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Demand of village 2070 BS, base period 5 years

Short answer

Design year 2095 BS, n=25, population 4312. Net demand 210128 l/d; adding 15% losses gives 241648 l/d = 241.65 m3/d = 2.80 l/s; maximum day demand 4.20 l/s.

Data obtained from a baseline survey of a village in year 2070 B.S. are as follows. Population of village = 2700, No. of day scholar students = 220, No. of big animals = 240, No. of small animals = 460. There are two offices, one health post. Calculate the water demand of that village in design year with base period of 5 years and design period of 20 years. Assume the population growth rate of that village as 1.89% per annum.

Survey year 2070 BS; base period 5 years and design period 20 years: design year 2095 BS, n=25n = 25 years.

Design population

P2095=2700×(1.0189)25=4,312P_{2095} = 2700\times(1.0189)^{25} = 4{,}312

Water demand

Design values assumed (Nepal rural water supply practice): domestic 45 lpcd (public taps); day scholar 5 l/d; boarder 45 l/d; office staff 45 l/d with 10 staff per office; big animal (cow, buffalo) 45 l/d; small animal (goat, pig) 5 l/d; poultry 0.2 l/d; health post 1000 l/d; tea shop 500 l/d; losses and wastage 15% of net demand; maximum day factor 1.5 and peak hour factor 3. Fire demand is neglected.

DemandQuantityRateDemand (l/day)
Domestic (45 lpcd)4,31245 l/c/d194,028
Day scholars2205 l/c/d1,100
Big animals24045 l/d10,800
Small animals4605 l/d2,300
Offices (10 staff each)2045 l/staff900
Health post11000 l/d1,000
Net demand210,128

Losses and wastage =15%= 15\% of net demand =31,519= 31{,}519 l/day

Qavg=210,128+31,519=241,648 l/day=241.65 m3/day=2.80 l/sQ_{avg} = 210{,}128 + 31{,}519 = 241{,}648\ \text{l/day} = 241.65\ \mathrm{m^3/day} = 2.80\ \text{l/s}

Answer: Design population = 4,312; average water demand = 242 m3^3/day (2.80 l/s); maximum day demand = 4.20 l/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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