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fluid mechanics
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Monroe L. Weber-Shirk
School of Civil and Environmental Engineering
Elementary Fluid Dynamics:The Bernoulli Equation
Elementary Fluid Dynamics:The Bernoulli Equation
CEE 331
April 21, 2023
BernoulliAlong a Streamline
ji
z
y
x
ks
n
ˆp g a kSeparate acceleration due to gravity. Coordinate system may be in any orientation!k is vertical, s is in direction of flow, n is normal.
s
zp
ss
d
da g
Component of g in s direction
Note: No shear forces! Therefore flow must be frictionless.
Steady state (no change in p wrt time)
BernoulliAlong a Streamline
s
dV Va
dt s
s
p dza
s ds
p pdp ds dn
s n
0 (n is constant along streamline, dn=0)
dp dV dzV
ds ds ds
Write acceleration as derivative wrt s
Chain rule
ds
dt
dp ds p s dV ds V s and
Can we eliminate the partial derivative?
VVs
Integrate F=ma Along a Streamline
0dp VdV dz
0dp
VdV g dz
21
2 p
dpV gz C
2'
12 pp V z C
If density is constant…
But density is a function of ________.pressure
Eliminate ds
Now let’s integrate…
dp dV dzV
ds ds ds
Assumptions needed for Bernoulli Equation
Eliminate the constant in the Bernoulli equation? _______________________________________
Bernoulli equation does not include ___________________________ ___________________________
Bernoulli Equation
Apply at two points along a streamline.
Mechanical energy to thermal energyHeat transfer, Shaft Work
FrictionlessSteadyConstant density (incompressible)Along a streamline
Bernoulli Equation
The Bernoulli Equation is a statement of the conservation of ____________________Mechanical Energy p.e. k.e.
212 p
pgz V C
2
"2 p
p Vz C
g
Pressure headp
z
pz
2
2
V
g
Elevation head
Velocity head
Piezometric head
2
2
p Vz
g
Total head
Energy Grade Line
Hydraulic Grade Line
Bernoulli Equation: Simple Case (V = 0)
Reservoir (V = 0)Put one point on the surface,
one point anywhere else
z
Elevation datum
21 2
pz z
g- =
Pressure datum 1
2
Same as we found using statics
2
"2 p
p Vz C
g
1 21 2
p pz z
We didn’t cross any streamlines
so this analysis is okay!
Mechanical Energy Conserved
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)?
p
z
2
2
V
g
Elevation datum
z
Pressure datum? __________________Atmospheric pressure
Teams
z
2
2
V
g
2
"2 p
p Vz C
g
Mechanical energy
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? ________
2 21 1 2 2
1 22 2
p V p Vz z
g g
( ) 22 1 2 12V g z z V= - +
2 21 2
1 22 2
V Vz z
g g
Free jet
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?
2
p"C2
p Vz
g
Stagnation point
Yes!
2
p"C2
p Vz
g
2a1a
2b
1b
Bernoulli Equation Application:Stagnation Tube
1a-2a_______________
1b-2a_______________
1a-2b_______________
_____________
Same streamline
Crosses || streamlines
Doesn’t cross streamlines
2 21 1 2 2
1 22 2
p V p Vz z
g g
z
21
22
Vz
g 21 2V gz
V = f(p)
V = f(z2)
V = f(p)
In all cases we don’t know p1
Stagnation 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?
EGL (defined for a point)Q V dA
Pitot Tubes
Used to measure air speed on airplanesCan connect a differential pressure
transducer to directly measure V2/2gCan be used to measure the flow of water
in pipelines Point measurement!
Pitot 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
Static pressure tapStagnation pressure tap
0
2 21 1 2 2
1 22 2
p V p Vz z
g g
Relaxed Assumptions for Bernoulli Equation
Frictionless (velocity not influenced by viscosity)
Steady
Constant density (incompressible)
Along a streamline
Small energy loss (accelerating flow, short distances)
Or gradually varying
Small changes in density
Don’t cross streamlines
Bernoulli Normal to the Streamlines
ks
n
ˆp g a k Separate acceleration due to gravity. Coordinate system may be in any orientation!n
zp
nn
d
da g
Component of g in n direction
Bernoulli Normal to the Streamlines
2
n
Va
R
n
p dza g
n dn
p pdp ds dn
s n
0 (s is constant normal to streamline)
2dp V dzg
dn R dn
R is local radius of curvature
dp dn p n
n is toward the center of the radius of curvature
Integrate F=ma Normal to the Streamlines
2
n
dp Vdn gdz C
R
(If density is constant)
2dp V dzg
dn R dn
2
n
p Vdn gz C
R
2
"n
Vp dn gz C
R
Multiply by dn
Integrate
n
Pressure Change Across Streamlines
1( )V r C r
If you cross streamlines that are straight and parallel, then ___________ and the pressure is ____________.
p gz Cr+ =hydrostatic
2
"n
Vp dn gz C
R
21 "np C rdr gz C
221
"2 n
Cp r gz C
dn dr
r
As r decreases p ______________decreases
2
'
1n
p Vdn z C
g R
End of pipeline?End of pipeline?
What must be happening when a horizontal pipe discharges to the atmosphere?
2
"n
Vp dn gz C
R
Try applying statics…
Streamlines must be curved!
(assume straight streamlines)
Nozzle Flow Rate: Find Q
D1=30 cm
D2=10 cm
Q
90 cm
1
2 3
1 3, 0p p
2 3, 0z z
1z h
z
Coordinate system
Pressure datum____________
Crossing streamlines
Along streamlineh
h=105 cm
gage pressure
Solution to Nozzle Flow
2
p"C2
p Vz
g
2
'
1n
p Vdn z C
g R
2ph
D1=30 cm
D2=10 cm
Q
h=105 cm
1
2
3
z
2 2 3 3Q V A V A 2 22
4QV
d
1 21 2
p pz z
Now along the streamline
223 32 2
2 32 2
p Vp Vz z
g g
h
2232
2 2
VVh
g g
Two unknowns…_______________Mass conservation
Solution to Nozzle Flow (continued)
2 2
2 4 2 42 3
8 8Q Qhg d g d
22 4 4
3 2
8 1 1h Q
g d d
2
4 43 2
1 18
hgQ
d d
D1=30 cm
D2=10 cm
Q
h=105 cm
1
2
3
z
Incorrect technique…
2
p"C2
p Vz
g
2
'
1n
p Vdn z C
g R
23
2
Vh
g p" 'C nC
223 31
1 3
12
p Vp Vdn z z
g R g
D1=30 cm
D2=10 cm
Q
h=105 cm
1
2
3
z
The constants of integration are not equal!
Bernoulli Equation Applications
Stagnation tubePitot tubeFree JetsOrificeVenturiSluice gateSharp-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?
How does the ball choose the distance above the source of the jet?
n r
Teams
2
p"C2
p Vz
g
2
'
1n
p Vdn z C
g R
Summary
By integrating F=ma along a streamline we found…That energy can be converted between pressure,
elevation, and velocityThat we can understand many simple flows by
applying the Bernoulli equationHowever, 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.
mechanical
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 reservoir. What is the flow rate (Q)?
z
p
z
2
2
V
g
Elevation datum
Are the 2 points on the same streamline?
2 5 mz =-
2 21 1 2 2
1 22 2
p V p Vz z
g g
2 22V g z
2 22 2
2 224 4
d dQ V g z
What is the radius of curvature at the end of the pipe?
2
n
p Vdn gz C
R
2
n
p Vdn gz C
R
dn dz
2
n
p Vz gz C
R
2
n
p Vz g C
R
2Vg
R
D1=30 cm
D2=10 cm
Q
h=105 cm
1
2
3
z
2
1
2VR
g
R1 2 0p p
Uniform velocityAssume R>>D2
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.
1 2
h
How many unknowns?What equations will you use?
Example Venturi
2 21 1 2 2
1 21 22 2
p V p Vz z
g g
gV
gVpp
22
21
2221
4
1
22
221 12 d
d
g
Vpp
412
212
1
)(2
dd
ppgV
412
212
1
)(2
dd
ppgACQ v
VAQ
2211 AVAV
44
22
2
21
1
dV
dV
222
211 dVdV
21
22
21
d
dVV