CE C212: By Dr. Ajit Pratap Singh 1 - Civil Engineering...

CE C212: By Dr. Ajit Pratap Singh 1

APPLICATIONS OF BERNOULLI’S EQUATION

Fluid flow measurement

By

Prof. A.P. Singh

Civil Engineering Department

BITS-Pilani

Applications: Flow Measurement

The Bernoulli equation can be applied to several commonly occurring

situations in which useful relations involving pressures, velocities and

elevations may be obtained.

A very important application in engineering is fluid flow measurement

� Measurement of flow rate: Orifices and Mouthpieces

Weirs and Notches

Venturi meter

Orifice meter

� Measurement of velocity : Pitot-static tube

Flow Through Orifices and Mouthpieces

Orifice

It is an opening having a closed perimeter, made in walls or bottom of a tank or a vessel containing fluid through which the fluid may be discharged.

Both orifices and mouthpieces are usually used for measuring the rate of flow of fluid.

MouthpieceIt is a short tube of length not more than two to three times its diameter, provided in a tank or a vessel containing fluid such that it is an extension of the orifice and through which also the fluid may be discharged.

Classification

�Small orifice• Large Orifice• Circular• Rectangular• Square• Triangular• Sharp edged orifice (Bevelled side facing the d/s)

• Rounded or bell mouthed orifices• Orifices discharging free• Drowned or submerged orifices

– Fully submerged– Partially submerged

Mouthpiece

• Cylindrical

• Convergent

• Divergent

• Convergent-divergent

• External mouthpieces

• Internal mouthpieces

• Running full mouthpieces

• Running free mouthpieces

• According to the shape of the upstream edge the orifices may be classifiedas sharp edged orifices and bell mouthed orifices or orifices with the round

corner.

Fig.(a) Sharp-edged orifice discharging free (b) Bell-mouth orifice


• As shown in previous Fig. a sharp edged orifice has the bevelled side facingthe downstream so that there is a minimum contact with the fluid flowingthrough the orifices and consequently the minimum frictional effects.

• According to discharge condition, the orifices may be classified as orificesdischarging free and drowned or submerged orifices.

Flow Through Small Orifice

• Figure shows a sharp edged small orifice in one

side of a reservoir containing liquid.

• The liquid will emerge from the orifice as a free jet,

that is, a jet discharged in the atmosphere and will

therefore be under the influence of gravity only.

• The area of the orifice is the area of the opening. The portion of the flow that approaches along the wall cannot make a right-angled turn at the opening and therefore maintains a radial velocity component that reduces the jet area. Since an abrupt change of direction of motion is impossible, the streamlines continue to converge beyond the orifice up to a certain distance until they become parallel at the section CC

• The cross section where the contraction is greatest is called the vena contracta. At Vena contrcta i.e. at the section CC of jet, the streamlines are straight and parallel to each other and perpendicular to the plane of orifice, and jet has the minimum x-sectional area.

• The streamlines are parallel throughout the jet at this section, and the pressure is atmospheric.

Figure: Figure: Orifice in a reservoir.

• The head H on the orifice is measured from the center of the orifice to the free surface. The head is assumed to be held constant.

• Bernoulli's equation applied from a point 1 on the free surface to the center of the vena contracta, point 2, with local atmospheric pressure as datum and point 2 as elevation datum, neglecting losses, is written

• Inserting the values gives

• The ratio of the actual velocity to the theoretical velocity is called the velocity coefficient; that is,

and hence

This Equation is known as

Torricelli's theorem and represents

theoretical velocity or ideal

velocity of the jet

• The ratio of jet area at vena contracta to area orifice is symbolized by another coefficient called the coefficient of contraction:

• The area at the vena contracta is CcA0. The actual discharge is thus

• It is customary to combine the two coefficients into a discharge coefficient Cd,

• From which

• There is no way to compute the losses between points 1 and 2; hence, must be determined experimentally. It varies from 0.95 to 0.99 for the square-edged or rounded orifice.


gHCv vactual 2=

The term is called a velocity coefficient. Then

The above equation gives a velocity of jet at vena-

contracta which always more than the average

velocity of jet at a section in the plane of the orifice

itself

Cv varies from 0.95 to 0.99 for different types of

orifices depending upon the shape and size of the

orifice

Sharp-Edged Orifice Discharging Free

• Sharp-edged small orifice in one side of a reservor

• Vena-Contacta: Min x-sectional area and maximum velocity of flow

• Torricelli’s formula for Velocity of formula

• Coefficient of velocity (0.95 to 0.99)

• Coefficient of contraction (π/(π+2) = 0.611) 0.61 to 0.69 or 0.64

• Coefficient of Discharge (0.61 to 0.65)

Applications: Flow Measurement Chee 223

•Attention to be drawn to the fact, that velocity does not depend

on dimensions of the orifice. They influence slightly to velocity

coefficient Cv only. Flow rate may be expressed as

Q=Acv

•Cross section of the flow at vena contracts Ac may be expressed

Ac = CcA.

• Actual Flow rate

gHACCQ vcActual 2=

Let us denote CcCv= Cd

name it flow rate coefficient. Now flow rate obtains expression

TheoreticdActualdActual QCQgHACQ =⇒= 2

•• Trajectory methodTrajectory method..

• By measuring the position of a point on the trajectory of the free jet downstream from the vena contracta (earlier Fig) experimentally, the actual velocity Va can be determined if air resistance is neglected. The x component of velocity does not change; therefore, Vat = x0, in which t is the time for a fluid particle to travel from the vena contracta to point 3.

• The time for a particle to drop a distance y0 under the action of gravity when it has no initial velocity in that direction is expressed by y0=gt2/2 . After t is eliminated in the two relations,

• With V2t determined by Eq. mentioned earlier, the ratio VaVt = Cv is known.

g

y

gHC

x

v

2

2=

from where

Hy

xCv

2=

Minor loss coefficient ζ may by recorded in such way

.11

2−=

vC

ζ

Flow rate coefficient Cd may be computed from measured flow

rate Q, hydrodynamic head H and area of orifice A as

gHA

QCd

2=

When Cv and Cd are known, contraction coefficient Cc may be

easily computed as

v

dc

C

CC =

• According to expression of Cv and physical sense of contraction

coefficient Cc (dimensions of jet cross section always are less

dimensions of an orifice) both of them are smaller than unit.

therefore product of them CcCv ==== Cd is also smaller than unit,

what should be kept in mind, when experimental results are treated

and used for practical computations.


ExampleExample

• A 75-mm-diameter orifice under a head of 4.88 m discharges

907.6 kg water in 32.6 s. The trajectory was determined by

measuring x0=4.76m for a drop of 1.22 m. Determine Cv, Cc,

Cd, the head loss per unit gravity force, and the power loss.

SolutionSolution

• The theoretical velocity V2t is

• The actual velocity is determined from the trajectory. The time

to drop 1.22 m is

• and the velocity is expressed by

• Then

• The actual discharge Qa is

• With Eq.

• Hence, from Eq.,

• The head loss is

• The power loss is

• A jet of water discharges horizontally into the atmosphere from a small orifice in the vertical side of a large open tank (filled with water head H above orifice). Derive an expression for actual velocity v of a jet at the vena contracta if the jet falls a distance y vertically in a horizontal distance x, measured from the vena contracta. Also determine the coefficient of velocity. If the orifice has an area of 650 mm2 and the jet falls a distance y of 0.5m in a horizontal distance x of 1.5m from the vena contracta, calculate the values of the coefficients of velocity, discharge and contraction, given that the volume rate of flow is 0.117 m3 and the head H above the orifice is 1.2m.

• A tank has two identical orifices in its of its vertical sides. The upper orifice is 3.0 m below the water surface and the lower one is 5.0 m below the water surface. If the value of coefficient of velocity Cv for each orifice is 0.96, find the point of intersection of the two jets.

Flow through a large orifice

• Velocity of liquid flow within limits of large orifice varies

significantly, what may be estimated computing flow rate as sum

flow rates passing small elements of the orifice

•

h1h2

dh

H h

b

a C

Fig. 7.6 Liquid flow through

large orifice

Let as analyze peculiarities of liquid flow from open reservoir

(H=h) through narrow horizontal belt of large rectangular orifice .

According to equation flow rate of liquid passing this elementary

belt may be expressed as

ghbdhCghdACdQdd

22 ==

Total flow rate Q may be received integrating within limits of h

from h1 to h2, and assuming Cd =const, i.e.

∫=2

1h

h

2ghbdhCQd

Solution of this equation has shape

−= 2

3

12

3

23

22 hhgbCQ d


Let us substitute h1 = H –a/2 and h2 = H +a/2 and rewrite (7.26),

where H is distance from free surface of liquid to the center of

orifice

−−

+=

2

3

2

3

223

22

aH

aHgbCQ d

Expanding difference of two terms in brackets by Newton

series leads to

−

−= ...

2048

1

96

112

3

242

H

a

H

agHabCQ

d

The terms are small enough to be

neglected and omitted. Thus

... ,2048

1 ,

96

142

H

a

H

a

2gHACQ do=

Where is large orifice flow rate coefficientdd0 C

3

2C =

Flow under Pressure through Orifices

Partially Submerged Orifice Pipe Flow-Major and Minor losses ( review)

• The goal is to study pressure losses due to viscous ( frictional) effects in fluid flows through pipes

Flow

meter

Differential Pressure

Gauge- measure ∆P

LValve

H

Reservoir

PipeD

Schematic of experimental Apparatus

• Pipes with different Diameter, Length, and surface characteristics will be used for the

experiments

Major and Minor losses

Total Head Loss( hLT) = Major Loss (hL)+ Minor Loss (hLM)

g

V

D

LfhEquationsDarcy l

2'

2

=⇒

Due to wall

friction

Due to sudden expansion,

contraction, fittings etc

g

VKhlm

2

2

=

K is loss coefficient must

be determined for each

situation

In this experiment you

will find friction factor for

various pipesFor Short pipes with multiple fittings, the minor losses are no

longer “minor”!!

Minor Losses

• Flow separation and associated viscous effects will tend to decrease the flow energy and hence the losses

• The phenomenon is fairly complicated. Loss coefficient ‘K’ will take care of this complicities

Valve

s

Bend

s

T joints Expansions Contraction

s

g

VKhlm

2

2

=


31

Minor Losses

• In addition to head loss due to friction, there are always other head losses due to pipe expansions and contractions, bends, valves, and other pipe fittings. These losses are usually known as minor losses (hLm).

• In case of a long pipeline, the minor losses maybe negligible compared to the friction losses, however, in the case of short pipelines, their contribution may be significant.

32

Losses due to pipe fittings

• where hLm= minor loss

K = minor loss coefficient

V = mean flow velocity

g2

VKh

2

Lm =

Table : Typical K values

Type K

Exit (pipe to tank) 1.0

Entrance (tank to pipe) 0.5

90° elbow 0.9

45° elbow 0.4

T-junction 1.8

Gate valve 0.25 - 25

33

Sudden Enlargement

� As fluid flows from a smaller pipe into a larger pipe through sudden enlargement, its velocity abruptly decreases; causing turbulence that generates an energy loss.

� The amount of turbulence, and therefore the amount of energy, is dependent on the ratio of the sizes of the two pipes.

� The minor loss (hLm)is calculated from;

• where is KE is the coefficient of expansion, and the values dependson the ratio of the pipe diameters (Da/Db) as shown below.

g2

VKh

2a

ELm =

Da/Db 0.0 0.2 0.4 0.6 0.8

K 1.00 0.87 0.70 0.41 0.15

Table 6.3: Values of KE vs. Da/Db

Sudden Enlargement

Frictional losses occur as result of turbulence generated

immediately downstream of the expansion

35 Copyright © ODL Jan

2005 Open University

Figure 6.9: Flow at Sudden Enlargement

Apply in direction of flow

Neglect surface shear

Divide by (A2 w)

Head Loss due to Sudden Expansion:Conservation of Momentum

Pressure is applied over all of section 1.Momentum is transferred over area corresponding to upstream pipe diameter.V1 is velocity upstream.

sspp FFFWMM +++=+2121

1 2

xx ppxx FFMM2121 +=+

1

2

11 AVM x ρ−= 2

2

22 AVM x ρ=

22212

2

21

2

1 ApApAVAV −=+− ρρ

g

A

AVV

gw

pp2

12

1

2

2

21

−

==

−

ρ

AA11

AA22

xx


Energy

Head Loss due to Sudden Expansion

g

VV

w

pphl

2

22

2121 −

+−

=

g

A

AVV

w

pp 2

12

1

2

2

21

−

=−

1

2

2

1

V

V

A

A=

g

VV

g

V

VVV

hl2

2

2

2

11

22

1

2

2−

+

−

=g

VVVVhl

2

2 2

121

2

2 +−=

( )g

VVhl

2

2

21 −=

2

2

1

2

1 12

−=

A

A

g

Vhl

2

2

11

−=

A

AK

Momentum

Mass

Head Loss due to Sudden Changes in Area of Flow

Sudden Expansion: The head loss at a sudden expansion is givenby the expression:

Portion Covered as on 19/04/2012

Sudden Contractions

Geometrically a sudden contraction is reverse of the sudden

enlargement but it is not possible to apply impulse momentum

equation to a control volume between section 1-1 and 2-2 in the

case of sudden contraction.

This is because just upstream of the junction between the two

pipes, in the wider pipe the stream lines are curved and the liquid

is accelerated, due to which the pressure at the annular face is

varies in unknown manner which can not be determined easily.

Sudden Contractions

At sudden contractions, flow streamlines converge causing the

downstream developed flow to have an area smaller than the

downstream pipe diameter. This flow constriction is called the

vena contracta. Viscous dissipation occurs in the vortices

developed in this area.

41

Sudden Contraction

•The energy loss due to a sudden contraction can be calculated using the following;

•

•The KC is the coefficient of contraction and the values depends on the ratio of the pipe diameter (Db/Da) as shown below.

g2

VKh

2b

CLm =

Db/Da 0.0 0.2 0.4 0.6 0.8 1.0

K 0.5 0.49 0.42 0.27 0.20 0.0

Table 6.4: Values of KC vs. Db/Da

Figure: Flow at sudden contraction42

� Determine the energy loss that will occur as 0.06 m3/s water flows from a 40-mm pipe diameter into a 100-mm pipe diameter through a sudden expansion.

• Solution:

� The head loss through a sudden enlargement is given by;

• Da/Db = 40/100 = 0.4

• From Table : K = 0.70

• Thus, the head loss is

g2

VKh

2

am =

s/m58.3)2/04.0(

0045.0

A

QV

2a

a =π

==

m47.081.9x2

58.3x70.0h

2

Lm ==


Flow Through Mouthpieces

• Flow through external cylindrical mouthpiece

• Flow through a convergent-divergent external mouthpiece

• Flow through internal or reentrant or Borda’s mouthpiece


Problems

1. Find the discharge from a 80mm diameter external

mouthpiece, fitted to a side of a large vessel if the head over the

mouth piece is 6m.

Solution.

For a cylindrical mouth piece Cd=0.853

gHaCQ d 2=

( )2310804

80

−××=∴

=

πa

mmd

2310026.5 m−×=

lps 46.52Q

/s0.04652m2x9.81x66x100.853x5.02Q 33

=⇒

=×=∴ −

2. An external cylindrical mouthpiece of 100mm diameter is

discharging water under a constant head of 8m. Determine the

discharge and absolute pressure head of water at Vena contracta.

Take Cd=0.855 and CC for Vena contracta =0.62. Take atmospheric

pressure head =10.3m of water

Solution.

H=8m, Q=?, Cd=0.855, Cc=0.62

232

m7.854x104

πx0.10.1m100mmd −=∴==

waterof m 10.3γ

pa =

2x9.81x8x4x100.855x7.852gHaCQ3

d−==

84.13lps/sec0.08413mQ3 ==

We know

(Gauge) water of m 7.1440.893x8γ

pc −=−=

−

=

∴ gauge

ppabsolute

p ccc

γγγ

)144.73.10( −= )(156.3 Absm=

)62.0 ( 893.0

)62.0 ( 893.0

=−=⇒

=−=

cc

cac

CwhenHg

p

CwhenHHH

ρ

3. An external cylindrical mouth piece 60mm diameter fitted in

the side of a tank discharges under a constant head of 3m, for

which CV=0.82

Determine i) the discharge in lps

ii) absolute pressure at Vena contracta

iii) Maximum head for steady.

Flow assuming that separation occurs at 2.5m of water

absolute. Local barometer reads 760mm Hg.

Solution.

(i)Discharge(Q)

At the exit of the mouth piece CC=1

1x0.82xCCC vcd ==∴

gHaCQ d 2=

( )

=−

2x9.81x3x4

60x10πx0.82x

23

lpssm 8.17/0178.03 ==


(ii)Absolute Pressure head at Vena contracta

Applying Bernoulli’s equation between (A)

& © © with the centre line of the mouth

piece as datum and neglecting losses hL

LCCAA

A hg

VpZ

g

VpZ +++=++

22

2

1

2

γγ

02

0{}002

+++=++g

VpH Cc

γ

)1(2

2

−−−+=g

VpH CC

γ

From Continuity equation Q=aV

a

QV =

( )

==− 231060

462.0

0178.0

xxxxaC

QV

C

C π

)1(FromEq∴

smVc /154.10=∴

g

VH

pcc

2

2

−=∴γ

−=

81.92

154.103

2

x=2.255m of water (Gauge)

Gauge

catimc ppp

−=

∴

γγγ

ymmofmercurp

atm 760=γ

)(336.10 2211 HSHSmofwater == Q

( ) )(781.7555.2336.10 Absolutep

abs

c =−=

∴

γ

(iii) Hmax for steady flow

Applying Bernoulli’s equation between Vena

contracta and exit of the mouth piece with the

centre line of the mouth piece as datum &

considering head loss hL due to sudden

expansion of flow.

LCc

C hg

VpZ

g

VpZ +++=++

22

2

111

2

γγ

g

VhgCAssu LC

2375.0,62.0min

2

==

( )g

V

g

VV

2

375.0

200

6219

62.0/836.70

222

+++=−

+−

g

V

22265.1836.7

2

−=−

mofwaterg

V389.6

226.1

836.7

2

2

=

=

gHCweknowV V 2=

2

21

2 VCx

g

VH =

2max82.0

1389.6 xH =∴

mofwater5.9=

Alternatively, Hpp

ac 82.0−=γγ

-7.781=0-0.82xHmax

Hmax=9.49m of water

Flow through a convergent and Divergent Mouthpiece

• In the case of a cylindrical mouthpiece, on account of the formation of vena-contracta and the subsequent enlargement of the jet there occurs a loss of energy which results in reducing the coefficient of discharge of the mouthpiece.

• If the mouthpiece is made to conform to the shape of the jet up to vena-contacta then loss of energy can be eliminated. Such mouthpiece is known as convergent mouthpiece.

• Theoretically cd for a convergent mouthpiece = 1 but actually on account of frictional resistance some energy is lost and hence cd for a convergent mouthpiece = 0.975

• The loss of energy owing to the sudden enlargement of a jet in a mouthpiece can however be eliminated by making the mouthpiece gradually diverging.



Flow through internal or reentrant or Borda’s mouthpiece

• An internal mouthpiece is a short cylindrical tube attached to a circular

orifice in the side of reservoir or tank such that it projects inward.

• The Borda mouthpiece, a short, thin-walled tube about one diameter

long that projects into the reservoir (re-entrant), permits application of

the momentum equation, which yields, one relation between Cv and Cd.


FigureFigure The Borda mouthpiece.

• The velocity along the wall of the tank is almost zero at all points; hence, the pressure distribution is hydrostatic.

• The final velocity is V2a; the initial velocity is zero; and Qa is the actual discharge. Then

• and

• Substituting for Qa and V2a and simplifying lead to


The Pitot-Static Tube

P1,V1 Stagnation

Point V2=0

1 2

�P1 is a Static pressure: It is

measured by a device (static

tube) that causes no velocity

change to the flow. This is

usually accomplished by drilling

a small hole normal to a wall

along which the fluid is flowing.

�P2 is a Stagnation pressure:

It is the pressure measured by

an open-ended tube facing the

flow direction. Such a device is

called a Pitot tube.

P2


Pitot-Static tube

Bernoulli equation between 1 and 2:0

2

)VV()PP(2

1

2

212 =−

+ρ

−

(Recall that position 2 is a stagnation point: V2= 0)

2/1

121

)PP(2V

ρ

−=

We can measure pressures P1 and P2 using hydrostatics:

P1=Patm + ρgh1, P2=Patm + ρgh2

or using a Pressure Gauge

2

112 V2

1PP ρ+=

(I)

�Stagnation Pressure is higher than

Static Pressure


Pitot-static tubeThe static and Pitot tube are often combined into the one-piece Pitot-static

tube.

In this, static tube surrounds the total head tube and two or more small

holes are drilled radially through the outer wall into the annular space

Alternatively, the dynamic pressure head can be determined directly by

connecting a suitable differential manometer between the Pitot tube and

the pressure tap


Problem

• A piezometer and a Pitot tube are tapped into a horizontal water pipe, as shown in Fig., to measure static and stagnation (static dynamic) pressures. For the indicated water column heights, determine the velocity at the center of the pipe.


Example: Measurement of Velocity

A Pitot static probe connected to a water manometer is used to measure the velocity of air. If the deflection (the vertical distance between the fluid levels in the two arms ) is 7.3 cm, dwtemine the air velocity . Take the density of air to be 1.25 Kg/m3


Example: Measurement of Velocity

Water flows in a 300 mm pipe. Two Pitot tubes are installed in the pipe, one on the centerline and the other 75 mm from the centerline. If the velocities at the two points are 3 m/sec and 2 m/sec respectively, Calculate the reading on the differential manometer connected to the two tubes.


Venturi, Orifice, and Nozzle meters

• Fluid is accelerated by forcing it to flow through a constriction, thereby

increasing kinetic energy and decreasing pressure energy. The flow

rate is determined by measuring the pressure difference between the

meter inlet and a point of reduced pressure.

• Desirable characteristics of flow meters:

– Reliable, repeatable calibration

– Introduction of small energy loss into the system

– Inexpensive

– Minimum space requirements

Basic principle: Increase in velocity causes a decrease

in pressure.

5.14


Venturi MeterThis device consists of a conical contraction, a short cylindrical throat

and a conical expansion. The fluid is accelerated by being passed

through the converging cone. The velocity at the “throat” is assumed to

be constant and an average velocity is used. The venturi tube is a

reliable flow measuring device that causes little pressure drop. It is used

widely particularly for large liquid and gas flows.

P1

P2

])A/(A-[1

)PP(2C V

2

12

212

ρ

−= υ

Where the discharge

coefficient, Cυ =f(Re), can

be found in Figure 5.11,

textbook (5.9 2nd edition)

Flow through a Venturi Meter

In a venturi, 0.95 < C < 0.98

Advantage:

Pressure recovery

Uses little power


Flow through a Venturi Meter

C

Re

2 x 105

µ

ρ= 11vd

eR

0.98

Based upon the conditions in the

pipe approaching the meter Applications: Flow Measurement

Example 1: Flow through an orifice meter

A Venturimeter has its axis vertical, the inlet and throat diameters being 150 mm and 75 mm respectively. The throat is 225 mm above inlet and Cd = 0.96. Petrol of sp.gravity 0.78 flows up through the Venturimeter at a rate of 0.029 m3/sec. Determine the pressure difference between the inlet and the throat.

Air at 110 kPa and 50°C flows upward through a 6-cm-

diameter inclined duct at a rate of 45 L/s. The duct diameter

is then reduced to 4 cm through a reducer. The pressure

change across the reducer is measured by a water

manometer. The elevation difference between the two points

on the pipe where the two arms of the manometer are

attached is 0.20 m. Determine the differential height between

the fluid levels of the two arms of the manometer if the

density of water is 1000 kg/m3 and the gas constant of air is

R = 0.287 kPa-m3/kg.K. Clearly write all assumptions and

give your comments on the result.

Thank You

CE C212: By Dr. Ajit Pratap Singh 1 - Civil Engineering...

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