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
CE C212: By Dr. Ajit Pratap Singh 2
• 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.
CE C212: By Dr. Ajit Pratap Singh 3
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.
CE C212: By Dr. Ajit Pratap Singh 4
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
CE C212: By Dr. Ajit Pratap Singh 5
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
=
CE C212: By Dr. Ajit Pratap Singh 6
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
CE C212: By Dr. Ajit Pratap Singh 7
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 ==
CE C212: By Dr. Ajit Pratap Singh 8
Flow Through Mouthpieces
• Flow through external cylindrical mouthpiece
• Flow through a convergent-divergent external mouthpiece
• Flow through internal or reentrant or Borda’s mouthpiece
Applications: Flow Measurement
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 ==
CE C212: By Dr. Ajit Pratap Singh 9
(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.
Applications: Flow Measurement
CE C212: By Dr. Ajit Pratap Singh 10
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.
Applications: Flow Measurement
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
Applications: Flow Measurement
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
Applications: Flow Measurement
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
Applications: Flow Measurement
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
CE C212: By Dr. Ajit Pratap Singh 11
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.
Applications: Flow Measurement
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
Applications: Flow Measurement
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.
Applications: Flow Measurement
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
Applications: Flow Measurement
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
CE C212: By Dr. Ajit Pratap Singh 12
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