27112846 8 in compressible flow

7/29/2019 27112846 8 in Compressible Flow

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FUNDAMENTALS OFFUNDAMENTALS OFFLUID MECHANICSFLUID MECHANICS

Chapter 8 Pipe FlowChapter 8 Pipe Flow


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MAIN TOPICSMAIN TOPICS

General Characteristics of Pipe FlowGeneral Characteristics of Pipe Flow

Fully Developed Laminar FlowFully Developed Laminar FlowFully Developed Turbulent FlowFully Developed Turbulent Flow

Dimensional Analysis of Pipe FlowDimensional Analysis of Pipe Flow

Pipe Flow ExamplesPipe Flow Examples

PipePipe FlowrateFlowrate MeasurementMeasurement


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IntroductionIntroduction

Flows completely bounded by solid surfaces are calledFlows completely bounded by solid surfaces are called INTERNALINTERNALFLOWSFLOWS which include flows throughwhich include flows through pipespipes ((Round cross sectionRound cross section ),),ductsducts (NOT(NOT Round cross sectionRound cross section )) , nozzles, diffusers, sudden, nozzles, diffusers, suddencontractions and expansions, valves, and fittings.contractions and expansions, valves, and fittings.

The basic principles involved are independent of the crossThe basic principles involved are independent of the cross --sectionalsectionalshape, although the details of the flow may be dependent on it.shape, although the details of the flow may be dependent on it.

The flow regime (laminar or turbulent) of internal flows is primThe flow regime (laminar or turbulent) of internal flows is prim arilyarilya function of the Reynolds number.a function of the Reynolds number.

BB Laminar flow: Can be solved analytically.Laminar flow: Can be solved analytically.BB Turbulent flow: Rely Heavily on semiTurbulent flow: Rely Heavily on semi --empirical theories andempirical theories and

experimental data.experimental data.


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General Characteristics of General Characteristics of Pipe FlowPipe Flow


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Pipe SystemPipe System

A pipe system include the pipes themselvesA pipe system include the pipes themselves

(perhaps of more than one diameter), the various(perhaps of more than one diameter), the variousfittings, thefittings, the flowrateflowrate control devices valves) , andcontrol devices valves) , andthe pumps or turbines.the pumps or turbines.


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Pipe Flow vs. Open Channel FlowPipe Flow vs. Open Channel Flow

Pipe flow: Flows completely filling the pipe. (a)Pipe flow: Flows completely filling the pipe. (a)

The pressure gradient along the pipe is main driving force.The pressure gradient along the pipe is main driving force.Open channel flow: Flows without completely filling theOpen channel flow: Flows without completely filling the

pipe. (b) pipe. (b)

The gravity alone is the driving force.The gravity alone is the driving force.


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Laminar or Turbulent FlowLaminar or Turbulent Flow 1/21/2

The flow of a fluid in a pipe may beThe flow of a fluid in a pipe may be Laminar ?Laminar ?

Turbulent ?Turbulent ?Osborne ReynoldsOsborne Reynolds , a British scientist and mathematician,, a British scientist and mathematician,was the first to distinguish the difference between thesewas the first to distinguish the difference between theseclassification of flow by using aclassification of flow by using a simple apparatussimple apparatus asasshown.shown.


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Laminar or Turbulent FlowLaminar or Turbulent Flow 2/22/2

>> For For small enoughsmall enough flowrateflowrate the dye streak will remain as athe dye streak will remain as a

wellwell --defined line as it flows along, with only slight blurring duedefined line as it flows along, with only slight blurring dueto molecular diffusion of the dye into the surrounding water.to molecular diffusion of the dye into the surrounding water.

>> For a somewhat larger For a somewhat larger intermediateintermediate flowrateflowrate the dyethe dye

fluctuates in time and space, and intermittent bursts of irregulfluctuates in time and space, and intermittent bursts of irregul ar ar behavior appear along the streak. behavior appear along the streak.

>> For For large enoughlarge enough flowrateflowrate the dye streak almostthe dye streak almost

immediately become blurred and spreads across the entire pipe inimmediately become blurred and spreads across the entire pipe ina random fashion.a random fashion.


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Time Dependence of Time Dependence of

Fluid Velocity at a PointFluid Velocity at a Point


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Indication of Indication of

Laminar or Turbulent FlowLaminar or Turbulent FlowThe termThe term flowrateflowrate should be replaced by Reynoldsshould be replaced by Reynolds

number, ,where V is the average velocitynumber, ,where V is the average velocity

inin

the pipe.the pipe.It isIt is not only the fluid velocitynot only the fluid velocity that determines thethat determines the

character of the flowcharacter of the flow its density, viscosity, and the pipeits density, viscosity, and the pipesize are of equal importance.size are of equal importance.For general engineering purpose, the flow in a round pipeFor general engineering purpose, the flow in a round pipe

>> LaminarLaminar>> TransitionalTransitional>>

TurbulentTurbulent

= /VDR e

2100R e 0 if the flow is uphill>0 if the flow is uphill


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Example 8.2 Laminar Pipe Flow Example 8.2 Laminar Pipe Flow

zz An oil with a viscosity of An oil with a viscosity of = 0.40 N= 0.40 N s/ms/m 22 and densityand density = 900= 900kg/mkg/m 33 flows in a pipe of diameter D= 0.20m . (a) What pressureflows in a pipe of diameter D= 0.20m . (a) What pressuredrop, pdrop, p 11-- p p 22 , is needed to produce a, is needed to produce a flowrateflowrate of Q=2.0of Q=2.0 1010 --55 mm 33/s if /s if the pipe is horizontal with xthe pipe is horizontal with x 11=0 and x=0 and x 22=10 m? (b) How steep a hill,=10 m? (b) How steep a hill, , must the pipe be on if the oil is to flow through the pipe at t, must the pipe be on if the oil is to flow through the pipe at t hehe

same rate as in part (a), but with psame rate as in part (a), but with p 11=p=p 22? (c) For the conditions of ? (c) For the conditions of part (b), if p part (b), if p 11=200=200 kPakPa , what is the pressure at section, x, what is the pressure at section, x 33=5 m,=5 m,where x is measured along the pipe?where x is measured along the pipe?


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Example 8.2Example 8.2 SolutionSolution 1/21/2

210087.2/VDR e


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With pWith p 11=p=p 22 the length of the pipe,the length of the pipe, AA, does not appear in the, does not appear in the flowrateflowrateequationequation

kPa200 p p p 321 ===

p=0 for all p=0 for all AA


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From theFrom the Navier Navier --Stokes EquationsStokes Equations 1/31/3

General motion of an incompressible Newtonian fluid isGeneral motion of an incompressible Newtonian fluid is

governed by the continuity equation and the momentumgoverned by the continuity equation and the momentumequationequation

0V =G

Vg p

VVtV 2 GGGGG

++=+

Steady flowSteady flow

k ggGG =

For steady, fully developed flow in a pipe, the velocityFor steady, fully developed flow in a pipe, the velocitycontains only an axial component, which is a function of contains only an axial component, which is a function of only the radial coordinateonly the radial coordinate i)r (uV GG =


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The flow is governed by a balance of pressure, weight, andThe flow is governed by a balance of pressure, weight, and

viscous forces in the flow direction.viscous forces in the flow direction.

Vk g p2

GG

=+

0V = GSimplify theSimplify the Navier Navier --Stokes equationStokes equation


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=+

r

ur

r r

1sing

x

p

A

px

p.const

x p =

=

i)r (uVGG

=

Function of, at most, only xFunction of, at most, only x Function of ,at most, only rFunction of ,at most, only r

IntegratingIntegrating Velocity profile u(r)=Velocity profile u(r)=

B.C. (1) r = R , u = 0 ;B.C. (1) r = R , u = 0 ;

(2) r = 0 , u


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From Dimensional AnalysisFrom Dimensional Analysis 1/31/3

Assume that the pressure drop in the horizontal pie,Assume that the pressure drop in the horizontal pie, p, is p, is

a function of the average velocity of the fluid in the pipe,a function of the average velocity of the fluid in the pipe,V, the length of the pipe,V, the length of the pipe, AA, the pipe diameter, D, and the, the pipe diameter, D, and theviscosity of the fluid,viscosity of the fluid, ..

),D,,V(F p = A Dimensional analysisDimensional analysis

= DV pD A an unknown function of the length toan unknown function of the length todiameter ratio of the pipe.diameter ratio of the pipe.


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D

C

V

pD A=

where C is a constant.where C is a constant.

2DVC p =

A A==

4 pD)C4/(AVQ

The value of C must be determined by theory or experiment.The value of C must be determined by theory or experiment.For a round pipe, C=32. For duct of other crossFor a round pipe, C=32. For duct of other cross --sectionalsectionalshapes, the value of C is different.shapes, the value of C is different.

2DV32

pA=For a round pipeFor a round pipe


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f is termed the friction factor, or f is termed the friction factor, or sometimes the Darcy friction factor.sometimes the Darcy friction factor.

For a round pipeFor a round pipeDRe

64DVD

64V

D/V32V p

2

2

1

2

2

2

1

AAA =

=

=

2V

Df p

2= A

2V

D p

f 2

= A

D4 p w= A

2w

V8

Re64f

==

For laminar flowFor laminar flow


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Energy ConsiderationEnergy Consideration 1/31/3

The energy equation for incompressible, steady flowThe energy equation for incompressible, steady flow

between two locations between two locations

L2

2222

1

2111 hz

g2V p

zg2V p +++

=++

g2V

g2V 222

211 =

D4

r 2hz pz p wL2

21

1

===

+

+

AA

r 2sin p = AA

Dr 2 w=

The head loss in aThe head loss in a pipe is a result of pipe is a result of the viscous shear the viscous shear stress on the wall.stress on the wall.


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Example 8.3 Laminar Pipe Flow PropertiesExample 8.3 Laminar Pipe Flow Properties 1/2 1/2

zz TheThe flowrateflowrate , Q, of corn syrup through the horizontal pipe shown in, Q, of corn syrup through the horizontal pipe shown inFigure E8.3 is to be monitored by measuring the pressure differeFigure E8.3 is to be monitored by measuring the pressure differe ncence

between sections (1) and (2). It is proposed that Q=K between sections (1) and (2). It is proposed that Q=K p, where the p, where thecalibration constant, K, is a function of temperature, T, becauscalibration constant, K, is a function of temperature, T, becaus e of e of the variation of the syrupthe variation of the syrup s viscosity and density with temperature.s viscosity and density with temperature.

These variations are given in Table E8.3. (a) Plot K(T) versus TThese variations are given in Table E8.3. (a) Plot K(T) versus T for for 6060FFTT 160160F. (b) Determine the wall shear stress and the pressureF. (b) Determine the wall shear stress and the pressuredrop,drop, p=p p=p 11-- p p 22 , for Q=0.5 ft, for Q=0.5 ft 33/s and T=100/s and T=100 F. (c) For the conditionsF. (c) For the conditions

of part (b), determine the nest pressure force.(of part (b), determine the nest pressure force.( DD22

/4)/4) p, and the p, and thenest shear force,nest shear force, DD AA ww , on the fluid within the pipe between the, on the fluid within the pipe between thesections (1) and (2).sections (1) and (2).


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Example 8.3 Laminar Pipe Flow PropertiesExample 8.3 Laminar Pipe Flow Properties 1/2 1/2


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21001380.../VDR e


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The new pressure force and viscous force on the fluid within theThe new pressure force and viscous force on the fluid within the pipe pipe between sections (1) and (2) is between sections (1) and (2) is

lb84.5...2D2F

lb84.5... p4D

F

wv

2

p

===

===

A

The values of these two forces are the same. The net force is zeThe values of these two forces are the same. The net force is ze ro;ro;there is no acceleration.there is no acceleration.


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Fully Developed Turbulent FlowFully Developed Turbulent Flow

Turbulent pipe flow is actually more likely to occur thanTurbulent pipe flow is actually more likely to occur than

laminar flow in practical situations.laminar flow in practical situations.Turbulent flow is a very complex process.Turbulent flow is a very complex process.

Numerous persons have devoted considerable effort in an Numerous persons have devoted considerable effort in anattempting to understand the variety of baffling aspects of attempting to understand the variety of baffling aspects of turbulence. Although a considerable amount if knowledgeturbulence. Although a considerable amount if knowledgeabout the topics has been developed,about the topics has been developed, the field of turbulentthe field of turbulentflow still remains the least understood area of fluidflow still remains the least understood area of fluidmechanics.mechanics.

Much remains to be learned about the nature of turbulent flow.Much remains to be learned about the nature of turbulent flow.


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Transition from Laminar to TurbulentTransition from Laminar to Turbulent

Flow in a PipeFlow in a Pipe1/21/2

For any flow geometry, there is one (or more)For any flow geometry, there is one (or more)

dimensionless parameters such as with this parameter dimensionless parameters such as with this parameter value below a particular value the flow is laminar, whereasvalue below a particular value the flow is laminar, whereaswith the parameter value larger than a certain value thewith the parameter value larger than a certain value the

flow is turbulent.flow is turbulent. The important parameters involved and their criticalThe important parameters involved and their critical

values depend on the specific flow situation involved.values depend on the specific flow situation involved.

Consider a long section of pipe that is Consider a long section of pipe that is initially filled with a fluid at rest.initially filled with a fluid at rest.

For flow in pipe : 21004000For flow along a plate Re x~5000


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Transition from Laminar to TurbulentTransition from Laminar to Turbulent

Flow in a PipeFlow in a Pipe2/22/2

As the valve is opened to start the flow, the flow velocity and,As the valve is opened to start the flow, the flow velocity and, hence,hence,the Reynolds number increase from zero (no flow) to their the Reynolds number increase from zero (no flow) to their maximum steady flow values.maximum steady flow values.For the initial time period the Reynolds number For the initial time period the Reynolds number osos small enough for small enough for laminar flow to occur.laminar flow to occur.

At some time the ReynoldsAt some time the Reynoldsnumber reaches 2100, and thenumber reaches 2100, and theflow begins its transition toflow begins its transition toturbulent conditions.turbulent conditions.Intermittent spots or burstIntermittent spots or burstappear appear ....


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Description for Turbulent FlowDescription for Turbulent Flow 1/51/5

Turbulent flows involveTurbulent flows involverandomly fluctuatingrandomly fluctuating

parameters. parameters.

The character of many of theThe character of many of the

important properties of theimportant properties of theflow (pressure drop, heatflow (pressure drop, heattransfer, etc.) depends stronglytransfer, etc.) depends strongly

on the existence and nature of on the existence and nature of the turbulent fluctuations or the turbulent fluctuations or randomness.randomness.

The timeThe time --averaged,averaged, , and, andfluctuating,fluctuating, description of adescription of a

parameter for tubular flow. parameter for tubular flow.

A typical trace of the axial component of A typical trace of the axial component of velocity measured at a given location invelocity measured at a given location inthe flow, u=u(t).the flow, u=u(t).


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Due to the macroscopic scale of the randomness inDue to the macroscopic scale of the randomness inturbulent flow, mixing processes and heat transfer turbulent flow, mixing processes and heat transfer

processes are considerably enhanced in turbulent flow processes are considerably enhanced in turbulent flowcompared to laminar flow.compared to laminar flow.Such finiteSuch finite --sized random mixing is very effective insized random mixing is very effective intransporting energy and mass throughout the flow field.transporting energy and mass throughout the flow field.[Laminar flow can be thought of as very small but finite[Laminar flow can be thought of as very small but finite --

sized fluid particles flowing smoothly in layer, one over sized fluid particles flowing smoothly in layer, one over another. The only randomness and mixing take place onanother. The only randomness and mixing take place onthe molecular scale and result in relatively small heat,the molecular scale and result in relatively small heat,mass, and momentum transfer rates.]mass, and momentum transfer rates.]


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In some situations, turbulent flow characteristics areIn some situations, turbulent flow characteristics are

advantages. In other situations, laminar flow is desirable.advantages. In other situations, laminar flow is desirable. Turbulence: mixing of fluids.Turbulence: mixing of fluids. Laminar: pressure drop in pipe, aerodynamic drag onLaminar: pressure drop in pipe, aerodynamic drag on

airplane.airplane.


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Turbulent flows are characterized by random, threeTurbulent flows are characterized by random, three --

dimensionaldimensional vorticityvorticity ..Turbulent flows can be described in terms of their meanTurbulent flows can be described in terms of their meanvalues on which are superimposed the fluctuations.values on which are superimposed the fluctuations.

( )+

=Tt

t

O

O

dtt,z,y,xuT1

u

'uuu +=

uu'u =


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The time average of the fluctuations is zero.The time average of the fluctuations is zero.

The square of a fluctuation quantity is positive.The square of a fluctuation quantity is positive.

Turbulence intensity or the level of the turbulenceTurbulence intensity or the level of the turbulence

( ) ( ) 0uTuTT1

dtuuT1

'uTt

t

O

O===

+

( ) 0dt'uT1)'u( Tt

t

22 O

O

>= +

( )

u

dt'uT1

u

)'u(

2Tt

t

22

O

O==

+ The larger the turbulence intensity, the larger The larger the turbulence intensity, the larger the fluctuations of the velocity. Wellthe fluctuations of the velocity. Well --designed wind tunnels have typical value of designed wind tunnels have typical value of =0.01, although with extreme care, values=0.01, although with extreme care, valuesas low asas low as =0.0002 have been obtained.=0.0002 have been obtained.


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Shear Stress for Laminar FlowShear Stress for Laminar Flow 1/21/2

Laminar flow is modeled as fluid particles that flow smoothly alongin layers, gliding past the slightly slower or faster ones on either side.The fluid actually consists of numerous molecules darting about inan almost random fashion. The motion is not entirely random aslight bias in one direction.

As the molecules dart across a given plane (plane A-A, for example), theones moving upward have come froman area of smaller average xcomponent of velocity than the onesmoving downward, which have come

from an area of large velocity.


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Shear Stress for Laminar FlowShear Stress for Laminar Flow 2/22/2

The momentum flux in the x direction across plane A-A give rise toa drag of the lower fluid on the upper fluid and an equal but oppositeeffect of the upper fluid on the lower fluid. The sluggish moleculesmoving upward across plane A-A must accelerated by the fluidabove this plane. The rate of change of momentum in this process

produces a shear force. Similarly, the more energetic moleculesmoving down across plane A-A must be slowed down by the fluid

below that plane.

BY combining these effects, we obtain the well-known Newtonviscosity law

dy

duyx

= Shear stress is present only if there is aShear stress is present only if there is agradient in u=gradient in u= u(yu(y ).).


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Shear Stress for Turbulent FlowShear Stress for Turbulent Flow 1/21/2

The turbulent flow is thought as aThe turbulent flow is thought as a

series of random, threeseries of random, three --dimensional eddy type motions.dimensional eddy type motions.

These eddies range in size fromThese eddies range in size from

very small diameter to fairly largevery small diameter to fairly largediameter.diameter.

This eddy structure greatlyThis eddy structure greatly promotes mixing within the fluid. promotes mixing within the fluid.


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Shear Stress for Turbulent FlowShear Stress for Turbulent Flow 2/22/2

The flow is represented by (timeThe flow is represented by (time --mean velocity ) plus umean velocity ) plus u and vand v (time randomly fluctuating velocity components in the x and y(time randomly fluctuating velocity components in the x and ydirection).direction).The shear stress on the plane AThe shear stress on the plane A --AA

u

turbulentar minla'v'udyud +==

'v'uis called Reynolds stress introduced byis called Reynolds stress introduced by

Osborne Reynolds.Osborne Reynolds.

0'v'u As we approach wall, and is zero at the wallAs we approach wall, and is zero at the wall(the wall tends to suppress the fluctuations.)(the wall tends to suppress the fluctuations.)

The shear stress is not merely proportional to the gradient of the

time-averaged velocity, .)y(u


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Structure of Turbulent Flow in a PipeStructure of Turbulent Flow in a Pipe 1/21/2

Near the wallNear the wall (the viscous(the viscous sublayer sublayer ), the), the laminar shearlaminar shear

stressstress lamlam is dominant.is dominant.Away fromAway from the wall (in the outer layer) ,the wall (in the outer layer) , the turbulentthe turbulentshear stressshear stress turbturb isis dominantdominant ..

The transition between these two regions occurs in theThe transition between these two regions occurs in theoverlap layer.overlap layer.


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Structure of Turbulent Flow in a PipeStructure of Turbulent Flow in a Pipe 2/22/2

The relative magnitude of The relative magnitude of lamlam compared tocompared to turbturb is ais a

complex function dependent on the specific flow involved.complex function dependent on the specific flow involved.

Typically the value of Typically the value of turbturb is 100 to 1000 times greater is 100 to 1000 times greater thanthan lamlam inin the outer region.the outer region.


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Alternative Form of Shear StressAlternative Form of Shear Stress 1/21/2

turbturb : requiring an accurate knowledge of the fluctuations: requiring an accurate knowledge of the fluctuations

uu

and vand v

, or , or

The shear stress for turbulent flow is given in terms of theThe shear stress for turbulent flow is given in terms of theeddy viscosityeddy viscosity ..

dyud

turb =This extension of of laminar flow terminologyThis extension of of laminar flow terminologywas introduced by J.was introduced by J. Boussubesq Boussubesq , a French , a French

scientist, in 1877. scientist, in 1877.

?? A semiempirical theory was proposed by L.L. PrandtlPrandtl todetermine the value of

'v'u


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Alternative Form of Shear StressAlternative Form of Shear Stress 2/22/2

dy

ud2mA=

2

2mturb

dy

ud

= A

mixing length, is not constantthroughout the flow field.

There is no general, allThere is no general, all -- encompassing,encompassing,

useful model that can accurately predictuseful model that can accurately predictthe shear stress throughout a generalthe shear stress throughout a generalincompressible, viscous turbulent flow.incompressible, viscous turbulent flow.


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Turbulent Velocity ProfileTurbulent Velocity Profile 1/51/5

Fully developed turbulent flow in a pipe can be broken into threFully developed turbulent flow in a pipe can be broken into thre eeregion: the viscousregion: the viscous sublayer sublayer , the overlap region, and the outer , the overlap region, and the outer turbulentturbulent sublayer sublayer ..Within the viscousWithin the viscous sublayer sublayer the shear stress is dominant comparedthe shear stress is dominant comparedwith the turbulent stress, and the random, eddying nature of thewith the turbulent stress, and the random, eddying nature of the flowflowis essentially absent.is essentially absent.In the outer turbulent layer the Reynolds stress is dominant, anIn the outer turbulent layer the Reynolds stress is dominant, an ddthere is considerable mixing and randomness to the flow.there is considerable mixing and randomness to the flow.

Within the viscousWithin the viscous sublayer sublayer the fluid viscosity is an importantthe fluid viscosity is an important parameter; the density is unimportant. In the outer layer the op parameter; the density is unimportant. In the outer layer the op posite positeis true.is true.


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Considerable information concerning turbulent velocity profilesConsiderable information concerning turbulent velocity profiles hashas been obtained through the use of been obtained through the use of dimensional analysis, and semidimensional analysis, and semi --empirical theoretical effortsempirical theoretical efforts ..In the viscousIn the viscous sublayer sublayer the velocity profile can be written inthe velocity profile can be written indimensionless form asdimensionless form as

++ =

== yyuuuu

*

*

( ) 2/1w* /u =Where y is the distance measured from the wall y=R Where y is the distance measured from the wall y=R --r.r.

is called the friction velocity.is called the friction velocity.

Law of the wallLaw of the wall

Is valid very near the smooth wall, for 5yu

0*


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In the overlap region the velocity should vary as theIn the overlap region the velocity should vary as the

logarithm of ylogarithm of y

In transition region or buffer layer In transition region or buffer layer

for 30yu * >

0.5 y

yuln5.2

uu +

=

30yu

7-5*

=

y R

ln5.2u

uU forfor

Determined experimentally


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0.5y

yuln5.2

uu +

=

=

*

*

yuuu


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The velocity profile for turbulentflow through a smooth pipe may

also be approximated by theempirical power power --law equationlaw equation

The power The power --law profile is notlaw profile is notapplicable close to the wall.applicable close to the wall.

n/1n/1

R

r 1

R

y

U

u

=

=

Where the exponent, n, variesWhere the exponent, n, varieswith the Reynolds number.with the Reynolds number.

l 8 b l i lE l 8 4 T b l Pi Fl


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Example 8.4 Turbulent Pipe Flow Example 8.4 Turbulent Pipe Flow

PropertiesPropertieszz Water at 20Water at 20 (( =998kg/m=998kg/m 33 andand =1.004=1.004 1010 --66mm 22/s) flows through/s) flows through

a horizontal pipe of 0.1a horizontal pipe of 0.1 --m diameter with am diameter with a flowrateflowrate of Q=4of Q=4 1010 --22mm 33/s/sand a pressure gradient of 2.59and a pressure gradient of 2.59 kPa/mkPa/m . (a) Determine the. (a) Determine theapproximate thickness of the viscousapproximate thickness of the viscous sublayer sublayer . (b) Determine the. (b) Determine theapproximate centerline velocity,approximate centerline velocity, VV cc. (c) Determine the ration of the. (c) Determine the ration of the

turbulent to laminar shear stress,turbulent to laminar shear stress, t urbt urb // l aml am at a point midwayat a point midway between the centerline and the pipe wall (i.e., at r=0.025m) between the centerline and the pipe wall (i.e., at r=0.025m)


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The thickness of viscousThe thickness of viscous sublayer sublayer ,, ss , is approximately, is approximately

5u*

s =

*s u

5 =( ) s/m255.0.../u 2/1w* ===

2w m/ N8.64...4

pD==

= A

mm02.0m1097.1...u55*s ===

=

The centerline velocity can be obtained from the average velocitThe centerline velocity can be obtained from the average velocit y andy andthe assumption of a power the assumption of a power --law velocity profilelaw velocity profile

s/m09.5

4/)m1.0(

s/m04.0

A

QV 2

3

=

== 5e 1007.5.../VDR ===


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n/1n/1

R r

1R y

Uu

=

=

VR )1n2)(1n(n

VR 2...dAuAVQ 22

c2 =++====

n=8.4n=8.4

e 107.5.../VDR ===

s/m04.6...V)1n2)(1n(

n2

V

Vc

2

c

==++

=

D

r 2 w= Valid for laminar or turbulent flow

2turblam

2w

m/ N4.32

)m1.0()m025.0)(m/ N8.64(2

Dr 2

=+=

==


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2turblam

2w

m/ N4.32

)m1.0()m025.0)(m/ N8.64(2

Dr 2

=+=

==

2

n/)n1(

clam m/ N0266.0R r

1nR V

dr ud

= ==

12200266.00266.04.32

lam

lam

lam

turb =

=

=


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Dimensional Analysis of Dimensional Analysis of Pipe FlowPipe Flow


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Energy ConsiderationsEnergy Considerations 1/81/8

Considering the steady flow through the piping system, includingConsidering the steady flow through the piping system, including aareducing elbow. The basic equation for conservation of energyreducing elbow. The basic equation for conservation of energy thethefirst law of thermodynamicsfirst law of thermodynamics

inShaftinnetCS

2

CVWQAdnV)gz

2V p

u(Vdet

GGG +=++

++

+=+

+=++

CS nnCSCVinShaftinnet

CSCVCS nninShaftinnet

dAnVdAnVeVdet

WQ

dAnVeVdet

dAnVWQ

GGGG

GGGG

Energy equationEnergy equation

gz2

Vue2

++=


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69


=

0Vdet CV

gz2

V pumgz

2V p

udAnVgz2

V pu

in

2

out

22

CS

GG

++

+

++

+=++

+

in

in

2

out

out

2

2

CS

mgz2

V pumgz

2

V pu

dAnVgz2

V pu

GG

++

+

++

+=

++

+

When the flow is steadyWhen the flow is steady

The integral of

dAnVgz2

V pu

2

CS

GG++

+

??????

Uniformly distribution

Only one streamentering and leavingOnly one streamentering and leaving


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70


( )

innetshaftinnet

inout

2

in

2

out

inout

inout

WQ

zzg2

VV p puum

+=

++ +

+= puh

( ) in/netshaftin/netinout2in

2out

inout WQzzg2VV

hhm +=++

If shaft work is involvedIf shaft work is involved ..

OneOne --dimensional energy equationdimensional energy equationfor steadyfor steady --inin --thethe --mean flowmean flow

EnthalpyEnthalpy The energy equation is written in termsThe energy equation is written in terms

of enthalpy.of enthalpy.


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71


( )innetinoutin2inin

out

2outout quugz

2V p

gz2

V p ++

=++

( )innetinout

2in

2outinout

inoutQzzg

2

VV p puum =++

+

m

For steady, incompressible flowFor steady, incompressible flow OneOne --dimensional energy equationdimensional energy equation

m/Qq innetinnet =

in

2in

inout

2out

out z2V

pz2

V p ++=++

0quu innetinout =

wherewhere

For steady, incompressible,For steady, incompressible, frictionless flowfrictionless flow

Bernoulli equationBernoulli equation

Frictionless flowFrictionless flow


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For steady, incompressible,For steady, incompressible, frictional flowfrictional flow

0quu innetinout >

lossquu innetinout =

lossgz2V p

gz2V p

in

2

ininout

2

outout ++=++

Defining useful or available energy gz2

V p 2 ++

Defining loss of useful or available energy

Frictional flowFrictional flow


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73


( ) innetshaftinnetinout2

in

2

outinoutinout WQzzg2VV p p

uum +=+

+

+

m )quu(wgz2

V pgz

2

V pinnetinoutinnetshaftin

2inin

out

2outout +++

=++

For steady, incompressible flow with friction and shaft work For steady, incompressible flow with friction and shaft work

losswgz2

V pgz

2V p

innetshaftin

2inin

out

2outout +++

=++

g Lsin

2inin

out

2outout hhz

g2V p

zg2

V p +++

=++

QW

gmW

gwh innetshaftinnetshaftinnetshaftS == g

lossh L =Head lossHead lossShaft headShaft head


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For turbineFor turbine

For pumpFor pumpThe actual head drop across the turbineThe actual head drop across the turbine

The actual head drop across the pumpThe actual head drop across the pump

)0h(hh TTs >=Ps hh = hh pp is pump headis pump head

hh TT is turbine headis turbine head

TLsT)hh(h +=

pLs p )hh(h =

Lsin

2inin

out

2outout hhz

g2

V pz

g2

V p +++

=++


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75


Total head loss ,Total head loss , hhLL , is regarded as the sum of major losses,, is regarded as the sum of major losses,

hh LL majormajor , due to frictional effects in fully developed flow, due to frictional effects in fully developed flowin constant area tubes, and minor losses,in constant area tubes, and minor losses, hhLL minor minor , resulting, resultingfrom entrance, fitting, area changes, and so on.from entrance, fitting, area changes, and so on.

or minmajor LLLhhh +=


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76

Major Losses: Friction Factor Major Losses: Friction Factor

The energy equation for steady and incompressible flowThe energy equation for steady and incompressible flowwith zero shaft work with zero shaft work

L2

2222

1

2111 hgz

2V p

gz2V p =

++

++

L1221 h)zz(g p p +=>>>

For fully developed flow through a constant area pipe ,For fully developed flow through a constant area pipe ,hhLL=0=0

For horizontal pipe, zFor horizontal pipe, z 22=z=z 11L

21 h p p p ==

>>>


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77

Major Losses: Laminar FlowMajor Losses: Laminar Flow

In fully developed laminar flow in a horizontal pipe, theIn fully developed laminar flow in a horizontal pipe, the pressure drop pressure drop

( )

2V

DR 64

VD64

2V

DDV

D32h

2V

Df p

DRe64

DVD64V21

pDV

D32

D4/DV128

DQ128

p

2

e

2

L

2

2

4

2

4

AAAA

AA

AAA

=

==>>=

=

=

=

=

=

Re64

f ar minla =

Friction FactorFriction Factor ( ) ( )2/V//D pf 2= A


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78

Major Losses: Turbulent FlowMajor Losses: Turbulent Flow 1/31/3

In turbulent flow we cannot evaluate the pressure drop analyticaIn turbulent flow we cannot evaluate the pressure drop analytica lly;lly;we must resort to experimental results and use dimensional analywe must resort to experimental results and use dimensional analy sissisto correlate the experimental data.to correlate the experimental data.

( )= ,,,,D,VF p A

In fully developed turbulent flow theIn fully developed turbulent flow the pressure drop, pressure drop, p p , caused by friction, caused by friction

in a horizontal constantin a horizontal constant --area pipe isarea pipe isknown toknown to depend on pipe diameter,D,depend on pipe diameter,D,

pipe length, pipe length, AA, pipe roughness,e,, pipe roughness,e,

average flow velocity,average flow velocity, V, fluidV, fluiddensitydensity , and fluid viscosity,, and fluid viscosity, ..


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79


Applying dimensional analysis, the result were a correlation of theform

Experiments show that the nondimensional head loss is directly proportional to A/D. Hence we can write

=

DRe,

DV

p2

21

A

D

Re,f g2

VDf h

2

L major

A

2

V

D

f p2= A

DarcyDarcy --WeisbachWeisbach equationequation

= D

,D

,VDV p 2

21

A


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80

Roughness for PipesRoughness for Pipes


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81

Friction Factor by L. F. MoodyFriction Factor by L. F. Moody

Depending on the specificDepending on the specificcircumstances involved.circumstances involved.


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82

About Moody ChartAbout Moody Chart

For laminar flow, f=64/Re, which is independent of theFor laminar flow, f=64/Re, which is independent of therelative roughness.relative roughness.For very large Reynolds numbers, f=For very large Reynolds numbers, f= (( / D), which is/ D), which isindependent of the Reynolds numbers.independent of the Reynolds numbers.For flows withFor flows with very large value of Revery large value of Re , commonly termed, commonly termedcompletely turbulent flow (or wholly turbulent flow), thecompletely turbulent flow (or wholly turbulent flow), thelaminar laminar sublayer sublayer is so thin (its thickness decrease withis so thin (its thickness decrease withincreasing Re) that the surface roughness completelyincreasing Re) that the surface roughness completelydominates the character of the flow near the wall.dominates the character of the flow near the wall.For flows with moderate value of Re, the friction factor For flows with moderate value of Re, the friction factor f=f= (Re,(Re, /D) ./D) .


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Colebrook To avoid having to use a graphical method for obtaining f for turbulent flows.

Miler suggests that a single iteration will produce a result within1 percent if the initial estimate is calculated from

+=f Re

51.27.3D/

log0.2f

1

2

9.00 Re74.5

7.3D/log25.0f

+=

Valid for the entireValid for the entire nonlaminar nonlaminar range of the Moody chart.range of the Moody chart.

Colebrook formulaColebrook formula

Example 8 5 Comparison of Laminar orExample 8 5 Comparison of Laminar or


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Example 8.5 Comparison of Laminar or Example 8.5 Comparison of Laminar or

Turbulent pressure DropTurbulent pressure Dropzz Air under standard conditions flows through a 4.0Air under standard conditions flows through a 4.0 --mmmm --diameter diameter

drawn tubing with an average velocity of V = 50 m/s. For suchdrawn tubing with an average velocity of V = 50 m/s. For suchconditions the flow would normally be turbulent. However, if conditions the flow would normally be turbulent. However, if

precautions are taken to eliminate disturbances to the flow (the precautions are taken to eliminate disturbances to the flow (theentrance to the tube is very smooth, the air is dust free, the tentrance to the tube is very smooth, the air is dust free, the t ube doesube does

not vibrate, etc.), it may be possible to maintain laminar flow.not vibrate, etc.), it may be possible to maintain laminar flow. (a)(a)Determine the pressure drop in a 0.1Determine the pressure drop in a 0.1 --m section of the tube if them section of the tube if theflow is laminar. (b) Repeat the calculations if the flow is turbflow is laminar. (b) Repeat the calculations if the flow is turb ulent.ulent.

1/2


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85


flowTurbulent700,13.../VDR e ===

Under standard temperature and pressure conditionsUnder standard temperature and pressure conditions

=1.23kg/m=1.23kg/m 33,, =1.79=1.79 1010 --55 N N s/ms/mThe Reynolds number The Reynolds number

kPa179.0...V21

Df p 2 === A

If the flow were laminar If the flow were laminar

f=64/Re=f=64/Re= =0.0467=0.0467

2 22/2


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86


kPa076.1...V21

Df p 2 ===

A

If the flow were turbulentIf the flow were turbulent

From Moody chartFrom Moody chart f=f= (Re,(Re, /D) =/D) = 0.0280.028

1/51/5


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87

Minor LossesMinor Losses 1/51/5

Most pipe systems consist of Most pipe systems consist of considerably more than straightconsiderably more than straight

pipes. These additional pipes. These additionalcomponents (valves, bends, tees,components (valves, bends, tees,

and the like) add to the overalland the like) add to the overallhead loss of the system.head loss of the system.

Such losses are termed MINOR Such losses are termed MINOR

LOSS.LOSS.

The flow pattern through a valvThe flow pattern through a valv

2/52/5


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88


The theoretical analysis to predict the details of flowThe theoretical analysis to predict the details of flow pattern (through these additional components) is not, as pattern (through these additional components) is not, asyet, possible.yet, possible.

The head loss information for essentially all components isThe head loss information for essentially all components is

given in dimensionless form and based on experimentalgiven in dimensionless form and based on experimentaldata. The most common method used to determine thesedata. The most common method used to determine thesehead losses or pressure drops is to specify the losshead losses or pressure drops is to specify the loss

coefficient, K coefficient, K LL

3/53/5


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89


2L

22L

L V21

K pV

2

1 p

g2/V

hK or min =

==

Re),geometry(K L =

f DK

g2V

Df

g2V

K h

Leq

2eq

2

LL or min

=

==

A

AMinor losses are sometimesgiven in terms of an equivalentlength Aeq

The actual value of K L is strongly dependent on the geometry of

the component considered. It may also dependent on the fluid properties. That is

4/54/5


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90


For many practical applications the Reynolds number isFor many practical applications the Reynolds number islarge enough so that the flow through the component islarge enough so that the flow through the component isdominated by inertial effects, with viscous effects being of dominated by inertial effects, with viscous effects being of secondary importance.secondary importance.In a flow that is dominated by inertia effects rather thanIn a flow that is dominated by inertia effects rather thanviscous effects, it is usually found that pressure drops andviscous effects, it is usually found that pressure drops andhead losses correlate directly with the dynamic pressure.head losses correlate directly with the dynamic pressure.This is the reason why the friction factor for very largeThis is the reason why the friction factor for very largeReynolds number, fully developed pipe flow isReynolds number, fully developed pipe flow isindependent of the Reynolds number.independent of the Reynolds number.

Mi L 5/55/5


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91


This is true for flow through pipe components.This is true for flow through pipe components.

Thus, in most cases of practical interest the lossThus, in most cases of practical interest the losscoefficients for components are a function of geometrycoefficients for components are a function of geometryonly,only,

)geometry(K L =

ffMi L C ffi i E t fl 1/3Entrance flow 1/3


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92

Minor Losses CoefficientMinor Losses Coefficient Entrance flow 1/3Entrance flow 1/3

Entrance flow conditionEntrance flow conditionand loss coefficientand loss coefficient

((aa ) Reentrant,) Reentrant, K K L L = 0.8= 0.8

((bb) sharp) sharp --edged,edged, K K L L = 0.5= 0.5

((cc) slightly rounded,) slightly rounded, K K L L = 0.2= 0.2((d d ) well) well --rounded,rounded, K K L L = 0.04= 0.04

K K L L = function of rounding of = function of rounding of the inlet edge.the inlet edge.

Mi L C ffi iMi L C ffi i Entrance flow 2/3Entrance flow 2/3


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93


A venaA vena contractacontracta region may result because the fluidregion may result because the fluidcannot turn a sharp rightcannot turn a sharp right --angle corner. The flow is said toangle corner. The flow is said toseparate from the sharp corner.separate from the sharp corner.

The maximum velocity velocity at section (2) is greater The maximum velocity velocity at section (2) is greater

than that in the pipe section (3), and the pressure there isthan that in the pipe section (3), and the pressure there islower.lower.

If this high speed fluid could slow down efficiently, theIf this high speed fluid could slow down efficiently, thekinetic energy could be converted into pressure.kinetic energy could be converted into pressure.

Mi L C ffi iMi L C ffi i Entrance flow 3/3Entrance flow 3/3


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94


Such is not the case. AlthoughSuch is not the case. Althoughthe fluid may be acceleratedthe fluid may be accelerated

very efficiently, it is veryvery efficiently, it is verydifficult to slow downdifficult to slow down(decelerate) the fluid(decelerate) the fluid efficentlyefficently ..

(2)(2) (3) The extra kinetic(3) The extra kineticenergy of the fluid is partiallyenergy of the fluid is partiallylost because of viscouslost because of viscous

dissipation, so that the pressuredissipation, so that the pressuredoes not return to the idealdoes not return to the idealvalue.value.

Flow pattern and pressure distributionFlow pattern and pressure distribution

for a sharpfor a sharp --edged entranceedged entrance

Mi L C ffi iMi L C ffi i Exit flowExit flow


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95

Minor Losses CoefficientMinor Losses Coefficient Exit flowExit flow

Exit flow condition andExit flow condition andloss coefficientloss coefficient

((aa ) Reentrant,) Reentrant, K K L L = 1.0= 1.0

((bb) sharp) sharp --edged,edged, K K L L = 1.0= 1.0

((cc) slightly rounded,) slightly rounded, K K L L = 1.0= 1.0((d d ) well) well --rounded,rounded, K K L L = 1.0= 1.0

Mi L C ffi iMi L C ffi i t varied diametervaried diameter


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96

Minor Losses CoefficientMinor Losses Coefficient varied diameter varied diameter

Loss coefficient for suddenLoss coefficient for suddencontraction, expansion,typicalcontraction, expansion,typical

conical diffuser.conical diffuser.

2

2

1L A

A1K

=

Mi L C ffi iMi L C ffi i t BendBend


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97

Minor Losses CoefficientMinor Losses Coefficient BendBend

Carefully designed guide vanesCarefully designed guide vaneshelp direct the flow with lesshelp direct the flow with lessunwanted swirl and disturbances.unwanted swirl and disturbances.

Character of the flow in bendCharacter of the flow in bendand the associated lossand the associated loss

coefficient.coefficient.

I t l St t f V lI t l St t f V l


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98

Internal Structure of ValvesInternal Structure of Valves

((aa ) globe valve) globe valve((bb) gate valve) gate valve((cc) swing check valve) swing check valve

((d d ) stop check valve) stop check valve

Loss Coefficients for PipeLoss Coefficients for Pipe


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99

pp

ComponentsComponents

E l 8 6 Mi LE mple 8 6 Minor Loss 1/21/2


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100

Example 8.6 Minor LossExample 8.6 Minor Loss 1/2 1/2

zz Air at standard conditions is to flow through the test sectionAir at standard conditions is to flow through the test section[between sections (5) and (6)] of the closed[between sections (5) and (6)] of the closed --circuit wind tunnelcircuit wind tunnel

shown if Figure E8.6 with a velocity of 200 ft/s. The flow is dr shown if Figure E8.6 with a velocity of 200 ft/s. The flow is dr iveniven by a fan that essentially increase the static pressure by the am by a fan that essentially increase the static pressure by the am ountount p p11-- p p 99 that is needed to overcome the head losses experienced by thethat is needed to overcome the head losses experienced by the

fluid as it flows around the circuit. Estimate the value of pfluid as it flows around the circuit. Estimate the value of p 11-- p p 99 andandthe horsepower supplied to the fluid by the fan.the horsepower supplied to the fluid by the fan.

E l 8 6 Mi LExample 8 6 Minor Loss 2/22/2


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101

Example 8.6 Minor LossExample 8.6 Minor Loss 2/2 2/2

E l 8 6Example 8 6 S l tiSolution 1/31/3


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102


The maximum velocity within the wind tunnel occurs in theThe maximum velocity within the wind tunnel occurs in thetest section (smallest area). Thus, the maximum Mach number test section (smallest area). Thus, the maximum Mach number of the flow is Maof the flow is Ma 55=V=V 55/c/c55

91L9

299

1

211 hz

g2V p

zg2

V p+++

=++

s/ft1117)KRT(cs/ft200V 2/1555 ===

The energy equation between points (1) and (9)The energy equation between points (1) and (9)

= 9191L

p ph The total head loss from (1) to (9).The total head loss from (1) to (9).

Example 8 6Example 8 6 SolutionSolution 2/32/3


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103


The energy across the fan, from (9) to (1)The energy across the fan, from (9) to (1)

91L55 p55 pa hVAhVAQhP ===

91L91

p h p p

h ==

1

211 p9

299 z

g2V phz

g2V p ++=+++

HH p p is the actual head rise suppliedis the actual head rise supplied by the pump (fan) to the air. by the pump (fan) to the air.

The actual power supplied to the air (horsepower, PThe actual power supplied to the air (horsepower, P aa) is obtained) is obtained

from the fan head byfrom the fan head by



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104


The total head lossThe total head loss

hp3.62s/lbft34300...P

psi298.0...)ft560)(ft/lb765.0(h p p

a

291L91

===

====

scr nozdif 3corner 2corner 8corner 7corner LLLLLLL91L hhhhhhhh ++++++=

0.4K 2.0K

g2V

6.0g2

VK h

g2V

2.0g2

VK h

scr noz

dif dif corner

LL

22

LL

22

LL

==

====

Noncircular DuctsNoncircular Ducts 1/41/4


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105


The empirical correlations for pipe flow may be used for The empirical correlations for pipe flow may be used for computations involving noncircular ducts, provided their computations involving noncircular ducts, provided their cross sections are not too exaggerated.cross sections are not too exaggerated.The correlation for turbulent pipe flow are extended for The correlation for turbulent pipe flow are extended for use with noncircular geometries by introducing theuse with noncircular geometries by introducing thehydraulic diameter, defined ashydraulic diameter, defined as

P

A4D h

Where A is crossWhere A is cross --sectional area, and Psectional area, and P

is wetted perimeter.is wetted perimeter.



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106


For a circular ductFor a circular duct

For a rectangular duct of width b and height hFor a rectangular duct of width b and height h

The hydraulic diameter concept can be applied in theThe hydraulic diameter concept can be applied in theapproximate rangeapproximate range


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107


The friction factor can be written asThe friction factor can be written as f=C/f=C/ ReRe hh , where the, where theconstant C depends on the particular shape of the duct, andconstant C depends on the particular shape of the duct, andReRe hh is the Reynolds number based on the hydraulicis the Reynolds number based on the hydraulicdiameter.diameter.

The hydraulic diameter is also used in the definition of theThe hydraulic diameter is also used in the definition of thefriction factor,friction factor, , and the relative, and the relativeroughnessroughness /D/D hh ..

)g2/V)(D/(f h 2hL A=



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108

Noncircular DuctsNoncircular Ducts 4/4

For Laminar flow, the value of C=fRe h have been obtainedfrom theory and/or experiment for various shapes.

For turbulent flow in ducts of noncircular cross section,calculations are carried out by using the Moody chart data

for round pipes with the diameter replaced by the hydraulicdiameter and the Reynolds number based on the hydraulicdiameter.

The Moody chart, developed for round pipes, can alsoThe Moody chart, developed for round pipes, can also be used for noncircular ducts. be used for noncircular ducts.

Friction Factor for Laminar Flow inFriction Factor for Laminar Flow in


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109

Noncircular DuctsNoncircular Ducts

Example 8 7 Noncircular DuctExample 8 7 Noncircular Duct


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110

Example 8.7 Noncircular Duct Example 8.7 Noncircular Duct

zz Air at temperature of 120Air at temperature of 120 F and standard pressure flows from aF and standard pressure flows from afurnace through an 8furnace through an 8 --in.in. --diameter pipe with an average velocity of diameter pipe with an average velocity of

10ft/s. It then passes through a transition section and into a s10ft/s. It then passes through a transition section and into a s quarequareduct whose side is of length a. The pipe and duct surfaces areduct whose side is of length a. The pipe and duct surfaces aresmooth (smooth ( =0). Determine the duct size, a, if the head loss per foot=0). Determine the duct size, a, if the head loss per foot

is to be the same for the pipe and the duct.is to be the same for the pipe and the duct.



111/160

111

Example 8.7Example 8.7 SolutionSolution

The head loss per foot for the pipeThe head loss per foot for the pipe

2sh a

49.3

A

QVa

P

A4D ====

For given pressure and temperatureFor given pressure and temperature =1.89=1.89 1010 --44ftft22/s/s

g2V

Df h 2

L =A

35300VD

Re =

=

0512.0g2

V

D

f h 2sh

L ==AFor the square ductFor the square duct



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112


Have three unknown ( Have three unknown ( a,f a,f , and , and Re Rehh ) and three equation ) and three equation Eqs Eqs . 1, 2, and either in graphical form the Moody chart or . 1, 2, and either in graphical form the Moody chart or the Colebrook equationthe Colebrook equation

The Reynolds number based on the hydraulic diameter The Reynolds number based on the hydraulic diameter

a

1089.1

1089.1

a)a/49.3(DVRe

4

4

2hs

h

=

=

=

5/1222

s

h

L f 30.1a)2.32(2)a/49.3(

af

0512.0g2

VDf h ====

A(1)(1)

(2)(2)

Find aFind a



113/160

113


Use the Moody chartUse the Moody chart

Assume the friction factor for the duct is the same as for the pAssume the friction factor for the duct is the same as for the p ipe.ipe.That is, assume f=0.022.That is, assume f=0.022.

FromFrom EqEq . 1 we obtain a=0.606 ft.. 1 we obtain a=0.606 ft.

FromFrom EqEq . 2 we have. 2 we have ReRe hh=3.05=3.05 1010 44From Moody chart we find f=0.023, which does not quite agree theFrom Moody chart we find f=0.023, which does not quite agree the

assumed value of f.assumed value of f.Try again, using the latest calculated value of f=0.023 as our gTry again, using the latest calculated value of f=0.023 as our g uess.uess.

The final result is f=0.023,The final result is f=0.023, ReRe hh =3.05=3.05 1010 44 , and a=0.611ft., and a=0.611ft.

Pipe Flow ExamplesPipe Flow Examples 1/21/2


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114


The energy equation, relating the conditions at any twoThe energy equation, relating the conditions at any two points 1 and 2 for a single points 1 and 2 for a single -- path pipe system path pipe system

by judicious choice of points 1 and 2 we can analyze not by judicious choice of points 1 and 2 we can analyze notonly the entire pipe system, but also just a certain sectiononly the entire pipe system, but also just a certain sectionof it that we may be interested in.of it that we may be interested in.

g2V

Df h

2

Lmajor

A

+==

++

++ or minmajor LLL2

2222

1

2111 hhhz

g2V

g p

zg2V

g p

Major lossMajor loss Minor lossMinor lossg2

VK h2

LL or min=

Pipe Flow ExamplesPipe Flow Examples 2/22/2


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115


Single pipe whose length may be interrupted by variousSingle pipe whose length may be interrupted by variouscomponents.components.

Multiple pipes in different configurationMultiple pipes in different configuration ParallelParallel

SeriesSeries Network Network

SingleSingle --Path SystemsPath Systems 1/21/2


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116

SingleSingle -Path SystemsPath Systems

Pipe flow problems can be categorized by what parametersPipe flow problems can be categorized by what parametersare given and what is to be calculated.are given and what is to be calculated.

SingleSingle --Path SystemsPath Systems 2/22/2


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117

SingleSingle Path SystemsPath Systems

Given pipe (L and D), and flow rate, and Q, find pressureGiven pipe (L and D), and flow rate, and Q, find pressuredropdrop p p

GivenGiven p, D, and Q, find L. p, D, and Q, find L.

GivenGiven p, L, and D, find Q. p, L, and D, find Q.

GivenGiven p p , L, and Q, find D., L, and Q, find D.

Given L D and QGiven L D and Q findfind pp


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118

Given L , D, and Q,Given L , D, and Q, findfind pp

The energy equationThe energy equation

The flow rate leads to the Reynolds number and hence theThe flow rate leads to the Reynolds number and hence thefriction factor for the flow.friction factor for the flow.Tabulated data can be used for minor loss coefficients andTabulated data can be used for minor loss coefficients and

equivalent lengths.equivalent lengths.The energy equation can then be used to directly to obtainThe energy equation can then be used to directly to obtainthe pressure drop.the pressure drop.

+==

++

++ or minmajor LLL2

2222

1

2111 hhhz

g2V

g pz

g2V

g p

GivenGiven p D and Qp, D, and Q, find Lfind L


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119

GivenGiven p, D, and Q,p, D, and Q, find Lfind L


The flow rate leads to the Reynolds number and hence theThe flow rate leads to the Reynolds number and hence thefriction factor for the flow.friction factor for the flow.Tabulated data can be used for minor loss coefficients andTabulated data can be used for minor loss coefficients and

equivalent lengths.equivalent lengths.The energy equation can then be rearranged and solvedThe energy equation can then be rearranged and solveddirectly for the pipe length.directly for the pipe length.

+==

++

++ or minmajor LLL22

2221

2

111 hhhzg2Vg pzg2Vg p

GivenGiven p L and Dp, L, and D, find Qfind Q 1/21/2


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GivenGiven p, L, and D,p, L, and D, find Qfind Q

These types of problems required either manual iterationThese types of problems required either manual iterationor use of a computer application.or use of a computer application.

The unknown flow rate or velocity is needed before theThe unknown flow rate or velocity is needed before theReynolds number and hence the friction factor can beReynolds number and hence the friction factor can be

found.found.

Repeat the iteration processRepeat the iteration processf f VV ReRe f until convergencef until convergence

GivenGiven pp L and D, L, and D, find Qfind Q 2/22/2


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GivenGiven pp , L, and D,, L, and D, find Qfind Q

First, we make a guess for f and solve the energy equationFirst, we make a guess for f and solve the energy equationfor V in terms of known quantities and the guessedfor V in terms of known quantities and the guessedfriction factor f.friction factor f.

Then we can compute a Reynolds number and henceThen we can compute a Reynolds number and hence

obtain a new value for f.obtain a new value for f.

Repeat the iteration processRepeat the iteration process

f f VV ReRe f until convergencef until convergence

GivenGiven p, L, and Q,p, L, and Q, find Dfind D 1/21/2


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GivenGiven p, L, and Q,p, L, and Q, find Dfind D

These types of problems required either manual iterationThese types of problems required either manual iterationor use of a computer application.or use of a computer application.

The unknown diameter is needed before the ReynoldsThe unknown diameter is needed before the Reynoldsnumber and relative roughness, and hence the frictionnumber and relative roughness, and hence the friction

factor can be found.factor can be found.

GivenGiven p, L, and Q,p, L, and Q, find Dfind D 2/22/2


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GivenGiven p, L, and Q,p, L, and Q, find Dfind D

First, we make a guess for f and solve the energy equationFirst, we make a guess for f and solve the energy equationfor D in terms of known quantities and the guessedfor D in terms of known quantities and the guessedfriction factor f.friction factor f.

Then we can compute a Reynolds number and henceThen we can compute a Reynolds number and hence

obtain a new value for f.obtain a new value for f.

Repeat the iteration processRepeat the iteration process

f f DD Re andRe and /D/D f until convergencef until convergence

Example 8.8 Type I Determine PressureExample 8.8 Type I Determine PressureDD


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DropDropzz Water at 60Water at 60 F flows from the basement to the second floor through theF flows from the basement to the second floor through the

0.750.75 --in. (0.0625in. (0.0625 --fy)fy) --diameter copper pipe (a drawn tubing) at a rate odiameter copper pipe (a drawn tubing) at a rate of

Q = 12.0 gal/min = 0.0267 ftQ = 12.0 gal/min = 0.0267 ft 33/s and exits through a faucet of diameter /s and exits through a faucet of diameter 0.50 in. as shown in Figure E8.8.0.50 in. as shown in Figure E8.8.

Determine the pressure atDetermine the pressure at

point (1) point (1) if: (a) all lossesif: (a) all lossesare neglected, (b) the onlyare neglected, (b) the onlylosses included are major losses included are major

losses, or (c) all losses arelosses, or (c) all losses areincluded.included.



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Example 8.8a p e 8.8 SolutionSo ut o


45000/VDReft/slb1034.2

ft/slug94.1s/ft70.8...AQ

V

25

3

11

===

====

s/ft6.19...A/QV

) jetfree(0 p,ft20z,0z

22

221

======

L2

2222

1

2111

hzg2V

g p

zg2V

g p

++

+=+

+

The flow is turbulentThe flow is turbulent

L2

12

221

21 h)VV(z p ++=

Head loss is different for Head loss is different for each of the three cases.each of the three cases.



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126

Example 8.8p Solution

(a) If all losses are neglected ((a) If all losses are neglected ( hhLL=0)=0)

45000Re108D/000005.0 5 ===

psi7.10ft/lb1547...)VV(z p 2212

221

21 ===+=

F=0.0215F=0.0215

(b) If the only losses included are the major losses, the head l(b) If the only losses included are the major losses, the head l oss isoss is

g2

V

Df h

21

L

A= Moody chartMoody chart

p3.21ft/lb3062...2

VD

)ft60(f )VV(z p 22

121

222

121 ====++= A



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a p e 8.8p So ut o

(c) If major and minor losses are included(c) If major and minor losses are included

]2)5.1(410[2

)s/ft70.8()ft/slugs94.1( psi3.21

2

VK psi3.21 p

2V

K g2V

Df )VV(z p

23

2

L1

2

L

212

12

221

21

+++=

+=

+++=

A

psi5.30 psi17.9 psi3.21 p1 =+=



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pp

Example 8.9 Type I, Determine Head LossExample 8.9 Type I, Determine Head Loss


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p yp ,p yp

zz Crude oil at 140Crude oil at 140 F withF with = 5 3.7 lb/ft= 5 3.7 lb/ft 33 andand = 8= 8 1010 --5 lb5 lb s/fts/ft 22 (about(aboutfour times the viscosity of water) is pumped across Alaska throufour times the viscosity of water) is pumped across Alaska throu ghgh

the Alaska pipeline, a 799the Alaska pipeline, a 799 --milemile --along, 4along, 4 --ftft --diameter steel pipe, at adiameter steel pipe, at amaximum rate of Q = 2.4 million barrel/day = 117ftmaximum rate of Q = 2.4 million barrel/day = 117ft 33/s, or /s, or V=Q/A=9.31 ft/s. Determine the horsepower needed for the pumpsV=Q/A=9.31 ft/s. Determine the horsepower needed for the pumps

that drive this large system.that drive this large system.



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pp

The energy equation between points (1) and (2)The energy equation between points (1) and (2)

Assume that zAssume that z 11=z=z 22, p, p 11=p=p 22=V=V 11=V=V 22=0 (large, open tank)=0 (large, open tank)

ft17700...g2VDf hh2

PL ====A

L2

2

22P1

2

11 hzg2

V phzg2

V p +++=+++hh

PPis the head provided to the oilis the head provided to the oil

by the pump. by the pump.

Minor losses are negligible because of the large lengthMinor losses are negligible because of the large length --toto--diameter ratio of the relatively straight, uninterrupted pipe.diameter ratio of the relatively straight, uninterrupted pipe.

f=0.0124 from Moody chartf=0.0124 from Moody chart /D=(0.00015ft)/(4ft), Re=/D=(0.00015ft)/(4ft), Re= ....



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pp

The actual power supplied to the fluid.The actual power supplied to the fluid.

hp202000s/lbft550

hp1...QhP Pa = ==

Example 8.10 Type II, DetermineExample 8.10 Type II, Determine FlowrateFlowrate


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p yp

zz According to an appliance manufacturer, the 4According to an appliance manufacturer, the 4 --inin --diameter diameter galvanized iron vent on a clothes dryer is not to contain more tgalvanized iron vent on a clothes dryer is not to contain more t hanhan

20 ft of pipe and four 9020 ft of pipe and four 90 elbows. Under these conditions determineelbows. Under these conditions determinethe air the air flowrateflowrate if the pressure within the dryer is 0.20 inches of if the pressure within the dryer is 0.20 inches of water. Assume a temperature of 100water. Assume a temperature of 100 and standard pressure.and standard pressure.



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pp

Application of the energy equation between the inside of the dryApplication of the energy equation between the inside of the dry er,er, point (1), and the exit of the vent pipe, point (2) gives point (1), and the exit of the vent pipe, point (2) gives

++++=++ g2V

K g2

VD

f zg2

V pz

g2V p 2

L

2

2

222

1

211

A

Assume that zAssume that z

11=z=z

22, p, p

22=0, V=0, V

11=0=0

231

OH

1 ft/lb04.1)ft/lb4.62(.in12

ft1.)in2.0( pin2.0

p

2

=

==

WithWith =0.0709lb/ft=0.0709lb/ft 33, V, V 22=V, and=V, and =1.79=1.79 1010 --44 ftft22/s./s.

2V)f 605.7(945 += (1)(1)f is dependent on Re, which is dependent on V, and unknown.f is dependent on Re, which is dependent on V, and unknown.



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pp

V1860...VD

Re === (2)(2)

We have three relationships (We have three relationships ( EqEq . 1, 2, and the. 1, 2, and the / D =0.0015 curve of / D =0.0015 curve of the Moody chart) from which we can solve for the three unknownsthe Moody chart) from which we can solve for the three unknowns f,f,Re, and V.Re, and V.

This is done easily by iterative scheme as follows.This is done easily by iterative scheme as follows.Assume f=0.022Assume f=0.022 V=10.4ft/s (V=10.4ft/s ( EqEq . 1). 1) Re=19,300 (Eq.2) f=0.029f=0.029

Assume f=0.029Assume f=0.029 V10.1ft/sV10.1ft/s Re=18,800Re=18,800 f=0.029f=0.029

/s0.881ft...AVQ 3===

Example 8.11 Type II, DetermineExample 8.11 Type II, Determine FlowrateFlowrate


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135

zz The turbine shown in Figure E8.11 extracts 50 hp from the water The turbine shown in Figure E8.11 extracts 50 hp from the water flowing through it. The 1flowing through it. The 1 --ftft --diameter, 300diameter, 300 --ftft --long pipe is assumed tolong pipe is assumed to

have a friction factor of 0.02. Minor losses are negligible. Dethave a friction factor of 0.02. Minor losses are negligible. Det ermineerminethethe flowrateflowrate through the pipe and turbine.through the pipe and turbine.



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136

p

The energy equation can be applied between the surface of the laThe energy equation can be applied between the surface of the la kekeand the outlet of the pipe asand the outlet of the pipe as

TL2

222

1

211 hhz

g2V p

zg2

V p ++++

=++

Where pWhere p11=V=V

11= p= p

22=z=z

22=0, z=0, z

11=90ft, and V=90ft, and V

22=V, the fluid velocity in=V, the fluid velocity in

the pipethe pipe

0561V90V107.0 3 =+

ftV0932.0

g2

V

D

f h 22

L ==A

ft

V

561...

Q

Ph aT ==

=

There are two real, positive roots: V=6.58 ft/s or V=24.9 ft/s. The third

root is negative (V=-31.4ft/s) and has no physical meaning for this flow.



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p

Two acceptableTwo acceptable flowratesflowrates areare

s/ft6.19...VD4

Q

s/ft17.5...VD4Q

32

32

=====

=

Example 8.12 Type III Without Minor Example 8.12 Type III Without Minor Losses Determine DiameterLosses Determine Diameter


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138

Losses, Determine Diameter Losses, Determine Diameter zz Air at standard temperature and pressure flows through a horizonAir at standard temperature and pressure flows through a horizon tal,tal,

galvanized iron pipe (galvanized iron pipe ( = 0.0005 ft) at a rate of 2.0ft= 0.0005 ft) at a rate of 2.0ft 33/s. Determine/s. Determine

the minimum pipe diameter if the pressure drop is to be no morethe minimum pipe diameter if the pressure drop is to be no morethan 0.50than 0.50 psi psi per 100 ft of pipe per 100 ft of pipe ..



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139

Assume the flow to be incompressible withAssume the flow to be incompressible with =0.00238 slugs/ft=0.00238 slugs/ft 33 andand =3.74=3.74 1010 --77 lblb s/fts/ft 2.2.

If the pipe were too long, the pressure drop from one end to theIf the pipe were too long, the pressure drop from one end to the other,other, p p11-- p p22, would not be small relative to the pressure at the beginning,, would not be small relative to the pressure at the beginning, andandcompressible flow considerations would be required.compressible flow considerations would be required.

With zWith z 11=z=z 22, V, V 11=V=V 22 , The energy equation becomes, The energy equation becomes gVDf p p2

21 +=A

2

D

55.2

A

QV ==

g2

V

)ft/slugs00238.0(D

)ft100(

f ft/lb)144)(5.0( p p

232

21 ==1/50.404f D = (1)(1)



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140

D1062.1

...VD

Re4==

= (2)(2)

(3)(3)D

0005.0D

=

We have four equations (We have four equations ( EqEq . 1, 2, 3, and either the Moody chart or . 1, 2, 3, and either the Moody chart or the Colebrook equation) and four unknowns (f, D,the Colebrook equation) and four unknowns (f, D, / D, and Re)/ D, and Re)from whichfrom which the solution can be obtained by trialthe solution can be obtained by trial --andand --error methods.error methods.

Repeat the iteration processRepeat the iteration processf f DD Re andRe and /D/D f until convergencef until convergence

(1)(1) (2)(2) (3)(3)

Example 8.13 Type III With Minor Losses,Example 8.13 Type III With Minor Losses,Determine DiameterDetermine Diameter


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141

Determine Diameter Determine Diameter zz Water at 60Water at 60 F (F ( =1.21=1.21 1010 --55 ftft22/s) is to flow from reservoir A to/s) is to flow from reservoir A to

reservoir B through a pipe of length 1700 ft and roughness 0.000reservoir B through a pipe of length 1700 ft and roughness 0.000 5 ft5 ft

at a rate of Q= 26 ftat a rate of Q= 26 ft 33/s as shown in Figure E8.13. The system/s as shown in Figure E8.13. The systemcontains a sharpcontains a sharp --edged entrance and four flanged 45edged entrance and four flanged 45 elbow.elbow.Determine the pipe diameter needed.Determine the pipe diameter needed.



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142

The energy equation can be applied between two points on theThe energy equation can be applied between two points on thesurfaces of the reservoirs (psurfaces of the reservoirs (p 11=V=V 11= p= p 22=z=z 22=V=V 22=0)=0)

+=

+++

=++

L2

1

L2

2221

211

K D

f g2

Vz

hzg2

V pzg2

V p

A

2D1.33

AQ

V == K Lent =0.5, K Lelbow =0.2, and K Lexit =1

+++= ]15.0)2.0(4[

D1700

f )s/ft2.32(2

Vft44 2

2

(1)(1)D00135.0D00152.0f 5 =



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143

D1074.2

...VD

Re6==

= (2)(2)

(3)(3)D

0005.0D

=

We have four equations (We have four equations ( EqEq . 1, 2, 3, and either the Moody chart or . 1, 2, 3, and either the Moody chart or the Colebrook equation) and four unknowns (f, D,the Colebrook equation) and four unknowns (f, D, / D, and Re)/ D, and Re)from whichfrom which the solution can be obtained by trialthe solution can be obtained by trial --andand --error methods.error methods.

Repeat the iteration processRepeat the iteration processDD f f Re andRe and /D/D f until convergencef until convergence

(1)(1) (2)(2) (3)(3)

MultipleMultiple --Path SystemsPath SystemsSeries and Parallel Pipe SystemSeries and Parallel Pipe System


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Series and Parallel Pipe SystemSeries and Parallel Pipe System

321 LLL321 hhhQQQQ ==++=

321BA LLLL321 hhhhQQQ ++===

MultipleMultiple --Path SystemsPath SystemsMultiple Pipe Loop SystemMultiple Pipe Loop System


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Multiple Pipe Loop SystemMultiple Pipe Loop System

( )

( )

3L2L

3L1LB

2BB

A

2AA

2L1LB

2BB

A

2AA

321

hh

31hhzg2

V pz

g2V p

21hhzg2

V pz

g2

V p

QQQ

=

++++

=++

++++

=++

+=

Multiple-Path SystemsThree-Reservoir System


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Three Reservoir SystemIf valve (1) was closed, reservoir BIf valve (1) was closed, reservoir B reservoir Creservoir CIf valve (2) was closed, reservoir AIf valve (2) was closed, reservoir A reservoir Creservoir C

If valve (3) was closed, reservoir AIf valve (3) was closed, reservoir A reservoir Breservoir BWith all valves openWith all valves open ..

( )

( )CBhhzg2

V pz

g2V p

BAhhz

g2

V pz

g2

V p

QQQ

3L2LC

2CC

B

2BB

2L1LB

2BB

A

2AA

321

++++

=++

++++

=++

+=

Example 8.14 Three reservoir, MultipleExample 8.14 Three reservoir, MultiplePipe SystemPipe System

7/29/2019 27112846 8 in Compressibl

27112846 8 in compressible flow

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