unit ii 2 environmental engineering i

7/29/2019 Unit II 2 Environmental Engineering I

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Unit II-2

Environmental Engineering-I

Water Demand


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Story so far

Water Supply Sources

Quality Criteria

Water Demand

Water Treatment

Conveyance


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Water Demand

Demands

Domestic

Industrial (automobile, distillery, textiles, etc.)

Institution & commercial (school, hospital, etc.) Public use (road cleaning, gardening, etc.)

Fire Demand

Loss & theft compensation


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Domestic demandMinimum domestic water consumption-India (IS 1172)

Use Consumption-Town & city

In l/h/d (liter/head/day)

Consumption-smalltown & city

Drinking 5 5

Cooking 5 5

Bathing 75 55

Cloth washing 25 20

Utensil washing 15 10

House cleaning 15 10

Gardening/lawns 15 -

WC flushing 45 30

200 l/h/d 135 l/h/d


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Domestic demand

Average demand: 135-225 l/h/d.

Demand varies with level of development &industrialization

Demand higher in metropolitan cities.

USA: ~340 l/h/d (& other developed &industrialized nations)

Air-conditional, cooling, heating , carwashing,


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Industrial demand

For specific industries

Automobiles: 40 klt./vehicle

Distillery: 122-170 klt./klt.

Fertilizer: 80-200 klt./tonne

Paper: 200-400 klt./tonne

Steel: 200-250 klt./tonne

Textile: 80-140 klt./tonne

~50 l/p/d as average assumption (no provision for large

industries)


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Institution & commercial demand

Office: 45-90 l/h/d

Factories: 45-90 l/h/d

School

Day scholar: 45-90 l/h/d Residential: 135-225 l/h/d

Hotels: 180/bed

Hospitals

< 100 bed: 340/bed > 100 bed: 450/bed

Airport: 70 l/h/d

~20-50 l/p/d depending on level of commercialisation


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Public Use

Watering of public parks, gardening, washing& sprinkling on roads, public fountains, etc.

Normally 10 l/h/d

Not necessary for a developing country

Gen.,


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Fire demand

Short duration, high discharge demand, lowper head demand.

Fire hydrants: 100-150 m spacing, 10-15 m

water head


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Fire demand: example

3 water streams of 1000 l/min for 6 fires/day,each of 3 hour duration in a town of 60 lakh.

Total water required 6x3x3x1000x60 l/day= 32,40,000 l/day

Per capita demand= 32,40,000/60,00,000 l/p/d

~


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Fire demand calculations

Kuichlings formula: Q=3182 P0.5

Q: water (l/min), P: population in 1000s

Freeman Formula: Q= 1136 (0.1P+10)

Bustons Formula: Q= 5663 P0.5

National Board of Fire Underwriters Formula

Insurance Companies

Q= 4637 P0.5(1-0.01 P0.5), Population< 2 lakh Q= 54,600 l/min, population > 2 lakh


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Loss & Theft compensation

Bad plumbing, damaged meters,unauthorized connections, wastes &leakages.

Could be up to 15% of total consumption.


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Factors affecting losses & thefts

Water tight joints: better joints~ less loss

Pressure in distribution network

Higher pressure~ more loss

System of supply

Intermittent vs. continuous

Metering

Discourages losses Unauthorized connections

Increases losses/theft.


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Per capita total demand

Includes all the possible uses: domestic,industrial,.

India: Domestic (200) + Industrial (50) +

commercial (20)+ Civil (10) + Losses (55) =335 l/h/d

Annual average amount of daily waterrequired per person.

Per capita demand (l/p/d) =

= total yearly requirements of city (liters)/ (365 xpopulation)


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Factors affecting per capita daily

demand

1. City size: more for larger cities2. Climatic conditions: more for hot climate3. Gentry type & peoples habits: rich-more, poor-less4. Industrial & commercial activities:

5. Quality of supply: more for better quality.6. Pressure in distribution system: more for high

pressure.7. Sewerage facility8. System of supply: intermittent vs. continuous9. Water cost: free vs. cheap vs. expensive10. Policy of metering & method of charging: reading vs

fixed


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Variations in demand

Seasonal Diurnal hourly

Max. daily consumption = 1.8 x avg. daily demand

Max. hourly consumption = 1.5 x avg. hourly demand Max hr. consumption of max day= peak demand =

1.8 x 1.5 x avg. demand = 2.7 x avg. demand


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Design capacities of supply scheme

Supply Source, Pipe Mains

~ Maximum daily consumption

Filters, Pump

~Maximum daily consumption + additional capacity for

break-down/repairs ~2x avg. demand (instead of 1.8x)

Distribution system

~Max. hourly consumption for max. day

Service reservoir ~for hourly fluctuations, fire demands, emergency

reserve,..

~daily consumption. (unless pumping is done< 24 hrs)


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Design supply or total city supply

= per capita demand x (design) population

Design population

Projected population over design period.

~ Future population for project life.

Population forecasting methods


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Design Period

Supply scheme: costly & complex to install, tough tomake changes in short term.

Designed for a long life.

Too short: not too useful for future needs.

Too long: unnecessary financial burden.

Factors:

Ease of expansion/modification, useful life ofcomponents, economics (additional investment,loan/interest rates), population growth rate.


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Design periods

Dam storage: 50 yrs

Pump House: 30 yrs

Water treatment unit: 15 yrs

Raw/clean water conveyance network: 30 yrs

Clear water/storage reservoirs: 15 yrs


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Population Forecasting

Census (every 10 yrs for India)

~1.241 billion (2011 WB estimate)

Variations in population

Birth

Death

Migration (emigration & immigration) Diseases/natural hazards


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Projection/Forecasting

A statement of the most likely future.


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Population growth curve


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Population forecasting

1. Arithmetic Increase

2. Geometric Increase

3. Incremental Increase

4. Decreasing rate

5. Simple graphical

6. Graphical comparison

7. Master Plan

8. Logistic Curve

NUMERICALS


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1. Arithmetic Increase

Assumption: Population increase is constant.

Difference between successive terms issame.: Arithmetic mean (m)

A1 = A0 + m A2 = A1 + m = A0 + 2m

A3 = A2 + m = A1 + 2m = A0 + 3m ..

An = A0+ n.m

m: average increase (or decrease) per unit time(gen., decades)


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1. Example

Year Population

1950 25000

1960 30000

1970 330001980 37000

1990 40000

Population of 2000, 2010, 2020 ????


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1. How to solve?

1. Find increase in population in each decade

2. Find average increase (m)

3. Use An = A0+ n.m

where, n is the time duration (generally,number of decades)


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1. Example

Year Population

1950 25000

1960 30000

1970 33000

1980 37000

1990 40000

Population of 2000, 2010, 2020 ????

m = 3,750/decade

A2000= 43,750, A2010= 47,500, A2020= 51,125


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2. Geometric Increase

Rate of increase is constant.

Ratio between successive terms is same: GeometricMean (r)

A1 = A0 + A0 x r = A0 (1+r)

A2 = A1 (1+r) = A0 (1+r)2

A3 = A2 (1+r) = A0 (1+r)3 ..

An = A0 x (1+r)n

r is a fraction here. If r is given as percentage:

An = A0 x (1+(r/100))n


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2. Example

Year Population

1950 25000

1960 30000

1970 330001980 37000

1990 40000

Population of 2000, 2010, 2020 ????


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2. How to solve?

Find the rate of increase for each decade.

Find average rate of increase (r)

r = (r1 x r2 x rn)1/n

Use:

An = A0 x (1+r)n

An = A0 x (1+(r/100))n


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2. Example

Year Population

1950 25000

1960 30000

1970 33000

1980 37000

1990 40000

Population of 2000, 2010, 2020 ????

r = 11.83%/decade

A2000 = 44,732; A2010= 50,024; A2020= 55942


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3. Incremental Increase

Growth rate is either increasingordecreasing.

Uses m (arithmetic mean) and n(incremental increase).

n: 1 for decade 1, 2 for 2nddecade,.

A1 = A0 + m + n

A2 = A1 + m + 2n = A0 + 2m + 3n = A0 + 2m+ 2.(2+1)n/2, .

Ai = A0 + i.m + i.(i+1)n/2


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4. Example

Year Population

1950 25000

1960 30000

1970 330001980 37000

1990 40000

Population of 2000, 2010, 2020 ????


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3. How to solve Incremental increase

1. Calculate the increase for each decade.

2. Calculate the incremental increase for eachdecade (careful: it could be


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3. Example

Year Population

1950 25000

1960 30000

1970 33000

1980 37000

1990 40000

Population of 2000, 2010, 2020 ????

m = 3,750/decade; n = -667/decade

A2000 = 43,083; A2010= 45,499; A2020= 47,248


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4. Decreasing rate

Saturation kinetics: Growth rate decreaseswith time.

Used for cases when growth rate is

decreasing.

Decrease in % increase is calculated.

Subtracted from the latest % increase for eachdecade.


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4. Example

Year Population

1950 25000

1960 30000

1970 330001980 37000

1990 40000

Population of 2000, 2010, 2020 ????


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4. Example

Year Population

1950 25000

1960 30000

1970 33000

1980 37000

1990 40000

Population of 2000, 2010, 2020 ????

r = 11.83%/decade, (20, 10, 12.1, 8.1; 10, -2.1, 4),rate~3.97% (arithematic mean)

A2000 = 41,652; A2010= 41,719; A2020= 40,130

A2000 = A1990 + A1990 x (8.1-3.97)/100


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5. Simple Graphical Method

Plot a graph using givenpopulation data

Extrapolate the graph.

40000

60000

80000

100000

120000

1940 1960 1980 2000 2020 2040 2060

Year

Population


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Last 5 methods

Simples and less time-intensive

Assumes that past conditions will continue infuture as well.

May not be most accurate


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6. Graphical Comparison Method

Comparison with similar cities, at similarpopulation level.

Population data for other cities is given.

Population curves of other cities is plotted &for subject city is extrapolated.


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7. Master Plan Method

Use of master plan for forecasting

More precise: citys growth is restricted.

An accurate forecast is already planned in the

design.


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8. Logistic Curve Method

Using the equation ofPopulation growth

curve.

Saturation kinetics..

Complex/messyequations

s

s

e

s

PKn

P

PPm

ntm

PP

.

)(log1

0

0

1

Where,

m, n and K are constants, P0 is population at start of curve, Ps issaturation population, P is population at any time t from start


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Logistic Curve

)(

)(log

1

)(2

01

10

1

0

0

2

120

20

2

1210

PPP

PPP

tn

P

PPm

PPP

PPPPPPP

s

s

e

s

s

Using only 3 population data fortime t0, t1 and t2 (=2t1),parameters are calculated for thelogistic equation.


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Numerical: Logistic Curve

Q: Assuming Delhis population grew from 80,00,000 in 1990 to

98,00,000 in 2000 and further to 1,12,00,000 in 2010, determineits population in 2030 and 2050 using Logistic Curve method.Also determine the year when the Saturation population will bereached.

Solution:

Using the equations given previously, Ps= 1,36,34,783

m = 0.704 and n = -0.0588

P2030=1, 27,78,547 (t = 40 yrs) and P2050=1,33,58,634 (t = 60 yrs)

For Saturation population time, use P = 1,36,34,783 (slightly less than Ps).

Ts = 254.74 yrs, i.e. saturation population will be reached by about 2244.74 A.D.


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Now to

Water Treatment

unit ii 2 environmental engineering i

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