Bernouli's Equation

8/2/2019 Bernouli's Equation

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Bernoulli equation

Daniel Bernoulli (1710-1782)


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Bernoulli: preamble

Want to discuss the properties of a moving uid.Will do this initially under the simplest possibleconditions, leading to Bernoullis equation. Thefollowing restrictions apply.

Flow is inviscid, there are no viscous drag forces Heat conduction is not possible for an inviscid

ow

The uid is incompressible . The ow is steady (velocity pattern constant). The paths traveled by small sections of the uid

are well dened.

Will be implicitly using the Euler equations of motion (discussed later)


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Coordinates and streamlines

Each piece of uid has velocity v . Steady ow, nothing changes with time at given

location. All particles passing through (1) end

up at (2) with velocity v

The trajectories followed by the particles arecalled streamlines.

Describe motion is terms of distance traveledalong streamline, s .

Velocity given by |v | = dsdt . Normal to velocityis n .

Stream-line can bend, R is radius of curvature.


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Coordinates and streamlines

Using body xed coordinates. If the particles changespeed along stream-line, or if stream-line bends, thenaccelerations must be present.The tangential acceleration

a s =dvdt

=vs

dsdt

= vvs

The normal acceleration

an =v2

R

The radius of curvature R changes along thestreamline.


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Streamline coordinates

It is convenient to use a coordinate system dened in

terms of the ow streamlines. The coordinate alongthe streamline is s and the coordinate normal to thestreamline is n . The unit vectors for the streamlinecoordinates are s and n .

The direction of s will be chosen to be in the samedirection as the velocity. So v = vs .

s

n^

s^

V

s = 0

s = s 1

s = s 2n = n 2

n = n 1

n = 0

Streamlines

y

x

The ow plane is covered with an orthogonal curvednet of coordinate lines and v = v(s, n ) s ands = s (s, n ) for steady ow.


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Forces on streamlines

Any particle travelling along the streamline will besubjected to a number of forces.

The relevant Forces for Bernoullis equation are

gravity and pressure.


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Streamlines F = ma

Will resolve forces in directions parallel s andperpendicular n to particles motions. y is out of

page, z is down, x is horizontal.

F s = m a s = m vvs

= V vvs


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Streamlines F = ma

Resolve forces

W s = W sin = V sin

W s would be zero for horizontal motion.

The pressure changes with height. Let p be pressurein middle of uid slab. Let p + ps be pressure infront of slab and p ps be pressure behind slab.From Taylors series

ps =ps

s2

The net pressure forceF ps = ( p ps )ny ( p + ps )ny

= 2ps ny = ps

sny = ps

V

Net forceF s = W s + F ps

= sin ps

V


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Bernoulli equation

Equate two expressions for F s

F s = sin ps

V = V vvs

sin ps

= vvs

The change in uid particle speed along a streamlineis accomplished by a combination of pressure andgravity forces.

Now use sin =dz

ds

And v dvds =12

dv 2

ds

And dp = ps ds + pn dn

Along streamline dn = 0

dzds

dpds

=12

dv2

ds


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Bernoulli, compact expression

Now making the assumption that density is constant,Bernoullis equation is obtained

dzds

+dpds

+12

dv2

ds= 0

dds

z + p + 12

v2 = 0

z + p +12

v2 = Constant

The constant density assumption (incompressibleow) is good for liquids (sometimes gases at lowspeed). Bernoullis equation presented in 1738monograph Hydrodynamics by Daniel Bernoulli.

If one has compressible uid

dp + 12v2 + gz = Constantand knowledge of how varies with p .


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Forces normal to streamline

The acceleration normal to the streamline is an = v2

R

where R is the local radius of curvature of thestreamline.

F n =(m)v2

R=

V v2

R


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A change in stream direction occurs from pressureand/or and gravity forces. Resolve forces

W n = W cos = V cos

W n would be zero for vertical motion.

The pressure changes with height. Let p be pressurein middle of uid slab, p + pn is pressure at top of slab and p pn be pressure at bottom of slab.From Taylors series

pn =pn

n2

The net pressure force, F pn

F pn = ( p

pn )s y

( p + pn )s y

= 2pn n y = pn

n s y = pn

V

Need to combine pressure and weight forces to getnet Force


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Combine weight and pressure forces

F n = W n + F pn

= cos pn

V =V v2

R

Pressure and weight forces imbalance produces thecurvature. For gas ows it is common to use

pn

= v2

R

The pressures increaseswith distance away fromthe center of curvature( pn is negative sincev2 /R is positive).

For straight parallel streamlines (in gases), pn = 0 .No pressure change across streamlines


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Will consider uid parameters normal to stream line

cos + pn

+ v2

R= 0

pn = dpdn since s is constant.

cos = dzdnand so for incompressible ows

dpdn

+ dzdn

+v2

R= 0

dpdn

+ dzdn

+ v2

R= 0

ddn

( p + z ) +v2

R= 0

p + z +

v2

Rdn = Constant

For a compressible substance, the best reduction is

dp + v2

R+ gz = Constant


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Interpretation for incompressible ows

Along the streamline

z + p +12

v2 = Constant

Across the streamline

p + z + v2

Rdn = Constant

The units of Bernoullis equations are J m 3 . This

is not surprising since both equations arose from anintegration of the equation of motion for the forcealong the s and n directions.

The Bernoulli equation along the stream-line is a

statement of the work energy theorem. As theparticle moves, the pressure and gravitational forcescan do work, resulting in a change in the kineticenergy.


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Dynamic and static pressures

p +12

v2 + gz = constant

Static pressure is the pressure as measured movingwith the uid. (e.g. static with uid). This is the pterm in Bernoullis equation. Imagine moving alongthe uid with a pressure gauge.

Some times the gz term in Bernoullis equation iscalled the hydrostatic pressure. (e.g. it is the changein pressure due to change in elevation.)

Dynamic pressure is a pressure that occurs when

kinetic energy of the owing uid is converted intopressure rise. This is the pressure associated withthe 12 v

2 term in Bernoullis equation.


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Dynamic and static pressuresThe static pressure at 1 can be estimated by theheight of the column.

p1 = h 3 1 + p3

= h 3 1 + h 4 3

= h

(1) (2)

(3)

(4)

h 3-1

h h 4-3

Open

H

V

V 1

= V V 2

= 0

The dynamic pressure at 2 is estimated by

p2 = p1 +12

v21 (v2 = 0)

The additional pressure due to the dynamic pressurewill cause the uid to rise a height of H > h .The point (2) is called a stagnation point.


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The stagnation point

Stagnation point

(a )

Stagnation streamline

Stagnation point

(b )

When uid ows around any stationary body, someof the streamlines pass over and some pass under theobject. But there is always a stagnation point wherethe stagnation streamline terminates. The

stagnation pressure is pstagnation = p +

12

v2

v is velocity at some point on stream-line away fromobstruction.The total pressure, pT

pT = p +12

v2 + z

is sum of static, dynamic and hydrostatic pressures.


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The pitot tube

Knowledge of static and stagnation pressures makes

it possible to determine the uid velocity. Geometryarranged so that elevation differences have littleimpact. The free stream pressure is p .

Pressure measured at

points 3 and 4Stagnation pressure

p2 = p3 = p +12

v2

V

p

(1)

(2)

(4)

(3)

Static pressure is just p1 = p p4 . Combiningequations

p = p4 = p3 12

v2

Rearranging leads to

v = 2( p3 p4 )


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The pitot tube: complications

The main question is de-sign of pitot tubes is whereto place the orice to mea-

sure the static pressure.

V American Blower company

National Physical laboratory (England)

American Society of Heating & Ventilating Engineers

The static pressure doesvary along the length of the tube. More com-

plicated analysis thanBernoulli required here.

Also make mouth of tube smooth.


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The free jet

The free jet result was rst obtained in 1643 by

Evangelista Torricelli.

p1 +12

v21 + z 1 = p2 +12

v22 + z 2

z 1 = 12

v22

p1 = patm = 0 ; gauge pressure

v1 0 ; large surface, so v1


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The free jet

z 1 =12

v22

v2 = 2h

= 2ghOutside nozzle, stream continues to fall and at (5)

v2 = 2ghv5 = 2g(h + H )Result v = 2gh is speed of freely falling bodystarting from rest.For the uid, all the potential energy is converted tokinetic energy when jet leaves tank. (assume noviscous forces).


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Free jet, ne details

Horizontal nozzle, veloc-ity at center line v2 isslightly smaller than v3and slightly larger thanv1 .

For d h , OK to usev2 as average velocity.

Streamlines cannot follow sharp corner exactly.Would take an innite pressure gradient to achievezero radius of curvature(i.e. R = 0 ). Uniform velocity only occurs at a-a line.

Vena Contracta effect.Jet diameter, d j isslightly smaller thanhole diameter dh .


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Free jet, Vena contracta effect

The contraction coefficient, C d = A j /A h is the ratioof the jet area A j , and hole area Ah .


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Flow rates

How much water ows down a channel or through apipe?

The volume ow rate,Q1 is dened as the vol-ume of uid that owspast an imaginary (orreal) interface.

Volume of uid leaving V = v1 tA 1

Rate of volume changeV t

= v1 A1

The volume ow rate Q = v1 A1 Mass of uid leaving m = v1 tA 1

Rate uid leaving m =dmdt

= v1 A1


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Equation of continuity

For a steady state situation, the mass of uid goinginto the tank must be the same as the mass of uidleaving the tank.

Mass of water in = Mass of water out

1 A1 v1 = 2 A2 v2

This is the continuity equation and forincompressible ow

A1 v1 = A2 v2 or Q1 = Q2

The equation of continuity and the Bernoullisequation are used into conjunction to analyze manyow situations.


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Flow rate: Example 1

Given the water velocityat (2) is 8.0 m/s and

the pipe diameter is 0.10m , what are the volumeand mass ow rates?

Q = vA = v(d/ 2)2

= 8 .0 0.0502 = 0 .06283 m3 /s

The mass ow is just Q sodm

dt= 1000

0.06283 = 62 .83 kg/s


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Flow rate: Example 2A stream of water d =0.10 m ows steadilyfrom a tank of diameterD = 1 .0 m as shownin the gure. What ow-

rate is needed from theinlet to maintain a con-stant water volume in theheader tank depth? Thedepth of water at the out-let is 2.0 m .

Can regard outlet as a free jet (note water level at(1) is not going down).

v2 =

2gh = 2

9.8

2.0 = 6 .26 m/s

Q2 = A2 v2 = (0.050)2 6.26 = 0 .0492 m3 /s

= Q1


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Flow rate measurement

One way to measure ow-rate is to place a

constriction in a pipe. The resulting change invelocity (continuity equation), leads to a pressuredifference. The absolute uid velocity can bedetermined from pressure difference between (1) and(2) .

The Orice, Nozzle and Venturi meters analysis hereignores viscous, compressibility and other real-world

effects.


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Flow rate measurement: 2

Want to determine ow rate, need v2

v1 A1 = v2 A2

v1 =A2A1

v2

p2 +12

v22 = p1 +12

v21

12

v22 12

v21 = p1 p2

12v

2

2 12

A22A21 v

2

2 = p1 p2

v22 =

2( p1 p2 ) 1

A 22A 21

So the ow rate is

Q = A2 2( p2 p1 ) 1 A 22A 21The pressure differences give the ow rate. Real

world ows are 1% to 40% smaller.


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Bernoulli and Cavitation

The temperature at which water boils depends onpressure.

T (oC) pvap (kPa)

10 1.23

20 2.34

30 4.24

40 7.34

Q

p

(Absolutepressure)

(1) (2) (3)

Small Q

Moderate Q

Large Q Incipient cavitation

pv

0 x

The process of cavitation involves

Fluid velocity increases

Pressure reduction

If p < p vap , water boils Bubbles collapse when reach high pressure part

of uid


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Bernoulli and Cavitation

Pressure transients exceeding 100 MPa can be

produced. These transients can produce structuraldamage to surfaces.


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Sluice gate ow rate

The height of water in the channel can be used todetermine the ow rate of water out the reservoir.

Q = z2 b 2g(z1 z2 )1

z 22

z21

z2 b

2gz1

(b) is the width of the reservoir.


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Venturi meter problem

Determine owrate as a functionof the diameter of the tube.

0.2 m

Q0.1 m D

Use venturi meter equation

Q = A2 2( p2 p1 ) 1 A 22A 21 A2 = D

2 / 4 m2

p2 p1 = 0.20 = 98000.20 = 1960 Pa = 1000 kgm

3

1 A22 /A 21 = 1 D 2 / 0.102 = 1 100D 2 m2

Q = D 2

4 39201000(1 100D 2 ) = D 2 0.245

(100D 2 1)


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Sharp crested weir

Between (1) and (2) pressure and gravitationalforces cause uid to accelerate from v1 v2 .

p1 = h and p2 0 forces cause uid toaccelerate from v1 v2 .

Assume ow is like free jet. Average velocityacross weir is C 1 2gH , C 1 = constant.

Flow rate is

Q = ( Hb) C 2gH = C 1 b 2gH 3The parameter C 1 is determined empirically.


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The energy line and hydraulic grade line

Consider Bernoulli equation divided by = g

p

+ 12

v2g

+ z = H = Constant on streamline

The dimensions of the equation are in length. Thereis the pressure head, velocity head, and elevation

head. The sum, H is called the total head.The energy line gives the total head available to auid. It can be measured by measuring thestagnation pressure with a pitot tube.The Hydraulic grade line is the line produced from

the pressure and elevation heads. It is measured witha static pitot tube.


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The energy line and hydraulic grade line

The energy line will be horizontal along thestream line as long as Bernoulli assumptions arevalid.

The hydraulic grade line will not be horizontal if the uid velocity changes along the stream line.

If forces are present (this does occur in pipeows), then there will be a loss in energy andthe energy line will not be constant.


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Example: EL and HGL

A scale drawing can be used to depict the pressure inthe tank/pipe system.

The energy line is horizontal The elevation head at (2) is converted into

increased pressure head p2 / and velocity headv22 / (2g) . The HGL decreases.

At (3) , pressure is atmospheric. So the HGL tothe level of the pipe and the elevation head hasbeen converted entirely into a velocity headv23 / (2g) .


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EL and HGLThe EL and HGL can depict whether there ispositive pressure p > p atm or negative pressure p < patm .

The water velocity will be constant in curvedpipe (equation of continuity).

The pressure head will increase or decrease asthe elevation head changes. Useful to know forleaking pipes.


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Limitations on Bernoulli equation

A number of problems can invalidate the use of theBernoulli equation, these are compressibility effects ,rotational effects , unsteady effects .

Compressibility effects

When can compressibility effects impact on gasows? Consider stagnation point

Stagnation pressure is greater than staticpressure by 12 v

2 (dynamic pressure), provided

constant.

will not changes too much as long as dynamicpressure is not too large when compared tostatic pressure.

So ows at low v will be incompressible But dynamic pressure increases as v

2 , socompressibility effects most likely at high speed.


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Compressibility effects

The isothermal model for an ideal gas, p = R T

C = dp + 12v21 + gz1

C = RT dp p + 12v21 + gz1C = RT ln p1 +

1

2v21 + gz1

This can be used to get

v212g

+ z1 +RT g

ln( p1 /p 2 ) =v222g

+ z2

Now write as p1 /p 2 = 1 + ( p1 p2 )/p 2 = 1 + p/p 2and use ln(1 + x) = x for x1 .v212g

+ z1 +RT g

ln(1 + p/p 2 ) =v222g

+ z2

v21

2g+ z1 + RT

g( p/p 2 ) v

22

2g+ z2

This can be reduced to the standard Bernoulliequation. Bernoulli recovered as long as pressuredifferences are not large.


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Compressibility effects: Isentropic ow

This is the situation that applies when there is no

heat transfer or friction during the ow (reasonablefor many gases). This gas law is p = k D where kdepends on specic heat capacities. Introduce theMach number, Ma = v/c (ratio of the ow speed tothe speed of sound). Consideration of the pressureratio between free stream and stagnation pointsleads to

p2 p1 p1

=kMa 21

2incompressible

p2 p1 p1

= 1 +k 1Ma 21

+k

k 1

1 compressible

The compressible andincompressible expres-sions agree to 2% forMa < 0.3 .

Compressible(Eq. 3.25)

Incompressible(Eq. 3.26)

k = 1.4

0.3

0.2

0.1

00 0.2 0.4 0.6 0.8

Ma1

p 2

p

1

______

p 1


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Unsteady effectsImplicit in the discussion was an assumption thatthe uid ows along steady state streamlines, sov = v(s) is a function of position along the stream

and does not contain any explicit time dependence.If v = v(s, t ) then then it would be necessary toinclude this when integrating along the streamline.

p1 +1

2v21 + z 1 = p2 +

1

2v22 + z 2 +

t 2

t 1

p

sds

The additional term does complicate matters andcan only be easily handled under restrictedcircumstances. There are quasi-steady ows wheresome time dependence exists, but Bernoullisequations could be applied as if the ow were steady(e.g. the draining of a tank).


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Rotational effects

Model of wake behind insect

Y. D. Afanasyev, Memorial University of Newfoundland

Bernoulli equation describes motion of uid particlesalong streamline. If particles spin about the

streamline then Bernoulli is no longer valid.Need to characterize irrotational and rotational ows.

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Bernouli's Equation

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