rate of water consumption: water distribution systems (wds)

Section III Water Demand Estimation

Rate of Water Consumption:

Water Distribution Systems (WDS)

The objective of WDS is to deliver water to individual consumers with

appropriate quality, quantity and pressure.

The distribution system describes collectively the facilities used to supply

water from its source to the point of usage.

This may include extensive system of pipes, storage reservoirs, pumps and

related appurtenances.

The proper functioning of a water distribution system is critical to providing

sufficient drinking water to consumers as well as providing sufficient water

for fire protection

Average day demand. The total annual quantity of water production for an

agency or municipality divided by 365.Used for pumping system design

Example:

Assume that the municipality has a population of 22,570. It was determined from

water meter readings that the average daily consumption was 137 gallons per person

per day. Calculate the required water to be pumped to the system??

22,570 × 137 gallons = 3,092,090 gallons =3.1 million gallons

( This means that on an average day, the water supply works need to have

available nearly 3.1 million gallons of water over a 24-hour period)

3,092,090 / (24*60) = 2,147 gpm (gallon per minute) of finished water delivered

into the water supply distribution system

Maximum day demand. The highest water demand of the year during any 24h

period. (it is around 160%-190% 0f the ADD).

It is used in the design of ground storage tanks.

137* 1.8 =246.6 gpd

Maximum monthly demand = 1.2 : 1.6 of Qav , and used in plants design

Peak hour demand. The highest water demand of the year during any 1-h

period.

Usually equals to 2.5-3.5Qav, it is used to design the network and high storage

tanks.


There are generally two peak periods in the day when consumption is greatest:

between (7-9am) .and between (6-8 pm)?? Why it is needed?

Minimum day demand = 0.7 Qaverage

Max daily demand – demand on the day of the year that uses the most water = 1.8 x

Avg daily demand

Max hourly demand – the demand during the hour that uses the most water = 3.25

x avg daily demand (23/10)

Peaking factors. The increase above average annual demand, experienced

during a specified time period. Peaking factors are customarily used as

multipliers of average day demand to express maximum day and peak hour

demands.

Distribution pipeline or main. A smaller diameter water distribution pipeline

that serves a relatively small area.

Water services to individual consumers are normally placed on distribution

pipelines.

Distribution system pipelines are normally between 150 and 400 mm (6-16

in.).

Transmission pipeline or main. A larger-diameter pipeline, designed to

transport larger quantities of water during peak demand-periods.

Water services for small individual consumers are normally not placed on

transmission pipelines.

Transmission mains are normally pipelines larger than 400 mm (16 in.).

50L/C/d is the minimum

recommended daily water

demand

The water is

pumped direct

without tank at roof


Capacity Requirements - Water Demand

Demand vary throughout the 24hr period and can range

- from 25-40% of the average daily demand between 12.00 to 6.00am

- to as high as 150 to 175% during morning and evening peak

Demand also vary from year to year; season to season; day to day and house to house

Fire Water Demand:(24/10)

Generally, the amount of water supply for fire fighting is low but the annual average

consumption is high.

It depend on: population number, construction type (single floor, building, towers),

floor area, connection to other construction and occupancy of a building.

- Demand based on number of population

Q= 𝑝× (1 − 0.01 𝑝)…..for community less than 200,000 people

Q: water demand in gpm

P: population in thousands

- Demand based on building type

Q= 6.8𝐶 × 10.76 ∗ 𝐴

Q: fire fighting water (l/min)

1US gallon = 3.78L

Make question to

converse 2gpm to

m3/day??


A: Total floors Area excluding the basement in m2

C: construction type Coefficient ( for wood C= 1.5, C= 1 for ordinary reinforced

construction, C= 0.6 for fire resistant construction)

Example:

determine the amount of water required for fire fighting that extended to 3 hours in a

concrete building contains five stories, each with an area of 1500 m2, if the building is

located in residential area with a population about 3000 capita and average water

consumption (Qav = 670lcd) and find the maximum water consumption at that day?

Daily Qav = 670 * 3000 = 2010 000 l/day

Daily Qmax = 1.8 * 2010× 103 = 3,618 × 103 l/day

Fire fighting water demand = 6.8*1* 10.76 ∗ 5 ∗ 1500 = 1931.7 l/min

= 1931.7 *3*60 = 347700 l

Maximum consumption = 3,618 × 103 + 347700 = 3,965.7 × 103 l

Example:

if the maximum capacity of the current wastewater treatment plant is 15,000 m3/day,

and the current population is 40,000 and the maximum monthly consumption is 5,000

m3/day. If the expected population after 25 years is 100,000 capita .find when the city

council needs to construct new plant assuming the population increase is arithmetic?

- First determine the maximum daily consumption (cmcd)

Maximum monthly consumption =5000

40,𝑜𝑜𝑜= 0.125 m3/capita. Day

- Determine the maximum population can be served by the plant

15000 m3/day = Qmax× 𝑝𝑜𝑝𝑓𝑢𝑡𝑢𝑟𝑒 popfuture= 15000/0.125 = 120,000 capita

- Determine the arithmetic increase rate

Ka = 100,000 – 40,000

25 = 2400 capita/ year

Thus

Popfuture =Popcurrent + r* n

120,000 = 40,000 + 2400 * n n = 33.33 years


Water Distribution methods:

consumer the following may be used to transport water to consumers with adequate

pressure:

Gravity – when the source is at a sufficient elevation above the consumer to

produce the desired pressure. Highly economical

Pumping – Pumps are used to develop the necessary head (pressure) to

distribute to consumers and storage reservoirs

Pump-Storage System- storage reservoirs are used to maintain adequate

pressure during periods of high demand and emergency (fires & power

failures). During low consumption, water is pumped and stored in the storage

reservoir.

Water Distribution system (WDS) consists of

1- Storage and distribution reservoir

2- Pumping station and boosters

3- Pipes network

4- Valve and fittings

Purposes of WDS :

1- Distribute water for consumption all over the served city in required

quantities, pressure and quality.

2- Save water for fire fighting at high rates

Basic requirements for WDS

1- Use pipe that made from durable materials

2- Use more than one feed source to guarantee flow continuity

3- Avoid the dead zone( points at the end of the pipes where water stops, and

chlore dosage finish, thus the environment becomes suitable for bacteria and

microorganisms grow) which result in reduction the water quality and network

pollution.

4- Control and measure water flow using the valve and meters.

5- Protect the water from pollution

6- Optimum pump and capacity to meet the maximum demands


Storage and distribution reservoirs: (26/10)

it is the first part of the drinking water reservoirs and mainly includes the

a- Ground Reservoir: it is usually exist in water treatment plants and used to

pump water from it by H.L.P( high lifting pumps). It used for:

- Storage for disinfection (Vdisinfection) with retention time (𝜏 = 0.5-1hr)

- storage for emergency (Vemergency) with retention time (𝜏 = 4-10hr)

- Storage for difference between maximum daily and maximum monthly

consumption (Vdiff)

- Storage for fire demand (80% 0f the fire demand)

Fire demand (Qfire)=60m2/2hr per 10,000 capita

Thus theVfire = 0.8 * Qfire = 0.8* 120 * population in (10,000) [m3]

Thus the Tank Volume = max of (Vdisinfection;Vemergency; Vdiff) + Vfire

Recommended dimension

l = (1.2 – 1.5) b

l ≤ 50 𝑚; it is recommended to be divided by 5 for structural consideration

d = 3-5 m

Area ≤ 2000 m2

n≥ 2


Example:

For a water treatment plant (WTP) determine the required number of ground tanks if

the plant serves 300,000 capita with average summer water consumption = 420 lcd

Solution:

Maximum (monthly) daily demand = 420𝑙

1000 𝑙/𝑚3 * 300,000 = 126,000 m3/day

Average daily demand = 420𝑙

1.4 ×1000 𝑙/𝑚3 * 300,000 = 90,000 m3/day

Maximum daily demand = 90,000 * 1.8 = 162,000 m3/day

- Thus the required tank volume for disinfection 𝜏 = 1hr

Vtank= Qm.m*𝜏 126,000 ×1ℎ𝑟

24ℎ𝑟/𝑑𝑎𝑦= 5,250 m3

- The required tank volume for emergence case 𝜏 = 4-10hr

Vtank= Qm.m *𝜏 126,000 ×6ℎ𝑟

24ℎ𝑟 /𝑑𝑎𝑦 = 31,500 m3

- The required storage tank volume for difference between maximum daily

demand and maximum monthly demand

Vdiff = 162,000 – 126,000 = 36,000 m3

- The required tank volume for fire demand

Qfire = 120∗300,000

10,000= 3600 m3

Vfire = 0.8 ∗ 120∗300,000

10,000 = 2880 m3

Finally the required tank volume = max of (Vdisinfection;Vemergency; Vdiff) + Vfire

= 36,000 + 2,880 = 38,880 m3

Let the water depth inside the tank = 5 m

Thus the required area = 38,880 / 5 = 7776 m2

assuming 4 tanks each with area of 2,000 m2 (and dimension 50 × 40 m)


Water stop


b- Elevated Tanks:

the idea of the tank is to preserve a certain amount of water in the network as

the following:

- When water consumption is less than water pumping ( at night or times rather

than peak consumption) the inlet valve is opened and allows water to be stored

inside the tank.

- At peak times when water consumption is higher than water pumping the

outlet valve is opened to compensate the water deficit

- The tank is provided by float valve to keep the water level

HW: how water stop

installation and work

principles?


- The tank is provided with overflow drainage for emergency

it is raised tanks filled for various purposes

1- For quantity (discharge)

- Cover fluctuation in water consumption through a day

- Cover difference between max hourly consumption and max daily production

- Compensate for 20% of fire demand

2- For pressure

- Fix head on H.L.P so it can work at max efficiency

- Enhance water pressure in the network

- Protect the network against water hammer.


Water hammer:

it is a phenomena in a fluid when a valve or a raising pump is suddenly closed or

stopped working. Creating a sort of shock wave and a hammering noise. Thus a

reverse wave of a fluid is accrued that may explode the pipes.

Figure 1 how to accommodate the water hummer

The elevation tanks may be suited at (location)

1- Just after the H.L.P : this location raise the water pressure in the network and

protect from water hummer

2- At mid of the city: same as (1)

3- At the end of the city: it enhances the water pressure , enhances for future

expansions, but dose not protect from water hummer.

The best system is a combination of the three locations

Design of elevation tank:

- Tank Capacity

1- From cumulative curve (total mass curve)

Vtank =1.75 ×𝑎+𝑏 (𝑙𝑐𝑑 )

1000× 𝑝𝑜𝑝 + 0.2 𝑓𝑖𝑟𝑒𝑑𝑒𝑚𝑎𝑛𝑑

Where

a& b are max. Vertical distance below and above pumping line


2- From difference between max hourly and max daily discharge for 1 hr

Vtank = (Qmax. daily - Qmax. monthly ) (m3/hr) × 1ℎ𝑟+ 0.2 𝑓𝑖𝑟𝑒𝑑𝑒𝑚𝑎𝑛𝑑

Tank dimension:

d = 6-10 m

ϕo≤ 35𝑚(over alldiameter)ϕi = 2-3m for pipes (inside diameter for services)

n≥ 2

volume of tanks = n×π

4(φ

02 - φ

12) .d

where n: number of tanks

d: water depth in the tank


Cumulative mass :

assuming a community with the following consumption data

time

water consumption (

m3/min)

cumulative consumption

(m3)

12am - 1am 7 7*60 = 420

1am-2am 7.2 432+420 = 852

2am-3am 7.2 852+432 = 1284

3am-4am 7.5

10 pm-11pm

11pm-12

midnight

Q max daily

However, the pumping may work for 24 hr... thus the pumping rate = Q max daily

24

Or it works for 6 hrs only:

thus the pumping rate = Q max daily

6 in this case the pumping rate should increased but

there is a demand for a tank.

Example:

For a city of population 200,000 capita and average water consumption (WC)

=240lcdFind the required elevated tank volume if the daily WC as follows:

T(2hr) 12-

2am

2-4 4-6 6-8 8-10 10-12 12-2 2-4 4-6 6-8 8-10 10-

12

WC 3 7 10 20 30 40 40 30 30 15 10 5

Solution:

If Water pumping (WP) is for 24 hr:

Total pumped demand = 240 / 24 hr = 10 m3/hour .since the table is represented by

2hr step then each step = 20 m3

T(2hr) 12-

2am

2-4 4-6 6-8 8-10 10-12 12-2 2-4 4-6 6-8 8-10 10-

12

WC 3 7 10 20 30 40 40 30 30 15 10 5

Accum 3 10 20 40 70 110 150 180 210 225 235 240

Accum.pump 20 40 60 80 100 120 140 160 180 200 220 240

Pump-accum 17 30 40 40 30 10 -10 -20 -30 -25 -15 0


From the chart or table (amax= 40; bmax= 30)

Vtank =1.75 ×40+30 (𝑙𝑐𝑑 )

1000× 200,000 + 0.2

200,000

10,000× 120 = 24,500 + 480 =

24,980 𝑚3

Thus tank dimension and design:

Assume water depth in the tank = 8m Surface area (S.A) = 24,980 / 8 = 3,122.5

m2

Assume: ϕo= 35𝑚 (over all diametr) ϕi = 2m


4(φ

02 - φ

12) .d

Thus number of tanks = 24,980

π

4 φ0

2−φ12 ×water depth

n= 3.25 tank

T(2hr) 12-2am 2-4 4-6 6-8 8-10 10-12 12-2 2-4 4-6 6-8 8-10 10-12

WC 3 7 10 20 30 40 40 30 30 15 10 5

Accum 3 10 20 40 70 110 150 180 210 225 235 240

pump 20 40 60 80 100 120 140 160 180 200 220 240

Pump-accum 17 30 40 40 30 10 -10 -20 -30 -25 -15 0

For 16 hr water pump:

0

50

100

150

200

250

300

1 2 3 4 5 6 7 8 9 10 11 12

accumulative water consumption

water pump

a max

bmax


T(2hr) 1 2 3 4 5 6 7 8 9 10 11 12

WC 3 7 10 20 30 40 40 30 30 15 10 5

Accum 3 10 20 40 70 110 150 180 210 225 235 240

pump 30 60 90 120 150 180 210 240 No pump

Pump-accm 27 50 70 80 80 70 60 60 30 15 5 0

From the chart or table (amax= 80; bmax= 60)

Tank volume =

Vtank =1.75 ×80+60 (𝑙𝑐𝑑 )

1000× 200,000 + 0.2

200,000

10,000× 120 = 49,000 + 480 =

49,480 𝑚3

Assume water depth in the tank = 8m Surface area (S.A) = 49,480 / 8 = 6,185 m2

Assume: ϕo= 35𝑚 (over all diametr) ϕi = 2m


4(φ0

2 - φ12) .d

Thus number of tanks = 24,980

π

4 φ0

2−φ12 ×water depth

n= 6.45 tanks

0

50

100

150

200

250

300

1 2 3 4 5 6 7 8 9 10 11 12 13

water consumption

water pump


System Losses:

Leaking and overflow from reservoir

Leaking from main and service pipelines

Leaking and losses on premises

Leaking from public taps

In a well maintained system losses are about 20%. Partially maintained systems are

about 50%

In Amman the water loss was about 65% and now is it around 30%

In Madaba in 2014 water loss was 64% , after Miyahuna management the

losses became 34%

Un-collected

bills due to

falsification

address ....

rate of water consumption: water distribution systems (wds)

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