water supply components

8/10/2019 Water Supply Components

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Monroe L. Weber-Shirk School of Civil and

Environmental Engineering

Gravity Water Supply Design
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2/32

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


5/32

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


6/32

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 _____ ____ __________ ______


8/32

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 = +


10/32


11/32

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


17/32

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


18/32

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


20/32

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
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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


23/32

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


28/32

Surveying

qVertical angle

rx

z

cos2

x r pq

D = -

sin2

z r pq

D = -


29/32

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


31/32

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


32/32

Pipe Length (along the slope)

r Mb= sinb c=

sinr Mc=

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