli Equation
Elementary Fluid Dynamics:The Bernoulli Equation
CVEN 311 Fluid DynamicsCVEN 311 Fluid Dynamics
StreamlinesStreamlines
Steady State
BernoulliAlong a Streamline
BernoulliAlong a Streamline
ji
zz
yy
xx
ks
n
ˆp r g- Ñ = +a k Separate acceleration due to gravity. Coordinate system may be in any orientation!s
pa
ssdzd
r g¶
- = +¶
Component of g in s directionNote: No shear forces! Therefore flow must be frictionless.
Steady state (no change in p wrt time)
(eqn 2.2)
BernoulliAlong a Streamline
BernoulliAlong a Streamline
s
dV Va
dt s¶
= =¶s
dV Va
dt s¶
= =¶
s
p dza
s dsr g
¶- = +¶
p pdp ds dn
s n¶ ¶
= +¶ ¶
0 (n is constant along streamline)
dp dV dzV
ds ds dsr g- = +
( )212
d VdVVds ds
=
Write acceleration as derivative wrt s
chain rule
dsdt=
dsdt=
dp ds p s\ =¶ ¶ dV ds V s=¶ ¶and
Can we eliminate the partial derivative?
VVs
¶¶VVs
¶¶
Integrate F=ma Along a Streamline
Integrate F=ma Along a Streamline
( )212
d Vdp dzds ds ds
r g- = +
( )210
2dp d V dzr g+ + =
( )210
2dp
d V g dzr+ + =ò ò ò
212
dpV gz C
r+ + =ò
212
p V z Cr g+ + =
If density is constant…
Along a streamline
But density is a function of ________.pressure
Eliminate ds
Now let’s integrate…
Inviscid (frictionless)
Bernoulli EquationBernoulli Equation
Assumptions needed for Bernoulli Equation
Eliminate the constant in the Bernoulli equation? _______________________________________
Bernoulli equation does not include ___________________________ ___________________________
Assumptions needed for Bernoulli Equation
Eliminate the constant in the Bernoulli equation? _______________________________________
Bernoulli equation does not include ___________________________ ___________________________
Apply at two points along a streamline.
Mechanical energy to thermal energyHeat transfer, shaft work
SteadyConstant density (incompressible)Along a streamline
Bernoulli EquationBernoulli Equation
The Bernoulli Equation is a statement of the conservation of ____________________Mechanical Energy p.e. k.e.
212
pgz V C
r+ + =
2
2p Vz C
gg+ + =
Pressure headpg=
z =
pz
g+ =
2
2Vg=
Elevation head
Velocity head
Piezometric head
2
2p Vz
gg+ + = Total head
Hydraulic and Energy Grade Lines (neglecting losses for now)
Hydraulic and Energy Grade Lines (neglecting losses for now)
The 2 cm diameter jet is 5 m lower than the surface of the reservoir. What is the flow rate (Q)?
z
z
2
2Vg
Elevation datum
Pressure datum? __________________Atmospheric pressure
What about the free jet?
p
Bernoulli Equation: Simple Case (V = 0)
Bernoulli Equation: Simple Case (V = 0)
Reservoir (V = 0) Put one point on the surface,
one point anywhere else
Reservoir (V = 0) Put one point on the surface,
one point anywhere else
h
Elevation datum
21 2
pz z
g- = 2
1 2
pz z
g- =
Pressure datum 1
2
Same as we found using statics
2
2p Vz C
gg+ + =
1 21 2
p pz z
g g+ = + We didn’t cross any streamlines
so this analysis is okay!
Bernoulli Equation: Simple Case (p = 0 or constant)
Bernoulli Equation: Simple Case (p = 0 or constant)
What is an example of a fluid experiencing a change in elevation, but remaining at a constant pressure? ________
What is an example of a fluid experiencing a change in elevation, but remaining at a constant pressure? ________
2 21 1 2 2
1 22 2p V p V
z zg gg g
+ + = + +2 2
1 1 2 21 22 2
p V p Vz z
g gg g+ + = + +
( ) 22 1 2 12V g z z V= - +( ) 22 1 2 12V g z z V= - +
2 21 2
1 22 2V V
z zg g
+ = +2 2
1 21 22 2V V
z zg g
+ = +
Free jet
Bernoulli Equation Application:Stagnation Tube
Bernoulli Equation Application:Stagnation Tube
What happens when the water starts flowing in the channel?
Does the orientation of the tube matter? _______
How high does the water rise in the stagnation tube?
How do we choose the points on the streamline?
What happens when the water starts flowing in the channel?
Does the orientation of the tube matter? _______
How high does the water rise in the stagnation tube?
How do we choose the points on the streamline?
2
C2
p Vz
gg+ + =
2
C2
p Vz
gg+ + =
Stagnation point
Yes!
2 21 1 2 2
1 22 2p V p V
z zg gg g
+ + = + +
x2
C2
p Vz
gg+ + =
2
C2
p Vz
gg+ + =
2a1a
2b
1b
Bernoulli Equation Application:Stagnation Tube
Bernoulli Equation Application:Stagnation Tube
1a-2a _______________
1b-2a _______________
1a-2b _______________
_____________
1a-2a _______________
1b-2a _______________
1a-2b _______________
_____________
Same streamline
Crosses streamlines
Doesn’t cross streamlines
z
1. We can obtain V1 if p1 and (z2-z1) are known2. z2 is the total energy!
Stagnation TubeStagnation Tube
Great for measuring __________________ 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?
Great for measuring __________________ 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?
EGL (defined for a point)EGL (defined for a point)Q V dA= ×òQ V dA= ×ò
Bernoulli Normal to the Streamlines
Bernoulli Normal to the Streamlines
ks
n
ˆp r g- Ñ = +a k Separate acceleration due to gravity. Coordinate system may be in any orientation!n
pa
nndzd
r g¶
- = +¶
Component of g in n direction
Bernoulli Normal to the Streamlines
Bernoulli Normal to the Streamlines
2
n
Va =
R
2
n
Va =
R
n
p dza
n dnr g
¶- = +¶
p pdp ds dn
s n¶ ¶
= +¶ ¶
0 (s is constant along streamline)
2dp V dzdn dn
r g- = +R
centrifugal force. R is local radius of curvature
dp dn p n\ =¶ ¶ dV dn V n=¶ ¶and
n is toward the center of the radius of curvature
Integrate F=ma Normal to the Streamlines
Integrate F=ma Normal to the Streamlines
2dp Vdn gdz C
r+ + =óó
ô ôõ õ òR
If density is constant…
Normal to streamline
2dp V dzdn dn
r g- = +R
2p Vdn gz C
r+ + =óôõ R
2Vp dn z Cr g+ + =ó
ôõ R
Multiply by dn
Integrate
n
Pressure Change Across Streamlines
Pressure Change Across Streamlines
1( )V r C r=
If you cross streamlines that are straight and parallel, then ___________ and the pressure is ____________.
p z Cg+ =hydrostatic
2Vp dn z Cr g+ + =ó
ôõ R
21p C rdr z Cr g- + =ò
221
2C
p r z Cr
g- + =
dn dr=-
r
As r decreases p ______________decreases
Pitot TubesPitot 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
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 TubePitot Tube
VV1 =
12
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
z1 = z2( )1 2
2V p p
r= -( )1 2
2V p p
r= -
Static pressure tapStagnation pressure tap
0
2 21 1 2 2
1 22 2p V p V
z zg gg g
+ + = + +2 2
1 1 2 21 22 2
p V p Vz z
g gg g+ + = + +
Relaxed Assumptions for Bernoulli Equation
Relaxed Assumptions for Bernoulli Equation
Frictionless
Steady
Constant density (incompressible)
Along a streamline
Frictionless
Steady
Constant density (incompressible)
Along a streamline
Viscous energy loss must be small
Or gradually varying
Small changes in density
Don’t cross streamlines
Bernoulli Equation ApplicationsBernoulli Equation Applications
Stagnation tube Pitot tube Free Jets Orifice Venturi Sluice gate Sharp-crested weir
Stagnation tube Pitot tube Free Jets Orifice Venturi Sluice gate Sharp-crested weir
Applicable to contracting streamlines (accelerating flow).
Teams
Ping Pong BallPing 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?
How does the ball choose the distance above the source of the jet?
n r
Teams
SummarySummary
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 ProblemJet 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!)
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 SolutionJet Solution
The 2 cm diameter jet is 5 m lower than the surface of the reservoir. What is the flow rate (Q)?
z
pgz
2
2Vg
Elevation datum
What about the free jet?
2 5 mz =-2 2
1 1 2 21 22 2
p V p Vz z
g gg g+ + = + +
Example: VenturiExample: Venturi
Example: VenturiExample: Venturi
Find the flow (Find the flow (QQ) given the pressure drop between point 1 and 2 ) given the pressure drop between point 1 and 2 and the diameters of the two sections. You may assume the head and the diameters of the two sections. You may assume the head loss is negligible. Draw the EGL and the HGL. loss is negligible. Draw the EGL and the HGL.
11 22
hh
Example VenturiExample Venturi
2 21 1 2 2
1 21 22 2p V p V
z zg gg g
+ + = + +2 2
1 1 2 21 2
1 22 2p V p V
z zg gg g
+ + = + +
gV
gVpp
22
21
2221
gV
gVpp
22
21
2221
4
1
22
221 12 d
d
g
Vpp
4
1
22
221 12 d
d
g
Vpp
412
212
1
)(2
dd
ppgV
4
12
212
1
)(2
dd
ppgV
412
212
1
)(2
dd
ppgACQ v
4
12
212
1
)(2
dd
ppgACQ v
VAQ VAQ
2211 AVAV 2211 AVAV
44
22
2
21
1
dV
dV
44
22
2
21
1
dV
dV
222
211 dVdV 2
222
11 dVdV
21
22
21
d
dVV
21
22
21
d
dVV