introduction - ipb university

WaterSuplly Asep Sapei

2006 Edition 1

INTRODUCTION

WATER SUPPLY

Is a matter of concern since the beginning of civilization Early water supply, just brought water from distance source

to a few central locations At the middle of 17 century, water supply used pipe made of

wood, clay or lead Since the development of cast iron pipe and pumps, water

delivered to individual residence Coagulation and filtration have been used since 2000 BC in

water treatment, but the application in municipal treatment

started on 1900 Chlorine disinfection was introduced in 1913

WATER DEMANDS

CONSUMPTION CATEGORIES

1. Domestic purposes

Categories:

- In house use: for drinking, cooking, ablution, sanitation, house

cleaning, laundry, patio and car washing - Out of house use: for garden watering, lawn sprinkling and

bathing pools

- Standpipe use : for standpipes and public fountains

Amount

- In house demand : in table

- Out of house demand :

Depends 0n whether a dry or wet climate

- For garden and lawn irrigation : 70 – 800 lcd


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- For swimming pools and others : 15 – 20 lcd

- Total domestic demand : 75 – 380 lcd

- on the average in Malaysia 230 lcd


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2. Agriculture for crops, livestock, horticulture, dairies, greenhouse, farmsteads

Rice : - Land preparation : 150 – 250 mm

- Nursery : 50 mm

- Growing : 500 – 1200 mm

3. Trade and industrial

- Industrial : for factories, industries, power station, dock etc.

- Commercial : for shops, offices, restaurants, hotels, railway

stations, airports, small trades, workshops, etc - Institutional : for hospitals, schools, universities, government

offices, military establishment, etc. Divided into 4 categories

- Cooling water demand : none is use from the public supply


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- Major industrial demand : use > 1000 m3/day, from private

source or non-potable source - Large industrial demand : use 100 – 500 m3/day

- Medium to small industrial demand : use < 50 m3/day

- About 15 % of the total consumption


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4. Public

for fire-fighting, street watering, public gardens, public parks and

sewer flushing

From 50 – 75 lcd

5. Losses

- Distribution losses : leakage from mains and connections, leakage

and overflow from service reservoirs,


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- Consumer wastages : leakage from their supply pipes , misuse and

unnecessary use - Metering and other losses : meter errors, unauthorized or

unrecorded consumptions

TOTAL CONSUMPTION

Is the amount supplied water per head of population

Is total legitimate potential demand + consumer wastage and

distribution losses – unsatisfied demand

Expressed in liters per head or capita per day (lhd or lcd)

The general range of total supply:

- For big industrial cities in USA : 600 – 800 lcd

- For many major cities and urban area throughout the

world: 300 – 550 lcd

- For areas where supplies are short or there are many

street standpipes, or many of population have private wells:

90 – 150 lcd


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FACTORS AFFECTING WATER USE

Size of the city. Small communities tend to have more limited uses of water

Industry and commerce. - Industrial use has no direct relation to the population, but great

care must be taken.

- Commercial consumption is largely dependent on the number of

population.

- In highly developed district, water used sometimes made upon

the basis of floor area (10-15 l/m2/day) or ground area (up to 95

l/m2/day)

Characteristics of population Economic level. More water needed for highest economic level

people

Metering of water supplied

Miscellaneous factors - Climate. more water needed at warm climate

- Water quality. Quality water used

- Water pressure. Pressure water used

VARIATION IN WATER USE

Variation during hour to hour during a day, day to day during a

week , week to week during a month and month to month

during a year

Beside average consumption Maximum or peak

consumption: maximum hour’s consumption, maximum day’s

consumption, maximum week’s consumption, maximum month’s

consumption

Calculated by Goodrich formula

1.0180

tP

Where P is the percentage of average rate which occurs

during peak period, t is the length of the period in day


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Maximum daily rate : 1.01180 xP = 180 %

Maximum weekly rate : 1.07180 xP = 148 %

Maximum monthly rate : 128 %

POPULATION PREDICTION

If the design life of the scheme is 25 years, thus we have

estimate for the design period up to 25 years ahead

a) Arithmetic methods

Assume the rate of growth is constant, dP/dt=K

Future population : Pt=P0 + Kt

Example :

Year population increase

1930

1940

1950

1960

1970

25000

28000

34000

42000

47000

3000

6000

8000

5000

: 22000

K = 22000/4 = 5500/10 years

Population in 1990: P = 47000 + 5500x2 = 58000

b) Geometric increase method

Assume the percentage of growth per decade is constant

non

rPP )

1001( , where n : numbers of decade

1100 0

tt

P

Pr

From the example :

17.0125000

47000

1004

r

Population in 1990: 64340)17.01(470002 nP

c) Curvilinear method

- as a graphical projection from the historical data


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- by make comparison of projected growth to the

recorded growth of other larger city which have

geographical proximity.

- Example: City A, the city being study, has historical data

from 1950 to 1990. the comparison cities are B, C, D and

E. The projected curve for city A is the dash line

FIRE DEMAND

- The actual amount of water used in a year is small but the

rate of use is high.

- The required flow : F=18C(A)0.5

F is the required flow in gal/min (:3.78 l/min), C is the

coefficient and A is the total floor area in ft2 (10.76 m2)

C = 1.5 for wood frame construction, 1.0 for ordinary

construction, 0.8 for noncombustible construction and 0.6 for

resistive construction

- For a single fire between 1890 – 45360 l/min

- The fire flow must be added to the daily consumption rate

- Fire flow duration


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- Example: city with 22000 population. Average consumption 600 lcd.

Fire flow for ordinary construction, 1000 m2 floor area and six stories.

Average domestic demand: 22000x600 = 13.2 x 106 l/day

Max. daily demand = 1.8 x 13.2 x 106 = 23.76 x 106 l/day

Fire flow = 18(1)(1000x10.76x6)0.5=4574 gal/min=24.89 x 106 l/day

Max. rate = 23.76 x 106 + 24.89 x 106= 48.65 x 106 l/day

The flow should be maintain for 10 hrs, so total flow required during

this day = 23.76 x 106 + 24.89 x 106 (10/24) = 34.13 x 106 l/day


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SOURCE OF WATER

The types of source are usually used for water supply:

1. Surface water

2. Groundwater – Subsurface water

3. Water reclamation – desalination water, re-use water,

demineralisation water

4. Other types – integrated sources

SURFACE WATER

May be derived from:

- River/stream flow

- Lakes or reservoir

- Runoff collection

Are fed directly by rainfall Runoff (Surface run off is a

portion of rainfall that flows on the surface of the ground and

eventually collects in the rivers, streams, the lakes and oceans)

Rainfall

- Rainfall parameters:

Amount

Intensity

Time of beginning and ending of precipitation

Raindrop-size distribution (rare)

- On the basis of the vertical depth of water (mm of hourly,

daily, weekly, monthly or annually) that would accumulate

on a level surface if the precipitation remained where it

fell


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- Precipitation gages

Non-recording type: for daily rainfall (at 7:00 AM)

Recording type: record by a pen trace on a chart,

punched tape recorder or electronic (data logger)

a. Tipping bucket type

b. Weighing type

c. Float recording

(a) (b) (c)


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- Precipitation gage network

Minimum densities of precipitation network: o For flat regions of temperate, Mediterranean and

tropical zone: 600 – 900 km2 per station

o For mountainous regions of temperate, Mediterranean

and tropical zones : 100 – 250 km2 per station

o For small mountainous islands with irregular precipitation

: 25 km2 per station

o For arid and polar zones : 1500 – 10000 km2 per station

- Other instrument:

Radar measurement: use electromagnetic energy

Satellite : for area where gage network are inadequate

or nonexistent, such as over the ocean.

Runoff Estimation

- Rational Method

Predicting a design peak runoff from a catchment (Is

the area where runoff from rainfall on that area flow out

through an outlet )


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Assumption: uniform intensity rainfall for a duration at

least equal to time of concentration (Tc) and for over

the entire area CiAq 0028.0

q : the design peak runoff, m3/s

C : the runoff coefficient

i : rainfall intensity in mm/h for the design return period and

for a duration equal to the time of concentration (Tc) of the

watershed

A : the watershed area, ha 385.077.0

0195.0

gc SLT

Tc : time of concentration, min

L : maximum length of flow, m

Sg : the watershed gradient, m/m, or the difference in

elevation between the outlet and the most remote point

divided by the length (L)


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- Soil Conservation Service Method

The peak flow AQqq u

q : peak runoff rate, m3/s

qu: unit peak flow rate, m3/s per ha/mm of runoff (from graph)

A : watershed area, ha

Q : runoff depth, mm


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5.07.0

8.0)(4407/9100

gc SN

LT

Tc : time of concentration, hrs

L : longest flow length, m

N : runoff curve number

Sg : average watershed gradient, m/m

SI

SIQ

8.0

2.02

254/25400 NS

I : rainfall, mm

S : max. potential difference between rainfall and runoff, mm

- Empirical Formula x

KAq

q : the magnitude of the peak runoff , L3/T

K : a coefficient dependent on various characteristics of the

watershed

A : watershed area, L2

x : a constant for a given location


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- Infiltration indexes (

index)

As the rate of rainfall

above which the rainfall


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volume equals the runoff volume

o Average precipitation over area

- Arithmetically mean Good for flat country and the gages are uniformly distributed

- Thiesen method For nonuniform distribution of gage

By providing a weighting factor for each gage

The stations are plotted on a map, and the effective area are

determined


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- Isohyetal method The most accurate method Use contours of equal precipitation (isohyets)

River/Stream flow

- River/stream flow measurement :

Stage level : The elevation of the water

surface above the arbitrary

zero datum

Use stage level : manual or

recording

Discharge The discharge at a section is derived from point

measurement of velocity

Q = AV


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Q : discharge, A: cross section area and V: flow velocity

Velocity measured by a current meter

o Type of current meter : cup type and propeller type

Cup type propeller type

o Velocity (V) is a function of propeller revolution speed

(N)

bNaV ; a and b : constants

o Velocity may be measured at:

One depth : at the depth of 0.6D for shallow flow

Two depths : at the depth of 0.2D and 0.8D

2/8.02.0 vvvaverage

Three depths : at the depth of 0.2D, 0.6D and 0.8D.

For deep flow

3/8.05.02.0 vvvvaverage

Measurement of wide stream or river

o Divide the stream into 20 to 30 vertical sections. No

section include more than 10 % of the total flow

o Measure the velocity of each section at 1, 2 or 3 depths,

then determine the average velocity of each section

o Measure the area of section

o Calculate the discharge of each section

o Add the increment of discharge


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Rating curve Periodic measurements of flow and simultaneous stage

observation provide data for making a rating curve

such a curve is approximately parabolic

Lake

- A small to moderately inland bodies of water with surfaces

exposed to the atmosphere

- They occupy depressions into the zone of saturation in the

environmental soil and rock

- Derive their water from rain, melted snow, ice, etc.

- Area : less than 250 km2 – larger than 80,000 km2

- Depth : about 30 m – 1700 m

- Classification:

Tectonic lakes : resulting from differential movement of the planetary crust


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usually have elongated shape, steep side and great depth such as: lake baikal, lake albert

Glacial lakes: Derive their water from ice Such as : lake superior

Volcanic lakes: Result when a lava flows dams a valley and impounds water

to form a lake or in cavities caused by subsidence after

subterranean volcanism

Such as: lake chambon, lake chala

Reservoir

o A stream is dammed and water is collected in a lake behind

the dam

o Types of reservoir : - Direct supply


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- Pump storage :

- Regulating reservoir

o Other reservoir classification: - Dugout reservoir fed by ground water

- On-stream reservoir fed by surface runoff, stream or spring

- Off-stream reservoir. It is constructed adjacent stream


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Runoff collection

o Collection of rainfall runoff from the roofs of buildings or

bare catchments and feeding to storage tanks

o three principal components; the catchment area, the

collection device, and the conveyance system


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GROUNDWATER

Groundwater

o Is water contained by an saturated groundwater bearing

formation and which will flow to wells, springs or other

points of recovery

o Often superior in quality to surface water

o Generally less expensive to develop for use

o Usually provide a more certain supply

Types of ground water based on the origin :

1. Connate water : - water that has been out of contact with the

atmosphere for at least an appreciable part

of a geologic period.

- Consists of fossil interstitial water that

migrated from its burial location

- May have been derived from oceanic or

freshwater sources

- Highly mineralized 2. Magmatic water: - water is derived from magma

- if the separation is deep plutonic water


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- if shallow volcanic water 3. Juvenile water : - New water of magmatic or cosmic origin

that has not previously been a part of the

hydrosphere 4. Metamorphic water : - water that is or has been associated

with rock during their metamorphism

Groundwater bearing formation:

o Aquifer: groundwater bearing formation which are sufficiently

permeable to yield usable quantities of water, such as :

unconsilidated sands and gravels o Aquiclude: ground water bearing formation which are not

sufficiently permeable to yield usable quantities of water, such as :

clay o Aquifuge : A relatively impermeable formation neither

containing nor transmiting water, such as : solid granite

o Aquitard : A saturated but poorly permeable stratum that

impedes groundwater movement and does not yield water freely to

wells, such as sandy clay

Type of aquifer:

o Confined : is aquifer which is located between two less permeable

layers, and usually under pressure artesian o Unconfined: is aquifer which is not overlain by an aquiclude. The

top of the saturated zone water table or phreatic surface o Leaky aquifer : where a permeable stratum is overlain by a

semipervious aquitard or semiconfining layer o Idealized aquifer : an aquifer that is assumed to be homogeneous

and isotropic


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Occurrence of aquifer

o The value of aquifers depends upon their ability to provide

usable quantities of water as a function of porosity

and particle size

o Good : sand, gravel and sandstone

Groundwater flow

o Darcy’s law: ksv v : flow velocity, k: coefficient having the unit of v, and s : slope of

hydraulic gradient


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o Discharge : KAskpAsq p : porosity, A : area, pA : effective area, and K : coefficient of

permeability

o Transmissivity, T : the flow rate per day through a section

1 ft wide and the thickness of the aquifer under a unit

head (slope of 1 ft/ft)

KYT Y is the saturated thickness of the aquifer

o Analysis Pumping test

Equilibrium Analysis

Asumption:

Homogenous and infinite aquifer

Initial water table horizontal

Well penetrates aquifer completely

For flow to occur to the well there must be a gradient

toward the well cone of depression

Flow toward the well through a cylindrical surface at radius

x must equal the discharge of the well

dx

dyxyKq 2


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2 xy : the area of the cylinder, dy/dx : the slope of the water

table

Integrating with respect to x from r1 to r2 and y from h1 to

h2 :

)ln(

)(

2

1

22

21

rr

hhKq

h : the height of water table above the base of aquifer at distance

r from the pump well

derived by Dupuit and modified by Thiem

For confined aquifer in which Y is constant, them

)ln(

)(2

2

1

21

rr

hhkYq

Since we have assumed the drawdown Z to be small

compared with the saturated thickness, so:

)(2

)ln(

12

2

1

ZZ

rr

q

T

Example:

A well in an unconfined aquifer is pumped at a rate of 25 l/s. The

thickness of the aquifer is 15 m and the elevation of the phreatic surface

is 12.5 m above the underlying aquiclude at an observation well 20 m away

from the well and 14.6 above at a well 50 m away. What is the value of K

for this aquifer?

)ln(

)(

2

1

22

21

rr

hhKq


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KK

195)

2050ln(

)5.126.14(025.0

22

K = 1.28 10-4 m/s = 11.1 m/day

A well in a confined aquifer with a thickness of 15 m produces a flow of

25 l/s. The height of the phreatic surface is at an elevation of 114.6 m

at an observation well 50 m away and at 112.5 m at an observation well 20

m away. Find K and T for the aquifer and estimate the height of the

phreatic surface at the 0.5 m diameter well

)ln(

)(2

2

1

21

rr

hhkYq

)20

50ln(

)5.1126.114)(15(2025.0

k=216K K=1.16 10-4 m/s = 10.0 m/day

T=KY= 10.0 x 15 = 150 m2/day

At the well

)25.0

20ln(

)5.112)(15)(1016.1(2025.0

4hx

h = 102.5 m

Nonequilibrium analysis

In general, as elapse time (t) of pumping increases, the

drawndown (s) increases at a decreasing rate

Theis method (1935)

Take account the effect of time and storage

characteristics of the aquifer

)(44

uWT

qdu

u

e

T

qZ

u

u

r

Where: Zr is drawndown in an observation well at distance r

from the pumped well (ft), q is the flow (ft3/day), T is

transmissibility (ft3/day.ft) and u is given by:

Tt

Sru c

4

2


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t is the time since pumping began (days) and Sc is the

storage constant of the aquifer (that is the volume of

water removed from a column of aquifer 1 ft2 when the

water table or piezometric surface is lowered 1 ft)

duu

e

u

u

is as well function of u , = W(u)

Solution steps

1. Make W(u) vs u curve (type curve) on

transparence logarithmic paper. Data from the

table above

2. Make Zr vs r2/t curve (data curve) on logarithmic

paper. Data from file observation

3. Superimpose chart (1) and chart (2), determine

the match point


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4. according to the match point, determine the

values of u, W(u), Z and r2/t

5. Substitute the value into eq. )(4

uWT

qZ r

and

Tt

Sru c

4

2

Example:

A 12 in diameter well is pumped at a uniform rate of 1.5 ft3/s while

observations of drawndown are made in a well 100 ft distant.

Values of t and Z as observed and computed values of r2/t are

given below. Find T and Sc for the aquifer and estimate the

drawndown in the observation well at the end of 30 days of pumping

From curve, the match point are :from type curve u = 0.4 and W(u)

= 0.7, and from data curve Z = 3.4 ft and r2/t = 5.3 x 104 ft2/day


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q in ft3/day = 1.5 x 86400 = 129600

T = (qW(u))/(4Z) = (129600x0.7)/(12.56x3.4)=2124 ft2/hr

Sc=(4uT)/(r2/t)=(4x0.4x2124)/(5.3x104)=0.064

After 30 days

u=(r2Sc)/(4Tt)=(10000x0.064)/(4x2124x30)=0.0025

from table W(u)=5.44 so:

Z=(qW(u))/(4T)=(129600x5.44)/(12.57x2124)=26.4 ft

o Modified Theis (Yacob)

Since u small, so 1

2

4

32

t

tlog

Z

q.T

Where Z is the change in drawndown between time t1 and t2.

The drawdown Z is ploted on an arithmetic scale against time t on

logarithmic scale. If Z is taken as the change in drawdown during

one log cycle, log(t2/t1)=1. When Z=0,

2

0252

r

Tt.Sc

to is intercept obtained if straight line portion of the curve is

extended to Z =0

Example:

From the previous example, find T and Sc

Plot the data. The drawdown Z is ploted on an arithmetic scale

and time t on logarithmic scale


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Draw a straight line portion of the curve

Between t=3 hrs and t = 30 hrs, we get Z=11 ft

So T=(2.3x129600)/(12.57x11)=2156 ft2/days

From the curve, to=2.7 hrs = 0.112 days

so Sc=(2.25x2156x0.112)/(10000)=0.0546


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YIELD ESTIMATION AND PREDICTION

SURFACE SOURCES

o Yield definitions :

Yield of experience : the abstraction taken from a sources

over a number of years Historic yield : the steady supply that could just maintained

through a repetition of the worst drought on record

Probability yield : the steady supply that could just

maintained through a drought of specified severity Reliability (safety) or risk

Exp.: once in 100 years (99 % reliability) = 1 % yield for short or

risk

Operational yield : the supply that could be taken into a

waterworks system under a fixed set of operating rules Gross yield : the total available resources

Net yield : the water remaining for supply after any

compensation water or residual flow has been left for other

riparian interest Many countries have adopted probability yield when

planning a new source, and then switch to operational yield

for the final design

In Malaysia, domestic water supply especially in cities are

adequate through a 50 years drought

o River intake yield

Many water supply system abstract water directly from

the river

Only a small proportion of the average flow of river can

often abstracted straight to treatment from a river

intake without affecting other uses

Yield estimation may be obtained by analysis of a

reliable long series of flow measurement at or near the

abstraction point. Usually weekly flow at least 20 years


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If the period of records is short, maybe possible to

extent it by correlation with a longer record of flows

of nearby catchment with similar characteristics

Flow record can be generate from rainfall data by using

programs (HYSIM etc)

This analysis are looking the probability of drought

rather than the probability of flood

Procedure recommended to estimate an intake yield - Locate a flow record of adequate length (20 years or over)

for the river concerned

- Adjust this record back to natural conditions, if necessary;

for example add back exported water

- Locate the lowest weekly flow each year and rank the

resulting series of extreme values from the smallest

upward

- Plot these on an appropriate probability paper. Draw a

straight line

- From the drawn line the flow for varying risk (probability of

failure) can be determined

Total flow in one week

with 2 % risk (1 drought in

50 years) = 30.2 Ml

-


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o Yield from impounding reservoir

Assessing reservoir storage volume

Reservoir is used to collect water in the rainy

season, so that there is adequate supply during the

dry months

To evaluate the reservoir capacity by

computing and plotting the cumulative inflow into

the reservoir

Type of storage:

o Within year storage. When the demands for water

can be satisfied by storing some of the high flow each

year for release during a later period of low flow during

the year o Carry over storage. If there not enough high flow

every year to raise storage to the desire level, extra

water must be stored during wetter years to release

during dry years Approach for determining the size of reservoir

o Mass curve analysis: - Accumulated inflow (Ii) can be plot against time (t)


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- If the total amount of Yi taken out of the reservoir

by a time t in order to satisfy a yield Y (=Yi/T,

where T is the length of the drought) is then

subtracted from this curve

- The depletion in the reservoir will be Yi-Ii

- The maximum value of difference (x) give the size of

reservoir needed

- Some adjustment is necessary if any water is lost by

spillage

- If T=36 months, then Yi = Ii over 36 months, and

Ysafe = Ii/T

o Computer simulation - Depend of type of reservoir :

o Direct supply :

the yield over a given periods is equal to the

amount of entering the reservoir that time

o Pump storage :

The minimum acceptable flow must always be

maintained in the stream decrease the

amount.

The maximum allowable flow (MAF) is a constant

value The total yield = the sum of MAF and

the drinkable supply


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o Regulating reservoir

At the time of low flow in the main stream, water

is released from behind the dam and enter the

main river

The MAF must be maintained at the point of

abstraction


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YIELD ESTIMATION AND FORECASTING -FREQUENCY AND PROBABILITY ANALYSIS-

Frequency and Probability

o Frequency : Different values have occur over the range of

occurrence o Probability : Different values might occur over the range of

occurrence

Concept of probability: NxnxP /)()(

P(x) : relative frequency of event x

n : numbers of event x

N : total amount of events

- 0 ≦ P(x) ≦ 1

- ΣP(xi) = 1

- P(x1∪x2) = P(x1) + P(x2)

- P(x1∩x2) = P(x1) x P(x2)

Probability:

The probability of event that is expected to be equaled or

exceeded, on the average, a particular level

Recurrence interval (Return period, Tr)

- Tr is the time that, on the average, elapses between two

events that equal or exceed a particular level

- )(

1

xPTr

General probability relation :

- The probability that x will be equaled or exceeded in any

year : rT

xP 1)(


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- The probability that x will not be exceeded in any year:

rTxPxP 11)(1)(

- The probability that x will not be equaled or exceeded in

any of n successive years: n

r

n

TxP )11()(

- Risk that x will be equaled or exceeded at least once in n

successive years : n

rTR )11(1

Probability distribution

- Discrete bar diagram

- Continuous histogram

- Cumulative distribution function (CDF) : as the probability

that any outcome in x is less than or equal to a state, limiting value

x )()( xXPxF


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- Some common distributions used

Normal Gumbel

Log-normal Gamma

Weibull Log-Pearson III

PROBABILITY ANALYSIS

- For determining magnitude of the N year event or the

event with P probability

- Such as: rainfall depths and intensities, peak annual

discharge, flood flow, low-flow durations

- Data requirement :

Adequacy : especially the time length of recording.

> 30 years

Accuracy : especially the homogeneity of data

1. Graphical method

- Rank the magnitude in descending order and calculate the

return period (Tr)

- Tr formula:

o Weibull : m

nTr

1

o Gringorten : 440

120

.m

.nTr

o Hazen : 12

2

m

nTr

o Cunnane : 4.0

2.0

m

nTr

Where m is event ranking and n is number of event - Calculate the probability, P, of an N year even of return

period Tr

rT

P 1

- Plot on graphs (log normal probability paper or Gumbel

extreme value paper) the variable and either Tr or P


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- Draw a straight best fit line

- Using this graph, the magnitude of a event can be

determined


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Event with 20 years return periode : 600 unit

Event with 80% probability : 230 unit


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2. Computational determination

a. Normal distribution

Formula: KCXX vt

Xt : magnitude of event with a return period t

Cv : coefficient of variation

K : frequency factor for return period t or probability P

b. Log Pearson Type III Distribution

xKSXX logloglog

K : Coefficient depend on skewness (g)

n/)Xlog(Xlog

2

22

log )log()(log

11

loglogX

n

X

n

n

n

XXS X

3log

3232

3log

3

))(2)(1(

))(log(2))(log)((log(3))(log(

21

loglog

X

X

Snnn

XXXnXn

Snn

XXng


2006 Edition 52

Gumber extreem value (tipe I)

X = X + KS Where X : the magnitude of event

K : coefficient

S : standard deviation


2006 Edition 53

Example

Determine the annual maximum mean daily discharge of the River Thames

at Teddington with a 100 year period using:

a. Normal distribution

b. Log Pearson type III

c. Gumbel extreme value

From the annual maximum mean daily discharge data:

(logQmax)

3 = 1651.799

(logQmax)3 = 18320613.4

a. A From table K for Tr 100 years = 2.33

4.6418.133)33.2(7.329 xKCXX vt m3/s

b.

3log

3232

))(2)(1(

))(log(2))(log)((log(3))(log(

XSnnn

XXXnXng

= -0.066


2006 Edition 54

From table K for g=-0.066 and Tr=100 years is 2.278

862.2165.0)278.2(487.2loglog log xKSXX x

X = 728 m3/s

c. From table K for 100 years record length and Tr = 100 years is 3.35

d. X = X + KS = 329.7+(3.32)x133.8 = 773.9 m3/s


2006 Edition 55

DAM AND RESERVOIR

To maintain water supply when natural sources fail, water

must be collected wherever and whenever there is excess

Storage Reservoir i.e.: Impounding reservoir

Impounding reservoir :

o A stream is dammed and water is collected in a lake behind

the dam

o Types of impounding reservoir : - Direct supply

- Pump storage :


2006 Edition 56

- Regulating reservoir

o Other reservoir classification: - Dugout reservoir fed by ground water

- On-stream reservoir fed by surface runoff, stream or spring

- Off-stream reservoir. It is constructed adjacent stream

Storage capacity of reservoir

Capacity of reservoir on natural sites must usually

determined from topographic survey

The increment of storage between two elevation is

computed by : ZAAVolume )(2

121


2006 Edition 57

Zones of storage in a reservoir

Reservoir capacity for a given yield

o Constant inflow (such as uniform pump rate)

Example:

The water supply for a city is pumped from wells to a

distribution reservoir. The estimated hourly water

requirements for the maximum day are as in table. The

pumps are to operate at a uniform rate. The capacity

calculation is as follows


2006 Edition 58


2006 Edition 59

Selection of capacity for a river reservoir

o May be performed with annual, monthly or daily time

interval. Monthly data are most commonly used

o Methods:

Sequent peak algorithm - Calculate the cumulative sum of inflow minus withdrawals

(including average evaporation and seepage)

- Identify the first peak and the sequent peak (next following

peak that is greater than the first

- The required storage for the interval is the difference

between initial peak and the lowest through in the interval

Mass curve (Rippl diagram) - Plot the cumulative net reservoir inflow. The slope of the

mass curve at any time is a measure of the inflow at that

time

- Demands curve representing a uniform rate of demand are

straight lines

- Draw tangent lines parallel to the demand curve at the high

points of the mass curve (at A , B) represent rate of

withdrawal


2006 Edition 60

- The maximum departure between the demand line and the

mass curve represents the reservoir capacity required to

satisfy the demand

Reservoir reliability

o Is as the probability that it will deliver the expected

demand throughout its lifetime (usually 50 to 100 years)

without incurring a deficiency

o Can be use Gumbel extreme value probability graph to plot

the probability


2006 Edition 61

Reservoir sedimentation

o The ultimate destiny of all reservoir is to be filled with

sediment

o Important data :

specific weight of the settled sediments

)log(100

%

)log(100

%)log(

100

%

33

2211

TBWclay

TBWsilt

TBWsand

W

W: the specific weight (dry) of a deposit with an age of T years

The percent of sand, silt and clay is on weight basis

W1,W2 and W3 : represent the specific weight of sand, silt and

clay

B1,B2 and B3 : constants


2006 Edition 62

the percentage of entering sediment that is deposited

in the reservoir/retained in reservoir (trap efficiency)

= a function of the ratio of reservoir capacity to

total inflow

The trap efficiency decrease with age

Reservoir sedimentation cannot be prevented, but can

be control by reduction in sediment inflow (through soil

conservation)

Reservoir site selection

General rule for choice of reservoir sites:

A suitable dam site must exist. The cost of dam is

often controlling factor

The cost of real estate for the reservoir (road,

railroad, dwelling location, etc.) must not be excessive

Must have adequate capacity

A deep reservoir is preferable than a shallow one

Tributary area that are unusually productive of

sediment should be avoided

The quality of the stored water must be satisfactory

No or less environmental impact


2006 Edition 63

DAMS

Type of dams:

1. Gravity dam

2. Arch dam

3. Buttress dam

4. Embankment dam

Gravity dam

o Depends on its own weight for stability

o Usually straight in plan, sometime slightly curve

o Force on gravity dam :

- Weight of dam (W)


2006 Edition 64

- Hydrostatic pressure (Hh and Hv)

- Uplift (U)

- Ice pressure

- Earthquake forces

(Ew)

2

2h

Hh

thh

U2

21

t is base thickness of the

dam, h1 and h2 are water

depths at the heel

(upstream face) and toe

(downstream face)

o Analysis of gravity dam

- To analysis the stability of dam against overturning at

toe and against sliding

- Factor of safety against overturning is about 2 and

against sliding is 1 – 1.5

- Top width dams vary from about 0.15 times the height


2006 Edition 65

Arch dam

o Arch dams transmit most horizontal thrust of the water

behind them to the abutments by arch action

o Have thinner cross sections than comparable gravity dams

o Use only in narrow canyons where the walls are capable of

withstanding the thrust

o Type of arch dams : constant center and variable center

o Simplified force 2/sin2 rhHh

This force balance by Ry=2Rsin/2.

since Fy=0

2/sin22/sin2 hrR or hrR

If t << r, so =R/t, then whrt /

Where w is the allowable working


2006 Edition 66

stress for concrete in compression

This indicates that the thickness of the ribs should

increase linearly with distance below the water surface

Buttress dams

o Consists of sloping membrane that transmit the water load

to a series of buttresses at right angles to the axins of

the dam

o Buttress type: flat slab and multiple arch

o Require only 1/3 to ½ as much concrete as gravity dams of

similar height

o Power house or water treatment plants can be placed

between the buttresses


2006 Edition 67

Embankment Dams

o Consists of earth-fill and rock-fill types

o Utilize natural materials with a minimum of processing

o May be built with primitive equipment

o Adapted to earth foundations

o Earth-fill embankment dams

Types of earth-fill embankment dams:

Simple type. It is constructed of relatively homogenous

soil material and either is keyed into a impervious

foundation stratum

Core or Zone type. A central section (core) of highly

impermeable soil is placed within the dam


2006 Edition 68

Diaphragm type. It uses a thin wall of plastic, butyl,

concrete, steel or wood to form a barrier against seepage

through the fill

General requirements for earth embankment 1. The topographic condition at dam site must allow economical

construction

2. Soil material must be available to provide a stable,

impervious fill

3. Must have adequate mechanical and flood spillway

4. Large storage embankment should be equipped with a

bottom drain pipe for maintenance and fish management

5. Should be provided by safety equipment

6. Provided by maintenance program

Height of dam - Is the distance from the foundation to the water surface in

the reservoir when the spillway is discharging, plus

freeboard

- Gross freeboard is the distance between the crest of the

mechanical spillway and the top of the dam

- Net freeboard is the distance between the maximum

designed high water level or flood peak and the top of

settled dam

- Net freeboard should sufficient to prevent wave

- Wave height for moderate reservoir : 2/1)(014.0 fDh

H : height of wave, m and Df : length of exposure, m.


2006 Edition 69

Top width - Should be sufficient to keep phreatic line when the

reservoir full

- For dams height 3.5 m, minimum top width is 2.4 m

- For dam height > 3.5 m, 14.0 HW

W : top width in m, H : maximum height of dam in m

Side slope - Depend on the height, the shearing resistance of the

foundation and the duration of inundation

- On structure < 15 m height with average material, side slope

3 : 1 on the upstream face and 2 : 1 on the downstream

face

- Coarse, uncompactable soil, side slope 3 : 1 or 4 : 1

- For structure > 15 m height complete analysis

Seepage - No earth fill dam can be considered impervious seepage

- If rate of seepage height enough soil particles are

moved called piping dam failure


2006 Edition 70

- Seepage through earth fill dam reduced by:

o Placing impervious blanket on upstream face

o Using a clay core or diaphragm

o Placing drain blanket and filters at the toe

- Seepage through permeable foundation reduced by

extending blanket to upstream

- Can be analyzed by flownet

Total head lost: H=H x no. of equipotential drops

=HxNd

Total seepage flow: q=q x no. of flow intervals=qxNf

Darcy: total flow, q = kHNf =d

f

N

NkH

Slope stability - Can use method of slices

- By comparing the moment tending to rotate the soil (M) with

the resisting moment (Mr) WxM

W : weight and x : moment arm

LrsM sr

Ss : shear strength, L: length of the failure and r : radius

of failure

Ss = c + tan

c: soil cohesion, : effective pressure and : internal

friction angle


2006 Edition 71

wuL

W

cos

uw : pore pressure


2006 Edition 72

o Rock-Fill dams

Have characteristics midway between gravity dams and earth fill

dam

Rock is as the main structural element

Two types: impervious face and impervious earth core

introduction - ipb university

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