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