Problems and solutions.pptx

7/29/2019 Problems and solutions.pptx

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Dsgn.pop

ulation

Per capita

demand

(lpcd)

Average

water

demand

(m3/d)

Max. daily

dema

(m3/d)

nd

Peak

hourly

demand

(m3/d)

Chx for fire

demand

Fire demand

(m3/d)

Dsn flow

(water

distributi

on

system)

22000 600 6 storey

building

with

a)Ordinary

constructio

n b)wood

constructio

n (Each

floor

area=1000

m2. )

55000 170

120000 610

Q.1 Estimate the water requirement for the following communities

Numerical 2

(Water Demand)


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Design

population

Per

capita

demand(lpcd)

Average

water

demand(m3/d)

Max.

daily

demand(m3/d)

Peak

hourly

demand(m3/d)

Chx for fire

demand

Max

daily+

Firedemand

(m3/d)

ordinary

Max

daily+

Firedeman

d

(m3/d)

wood

22000 600 13200 19800 297006 storey

building with

a)Ordinary

b)wood

construction

(Each floorarea=1000

m2)

30172.47 35358.7

55000 170 9350 14025 21037.5 24397.47 29583.7

120000 610

73200 109800 164700 120172.5 125358.7

Q.1Estimate the water requirement for the following communities

Fire demand =17287.45 l/min=10372.5 m3/day(ordinary)

=25931.17l/min= 15558.7 m3/day (wood)


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Numerical 3

(Water demand)

3 storey wooden-frame building has each floor area

a)900 m2

b)700m2

c)400m2

Determine fire flow and total daily amount of fire flow for

maximum duration of fire flow .


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Solution problem 3

floor area

per

storey(m2) floors

total

area(m2)

fire demand

wood(litres/

min)

daily fire

demand(m3

/day)

900 3 2700 17395.16 10437.1

700 3 2100 15341.09 9204.653

400 3 1200 11596.77 6958.064


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Predict the population for the years 1981, 1991,1994, and 2001 from the following census figures of

a town by different methods.

Year Population:

(thousands)

Arithmatic

Increment per year

Geometric Increment per

year

1901 60 - -

1911 65

1921 63

1931 72

1941 791951 89

1961 97

1971 120

Averages -

Numerical 4

(Population Forecasting)


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Problem: Predict the population for the years 1981,

1991, 1994, and 2001 from the following census

figures of a town by different methods.Year Population:

(thousands)

Arithmatic

Increment per year

geometric Increment per

year

1901 60 - -

1911 65 0.5 0.008004

1921 63 -0.2 -0.00313

1931 72 0.9 0.013353

1941 79 0.7 0.009278

1951 89 1 0.011919

1961 97 0.8 0.008607

1971 120 2.3 0.021278

Averages 0.857143 0.009902

-


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Solution problem 1

year PO tf-to Kar

Pf

(arithmet

ic) KgeoP

f(Geometric)

1981 120 10 0.857 129 0.0099 132.491

1991 120 20 0.857 137 0.0099 146.282

2001 120 30 0.857 146 0.0099 161.508

1994 120 23 0.857 140 0.0099 150.692

For arithmetic growth method : Pf= PO + K(tf-to)

Average increases per year =K = 0.857Population for the years,

1981= population 1971 + K(tf-to) = 120 + 8.57 = 128.57

For geometric growth method : Pf= POeKn; Where n=(tf-to)

Average increases per year =K = 0.0099

Population for the years,

1981= Population 1971 x e Kn = 120 x e Kn =132.5 (n = 10)


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Numerical 5

(Population forecasting)

A city had a population of 210000 in 1991 and

240000 in 2001.If the city is assumed to follow

arithmetic rate of growth find the population

of the city in 2018

Present (2009) population of city is 1350000

and it is expected to grow at a uniform rate of

3% per annum. Find its population in 2033


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umer caPopulation Forecasting

Year Populatio

n

Average

Increme

nt /yr

Geometr

ic

Increme

nt /yr1970 30000

1980 40000

1990 62000

2000 67000

Avg

Year Population Average

Increme

nt /yr

Geometr

ic

Increme

nt /yr

1950 8000

1960 8990

1970 11300

1980 14600

1990 18400

Avg

Estimate population for 2015 and 2048 in both of the cases

A B


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Population estimate

Year Populatio

n

Average

Increme

nt /yr

Geometr

ic

Increme

nt /yr

1970 30000

1980 40000 1000 0.028768

1990 620002200 0.043825

2000 67000 500 0.007756

Avg 1233.333 0.026783

Year Population Average

Increme

nt /yr

Geometr

ic

Increme

nt /yr

1950 8000

1960 8990 99 0.011667

1970 11300231 0.022869

1980 14600 330 0.025622

1990 18400 380 0.023133

Avg 313.6667 0.023875

Problem 1 Problem 2

year

number of

years

arithmetic

growth

geomtric

growth

solution problem 1 2015 15 85500 100127.1

2048 48 126200 242326.3

solution problem 2 2015 25 26241.7 33422.04

2048 58 36592.7 73484.69


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Problem 2

(Sources of water)

A small community had a population of 65000

and 85000 in the year of 1995 and 2005

respectively. Assuming a geometric growthrate and an average WC of 300lit/cap/day.

Calculate the design flow for the treatment

plant and the transmission main fromcurrent year. Select an appropriate value for

design period.


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P1995 =65000 , P2005=85000, Pf= POe

Kn =85000= 65000x(e10K)

K=0.0268person/yr

For transmission main design period=25 yrs(design yr=2037)

Treatment plant =15 yrs(design yr=2027)

Pf(transmission main)= 85000x(e0.0268x(2037-2005))=200387.15

Pf(treatment plant)= 85000x(e0.0268x(2027-2005))=153277.7

Max daily WC=1.5x300=450Lit/capita/day

Capacity for transmission mains= 200387.15 x450=90174.2m3/day

Capacity for treatment plant = 153277.7 x450=68974.96m3/day

Solution Problem 2


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Problem 3

(Sources of wtaer) The present population of a community is

160000 increasing at a geometric growthrate of 0.043 per yr. The present water

requirement of the community are fullymet by a number of tube wells installed inthe city. The average WC is 350l/c/d using adesign period of 15 yrs. Calculate the

number of additional tube-wells of3.4m3/min capacity to meet the demand ofdesign period.


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Solution problem 3Avg WC=350l/c/d; design period=15yrs

Present population=Po=160000;K=0.043

Pf= POeKn =160000X(e0.043x15)=304957.92

Additional poulation=304957.92-160000=144957.92

Total WC= 350x144957.92=50735272l/d=50735.2m3/day

Tubewell capacity=3.4x60x24=4896m3/day

With storage/ overhead reservoir(OHR)

max. Daily WC=1.5x50735.2=76102.9m3/day

Total no. of tube wells=76102.9/4896=15.516

No overhead reservoir(OHR)

Peak hourly flow=2.25x50735.2=114154.4m3/day

Total no. of tubewells wells=114154.4/4896=23.324

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