water supply components
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Monroe L. Weber-Shirk School of Civil and
Environmental Engineering
Gravity Water Supply Design
http://ceeserver.cee.cornell.edu/mw24/Default.htmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/faculty/info.cfm?abbrev=faculty&shorttitle=bio&netid=mw24http://ceeserver.cee.cornell.edu/mw24/Default.htmhttp://ceeserver.cee.cornell.edu/mw24/Default.htmhttp://ceeserver.cee.cornell.edu/mw24/Default.htmhttp://www.cornell.edu/ -
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Population Projection
Example from Agua Para el Pueblo(Honduras)Count the housesAssume 6 people per houseAssume linear growth for design period
N = design periodK = growth rate
Pob lacin fu tura ( Pf ) = Pa(1+N*K /100)
K = Tasa de crecimiento ( 3.5% ) N = Per o do de di seo ( 22 ao s )
( )1 future present P P NK = +
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Water Demand
Assume a per capita demand (this might be based on a governmental regulation)Multiply per capita demand by the future
population to get design average demandMultiply average demand by scaling factorsto get maximum day demand and maximumhour demand
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Distribution Storage Tank Size
Based on 8 hours of storage at averagedemandThese systems arent designed for fire
protection
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Design Flows
Transmission Line Design flowPerhaps based on maximum daily demand or onmaximum hourly demand
Distribution system design flowsTake peak hourly flow at the end of the system designlifeDivide that flow by the current number of houses to geta flow per houseThe flow in each pipe is calculated based on thenumber of houses downstream
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Pipe Diameters
How are pipe sizes chosen?Energy Equation
An equation for head lossRequirement of minimum pressure in thesystem
Lt p hh z g V p
h z g
V p2
22
22
1
21
11
22
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EGL (or TEL) and HGL
velocity
head
elevationhead (w.r.t.
datum)
pressurehead (w.r.t.
reference pressure)
z g
V p EGL
2
2
z
p HGL
downward
lower than reference pressure
The energy grade line must always slope ___________ (indirection of flow) unless energy is added (pump)The decrease in total energy represents the head loss or energy
dissipation per unit weightEGL and HGL are coincident and lie at the free surface for waterat rest (reservoir)If the HGL falls below the point in the system for which it is
plotted, the local pressures are _____ ____ __________ ______
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Energy equation
z = 0
pump
Energy Grade LineHydraulic G Lvelocity head
pressurehead
elevation
datum
z
2g
V 2
p
Lt p hh z g V p
h z g V p
2
2
2221
2
111 22
static head
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Transmission Line Design
Air release valves
HGL
EGL
Spring
box
DistributionTank
2
2 5
8f f
LQh
g D p=
( )1 future present P P NK = +
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Methods to Calculate Head Loss(Mechanical Energy Loss)
Moody DiagramSwamee-JainHazen-Williams
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Moody Diagram
0.01
0.10
1E+03 1E+04 1E+05 1E+06 1E+07 1E+08Re
f r
i c t i
o n
f a c
t o r
laminar
0.050.04
0.03
0.020.015
0.010.0080.0060.004
0.002
0.0010.0008
0.0004
0.0002
0.0001
0.00005
smooth
l D
C pf
D
0.02
0.03
0.040.05
0.06
0.08
2
2 5
8f f
LQh
g D p=
Re
4Q D
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Swamee-Jain
1976limitations
/D < 2 x 10 -2
Re >3 x 10 3 less than 3% deviationfrom results obtainedwith Moody diagram
easy to program forcomputer or calculatoruse
0.044.75 5.221.25 9.4
f f
0.66 LQ L
D Q gh gh
e n
= +
2
0.9
0.25f
5.74log
3.7 Re D
e=
+
Each equation has two terms. Why?
2 f
f
1.7840.965 ln
3.7 gDh
Q D L D gDh
D L
e n
= - +
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Pipe roughness
pipe material pipe roughness (mm)
glass, drawn brass, copper 0.0015
commercial steel or wrought iron 0.045asphalted cast iron 0.12
galvanized iron 0.15
cast iron 0.26
concrete 0.18-0.6rivet steel 0.9-9.0
corrugated metal 45PVC 0.12
d Must bedimensionless!
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Pipeline Design Steps
Find the minimum pipe diameter that will keep theHGL above the pipeline
Round up to the next real pipe sizeCalculate the location of the HGL given the real
pipe size
If an intermediate high point constrained thedesign then investigate if a smaller size pipe could
be used downstream from the high point.
0.044.75 5.221.25 9.4
f f
0.66 LQ L
D Q gh gh
e n
= +
2
f 2 5
8f
LQh
g D p=
2
0.9
0.25f
5.74log 3.7 Re D
e=
+
Re 4Q
D
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Minor Losses
Most minor losses (with the exception ofexpansions) can not be obtained
analytically, so they must be measuredMinor losses are often expressed as a losscoefficient, K, times the velocity head.
2
2l V
h K g
=
( )geometry,Re p
C f =2
2C
V
p p 2
2C
V
ghl p
g
V h pl
2C
2High Re
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g
V K h ee
2
2
0.1e K
5.0e K
04.0e K
Entrance Losses
Losses can bereduced by
accelerating the flowgradually andeliminating thevena contracta
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Head Loss in Valves
Function of valve type andvalve position
The complex flow path throughvalves can result in high headloss (of course, one of the
purposes of a valve is to createhead loss when it is not fullyopen)
g
V K h vv 2
2
What is the maximum value of K v? ______
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Solution Technique: Head Loss
Can be solved explicitly
f l minor h h h= +
2
2minor V h K
g =
2
f 2 5
8f
LQh g D p
=2
0.9
0.25f
5.74log3.7 Re D
e
=
+
2
2 48
minor Q K h
g D p=
D
Q4Re
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Solution Technique 1:Find D
Assume all head loss is major head lossCalculate D using Swamee-Jain equationCalculate minor lossesFind new major losses by subtracting minorlosses from total head loss
0.044.75 5.221.25 9.4
f f
0.66 LQ L
D Q gh gh
e n
= +
42
28
D g
Q K h minor
f l minor h h h= -
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Solution Technique 2:Find D using Solver
Iterative techniqueSolve these equations
f l minor h h h= +
42
2
8 D g
Q K h minor
2
f 2 5
8f
LQh
g D p=2
0.9
0.25f
5.74log
3.7 Re De
=
+
D
Q4Re
Use goal seek or Solver tofind diameter that makes thecalculated head loss equalthe given head loss.
Spreadsheet
http://ceeserver.cee.cornell.edu/mw24/cee331/tools/Swamee-Jain.xlshttp://www.cee.cornell.edu/cee331summer/http://ceeserver.cee.cornell.edu/mw24/cee331/tools/Swamee-Jain.xlshttp://www.cee.cornell.edu/cee331summer/ -
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Exponential Friction Formulas
f
n
m
RLQh
D=
unitsSI675.10
unitsUSC727.4
n
n
C
C R
1.852
f 4.8704
10.675 SI units
L Qh
D C =
C = Hazen-Williams coefficient
range of data
Commonly used in commercial andindustrial settings
Only applicable over _____ __ ____collectedHazen-Williams exponential frictionformula
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Head loss:Hazen-Williams Coefficient
C Condition150 PVC
140 Extremely smooth, straight pipes; asbestos cement130 Very smooth pipes; concrete; new cast iron120 Wood stave; new welded steel110 Vitrified clay; new riveted steel
100 Cast iron after years of use95 Riveted steel after years of use60-80 Old pipes in bad condition
1.85210 6 Q
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Hazen-Williamsvs
Darcy-Weisbach
1.852
f 4.8704
10.675 SI units
L Qh
D C =
2
f 2 5
8f
LQh
g D p=
preferred
Both equations are empiricalDarcy-Weisbach is dimensionally correct,and ________.Hazen-Williams can be considered validonly over the range of gathered data.Hazen- Williams cant be extended to otherfluids without further experimentation.
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Air Release Valve
http://www.ipexinc.com/industrial/airreleasevalves.html
http://www.apcovalves.com/airvalve.htm
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Pipes
http://www.ipexinc.com/industrial/4080_pipe.html
Diameter O.D. Wall
Thickness
I.D. Pressure
73F
Wall
Thickness
I.D. Pressure
73F
(inches) (inches) (inches) (inches) (psi) (inches) (inches) (psi) 1/2 0.84 0.109 0.602 600 0.147 0.526 850 3/4 1.05 0.113 0.804 480 0.154 0.722 690
1 1.315 0.133 1.029 450 0.179 0.936 6301 1/4 1.66 0.141 1.36 370 0.191 1.255 5201 1/2 1.9 0.145 1.59 330 0.2 1.476 470
2 2.375 0.154 2.047 280 0.218 1.913 4002 1/2 2.875 0.203 2.445 300 0.276 2.29 420
3 3.5 0.216 3.042 260 0.3 2.864 370
Schedule 40
PVC
Schedule 80
PVC
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Additional PVC Pipe Schedules
http://www.prodigyweb.net.mx/pofluisa/pvc.htm#tubocementar
Presin de Trabajo
RD-13.5 22.4 kg/cm 2 315 psi
RD-21 14.0 kg/cm 2 200 psiRD-26 11.1 kg/cm 2 160 psi
RD-32.5 8.6 kg/cm 2 125 psi
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Surveying
qVertical angle
rx
z
cos2
x r pq
D = -
sin2
z r pq
D = -
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Surveying using Stadia
q
Vertical angle
rx
cos2
x r pq
D = - sin
2 z r
pq
D = - -
The reading is on a vertical rod, so it needs to becorrected to the smaller distance measured
perpendicular to a straight line connecting thetheodolite to the rod.
a
z
b c
cos
2
b c pq
= -
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Horizontal Distance
cos 2 x r pq D = -
cos2
b c pq
= -
sin cos2
pq q
= -
Trig identity
sin x r D =
sinb c=
r Mb= M is the Stadia multiplier (often 100)
( )2sin x Mc qD = c is the Stadia reading
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Vertical Distance
sin2
z r pq
D = - -
( )sin cos cos2 pq q p q - = - = -
sin cos2
pq q
= -
sinb c=r Mb
=
sin cos z McD =
( )1 sin 2 sin cos2
q q q=
cos z r D =
sin2
2
Mc z qD =
Trig identities
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Pipe Length (along the slope)
r Mb= sinb c=
sinr Mc=
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