Elementary Mechanics of Fluids

CE 319 FDaene McKinney

Flow inPipes

Reynolds Experiment• Reynolds Number

• Laminar flow: Fluid moves in smooth streamlines

• Turbulent flow: Violent mixing, fluid velocity at a point varies randomly with time

• Transition to turbulence in a 2 in. pipe is at V=2 ft/s, so most pipe flows are turbulent

2lowfTurbulent4000flowTransition40002000

flowLaminar2000Re

Vh

VhVD

f

f

Laminar Turbulent

Shear Stress in Pipes• Steady, uniform flow in a pipe: momentum

flux is zero and pressure distribution across pipe is hydrostatic, equilibrium exists between pressure, gravity and shear forces

DL

hhh

dsdhD

zpdsdD

sDdsdzsAsA

dsdp

sDWAsdsdpppAF

f

s

021

0

0

0

0

44

)]([4

)(0

)(sin)(0

• Since h is constant across the cross-section of the pipe (hydrostatic), and –dh/ds>0, then the shear stress will be zero at the center (r = 0) and increase linearly to a maximum at the wall.

• Head loss is due to the shear stress.

• Applicable to either laminar or turbulent flow• Now we need a relationship for the shear

stress in terms of the Re and pipe roughness

Darcy-Weisbach Equation

)(Re,

)(Re,

;Re;

,,:variablesRepeating),(

),,,,(

20

20

20

321

214

0

DeFV

DeF

V

VDe

DVF

eDVF

V D e

ML-1T-2 ML-

3LT-1 ML-1T-1 L L

)(Re,82

)(Re,82

)(Re,4

4

2

2

2

0

DeFf

gV

DLfh

DeF

gV

DL

DeFV

DL

DLh

f

f

Darcy-Weisbach Eq. Friction factor

Laminar Flow in Pipes• Laminar flow -- Newton’s law of viscosity is valid:

2

0

20

20

2

14

44

2

2

2

rr

dsdhrV

dsdhrCC

dsdhrV

drdsdhrdV

dsdhr

drdV

drdV

dydV

dsdhr

dydV

• Velocity distribution in a pipe (laminar flow) is parabolic with maximum at center.

2

0max 1

rrVV

Discharge in Laminar Flow

dsdhD

dsdhr

Q

rrdsdh

rdrrrdsdhVdAQ

rrdsdhV

r

r

128

8

2)(

4

)2()(4

)(4

4

40

0

220

2

022

0

220

0

0

dsdhDV

AQV

32

2

Head Loss in Laminar Flow

2

21

12212

2

2

2

32

)(32

32

3232

DVLh

hhh

ssD

Vhh

dsD

Vdh

DV

dsdh

dsdhDV

f

f

Re64

2

2/)(Re64

2/))((64

2/2/32

32

2

2

2

2

2

2

2

fVDLfh

VDL

VDL

DV

VV

DVL

DVLh

f

f

Nikuradse’s Experiments• In general, friction factor

– Function of Re and roughness• Laminar region

– Independent of roughness• Turbulent region

– Smooth pipe curve• All curves coincide @

~Re=2300

– Rough pipe zone• All rough pipe curves flatten

out and become independent of Re

Re64

f

Blausius

Re 4/1kf

Rough

Smooth

Laminar Transition Turbulent

Blausius OK for smooth pipe

)(Re,DeFf

Re64

f

2

9.010Re

74.57.3

log

25.0

De

f

Moody Diagram

Pipe Entrance• Developing flow

– Includes boundary layer and core, – viscous effects grow inward from the

wall• Fully developed flow

– Shape of velocity profile is same at all points along pipe

flowTurbulent 4.4Re

flowLaminar Re06.01/6D

Le

eL

Entrance length LeFully developed

flow region

Region of linear pressure drop

Entrance pressure drop

Pressure

x

Entrance Loss in a Pipe• In addition to frictional losses, there are minor

losses due to – Entrances or exits– Expansions or contractions– Bends, elbows, tees, and other fittings– Valves

• Losses generally determined by experiment and then corellated with pipe flow characteristics

• Loss coefficients are generally given as the ratio of head loss to velocity head

• K – loss coefficent– K ~ 0.1 for well-rounded inlet (high Re)– K ~ 1.0 abrupt pipe outlet– K ~ 0.5 abrupt pipe inlet

Abrupt inlet, K ~ 0.5

gVKh

gVhK L

L2

or

2

2

2

Elbow Loss in a Pipe• A piping system may have many minor

losses which are all correlated to V2/2g • Sum them up to a total system loss for pipes

of the same diameter

• Where,

m

mm

mfL KDLf

gVhhh2

2

lossheadTotalLh

lossheadFrictionalfh

mhm fittingfor lossheadMinor

mKm fittingfor tcoefficienlossheadMinor

EGL & HGL for Losses in a Pipe• Entrances, bends, and other flow transitions

cause the EGL to drop an amount equal to the head loss produced by the transition.

• EGL is steeper at entrance than it is downstream of there where the slope is equal the frictional head loss in the pipe.

• The HGL also drops sharply downstream of an entrance

Ex(10.2)Given: Liquid in pipe has = 8 kN/m3. Acceleration = 0.

D = 1 cm, = 3x10-3 N-m/s2. Find: Is fluid stationary, moving up, or moving down?

What is the mean velocity?Solution: Energy eq. from z = 0 to z = 10 m

smV*x*

).*(.V

μLγDhV

γDμLVh

mh

h

h

zpg

Vhzpg

V

-

L

L

L

L

L

L

/04.11010332

0108000251

32

32

upward) (moving25.1

10890

108000

000,1108000

000,20022

3

2

2

2

22

22

211

21

1

1

2

Ex (10.4)• Given: Oil (S = 0.97, m = 10-2 lbf-s/ft2) in 2

in pipe, Q = 0.25 cfs.• Find: Pressure drop per 100 ft of horizontal

pipe.• Solution:

ftpsi/100791122

46111002103232

(laminar)36010

)12/2(*46.11*94/1*97.0Re

/46.114/)12/2(

25.0

22

2

2

.)/(

.**-*DμLVΔp

VD

sftAQV

Ex. (10.8)Given: Kerosene (S=0.94, =0.048 N-s/m2). Horizontal 5-

cm pipe. Q=2x10-3 m3/s.Find: Pressure drop per 10 m of pipe.Solution:

cfs

sftVVV

VVg

VγD

μLVg

α

gVα

γDμLV

γDμLVh

zγp

gVαhz

γp

gVα

L

L

32

5

22

2

52

2

22

22

22

22

2

22

22

211

21

1

10*23.14/(0.25/12)**1.6A*VQ

(laminar)129310*4

)12/25.0(*6.1*94.1*8.0Re

/60.101.1645.8

05.0)32/1(*4.62*8.0

10*10*4*3222

05.0322

002

325.000

3222

Ex. (10.34)

2/62.0,/300,12 mNsmN

Given: Glycerin@ 20oC flows commercial steel pipe.

Find: hSolution:

mγDμLVhh

VDVD

hzγpz

γph

zγphz

γp

zγ

pg

Vαhzγp

gVα

L

L

L

L

42.2)02.0(*300,12

)6.0)(1)(62.0(3232

(laminar)5.2310*1.5

02.0*6.0Re

)(

22

22

4

22

11

22

11

22

22

211

21

1

Ex. (10.43)Given: FigureFind: Estimate the elevation required in the upper

reservoir to produce a water discharge of 10 cfs in the system. What is the minimum pressure in the pipeline and what is the pressure there?

Solution:

ftz

sftAQV

DLfKKK

gV

DLfKKKh

zhz

zγ

pg

Vαhzγp

gVα

Ebe

EbeL

L

L

1332.32*2

73.1275.100.14.0*25.0100

/73.121*4/

10

75.101

430*025.0;0.1;(assumed)4.0;5.0

22

000022

21

2

221

22

22

211

21

1

55

2

22

1

2

1

21

12

11

10*910*14.1

1*73.12Re

59.0)53.1(*4.6235.1

2.32*273.12

1300025.04.05.00.17.110133

22

2*100

22

VD

psigpft

gV

DLfKK

gVzz

γp

zγp

gV

hz

zγp

gVαhz

γp

gVα

b

beb

bb

bbb

L

bbb

bL

Ex. (10.68)

ftxksftx s425 105.1;/1022.1

ftDD

gDQ

D

gV

DLfh f

06.8

984,331

2))4//((1000*015.01

2

5

22

2

Given: Commercial steel pipe to carry 300 cfs of water at 60oF with a head loss of 1 ft per 1000 ft of pipe. Assume pipe sizes are available in even sizes when the diameters are expressed in inches (i.e., 10 in, 12 in, etc.).

Find: Diameter.Solution:

Assume f = 0.015

Relative roughness: 00002.006.8105.1 4

x

Dks

Get better estimate of f

65

2

109.31022.1)06.8)(4/(

300Re

)4/()4/(

Re

xx

DQ

DD

Q

VD

f=0.010

.8943.7

656,221 5

inftDD

Use a 90 in pipe

Ex. (10.81)

mh

h

zγ

pg

Vαhzγp

gVα

gQ

hfLD

gDQ

DLf

gV

DLfh

f

f

L

f

f

45.14

109810

000,609810

000,30022

8

2))4//((

2

22

22

211

21

1

5/1

2

2

222

Given: The pressure at a water main is 300 kPa gage. What size pipe is needed to carry water from the main at a rate of 0.025 m3/s to a factory that is 140 m from the main? Assume galvanized-steel pipe is to be used and that the pressure required at the factory is 60 kPa gage at a point 10 m above the main connection.

Find: Size of pipe.Solution:

Assume f = 0.020

m

gQ

hfLDf

100.081.9)025.0(140

45.1402.08

8

5/1

2

2

5/1

2

2

Relative roughness:

022.0

0015.0100

15.0

f

Dks

Friction factor:

mD 102.0020.0022.0100.0

5/1

Use 12 cm pipe

Ex. (10.83)Given: The 10-cm galvanized-steel pipe is 1000 m long

and discharges water into the atmosphere. The pipeline has an open globe valve and 4 threaded elbows; h1=3 m and h2 = 15 m.

Find: What is the discharge, and what is the pressure at A, the midpoint of the line?

Solution:

002

)41(1200

222

22

22

211

21

1

gV

DLfKKK

zγp

gVαhz

γp

gVα

bve

L

D = 10-cm and assume f = 0.025

46

32

2

2

1071031.1

1.0*942.0Re

/0074.0)10.0)(4/(942.0

/942.01.265

24

)1.0

1000025.09.0*4105.01(24

xx

VDsmVAQ

smV

gV

Vg

So f = 0.025

kPap

mgγ

p

gV

DLfK

γp

zγp

gVαhz

γp

gVα

A

A

bA

LAAA

A

8.90)26.9(*9810

6.9152

)942.0()1.0

500025.09.0*2(

2)2(15

22

2

2

22

22

22

Near cavitation pressure, not good!

Ex. (10.95)Given: If the deluge through the system shown is 2 cfs,

what horsepower is the pump supplying to the water? The 4 bends have a radius of 12 in and the 6-in pipe is smooth.

Find: HorsepowerSolution:

55

22

2

22

22

22

211

21

1

1017.41022.1

)2/1(*18.10Re

611.12

/18.10)2/1)(4/(

2

)45.01(2

6003000

22

xx

VD

ftg

V

sftAQV

DLfK

gVh

hzγp

gVαhz

γp

gVα

bp

Lp

hphQ

p

ft

h

p

p

4.24550

6.107

))2/1(

17000135.019.0*45.01(611.13060

So f = 0.0135