7/30/2019 Chapter 8 SU13

1/106

Figure8.1

Objec/ve:todesignapipingsystemhowtoes/matethepumpsize

Chapter8ViscousflowinPipes

WeneedtohaveapumptocompensateHeadlossduetofric/onHeadlossduetoelbows,valves,andEleva/onchange

7/30/2019 Chapter 8 SU13

2/106

Gas Turbine to Produce Electricity

7/30/2019 Chapter 8 SU13

3/106

Cogeneration with Gas Turbine

7/30/2019 Chapter 8 SU13

4/106

Gas Turbine Blade

7/30/2019 Chapter 8 SU13

5/106

Cooling of Gas Turbine Blade

7/30/2019 Chapter 8 SU13

6/106

6.9.3.Viscousflowinacirculartube

Figure6.34

Navier-Stokesequa/onincylindricalcoordinates

7/30/2019 Chapter 8 SU13

7/106

8.1.1LaminarandTurbulentFlows

Figure8.3FlowislaminarifRe3,000

Isthislaminarorturbulentflow?

ReynoldsDyeExperiment

7/30/2019 Chapter 8 SU13

8/106

Figure8.4

Timedependenceoffluidvelocityatapoint

Fluctua/oncomponentinthevelocityofaturbulentflow

7/30/2019 Chapter 8 SU13

9/106

fig_08_18

VelocityProfiles

ParabolicprofileForLaminarFlow

Power-lawprofileForTurbulentFlow

7/30/2019 Chapter 8 SU13

10/106

B

7/30/2019 Chapter 8 SU13

11/106

fig_08_19

Fric/oninturbulentflowisgreatlyaffectedbyaroughsurface

Inlaminarflow,

whyisfric/onnotaffectedbysurfaceroughness?

7/30/2019 Chapter 8 SU13

12/106

W

A. 28,158;B.31,424;C.35,597;D.42,632

Y

7/30/2019 Chapter 8 SU13

13/106

8.1.2Entranceregion,developingflow,andfully

developedflow

Figure8.5

7/30/2019 Chapter 8 SU13

14/106

8.1.3Pressurevaria/onalongahorizontalpipe,p(x)

Figure8.6

dPdzisconstantinafullydevelopedflow.

Whyisdpdxgreater

attheentranceregionthan

atthefullydevelopedregion?

Why?

7/30/2019 Chapter 8 SU13

15/106

8.2FullyDevelopedLaminarFlow

Rela/onshipbetweenwallshearstressandpressuredrop

Figure8.7

Objec/ve:toderivetheaboverela/onfromtheabovesketchAssume:steadystateflow,constantdiameterpipe

accelera/oniszero.

Thiswasderivedfrom

Navier-Stokeseq.(Ch6)

7/30/2019 Chapter 8 SU13

16/106

8.2FullyDevelopedLaminarFlow

Rela/onshipbetweenwallshearstressandpressuredrop

Figure8.8

Free-bodydiagramofacylinderoffluid

7/30/2019 Chapter 8 SU13

17/106

8.2FullyDevelopedLaminarFlow

Rela/onshipbetweenwallshearstressandpressuredrop

Alterna/vely,wecanderiveitfromvelocityprofile.(Chapter6)

7/30/2019 Chapter 8 SU13

18/106

8.2FullyDevelopedLaminarFlow

Rela/onshipbetweenwallshearstressandpressuredrop

7/30/2019 Chapter 8 SU13

19/106

CapillaryTubeViscometer

Capillarytube

d=3mm

Balancetomeasure

Massflowrate

Reservoir

tank

OpentoAir

Liquid

L=15cm

H=7cmObjec/ve:todetermineliquidviscosity

Given:Wecollected55gper1min.

WhatisP[Pa]acrosscapillarytube?

WhatisthevalueofvolumeflowrateQ[m3s]?

Whatistheviscosityofliquid?

7/30/2019 Chapter 8 SU13

20/106

H=7cm

L=15cm

Wecollected55gper1min.

CapillaryTubeViscometer

7/30/2019 Chapter 8 SU13

21/106

7/30/2019 Chapter 8 SU13

22/106

CapillaryTubeViscometer

7/30/2019 Chapter 8 SU13

23/106

W

A. 8;B.10;C.12;D.15

WhydoesEGLdecrease

alongxdirec/on?

Answer:duetopressuredrop

7/30/2019 Chapter 8 SU13

24/106

W

7/30/2019 Chapter 8 SU13

25/106

8.2.3PipingSystemDesign

(Es/ma/onofthelossduetofric/on)

Deriva/onoffric/onfactorDefini/on

Recall

64/

7/30/2019 Chapter 8 SU13

26/106

8.2.3Es/ma/onoffric/onalpressuredropusingfric/onfactor

Recall

From

WhereForlaminarflow:

Forturbulentflow:

7/30/2019 Chapter 8 SU13

27/106

Calcula/onprocedureofpressuredropduetofric/on

Step1:CalculateReynoldsnumber

Step2:Determineifflowislaminarorturbulent.

Step3:Determinefric/onfactor

Step4:CalculateP

7/30/2019 Chapter 8 SU13

28/106

Example(Laminarflow)

Objec/ve:toes/matethefric/onalpressureloss

Given:Water,pipediameterd=4mm;Length=19m

Averagevelocity=0.35msDensity=1,000kgm3;Viscosity=1mPa.s

10m

7m

2m

Valve

Step1:CalculateReynoldsnumber

Step2:Determineifflowislaminarorturbulent.

Step3:Determinefric/onfactor(f=0.04)

Step4:Calculatefric/onalpressuredrop(Answer:P=11,638Pa)

=0.04 19 /0.004 0.5(1000)0.352=11,638

7/30/2019 Chapter 8 SU13

29/106

Example(turbulentflow)

Objec/ve:toes/matethefric/onalpressureloss

Given:Water,pipediameterd=4mm;Length=19mAveragevelocity=15ms

Density=1,000kgm3;Viscosity=1mPa.s

10m

7m

2m

Valve

Step1:CalculateReynoldsnumber

Step2:Determineifflowislaminarorturbulent.

Step3:Determinefric/onfactor(f=0.0202)

f=0.316(60,000)0.25=0.31615.65=0.0202

Step4:Calculatefric/onalpressuredrop(P=10,790,000Pa)

Pump

=0.0202 19 /0.004 0.5(1000)152=10,790,000

7/30/2019 Chapter 8 SU13

30/106

Page412

8.4.1MajorLossesSurfaceRoughnessEffectonFric/onalPressureDrop

Forasmoothpipe(inturbulentflow)

Foraroughsurfacepipe

UseMoodychart

7/30/2019 Chapter 8 SU13

31/106

Fric/onfactorasafunc/onofReynoldsnumberandrela/veroughnessfor

roundpipestheMoodychartWhatisfforRe=60,000?Forasmoothpipef=0.02Foraroughpipe(/D=0.002)f=0.0253

Figure8.20

7/30/2019 Chapter 8 SU13

32/106

B

V1=3msV2=2.1ms

7/30/2019 Chapter 8 SU13

33/106

7/30/2019 Chapter 8 SU13

34/106

fig_03_21

Thelocusofeleva/onsprovidedbyPitottubesisEnergyline.

Thelocusofeleva/onsprovidedbysta/ctapishydraulicgradeline

Ifweassumenofric/onallossandnominorloss,(Figfromchapter3)

Pressurehead

WithPitottube

7/30/2019 Chapter 8 SU13

35/106

fig_03_22

Ifweassumenofric/onallossandnominorloss,(Figfromchapter3)

Pressure

head

WithPitottube

7/30/2019 Chapter 8 SU13

36/106

Mc

Thesearesta/cpressuretubes,

measuringsta/cpressures.

?

7/30/2019 Chapter 8 SU13

37/106

7/30/2019 Chapter 8 SU13

38/106

Mc

FromB.Eq.,wehave

Vexit=24.3ms

Specificenergyisgivenas[Jkg].Thetotalenergyisconstanteverywhere.

Answer:totalenergyintermsofpressure;P=(1000)(9.8)(30m)=294.3kPa.

Totalenergydoesnotchange(duetofric/onlessassump/on).

Totalspecificenergy=/=294.3/

7/30/2019 Chapter 8 SU13

39/106

Y

Disthepipediameter.

7/30/2019 Chapter 8 SU13

40/106

7/30/2019 Chapter 8 SU13

41/106

7/30/2019 Chapter 8 SU13

42/106

fig_08_32

8.4.2MinorLossinaSuddenExpansion,Fig.8.32

GlobeValve

SwingCheckValve

GateValve

StopCheckValve

fig_08_33

Notethatinletvelocityisequaltooutletvelocity.

Whataboutinletpressureandoutletpressure?

Sameornot?Why?

Answer:Thereisalossofenergyintheformofpressuredropacrossavalve.

Whydowelosesomeenergyorpressure?

7/30/2019 Chapter 8 SU13

43/106

)2

(2

g

VKh LL =

LossCoefficients

forPipeComponents

Table8.2

7/30/2019 Chapter 8 SU13

44/106

Example

Objec/ve:toes/matethedischargepressureofpump

Given:Water,pipediameterd=2cm;Length=23m

Averagevelocity=2ms

Density=1,000kgm3;Viscosity(water)=1mPa.s

15m

5m3m

Valve

Kforelbow=0.7

Kforvalve=1.2

Pump

Duetofric/on

Duetominorloss

7/30/2019 Chapter 8 SU13

45/106

7/30/2019 Chapter 8 SU13

46/106

7/30/2019 Chapter 8 SU13

47/106

7/30/2019 Chapter 8 SU13

48/106

7/30/2019 Chapter 8 SU13

49/106

7/30/2019 Chapter 8 SU13

50/106

Example8.8

7/30/2019 Chapter 8 SU13

51/106

Example8.8

7/30/2019 Chapter 8 SU13

52/106

2.Considerawaterpumpingsystemshowninthefigurebelow.Waterflowsat12gpm.

Insidediameterofthepipeis0.75in,lengthsaregivenin,losscoefficientsfor90-degree

elbows=0.2,losscoefficientforwideopenvalve=0.4.1gal=0.133683;

7/30/2019 Chapter 8 SU13

53/106

7/30/2019 Chapter 8 SU13

54/106

2.Considerawaterpumpingsystemshowninthefigurebelow.Waterflowsat12gpm.

Insidediameterofthepipeis0.75in,lengthsaregivenin,losscoefficientsfor90-degree

elbows=0.2,losscoefficientforwideopenvalve=0.4.1gal=0.133683;

7/30/2019 Chapter 8 SU13

55/106

2.Considerawaterpumpingsystemshowninthefigurebelow.Waterflowsat12gpm.

Insidediameterofthepipeis0.75in,lengthsaregivenin,losscoefficientsfor90-degree

elbows=0.2,losscoefficientforwideopenvalve=0.4.1gal=0.133683;

7/30/2019 Chapter 8 SU13

56/106

Powerthatshouldbedeliveredtopump

=[]=Where

Q=volumeflowrate[m3s]or[3s]

P=totalpressureliinpump[Pa]or[psf]

Conversion:

1horsepower=1hp=550.lbs=745.7W

Ifpumphas83%efficiency,=[]= /0.83

7/30/2019 Chapter 8 SU13

57/106

7/30/2019 Chapter 8 SU13

58/106

Y

Eleva/onfromEtoB

=2.1hp

7/30/2019 Chapter 8 SU13

59/106

Y

7/30/2019 Chapter 8 SU13

60/106

Y

7/30/2019 Chapter 8 SU13

61/106

fig_08_e11

Example8.11

Powerthatshouldbedeliveredtoturbine

with100%efficiency:givenV=6.58s

7/30/2019 Chapter 8 SU13

62/106

Example8.11

7/30/2019 Chapter 8 SU13

63/106

W

Conversion:1horsepower=1hp=550.lbs=745.7W

A.1.0;B.1.5;C.2.0;D.3.0

7/30/2019 Chapter 8 SU13

64/106

W

A.0.2;B.0.6;C.1.0;D.2.0

Conversion:1horsepower=1hp=550.lbs=745.7W

85

.85

.85

7/30/2019 Chapter 8 SU13

65/106

W

Conversion:1horsepower=1hp=550.lbs=745.7W

7/30/2019 Chapter 8 SU13

66/106

W

Thenconsidering80%efficiency,

Weget3.2650.8=4.08.

Conversion:1horsepower=1hp=550.lbs=745.7W

7/30/2019 Chapter 8 SU13

67/106

Fric/onFactorsforLaminarFlowinNoncircularDucts

Usehydraulicdiameter,definedas

Table8.3

1

2cm

7/30/2019 Chapter 8 SU13

68/106

7/30/2019 Chapter 8 SU13

69/106

B

V1A1=V2A2

7/30/2019 Chapter 8 SU13

70/106

CoolingofAegisRadarSystem(phased-arraysystem)

7/30/2019 Chapter 8 SU13

71/106

AS-17"Krypton"(Kh-31P)

AegisRadarSystem(phased-arraysystem)

OldRadar

7/30/2019 Chapter 8 SU13

72/106

p_08_102

MicrochannelcoolingsystemforVLSIchip

7/30/2019 Chapter 8 SU13

73/106

7/30/2019 Chapter 8 SU13

74/106

7/30/2019 Chapter 8 SU13

75/106

7/30/2019 Chapter 8 SU13

76/106

Figure8.35Series(a)andparallel(b)pipesystems.

8.5.2Mul/plePipeSystem

Whatisconstant?

Whatisconstant?

7/30/2019 Chapter 8 SU13

77/106

Y

7/30/2019 Chapter 8 SU13

78/106

W

7/30/2019 Chapter 8 SU13

79/106

W

Notethatf.V2=constant

1

2

V2=2V1

7/30/2019 Chapter 8 SU13

80/106

B

V1=Q1A1=101.3963=7.1718s

CalculateDP1

DP1=2,238.9psf

DP1=DP2

1

2

7/30/2019 Chapter 8 SU13

81/106

TypicalVenturimeterconstruc/on.Figure8.44

Whatistheopera/ngprinciple?

8.6PipeFlowrateMeasurement

?

7/30/2019 Chapter 8 SU13

82/106

Rotameter-typeflowmeter.(CourtesyofFischer&PorterCo.)

Figure8.46

8.6PipeFlowrateMeasurement

Whatistheopera/ngprinciple?

7/30/2019 Chapter 8 SU13

83/106

Turbine-typeflowmeter.(CourtesyofE.G.&G.FlowTechnology,Inc.)

Figure8.47

8.6PipeFlowrateMeasurement

Whatistheopera/ngprinciple?

7/30/2019 Chapter 8 SU13

84/106

FigureP8.84

HWProblem8.84

Waterat40oF,Flowrate=0.9galmin

(gpmconversion:1USgallon=0.003786m3=0.133683)

(Note1Bri/shimperialgallon=1.2USgallon)

Determinepressuredrop.

7/30/2019 Chapter 8 SU13

85/106

Chapter12Turbomachines

Typicalposi/vedisplacementpumps:(a)/repump,(b)

humanheart,(c)gearpump.

Figure12.1

7/30/2019 Chapter 8 SU13

86/106

12.4.2PumpPerformanceCharacteris/cs

Typicalperformancecharacteris/csforacentrifugalpumpofa

givensizeopera/ngataconstantimpellerspeed.

Figure12.11

7/30/2019 Chapter 8 SU13

87/106

Schema/cofapumpinstalla/oninwhichthepumpmustli

fluidfromoneleveltoanother.

Figure12.13

7/30/2019 Chapter 8 SU13

88/106

.

Typicalflowsystem

WithReservoirtankabovepump

Figure12.14

7/30/2019 Chapter 8 SU13

89/106

12.4.2PumpPerformanceCharacteris/csFigure12.10(p.658)

Units:

1hP=745.7W

1hP=550

IfQ=0.5m3sandisP=20,000Pa,whatshouldbethesizeofmotor?(100%efficiency)

Pump efficiency is not 100%

7/30/2019 Chapter 8 SU13

90/106

Pumpefficiencyisnot100%.

Figure12.11

84%

Ifpumpefficiencyis84%,13.4shouldbechangedto

7/30/2019 Chapter 8 SU13

91/106

Figureonp.684(7thed)

Systemcurve

7/30/2019 Chapter 8 SU13

92/106

U/liza/onofthesystemcurveandthepumpperformancecurve

toobtaintheopera/ngpointforthesystem.

Figure12.15

7/30/2019 Chapter 8 SU13

93/106

Example12.4FigureE12.4

Objec/ve:todetermineQflowrate

Given:totallengthofpipe=200

Fric/onfactorf=0.02

Pipediameter=6in

Pumpcurveisgiven

Solu/on:

Weneedtodevelopasystemcurve,

whichlookslikethis.

7/30/2019 Chapter 8 SU13

94/106

FigureE12.4con7nued 1gal=0.133683

7/30/2019 Chapter 8 SU13

95/106

Effectofopera/ngpumpsin(a)seriesand(b)inparallel.

Figure12.16

W

7/30/2019 Chapter 8 SU13

96/106

W

Conversion:1horsepower=1hp=550.lbs=745.7W

Answerseenextpage

7/30/2019 Chapter 8 SU13

97/106

AnswerforW49.

W

7/30/2019 Chapter 8 SU13

98/106

W

Conversion:1horsepower=1hp=550.lbs=745.7W

Answerseenextpage

7/30/2019 Chapter 8 SU13

99/106

Answer

W

7/30/2019 Chapter 8 SU13

100/106

W

Answerseenextpage

PumpCurve

SystemCurve

7/30/2019 Chapter 8 SU13

101/106

W

7/30/2019 Chapter 8 SU13

102/106

W

A.0.55;B.2.15;C.4.35;D.5.40

W

7/30/2019 Chapter 8 SU13

103/106

W

W

7/30/2019 Chapter 8 SU13

104/106

W

Answer:P=15+4+40=59m.Thus,ModelB.

7/30/2019 Chapter 8 SU13

105/106

Answer:PumpC

W

7/30/2019 Chapter 8 SU13

106/106

W