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Design discharge 2040, Ilam rural municipality

Short answer

Design population in 2040 is 2,521 by geometric growth. Average demand is 193.4 m3/day or 2.24 l/s; maximum day demand is 3.36 l/s and peak hour demand is 6.72 l/s.

Calculate the design discharge for design year 2040 for a Rural Municipality in Ilam District. The data collected in survey year 2020 is as below:

Survey year population = 1600; Population growth rate = 2.3% per year; number of buffalos = 350; Number of cows = 500; Number of goats = 900; Number of chickens = 2500; Number of boarder students = 100; Number of day scholar students = 550; Number of offices = 5; Health post = 2 nos.

Survey year 2020, design year 2040, so n=20n = 20 years.

Design population

P2040=1600×(1.023)20=2,521P_{2040} = 1600\times(1.023)^{20} = 2{,}521

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 for a rural area.

DemandQuantityRateDemand (l/day)
Domestic (45 lpcd)2,52145 l/c/d113,461
Boarders10045 l/c/d4,500
Day scholars5505 l/c/d2,750
Buffaloes + cows85045 l/d38,250
Goats9005 l/d4,500
Chickens2,5000.2 l/d500
Offices (10 staff each)5045 l/staff2,250
Health posts21000 l/d2,000
Net demand168,211

Losses and wastage =15%= 15\% of net demand =25,232= 25{,}232 l/day

Qavg=168,211+25,232=193,442 l/day=193.44 m3/day=2.24 l/sQ_{avg} = 168{,}211 + 25{,}232 = 193{,}442\ \text{l/day} = 193.44\ \mathrm{m^3/day} = 2.24\ \text{l/s}

Design discharge

  • Maximum day demand =1.5×193,442=290,163= 1.5\times193{,}442 = 290{,}163 l/day =3.36= 3.36 l/s (for source, intake and transmission main).
  • Peak hour demand =3×2.24=6.72= 3\times2.24 = 6.72 l/s (for the distribution system).

Answer: Design discharge (maximum day) = 3.36 l/s; average demand = 2.24 l/s (193.4 m3^3/day); peak hour = 6.72 l/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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