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Demand of rural area survey 2019, 1.77% growth

Short answer

Population grows geometrically at 1.77% for 22 years to 6,561. Applying rural unit demands and 15% losses gives total water demand 382 m3/day, about 4.42 l/s.

A survey was carried out in 2019 in a rural area of Nepal and the following data were obtained: Population = 4460, Offices = 3 nos, Day students = 654, boarding students = 145, Cows and buffaloes = 480, goats and pigs = 855. A 15% of net water demand is considered as compensation for losses and wastage. Estimate the total water demand for the scheme if base period is 2 years and design period is 20 years. Consider the annual population growth rate as 1.77%.

Survey year 2019; base period 2 years and design period 20 years, so the design year is 2041, i.e. n=22n = 22 years after the survey.

Design population

P2041=4460×(1.0177)22=6,561P_{2041} = 4460\times(1.0177)^{22} = 6{,}561

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. Losses and wastage are 15% of net demand as given. Student and livestock numbers are taken as constant.

DemandQuantityRateDemand (l/day)
Domestic (45 lpcd)6,56145 l/c/d295,245
Day students6545 l/c/d3,270
Boarding students14545 l/c/d6,525
Offices (10 staff each)3045 l/staff1,350
Cows + buffaloes48045 l/d21,600
Goats + pigs8555 l/d4,275
Net demand332,265

Losses and wastage =15%= 15\% of net demand =49,840= 49{,}840 l/day

Qavg=332,265+49,840=382,105 l/day=382.10 m3/day=4.42 l/sQ_{avg} = 332{,}265 + 49{,}840 = 382{,}105\ \text{l/day} = 382.10\ \mathrm{m^3/day} = 4.42\ \text{l/s}

Answer: Design population = 6,561; total water demand = 382 m3^3/day (4.42 l/s).

✓ Answer checked by Claude Opus 5.5 (high effort)

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