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 18, 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