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

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 zg gγ γ

+ + = + +VAQ =

gV

gVpp

22

21

2221 −=−

γγ2211 AVAV =

44

22

2

21

1dVdV ππ

=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−=−

4

1

22

221 12 d

d

g

Vpp

γγ 222

211 dVdV =

21

22

21d

dVV =( )[ ]4

12

212

1

)(2

dd

ppgV−

−=

γ

( )[ ]412

212

1

)(2

dd

ppgACQ v−

−=

γ