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Monroe L. Weber-Shirk School of Civil and
Environmental Engineering
Elementary Fluid Dynamics:The Bernoulli Equation
CEE 331June 8, 2006
BernoulliAlong a Streamline
ji
z
y
x
ks
n
ˆp ga kρ ρ−∇ = +Separate acceleration due to gravity. Coordinate system may be in any orientation!k is vertical, s is in direction of flow, n is normal.
szpss
dd
a gρ ρ∂− = +∂ 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
sdV Vadt s
∂= =
∂
sp dzas ds
ρ γ∂− = +∂
Can we eliminate the partial derivative?
p pdp ds dns n
∂ ∂= +∂ ∂
0 (n is constant along streamline, dn=0)
dp dV dzVds ds ds
ρ γ− = +
Write acceleration as derivative wrt s
Chain ruledsdt
=
dp ds p s∴ = ∂ ∂ dV ds V s= ∂ ∂and
V Vs
∂∂
Integrate F=ma Along a Streamline
dp dV dzVds ds ds
ρ γ− = + Eliminate ds
0dp VdV dzρ γ+ + =But density is a function of ________.pressure
Now let’s integrate…
0dp VdV g dzρ+ + =∫ ∫ ∫
212 p
dp V gz Cρ+ + =∫ If density is constant…
2'
12 pp V z Cρ γ+ + =
Bernoulli Equation
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
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
p gz V Cρ+ + =
2
"2 pp Vz C
gγ+ + =
Pressure headpγ=
z =
p zγ+ =
2
2V
g=
Elevation head
Velocity head
Piezometric head
2
2p Vz
gγ+ + =
Total headEnergy Grade Line
Hydraulic Grade Line
Bernoulli Equation: Simple Case (V = 0)
zReservoir (V = 0)
Put one point on the surface, one point anywhere else
1
2
Pressure datum
Elevation datum2
"2 pp Vz C
gγ+ + =
1 21 2
p pz zγ γ+ = + We didn’t cross any streamlines
so this analysis is okay!
21 2
pz zγ
− = Same as we found using statics
Hydraulic and Energy Grade Lines (neglecting losses for now)
Mechanical energy
Mechanical Energy Conserved
The 2 cm diameter jet is 5 m lower than the surface of the reservoir. What is the flow rate (Q)?p
γ
z
2
2V
g
Elevation datum
z
Pressure datum? __________________Atmospheric pressure
z
2
2V
g
2
"2 pp Vz C
gγ+ + =
Teams
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 2p V p Vz z
g gγ γ+ + = + +
( ) 22 1 2 12V g z z V= − +
2 21 2
1 22 2V 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 Vzgγ
+ + =
Stagnation point
Yes!
Bernoulli Equation Application:Stagnation Tube
2
p"C2
p Vzgγ
+ + =
2a1a
2b
1b
1a-2a_______________
1b-2a_______________
1a-2b____________________________
Same streamline
Crosses || streamlines
Doesn’t cross streamlines
2 21 1 2 2
1 22 2p V p Vz z
g gγ γ+ + = + +
z
21
22V z
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
z1 = z2( )1 2
2V p pρ
= −
Static pressure tapStagnation pressure tap
0
2 21 1 2 2
1 22 2p V p Vz z
g gγ γ+ + = + +
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)
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
BernoulliNormal to the Streamlines
Separate acceleration due to gravity. Coordinate system may be in any orientation!
ˆp ga kρ ρ−∇ = +
nzpnn
dd
a gρ ρ∂− = +∂
Component of g in n direction
ks
n
BernoulliNormal to the Streamlines
np dza gn dn
ρ ρ∂− = +∂
R is local radius of curvature2
nVaR
=
p pdp ds dns n
∂ ∂= +∂ ∂
0 (s is constant normal to streamline)
dp dn p n∴ = ∂ ∂
n is toward the center of the radius of curvature
2dp V dzgdn R dn
ρ ρ− = +
Integrate F=ma Normal to the Streamlines
2dp V dzgdn R dn
ρ ρ− = + Multiply by dn
2
ndp V dn gdz C
Rρ+ + =⌠ ⌠⎮⎮⌡⌡ ∫ Integrate
2
np V dn gz C
Rρ+ + =⌠⎮⌡
(If density is constant)
2
"nVp dn gz CR
ρ ρ+ + =⌠⎮⌡
Pressure Change Across Streamlines
2
'1
np V dn z C
g Rγ+ + =⌠⎮
⌡
2
"nVp dn gz CR
ρ ρ+ + =⌠⎮⌡
If you cross streamlines that are straight and parallel, then ___________ and the pressure is ____________.
p gz Cρ+ =hydrostatic n r
1( )V r C r=21 "np C rdr gz Cρ ρ− + =∫ dn dr= −
221
"2 nCp r gz Cρ ρ− + =
As r decreases p ______________decreases
End of pipeline?End of pipeline?
What must be happening when a horizontal pipe discharges to the atmosphere?
2
"nVp dn gz CR
ρ ρ+ + =⌠⎮⌡
(assume straight streamlines)Try applying statics…
Streamlines must be curved!
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
Pressure datum____________
Crossing streamlines
Coordinate system
h
h=105 cm
gage pressure
Along streamline
Solution to Nozzle Flow
D1=30 cm
D2=10 cm
Q
h=105 cm
1
2
3
z2
'1
np V dn z C
g Rγ+ + =⌠⎮
⌡
1 21 2
p pz zγ γ+ = + 2ph
γ=
2
p"C2
p Vzgγ
+ + =
2 2 3 3Q V A V A= = 2 22
4QVdπ
=
Now along the streamline
223 32 2
2 32 2p Vp Vz z
g gγ γ+ + = + +
h
2232
2 2VVh
g g+ =
Two unknowns…_______________Mass conservation
Solution to Nozzle Flow (continued)
D1=30 cm
D2=10 cm
Q
h=105 cm
1
2
3
z2 2
2 4 2 42 3
8 8Q Qhg d g dπ π
+ =
22 4 4
3 2
8 1 1h Qg d dπ
⎡ ⎤= −⎢ ⎥
⎣ ⎦
2
4 43 2
1 18
hgQ
d d
π=
⎡ ⎤−⎢ ⎥
⎣ ⎦
Incorrect technique…
D1=30 cm
D2=10 cm
Q
h=105 cm
1
2
3
z2
'1
np V dn z C
g Rγ+ + =⌠⎮
⌡
2
p"C2
p Vzgγ
+ + =
223 31
1 31
2p Vp V dn z z
g R gγ γ+ + = + +⌠⎮
⌡
23
2Vh
g= p" 'C nC= The constants of integration
are not equal!
Bernoulli Equation Applications
Stagnation tubePitot tubeFree JetsOrificeVenturiSluice gateSharp-crested weir
Applicable to contractingstreamlines
(acceleratingflow).
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 Vzgγ
+ + =2
'1
np V dn z C
g Rγ+ + =⌠⎮
⌡
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 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.
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
pγ
z2
2V
g
Elevation datum
Are the 2 points on the same streamline?
2 5 mz = −
2 21 1 2 2
1 22 2p V p Vz z
g gγ γ+ + = + +
( )2 22V g z= −
( )2 22 2
2 224 4d dQ V g zπ π
= = −
What is the radius of curvature at the end of the pipe?
D1=30 cm
D2=10 cm
Q
h=105 cm
1
2
3
z
2
1
2
np V dn gz C
Rρ+ + =⌠⎮⌡
2
np V dn gz C
Rρ+ + =∫
dn dz= −
R1 2 0p p= =
Uniform velocityAssume R>>D2
2
np V z gz C
Rρ− + =
2
np Vz g C
Rρ⎛ ⎞
+ − =⎜ ⎟⎝ ⎠
2VRg
=2Vg
R=
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?