9 flow in pipes_with figures
TRANSCRIPT
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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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):
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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( )