What rates are used for animals?

Cows and buffaloes 45 l/d, goats 5 l/d, poultry 0.2 l/d

Cows and buffaloes 5 l/d, goats 45 l/d

How are losses and max day demand applied?

Losses 15% of net demand; maximum day = 1.5 x average

Losses 15% of domestic only; max day = 3 x average

2079 Baisakh · 8 marks

Demand of rural municipality, base period 3 years

Short answer

With 2% annual growth, design population in 2044 is 9,461. Summing domestic (45 lpcd), livestock, school, and office demands, then adding 15% losses, average daily demand = 784 m³/day (9.1 L/s) and maximum day demand = 13.6 L/s.

Baseline survey of a newly developing rural municipality in 2021 AD has the following information:

Population = 6000 with annual growth rate of 2%; No. of cows = 3050; No. of buffaloes = 2000; No. of goats = 2500; No. of ducks = 1000; No. of chickens = 6000; No. of rural municipality = 1 and no. of students in a school = 150 boarders and 1500 day scholars.

Neglecting the fire demand, calculate the water demand of the area considering a base period of 3 and design period of 20 years.

Survey year 2021; base period 3 years and design period 20 years, so the design year is $2021+3+20 = 2044$ , i.e. 23 years after the survey.

Population forecast (geometrical, r = 2%)

P_{2044} = P_0(1+r)^{n} = 6000\times(1.02)^{23} = 9{,}461

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.

Livestock and school numbers are taken as constant. Fire demand is neglected.

Demand	Quantity	Rate	Demand (l/day)
Domestic (45 lpcd)	9,461	45 l/c/d	425,763
Boarders	150	45 l/c/d	6,750
Day scholars	1,500	5 l/c/d	7,500
Cows + buffaloes	5,050	45 l/d	227,250
Goats	2,500	5 l/d	12,500
Ducks + chickens	7,000	0.2 l/d	1,400
Rural municipality office (10 staff)	10	45 l/staff	450
Net demand			681,613

Losses and wastage $= 15\%$ of net demand $= 102{,}242$ l/day

Q_{avg} = 681{,}613 + 102{,}242 = 783{,}855\ \text{l/day} = 783.85\ \mathrm{m^3/day} = 9.07\ \text{l/s}

Peak (maximum day) demand $= 1.5\times783.9 = 1{,}175.8\ \mathrm{m^3/day}$ .

Answer: Design population = 9,461; average daily water demand = 784 m $^3$ /day (9.1 l/s); maximum day demand = 13.6 l/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

Demand of rural area survey 2019, 1.77% growth · Water Supply Engineering, chapter 3
Rural village demand, survey 2070 BS, base year 2073 · Water Supply Engineering, chapter 3
Demand of rural municipality 2018, 15,000 pop · Water Supply Engineering, chapter 3
Demand of village 2070 BS, base period 5 years · Water Supply Engineering, chapter 3
Population in survey year from design demand 800 m3 · Water Supply Engineering, chapter 3 · asked 2×
Average demand, village survey 2013, design 25 years · Water Supply Engineering, chapter 3 · asked 2×
Prove mean annual flood has 2.33-year return period · Engineering Hydrology, chapter 6
5th year peak load with population growth · Transmission and Distribution System Design, chapter 6 · asked 7×
Asphalt Institute design: 1.888×10³ ESAL/day, 10 years · Transportation Engineering II, chapter 2
5th year peak load, incremental growth rate · Transmission and Distribution System Design, chapter 6 · asked 7×

All Water Supply Engineering chapter 3 questions →

Demand of rural municipality, base period 3 years

Population forecast (geometrical, r = 2%)

Water demand

Related questions