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Fluid Flow of Food Processing

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Title: Fluid Flow of Food Processing


1
Fluid Flow of Food Processing
2
Production Line for milk processing
3
Fluid Mechanics Overview
Fluid
Mechanics
Gas
Liquids
Statics
Dynamics
, Flows
Water, Oils, Alcohols, etc.
Stability
Air, He, Ar, N2, etc.
Buoyancy
Compressible/ Incompressible
Pressure
Laminar/ Turbulent
Surface Tension
Steady/Unsteady
Compressibility
Viscosity
Density
Vapor Pressure
Viscous/Inviscid
Fluid Dynamics
Introduction
4
Characteristics of Fluids
  • Gas or liquid state
  • Large molecular spacing relative to a solid
  • Weak intermolecular cohesive forces
  • Can not resist a shear stress in a stationary
    state
  • Will take the shape of its container
  • Generally considered a continuum
  • Viscosity distinguishes different types of fluids

5
Measures of Fluid Mass and Weight Density
The density of a fluid is defined as mass per
unit volume.
m mass, and v volume.
  • Different fluids can vary greatly in density
  • Liquids densities do not vary much with pressure
    and temperature
  • Gas densities can vary quite a bit with pressure
    and temperature
  • Density of water at 4 C 1000 kg/m3
  • Density of Air at 4 C 1.20 kg/m3

Alternatively, Specific Volume
6
Measures of Fluid Mass and Weight Specific
Weight
The specific weight of fluid is its weight per
unit volume.
g local acceleration of gravity, 9.807 m/s2
  • Specific weight characterizes the weight of the
    fluid system
  • Specific weight of water at 4 C 9.80 kN/m3
  • Specific weight of air at 4 C 11.9 N/m3

7
Measures of Fluid Mass and Weight Specific
Gravity
The specific gravity of fluid is the ratio of the
density of the fluid to the density of water _at_ 4
C.
  • Gases have low specific gravities
  • A liquid such as Mercury has a high specific
    gravity, 13.2
  • The ratio is unitless.
  • Density of water at 4 C 1000 kg/m3

8
Viscosity Kinematic Viscosity
  • Kinematic viscosity is another way of
    representing viscosity
  • Used in the flow equations
  • The units are of L2/T or m2/s and ft2/s

9
Surface Tension
At the interface between a liquid and a gas or
two immiscible liquids, forces develop forming
an analogous skin or membrane stretched over
the fluid mass which can support weight. This
skin is due to an imbalance of cohesive forces.
The interior of the fluid is in balance as
molecules of the like fluid are attracting each
other while on the interface there is a net
inward pulling force. Surface tension is the
intensity of the molecular attraction per unit
length along any line in the surface. Surface
tension is a property of the liquid type, the
temperature, and the other fluid at the
interface. This membrane can be broken with a
surfactant which reduces the surface tension.
10
Surface Tension Liquid Drop
The pressure inside a drop of fluid can be
calculated using a free-body diagram
Real Fluid Drops
Mathematical Model
R is the radius of the droplet, s is the surface
tension, Dp is the pressure difference between
the inside and outside pressure.
The force developed around the edge due to
surface tension along the line
This force is balanced by the pressure difference
Dp
11
Surface Tension Liquid Drop
Now, equating the Surface Tension Force to the
Pressure Force, we can estimate Dp pi pe
This indicates that the internal pressure in the
droplet is greater that the external pressure
since the right hand side is entirely positive.
12
Surface Tension Capillary Action
Capillary action in small tubes which involve a
liquid-gas-solid interface is caused by surface
tension. The fluid is either drawn up the tube
or pushed down.
Wetted
Non-Wetted
Cohesion gt Adhesion
Adhesion gt Cohesion
h is the height, R is the radius of the tube, q
is the angle of contact.
The weight of the fluid is balanced with the
vertical force caused by surface tension.
13
Surface Tension Capillary Action
Free Body Diagram for Capillary Action for a
Wetted Surface
Equating the two and solving for h
For clean glass in contact with water, q ? 0,
and thus as R decreases, h increases, giving a
higher rise. For a clean glass in contact with
Mercury, q ? 130, and thus h is negative or
there is a push down of the fluid.
14
Pressure
  • Pressure is the force on an object that is spread
    over a surface area. The
  • equation for pressure is the force divided by the
    area where the force is applied.
  • Although this measurement is straightforward when
    a solid is pushing on a
  • solid, the case of a solid pushing on a liquid or
    gas requires that the fluid be
  • confined in a container. The force can also be
    created by the weight of an
  • object.

Unit of pressure is Pa
15
Force Equilibrium of a Fluid Element
  • Fluid static is a term that is referred to the
    state of a fluid where its velocity is zero and
    this condition is also called hydrostatic.
  • So, in fluid static, which is the state of fluid
    in which the shear stress is zero throughout the
    fluid volume.
  • In a stationary fluid, the most important
    variable is pressure.
  • For any fluid, the pressure is the same
    regardless its direction. As long as there is no
    shear stress, the pressure is independent of
    direction. This statement is known as Pascals
    law

16
Force Equilibrium of a Fluid Element
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Standard Atmosphere
1 atm 101325 Pa
760 mmHg 760 torr
1 bar
h 76 cm
Mercury
Mercury barometer
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  • A pressure is quoted in its gauge value, it
    usually refers to a standard atmospheric pressure
    p0. A standard atmosphere is an idealised
    representation of mean conditions in the earths
    atmosphere.
  • Pressure can be read in two different ways the
    first is to quote the value in form of absolute
    pressure, and the second to quote relative to the
    local atmospheric pressure as reference.
  • The relationship between the absolute pressure
    and the gauge pressure is illustrated in Figure
    2.6.

23
  • The pressure quoted by the latter approach
    (relative to the local atmospheric pressure) is
    called gauge pressure, which indicates the
    sensible pressure since this is the amount of
    pressure experienced by our senses or sensed by
    many pressure transducers.
  • If the gauge pressure is negative, it usually
    represent suction or partially vacuum. The
    condition of absolute vacuum is reached when only
    the pressure reduces to absolute zero.

24
Pressure Measurement
  • Based on the principle of hydrostatic pressure
    distribution, we can develop an apparatus that
    can measure pressure through a column of fluid
    (Fig. 2.7)

25
Pressure Measurement
We can calculate the pressure at the bottom
surface which has to withstand the weight of four
fluid columns as well as the atmospheric
pressure, or any additional pressure, at the free
surface. Thus, to find p5, Total fluid
columns (p2 p1) (p3 p2) (p4 p3) (p5
p4) p5 p1 ?og (h2 h1) ?wg (h3 h2)
?gg (h4 h3) ?mg (h5 h4) The p1 can be the
atmospheric pressure p0 if the free surface at z1
is exposed to atmosphere. Hence, for this case,
if we want the value in gauge pressure (taking
p1p00), the formula for p5 becomes p5 ?og
(h2 h1) ?wg (h3 h2) ?gg (h4 h3) ?mg
(h5 h4) The apparatus which can measure the
atmospheric pressure is called barometer (Fig
2.8).
26
  • For mercury (or Hg the chemical symbol for
    mercury), the height formed is 760 mm and for
    water 10.3 m.
  • patm 760 mm Hg (abs) 10.3 m water (abs)
  • By comparing point A and point B, the atmospheric
    pressure in the SI unit, Pascal,
  • pB pA ?gh
  • pacm pv ?gh
  • 0.1586 13550 (9.807)(0.760)
  • ? 101 kPa

27
  • This concept can be extended to general pressure
    measurement using an apparatus known as
    manometer. Several common manometers are given
    in Fig. 2.9. The simplest type of manometer is
    the piezometer tube, which is also known as
    open manometer as shown in Fig. 2.9(a). For
    this apparatus, the pressure in bulb A can be
    calculated as
  • pA p1 p0
  • ?1gh1 p0
  • Here, p0 is the atmospheric pressure. If a known
    local atmospheric pressure value is used for p0,
    the reading for pA is in absolute pressure. If
    only the gauge pressure is required, then p0 can
    be taken as zero.

28
Although this apparatus (Piezometer) is simple,
it has limitations, i.e. It cannot measure
suction pressure which is lower than the
atmospheric pressure, The pressure measured is
limited by available column height, It can only
deal with liquids, not gases. The restriction
possessed by the piezometer tube can be overcome
by the U-tube manometer, as shown in Fig. 2.9(b).
The U-tube manometer is also an open manometer
and the pressure pA can be calculated as
followed p2 p3 pA ?1gh1 ?2gh2
p0 ? pA ?2gh2 - ?1gh1 p0
29
If fluid 1 is gas, further simplification can be
made since it can be assumed that ?1 ?  ? 2, thus
the term ?1gh1 is relatively very small compared
to ?2gh2 and can be omitted with negligible
error. Hence, the gas pressure is
pA ? p2 ?2gh2 - p0 There is also a closed
type of manometer as shown in Fig. 2.9(c), which
can measure pressure difference between two
points, A and B. This apparatus is known as the
differential U-tube manometer. For this case, the
formula for pressure difference can be derived as
followed p2 p3 pA ?1gh1
pB ?3gh3 ?2gh2 ? pA - pB ?3gh3
?2gh2 - ?1gh1
30
Piezometer tube
Open
h
31
U-tube manometer
Pressure is defined as a force per unit area -
and the most accurate way to measure low air
pressure is to balance a column of liquid of
known weight against it and measure the height of
the liquid column so balanced. The units of
measure commonly used are inches of mercury (in.
Hg), using mercury as the fluid and inches of
water (in. w.c.), using water or oil as the fluid
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Example
  • An underground gasoline tank is accidentally
    opened during raining causing the water to seep
    in and occupying the bottom part of the tank as
    shown in Fig. E2.1. If the specific gravity for
    gasoline 0.68, calculate the gauge pressure at
    the interface of the gasoline and water and at
    the bottom of the tank. Express the pressure in
    Pascal and as a pressure head in metres of water.
    Use ?water  998 kg/m3 and g  9.81 m/s2.

35
  • For gasoline
  • ?g 0.68(998) 678.64kg/m3
  • At the free surface, take the atmospheric
    pressure to be zero, or p0  0 (gauge pressure).
  • p1 p0 pgghg 0 (678.64)(9.81)(5.5)
  • 36616.02 N/m2 36.6 kPa
  • The pressure head in metres of water is
  • h1 p1 p0 36616.02 - 0
  • pwg (998)(9.81)
  • 3.74 m of water
  • At the bottom of the tank, the pressure
  • p2 p1 pgghg 36616.02 (998)(9.81)(1)
  • 46406.4 N/m2 46.6 kPa
  • And, the pressure head in meters of water is
  • h2 p1 p0 46406.4 - 0
  • pwg (998)(9.81)
  • 4.74 m of water

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Strain and Strain (Shear) Rate
  • Strain
  • a dimensionless quantity representing the
    relative deformation of a material
  • Normal Strain Shear Strain

39
Shear Stress is the intensity of force per unit
area, acting tang
40
Simple Shear Flow
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Solids Elastic and Shear Moduli
  • When a solid material is exposed to a stress, it
    experiences an amount of deformation or strain
    proportional to the magnitude of the stress
  • Stress (?) ? Strain (? or ?)
  • Stress Modulus ? Strain
  • Normal stress elastic modulus (E)
  • Shear stress shear modulus (G)

43
Fluid Viscosity
  • Newtonian fluids
  • viscosity is constant (Newtonian viscosity, ?)
  • Non-Newtonian fluids
  • shear-dependent viscosity (apparent viscosity, ?
    or ?a)

44
Viscosity Introduction
The viscosity is measure of the fluidity of the
fluid which is not captured simply by density or
specific weight. A fluid can not resist a shear
and under shear begins to flow. The shearing
stress and shearing strain can be related with a
relationship of the following form for common
fluids such as water, air, oil, and gasoline
  • is the absolute viscosity or dynamics viscosity
    of the fluid, u is the velocity of the fluid and
    y is the vertical coordinate as shown in the
    schematic below

No Slip Condition
45
Viscosity
Viscosity is a property of fluids that indicates
resistance to flow. When a force is applied to a
volume of material then a displacement
(deformation) occurs. If two plates (area, A),
separated by fluid distance apart, are moved (at
velocity V by a force, F) relative to each
other,
Newton's law states that the shear stress (the
force divided by area parallel to the force, F/A)
is proportional to the shear strain rate . The
proportionality constant is known as the
(dynamic) viscosity
46
Shear stress
The unit of viscosity in the SI system of units
is pascal-second (Pa s)
In cgs unit , the unit of viscosity is expressed
as poise
Shear rate
1 poise 0.1 Pa s
1 cP 1 m Pa s
47
Example Shear stress in soybean oil
  • The distance between the two parallel plates is
    0.00914 m and the lower plate is being pulled at
    a relative velocity of 0.366 m/s greater than the
    top plate. The fluid used is soybean oil with
    viscosity of 0.004 Pa.s at 303 K
  • Calculate the shear stress and the shear rate
  • If water having a viscosity of 880x10-6 Pa.s is
    used instead of soybean oil, what relative
    velocity in m/s needed using the same distance
    between plates so that the same shear stress is
    obtained? Also, what is the new shear rate?

48
Kinematics Viscosity
49
Reynolds Number
50
Experiment for find Re
Laminar flow
Turbulent flow
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52
Example At what velocity does water flow
convert from laminar to transition flow in a 5
cm diameter pipe at 20 C
Properties of water _at_ 20 C
53
Milk is flowing at 0.12 m min in a 2.5-cm
diameter pipe. If the temperature of the milk is
21C, is the flow turbulent or streamline
3
-1
Given
Density of milk 1030
kg /m 3 Viscosity
2.12 cP
54
Velocity Profile in a Liquid Flowing under Fully
Developed Flow Condition
The velocity profile for a larminar , fully
develop flow in horizontal pipe is
55
Measurement of Viscosity
Viscosity of a liquid can be measurement
56
Viscosity Measurements
A Capillary Tube Viscosimeter is one method of
measuring the viscosity of the fluid. Viscosity
Varies from Fluid to Fluid and is dependent on
temperature, thus temperature is measured as
well. Units of Viscosity are Ns/m2 or lbs/ft2
Movie Example using a Viscosimeter
57
Capillary Tube Viscometer
58
Rotational Viscometer
59
Influence of Temperature on Viscosity
There is considerable evidence that the influence
of temperature on viscosity for liquid food may
be described by an Arrhenius type relationship
60
Energy Balance
In addition to the mass balance, the other
important quantity we must consider in the
analysis of fluid flow, is the energy balance.
Referring again to this picture , we shall
consider the changes in the total energy of unit
mass of fluid, one kilogram, between Section 1
and Section 2
There may be energy interchange with the
surroundings including      (4) Energy lost to
the surroundings due to friction.      (5)
Mechanical energy added by pumps.      (6) Heat
energy in heating or cooling the fluid.
Firstly, there are the changes in the intrinsic
energy of the fluid itself which include changes
in      (1) Potential energy.      (2) Kinetic
energy.      (3) Pressure energy.
61
Bernoulli Equation
Bernoullis equation is a consequence of
Conservation of Energy applied to an ideal fluid
Assumes the fluid is incompressible and
non-viscous, and flows in a non-turbulent,
steady-state manner
States that the sum of the pressure, kinetic
energy per unit volume, and the potential energy
per unit volume has the same value at all points
along a streamline
62
A 3 m diameter stainless steel tank contains
wine. In the tank , the wine is filled to 5 cm
depth. A discharge port, 10 cm diameter , is
opened to drain the wine. Calculate the discharge
velocity of wine ,assuming the flow is steady and
frictionless and the time required in emptying it
5 m
Location 2
Then volumetric flow rate from the discharge port
using
63
Water flows at the rate of 0.4 m3 / min in a 7.5
cm diameter pipe at a pressure of 70 kPa. If the
pipe reduces to 5 cm diameter calculate the new
pressure in the pipe
P 70 kPa D 7.5 cm A 4.42 x 10 -3 m 2
P ? kPa D 5 cm A 1.90 x 10 -3 m 2
P2 65.3 k Pa.
64
Forces due to Friction
When a fluid moves through a pipe or through
fittings, it encounters frictional resistance and
energy can only come from energy contained in the
fluid and so frictional losses provide a drain on
the energy resources of the fluid. The actual
magnitude of the losses depends upon the nature
of the flow and of the system through which the
flow takes place. In the system, let the energy
lost by 1 kg fluid between section 1 and section
2, due to friction, be equal to Eƒ (J).
Common parameter used in laminar and turbulent
flow is Fanning friction factor (ƒ) ƒ is defined
as drag force per wetted surface unit area
product of density
times velocity head
? ½?v2
65
Major loss
f
66
Type of pipe Roughness (?) Roughness (?)
Ft m
Riveted steel 0.003-0.03 0.0009-0.009
Concrete 0.001-0.01 0.0003-0.003
Wood stave 0.0006-0.003 0.0002-0.0009
Cast iron 0.00085 0.00026
Galvanized iron 0.0005 0.00015
Asphalted cast iron 0.0004 0.0001
Commercial steel or wrought iron 0.00015 0.000046
Drawn brass or copper tubing 0.000005 0.0000015
Glass and plastic smooth smooth
67
Moody Diagram for the Fanning friction factor
68
Moody Diagram for the Moody friction factor
69
Laminar Zone
Turbulent Zone
70
Laminar
Lamina Zone
Turbulent Zone
Transition
71
Water at 30 C is being pumped through a 30 m
section of 2.5 cm. diameter steel pipe at a mass
flow rate 2.5 kg/s. Compute the pressure loss due
to friction in pipe section
Re
240.08 kPa
72
A liquid is flowing through a horizontal straight
commercial steel pipe at 4.57 m/s. The inside
diameter of the pipe is 2.067 in. The viscosity
of the liquid is 4.46 cP and the density 801
kg/m3. Calculate the mechanical-energy friction
loss in J/kg for a 36.6 m section of pipe. Given
roughness of commercial steel pipe is 4.6 X
10-5 m.
73
For Non-circular pipe
4(Cross-sectional area)
4 Hydraulic radius


Wetted Perimeter
74
0.875 m
75
Water at 30 C is being pumped through a 30 m
section of annular area of steel pipe at a mass
flow rate 2 kg/s. Compute the pressure loss due
to friction in pipe section
2.5 cm
5 cm
Determine Equivalent Diameter
0.025 m
76
Determine Reynolds Number and
Determine Friction factor
240.08 kPa
77
Energy Equation for Steady Flow of Fluids
78
Frictional Energy Loss
The frictional energy loss for a liquid flowing
in pipe is composed of major and minor loses
The minor losses are due to various components
used in pipeline system such as values , tees
and elbow and contraction of fluid
The major losses are due to the flow of viscous
liquid in the straight portions of a pipe
79
General Equation for Minor Losses
hLm minor loss K minor loss coefficient Le
equivalent length
80
Expansion and Contraction
81
Transition Loss Equations
  • Contraction
  • Expansion

82
Pipe Bends
83
Friction loss of turbulent flow through valves
and fittings
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Example Friction losses and mechanical-energy
balance
  • An elevated storage tank contains water at
    82.2ºC as shown in figure. It is desired to have
    a discharge rate at point 2 of 6.315 X 10-3 m3/s.
    What must be the height H in m of the surface of
    the water in the tank relative to the discharge
    point? The pipe used is commercial steel pipe.

89
Pump
  • Power and work
  • Using the total mechanical-energy-balance
    equation on pump and piping system, the actual or
    theoretical mechanical energy (Ws) added to fluid
    by the pump can be calculated.
  • If ? fractional efficiency and Wp the shaft
    work delivered to the pump
  • Wp Ws/ ?
  • The mechanical energy Ws in J/kg added to fluid
    often expressed as the developed head (H) of pump
    in m of fluid
  • Wp H.g.mº/ ?

90
Suction lift and cavitation
  • Power can be calculated from difference in
    pressure between discharge and suction.
  • Practically, lower limit of suction pressure is
    fixed by vapor pressure (corresponding to
    temperature at suction). If suction pressure is
    equal to vapor pressure, liquid flashes into
    vapor. This process call cavitation that can
    occur in suction line and no liquid can be drawn
    into pump. This process can cause severe erosion
    and mechanical damage to pump.
  • Therefore, it should have net positive suction
    head (NPSH)

91
Net positive suction head (NPSH)
  • NPSH positive difference between pressure at
    pump inlet and vapor pressure of liquid being
    pumped to prevent cavitation.
  • NPSH
  • required NPSH function of impeller design (its
    value is provided by manufacturer)
  • available NPSH function of suction system.
  • NPSHa ha hvp hs hf
  • Where ha absolute pressure
  • hvp vapor pressure
  • hs static head of liquid above center line
    of pump
  • hf friction loss

92
Centrifugal pump
  • Pressure developed by rotating impeller.
  • Impeller impact a centrifugal force on liquid
    entering the center of impeller.
  • Affinity laws govern performance of
    centrifugal pumps at various impeller speeds
  • Vº2 Vº1(N2 /N1)
  • h2 h1(N2 /N1)2
  • P2 P1(N2 /N1)3
  • Where Vº volumetric flow rate
  • h total head
  • P power

93
Positive displacement
  • Constant discharge pressure at difference flow
    rates.
  • Regulation of flow rate is done by changing
    displacement or capacity of intake chamber

94
Characteristic curve
  • plot head, power consumption and efficiency with
    respect to volumetric flow rate (capacity)
  • use for rating pumps

95
Pipe flow problems
  • Basically, there are 3 types of problems
  • Direct solution finding h or ?P for given L, D,
    v, f
  • Finding v or Q for given L, D, h or ?P, ?/D
  • Trial and error type solution
  • Use Re.f1/2 constant (not depend on v or Q)
  • Finding D for given L, Q, h or ?P, ?/D
  • Trial and error type solution
  • Use Re.f1/5 constant (not depend on D)

96
Assumption V
Trial and error type solution For finding v
Determine Re
Determine f
Determine V
YES
NO
97
Example Trial and error type solution For
finding v
2
Constant level
5 m
1
12 m
98
1 Equation 2 Variable cannot determine
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Whole milk flows through a horizontal stainless
steel pipe ( ? 4.2 x 10-6 m) , 30 m long and
having inside diameter D 2.54 cm . Determine
the flow rate of milk when a 2 HP motor with 75
efficiency is used.
Given
Density of whole milk 1030 kg /m
Viscosity of whole milk 0.00202 Pa s
1
2
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Flow Measurement
Flow measurement is essential in many industries
such as the oil, power, chemical, food, water,
and waste treatment industries. These industries
require the determination of the quantity of a
fluid, either gas, liquid, or steam, that passes
through a check point, either a closed conduit or
an open channel, in their daily processing or
operating. The quantity to be determined may be
volume flow rate, mass flow rate, flow velocity,
or other quantities related to the previous
three. These methods include (a) Pitot tube, (b)
Orifice plate and (c) Venturi tube are the
measurement involves pressure difference.
Differential pressure flow meters employ the
Bernoulli equation that describes the
relationship between pressure and velocity of a
flow. These devices guide the flow into a section
with difference cross section areas (different
pipe diameters) that causes variations in flow
velocity and pressure. By measuring the changes
in pressure, the flow velocity can then be
calculated. Many types of differential pressure
flow meters are used in the industry.
105
The Pitot Tube
The Pitot tube is a widely used sensor to measure
velocity of fluid
Total pressure
The principle is based on the Bernoulli Equation
where each term can be interpreted as a form of
pressure
Air
Static Pressure
Stagnation point (v 0)
Water
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The Orifice Meter
A flat plate with an opening is inserted into
the pipe and placed perpendicular to the flow
stream. As the flowing fluid passes through the
orifice plate, the restricted cross section area
causes an increase in velocity and decrease in
pressure. The pressure difference before and
after orifice plate is used to calculate the flow
velocity.
108
D1
D2
PA - PB
109
The Venturi Meter
To reduce energy loss due to friction created by
the sudden contraction in flow in an orifice
meter
110
Variable-Area Meter
Variable area flow meter 's cross section area
available to the flow varies with the flow rate.
Under a (nearly) constant pressure drop, the
higher the volume flow rate, the higher the flow
path area.
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