9 Flow in Pipes_with Figures

7/29/2019 9 Flow in Pipes_with Figures

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9Flowinpipes

o Sewer/Culverts--->Openchannelflowo Work-energyprincipleo ContinuityEquationo Principles&Equationoffluidresistance

9.1FundamentalEquations: Work-energyEquation:

1

2

HeadLoss

Engineering:

hL12

=

v1

2

2gn+

p1

+z

1

v2

2

2gn+

p2

+z

2

Exact:

hL12 = 1v1

2

2gn+

p1

+z

1

2

v2

2

2gn+

p2

+z

2

=1

v2

v3

A

dA

v

A

dA, =

1

v

v2

A

dA

v

A

dA


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, are dimensionless, coefficients that represent correction factors to the conventional

velocity headv2

2gn

and Momentum QV , respectively.

Uniform velocitiy = = 1

Non-uniform velocity > > 1 Total Kinetic energy (J/s) =

2v 3

A

dA (1)

using mean velocity Q(v

2gn) = gnQ

v 2

2gn

=1

v

2

v 3

A

dA

Q

=

1

v

2

v 3

A

dA

vA dA Momemtum flux (M) = v 2

A

dA (1)

Mean velocity = Qv

=1

v

v 2

A

dA

vA

dA

z1+

p1

+

1

v1

2

2gn

= z2+

p2

+

2

v2

2

2gn

+ hL

In most pipe flow problems is omitted for several reasons:

1. Most engineering pipe flow problems atre turbulent --> is only slightly biggerthan 1.

2. In laminar glow where is large, velocity head are usually negligible whencompared to to other terms.

3. Velocity heads in most pipe flows are usually so small compared to other terms.4. Engineering answers are not usually required to an accuracy which would justify the

inclusion of.

Including depens on:

+ an understanding of the factors which afect the head loss hL.+ the methods available for calculating this quantity.


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Darcy-Weiscbach Eq. (1850):

flows in long, straight, cylindrical pipes

hL = fl

d

v 2

2gn(1) , f : friction factor (dimensionless)

f depends on roughness, velocity & diameter

hL =

olRh

, Hydraulic radius :Rh =Ap=

d2

4d

=d4, then hL =

4old

(2)

(1)& (2)4ol

gn d= f

l

d

v 2

2gn o =

fv 2

8

o= v

f

8

friction velocity : v*=

o= v

f

8


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Example: Obtain v*

hL = fl

d

v2

2gn f =

hL d2gn

l v2

f =5.33 0.15 2 9.81

30 (4.5)2

= 0.026

v* =o

= v

f

8= 4.5

0.026

8= 0.26m /s

9.2Laminarflow:

Assumptions:

Symetrical distribution of shear stress & velocity.

Maximum velocity at the center of the pipe, no-slip condition on the pipe wall (v = 0 at the

wall).


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Linear shear-stress distribution in the fluid given by:

hL = l

Rh, Rh =

A

p=

d

4=

r

2 =

hL

2l

r

Shear stress in laminar flow =

dv

dy

Find velocity profile:

r = R y dr = dy

=hL

2l

r =

dv

dy=

dv

dr

At the wall : r = R ,y = 0 0 =hLR

2l

hL

2l

=

0

Rdv

dr=

1

hL

2l

r =

1

0

Rr v =

1

0

Rr dr =

1

0

R

1

2r2 +C

at r = R, v = 0 C=1

0

R

R2

2 v =

0

2R(R

2 r2) parabolic profile (9.5)

at r = 0 v = vc =0R

2

2R v = vc 1

r2

R2

(9.5)

v* =0

=

v =v*

2

2RR

2 r2( ) v

v*=

v*

2RR

2 r2( )

r2 = R y( )2

v

v*=

v*

y

y 2

2R

Velocity profile :v(y,v*), distance from wallv*: characteristic "velocity", y : laminar profile

wheny

2

2R


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Flow rate Q: Q = v A = v2r dr( )

Q = v 2r dr( )=

0

2RR2 r2( ) 2r( )dr

0

R

0

R

Q =

0

RR2 r2( )r dr

0

R

=

0

RR2

1

2r2

1

4r4

0

R

=

0

R

R4

2

R4

4

=

0R

3

4

0=

hLR

2l Q =

R4hL

8l=

d4hL

128l(Hagen Poiseuille Law)

Q = R2v v =R2hL

8l=

d2hL

32l

Head Loss hL =32l

d2(9.9)

In laminar flow, head loss varies with first power of the velocity:

Darcy Weisbach : hL = fl

d

v 2

2gn f =

64

v d=

64

Re(9.10)

In laminar flow, friction factor depends only on Reynold number.

Example 1:

400 l/minute of oil = 855.6Kg

m3, = 71.8 103

flow throught an 8cm diameter

pipeline. Calculate the centerline velocity, head loss in 300 m of pipe, and shear stress &

velocity at a point 2cm from the centerline.

d = 0.08m

l = 300m

Q


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Q = 400l

min=

0.4m3

60s= 0.0067

m3

s, l = 300m, d= 0.08

v = QA

= 0.067

d2

4

= 0.067

0.08( )

2

4

=1.33 ms

Re =Vd

=

855.6 1.33 0.08

71.8 103=1268 Re < 2100 lamin ar flow

hL =32 l v

d2=

32 71.8 103 300 1.33

855.6 9.81 0.08( )2

=17.065m

=hL

2l

r =

855.6 9.81 17.065

2 300 0.02

at centerline, r = 0 v = vc =

0R2

2R=

0

R

2=

hLR2

4l=

855.6 9.81 17 0.04( )2

4 71.8 103 300= 2.65

m

s

v =1

A

v dAA

=1

R2

vc 1r 2

R2

2r dr

0

R

=2vc

R2

1

2

r 2 1

4R2

r4

0

R

=

2vc

R2

R2

2

R2

4

=

1

2

vc vc = 2v

v = vc 1r2

R2

= 2.65 1

0.02( )2

0.08( )2

=


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Example 2:

100 mm

d = 4mm, Q = ?, hL = 1 m, v = ?

Q =Volume

time=

1300 106

600= 2.17 10

6 m3

s, Q =

d4 hL

128 l=

d4 g hL

128 l

= d4 g h

L

128Q l=

3.14 4 103

( )

4

9.81 1

128 2.17 106 0.1

= 28 10

5

m2

s

v =Q

A=

2.17 106

4 4 10

3( )2= 0.17

m

s, Re =

v d

=

0.17 4 103

28 105

= 2.4 Flow is La min ar

9.3Turbulentflow-Smoothpipes:

Chaotic / iregular moving in time & space

Diffusivity

High Re

3D vorticity fluctuation

Dissipation of KE of the turbulent by viscous shear stress

Boundary layer

Boundary layer: existence of a viscous sublayer near to the pipe walls

= +( ) dvdy

, : inludes viscous effect, : includes turbulent effect


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vx =Vx + vx

vy =Vy + vy = v ivj, Reynolds stress

vz =Vz + vz 2D : = vxvy

= dvdy

vxv (9.11), dvdy

: viscous stress, vxvy : turbulent stress

Pie de figura: Re >> 2100 Re~

5104

Laufer(1954)NACA

Re ~ 5105

(N.A.C.A:NationalAdvisoryCommiteeforAeronauticsNASA:Aeronautics&Space

Administration)

+Mostofflow,turbulentstressdominates:

Maximumstress=viscousstressatthewallwhereturbulentstressiszero.

torepresentReynoldsstress,Prandtl'smixinglengththeory:

= v ivj = l2

dv

dy

2

(9.11)


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as eq (9.12) says : 0

1y

R

= l2

dv

dy

2

ldv

dy= v* 1

y

R (

dv

dy~

1

y) l y 1

y

R

near wall, y 0 l = ky( )

0 l2

k2

y2= l

2 dv

dy

2

dv

dy

=

0

l

ky

=

v*

ky

Integrating : v =v*

kln y( ) +C, ifv = vc at y = R vc =

v*

kln R( )+C C= vc

v*

kln

v = vc +v*

kln y( ) ln R( )( )

vc v

v*= 2.5 ln

y

R

, proved!( )

Forsmoothpipes: turbulentvelocityprofile

v

v*

= 2.5 ln(v* y

) +5.5, or :v

v*

= 5.75 log(v* y

) +5.5, (9.17)

Near to the smooth wall in the viscous sublayer, the laminar shear stress:

v

v*

=

v* y

where y


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FromEq.(9.17)at

y = R, where v = vc:

FromEq (9.17) at y = R, where v = vc :vc

v*= 5.75 log(

v* R

) +5.5,

andEq (9.19) :

v

v*= 5.75log

v*R

+1.75

vc v

v*= 3.75

as v* = vf

8 vc v = 3.75vf

8 vc

v= 3.75 f 8 +1

adjusting this expression:vc

v= 4.07 f 8 +1 (9.20)

v* = vf

8 , Re =vd

=

2vR

,v

v*= 5.75log

v*R

+1.75,

v

v f= 8 5.75log

v R f

8

+1.75

adjusting this expression :1

f

= 2.0log Re f( ) 0.8, (Smooth pipes) (9.21)

v* = vf

8, v =11.6

v*

,

v

d=

11.6

v* d=

11.6

v df

8

=

32.8

Re f, Re f =

32.8

vd

(9.22)

v as Re , Sublayer thickness as Re

Substituting into Eq. (9.21):1

f= 2.0log

32.8

vd

0.8 (9.23)

Forturbulentflowoversmoothwalls,thefrictionfactorisafunctiononlyoftheratio

ofthesublayerthicknesstothepipediameter.


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Example 1:

d = 75mm

Smooth pipe

Knowing that at 20 C = 998 Kg/m3, 0 = 3.68 N/m. Calculate:

Thinkness of viscous sublayerv frition factor f Centerline velocity vc

Mean velocity v Flow rate Q hL in 1000m

v, at 25mm from centerline

=11.6

v*

, v* =

3.68 N/m2

998 Kg /m3= 0.061 m /s v =11.6

106 m2 /s

0.061m /s

=1.9 104 m

1

f= 2.0log

32.8

vd

0.8 = 7.42 f =

1

7.42( )2 = 0.018

v

v*= 5.75log

v* R

+1.75 = 21.1 R =

d

2

v = 21.1 v* = 21.1 0.061 =1.29m /s

Q = v A =1.29 4

0.075( )2 = 0.0057m3 /s,

vc

v=1+ 4.07

f8

=1.193 vc =1.193 1.29 =1.54m /s

v

v*= 5.75log

v* y

+ 5.5 y = 37.5 25 =12.5mm( )

v

v*= 5.75log

0.061 0.0125

106

+ 5.5 = 22

v = 22.1 0.061 =1.35m /s

Shear stress is linearly with radius :

25

37.5=

2

3, =

2

30 =

2

3 3.68 = 2.45N/m2 Head loss : hL = f

l

d

v 2

2gn= 0.018

1000

0.075

(1.29)2

2 9.81= 20.4m

Blasius (German, 1913): a) Empirical, b) Mathematical

Empirical work, representing the friction factor for 3.000 < Re < 100.000


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f =0.316

Re0.25

Blasius (9.24)

Substituting into Darcy- Weisbach: hL = fl

d

v 2

2gn hL v

1.75

for turbulent flow in smooth pipes with Re < 105

v* =

0

l= v

f

8 0 =

f v 2

8=

0.316

2Rv

0.25

v 2

8= 0.0332

14 R

14 v

74

34 (9.25)

Blasiusassumedthattheturbulentvelocityprofilecouldbeaproximatedbyapower

relationship:

Fig.

v

vc

=

y

R

m

v =1

Av dA =

1

R2

vcy

R

m

2r( )dr0

R

r = R yA

vR2 = vcy

R

m

2 R y( ) dy( )0

R

v =2vc

m +1( ) m + 2( )=

2

m +1( ) m + 2( )v

R

y

m

, Substituting into (9.25) :

0= 0.0332

2

m +1( ) m +2( )

7 4

1

4 R 1

4+

7m4 v

74 y

7m4

34

However, wall shear stress could depend only on velocity and fluid properties, not on the

Radius of the pipe R so that, 1

4+

7m

4= 0 m =

1

7

Seventh-root law for turbulent velocity distribution:

vvc

=yR

1

7 m = 1

7 v

vc=4960

v = 4960

vc

Substituting into (9.25) 0 = 0.0464

vcR

14vc

2

2


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Example 2:

Fig. (same as previous example)

Knowing that v = 1.29 m/s, d = 75mm and water at 20C. Calculate:

Wall shear stress 0

frition factor f Centerline velocity vc

v at 25mm from centerline usinf seventh-root law

f =0.316

Re=

0.316

v d

=

0.316

1.29 0.075

106

= 96.750 Re =v d

= 96750 in the range of Blasius 3

v

vc

=

49

60

vc =60

49

v =60

49

1.29 = 1.58m/s

0 = 0.0464

vcR

0.25

vc2

2= 0.0464

106

1.58 0.375

0.25

998 1.582

2= 3.70 Pa

Using seventh root law : (@25mm)

v25

vc

=

y

R

17

=

0.0125

0.0375

17

= 0.855 v25 =1.35 m /s


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9.4Turbulent-flow - Roughpipes

Nikuradse:

vc v

v*=

2.5ln

y

R

for all pipes,

for rough pipes :v

v*= 5.75log

y

e

+8.5 e : roughness height, Smooth pipe log

v*R

and flowrate Q : Q = R2v* 5.75logR

e

+ 4.75

v =

Q

A=

Q

R2= v* 5.75log

R

e

+ 4.75

for rough pipes :v

v*= 5.75log

R

e

+ 4.75

Substituting : v* = v f /8 1f

= 2.0log Re

+1.68

adjusting by Nikuradse's experiment:1

f= 2.0log

R

e

+1.75 or

1

f= 2.0log

d

e

+1.14 for rou

Example:

d = 300 mm

v = 3 m/s

Knowing that Relative roughness, e /d= 0.002 ,and kinematic viscosity of the water,

= 9 107m

2/s. Calculate:

frition factor f , Centerline velocity vc, v at 50mm from pipewall (y=50mm) and hL 300m


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1

f= 2.0log

d

e

+1.14 = 2.0log

1

0.002

+1.14 = 6.54 f =

1

6.542= 0.0234

v* = v f /8 = 30.0234

8

= 0.162m /s

v

v*= 5.75log

R

e

+ 4.75 = 5.75log

1

0.004

+ 4.75 v* = 0.162m /s

v

v*= 5.75log

y

e

+ 8.5

e

d= 0.002 e = 0.02 0.3 = 0.0006m

yc

e=

0.150

0.0006= 250

vc = 0.162 5.75log 250( )+ 8.5( ) = 3.61m /s

yc

e=

0.050

0.0006= 83.3 v50 = 0.162 5.75log 83.3( )+ 8.5( ) = 3.17m /s

head loss : hL = fl

d

v 2

2gn= 0.0234

300

0.3

32

2 9.81=10.7m

9.5ClassificationofSmoothless&Roughness

Friction factor Velocity profiles

Turbulent flows in smooth pipe :

1

f= 2.0log

32.8

vd

0.8, friction factor depends on :Sublayer thickness (v) and characteristic

Turbulent flows in rough pipe :1

f= 2.0log

d

e

+1.14, f = f(e,d)

In transition flow :e

must be significant parameter

In laminar flow : v = R b c the viscous effects dominate the whole flow

e

R


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In laminar sublayer :v

d=

32.8

Re f,

e

v

=

e d

v d=

e

d

Re f

32.8,

e

dRe f = 32.8

e

v

(9.32)

For rought pipe :1

f 2.0 log

d

e

=1.14 plot:

1

f 2.0 log

d

e

versus

e

dRe f

For smooth pipe :1

f= 2.0 log

32.8

v d

0.8 = 2.0 log Re f( ) 0.8

1

f 2.0 log

d

e

= 2.0 log

e

dRe f

0.8 plot in Fig. (9.7)

For smooth flow :e

d

Re f 10e

v

0.3

For transition flow : 10 maximum KL

better abrupt enlargement than = 60 (KL =1) => desing of connector of pipelines.

Example:

P300 = 140 kPa 300 mm 600mm 20

Flowrate: Q = 0.3 m3/s. Calculate: P600 in the larger pipe, neglecting pipe friction


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z300

+

P300

+

v300

2

2gn= z

600+

P600

+

v600

2

2gn+ hL ,

v300

=

Q

A300

=

Q

d

300

2

4

=

0.3 4

0.3( )2= 4.24 m /s v

600=

Q

A600

=1.06m /s,

From Fig. 9.15, = 20 andA

1

A2

=

d600

2

d300

2=

0.6

0.3

2

= 4, given KL 0.43

hL = KL v300

v600( )

2

2gn= 0.43

4.24 1.06( )2

2 9.81= 0.2216

Taking the datum at the CL

to e liminate z from the Eq :

140 103

9.81 103+

4.24( )2

2 9.81=

P600

9.81 103+

1.06( )2

2 9.81+ 0.2216 P

600= 143 kPa

Another abrupt contraction form:

Experimental measurements of KL 0.5, 1[ ],

A2

A1

= 0, 1[ ], Cc =A

c

A2


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A2

A1= 0 (Fig.) KL 0.5 for high Re

See Table 2 :

A2A1

=1 (Fig.) KL 0.8

Hamilton:


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r > 0.14d will prevent the formation of a vena contracta, and thus eliminate the head loss

due to flow deceleration. KL 0.1, exact value depends on the detailed geomtry of the

entrance & boundary layer.

For short well-streamlined contraction: KL0.04

For long contraction: KL > 0.04

Losses of head in smooth pipe bends due to effects of separation, wall friction & secondary

flow. Head loss coefficients for smooth pipe bends are shown in Fig 9.20. KL (Shape of

bend: R d)

R d= 0, KL 1.1, miter bend using in large duct as wind& water tunnels

The losses of head caused by commercial pipe fittings occur because of their rough &

irregular shapes. Values of KL in the Engineering Date Book are given in Table 3.


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Example:

A pipeline lenght = 3 km, diameter = 75mm carries water at v = 70 cm/s, has a pipe friction

loss of 25m, neglecting local losses. If the line contains a sharp-edged reservoir entrance, a

wide-open, screwed global valued, and four 90 regual, screwed elbow, and exits directly

into a reservoir.

Calculate: local losses in the line & the percent error incurred by neglecting them

hL = KLv2

2g n, Need KL = ?

Sharp edged entrance :KL = 0.5

Wide open, screw globe value Table 3, given KL =10

90 regular, screwed elbows Table 3, given KL =1.5

Exit into reservoir KL =1

Total local loss = KLv2

2gn= 0.5 +10 +1.5 +1( )

0.72

2 9.81= 0.325m

percent error =local Loss

pipe friction Loss 100% =

0.325

25 100% =1.3%


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9.10Pipelineproblems-Singlepipes

Engineeringpipe-flowproblemsusuallyconsistof:

1. CalculationofhLandpfromflowrateandpipelinecharacteristics.2. CalculationofflowrateQfrompipelinecharacteristicsandhL.3. Calculationof requiredpipe diameter topass a given flowrateQ between2

regionsofknownp.

P1. Can be solved directly.

P2 & Pe: using trial-and-error solution because friction factor f and loss coefficients KL

depend on Re, Re(flowrate [P.2], d [P.3]).

Construction of Energy & Hydraulic Grade lines for some typical pipelines problems mayfrequently be used in the solution of Engineeering problems.

Head loss due to friction hL f , abrupt entrance hLe , and abrupt exit to another tank hLx :

hLe

+ hL f + hLx = H 0.5+ fl

d+1

v

2

2gn= H

The effects of local losses in pipelines of common lengths is so small, which may often be

neglected entirely, increasing l /d also decreasesv2

2gn

.

EL HGL hL = fl

d

v 2

2gn= H


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Example: A clean cast iron pipeline has d=0.3m, length = 300m connects two

reservoirshavingsurfaceelevationsof60&75m.Calculatetheflowratethroughthe

line,assumingwaterat10Candsquared-edgedentrance.

water at10C, =1.306 106 m2 /s Re = vd

= v0.31.306 10

6= 229.000V

from Fig.9.11 for clean cast iron pipe, given e /d= 0.00083

Using iteration procedure, guess v 2m s . (Since in D-W formula hL = fl

dv 2

2g, need f,

in order to find f from figure 9.10, we need a Re number!) Re = 458000 .

From Fig 9.10, with e /d= 0.00083, Re = 458 103 given f 0.02, KLe = 0.5, KLx =1.

Chosen points 1&2 on the surfaces of 2 reservoirs work-energy Eq:

z1+

p1

+

v1

2

2g= z

2+

p2

+

v2

2

2g+ hL , p1 = p2 = 0, v1 = v2 = 0, 75 + 0 +0 = 60+ 0.5+ f

l

d+1

v 2

2gn

15 = 0.5 +0.02300

0.3+1

v 2

2 9.81 v = 3.7m /s

Recalculate Re to find f -value & v again.

Re =vd

=

3.70.3

1.306 106

= 847250

From Fig 9.10, with Re = 847250 & e /d= 0.00083 given f 0.0193:

75+0+0 = 60+ 0.5+0.0193300

0.3+1

v 2

2gn

v = 3.76m /s Q = v A = 3.76 0.3

2

4= 0.266 m

3


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Example: A smooth PVC pipeline 60m long carries a flowrate 0.003 m3/s between 2 water

tanks with a difference in water surface elevation of 1.5m. If there is a squared-edged

entrance and water at 10C, calculate the diameter of the required pipe.

water at10C, =

1.306 10

6

m

2

/s Re =

vd

=

Qd

A=

Qd

d2

4

=

4Q

d=

40.003

d

4 1.036 106

=

14700

d

Work energy equation : 0.5+ fl

d+1

v 2

2gn=1.5 0.5 + f

60

d+1

v 2

2gn=1.5

using Excel to find : d= 0.036m

In the main line velocity head is neglected EL HGL . Velocity head at nozzle exit

cannot be neglected work-done Energy Equation: (z0 at reservoir)


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40

z1 +p

1

+

v1

2

2g= z2 +

p2

+

v2

2

2g+ hL , hL = hLe + f

l

d1

v1

2

2gn

= KLe

v1

2

2gn+ f

l

d1

v1

2

2gn

Setting the datum at the CL of pipe (nozzle)

H+0+0 = 0 +0+v

2

2

2g

+ hL, H=v

2

2

2gn+ KLe

v1

2

2gn+ f

l

d1

v1

2

2gn

A1 v1 = A2 v2 d

1

2

4v1 =

d2

2

4v2 v2 =

d1

d2

2

v1 H=d1

d2

2

+KLe + fl

d1

v1

2

2gn

+ If we know the flowrate, we can solve this equation directly.+ If we know H, this equation can be solved by trial- and error.

desing: Sprinkler system, fire supression hoses,....

- Turbine, where the power in the jet is converted to electrical energy.


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41

power = Q N/ s( )v 2

2gnm

= Q

v2

2gn

Nm

s= Q

v2

2gn

J

s

neglecting local losses, we obtain :

v22

2gn= H f

l

d1

v12

2gn= H f

l

d1

Q2

2gn A12 power = Q H

f l Q2

2 d1 gn A12

To find maximum of jet power : dP

dQ= 0 H

f l

2 d1 gn A12 3Q2

= 0

f l Q2

2 d1 gn A12=

H

3

f l v12

2 d1 gn=

H

3

v22

2 gn=

2 H

3

itshowswhenthemaximum powermaybeexpected.

Work - energyequation: z1 +

p1

+ v12

2g+ EP = z2 + p

2

+ v2

2

2g+ ET

Horsepower of Machine =Q EP or ET( )

550(US) Kilowatts of Machine =

Q EP or ET( )1000

(SI)

1hp = 0.746kW (KyneticEnergyperunitweightv 2

2g)

z1 +

p1

+

v1

2

2gn+ EP = z2 +

p2

+

v2

2

2gn+ hL WHP =

QEP

550(US, 9.49a) WkW =

QEP

1000(SI, 9

+ Usually, we have been working with gravity-flow, however a more commonocurrence is the pumped pipeline (water supply).

+ Where the pumps are located at the upcstream end of the pipeline: sourcer pupms,the draw liquid from wells, reservoirs, tanks, etc.

+ Where the pumps are located at some intermediate point in the pipeline: boostenpumps.

Using formula (9.49), WHP (WkW) represent the power added to the fluid by the pump.


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42

Most engineering liquids contain dissolved gases, they move with liquids as large bubbles,

collected in high point of the pipeline, reducing the flow along a crosssection, and are sent

to disrupt the flow. In practice, large negative pressures in pipes should be avoided if

possible by design improvements. Where such negative pressures cannot be avoided they

should be prevented from exceeding about 2/3 of the different barometric and vapor

pressures. (

patm

pv

).

Example:

(1330m, ridge-1313m (distance from upstream), 1230m, pipeline 20km long)

There is concern that the ride is too high and will create an unacceptable low preassure in

the pipeline. What is your recommendation as to the feasibility of proposed location of the

pipeline?

Neglecting local losses & considering EL HGL . EL-HGL is falling

1330 m 1230 m =100 m over 20 km. the ridge distance from upstream 4km is about 1/5

of this falling.

Elevation of EL-HGL of ridge =1330 100 1/5 =1310m . It means that the preasure head

at ridge is -3m. The negative preasure head of water is approximately -10m, the preasure

haed at ridge = -3m is about 1/3 of this limitation.

Note: 0 35C 10m( at 20C), he = patm

pv

= 101Pa 2.3Pa

9.8110m


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43

Example: Calculate the horse power that the pump must supply to the water at 10C in

order to pump 0.07m3/s through a clean cust iron pipe from the lower reservoir to uuper

reservoir. Neglecting local loses and velocity heads.

water at10C, =1.306 106 m2 /s WkW =QEp

1000

we need to find Ep z1 +p

1

+

v1

2

2gn+ Ep = z2 +

p2

+

v2

2

2gn+ hL

v =Q

A: v0.2 =

0.07

0.2( )

2

4

= 2.22 m /s v0.15

=

0.07

0.15( )

2

4

= 3.96 m /s

R0.2 =

v0.2 d

=

2.22 0.2

1.306 106= 3.4 105 R

0.15=

v0.15

d

=

3.96 0.15

1.306 106= 454800

Fig 9.10 ed

0.2

= 0.012 Fig.9.11 f = 0.021

e

d

0.2

= 0.0018 Fig.9.11 f = 0.024

Head losses

hL0.2

= fl

d

v0.2

2

2gn= 0.021

300

0.2

2.22( )2

2 9.81= 7.91 m hL0.15 = 0.024

600

0.15

3.96( )2

2 9.81= 76.73 m

hL = 7.91 m + 76.73 m = 84.64 m

z1+

p1

+

v1

2

2gn+ Ep = z2 +

p2

+

v2

2

2gn+ hL 15+0+ 0+ Ep = 45+0+ 0+84.64 Ep =116.64 m

WkW =QEp

1000=

0.07 9.8 116.64

1000= 0.08kW WHP = 0.746 0.08 = 0.06hp


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44

9.11PipelineProblems-Multiplepipes

+ 2 pipes separating & rejoining

+ More complex system


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45

The basic principles of analysis are the same although the techniques of analysis vary

depending on the system complexity. Assumptions:

- Velocity heads & local lsses are neglected EL HGL -

Variation of the friction factor f with the Re are often neglected (unless handlingby a computer program).

One loop network:

+ Parallel laying is standard method: a) increasing capacity b) flexibility to repair.- The head loss through bouth the branches of the loop mush be the same, if EL-

EGL network above the pipes is to be continous throughout the system

+ Continuity equation: Q =QA +QB , hLA = hLB Darcy-Weisbach for head loss:

hL =

fl

d

v 2

2gn= f

l

2gnd

16 Q2

2 d4

=

16 f l

2 2 gn d5

Q

2

Generalwritingofthisequation: hL = k Qn kA QA

n= kB QB

n, Q =QA +QB


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46

Solution of this equation system allows prediction of the division of a flowrate Q into QA &

QB when the pipe characteristics are known.

Example:

d = 300mm, l = 1500m, Q = 0.15 m3/s. This pipe is looped with 600m pipe of the same

material & size parallel and connected to it. What % increase in maximum flowrate may be

expected?

Neglectic local losses and velocity head EL HGL

hL =16 f l

2 2 gn d5 Q2 k l( ) hL = k Q

2 k1500

=

hL

Q2=

24

0.15( )2=1067 1500mpipeline( )

Fortheloopsection: k600

=

600

1500 1067 = 427

Fortheunloopedsection: k900

=

900

1500 1067 = 640

Fortheoriginalpipe,theheadlossinthelooped & uploopedportions:

hL = 24 = k900 Q2+ k

600 QA

2= 640 Q2 + 427 QA

2

Forthenewpipe : hL = 24 = k900 Q2+ k

600 QB

2= 640 Q2 + 427 QB

2

Solvingthisequations,weobtain : QA =QB Q =QA +QB QA =Q

2

640 Q2 + 427Q

2

2

= 24 640 Q2 +106.75 Q2 = 24 Q = 0.18m3 s

percentage increasing =0.18 0.15

0.15 100% = 20%


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47

Multiple pipes system connects 3 reservoirs (or more)

Scenarios:

1. Flow from reservoir A into reservoirs B & C2. Flow from A to C without flow in & out from B3. Flow from A to B into C

Situation (1):

hA /C = kA QAn + kC QCn hA /B = kA QAn + kB QBn

Set(1)

zA kA QAn= kC QC

nzA zB = kA QA

n+ kB QB

n

zA kA QAn= zB + kB QB

n

QA =QB +QC

Situation (3):

zA kA QAn= kC QC

n

Set(2)

zB kB QBn= kC QC

n

zA kA QAn= zB kB QB

n

QA +QB =QC

Situation (2), the same as above, just set QB=0

In view of physical flow only one of these sets of equations can be satisfied:


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48

If QA > QC : using Set (1) If QA < QC: using Set (2)

After identification of the set of equations, these may be solved (by trial) to yield the

flowrates QA, QB and QC.

Multiple pipe systems are complicated problems of distribution of low in pipe networks

(e.g. city water supply system).

+ One Method is presented here to illustrate the bais principles: Network consistingof various sizes, geometry, hydraulic characteristics, pumps, valves, etc.

Assumptions:

o Flows are assumed positive in a clockwise direction around each loop.o The continuity principle states that the net flowrate into any pipe junction must be

zero: Qi = 0.

o The work-energy principle requires that at any junction there is only one position ofEL-HGL, which means the net head loss around any single loop of the network

must be zero.

Applying into the following network:


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49

EquationforloopI : (thesimilarwillbedoneineveryloop)

QA

=QA +Q2 Q1 = 0

QF

=Q1 +QF Q3 = 0

QE

=Q3 Q4 Q8 = 0

QB

= Q2 +Q4 +Q5 +Q7 = 0

hL = k1 Q13

I

+ k3 Q33 + k4 Q43 + k2 Q2n = 0

Assuming that: pipe sizes, lengths, hydraulic characteristics, Network in/out flows, pump

station & pump characteristics, and Network layout & elevations Qi i =1, ...,10 will be

solved.

There are several ways to solve for those flowrates, the simplest and easiest one is Hardy

Cross Method.

Hardy Cross Method (Iteration Method):

The essence of the method is to start with a best estimate of initial values, If the first

estimates are reaonably accurate, Q0i, the first iteration:

Qi =

Q0i i (9.52)

Sign (

) depends on the direction assumed for Q0,L , is the correction in a loop L.

For example: Q3= Q

03+ I, Q8 = Q08 + II, but Q4 = Q04 + I II

In general a head loss equation has a form: hL iL

= k Qin = 0L

(9.53)

The sign (

) depends on the flow direction, Qi is the magnitude of the flowrate, for

example:

Loop I k1Q

1

n+ k

3Q

3

n+ k

4Q

4

n+ k

2Q

2

n= 0

Loop II k8Q

8

n+ k

10Q

10

n+ k

7Q

7

n k

4Q

4

n= 0

Loop III k5Q

5

n k

7Q

7

n+ k

9Q

9

n+ k

8Q

8

n= 0

Substitute (9.52) into (9.53) we got: hLiL

= ki Q0i L( )n

= 0

L


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50

Expanding (using binomial theorem!)

1

1 1

1 2 1

1 3 3 1

neglecting higher order terms of L

hL iL

ki Q0i

n n Q

0i

n1L( ) + L( )2

= 0

L

L = ki Q

0i

n

L

nki Q0i

n1

L

This equation is used to calculate flowrate correction L

for each loop of the nerwork. The

iteration equation:

Lj+1( )

=

ki Q i( j)

( )n

L

nki Q i( j)( )

n1

L

, j: jth - iterationstepforloopL

When a pump is added to a pipe in the network, an expression representing the head

increase is a polynomial equation:

Ep i = a0 + a1 Qi + a2 Qi2+ a

3 Qi

3+ ...

with as many ai coefficients as necessary to provide a good representation of the pump

curve. For example if the pump added to the line 8 in the loop II:

hLiII

= k8Q8n a0 + a1 Q8 + a2 Q82 + ...( ) + k10Q10n + k7Q7n = 0


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51

Example:

o Paralel commercial steel pipeo By accident a valve in time in line BC was open.

What are the resulting flowrates in all the pipes, neglecting local losses, and assuming that

the flows are wholly rough:

Darcy-Weisbach: ki =16 fi li

2 2 gn di

5

pipe Nr. Length (m) Diameter (m) e/d f Ki

1 1000 0.5 910-5 0.012 31.7

2 1000 0.4 110-4 0.012 96.8

3 100 0.4 110-4

0.012 9.7

4 1000 0.5 910-5

0.012 31.7

5 1000 0.3 110

-4

0.013 442.0

Equation forL

for each Loop:

I = k1 Q01

2+ k2 Q02

2+ k3 Q03

2

2 k1 Q

01 +k2 Q

02 +k3 Q

03( )( first iteration step), II =

k3 Q032+ k5 Q05

2 k4 Q04

2

2 k3 Q

03 +k5 Q

05 +k4 Q

04( )


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52

The adjustment equations are:

Initial Calculation Subsequent Calculation

Loop I

Q1= Q

01+ I Q1

j+1= Q

1

j+ I

j

Q2= Q

02+ I Q2

j+1= Q

2

j+ I

j

Q3= Q

03+ I II Q3

j+1= Q

3

j+ I

j IIj

Loop II

Q3= Q

03+ II I Q3

j+1= Q

3

j+ II

j Ij

Q2= Q

04+ II Q4

j+1= Q

4

j+ II

j

Q3= Q

05+ II Q5

j+1= Q

5

j+ II

j

Take a look at problem 9.21

9.12Pipeflow:UnsteadyFlow

o Unsteady flow is important in engineering practice:- Cause escessive pressures, vibration, cavitation- Cause physical or performance failure of a System

o In many cases, the analysis of unsteady flow recurrin in pipeline system is based onsteady analysis flow recurring in pipeline system is base on steady analysis because

its transient nature and change with a small magnitude.

o We consider few cases, wherein significant changes in velocity cause large changesin pressure.

o Unsteady flows in pipeline system are far too complex and too uncertain to permitaccurate simulation by mathematical means. It requires considerable judgment to

simplify the problem to one that can be analyzed.

Two methods using to analyze unsteady flow in pipeline:

1. Rigid water column theory: treats the fluid as an inelastic substance obtain ODEequations and find a numerical solution.

2. Elastic or waterhammer theory, wherein the elasticity of both the fluid & the pipewalls is taken into account in the calculation. (Deforming)


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53

Action of Water hammer in a simple pipelne situation:

Friction loss is neglected, velocity head is small: EL HGL . Water hammer will be

introduced into de system by suddenly closing the valve. The activity will occur both

upstream & downstream of the valve, we will observe only what occurs on the upstream of

the valve.

+ Suddenly close the valve, velocity of the water at the valve is forced to be zero(B.E.) pressure head at the valve H, Pressure at the valve stretching of

the pipe and increasing of the density of fluid.

+ Preassure increase propagates upstream at a wave speed a.


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54


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55

9.13RigidwatercolumnTheory: Euler'sEquation

Rh = dy

Rh = AP

P = 4Ad

, Shear stress = P s

F = m a

pA p +p( ) A gnsAcos4A

d s = A s

v

tpA gnsA

z

s 4

A s

d= ...

... = A sv

t

1

p

sz

s4

d=

1

gn

v

t

Whenthecontrolvolumediameterisexpandedtothesizeofthepipediameter

1

p

sz

s4

d=

1

gn

v

t

1

p

sz

s

f v2

2 gn d=

1

gn

v

t =

r

R

0,

0=

r

R, =

h

2l

zisafunctionofonlys partialderivative totalderivative

1

p

sz

s

f v2

2 gn d=

1

gn

v

t9.56( )

Thisisunsteadyequationcanbeusedtosolveawiderangeofpipelineproblem.


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56

We will address some basic problems:

If the discharge in this pipeline is controlled by the value at the downstream end. The

preassure in the pipe = H0 when the valve is closed. When the valve is suddenly opened the

pressure at the valve drops instantly to zero and the fluid begins to accelerate.

Integrating (9.56) with respect to s from point 1 to 2.

1

p

s ds dz

ds ds f v

2

2gn d ds = 1

gn dv

dtds, horizontal

dz

ds= 0, v = v(t)only

Assumingthatthevalueoffinunsteady = finsteadyflow, weobtain : P1

P2

f L

2gn dv2 =

L

gn

dv

dt,

P1

= H0 = const. and P2 = 0 at t> 0 H0

f L

2gn dv2 =

L

gn

dv

dt

Intregratingbyseparatingvariables :

dt = Lgn

dv

H0 f L

2gn d

v

2

t= L d2gn f H0

ln2gn dH0 f L + v

2gn dH0 f L v

Ifthelocallossesareneglected,then 2gn dH0 f L = v0 steady flow velocity( ), then :

t=L v0

2gn f H0n

v0 + v

v0 v

Assteadyflowv v0 t ; unacceptable.

Weproposethatwhenv = 0.99v0 , wehaveessentiallysteadyflow, and : tgg = 2.65L v

gn H0


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57

9.14ElasticTheory(Waterhammer)

Using the impulse - momentum equation

We know that a change in velocity v will cause a pressure head H to propagate upstream

at apeed a. Calculate H & a.

Assume a = const, change reference frame mving with speed a unsteady steady

(important technique!)

One dimensional impulse-momentum equation (Chapter 6)


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58

Fext = Qv( )

out Qv( )

in, Q = discharge, densityofthefluid, Fext sumofexternalfo

Fext =

m vout v in( ), 9.67( ), m = Q= constant

using 9.67( )considerrigidpipeandthatF3isneglected

F1 F2 = m v v + a - v - a( ) = m v( )

where m = Qp = v + a( )Ap andvisreductioninvelocity

p A p+ p( ) A +A( ) = v + a( )Ap v( ) pA -pA p A = v + a( )Ap v( )

p = H, AisverysmallcomparedtoH, A &

wedropthesmall HA = v + a( )Ap v( ) H=p

v v + a( ) H=

av

gn1 +

Inengineeringpracticev

a< 0.01 H=

av

gn H= f(a)

weneedtofindabyusingthemassconservation

Duringdt :

M= vAt v - v( ) +( ) A +A( )t

expandingandneglectingsmallterms,weobtain : M= Avt= AvL

a

9.70( )

Because the pressure has increased, the volume of liquid in the section will compress

slightly to higher density. The hulk modulus of elasticity:

E =dp

dV VChapter 1( ), E:hulkmodulusofelasticityofthefluid, p,V : pressure & Volum

V p, (E : relativelyconstantoverawiderangeofpressureintheabsenceoffreeorentrain

V pL A

E, VischangeinvolumeoffluidinthepipesectionL

Because when the pressure increases stretching the pipe, the result of this stretching gives

(by evaluation):


59/59

V =

4 d

2 L

1 +2

2( )

1,

2representunitstraininlongitudinal & radialdirections,respectively

If the pipe is restrained from longitudinal streching then:

V =

4 d2 L

1 p2

Ep

p d

ep

, ep :pipewallthickness, Ep :Modulusofelasticity

p :Poisson'srationofpipematerial

From (9.70), Mass changes in the L pipe section:

M= p+p( ) A L +V( ) A L 9.73( )

Combine 9.70( ) & 9.73( )andsubstituting 9.71( ) & 9.72( )wegot :

a =E

1+E

Epd

ep 1 p

2( )

Documents

9 Flow in Pipes_with Figures