Bernoulli Equation

Elementary Fluid Dynamics – The Bernoulli Equation

• Consider inviscid, steady, two-dimensional flow in x-z plane

• Define streamlines

• Select coordinate systems based on streamlines

• Define acceleration

• Define forces

• Apply Newton’s second law of motion along and across streamline

Newton’s Second Law: F = ma

• Prior to apply Newton’s second law of motion to fluid particle:

• Consider motion of an inviscid fluid

• Assume that fluid motion is governed by pressure and gravity forces

• Select an appropriate coordinate system. Consider two dimensional motion (x-z plane)

Newton’s Second Law: F = ma

• Prior to apply Newton’s second law of motion to fluid particle:

• Consider motion of an inviscid fluid

• Assume that fluid motion is governed by pressure and gravity forces

• Select an appropriate coordinate system. Consider two dimensional motion (x-z plane)

Coordinate System

• Motion of a fluid particle is described by its velocity vector

Coordinate System


• As the particle moves, it follows a particular path, the shape of which is governed by velocity vector

Coordinate System


• As the particle moves, it follows a particular path, the shape of which is governed by velocity vector

• If flow is steady, each successive particle that passes through given point (1) will follow the same path. For such cases the path is a fixed line in the x-z plane. The entire x-z plane is filled with such paths.

Coordinate System

• For steady flow each particle slides along its path and its velocity vector is everywhere tangent to the path

Coordinate System


• The lines that are tangent to the velocity vectors throughout the flow field are called streamlines.

Streamlines


• The lines that are tangent to the velocity vectors throughout the flow field are called streamlines.

• We will use coordinates based on streamlines

Streamlines

• Particle motion is described in terms of its distance, s = s(t), along streamline, and local radius of curvature

Particle Motion

• Particle motion is described in terms of its distance, s = s(t), along streamline, and local radius of curvature

• Distance s is related to particle’s speed V = ds/dt, and radius of curvature is related to the shape of streamline

Particle Motion

• Acceleration:

Particle Acceleration

d dta V

• Acceleration:

• Components of acceleration in s and n directions:

Particle Acceleration

2

, s n

V Va V a

s

d dta V

• To determine forces consider free-body diagram of small fluid particle

Forces

F = ma along a Streamline

Free-body diagram of a fluid particle


Equation of motion along streamline (details)

sin s

p VV a

s s


Change in fluid particle speed is accomplished by combination of pressure gradient and particle weight along streamline

Example 3.1 Consider the inviscid, incompressible, steady flow along the horizontal streamline A–B in front of the sphere of radius a. From a more advanced theory of flow past a sphere, the fluid velocity along this streamline is

Determine the pressure variation along the streamline from point A far in front of the

sphere (xA = – and VA = V0) to point B on the sphere (xB = – a and VB = 0).

3

0 31

aV V

x




Solution Streamline is horizontal, then

Acceleration

3

0 31

aV V

x

p VV

s s

3 32

0 3 43 1

V V a aV V V

s x x x

3 2 3 30

4

3 1a V a xp

x x

632

0 2

a xap V

x




Solution Pressure gradient

Pressure distribution

3

0 31

aV V

x

20

2B

Vp

Bernoulli Equation

For incompressible fluid equation of motion along streamline reduces to Bernoulli equation (details)

Restricted to:

- inviscid flow

- steady flow

- incompressible flow

- along streamline

21constant along streamline

2p V z

Example 3.2 Consider the flow of air around a bicyclist moving through still air with velocity V0. Determine the difference in the pressure between points (1) and (2).

Example 3.2 Consider the flow of air around a bicyclist moving through still air with velocity V0. Determine the difference in the pressure between points (1) and (2).

Solution Apply Bernoulli equation between (1) and (2)

Pressure difference

2 21 1 1 2 2 2

1 1

2 2p V z p V z

2 22 1 1 0

1 1

2 2p p V V

F = ma Normal to a Streamline

Equation of motion along the normal direction (details)


Change in the direction of flow of a fluid particle is accomplished by combination of pressure gradient and particle weight normal to streamline

2dz p V

dn n

Example 3.3 Shown in Fig. a, b are two flow fields with circular streamlines. The velocity distributions are

where C1 and C2 are constant. Determine the pressure distributions, p = p(r), for each, given that p = p0 at r = r0.

1

2

for case ( )

for case ( )

V r C r a

CV r b

r



1

2

for case ( )

for case ( )

V r C r a

CV r b

r

Solution Assume steady, inviscid, and incompressible flow with streamlines in horizontal plane (dz/dn = 0)Since streamlines are circles, /n = - /r and = rThen equation of motion along the normal

becomes

2dz p V

dn n

2p V

r r



1

2

for case ( )

for case ( )

V r C r a

CV r b

r

Solution

For case (a)

For case (b)

Comments: (a) – free vortex, (b) forced vortex

2 2 2 21 1 0 0

1 and

2

pC r p C r r p

r

2222 03 2 2

0

1 1 1 and

2

p Cp C p

r r r r


For steady, inviscid, incompressible flow (details)

Restricted to:

- inviscid flow

- steady flow

- incompressible flow

- across streamline

Pressure variation across straight streamlines is hydrostatic

2

constant across streamlineV

p dn z

Physical Interpretation

Work done on a particle by all forces acting on the particle is equal to the change of the kinetic energy of the particle

Each term of Bernoulli equation can be interpreted as:

- head (elevation, pressure, velocity)

- form of pressure (static, hydrostatic, dynamic)


2p V z

Example 3.4

Pressure variation across straight streamlines is hydrostatic

Static, Stagnation, Dynamic, and Total Pressure

Useful concept associated with the Bernoulli equation deals with the stagnation and dynamic pressures.

As fluid is brought to rest its kinetic energy is converted to a pressure rise


Each term in Bernoulli equation can be interpreted as a form of pressure; static, p, hydrostatic, z, and dynamic, V 2/2 ,


2p V z

22 1 1

1

2p p V

Point (2) is a stagnation point

Pressure at the stagnation point is greater than static pressure by dynamic pressure


• There is a stagnation point on any stationary body that is placed onto a flowing fluid

• Some of the fluid flows “over” and some “under” the object. Dividing line is termed the stagnation streamline and terminates at the stagnation point on the body

• Location of the stagnation point is function of body shape.


• If elevation effect are neglected, stagnation pressure, p + V2/2, is the largest pressure obtainable along a given streamline. It represents the conversion of all of the kinetic energy into a pressure rise.

• Sum of the static pressure, hydrostatic pressure, and dynamic pressure is termed the total pressure, pT

• Bernoulli equation is a statement that total pressure remains constant along a streamline.


2 Tp V z p

Fluid Velocity Measurement

3 42 p pV

Pitot-static tubes measure fluid velocity by converting velocity into pressure

Pitot-static tube

Typical Pitot-Static Tube Designs

Measurement of Static Pressure

Incorrect and correct design of static pressure taps

Measurement of Static Pressure

Typical pressure distribution along a Pitot-static tube

Fluid Velocity Measurement

Many velocity measurement devices use Pitot-static tube principle

Cross section of a directional-finding Pitot-static tube

o

1 3

2

2

2 1

If 0 and 29.5

2

2

p p p

Vp p

p pV

Examples of Use of Bernoulli Equation

Free Jets

Exit pressure for an incompressible fluid jet is equal to the surrounding pressure

Velocity:

Vertical flow from a tank

2V gh

Free Jets

For horizontal nozzle velocity is not uniform

If d « h centerline velocity can be used as an “average velocity”

Horizontal flow from a tank

Free Jets

If exit is not smooth, diameter of the jet will be less than diameter of the hole.

This phenomena, called a vena contracta effect, is a result of the inability of the fluid to turn the sharp 90º corner indicated by dotted lines in the figure

Vena contracta effect for a sharp-edged orifice

Free Jets

Since streamlines in the exit plane are curved, the pressure across them is not constant.

The highest pressure occurs along the centerline at (2), and lowest pressure, p1 = p3 = 0


Free Jets

Assumption of uniform velocity with straight streamlines and constant pressure is not valid at the exit plane

It is valid in the plane of vena contracta, sectшon a-a, provided dj « h


Free Jets

Vena contracta effect is a function of the geometry of the ounlet.

Contraction coefficient:

Typical flow patterns and contraction coefficients for various round exit configurations

c j hC A A

Confined Flows

In nozzles and pipes of variable diameter velocity changes from one section to another

For such cases continuity equation must be used along with Bernoulli equation

Continuity equation states that mass cannot be created or destroyed

For incompressible fluid (details)

1 1 2 2 1 2 or AV A V Q Q

Example 3.7 A stream of water of diameter d = 0.1 m flows steadily from a tank of diameter D = 1.0 m. Determine the flowrate, Q , needed from the inflow pipe if the water depth remains constant, h = 2.0 m


SolutionAssume steady, inviscid, incompressible flow.Apply Bernoulli equation between points (1) and (2)

With p1 = p2 = 0, z1 = h and z2 = 0

From continuity equation

Exit velocity

and volume flowrate

2 21 1 1 2 2 2

1 1

2 2p V z p V z

2 21 2

1 1

2 2V gh V

2

1 2

dV V

D

2 4

26.26 m/s

1

ghV

d D

31 1 2 2 0.0492 m /sQ AV A V


SolutionIf D » d, then we can assume V1 ≈ 0. Error associated with this assumption:

4

2

40 2

2 1 1

2 1D

gh d DQ V

Q V gh d D

Example 3.8

Example 3.8

Answers: 13 2

3 22

269.0 m/s =7.67 m/s

0.00542 m / s 2963 N/m

pV V

Q p

Comments: V3 is determined strictly by the value of p1

In absence of viscous effect pressure throughout the hose is constant and equals to p2

Decrease in pressure from p1 to p3 accelerate the air and increase its kinetic energy

Pressure change (density change) is within 3%. Hence, incompressibility assumption is reasonable

Example 3.9

Answer:

Comments:For a given flowrate h does not depend on , but pressure difference, p1 – p2, as measured by pressure gage, does

2 2

2 1

2

1

2 1

A AQh

A g SG

Cavitation

Cavitation occurs when the pressure is reduced to the vapor pressure

Cavitation can cause damage to equipment

Pressure variation and cavitation in a variable area pipe

Cavitation

Tip cavitation from a propeller

Example 3.10

Answer:

Comments: Results are independent of diameter and length of the hose (provided viscous effects are not important

Proper design of hose is needed to ensure that it will not collapse due to the large pressure difference (vacuum) between the inside and outsides of the hose

28.2 ftH

Flowrate Measurement

Typical devices for measuring flowrate

Various flow meters are governed by the Bernoulli and continuity equations

We consider “ideal” flow meters – those devoid of viscous, compressibility, and other effects.

The flowrate is a function of the pressure difference across the flow meter

1 2

2 2

2 1

2

1

p pQ A

A A

Example 3.11

Answer:

Comments: These values represent “ideal” results, and these results are independent of flow meter geometry – an orifice, nozzle, or Venturi meter.

Tenfold increase in flowrate requires one-hundredfold increase in pressure difference. This nonlinear relationship can cause difficulties when measuring flowrates over a wide range of values. An alternative is to use two flow meters in parallel

1 21.16 kPa 116 kPap p

Flowrate Measurement. Sluice Gate

Sluice gate geometry

The flowrate under a sluice gate depends on the water depths on either side of the gate

1 22 2

2 1

2

1

g z zQ z b

z z

In the limit of z1»z2

A vena contracta occurs as water flows under a sluice gate

2 12Q z b gz

Flowrate Measurement. Sharp-crested Weir

Rectangular, sharp-crested weir geometry

Flowrate over a weir is a function of the head on the weir

3 21 12 2Q C Hb gH C b g H

Energy Line and Hydraulic Grade Line

Hydraulic grade line and energy line are graphical forms of the Bernoulli equation

Energy line represents the total head available to the fluid

Locus provided by a series of piezometric taps is termed the hydraulic grade line

Representation of the energy line and the hydraulic grade line


If the flow is steady, incompressible, and inviscid, the energy line is horizontal and at the elevation of the liquid in the tank.

Hydraulic grade line lies a distance of one velocity head below the energy line

At the pipe outlet the pressure head is zero (gage) so the pipe elevation and hydraulic grade line coincide

Energy line and hydraulic grade line for flow from a tank


The distance from pipe to hydraulic grade line indicates the pressure within the pipe

For flow below the hydraulic grade line, the pressure is positive

For flow above the hydraulic grade line, the pressure is negative

Use of the energy line and hydraulic grade line

Restriction on Use of the Bernoulli Equation

Restrictions on use for the Bernoulli equation are imposed by the assumptions used in its derivation.

To avoid incorrect use of Bernoulli equation one must take into account:

- Compressibility effects;

- Unsteady effects;

- Rotational effects;

- Viscosity effects;

- Presence of mechanical devices (pumps, turbines)

““Change of scene, and absence of the necessity for Change of scene, and absence of the necessity for thought, will restore the mental equilibrium”thought, will restore the mental equilibrium”

(Jerom K. Jerom, “Three Men In a Boat”)(Jerom K. Jerom, “Three Men In a Boat”)

END OF CHAPTER

Supplementary slides


s s

V VF ma mV V V

s s

Newton’s second law along streamline


sin sinsW W V Gravity force


ps s s

pF p p n y p p n y V

s

Pressure force


Net force sins s ps

pF W F V

s

back

back

Bernoulli Equation

Consider equation

Along streamline

Also

Finally, along streamline value of n is constant (dn = 0) so that

Hence, along streamline p/s = dp/ds . Then equation (a) becomes

Integration at constant density gives Bernoulli equation

sin (a)p V

Vs s

sindz

ds

21

2

d VVV

s ds

p p pdp ds dn ds

s n s

210 (along streamline)

2dp d V dz


Newton second law in normal direction2 2

n

mV V VF


Gravity force cos cosnW W V


Pressure force pn n n

pF p p s y p p s y V

n


Net force cosn n pn

pF W F V

n

back

Continuity EquationConsider a fluid flowing through a fixed volume. If the flow is steady, rate at which fluid flows into the volume must equal the rate at which it flows out of the volume (mass is conserved)

Mass flow rate is given by

Volume flow rate

Conservation of mass requires

If density remains constant

m QQ VA

1 1 1 2 2 2AV A V

1 1 2 2AV A V

back

Compressibility Effects

Bernoulli equation can be modified for compressible flows.

For compressible, inviscid, isothermal, steady flows:

Use of above equation is restricted by inviscid flow assumptions, since most isothermal flows are accompanied by viscous effects.

For compressible, isentropic (no friction or heat transfer), steady flow of a perfect gas:

2 21 1 2

1 22

ln2 2

V RT p Vz z

g g p g

2 21 1 2 2

1 21 21 2 1 2

k p V k p Vgz gz

k k

Compressibility EffectsBernoulli equation for compressible flow can be written for pressure ratio as

Where Ma = V/c is the Mach number; c is local speed of sound

122 11

1

11 Ma 1

2

k

kp p k

p

back

Pressure ratio as a function of Mach number for incompressible and compressible (isentropic) flow

A “rule of thumb” is that the flow of a perfect gas may be considered as incompressible provided the Mach number is less than about 0.3

Unsteady Effects

Bernoulli equation can be modified for unsteady flows.

For incompressible, inviscid, unsteady flows:

Use of this equation requires knowledge of variation of V/t along the streamline

2

1

2 21 2

1 1 2 22 2

s

s

V V Vp z ds p z

t

back

Example 3.12

Answers:

Comments: ?

24.83 m /sQ

b

Example 3.13

Answers:

Comments: ?

2 5 22 2tan 2 tan 2

2 2Q AV H C gh C ghH

0

0

5 2

3 2 05 2

2 0

tan 2 2 3

tan 2 2

H

H

Q C g H

Q C g H

Example 3.14

Answers:

Comments: ?

Example 3.16

Answers:

Comments: ?

2g

l

Example 3.17

Answers:

Comments: ?

21

2 518 kPa2

Vp h

Example 3.18

Bernoulli Equation

Documents

Transcript of Bernoulli Equation