FLUID FLOW THEORY.docx

8/20/2019 FLUID FLOW THEORY.docx

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FLUID FLOW THEORY

Many raw materials for foods and many finished foods are in the form of fluids. These fluids have to betransported and processed in the factory. Food technologists must be familiar with the principles that govern

the flow of fluids, and with the machinery and equipment that is used to handle fluids. In addition, there is anincreasing tendency to handle powdered and granular materials in a form in which they behave as fluids.Fluidization, as this is called, has been developed because of the relative simplicity of fluid handling comparedwith the handling of solids.

The engineering concept of a fluid is a wider one than that in general use, and it covers gases as well asliquids and fluidized solids. This is because liquids and gases obey many of the same laws so that it isconvenient to group them together under the general heading of fluids.

The study of fluids can be divided into the study of fluids at rest - fluid statics, and the study of fluids in motion- fluid dynamics. For some purposes, further subdivision into compressible fluids such as gases, andincompressible fluids such as liquids, is necessary. Fluids in the food industry vary considerably in theirproperties. They include such materials as

Thin liquids - mil!, water, fruit "uices,Thic! liquids - syrups, honey, oil, "am,#ases - air, nitrogen, carbon dio$ide,Fluidized solids - grains, flour, peas.

F%&I' (T)TI*(

) very important property of a fluid at rest is the pressure e$erted by that fluid on its surroundings, that is the

fluid pressure

+ressure is defined as force e$erted on an area. &nder the influence of gravity, a mass of any material e$ertsa force on whatever supports it. The magnitude of this force is equal to the mass of the material multiplied bythe acceleration due to gravity. The mass of a fluid can be calculated by multiplying its volume by its density,which is defined as its mass per unit volume. Thus the equation can be written

F = mg = V ρg

where F is the force e$erted, m is the mass, g the acceleration due to gravity, V the volume and ρ the #ree!letter rho the density. The units of force are ewtons, or !g m s-/. and of pressure +ascals, one +ascal beingone ewton m-/, and so one +ascal is also one !g m-0 s-/.

For a mass to remain in equilibrium, the force it e$erts due to gravity must be resisted by some supportingmedium. For a weight resting on a table, the table provides the supporting reaction1 for a multi-storey building,the upper floors must be supported by the lower ones so that as you descend the building the burden on thefloors increases until the foundations support the whole building. In a fluid, the same situation applies.

%ower levels of the fluid must provide the support for the fluid that lies above them. The fluid at any point mustsupport the fluid above. )lso, since fluids at rest are not able to sustain shearing forces, which are forcestending to move ad"acent layers in the fluid relative to one another, it can be shown that the forces at any point


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in a fluid at rest are equal in all directions. The force per unit area in a fluid is called the fluid pressure. It ise$erted equally in all directions.

*onsider a horizontal plane in a fluid at a depth Z below the surface, as illustrated in Fig. 3.1.

Figure 3.1 Pressure in a fluid

If the density of the fluid is ρ, then the volume of fluid lying above an area A on the plane is ZA and the weight

of this volume of fluid, which creates a force e$erted by it on the area A which supports it, is Z ρ Ag . 2ut thetotal force on the area A must also include any additional force on the surface of the liquid. If the force on thesurface is P s per unit area,

F = AP s + Z ρ Ag 3.0

where F is the total force e$erted on the area A and P s is the pressure above the surface of the fluid e.g. itmight be atmospheric pressure. Further, since total pressure P is the total force per unit area,

P = F/A = P s + Z ρg 3./

In general, we are interested in pressures above or below atmospheric. If referred to zero pressure as datum,the pressure of the atmosphere must be ta!en into account. 4therwise the atmospheric pressure represents adatum or reference level from which pressures are measured. In these circumstances we can write

P = Z ρg 3.3

This may be considered as the fundamental equation of fluid pressure. It states that the product of the densityof the fluid, acceleration due to gravity and the depth gives the pressure at any depth in a fluid.

EXAPLE 3.1. T!"al pressure in a "an# !f peanu" !il*alculate the greatest pressure in a spherical tan!, of / m diameter, filled with peanut oil of specific gravity5.6/, if the pressure measured at the highest point in the tan! is 75 !+a.


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'ensity of water 8 0555 !g m-3

'ensity of oil 8 5.6/ $ 0555 !g m-3 8 6/5 !g m-3

Z 8greatest depth 8 / mand g 8 6.90 m s-/

ow P = Z ρg 8 / $ 6/5 $ 6.90 !g m-0 s-/

8 09,5:5 +a 8 09.0 !+a.

To this must be added the pressure at the surface of 75 !+a.

Total pressure 8 75 ; 09.0 8 99.0 !+a.

ote in E$a%ple 3.1, the pressure depends upon the pressure at the top of the tan! added to the pressuredue to the depth of the liquid1 the fact that the tan! is spherical or any other shape ma!es no difference tothe pressure at the bottom of the tan!.

In the previous paragraph, we established that the pressure at a point in a liquid of a given density is solelydependent on the density of the liquid and on the height of the liquid above the point, plus any pressure whichmay e$ist at the surface of the liquid. 6.90 8 35 tonnes of water.

+ressures are sometimes quoted as a&s!lu"e pressures and this means the total pressure includingatmospheric pressure. More usually, pressures are given as gauge pressures, which implies the pressureabove atmospheric pressure as datum. For e$ample, if the absolute pressure is given as 3:5 !+a, the gauge

pressure is 3:5 - 055 8 /:5 !+a assuming that the atmospheric pressure is 055 !+a. These pressureconversions are illustrated in Fig. 3.'.


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Figure 3.' Pressure (!n)ersi!ns

(tandard atmospheric pressure is actually 050.3 !+a but for our practical purposes 055 !+a is sufficientlyclose and most convenient to use. )ny necessary ad"ustment can easily be made.

)nother commonly used method of e$pressing pressures is in terms of ?*ead? of a particular fluid. From eqn.3.3 it can be seen that there is a definite relationship between pressure and depth in a fluid of given density.Thus pressures can be e$pressed in terms of depths, or heads as they are usually called, of a given fluid.

The two fluids most commonly used, when e$pressing pressures in this way, are water and mercury. The mainreason for this method of e$pressing pressures is that the pressures themselves are often measured byobserving the height of the column of liquid that the pressure can support. It is straightforward to convertpressures e$pressed in terms of liquid heads to equivalent values in !+a by the use of eqn. 3.3..

EXAPLE 3.'. Head !f Wa"er *alculate the head of water equivalent to standard atmospheric pressure of 055 !+a.

'ensity of water 8 0555 !g m-3,g 8 6.90 m s-/

and pressure 8 055 !+a 8 055 $ 053 +a 8 055 $ 053 !g m-0s-/.but from eqn. 3.3

Z = P/ ρg 8 055 $ 053> 0555 $ 6.90 8 05./ m


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EXAPLE 3.3. Head !f %er(ur+ *alculate the head of mercury equivalent to a pressure of two atmospheres.

'ensity of mercury 8 03,@55 !g m-3

A 8 / $ 055 $ 053

> 03,@55 $ 6.90 8 0.:m

F%&I' 'B)MI*(

Mass balance

Cnergy balance

+otential energyDinetic energy+ressure energy Friction loss Mechanical energy 4ther effects 2ernouilliEs equation

In most processes fluids have to be moved so that the study of fluids in motion is important. +roblems on theflow of fluids are solved by applying the principles of conservation of mass and energy. In any system, or inany part of any system, it must always be possible to write a mass balance and an energy balance. Themotion of fluids can be described by writing appropriate mass and energy balances and these are the basesfor the design of fluid handling equipment.

ass ,alan(e

*onsider part of a flow system, such for e$ample as that shown in Fig. 3.3.

This consists of a continuous pipe that changes its diameter, passing into and out of a unit of processing plant,which is represented by a tan!. The processing equipment might be, for e$ample, a pasteurizing heate$changer. )lso in the system is a pump to provide the energy to move the f luid.

http://www.nzifst.org.nz/unitoperations/flfltheory2.htm#masshttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#energyhttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#energyhttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#potentialhttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#kinetichttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#pressurehttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#pressurehttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#frictionhttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#frictionhttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#mechanicalhttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#otherhttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#bernouillihttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#energyhttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#potentialhttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#kinetichttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#pressurehttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#frictionhttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#mechanicalhttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#otherhttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#bernouillihttp://www.nzifst.org.nz/unitoperations/flfltheory2.htm#mass


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Fiigure 3.3. ass and energ+ &alan(e in fluid fl!-

In the flow system of Fig. 3.3 we can apply the law of conservation of mass to obtain a mass balance. 4nce

the system is wor!ing steadily, and if there is no accumulation of fluid in any part the system, the quantity offluid that goes in at section 0 must come out at section /. If the area of the pipe at section 0 is A0 , the velocity

at this section, v 0 and the fluid density ρ0, and if the corresponding values at section / are A/, v /,

ρ/, the mass balance can be e$pressed as

ρ0 A0v 0 = ρ/ A/v / 3.=

If the fluid is incompressible ρ0 8 ρ/ so in this case

A0v 0 = A/v / 3.:

Cquation 3.: is !nown as the (!n"inui"+ eua"i!n for liquids and is frequently used in solving flow problems.It can also be used in many cases of gas flow in which the change in pressure is very small compared with thesystem pressure, such as in many air-ducting systems, without any serious error.

EXAPLE 3./. 0el!(i"ies !f fl!-


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This e$pression can be substituted for v 2 in the mass balance equation to give

ρ0 A0v 0 = ρ/ A/(A0v 0 – A3v 3 )/A/ + ρ3 A3v 3

ρ1 A0v 0 = ρ/ A0v 0 - ρ/ A3v 3 + ρ3 A3v 3.

So A0v 0( ρ0 - ρ/ ) = A3v 3( ρ3 - ρ/ ) b

From the !nown facts of the problem we have

A0 8 π>= $ 5.5:/ 8 0.6@ $ 05-3 m/

A/ = A3 8 π>= $ 5.5// 8 3.0= $ 05-= m/

v 0 8 5.// m s-0

ρ0 8 0.53: $ ρw, ρ/ 8 0.5= $ ρw , ρ3 8 0.50 $ ρw

where ρw is the density of water.

(ubstituting these values in eqn. b above we obtain

-0.6@ $ 05-3 $ 5.// 5.55: 8 -3.0= $ 05-= $ v 3 $ 5.53

so v 3 8 5./3 m s-0

)lso from eqn. a we then have, substituting 5./3 m s-0 for v 3,

v / 8 0.6@ $ 05-3 $ 5.// - 3.0= $ 05 -= $ 5./3G > 3.0= $ 05 -=

8 0.0m s-0

Energ+ ,alan(e

In addition to the mass balance, the other important quantity we must consider in the analysis of fluid flow, isthe energy balance. Heferring again to Fig. 3.3, we shall consider the changes in the total energy of unit massof fluid, one !ilogram, between (ection 0 and (ection /.

Firstly, there are the changes in the intrinsic energy of the fluid itself which include changes in 0 +otential energy. / Dinetic energy. 3 +ressure energy.

(econdly, there may be energy interchange with the surroundings including = Cnergy lost to the surroundings due to friction. : Mechanical energy added by pumps. @ eat energy in heating or cooling the fluid.

In the analysis of the energy balance, it must be remembered that energies are normally measured from adatum or reference level. 'atum levels may be selected arbitrarily, but in most cases the choice of a


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convenient datum can be made readily with regard to the circumstances.

P!"en"ial energ+

Fluid maintained above the datum level can perform wor! in returning to the datum level. The quantity of wor!it can perform is calculated from the product of the distance moved and the force resisting movement1 in thiscase the force of gravity. This quantity of wor! is called the potential energy of the fluid.Thus the potential energy of one !ilogram of fluid at a height of Z m above its datum is given by E p, where

E p = Zg J

ine"i( energ+

Fluid that is in motion can perform wor! in coming to rest. This is equal to the wor! required to bring a bodyfrom rest up to the same velocity, which can be calculated from the basic equation

v / = /as, therefore s = v / / /a,

where v m s-0 is the final velocity of the body, a m s-/ is the acceleration and s m is the distance the bodyhas moved.

)lso wor! done 8 W = F $ s, and from ewtonEs (econd %aw, for m !g of fluid

F = ma

and so E k = W = mas = mav / / /a = mv / / /

The energy of motion, or !inetic energy, for 0 !g of fluid is therefore given by E k where

E k = v / / / J.

Pressure energ+

Fluids e$ert a pressure on their surroundings. If the volume of a fluid is decreased, the pressure e$erts a forcethat must be overcome and so wor! must be done in compressing the fluid. *onversely, fluids under pressurecan do wor! as the pressure is released. If the fluid is considered as being in a cylinder of cross-sectional area

A m/ and a piston is moved a distance L m by the fluid against the pressure P +a the wor! done is PAL

"oules. The quantity of the fluid performing this wor! is ALρ !g. Therefore the pressure energy that can beobtained from one !g of fluid that is the wor! that can be done by this !g of fluid is given by E r where

E r = PAL / ALρ

= P/ ρ ( J )

Fri("i!n l!ss


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come from energy contained in the fluid and so frictional losses provide a drain on the energy resources of thefluid. The actual magnitude of the losses depends upon the nature of the flow and of the system throughwhich the flow ta!es place. In the system of Fig. 3.3, let the energy lost by 0 !g f luid between section 0 andsection /, due to friction, be equal to E ƒ J.

e(*ani(al energ+

If there is a machine putting energy into the fluid stream, such as a pump as in the system of Fig. 3.3, themechanical energy added by the pump per !g of fluid must be ta!en into account. %et the pump energy addedto 0 !g fluid be E J. In some cases a machine may e$tract energy from the fluid, such as in the case of awater turbine.

O"*er effe("s

eat might be added or subtracted in heating or cooling processes, in which case the mechanical equivalentof this heat would require to be included in the balance. *ompressibility terms might also occur, particularly

with gases, but when dealing with low pressures only they can usually be ignored.

For the present let us assume that the only energy terms to be considered are E p, E k , E r , E ! , E .

,ern!uilli2s Eua"i!n


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for fluids. In fact much of the remainder of this chapter will be concerned with applying one or another aspectof these equations.

The 2ernouilli equation is of sufficient importance to deserve some further discussion. In the form in which ithas been written in eqn. 3.6 it will be noticed that the various quantities are in terms of energies per unit

mass of the fluid flowing. If the density of the fluid flowing multiplies both sides of the equation, then we havepressure terms and the equation becomes

ρZg + ρv / / / + P = k" 3.05

and the respective terms are !nown as the potential head pressure, the velocity pressure and the staticpressure.

4n the other hand, if the equation is divided by the acceleration due to gravity, g , then we have an e$pressionin terms of the head of the fluid flowing and the equation becomes

Z + v / / /g + P/ ρg = k"" 3.00

and the respective terms are !nown as the potential head, the velocity head and the pressure head.The most convenient form for the equation is chosen for each particular case, but it is important to beconsistent having made a choice.

If there is a constriction in a pipe and the static pressures are measured upstream or downstream of theconstriction and in the constriction itself, then the 2ernouilli equation can be used to calculate the rate of flowof the fluid in the pipe. This assumes that the flow areas of the pipe and in the constriction are !nown.*onsider the case in which a fluid is flowing through a horizontal pipe of cross-sectional area A0 and then itpasses to a section of the pipe in which the area is reduced to A/. From the continuity equation eqn. 3.:Gassuming that the fluid is incompressible

A0v 0 = A/v /and so v / = v 0 A0 /A/

(ince the pipe is horizontal

Z 0 = Z /

(ubstituting in eqn. 3.9

v 0/ /2 + P 0 / ρ0 = v 0/ A0/ /( / A// ) + P / / ρ/

and since ρ0 8 ρ/ as it is the same fluid throughout and it is incompressible,

P 0 - P / = ρ0 v 0/((A0/ /A// - 0 )/ /. 3.0/

From eqn. 3.0/, !nowing P 0, P /, A0, A/, ρ0, the un!nown velocity in the pipe, v 0, can be calculated.

)nother application of the 2ernouilli equation is to calculate the rate of flow from a n!le with a !nownpressure differential. *onsider a nozzle placed in the side of a tan! in which the surface of the fluid in the tan!


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is A ft above the centre line of the nozzle as illustrated in Fig. 3./.

FI4. 3./. Fl!- fr!% a n!le.

Ta!e the datum as the centre of the nozzle. The velocity of the fluid entering the nozzle is appro$imately zero,as the tan! is large compared with the nozzle. The pressure of the fluid entering the nozzle is +0 and the

density of the fluid ρ0. The velocity of the fluid flowing from the nozzle is v / and the pressure at the nozzle e$itis 5 as the nozzle is discharging into air at the datum pressure. There is no change in potential energy as thefluid enters and leaves the nozzle at the same level. ρ0 8 gZ where Z is the head of fluid above the nozzle therefore

v / 8 3.03/ gZ

EXAPLE 3.5. Pressure in a pipe


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8 =.=/ $ 05-3 m/.(o velocity of flow in 7.: cm diameter pipe, v 0 8 5.=>@5>=.=/ $ 05

-3 8 0.:0 m s-0

)rea of : cm diameter pipe 8 π>=5.5:/

8 0.6@ $ 05-3 m/

and so velocity of flow in : cm diameter pipe, v / 8 5.=>@5>0.6@ $ 05

-3 8 3.= m s-0

ow

Z 0g + v 0/ / / + P 0 / ρ0 = Z /g + v // / / + P / / ρ/

and so5 ; 0.:0/>/ ; 75 $ 053>0555 8 5 ; 3.=/>/ ; P />0555

5 ; 0.0 ; 75 8 5 ; :.9 ; +/>0555 P />0555 8 70.0 - :.9 8 @:.3

P / 8 @:.3! +a.

EXAPLE 3.6. Fl!- ra"e !f !li)e !il4live oil of specific gravity 5.6/ is flowing in a pipe of / cm diameter. *alculate the flow rate of the olive oil, ifan orifice is placed in the pipe so that the diameter of the pipe in the constriction is reduced to 0./ cm, and ifthe measured pressure difference between the clear pipe and the most constricted part of the pipe is 9 cm ofwater.'iameter of pipe, in clear section, equals / cm and at constriction equals 0./ cm.

A0> A/ 8 #0>#// 8 />0.//

'ifferential head 8 9 cm water.

'ifferential pressure 8 €€€€ Z ρg 8 5.59 $ 0555 $ 6.90 8 79: +a.

substituting in eqn. 3.0/

79: 8 5.6/ $ 0555 $ v / />0./= - 0 G >/

v / 8 79:>3560 v 8 5.: m s-0

EXAPLE 3.7. ass fl!- ra"e fr!% a "an#

The level of water in a storage tan! is =.7 m above the e$it pipe. The tan! is at atmospheric pressure and thee$it pipe discharges into the air. If the diameter of the e$it pipe is 0./ cm what is the mass rate of flow through

this pipeK

From eqn. 3.03

v 8/ gZ

v 8/ $ 6.90 $ =.7

8 6.@ m s-0.


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ow area of pipe, A

8 π>=#/

8 π>= $ 5.50//

8 0.03 $ 05-= m/

Lolumetric flow rate, Av

8 0.03 $ 05-= m/ $ 6.@ m s-0

8 0.03 $ 05-= $ 6.@ m3 s-0

8 0.59 $ 05-3 m3 s-0

Mass flow rate, ρ Av

8 0555 !g m-3 $ 0.59 $ 05-3 m3 s-0

8 0.59 !g s-0

EXAPLE 3.8. Pu%p *!rsep!-er

= $ 5.57:/ 8 =.=/ $ 05-3 m/,

Lelocity in pipe, v

8 /.7 $ 05-/>=.=/ $ 05-3 8 @ m s-0,

)nd so applying eqn. 3.7 E = Z g ; v

/>/

E c 8 3: $ 6.90 ; @/>/

8 3=3.= ; 09 8 3@0.= J

Therefore total power required

8 E c $ mass rate of flow

8 E cV ρ 8 3@0.= $ /.7 $ 05-/ $ 0555 J s-0 8 67:9 J s-0


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and, since 0 h.p. 8 7.=@ $ 05/ J s-0,

required power 8 03 h.p.

LI(*4(ITB

ewtonian and on-ewtonian Fluids

Liscosity is that property of a fluid that gives rise to forces that resist the relative movement of ad"acent layersin the fluid. 0is(!us f!r(es are of the same character as s*ear f!r(es in solids and they arise from forcesthat e$ist between the molecules.

If two parallel plane elements in a fluid are moving relative to one another, it is found that a steady force mustbe applied to maintain a constant relative speed. This force is called the viscous drag because it arises fromthe action of viscous forces.*onsider the system shown in Fig. 3.5.

Figure 3.5. 0is(!us f!r(es in a fluid.

If the plane elements are at a distance Z apart, and if their relative velocity is v , then the force F required tomaintain the motion has been found, e$perimentally, to be proportional to v and inversely proportional to A formany fluids. The coefficient of proportionality is called the viscosity of the fluid, and it is denoted by the symbol

µ mu.

From the definition of viscosity we can write

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F/A = µv/Z 3.0=

where F is the force applied, A is the area over which force is applied, Z is the distance between planes, v is

the velocity of the planes relative to one another, and µ is the viscosity.

2y rearranging the eqn. 3.0=, the dimensions of viscosity can be found.

µG 8FZ

8

FG%GtG

8

FGtG

Av %/G%G %G/

= MG%G-0tG-0

There is some ambivalence about the writing and the naming of the unit of viscosity1 there is no doubt aboutthe unit itself which is the s m-/, which is also the +ascal second, +a s, and it can be converted to massunits using the basic mass>force equation. The older units, the poise and its sub-unit the centipoise, seem tobe obsolete, although the conversion is simple with 05 poises or 0555 centipoises being equal to 0 s m -/,and to 0 +a s.The new unit is rather large for many liquids, the viscosity of water at room temperature being around 0 $ 05 -3 s m-/ and for comparison, at the same temperature, the appro$imate viscosities of other liquids are acetone,5.3 $ 05-3 s m-/1 a tomato pulp, 3 $ 05 -31 olive oil, 055 $ 05-31 and molasses 7555 s m -/.

Liscosity is very dependent on temperature decreasing sharply as the temperature rises. For e$ample, theviscosity of golden syrup is about 055 s m -/ at 0@N*, =5 at //N* and /5 at /:N*. *are should be ta!en not to

confuse viscosity µ as defined in eqn. 3.0= which strictly is called the dynamic or absolute viscosity, with µ>ρwhich is called the !inematic viscosity and given another symbol. In technical literature, viscosities are oftengiven in terms of units that are derived from the equipment used to measure the viscosities e$perimentally.The fluid is passed through some form of capillary tube or constriction and the time for a given quantity topass through is ta!en and can be related to the viscosity of the fluid. Tables are available to convert thesearbitrary units, such as ?(aybolt (econds? or ?Hedwood (econds?, to poises.

The viscous properties of many of the fluids and plastic materials that must be handled in food processingoperations are more comple$ than can be e$pressed in terms of one simple number such as a coefficient ofviscosity.

9e-"!nian and 9!n:9e-"!nian Fluids

From the fundamental definition of viscosity in eqn. 3.0= we can write

F/A = µv /Z 8 µ dv >d$ 8 τ

where τ tau is called the shear stress in the fluid. This is an equation originally proposed by ewton andwhich is obeyed by fluids such as water. owever, for many of the actual fluids encountered in the foodindustry, measurements show deviations from this simple relationship, and lead towards a more generalequation

τ 8 k dv >d$ % 3.0:


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which can be called the p!-er:la- eua"i!n, and where k is a constant of proportionality.


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(THC)M%IC )' T&H2&%CT F%4<


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'iameter of pipe 8 5.5/: m.

*ross-sectional area of pipe 8 π>=#/

8 π>= $ 5.5/:/

8 =.6 $ 05-= m/

Hate of flow 8 5.0/ m3 min-0 8 5.0/>@5m3 s-0 8 / $ 05 m3 s-0

(o velocity of flow 8 / $ 05-3>=.6 $ 05-= 8 =.0 m s-0,

and so He 8 #v ρ / µ 8 5.5/: $ =.0 $ 05/6>/.0 $ 05-3

8 :5,/35and this is greater than =555 so that the flow is turbulent.

)s He is a dimensionless ratio, its numerical value will be the same whatever consistent units are used.owever, it is important that consistent units be used throughout, for e$ample the (I system of units as areused in this boo!. If1 for e$ample, cm were used instead of m "ust in the diameter or length term only, thenthe value of He so calculated would be greater by a factor of 055. This would ma!e nonsense of any

deductions from a particular numerical value of He.4n the other hand, if all of the length terms in He, and this includes not only # but also v m s-0, ρ

!g m-3 and µ s m-/, are in cm then the correct value of He will be obtained. It is convenient, but notnecessary to have one system of units such as (I. It is necessary, however, to be consistent throughout.

CCH#B %4((C( I F%4<

Friction in +ipes

Cnergy %osses in 2ends and Fittings+ressure 'rop through CquipmentCquivalent %engths of +ipe*ompressibility Cffects for #ases*alculation of +ressure 'rops in Flow (ystems

Cnergy losses can occur through friction in pipes, bends and fittings, and in equipment.

Fri("i!n in Pipes

In 2ernouilliEs equation the symbol E ƒ was used to denote the energy loss due to friction in the pipe. This lossof energy due to friction was shown, both theoretically and e$perimentally, to be related to the Heynoldsnumber for the flow. It has also been found to be proportional to the velocity pressure of the fluid and to afactor related to the smoothness of the surface over which the fluid is flowing.

If we define the wall friction in terms of velocity pressure of the fluid flowing, we can write

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F/A = ! ρv />/ 3.0@

where F is the friction force, A is the area over which the friction force acts, ρ is the density of the fluid, v is thevelocity of the fluid, and ! is a coefficient called the friction factor.

*onsider an energy balance over a differential length, dL, of a straight horizontal pipe of diameter #, as in Fig.3.7.

Figure 3.7. Energ+ &alan(e !)er a leng"* !f pipe.

*onsider the equilibrium of the element of fluid in the length dL. The total force required to overcome frictiondrag must be supplied by a pressure force giving rise to a pressure drop dP along the length dL.

The pressure drop force is

dP $ )rea of pipe 8 dP $ π#/>=The friction force is force>unit area $ wall area of pipe

8 F/A $ π# $ dL

so from eqn. 3.0@, 8 ! ρv />/ $ π# $ dL

Therefore equating prressure drop and friction force

π#/>= dP 8 ! ρv />/ π# $ dL,therefore

dP 8 =! ρv />/ $ dL/#

Integrating between L0 and L/, in which interval P goes from P 0 to P / we have

dP 8 =! ρv />/ $ dL/#

P 0 - P / 8 =! ρv/>/L0 - L/>#i.e.

∆P ! 8 =! ρv />/ $ % /# 3.07or

E ƒ 8 ∆P ! >ρ 8 /!v /L/# 3.09

where L = L1 - L2 8 length of pipe in which the pressure drop, ∆P ! 8 P 1 - P 2 is the frictional pressure drop, andE ƒ is the frictional loss of energy.

Cquation 3.07 is an important equation1 it is !nown as the Fanning eua"i!n, or sometimes the 'E)rcy orthe Fanning-'E)rcy equation. It is used to calculate the pressure drop that occurs when liquids flow in pipes.


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The factor ! in eqn.3.07 depends upon the Heynolds number for the flow, and upon the r!ug*ness of thepipe. In Fig. 3.8 e$perimental results are plotted, showing the relationship of these factors. If the Heynoldsnumber and the roughness factor are !nown, then ! can be read off from the graph.

Figure 3.8 Fri("i!n fa("!rs in pipe> )fter Moody,06==?

It has not been found possible to find a simple e$pression that gives analytical equations for the curve of Fig.

3.9, although the curve can be appro$imated by straight lines covering portions of the range. Cquations canbe written for these lines. (ome writers use values for ! which differ from that defined in eqn. 3.0@ bynumerical factors of / or =. The same symbol, ! , is used so that when reading off values for ! , its definition inthe particular conte$t should always be chec!ed. For e$ample, a new f 8 =! removes one numerical factorfrom eqn. 3.07.

Inspection of Fig. 3.9 shows that for low values of He, there appears to be a simple relationship between Qand He independent of the roughness of the pipe. This is perhaps not surprising, as in streamline flow thereis assumed to be a stationary boundary layer at the wall and if this is stationary there would be no liquidmovement over any roughness that might appear at the wall. )ctually, the friction factor f in streamline flowcan be predicted theoretically from the Hagen:P!iseuille eua"i!n, which gives

! 8 0@>He 3.06

and this applies in the region 5 O He O /055.

In a similar way, theoretical wor! has led to equations which fit other regions of the e$perimental curve, fore$ample the ,lasius eua"i!n which applies to smooth pipes in the range 3555 O He O 055,555 and inwhich

! Q 8 5.30@ He-5./: 3.06


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=

In the turbulent region, a number of curves are shown in Fig. 3.9. It would be e$pected that in this region, thesmooth pipes would give rise to lower friction factors than rough ones. The roughness can be e$pressed interms of a roughness ratio that is defined as the ratio of average height of the pro"ections, which ma!e up the?roughness? on the wall of the pipe, to the pipe diameter. Tabulated values are given showing the roughness

factors for the various types of pipe, based on the results of Moody 06==. These factors ε are then dividedby the pipe diameter # to give the roughness ratio to be used with the !!d+ grap*. The question of relativeroughness of the pipe is under some circumstances a difficult one to resolve. In most cases, reasonableaccuracy can be obtained by applying Ta&le 3.1 and Fig. 3.8.

T)2%C 3.0HC%)TILC H4C(( F)*T4H( F4H +I+C(

Material Houghness factor ε Material Houghness factor ε

Hiveted steel 5.550- 5.50 #alvanized iron 5.555/

*oncrete 5.5553 - 5.553 )sphalted cast iron 5.550

=#/

8 π>= $ 5.5:/

8 0.6@ $ 05-3 m/

From )ppendi$ =,

Liscosity of olive oil at /5N* 8 9= $ 05-3 s m-/ and density 8 605 !g m -3, and velocity 8 5.0 $ 0>@5>0.6@ $ 05 -3 8 5.9: m s-0,

ow He 8 #v ρ / µ

8 5.5: $ 5.9: $ 605>9= $ 05-3G 8 =@5

so that the flow is streamline, and from Fig. 3.9, for He 8 =@5

! 8 5.53.


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)lternatively for streamline flow from 3.09, ! 8 0@>He 8 0@>=@5 8 5.53 as before.

)nd so the pressure drop in 075 m, from eqn. 3.07

∆P ! 8 =! ρv />/ $ % /#

8 = $ 5.53 $ 605 $ 5.9:/ $ 0>/G $ 075 $ 0>5.5:G 8 0.3= $ 05: +a 8 03= !+a.

Energ+ L!sses in ,ends and Fi""ings

/ 3./5

where k has to be found for the particular fitting. Lalues of this constant k for some fittings are given in Ta&le3.'.

T)2%C 3./FHI*TI4 %4(( F)*T4H( I FITTI#(

k

Lalves, fully open

gate 5.03

globe @.5

angle 3.5

Clbows

65N standard 5.7=

medium sweep 5.:

long radius 5./:

square 0.:

Tee, used as elbow 0.:

Tee, straight through 5.:

Cntrance, large tan! to pipe


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sharp 5.:

rounded 5.5:

Cnergy is also lost at sudden changes in pipe cross-section. )t a sudden enlargement the loss has beenshown to be equal to

E ! = (v 1 - v 2 )2 >/ 3./0

For a sudden contraction E ! = kv 2

2 >/ 3.//

where v0 is the velocity upstream of the change in section and v/ is the velocity downstream of the change inpipe diameter from #0 to #/.

The coefficient ! in eqn. 3.// depends upon the ratio of the pipe diameters #/>#0 as given in Ta&le 3.3.

T)2%C 3.3

%4(( F)*T4H( I *4TH)*TI4(

#/>#0 5.0 5.3 5.: 5.7 5.6

! 5.3@ 5.30 5.// 5.00 5.5/

Pressure Dr!p "*r!ug* Euip%en"

Fluids sometimes have to be passed through beds of pac!ed solids1 for e$ample in the air drying of granularmaterials, hot air may be passed upward through a bed of the material. The pressure drop resulting is noteasy to calculate, even if the properties of the solids in the bed are well !nown. It is generally necessary, foraccurate pressure-drop information, to ma!e e$perimental measurements.

) similar difficulty arises in the calculation of pressure drops through equipment such as ban!s of tubes inheat e$changers. )n equation of the general form of eqn. 3./5 will hold in most cases, but values for ! willhave to be obtained from e$perimental results. &seful correlations for particular cases may be found in boo!son fluid flow and from wor!s such as +erry 0667 and Mc)dams 06:=.

Eui)alen" Leng"*s !f Pipe

In some applications it is convenient to clculate pressure drops in fittings from added equivalent lengths ofstraight pipe, rather than directly in terms of velocity heads or velocity pressures when ma!ing pipe-flowcalculations. This means that a fictitious length of straight pipe is added to the actual length, such that frictiondue to the fictitious pipe gives rise to the same loss as that which would arise from the fitting under

consideration. In this way various fittings, for e$ample bends and elbows, are simply equated to equivalentlengths of pipe and the total friction losses computed from the total pipe length, actual plus fictitious. )s E ƒ ineqn. 3./5 is equal to E ƒ in eqn. 3.07, k can therefore be replaced by =QL/# where L is the length of pipe ofdiameter # equivalent to the fitting.

!%pressi&ili"+ Effe("s f!r 4ases

The equations so far have all been applied on the assumption that the fluid flowing was incompressible, that is

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its density remained unchanged through the flow process. This is true for liquids under normal circumstancesand it is also frequently true for gases.

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