fundamentals of fluid mechanicsfluid mechanics chapter …cau.ac.kr/~jjang14/fme/chap8.pdf · fluid...

FUNDAMENTALS OFFUNDAMENTALS OFFLUID MECHANICSFLUID MECHANICSFLUID MECHANICSFLUID MECHANICS

Chapter 8 Pipe FlowChapter 8 Pipe FlowChapter 8 Pipe Flow Chapter 8 Pipe Flow

1

MAIN TOPICSMAIN TOPICSMAIN TOPICSMAIN TOPICS

G l Ch i i f Pi FlG l Ch i i f Pi FlGeneral Characteristics of Pipe FlowGeneral Characteristics of Pipe FlowFully Developed Laminar FlowFully Developed Laminar FlowFully Developed Turbulent FlowFully Developed Turbulent FlowDimensional Analysis of Pipe FlowDimensional Analysis of Pipe FlowDimensional Analysis of Pipe FlowDimensional Analysis of Pipe FlowPipe Flow ExamplesPipe Flow ExamplesPi Fl MPi Fl MPipe Flowrate MeasurementPipe Flowrate Measurement

2

IntroductionIntroductionIntroductionIntroduction

Flows completely bounded by solid surfacesFlows completely bounded by solid surfaces are called are called INTERNAL INTERNAL FLOWSFLOWS which include flows through which include flows through pipespipes ((Round cross sectionRound cross section), ), ductsducts (NOT(NOT Round cross sectionRound cross section)) nozzles diffusers suddennozzles diffusers suddenductsducts (NOT (NOT Round cross sectionRound cross section)), nozzles, diffusers, sudden , nozzles, diffusers, sudden contractions 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--sectionalsectional The basic principles involved are independent of the crossThe basic principles involved are independent of the cross sectional sectional shape, although the details of the flow may be dependent on it.shape, although the details of the flow may be dependent on it.

The The flow regime (laminar or turbulent)flow regime (laminar or turbulent) of internal flows is primarily of internal flows is primarily g ( )g ( ) p yp ya function of the Reynolds numbera function of the Reynolds number ((-->inertial force/viscous force).>inertial force/viscous force).Laminar flow: Can be solved analytically.Laminar flow: Can be solved analytically.y yy yTurbulent flow: Rely heavily on semiTurbulent flow: Rely heavily on semi--empirical theories and empirical theories and

experimental data.experimental data.3

pp

General Characteristics of General Characteristics of Pi FlPi FlPipe FlowPipe Flow

Laminar vs. TurbulentLaminar vs. TurbulentEntrance Region vs. Fully Developed FlowEntrance Region vs. Fully Developed Flow

4

Pipe SystemPipe SystemPipe SystemPipe System

A pipe system include the pipes themselves A pipe system include the pipes themselves (perhaps of more than one diameter), the various (perhaps of more than one diameter), the various (p p ),(p p ),fittings, the flowrate control devices valves) , and fittings, the flowrate control devices valves) , and the pumps or turbinesthe pumps or turbinesthe pumps or turbines.the pumps or turbines.

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

Pipe flow: Flows Pipe flow: Flows completely filling the pipecompletely filling the pipe. (a). (a)The The pressure gradientpressure gradient along the pipe is main driving force.along the pipe is main driving force.p gp g g p p gg p p g

Open channel flow: Flows Open channel flow: Flows without completely filling the without completely filling the pipepipe. (b). (b)pipepipe. (b). (b)The The gravitygravity alone is the driving force.alone is the driving force.

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

The flow of a fluid in a pipe may be The flow of a fluid in a pipe may be Laminar ? Or Laminar ? Or Turbulent ?Turbulent ?

Osborne ReynoldsOsborne Reynolds, a British scientist and mathematician, , a British scientist and mathematician, was the first to distinguish the difference between these was the first to distinguish the difference between these ggclassification of flow by using a classification of flow by using a simple apparatussimple apparatus as as shown.shown.

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

For “For “small enough flowratesmall enough flowrate” the dye streak will remain as a ” the dye streak will remain as a wellwell--defined linedefined line as it flows along, with only slight blurring due as it flows along, with only slight blurring due t l l diff i f th d i t th di tt l l diff i f th d i t th di tto 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 “intermediate flowrateintermediate flowrate” the dye ” the dye fl i i d d i i b f i lfl i i d d i i b f i lfluctuates in time and space, and intermittent bursts of irregular fluctuates in time and space, and intermittent bursts of irregular behavior appear along the streak.behavior appear along the streak.

F “F “l h fl tl h fl t ” h d k l” h d k lFor “For “large enough flowratelarge enough flowrate” the dye streak almost ” the dye streak almost immediately become blurred and spreads across the entire pipe in immediately become blurred and spreads across the entire pipe in aa randomrandom fashionfashiona a randomrandom fashion.fashion.

8

Time Dependence of Time Dependence of Fluid Velocity at a PointFluid Velocity at a Point

9

Indication of Indication of Laminar or Turbulent FlowLaminar or Turbulent Flow ThTh fl tfl t h ld bh ld b l d b R ldl d b R ld The term The term flowrateflowrate should be should be replaced by Reynolds replaced by Reynolds

numbernumber, ,where , ,where VV is the average velocity in the pipe, is the average velocity in the pipe, and and L L is the characteristic dimension of a flow. is the characteristic dimension of a flow. LL is usually is usually D D

/VLRe yy

(diameter)(diameter) in a pipe flow. in a pipe flow. --> a measure of inertial force to the > a measure of inertial force to the viscous force.viscous force.

I iI i l h fl id l il h fl id l i h d i h h f hh d i h h f h It is It is not only the fluid velocitynot only the fluid velocity that determines the character of the that determines the character of the flow flow –– its density, viscosity, and the pipe size are of equal its density, viscosity, and the pipe size are of equal importance.importance.pp

For general engineering purpose, the flow in a For general engineering purpose, the flow in a round piperound pipeLaminarLaminar 2100R e TransitionalTransitionalTurbulentTurbulent

2100R e

4000＞R e10

4000＞R e

Entrance Region and Entrance Region and Fully Developed Flow Fully Developed Flow 1/51/5

Any fluidAny fluid flowing in a pipeflowing in a pipe had to enter the pipe at some had to enter the pipe at some location.location.

The region of flow near where the fluid enters the pipe is The region of flow near where the fluid enters the pipe is termed the termed the entrance (entry) regionentrance (entry) region or developing flow or developing flow y gy g p g fp g fregionregion..

11


The fluid The fluid typicallytypically enters the pipe with a enters the pipe with a nearly uniform nearly uniform velocity profilevelocity profile at section (1).at section (1).

As the fluid moves through the pipe, viscous effects cause As the fluid moves through the pipe, viscous effects cause it to stick to the pipe wallit to stick to the pipe wall ((the no slip boundary the no slip boundary p pp p (( p yp yconditioncondition))..

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A A boundary layerboundary layer in which in which viscous effects are importantviscous effects are important is is produced along the pipe wall such that the produced along the pipe wall such that the initial velocity initial velocity profile changes with distance along the pipe, x , until the profile changes with distance along the pipe, x , until the fluid reaches the end of the fluid reaches the end of the entrance lengthentrance length, section (2), , section (2), beyond which the velocity profile does not vary with xbeyond which the velocity profile does not vary with x..

The boundary layer has grown in thickness to completely The boundary layer has grown in thickness to completely fill the pipe. fill the pipe.

13


Viscous effects are of considerable importance within the Viscous effects are of considerable importance within the boundary layer.boundary layer. Outside the boundary layer, the viscous Outside the boundary layer, the viscous effects are negligible.effects are negligible.

The The shape of the velocity profileshape of the velocity profile in the pipe depends on in the pipe depends on p y pp y p p p pp p pwhether the flow is laminar or turbulent, as does the length whether the flow is laminar or turbulent, as does the length of the entrance region, of the entrance region, ..

For laminar flowFor laminar flow For turbulent flowFor turbulent flow

eR06.0D 6/1

eR4.4D

14Dimensionless entrance length


Once the fluid reaches the end of the entrance region, Once the fluid reaches the end of the entrance region, section (2), the flow is simpler to describe because section (2), the flow is simpler to describe because the the velocity is a function of only the distance from the pipe velocity is a function of only the distance from the pipe centerline, r, and independent of xcenterline, r, and independent of x..

The flow between (2) and (3) is termedThe flow between (2) and (3) is termed fully developedfully developed..

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Pressure Distribution along PipePressure Distribution along PipePressure Distribution along PipePressure Distribution along PipeIn the entrance region of a pipe the fluidIn the entrance region of a pipe the fluidIn the entrance region of a pipe, the fluid In the entrance region of a pipe, the fluid accelerates or decelerates as it flows. There is accelerates or decelerates as it flows. There is a balance between pressure, viscous, and a balance between pressure, viscous, and inertia (acceleration) forceinertia (acceleration) force..

0

0constant

u

pxp

Th i d f hTh i d f h

0x

There is a balance between pressureThere is a balance between pressure andandviscousviscous forceforce

The magnitude of the The magnitude of the pressure gradient is constant.pressure gradient is constant.

The magnitude of the The magnitude of the pressure gradient is larger pressure gradient is larger than that in the fully than that in the fully

viscousviscous force.force.

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p gp gdeveloped region.developed region.

Fully DevelopedFully Developed LaminarLaminar FlowFlowThere are numerous ways to derive important There are numerous ways to derive important results pertaining to fully developed laminar flow:results pertaining to fully developed laminar flow:From F=ma applied directly to a fluid element.From F=ma applied directly to a fluid element.From the NavierFrom the Navier--Stokes equations of motionStokes equations of motionFrom dimensional analysis methodsFrom dimensional analysis methods

17

From F=maFrom F=ma 1/81/8From F=ma From F=ma 1/81/8

Considering a fully developed axisymmetric laminar flow in a long, straight, constant diameter section of a pipe.

The fluid element The fluid element is a circular cylinder of fluid of length l and radius r centered on the axis of a horizontal pipe of p pdiameter D.

18


Because the velocity is not uniform across the pipe, the initially flat end of the cylinder of fluid at time t become distorted at time t+t when the fluid element has moved to its new location along the pipe.

If the flow is fully developed and steady, the distortion on each end of the fluid element is the same, and no part of the fluid experiences any acceleration as it flows.

V

0tV

0ixuuVV

SteadySteady Fully developedFully developed

19


Apply the Newton’s second Law to the cylinder of fluidApply the Newton’s second Law to the cylinder of fluid

xx maF

22 202 p rx

xx

The force (pressure & friction) balance The force (pressure & friction) balance 21

21

21 02 ppp

rprrpprp rx

rx

Basic balance in forces needed to drive each fluid particle Basic balance in forces needed to drive each fluid particle l h i i h l il h i i h l ialong the pipe with constant velocityalong the pipe with constant velocity

Not function of rNot function of r Independent of rIndependent of r r2Crrx ?

rp rx2

B C r=0B C r=0 =0=0 D

rwrx

2

20Not function of rNot function of rB.C. r=0 B.C. r=0 rxrx=0=0

r=D/2 r=D/2 rxrx= = w < 0w < 0


Th d d ll h l d bTh d d ll h l d b

p w4

The pressure drop and wall shear stress are related byThe pressure drop and wall shear stress are related byrw

rx 2

p rx2

Dp

Valid for both laminar and turbulent flowValid for both laminar and turbulent flow..Drx r

LaminarLaminardu

21drrx


du pdu

p 2Since Since

drdu

2pp

r2

pdrdu

L iL i

rp rx2

With the boundary conditions:With the boundary conditions:

12 Cr

4purdr

2pdu

LaminarLaminar

With the boundary conditions: With the boundary conditions: u=0 at r=D/2 u=0 at r=D/2

2pD161

pDC

222 212116

)(DrV

DrpDru C

2

14

)(RrDru w

Velocity distributionVelocity distribution

p w4

22

4 R

Dp


The shear stress distributionThe shear stress distributionprdu

Volume flowrateVolume flowrate2drrx

222 212116

)(DrV

DrpDru C

2)(uQ2VRrdrruAd CR

2

.....2)(uQ

4

0A

pD

rdrruAd

128

pDQ Poiseuille’s LawPoiseuille’s Law

Valid for Laminar flow only

23


Average velocityAverage velocity

4 pD 2pDQQ

128

pDQ

322

pDRQ

AQVaverage

Point of maximum velocityPoint of maximum velocity

222

0drdu

at r=0at r=0

222 212116

)(DrV

DrpDru C

averageVpRUuu 22

max

24

average4max


Making adjustment to account Making adjustment to account for nonhorizontal pipesfor nonhorizontal pipes

sin pp>0 if the flow is uphill>0 if the flow is uphill 0 if th fl i d hill0 if th fl i d hill sin pp <0 if the flow is downhill<0 if the flow is downhill

2sin0sin2 222

rgprgrrpprp rx

rx

gDpV

sin 2 sin 4DpQ

r

Specific weightgVaverage

,

32 128Q

25

Example 8 2 Laminar Pipe FlowExample 8 2 Laminar Pipe FlowExample 8.2 Laminar Pipe FlowExample 8.2 Laminar Pipe Flow22 33 An oil with a viscosity of An oil with a viscosity of μμ= 0.40 N= 0.40 N··s/ms/m22 and density and density ρρ= 900 kg/m= 900 kg/m33

flows in a pipe of diameter D= 0.20m . flows in a pipe of diameter D= 0.20m . (a) What pressure drop p(a) What pressure drop p pp is needed to produce a flowrate of Q 2 0is needed to produce a flowrate of Q 2 0××1010--55 (a) What pressure drop, p(a) What pressure drop, p11--pp22, is needed to produce a flowrate of Q=2.0, is needed to produce a flowrate of Q=2.0××1010--55

mm33/s if the pipe is horizontal with x/s if the pipe is horizontal with x11=0 and x=0 and x22=10 m? =10 m? (b) How steep a hill, (b) How steep a hill, θθ,must the pipe be on if the oil is to flow through the pipe ,must the pipe be on if the oil is to flow through the pipe

at the same rate as in part (a), but with pat the same rate as in part (a), but with p11=p=p22? ? (c) For the conditions of part (b), if p(c) For the conditions of part (b), if p11=200 kPa, what is the pressure at section, =200 kPa, what is the pressure at section,

xx33=5 m where x is measured along the pipe?=5 m where x is measured along the pipe?xx33 5 m, where x is measured along the pipe?5 m, where x is measured along the pipe?

26

Example 8 2Example 8 2 SolutionSolution1/21/2Example 8.2 Example 8.2 SolutionSolution1/21/2

2100872/VDR 210087.2/VDR e

s/m0637.0QV A

The flow is laminar flowThe flow is laminar flowQ128 kPa4.20...

DQ128ppp 421

Q128

If the pipe is on the hill of angle If the pipe is on the hill of angle θ with Δp=0θ with Δp=0

34.13...

gDQ128sin 4

27


Wi hWi h h l h f h ih l h f h i d i h fld i h flWith pWith p11=p=p22 the length of the pipe, the length of the pipe, , does not appear in the flowrate , does not appear in the flowrate equationequation

Δp=0 for all Δp=0 for all

kPa200ppp 321

28

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

General motion of an General motion of an incompressible Newtonian fluidincompressible Newtonian fluid is governed by the is governed by the continuity equation and the momentum equationcontinuity equation and the momentum equation

Mass conservationMass conservation

Navier-Stokes EquationNavier-Stokes Equation in a cylindrical coordinate

Acceleration

29


Simplify the NavierSimplify the Navier--Stokes equationStokes equation

axial component: axial component: zz

Th fl i d b b l f i ht d iTh fl i d b b l f i ht d iThe flow is governed by a balance of pressure, weight, and viscous The flow is governed by a balance of pressure, weight, and viscous forces in the flow direction.forces in the flow direction.ForFor steady fully developed flow in a pipesteady fully developed flow in a pipe the velocitythe velocityFor For steady, fully developed flow in a pipesteady, fully developed flow in a pipe, , the velocity the velocity contains only an axial componentcontains only an axial component, which is , which is a function of a function of only the radial coordinateonly the radial coordinate i)r(uV

30

yy i)r(uV


1axial component: axial component: xx

rur

rr1sing

xp

i)r(uV

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

pxp.const

xp

xx

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 < ∞ or(2) r = 0 u < ∞ or u/u/r=0r=0

31

(2) r = 0 , u < ∞ or (2) r = 0 , u < ∞ or u/u/r=0r=0

From Dimensional AnalysisFrom Dimensional Analysis 1/31/3From Dimensional Analysis From Dimensional Analysis 1/31/3

Assume that the pressure drop in the horizontal pie, Assume that the pressure drop in the horizontal pie, Δp, is Δp, 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, , the pipe diameter, D, and the , the pipe diameter, D, and the viscosity of the fluid, viscosity of the fluid, μ. μ.

),D,,V(Fp Dimensional analysisDimensional analysis

pD an unknown function of the length to an unknown function of the length to diameter ratio of the pipediameter ratio of the pipe

DV

diameter ratio of the pipe.diameter ratio of the pipe.kk--r = 5 r = 5 ((총총변수변수) ) --33 ((reference reference dimension) dimension) ==2 dimensionless group2 dimensionless group

32

)) g pg p


DC

VpD

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

2DVCp

422 )4/(

4pDCD

CpDAVQ

D 4 C

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

DC 1)4/( 4 V

32 V

ppDCQ resistance Flow

1)4/(

RVi Analogy:Analogy:

332

32D

Vp For a round pipeFor a round pipe


DVp 64/32 2 For a round pipeFor a round pipeDDVDV

DVVp

Re6464/32

2212

21

f i t d thf i t d th f i ti f tf i ti f t2V

Dfp

2

Dynamic pressure Dynamic pressure --> Characteristic pressure > Characteristic pressure

f is termed the f is termed the friction factorfriction factor, or , or sometimes the sometimes the Darcy friction Darcy friction factorfactor --> dimensionless pressure > dimensionless pressure *

2

2 pVpDp

f

drop for internal flowsdrop for internal flows..*2

2

p

DV

f

w864f

2 D

4

2VRef

For laminar flowFor laminar flow

34D4p w

Energy ConsiderationEnergy Consideration 1/31/3Energy Consideration Energy Consideration 1/31/3

The energy equation for The energy equation for incompressible, steady flow between two incompressible, steady flow between two locationslocations

L2

2222

1

2111 hz

g2Vpz

g2Vp

g2V

g2V 2

222

11

zzpphzpzpL

1

212122

11

The The head losshead loss in a in a

pipe ispipe is a result ofa result of

Drgp w

42sin1

pipe is pipe is a result of a result of the viscous shear the viscous shear stressstress on the wall.on the wall.

35D

r2 w 2sinr

gp rx

Example 8 3 Laminar Pipe Flow PropertiesExample 8 3 Laminar Pipe Flow Properties 1/21/2Example 8.3 Laminar Pipe Flow Properties Example 8.3 Laminar Pipe Flow Properties 1/21/2

The flowrate, Q, of corn syrup through the horizontal pipe shown in The flowrate, Q, of corn syrup through the horizontal pipe shown in Figure E8.3 is to be monitored by measuring the pressure difference Figure E8.3 is to be monitored by measuring the pressure difference between sections (1) and (2) It is proposed that Q=Kbetween sections (1) and (2) It is proposed that Q=KΔΔp where thep where thebetween sections (1) and (2). It is proposed that Q=Kbetween sections (1) and (2). It is proposed that Q=KΔΔp, where the p, where the calibration constant, K, is a function of temperature, T, because of calibration constant, K, is a function of temperature, T, because of the variation of the syrup’s viscosity and density with temperature. the variation of the syrup’s viscosity and density with temperature. y p y y py p y y pThese variations are given in Table E8.3. These variations are given in Table E8.3. (a) Plot K(T) versus T for 60(a) Plot K(T) versus T for 60˚F≤T ≤ 160F≤T ≤ 160˚F. F. (b) Determine the wall shear stress and the pressure drop, (b) Determine the wall shear stress and the pressure drop, ΔΔp=pp=p11--pp22, for Q=0.5 , for Q=0.5

ftft33/s and T=100/s and T=100˚F. F. (c) For the conditions of part (b) determine the nest pressure force ((c) For the conditions of part (b) determine the nest pressure force (ππDD22/4)/4)ΔΔpp (c) For the conditions of part (b), determine the nest pressure force.((c) For the conditions of part (b), determine the nest pressure force.(ππDD /4)/4)ΔΔp, p,

and the nest shear force, and the nest shear force, ππDDττww, on the fluid within the pipe between the , on the fluid within the pipe between the sections (1) and (2).sections (1) and (2).

36

Example 8 3 Laminar Pipe Flow PropertiesExample 8 3 Laminar Pipe Flow Properties 1/21/2Example 8.3 Laminar Pipe Flow Properties Example 8.3 Laminar Pipe Flow Properties 1/21/2

37


If the flow is laminar (If the flow is laminar ( > should be verified)> should be verified)

51060.1K

If the flow is laminar (If the flow is laminar (--> should be verified)> should be verified)

pKpDQ4

For T=100For T=100˚F, F, μμ=3.8=3.8××1010--33 lb·s/ftlb·s/ft22, Q=0.5ft, Q=0.5ft33/s/s

pK128

Q

,, μμ , Q, Q

24 ft/lb119...

DQ128p

21001380.../VDR e

4D

s/ft2.10...AQV eA

2w ft/lb241Dp4p

LaminarLaminar

38

w ft/lb24.1...4D

p


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

D2

lb84.5...p4DF

2

p

lb84.5...2D2F wv

The values of these two forces are the same. The net The values of these two forces are the same. The net force is zero; there is no accelerationforce is zero; there is no accelerationforce is zero; there is no acceleration.force is zero; there is no acceleration.

39

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

TurbulentTurbulent pipe flow is actually pipe flow is actually more likely to occurmore likely to occur than 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 anNumerous persons have devoted considerable effort in anNumerous persons have devoted considerable effort in an Numerous persons have devoted considerable effort in an

attempting 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 knowledgeturbulence. Although a considerable amount if knowledge turbulence. Although a considerable amount if knowledge about the topics has been developed, about the topics has been developed, the field of turbulent the field of turbulent flow still remains the least understood area of fluidflow still remains the least understood area of fluidflow still remains the least understood area of fluid flow still remains the least understood area of fluid mechanics.mechanics.Much remains to be learned about the nature of turbulent flowMuch remains to be learned about the nature of turbulent flow

40

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

Transition from Laminar to Turbulent Transition from Laminar to Turbulent Flow in a Pipe Flow in a Pipe 1/21/2

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, whereas value below a particular value the flow is laminar, whereas with the parameter value larger than a certain value the with the parameter value larger than a certain value the flow is turbulent.flow is turbulent.

The important parameters involved and their critical The important parameters involved and their critical values depend on the specific flow situation involved.values depend on the specific flow situation involved.For flow in pipe : Re~4000 Turbulence initiatedTurbulence initiated

Consider a long section of pipe that isConsider a long section of pipe that is

p pFor flow along a plate Rex~500000

Turbulence initiated.Turbulence initiated.

41

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.

Energy ConsiderationsEnergy Considerations 1/1/7 (17 (1--6:6: option)option)Energy Considerations Energy Considerations 1/1/7 (17 (1 6: 6: option)option)

Considering the steady flow through the piping system, including a Considering the steady flow through the piping system, including a reducing elbow. The basic equation for conservation of energy reducing elbow. The basic equation for conservation of energy –– the the first law of thermodynamicsfirst law of thermodynamicsfirst law of thermodynamicsfirst law of thermodynamics

CSCVCS nninShaftinnet dAnVeVde

tdAnVWQ

CS nnCSCVinShaftinnet dAnVdAnVeVdet

WQ

t

Vp

2

Energy equationEnergy equation -p

inShaftinnetCSCVWQAdnVgzVpuVde

t

)2

ˆ(

V 2N Th h i li ibl llN Th h i li ibl ll

Kinetic energyKinetic energy

42gzVue

2ˆ Note: The shear stress power is negligibly small Note: The shear stress power is negligibly small

on a control surface.on a control surface.Internal energyInternal energy

Energy ConsiderationsEnergy Considerations 2/2/77Energy Considerations Energy Considerations 2/2/77

0VdeWhen the flow is steadyWhen the flow is steady

0Vdet CVWhen the flow is steadyWhen the flow is steady

The integral of

dAnVgz2

Vpu2

CS

??????

VpVpVp 222

Uniformly distribution

mgz2

Vpumgz2

VpudAnVgz2

Vpuinout

CS

2

dAVVpˆ

i

2

t

2

CS

mgzVpumgzVpu

dAnVgz2

pu

Only one stream

entering and leavingOnly one stream entering and leaving

43

inin

outout

mgz2

umgz2

u


If h ft k i i l dIf h ft k i i l d

2in

2out zzgVVppuum

If shaft work is involved….If shaft work is involved….

inoutinout

inoutinout

WQ

zzg2

ppuum

OneOne dimensional energy equationdimensional energy equationinnetshaftinnet WQ

pˆh

OneOne--dimensional energy equation dimensional energy equation for steadyfor steady--inin--thethe--mean flowmean flow

h lh l h i i i ih i i i i

puh

22

EnthalpyEnthalpy The energy equation is written in terms The energy equation is written in terms of enthalpy.of enthalpy.

in/netshaftin/netinout

2in

2out

inout WQzzg2

VVhhm

44


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

innetinout

2in

2outinout

inout Qzzg2

VVppuum

For steady, incompressible flow…For steady, incompressible flow…OneOne dimensional energy equationdimensional energy equation

innetinoutin

2inin

out

2outout quugz

2Vpgz

2Vp

2

m innetinoutinout 22

m/Qq innetinnet wherewhere

2i

2t VV

For steady, incompressible, For steady, incompressible, frictionless flowfrictionless flow…… Examples?Examples?

inin

inoutout

out z2Vpz

2Vp

0ˆˆ

Bernoulli equationBernoulli equation

450quu innetinout Frictionless flowFrictionless flow……


For steady incompressibleFor steady incompressible frictional flowfrictional flowFor steady, incompressible, For steady, incompressible, frictional flowfrictional flow……

0quu innetinout Frictional flowFrictional flow……q innetinout

Defining “useful or available energy”… gz2

Vp 2

lossquu innetinout

2

Defining “loss of useful or available energy”…

lossgz2

Vpgz2

Vpin

2inin

out

2outout

22 inout

46


FF t d i ibl fl ith f i ti d h ft kt d i ibl fl ith f i ti d h ft k

ih fii

2in

2outinout

i WQzzgVVppuum

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

innetshaftinnetinoutinout WQzzg2

uum

m )qˆˆ(gVpgVp 2inin

2outout m )quu(wgz

2pgz

2p

innetinoutinnetshaftininin

outoutout

lVpVp 2ii

2tt losswgz

2Vpgz

2Vp

innetshaftininin

outoutout

g 22 VpVpgLsin

ininout

outout hhzg2

Vpzg2

Vp

WW l47Q

Wgm

Wg

wh innetshaftinnetshaftinnetshaft

S

glosshL Head lossHead lossShaft headShaft head


22

Lsin

2inin

out

2outout hhz

g2Vpz

g2Vp

gg

Total head loss , Total head loss , hhLL, is regarded as the sum of major losses, , is regarded as the sum of major losses, hh d td t f i ti l ff t if i ti l ff t i f ll d l d flf ll d l d flhhL majorL major, due to , due to frictional effects in frictional effects in fully developed flow fully developed flow in constant area tubesin constant area tubes, and minor losses, , and minor losses, hhL minorL minor, , resulting from resulting from entrance, fitting, area changesentrance, fitting, area changes, and so on., and so on.

LLL hhh minormajor LLL

48

Major Losses: Friction FactorMajor Losses: Friction FactorMajor Losses: Friction FactorMajor Losses: Friction Factor

The energy equation for The energy equation for steady and incompressiblesteady and incompressible flow flow with with zero shaft workzero shaft work

dAVV2

LhzgV

gpz

gV

gp

2

2222

1

2111

22

1

2Vm

dAnV2

2A

For fully developed flow through a constant area pipe, α1= α2, V1= V2 , where is the kinetic energy coefficient and V is the

2

hpp )(21

V1 V2 , where is the kinetic energy coefficient and V is the average velocity. For a uniform velocity, α=1.

Lhzzg

pp )( 12

21

For hori ontal pipeFor hori ontal pipe hppp

21

49

For horizontal pipe, zFor horizontal pipe, z22=z=z11 Lhgg

Major Losses: Laminar FlowMajor Losses: Laminar FlowMajor Losses: Laminar FlowMajor Losses: Laminar Flow

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

64DV

D32

D4/DV128

DQ128p 4

2

4

DRe64

DVD64

V21

p2

2V

DR64

VD64

2V

DDV

D32h

2V

Dfp

2

e

2

L

2

64l i f

Friction FactorFriction Factor 2/V//Dpf 2

50Relaminarf

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

In turbulent flow, we cannot evaluate the pressure drop analytically; In turbulent flow, we cannot evaluate the pressure drop analytically; we must resort to experimental results and use dimensional analysis we must resort to experimental results and use dimensional analysis to correlate the experimental datato correlate the experimental datato correlate the experimental data.to correlate the experimental data.

In In fully developed turbulent flowfully developed turbulent flow thethepressure drop, pressure drop, △△pp , caused by friction , caused by friction p p,p p, △△pp , y, yin a horizontal constantin a horizontal constant--area pipe is area pipe is known to depend on pipe diameter,D, known to depend on pipe diameter,D, pipe length, pipe length, , pipe roughness,e, , pipe roughness,e, average flow velocity, V, fluid densityaverage flow velocity, V, fluid densityρρ, ,

d fl id i itd fl id i it

,,,,D,VFp

and fluid viscosity,and fluid viscosity,μ.μ.

51

,,,,D,VFp


Applying dimensional analysis, the result were a correlation of the form

VDp

E i h h h di i l h d l i di l

D,

D,

Vp

221

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

V2

DRe,

DVp

221

2V

Dfp

DRe,f Vfh

2

DarcyDarcy--Weisbach equationWeisbach equation

52

D g2D

fh L major

Roughness for PipesRoughness for PipesRoughness for PipesRoughness for Pipes

53

Moody chartMoody chartMoody chartMoody chart

Depending on the specificDepending on the specificDepending on the specific Depending on the specific circumstances involved.circumstances involved.

54

About Moody ChartAbout Moody ChartAbout Moody ChartAbout Moody Chart

FF l il i flfl f 64/R hi h i i d d t f thf 64/R hi h i i d d t f thFor For laminar laminar flow, flow, f=64/Re, which is independent of the f=64/Re, which is independent of the relative roughnessrelative roughness..

ForFor very large Reynoldsvery large Reynolds numbersnumbers f=f=ΦΦ((εε/D) which is/D) which isFor For very large Reynoldsvery large Reynolds numbers, numbers, f=f=ΦΦ((εε/D), which is /D), which is independent of the Reynolds numbersindependent of the Reynolds numbers. .

For flows withFor flows with very large value of Revery large value of Re, commonly termed, commonly termedFor flows with For flows with very large value of Revery large value of Re, commonly termed , commonly termed completely turbulent flow (or wholly turbulent flow), the completely turbulent flow (or wholly turbulent flow), the laminar sublayer is so thin (its thickness decrease with laminar sublayer is so thin (its thickness decrease with i i R ) th t th f h l t li i R ) th t th f h l t lincreasing Re) that the surface roughness completely increasing Re) that the surface roughness completely dominates the character of the flow near the wall.dominates the character of the flow near the wall.

For flows withFor flows with moderate value of Remoderate value of Re the friction factorthe friction factorFor flows with For flows with moderate value of Remoderate value of Re, the friction factor , the friction factor f=f=ΦΦ(Re,(Re,εε/D). /D).

55


Colebrook – To avoid having to use a graphical method for obtaining f for turbulent flows. Valid for the entire nonlaminar Valid for the entire nonlaminar

f h M d hf h M d h

fRe

51.27.3D/log0.2

f1 range of the Moody chart.range of the Moody chart.

Colebrook formulaColebrook formula --> needs > needs

Miler suggests that a single iteration will produce a result within 1 percent if the initial estimate is calc lated from

iteration.iteration.

1 percent if the initial estimate is calculated from274.5D/l250f

9.00 Re7.3log25.0f

56

Example 8.5 Comparison of Laminar or Example 8.5 Comparison of Laminar or Turbulent pressure DropTurbulent pressure Drop 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 such drawn tubing with an average velocity of V = 50 m/s. For such conditions the flow would normally be turbulent However ifconditions the flow would normally be turbulent However ifconditions 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 (the entrance to the tube is very smooth, the air is dust free, the tube does entrance to the tube is very smooth, the air is dust free, the tube does y , ,y , ,not vibrate, etc.), it may be possible to maintain laminar flow. not vibrate, etc.), it may be possible to maintain laminar flow. (a) Determine the pressure drop in a 0.1(a) Determine the pressure drop in a 0.1--m section of the tube if the flow is m section of the tube if the flow is

l il ilaminar. laminar. (b) Repeat the calculations if the flow is turbulent.(b) Repeat the calculations if the flow is turbulent.

57


Under standard temperature and pressure conditionsUnder standard temperature and pressure conditionsUnder standard temperature and pressure conditionsUnder standard temperature and pressure conditionsΡΡ=1.23kg/m=1.23kg/m33, μ=1.79, μ=1.791010--55NNs/ms/m

flowTurbulent700,13.../VDR e

The Reynolds numberThe Reynolds number

,e

If the flow were laminarIf the flow were laminar

f 64/R 0 0467f 64/R 0 0467

kPa1790V1fp 2

f=64/Re=…=0.0467f=64/Re=…=0.0467

kPa179.0...V2D

fp

58


If the flow were turbulentIf the flow were turbulentIf the flow were turbulentIf the flow were turbulent

From Moody chart From Moody chart f=f=ΦΦ(Re,(Re,εε/D) =…0.028/D) =…0.028

kPa076.1...V21

Dfp 2

59

Minor LossesMinor Losses 1/51/5Minor Losses Minor Losses 1/51/5

Most pipe systems consist of Most pipe systems consist of considerably considerably more than straight more than straight pipespipes. These additional . These additional components (valves, bends, tees, components (valves, bends, tees, and the like) add to the overall and the like) add to the overall head loss of the system.head loss of the system.

Such losses are termed MINOR Such losses are termed MINOR LOSS.LOSS. But, it is But, it is not minor at all not minor at all and it may be larger than the and it may be larger than the major losses.major losses.

The flow pattern through a valveThe flow pattern through a valve60

The flow pattern through a valveThe flow pattern through a valve


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

The head loss information for essentially all components is The head loss information for essentially all components is y py pgiven in dimensionless form and given in dimensionless form and based on experimental based on experimental datadata. The most common method used to determine these . The most common method used to determine these head losses or pressure drops is to specify the head losses or pressure drops is to specify the loss loss coefficientcoefficient, K, KLL..LL

61


1h 2L

22L

L V21Kp

V21

pg2/V

hK ormin

Large K : large pressure

drops for given velocities.2

g2V

Df

g2VKh

2eq

2

LL ormin

Minor losses are sometimes gi en in terms of an eq i alent

fDK

gg

Leq

given in terms of an equivalent length eq

The actual value of KL is strongly dependent on the geometry of the component considered. It may also dependent on the fluid

Re),(geometryKL properties. That is

62

Re),(geometryKL


F i l li i h R ld b iF i l li i h R ld b iFor many practical applications the Reynolds number is For many practical applications the Reynolds number is large enoughlarge enough so that so that the flow through the component is the flow through the component is dominated by inertial effectsdominated by inertial effects with viscous effects being ofwith viscous effects being ofdominated by inertial effectsdominated by inertial effects, with viscous effects being of , with viscous effects being of secondary importance. secondary importance.

In a flow that is dominated by inertia effectsIn a flow that is dominated by inertia effects rather thanrather thanIn a flow that is dominated by inertia effectsIn a flow that is dominated by inertia effects rather than rather than viscous effects, it is usually found that viscous effects, it is usually found that pressure drops and pressure drops and head losses correlate directly with the dynamic pressurehead losses correlate directly with the dynamic pressurehead losses correlate directly with the dynamic pressurehead losses correlate directly with the dynamic pressure..

This is the reason why This is the reason why the friction factor for very large the friction factor for very large Reynolds number fully developed pipe flow isReynolds number fully developed pipe flow isReynolds number, fully developed pipe flow is Reynolds number, fully developed pipe flow is independent of the Reynolds numberindependent of the Reynolds number..

63


This is true for flow through pipe components.This is true for flow through pipe components.Thus, Thus, in most cases of practical interest the loss in most cases of practical interest the loss pp

coefficients for components are a function of geometry coefficients for components are a function of geometry onlyonly,,

)t(K yy

)geometry(KL

64

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

Entrance flow condition Entrance flow condition and loss coefficientand loss coefficient((aa) Reentrant, ) Reentrant, KKLL = 0.8= 0.8((bb) sharp) sharp--edged, edged, KKLL = 0.5 = 0.5 (( ) p) p g ,g , LL

((cc) slightly rounded, ) slightly rounded, KKLL = 0.2= 0.2((dd) well) well--rounded, rounded, KKLL = 0.04= 0.04(( )) ,, LL

KKLL = function of rounding of = function of rounding of the inlet edge.the inlet edge.

65


A vena contract region may result because the fluid cannot A vena contract region may result because the fluid cannot turn a sharp rightturn a sharp right--angle corner. The flow is said to separate angle corner. The flow is said to separate from the sharp corner.from the sharp corner.

The maximum velocity velocity at section (2) is greater The maximum velocity velocity at section (2) is greater y y ( ) gy y ( ) gthan that in the pipe section (3), and the pressure there is than that in the pipe section (3), and the pressure there is lower.lower.

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

66


Such is not the case. Although Such is not the case. Although the fluid may be accelerated the fluid may be accelerated very efficiently it is veryvery efficiently it is veryvery efficiently, it is very very efficiently, it is very difficult to slow down difficult to slow down (decelerate) the fluid efficently.(decelerate) the fluid efficently.( ) y( ) y

(2)(2)(3) (3) The extra kinetic The extra kinetic energy of the fluid is partially energy of the fluid is partially lost because of viscous lost because of viscous dissipation, so that the pressure dissipation, so that the pressure d t t t th id ld t t t th id ldoes not return to the ideal does not return to the ideal value.value.

Flow pattern and pressure distribution Flow pattern and pressure distribution f hf h d dd d

67

for a sharpfor a sharp--edged entranceedged entrance

Minor Losses CoefficientMinor Losses Coefficient Exit flowExit flowMinor Losses Coefficient Minor Losses Coefficient Exit flowExit flow

Exit flow condition and Exit flow condition and loss coefficientloss coefficient((aa) Reentrant, ) Reentrant, KKLL = 1.0= 1.0((bb) sharp) sharp--edged, edged, KKLL = 1.0= 1.0(( ) p) p g ,g ,((cc) slightly rounded, ) slightly rounded, KKLL = 1.0= 1.0((dd) well) well--rounded, rounded, KKLL = 1.0= 1.0(( )) ,,

68

Minor Losses CoefficientMinor Losses Coefficient varied diametervaried diameterMinor Losses Coefficient Minor Losses Coefficient varied diametervaried diameter

Loss coefficient for sudden Loss coefficient for sudden contraction, expansion,typical contraction, expansion,typical conical diffuserconical diffuserconical diffuser.conical diffuser.

2

2

1L A

A1K

2A

69

Minor Losses CoefficientMinor Losses Coefficient BendBendMinor Losses Coefficient Minor Losses Coefficient BendBend

Character of the flow in bend Character of the flow in bend and the associated loss and the associated loss coefficientcoefficientcoefficient.coefficient.

Carefully designed guide vanes Carefully designed guide vanes help direct the flow with lesshelp direct the flow with less

70

help direct the flow with less help direct the flow with less unwanted swirl and disturbances.unwanted swirl and disturbances.

Internal Structure of ValvesInternal Structure of ValvesInternal Structure of ValvesInternal Structure of Valves

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

71((dd) stop check valve) stop check valve

Loss Coefficients for Pipe Loss Coefficients for Pipe ComponentsComponents

72

Example 8 6 Minor LossExample 8 6 Minor Loss 1/21/2Example 8.6 Minor Loss Example 8.6 Minor Loss 1/21/2

Air at standard conditions is to flow through the test section Air 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 tunnel circuit wind tunnel shown if Figure E8 6 with a velocity of 200 ft/s The flow is drivenshown if Figure E8 6 with a velocity of 200 ft/s The flow is drivenshown if Figure E8.6 with a velocity of 200 ft/s. The flow is driven shown if Figure E8.6 with a velocity of 200 ft/s. The flow is driven by a fan that essentially increase the static pressure by the amount by a fan that essentially increase the static pressure by the amount pp11--pp99 that is needed to overcome the head losses experienced by the that is needed to overcome the head losses experienced by the pp11 pp99 p yp yfluid as it flows around the circuit. Estimate the value of pfluid as it flows around the circuit. Estimate the value of p11--pp99 and and the horsepower supplied to the fluid by the fan.the horsepower supplied to the fluid by the fan.

73

Example 8 6 Minor LossExample 8 6 Minor Loss 2/22/2Example 8.6 Minor Loss Example 8.6 Minor Loss 2/22/2

74


The maximum velocity within the wind tunnel occurs in the The maximum velocity within the wind tunnel occurs in the test 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 =V=V /c/cof the flow is Maof the flow is Ma55=V=V55/c/c55

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

299

211 VpVp

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

91L999

111 hz

g2Vpz

g2Vp

9191L

pph The total head loss from (1) to (9).The total head loss from (1) to (9).

75


The energy across the fan, from (9) to (1)The energy across the fan, from (9) to (1)2

112

99 VphVp HHpp is the actual head rise supplied is the actual head rise supplied

pp

111

p999 z

g2phz

g2p

pp ppppby the pump (fan) to the air.by the pump (fan) to the air.

91L91

p hpph

The actual power supplied to the air (horsepower, PThe actual power supplied to the air (horsepower, Paa) is obtained ) is obtained from the fan head byfrom the fan head by

91L55p55pa hVAhVAQhP

76


The total head lossThe total head loss

LLLLLLL91L hhhhhhhh scrnozdif3corner2corner8corner7corner LLLLLLL91L hhhhhhhh

V6.0VKhV2.0VKh22

LL

22

LL

2

0.4K2.0Kg2g2g2g2

scrnoz

difdifcorner

LL

LLLL

hp3.62s/lbft34300...Ppsi298.0...)ft560)(ft/lb765.0(hpp

a

291L91

77

Noncircular DuctsNoncircular Ducts 1/41/4Noncircular Ducts Noncircular Ducts 1/41/4

Th i i l l i f i fl b d fTh i i l l i f i fl b d fThe 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 exaggeratedcross sections are not too exaggeratedcross 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 theuse with noncircular geometries by introducing the use with noncircular geometries by introducing the hydraulic diameterhydraulic diameter, defined as, defined as

A4PA4D h

wwhere A is crosshere A is cross--sectional area, and P sectional area, and P i tt d i ti tt d i t

78

is wetted perimeter.is wetted perimeter.


F i l d tF i l d tFor a circular ductFor a circular duct

DRADh 44 2

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

DRP

Dh 2

b/harar1h2

)hb(2bh4

PA4Dh

bbhh

The The hydraulic diameter concept can be applied in the hydraulic diameter concept can be applied in the approximate rangeapproximate range ¼<ar<4¼<ar<4 So the correlations for pipeSo the correlations for pipe

)( bb

approximate rangeapproximate range ¼<ar<4¼<ar<4. So the correlations for pipe . So the correlations for pipe flow give flow give acceptably accurate resultsacceptably accurate results for rectangular ducts.for rectangular ducts.

79


The friction factor can be written as The friction factor can be written as f=C/Ref=C/Rehh, where the , where the constant constant C depends on the particular shapeC depends on the particular shape of the duct, and of the duct, and ReRehh is the is the Reynolds number based on the hydraulic Reynolds number based on the hydraulic diameterdiameter.. Note: C = 64 for a circular tube.Note: C = 64 for a circular tube.

The hydraulic diameter is also used in the definition of the The hydraulic diameter is also used in the definition of the friction factor, ,friction factor, , and the relative and the relative )g2/V)(D/(fh 2

hL roughness roughness /D/Dhh..

)g)(( hL

80


For Laminar flow, the value of C=f·Reh have been obtained from 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 y g yfor round pipes with the diameter replaced by the hydraulic diameter and the Reynolds number based on the hydraulic diameter.

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

81

Friction Factor for Laminar Flow in Friction Factor for Laminar Flow in Noncircular DuctsNoncircular Ducts

Note:Note: CC = 64 for a circular tube= 64 for a circular tubeNote: Note: CC = 64 for a circular tube.= 64 for a circular tube.

82

Example 8 7 Noncircular DuctExample 8 7 Noncircular DuctExample 8.7 Noncircular DuctExample 8.7 Noncircular Duct

˚ Air at temperature of 120Air at temperature of 120˚F and standard pressure flows from a F and standard pressure flows from a furnace 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 square10ft/s It then passes through a transition section and into a square10ft/s. It then passes through a transition section and into a square 10ft/s. It then passes through a transition section and into a square duct whose side is of length a. The pipe and duct surfaces are duct whose side is of length a. The pipe and duct surfaces are smooth (smooth (εε=0). Determine the duct size, a, if the head loss per foot is =0). Determine the duct size, a, if the head loss per foot is (( ) , , p) , , pto be the same for the pipe and the duct.to be the same for the pipe and the duct.

83


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

Vfh 2L

For given pressure and temperatureFor given pressure and temperature νν=1 89=1 891010--44ftft22/s/s

g2D

For given pressure and temperature For given pressure and temperature νν 1.891.891010 ftft /s/s

35300VDRe

0512.0g2

VDfh 2

s

h

L

For the square ductFor the square duct

49.3QVaA4D

g2Dh

84

2sh aAQVa

PD


222 )a/493(fVfh 5/1s

h

L f30.1a)2.32(2)a/49.3(

af0512.0

g2V

Dfh

(1)(1)

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

1089.1a)a/49.3(DVRe42

hs

(2)(2)

Have three unknown (aHave three unknown (a f and Ref and Re ) and three equation) and three equation

a1089.1Re 4h

(2)(2)

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

Find aFind a

85


Use the Moody chartUse the Moody chart

A th f i ti f t f th d t i th f th iA th f i ti f t f th d t i th f th iAssume the friction factor for the duct is the same as for the pipe. Assume the friction factor for the duct is the same as for the pipe. That is, assume f=0.022.That is, assume f=0.022.F E 1 bt i 0 606 ftF E 1 bt i 0 606 ftFrom Eq. 1 we obtain a=0.606 ft.From Eq. 1 we obtain a=0.606 ft.From Eq. 2 we have ReFrom Eq. 2 we have Rehh=3.05=3.05101044

F M d h fi d f 0 023 hi h d i hF M d h fi d f 0 023 hi h d i hFrom Moody chart we find f=0.023, which does not quite agree the From 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 guess.Try again, using the latest calculated value of f=0.023 as our guess.…… The final result is f=0.023, Re…… The final result is f=0.023, Rehh=3.05=3.05101044, and a=0.611ft., and a=0.611ft.

86

Pipe Flow ExamplesPipe Flow Examples 1/21/2Pipe Flow Examples Pipe Flow Examples 1/21/2

The energy equation, relating the conditions at any two The energy equation, relating the conditions at any two points 1 and 2 for a singlepoints 1 and 2 for a single--path pipe system, for path pipe system, for steady and steady and incompressibleincompressible flow with flow with zero shaft workzero shaft work is given byis given by

2

2222

111 hhhVpVp

By judicious choice of points 1 and 2 we can analyze notBy judicious choice of points 1 and 2 we can analyze not

orminmajor LLL2

2221

111 hhhzg2g

pzg2g

p

By judicious choice of points 1 and 2, we can analyze not By judicious choice of points 1 and 2, we can analyze not only the entire pipe system, but also just a certain section only the entire pipe system, but also just a certain section of it that we may be interested inof it that we may be interested inof it that we may be interested in.of it that we may be interested in.

Vfh2

Major lossMajor loss Minor lossMinor loss VKh2

LL 87g2D

fhLmajorMajor lossMajor loss

g2Kh LL ormin

Pipe Flow ExamplesPipe Flow Examples 2/22/2Pipe Flow Examples Pipe Flow Examples 2/22/2

Single pipe whose length may be interrupted by various Single pipe whose length may be interrupted by various components.components.

Multiple pipes in different configurationMultiple pipes in different configurationParallelParallelParallelParallelSeriesSeriesN t kN t kNetwork Network

88

SingleSingle Path SystemsPath Systems 1/21/2SingleSingle--Path Systems Path Systems 1/21/2

Pipe flow problems can be categorized by what parameters Pipe flow problems can be categorized by what parameters are are givengiven and what is to be and what is to be calculatedcalculated..

89

SingleSingle Path SystemsPath Systems 2/22/2SingleSingle--Path Systems Path Systems 2/22/2

Type 1: Given pipe (L and D), and flow rate, and Q, find Type 1: Given pipe (L and D), and flow rate, and Q, find pressure drop Δppressure drop Δp

Type 1: Given Δp, D, and Q, find L. Type 1: Given Δp, D, and Q, find L. Type 2: Given Δp, L, and D, find Q.Type 2: Given Δp, L, and D, find Q.Type 2: Given Δp, L, and D, find Q.Type 2: Given Δp, L, and D, find Q.Type 3: Given Δp, L, and Q, find D.Type 3: Given Δp, L, and Q, find D.

90

Given L D and QGiven L D and Q findfind ΔΔppGiven L , D, and Q, Given L , D, and Q, find find ΔΔpp

Th iTh iThe energy equationThe energy equation

22 VpVp

orminmajor LLL2222

1111 hhhz

g2V

gpz

g2V

gp

The flow rate leads to the Reynolds number and hence the The flow rate leads to the Reynolds number and hence the friction factor for the flow.friction factor for the flow.

Tabulated data can be used for minor loss coefficients and Tabulated data can be used for minor loss coefficients and equivalent lengths.equivalent lengths.

The energy equation can then be used to directly to obtain The energy equation can then be used to directly to obtain the pressure drop.the pressure drop.

91

GivenGiven ΔΔp D and Qp D and Q find Lfind LGiven Given ΔΔp, D, and Q, p, D, and Q, find Lfind L

Th iTh iThe energy equationThe energy equation

2

2222

111 hhhVpVp

orminmajor LLL2

2221

111 hhhzg2g

pzg2g

p

The flow rate leads to the Reynolds number and hence the The flow rate leads to the Reynolds number and hence the friction factor for the flow.friction factor for the flow.

Tabulated data can be used for minor loss coefficients and Tabulated data can be used for minor loss coefficients and equivalent lengths.equivalent lengths.

The energy equation can then be rearranged and solved The energy equation can then be rearranged and solved directly for the pipe length.directly for the pipe length.

92

Given ∆p L and DGiven ∆p L and D find Qfind QGiven ∆p, L, and D, Given ∆p, L, and D, find Qfind QThese types of problems required either manual iteration These types of problems required either manual iteration yp p qyp p q

or use of a computer application.or use of a computer application.The unknown flow rate or velocity is needed before the The unknown flow rate or velocity is needed before the yy

Reynolds number and hence the friction factor can be Reynolds number and hence the friction factor can be found.found.

First, we make a First, we make a guess for fguess for f** and and solve the energy equation solve the energy equation for V in terms of known quantities and the guessed for V in terms of known quantities and the guessed

**friction factor ffriction factor f**..Then we can Then we can compute a Reynolds number and hence compute a Reynolds number and hence

obtain a new value for fobtain a new value for f..Repeat the iteration process Repeat the iteration process

93ff**→ V→ Re→ f until convergence (f→ V→ Re→ f until convergence (f**=f)=f)

GivenGiven ΔΔp L and Qp L and Q find Dfind DGiven Given ΔΔp, L, and Q, p, L, and Q, find Dfind DThese types of problems required either manual iteration These types of problems required either manual iteration yp p qyp p q

or use of a computer application.or use of a computer application.The unknown diameter is needed before the Reynolds The unknown diameter is needed before the Reynolds yy

number and relative roughness, and hence the friction number and relative roughness, and hence the friction factor can be found.factor can be found.

First, we make a guess for fFirst, we make a guess for f** and solve the energy equation and solve the energy equation for D in terms of known quantities and the guessed for D in terms of known quantities and the guessed friction factor ffriction factor f**..

Then we can compute a Reynolds number and hence Then we can compute a Reynolds number and hence obtain a new value for f.obtain a new value for f.

Repeat the iteration process Repeat the iteration process 94

f*→f*→ D→ (Re and D→ (Re and εε/D) → f until convergence /D) → f until convergence (f*=f)(f*=f)

Example 8.8 Type I Determine Pressure Example 8.8 Type I Determine Pressure DropDrop

˚ Water at 60Water at 60˚F flows from the basement to the second floor through the F 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 of diameter copper pipe (a drawn tubing) at a rate of Q = 12 0 gal/min = 0 0267 ftQ = 12 0 gal/min = 0 0267 ft33/s and exits through a faucet of diameter/s and exits through a faucet of diameterQ = 12.0 gal/min = 0.0267 ftQ = 12.0 gal/min = 0.0267 ft33/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 atDetermine the pressure at Determine the pressure at point (1)point (1) if: (a) all losses if: (a) all losses are neglected, (b) the only are neglected, (b) the only losses included are major losses included are major losses, or (c) all losses are losses, or (c) all losses are i l d di l d dincluded.included.

95


ft/l941/ft708QV 3

45000/VDReft/slb1034.2

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

25

3

11

The flow is The flow is turbulentturbulent

The energy equationThe energy equation22 VV

Nearly uniform velocity profileNearly uniform velocity profile

L2222

1111 hz

g2V

gpz

g2V

gp

121

ff )(0200

LhVVzp )( 21

222

121

sftAQV

jetfreepftzz

/6.19.../

)(0,20,0

22

221

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

96sftAQV /70.8/ 11

each of the three cases.each of the three cases.


(a) If all losses are neglected (h(a) If all losses are neglected (h =0)=0)(a) If all losses are neglected (h(a) If all losses are neglected (hLL=0)=0)

psi7.10ft/lb1547...)VV(zp 221

222

121

(b) If the only losses included are the major losses, the head loss is(b) If the only losses included are the major losses, the head loss isV2

45000R108/0000050 5D f=0 0215f=0 0215g2

VD

fh 1L

Moody chartMoody chart

45000Re,108/,000005.0 5 D f=0.0215f=0.0215

psi321ft/lb3062V)ft60(f)VV(p 22

1221

psi3.21ft/lb3062...2D

)(f)VV(zp 21122

121

97


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

VKVf)VV(zp22

1221

Vi

2K

g2Df)VV(zp

2

L12221

)s/ft708(2

VKpsi3.21p

2

L1 ValveValve (pp. 72)(pp. 72) FaucetFaucet

]2)5.1(410[2

)s/ft70.8()ft/slugs94.1(psi3.21 3 (pp )(pp )

ElbowElbow

psi5.30psi17.9psi3.21p1

98

Example 8 9 Type I Determine Head LossExample 8 9 Type I Determine Head LossExample 8.9 Type I, Determine Head LossExample 8.9 Type I, Determine Head Loss

˚ 33 22 Crude oil at 140Crude oil at 140˚F with F with γγ=53.7 lb/ft=53.7 lb/ft33 and and μμ= 8= 81010--5 lb·s/ft5 lb·s/ft22 (about (about four times the viscosity of water) is pumped across Alaska through four times the viscosity of water) is pumped across Alaska through the Alaskathe Alaska pipelinepipeline a 799a 799 milemile along 4along 4 ftft diameterdiameter steel pipesteel pipe at aat athe Alaska the Alaska pipelinepipeline, a 799, a 799--milemile--along, 4along, 4--ftft--diameter diameter steel pipesteel pipe, at a , at a maximum rate of Q = 2.4 million barrel/day = 117ftmaximum rate of Q = 2.4 million barrel/day = 117ft33/s, or /s, or V=Q/A=9.31 ft/s. Determine the horsepower needed for the pumps V=Q/A=9.31 ft/s. Determine the horsepower needed for the pumps Q p p pQ p p pthat drive this large system.that drive this large system.

99


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

hzVphzVp

2222

2111 hhPP is the head provided to the oil is the head provided to the oil

Assume that zAssume that z11=z=z22 pp11=p=p22=V=V11=V=V22=0 (<=0 (<-- large open tanklarge open tank))

LP hzg

phzg

p 2

2221

111

22 PP pp

by the pump.by the pump.

Assume that zAssume that z11 zz22, p, p11 pp22 VV11 VV22 0 (<0 (< large, open tanklarge, open tank))

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 pipediameter ratio of the relatively straight uninterrupted pipe

ft17700...Vfhh2

PL

diameter ratio of the relatively straight, uninterrupted pipe.diameter ratio of the relatively straight, uninterrupted pipe.

ft17700...g2D

fhh PL

f 0 0124 f M d h t /D (0 00015ft)/(4ft) R 7 76 10f 0 0124 f M d h t /D (0 00015ft)/(4ft) R 7 76 1055

Turbulent (Turbulent (αα=1)=1)Table 8.1Table 8.1

100

f=0.0124 from Moody chart ε/D=(0.00015ft)/(4ft), Re=7.76 x 10f=0.0124 from Moody chart ε/D=(0.00015ft)/(4ft), Re=7.76 x 1055

For steel pipeFor steel pipe


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

hp1

hpslbft

hpQhPa 202000/550

1...P

--> It requires many pump stations to be set up.> It requires many pump stations to be set up.

101

Example 8 10 Type II Determine FlowrateExample 8 10 Type II Determine FlowrateExample 8.10 Type II, Determine FlowrateExample 8.10 Type II, Determine Flowrate

According to an appliance manufacturer, the 4According to an appliance manufacturer, the 4--inin--diameter diameter galvanized irongalvanized iron vent on a clothes dryer is not to contain more than vent on a clothes dryer is not to contain more than 20 ft of pipe and four 9020 ft of pipe and four 90˚ elbows Under these conditions determineelbows Under these conditions determine20 ft of pipe and four 9020 ft of pipe and four 90 elbows. Under these conditions determine elbows. Under these conditions determine the the airair flowrate if the pressure within the dryer is 0.20 inches of flowrate 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. Kand standard pressure. KLL = = pp pp LL0.5 for an entrance and 1.5 for elbows.0.5 for an entrance and 1.5 for elbows.

102


Application of the energy equation betweenApplication of the energy equation between the inside of the dryerthe inside of the dryerApplication of the energy equation between Application of the energy equation between the inside of the dryerthe inside of the dryer, , point (1), and the point (1), and the exit of the vent pipeexit of the vent pipe, point (2) gives, point (2) gives

VVVpVp 2222

gVK

gV

Dfz

gVpz

gVp

L 2222 2222

1111

A th tA th t 0 V0 V 00 VKVfVpVp 222222

2111 Assume that zAssume that z11=z=z22, p, p22=0, V=0, V11=0 =0

231

1 ft/lb04.1)ft/lb4.62(i12ft1.)in2.0(pin2.0p

g

KgD

fzg

pzg

pL 2222 2

2221

111

1OH

)(.in12

)(p2

With With γγ=0.0709lb/ft=0.0709lb/ft33, V, V22=V, and =V, and νν=1.79=1.79××1010--44 ftft22/s. /s. 22

2

2

2

3

2

)6057(945

sft2.3225.145.0

12ft4ft201

ftlb0709.0ftlb04.1

Vf

Vf

(1)(1) Assume turbulent flow Assume turbulent flow ((αα=1)=1)

103

)605.7(945 Vf

f is dependent on Re, which is dependent on V, and unknown.f is dependent on Re, which is dependent on V, and unknown.


VD VVD 1860...Re

(2)(2)

We have three relationships (Eq 1 2 and the ε/D=0 0015 curve ofWe have three relationships (Eq 1 2 and the ε/D=0 0015 curve ofFor galvanized ironFor galvanized iron

We have three relationships (Eq. 1, 2, and the ε/D=0.0015 curve of We have three relationships (Eq. 1, 2, and the ε/D=0.0015 curve of the Moody chart) from which we can solve for the three unknowns f, the Moody chart) from which we can solve for the three unknowns 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=0 022→V=10 4ft/s (Eq 1)→Assume f=0 022→V=10 4ft/s (Eq 1)→Re=19 300 (Eq 2)→f=0 029→f=0 029Assume f=0.022→V=10.4ft/s (Eq. 1)→Assume f=0.022→V=10.4ft/s (Eq. 1)→Re=19,300 (Eq.2)→f=0.029→f=0.029

Assume f=0.029 →V10.1ft/s→Re=18,800 →f=0.029 (Final value) Assume f=0.029 →V10.1ft/s→Re=18,800 →f=0.029 (Final value)

/s0.881ft...AVQ 3 TurbulentTurbulent

104

Example 8 11 Type II Determine FlowrateExample 8 11 Type II Determine FlowrateExample 8.11 Type II, Determine FlowrateExample 8.11 Type II, Determine Flowrate

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 to long pipe is assumed to have a friction factor of 0 02 Minor losses are negligible Determinehave a friction factor of 0 02 Minor losses are negligible Determinehave a friction factor of 0.02. Minor losses are negligible. Determine have a friction factor of 0.02. Minor losses are negligible. Determine the flowrate through the pipe and turbine.the flowrate through the pipe and turbine.

105


The energy equation can be applied between the surface of the lakeThe energy equation can be applied between the surface of the lakeThe energy equation can be applied between the surface of the lake The energy equation can be applied between the surface of the lake and the outlet of the pipe asand the outlet of the pipe as

VpVp 22 VpVp 22 TL hhz

gVpz

gVp

2222

1111

22

Where pWhere p =V=V = p= p =z=z =0 z=0 z =90ft and V=90ft and V =V the fluid velocity in=V the fluid velocity in

TL hhzgVpz

gVp

2222

1111

22

ftV09320Vfh 22

Where pWhere p11=V=V11= p= p22=z=z22=0, z=0, z11=90ft, and V=90ft, and V22=V, the fluid velocity in =V, the fluid velocity in the pipethe pipe

ft561Ph a Assume turbulent flowAssume turbulent flowftV0932.0g2D

fh 2L

561 322V

ftV

...Q

h aT

Assume turbulent flow Assume turbulent flow

((αα=1)=1)

056190107.0 -ft 561ft0932.0sft2.322

ft90 322

VV

VVV

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

106

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


T bl flT bl flTwo acceptable flowrates areTwo acceptable flowrates are

s/ft175VDQ 32

At 60At 60ooF 1 21 x 10F 1 21 x 10--55 (ft(ft22/s) Re/s) Re ≈ 10≈ 1055

s/ft6.19...VDQ

s/ft17.5...VD4

Q

32

At 60At 60 F, 1.21 x 10F, 1.21 x 10 (ft(ft /s) Re /s) Re 10 10

s/t6.9...V4

Q

107

Example 8.12 Type III Without Minor Example 8.12 Type III Without Minor Losses, Determine DiameterLosses, Determine Diameter AirAir at standard temperature and pressure flows through a at standard temperature and pressure flows through a horizontalhorizontal, ,

galvanized irongalvanized iron pipe (ε=0.0005 ft) at a rate of 2.0ftpipe (ε=0.0005 ft) at a rate of 2.0ft33/s. Determine the /s. Determine the minimum pipe diameter if the pressure drop is to be no more thanminimum pipe diameter if the pressure drop is to be no more thanminimum pipe diameter if the pressure drop is to be no more than minimum pipe diameter if the pressure drop is to be no more than 0.50 psi per 100 ft of pipe0.50 psi per 100 ft of pipe..

108


Assume the flow to be incompressible with ρ=0.00238 slugs/ftAssume the flow to be incompressible with ρ=0.00238 slugs/ft33 and and μ=3.74μ=3.74××1010--77 lblb．．s/fts/ft2.2.

If the pipe were too long, the pressure drop from one end to the other, If the pipe were too long, the pressure drop from one end to the other, pp11--pp22, would not be small relative to the pressure at the beginning, and , would not be small relative to the pressure at the beginning, and compressible flow considerations would be required.compressible flow considerations would be required.

22

i hi h h i bh i b Vf2

g

VKg

VD

fzgVpz

gVp

L 2222

22

2

2222

1

2111

With zWith z11=z=z22, V, V11=V=V22 , The energy equation becomes, The energy equation becomesgV

Dfpp 21

V)ft/l002380()ft100(fft/lb)144)(50(2

32

552Q

g2)ft/slugs00238.0(

D)(fft/lb)144)(5.0(pp 32

21

1/50 404fD1092D

55.2AQV

1/50.404fD (1)(1)


D1062.1...VDRe

4

(2)(2)

(3)(3)D

0005.0D

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

Repeat the iteration process Repeat the iteration process ff**→ D→ Re and → D→ Re and εε/D→ f /D→ f until convergenceuntil convergence

110(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

˚ 55 22 Water at 60Water at 60˚F (F (νν=1.21=1.21××1010--5 5 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 0.0005 ft reservoir B through a pipe of length 1700 ft and roughness 0.0005 ft at a rate of Q= 26 ftat a rate of Q= 26 ft33/s as shown in Figure E8 13 The system/s as shown in Figure E8 13 The systemat a rate of Q= 26 ftat a rate of Q= 26 ft33/s as shown in Figure E8.13. The system /s as shown in Figure E8.13. The system contains a contains a sharpsharp--edged entrance and four flanged 45edged entrance and four flanged 45˚ elbowelbow. . Determine the pipe diameter needed.Determine the pipe diameter needed.p pp p

111


The energy equation can be applied between two points on the The energy equation can be applied between two points on the gy q pp pgy q pp psurfaces of the reservoirs (psurfaces of the reservoirs (p11=V=V11= p= p22=z=z22=V=V22=0)=0)

hzVpzVp 2222

2111

L

KfVz

hzg

pzg

p

222

2222

1111

LKD

fg

z21

1.33QV K =0 5 K lb =0 2 and K i =12DAV KLent 0.5, KLelbow 0.2, and KLexit 1

]150)20(4[1700fVft44

2

]15.0)2.0(4[D

f)s/ft2.32(2

ft44 2

(1)(1)D001350D001520f 5 112

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


D1074.2...VDRe

6

(2)(2)

(3)(3)D

0005.0D

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

Repeat the iteration process Repeat the iteration process D→ fD→ f** →Re and →Re and εε/D→ f until convergence/D→ f until convergence

113(1)(1) (2)(2) (3)(3)

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