The document contains information to estimate water requirements for different communities including design population, per capita demand, average water demand, maximum daily demand, peak hourly demand, and fire flow requirements. It provides the calculations to estimate the total water requirement considering both regular demand and fire demand for communities with populations of 22,000, 55,000 and 120,000. Fire flow rates and daily amounts are also estimated for different building types and floor areas.

1. 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)

2. 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+
Fire
demand
(m3/d)
ordinary
Max
daily+
Fire
deman
d
(m3/d)
wood
22000 600 13200 19800 29700 6 storey
building with
a)Ordinary
b)wood
construction
(Each floor
area=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)

3. 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 .

4. 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

5. 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 79
1951 89
1961 97
1971 120
Averages -
Numerical 4
(Population Forecasting)

6. 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
-

7. Solution problem 1
year PO tf-to Kar
Pf
(arithmet
ic) Kgeo
Pf
(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 = 0.857
Population for the years,
1981= population 1971 + K(tf-to) = 120 + 8.57 = 128.57
For geometric growth method : Pf = POe Kn; Where n=(tf-to)
Average increases per year =K = 0.0099
Population for the years,
1981= Population 1971 x eKn = 120 x eKn =132.5 (n = 10)

8. 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

9. Numerical 6
Population Forecasting
Year Populatio
n
Average
Increme
nt /yr
Geometr
ic
Increme
nt /yr
1970 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

10. Population estimate
Year Populatio
n
Average
Increme
nt /yr
Geometr
ic
Increme
nt /yr
1970 30000
1980 40000 1000 0.028768
1990 62000 2200 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 11300 231 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

11. 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 growth
rate and an average WC of 300lit/cap/day.
Calculate the design flow for the treatment
plant and the transmission main from
current year. Select an appropriate value for
design period.

12. • 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

13. Problem 3
(Sources of wtaer)
• The present population of a community is
160000 increasing at a geometric growth
rate of 0.043 per yr. The present water
requirement of the community are fully
met by a number of tube wells installed in
the city. The average WC is 350l/c/d using a
design period of 15 yrs. Calculate the
number of additional tube-wells of
3.4m3/min capacity to meet the demand of
design period.

14. Solution problem 3
Avg WC=350l/c/d; design period=15yrs
Present population=Po=160000;K=0.043
Pf = POe Kn =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=15.5≈16
No overhead reservoir(OHR)
Peak hourly flow=2.25x50735.2=114154.4m3/day
Total no. of tubewells wells=114154.4/4896=23.3≈24