Fluid Flow of Food Processing

Production Line for milk processingProduction Line for milk processing

Fluid Mechanics Overview

Gas Liquids Statics Dynamics

Air, He, Ar, N2, etc.

Water, Oils, Alcohols, etc.

0 iF

Viscous/Inviscid

Steady/Unsteady

Compressible/

Incompressible

0 iF

Laminar/

Turbulent

, Flows

Compressibility Viscosity Vapor

Pressure

Density

Pressure BuoyancyStability

Introduction Fluid StaticsFluid Dynamics

Surface

Tension

Fluid Mechanics

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

Measures of Fluid Mass and Weight: Density

v

m

The density of a fluid is defined as mass per unit 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:

1

m = mass, and v = volume.

Measures of Fluid Mass and Weight: Specific Weight

g The specific weight of fluid is its weight per unit volume.

•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

g = local acceleration of gravity, 9.807 m/s2

Measures of Fluid Mass and Weight: Specific Gravity

OH

SG2

The specific gravity of fluid is the ratio of the density of the fluid to the density of water @ 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

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

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 alongany 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.

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, is the surface tension, p is the pressure difference between the inside and outside pressure.

The force developed around the edge due to surface tension along the line:

RFsurface 2This force is balanced by the pressure difference p:

2RpFpressure Applied to Area

Applied to Circumference

Surface Tension: Liquid Drop

Now, equating the Surface Tension Force to the Pressure Force, we can estimate p = 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.

Rp

2

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.

h is the height, R is the radius of the tube, is the angle of contact.

“Wetted” “Non-Wetted”

The weight of the fluid is balanced with the vertical force caused by surfacetension.

Adhesion > Cohesion Cohesion > Adhesion

Adhesion

CohesionAdhesion

Cohesion

Surface Tension: Capillary Action

Rh

cos2

cos2 RFsurface

Free Body Diagram for Capillary Action for a Wetted Surface:

For clean glass in contact with water, 0°, and thus as R decreases, h increases, giving a higher rise.

For a clean glass in contact with Mercury, 130°, and thus h is negative or there is a push down of the fluid.

hRW 2

Equating the two and solving for h:

PressurePressurePressure 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.

FA

Pressure = Unit of pressure is Pa

mgF

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 Pascal’s law

Force Equilibrium of a Fluid Element

Fluid surfaces

Figure 2.1 Pressure acting uniformly in all directions

Figure 2.2: Direction of fluid pressures on boundaries

Standard AtmosphereStandard AtmosphereP

atmP

atm

P-vapor

Mercury

h = 76 cm

Mercury barometer

1 atm = 101325 Pa

= 760 mmHg

= 760 torr

= 1 bar

Pabs Pgages Patm+=

Pgages

Patm

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 earth’s 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.

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.

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)

Pressure Measurement

Pressure MeasurementWe 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 p1=p0=0), 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).

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

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.

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

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

Piezometer tube

Open

Pabs Pgages Patm+=

hghP

)(

/81.9

)/(2

3

mfluidoflevelhigthh

smg

mkgDensity

U-tube manometerPressure 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

ghP

)(

/81.9

)/(2

3

mfluidoflevelhigthh

smg

mkgDensity

FB = Wfluid

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.

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

Strain and Strain (Shear) Rate

• Strain– a dimensionless quantity representing the

relative deformation of a material

Normal Strain Shear Strain

Shear Stress is the intensity of force per unit area, acting tang

Simple Shear Flow

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)

Fluid Viscosity

• Newtonian fluids– viscosity is constant (Newtonian viscosity, )

• Non-Newtonian fluids– shear-dependent viscosity (apparent

viscosity, or a)

)(f

Viscosity: Introduction

dy

duor

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”

ViscosityViscosity

dy

dv

y = y-max

y = 0

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

dy

dv

= Coefficient Viscosity ( Pa s)

Shear rate

She

ar s

tres

s

dy

dv

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

1 poise = 0.1 Pa s

1 cP = 1 m Pa s

Example : Shear stress in soybean oilExample : 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?

Kinematics ViscosityKinematics Viscosity

Kinematics Viscosity = Dynamic viscosity

Density

The unit of kinematics viscosity are m /s2

Reynolds NumberReynolds Number

Kinematics = Inertial Forces

Viscous Forces

Dv

Re

D

mRe

o

4

= Density of liquid Pa s

= Density of liquid kg / m 3

V = Flow Velocity m/s

D = Diameter of the pipe m

m = mass flow rate kg /seco

Experiment for find Re

Laminar flow Turbulent flow

2100Re 000,10Re

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 @ 20 C

Milk is flowing at 0.12 m min in a 2.5-cm diameter pipe. If the temperature of the milk is 21°C, is the flow turbulent or streamline

3 -1

Given Density of milk = 1030 kg /m 3

Viscosity = 2.12 cP

)( 22 Pr r

)( 22 Pr)( 1

2 Pr

)2( rL

)2()( 122 rLPPr

)2(2 rLrxP

The velocity profile for a larminar , fully develop flow in horizontal pipe is

22

14

)(R

r

L

RPrv

Velocity Profile in a Liquid Flowing under Fully Developed Flow ConditionVelocity Profile in a Liquid Flowing under Fully Developed Flow Condition

Circumferential area rLA 2

Measurement of Viscosity

Capillary Tube ViscosityRotational Viscometer

Viscosity of a liquid can be measurement

Viscosity: MeasurementsA 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 N·s/m2 or lb·s/ft2

Movie Example using a Viscosimeter:

Capillary Tube Viscometer

Rotational Viscometer

Influence of Temperature on Viscosity

Ag

aA TR

EB lnln

There is considerable evidence that the influence of temperature on viscosity for liquid food may be described by an Arrhenius type relationship

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

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.

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.

Bernoulli EquationBernoulli Equation

Constant2

2

zg

v

g

P

Bernoulli’s 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

5 m

Location 2

2

222

1

211

22z

g

v

g

Pz

g

v

g

P

)(2 122 zzgv

sm

xxv

/9.9

)05(81.922

Then volumetric flow rate from the discharge port using

3 m

D = 10 cm

Location 1

Volume of tank

2

22

3.35

5344

m

xxLD

Time to empty the tank = 35.3 m

0.0078 m / s3

3

= 452.6 sec or 7.5 min

Q = A V = smxx /9.91.04

2= 0.0078 m /s3

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

2

222

1

211

22z

g

v

g

Pz

g

v

g

P

222 54.351.1

2

100070 kPaP

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

22

2112 2

vvPP

P2 = 65.3 k Pa.

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

Forces due to FrictionWhen 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

=

½v2

Major loss

DLu

2f P

E2

ff

Dg

Lh f

2u

2f

f = ReN16

Du

16

Type of pipe 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”

D

Moody Diagram for the Fanning friction factorMoody Diagram for the Fanning friction factor

Moody Diagram for the Moody friction factorMoody Diagram for the Moody friction factor

Laminar Zone

Turbulent Zone

Lamina Zone

Turbulent Zone

2

2Re74.5

7.3ln

33137.0

D

f

Re

16f

Laminar

Transition

m = 2.5 kg/so

30 m

Re = 550,128)025.0)(000792.0(

/2442

0

xsPa

skgx

D

m

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

Density and viscosity of water@30 C are 997.5 kg/m and 0.000792 Pa s3

36

10828.1025.0

107.45

xx

D

f = 0.006

128550

1.828 x 10-3

)092.4)(7.995)(025.0/30(006.022 2 xxvD

LfP

= 240.08 kPa

smmkg

skg

A

mv /092.4

)025.04

)(/7.995(

/5.2

23

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.

De4 Hydraulic radius=

De =

21

21

22 444

DD

DD

b

a

a x b

2 ( a + b)De = 4

4(Cross-sectional area)

Wetted Perimeter=

For Non-circular pipe

1212

21

22 )(

DDDD

DD

=

De =

12

21

22

4

)4

(444

dD

dD

d1 = 10 cm

D 2 = 1.2 m

De =

12

21

22

4

)(4

dD

dD

De =

)1.0(42.1

)1.0(42.1 22

= 0.875 m

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

D e =

12

21

22 )(

dD

dD

Determine Equivalent Diameter

5 cm2.5 cm

=

025.005.0

)025.0(05.0 22

= 0.025 m

550,128)025.0)(000792.0(

/24Re

2

xsPa

skgxDv

36

10828.1025.0

107.45

xx

D

f = 0.006

128550

1.828 x 10-3

)092.4)(7.995)(025.0/30(006.022 2 xxvD

LfP

e

= 240.08 kPa

Determine Reynolds Number and

Determine Friction factor

Energy Equation for Steady Flow of Fluids

212

22

221

11

21

22 L,hγ

pzy

g

v

γ

pzy

g

v

velocity head

elevation head

pressure head

Energy loss between sections 1 and 2

Frictional Energy LossThe frictional energy loss for a liquid flowing in pipe is composed of major and minor loses

EE f EE f , MinorEE f , Major+=

The major losses are due to the flow of viscous liquid in the straight portions of a pipe

The minor losses are due to various components used in pipeline system such as values , tees and elbow and contraction of fluid

EE fittingEE contraction +EE Minor EE expansion= +

General Equation for Minor Losses

g

VKh mL 2

2

Dg

fLeVh mL

22

hLm = minor loss

K = minor loss coefficient

Le = equivalent length

Expansion and Contraction

width 2

1 2

width 1b 1 b 2

Contraction b1 > b2

Expansion b1 < b2

Transition Loss Equations

• Contraction:

• Expansion:

g

V

A

A

g

VKh i 2

)1(55.02

22

1

22

2

g

V

A

A

g

VKh e 2

)1(2

212

2

12

1

Pipe Bends

Friction loss of turbulent flow through valves and fittings

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.

6.1 m

38.1 m

3 m

15.2 m

PumpPower 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/

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)

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

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

P`2 = P`1(N2 /N1)3

Where Vº = volumetric flow rate

h = total head

P` = power

Positive displacement

• Constant discharge pressure at difference flow rates.

• Regulation of flow rate is done by changing displacement or capacity of intake chamber

Characteristic curve• plot head, power consumption and efficiency with respect to volumetric flow rate (capacity)

• use for rating pumps

EfficiencyHead

Power requirement

Peak Efficiency

Volumetric flow rate (capacity)

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)

Assumption V

Determine Re

Determine “ f ”

Determine “ V ”

V new V old

V newV old

YES

NO

Trial and error type solution

For finding v

P

1

2

12 m

Constant level 5 m

Example :Trial and error type solution

For finding v

g

v

D

Lfz

g

v

g

Phz

g

v

g

PP

2

2

222

1

211 2

22

g

v

D

Lfz

g

vh P

2

2

22 2

2

22

22

81.905.0

3025

)81.9(212 v

xf

v

Constant level

2)5.12205.0(7 vf

1 Equation 2 Variable cannot determine

fv

5.12205.0

7

V newV old

Assumption V

Determine Re

Determine “ f ”

Determine “ V ”

V new V old YES

NO

fv

5.12205.0

7

Re =75361

YES

NO

Guess V = 1.5

f =0.0057

V new =3.059

1.5 =3.059

v = 3.059

Re =153678

YES

NO

Guess V =3.059

f =0.0053

V new =3.164

3.059 = 3.164

v = 3.164

Re = 158964

YES

NO

Guess V = 3.164

f = 0.0053

V new = 3.164

3.164= 3.164

V =3.164 m/s

30 m

P Diameter of pipe = 2.54 cm

Power = 2 HP

75 % efficiency

Whole Milk

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

g

v

D

Lfz

g

v

g

Phz

g

v

g

PP

2

2

222

1

211 2

22

g

v

D

LfhP

2

2Qx

hgHP p

746

0.75 x 2 HP = 1030 x 9.81 x h

746v)0254.0(

42

p

h p = 218.86v

218.86v

240.8 f=3

218.86v

2 x f x 30 x v

0.0254 x 9.81=

2

v = 0.91

f

1/3

Re = 12951.5

YES

NO

Guess V = 1

f = 0.007

V new = 5.06

1 = 5.06

v = 5.06

Re = 65534

YES

NO

Guess V = 5.06

f = 0.0052

V new = 5.59

5.06 = 5.59

v = 5.59

Re = 72399

YES

NO

Guess V = 5.59

f = 0.005

V new = 5.66

5.59 ~ 5.66

V = 5.66 m/s

Flow MeasurementFlow 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.

The Pitot TubeThe Pitot Tube

2

2

1

2

22z

g

v

g

Pz

g

v

g

P bbaa

air

aba

PPV

)(2

The principle is based on the Bernoulli Equation where each term can be interpreted as a form of pressure

The Pitot tube is a widely used sensor to measure velocity of fluid

Static Pressure

Total pressure

Water

Air

air

abd

PPC

)(2

C d = Discharge Coefficient

Stagnation point (v = 0)

waterd HC 11.129

air

waterwaterd

aird

HgC

PPCV

2)(2 12

1

The Orifice MeterThe 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.

4

1

21

)(2

DD

PPCV

f

BAd

D1 D2

PA - PB

The Venturi MeterThe Venturi Meter

2/1

4

1

21

12

DD

zg

CVf

f

d

To reduce energy loss due to friction created by the sudden contraction in flow in an orifice meter

Variable-Area MeterVariable-Area MeterVariable 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.