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Demand of rural municipality 2018, 15,000 pop

Short answer

Design population in 2040 is 22,210. Applying rural unit rates and 15% losses gives average demand 1,566 m3/day or 18.1 l/s; maximum day demand is about 27.2 l/s.

Data obtained from a baseline survey of a newly formed rural municipality in year 2018 A.D. are as follows. Population = 15,000, No. of day students in school = 1500, No. of big animals = 6500, No. of small animals = 8000. There are altogether 10 offices, one hospital with total 25 beds, No. of tea shops = 12, No. of health post = 2 and number of police check post = 2. Neglect fire fighting demand. Calculate the water demand of that rural municipality in design year with base period of 2 years and design period of 20 years. Assume the population growth rate of that community is 1.8 % per annum.

Survey year 2018; base period 2 years and design period 20 years, so the design year is 2040, n=22n = 22 years after the survey.

Design population

P2040=15000×(1.018)22=22,210P_{2040} = 15000\times(1.018)^{22} = 22{,}210

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. Hospital 340 l/bed/d (IS 1172). Fire demand is neglected.

DemandQuantityRateDemand (l/day)
Domestic (45 lpcd)22,21045 l/c/d999,436
Day students1,5005 l/c/d7,500
Big animals6,50045 l/d292,500
Small animals8,0005 l/d40,000
Offices (10 staff each)10045 l/staff4,500
Hospital (25 beds)25340 l/bed/d8,500
Tea shops12500 l/d6,000
Health posts21000 l/d2,000
Police check posts (10 staff each)2045 l/staff900
Net demand1,361,336

Losses and wastage =15%= 15\% of net demand =204,200= 204{,}200 l/day

Qavg=1,361,336+204,200=1,565,536 l/day=1,565.54 m3/day=18.12 l/sQ_{avg} = 1{,}361{,}336 + 204{,}200 = 1{,}565{,}536\ \text{l/day} = 1{,}565.54\ \mathrm{m^3/day} = 18.12\ \text{l/s}

Answer: Design population = 22,210; average daily water demand = 1,566 m3^3/day (18.1 l/s); maximum day demand = 27.2 l/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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