Quantity of Water

Factors affecting the rate of demand

1. Climate: Hot and dry weather increases the use of water for drinking, bathing, cooling and gardening.
2. Size and type of community: Large cities use more water per head than villages; industrial or commercial cities use more.
3. Standard of living: Rich households with flush toilets, baths and appliances use much more than poor households.
4. Industrial and commercial activities: Factories, hotels, and offices raise the demand.
5. Quality of water: Better and safer water encourages use.
6. Pressure in the distribution system: High pressure increases consumption and leakage.
7. Cost (tariff) and metering: Metered supply and higher tariff reduce wastage; flat rate and free taps increase it.
8. System of supply: Continuous supply uses more water than intermittent supply.
9. Water loss and wastage: Leaky pipes and careless use increase demand.
10. Other: Habits and culture, availability of sewerage, and fire demand.

Variation of demand

• Seasonal variation: Demand is highest in summer (about 1.1 to 1.2 times the average) and low in winter.
• Daily variation: Higher on holidays, hot days and during festivals; maximum daily demand is about 1.8 times the average daily demand.
• Hourly variation: Demand is low at night and high in the morning (6 to 9 am) and evening (5 to 8 pm). The peak hour demand is about 1.5 times the maximum daily demand, or 2.7 times the average.

 Demand
   |      /\        /\
   |     /  \      /  \
   |    /    \____/    \
   |___/                \___
   0   6   12   18   24  hour
      (morning & evening peaks)

Peak demand

Peak factors are applied to the average daily demand $Q_{avg}$:

• Maximum daily demand $= 1.8\,Q_{avg}$ (Nepal rural: about 1.5)
• Peak hourly demand $= 1.5\times$ maximum daily demand $\approx 2.7\,Q_{avg}$ (population 20,000 to 50,000: factor about 2.0 to 3.0 in Nepal practice).

The distribution mains are designed for the peak hourly demand, while the source and transmission main are designed for the maximum daily demand.

Census interval 10 years (B.S.), last census 2070; for 2090, $n = 2$ decades.

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{70{,}000-28{,}000}{4} = 10{,}500.0$

$$P_{2090} = P_n + n\bar x = 70{,}000 + 2\times 10{,}500.0 = 91{,}000$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{70{,}000}{28{,}000}\right)^{1/4}-1 = 25.743\%$

$$P_{2090} = P_n(1+r)^n = 70{,}000\times(1+0.25743)^{2} = 110{,}680$$

Decreasing rate of growth method

Average decrease in rate $= 5.008\%$ per decade.

$P_{2090} = 89{,}620$

Summary

Water demand in 2090

The decreasing rate of growth method follows the falling trend of the growth rate and is adopted for design. Per capita demand 110 lpcd.

$$Q = 89{,}620\times110 = 9{,}858{,}187\ \mathrm{l/day} = 9{,}858.2\ \mathrm{m^3/day}$$

Answer: Population in 2090: arithmetical 91,000, geometrical 110,680, decreasing rate 89,620. Total water demand = 9,858 m$^3$/day (114.1 l/s) on the adopted population.

Census interval 10 years (B.S.), last census 2050. For 2070, $n = 2$ decades.

Population in 2070

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{60{,}000-25{,}000}{4} = 8{,}750.0$

$$P_{2070} = P_n + n\bar x = 60{,}000 + 2\times 8{,}750.0 = 77{,}500$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{60{,}000}{25{,}000}\right)^{1/4}-1 = 24.467\%$

$$P_{2070} = P_n(1+r)^n = 60{,}000\times(1+0.24467)^{2} = 92{,}952$$

Decreasing rate of growth method

Average decrease in rate $= 10.303\%$ per decade.

$P_{2070} = 52{,}447$

Population in 2080 (n = 3)

Water demand in 2080

The per capita demand is not given; I assume 110 lpcd for a city. The last increment (5,000) is much smaller than before, so the growth is slowing; I adopt the average of the three methods as the design population.

$$P_{2080} = \frac{86{,}250+115{,}694+41{,}004}{3} = 80{,}983$$ $$Q = 80{,}983\times110 = 8{,}908{,}094\ \mathrm{l/day} = 8{,}908\ \mathrm{m^3/day}$$

Answer: Population in 2070: arithmetical 77,500, geometrical 92,952, decreasing rate 52,447. Water demand in 2080 = 8,908 m$^3$/day (103.1 l/s).

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

Design population

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

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.

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

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

Design discharge

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

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

Survey year 2016; base period 2 years, design period 20 years: design year 2038, so $n = 22$ years after the survey.

Design population

$$P_{2038} = 45{,}000\times(1.015)^{22} = 62{,}440$$

Water demand

Assumed unit rates for a municipality: domestic 110 lpcd; day scholar 5 l/d; big animal 45 l/d; small animal 5 l/d; office staff 45 l/d (10 staff per office); hospital 340 l/bed/d (IS 1172); losses 15%.

Losses and wastage $= 15\%$ of net demand $= 1{,}106{,}541$ l/day

$$Q_{avg} = 7{,}376{,}940 + 1{,}106{,}541 = 8{,}483{,}481\ \text{l/day} = 8{,}483.48\ \mathrm{m^3/day} = 98.19\ \text{l/s}$$

Fire demand (National Board of Fire Underwriters formula)

$P = 62.44$ thousand.

$$\begin{aligned} Q_f &= 4637\sqrt P\,(1-0.01\sqrt P)\ \text{l/min}\\ &= 4637\times7.902\times(1-0.0790) = 33{,}746\ \text{l/min} = 562.4\ \text{l/s} \end{aligned}$$

For a fire lasting 2 hours the volume is $Q_f\times120 = 4{,}050\ \mathrm{m^3}$. It is provided from storage and is not added to the daily average, but the system is checked for $Q_{avg}$-based peak flow plus the fire flow.

Total

Total design demand with fire (2 h) $= 8{,}483+4{,}050 = 12{,}533\ \mathrm{m^3/day}$.

Answer: Design population = 62,440; average daily demand = 8,483 m$^3$/day (98.2 l/s); fire demand = 33,746 l/min (562 l/s); total including 2 h fire = 12,533 m$^3$.

Survey year 2013; base period 3 years and design period 25 years: design year 2041, $n = 28$ years after the survey.

Design population

$$P_{2041} = 1250\times(1.015)^{28} = 1{,}897$$

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.

Losses and wastage $= 15\%$ of net demand $= 15{,}112$ l/day

$$Q_{avg} = 100{,}744 + 15{,}112 = 115{,}855\ \text{l/day} = 115.86\ \mathrm{m^3/day} = 1.34\ \text{l/s}$$

Answer: Design population = 1,897; average water demand = 116 m$^3$/day (1.34 l/s).

This is a reverse problem: the total design demand is known, so the design population is found first and then brought back to the survey year.

Assumed rates: domestic 110 lpcd (municipality), day scholar 5 l/d, big animal 45 l/d, small animal 5 l/d, office staff 45 l/d (10 staff per office), hospital 340 l/bed/d, losses and wastage 15% of net demand. The total demand of 800 m$^3$ is taken per day, with losses included.

Step 1: demand other than domestic

Step 2: domestic demand in the design year

Net demand $= \dfrac{800{,}000}{1.15} = 695{,}652$ l/day

Domestic demand $= 695{,}652 - 250{,}500 = 445{,}152$ l/day

$$P_{2038} = \frac{445{,}152}{110} = 4{,}047$$

Step 3: population in the survey year

The design year is $2016+2+20 = 2038$, so $n = 22$ years.

$$P_{2038} = P_{2016}(1+r)^{n} \Rightarrow P_{2016} = \frac{4{,}047}{(1.019)^{22}} = 2{,}675$$

Answer: Population in the survey year (2016) = 2,675 (about 2,670); design year population = 4,047.

Survey year 2015, design year 2030, so $n = 15$ years.

Design population

$$P_{2030} = 1500\times(1.02)^{15} = 2{,}019$$

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 $= 15\%$ of net demand $= 20{,}625$ l/day

$$Q_{avg} = 137{,}501 + 20{,}625 = 158{,}126\ \text{l/day} = 158.13\ \mathrm{m^3/day} = 1.83\ \text{l/s}$$

Design discharge

• Maximum day demand $= 1.5\times1.83 = 2.75$ l/s (source, intake and transmission main).
• Peak hour demand $= 3\times1.83 = 5.49$ l/s (distribution).

Answer: Design discharge (maximum day) = 2.75 l/s; average demand = 1.83 l/s (158.1 m$^3$/day); peak hour = 5.49 l/s.

Survey year 2013; base period 3 years and design period 15 years: design year 2031, $n = 18$ years.

Design population

$$P_{2031} = 250\times(1.015)^{18} = 327$$

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 $= 15\%$ of net demand $= 4{,}516$ l/day

$$Q_{avg} = 30{,}108 + 4{,}516 = 34{,}624\ \text{l/day} = 34.62\ \mathrm{m^3/day} = 0.40\ \text{l/s}$$

Answer: Design population = 327; total water demand = 34.6 m$^3$/day (0.40 l/s).

Census interval 10 years; last census 2011. For 2021, $n = 1$ decade; for 2026, $n = 1.5$ decades.

Population in 2021 (n = 1)

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{83{,}000-18{,}000}{5} = 13{,}000.0$

$$P_{2021} = P_n + n\bar x = 83{,}000 + 1\times 13{,}000.0 = 96{,}000$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{83{,}000}{18{,}000}\right)^{1/5}-1 = 35.757\%$

$$P_{2021} = P_n(1+r)^n = 83{,}000\times(1+0.35757)^{1} = 112{,}678$$

Decreasing rate of growth method

Average decrease in rate $= 6.061\%$ per decade.

$P_{2021} = 99{,}348$

Population in 2026 (n = 1.5)

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{83{,}000-18{,}000}{5} = 13{,}000.0$

$$P_{2026} = P_n + n\bar x = 83{,}000 + 1.5\times 13{,}000.0 = 102{,}500$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{83{,}000}{18{,}000}\right)^{1/5}-1 = 35.757\%$

$$P_{2026} = P_n(1+r)^n = 83{,}000\times(1+0.35757)^{1.5} = 131{,}286$$

Decreasing rate of growth method

Average decrease in rate $= 6.061\%$ per decade.

Population at 2026 is obtained by linear interpolation between the decade values: $P_{2026} = 106{,}122$.

Summary

Answer: 2021: arithmetical 96,000, geometrical 112,678, decreased rate 99,348. 2026: arithmetical 102,500, geometrical 131,286, decreased rate 106,122.

Census interval 10 years; last census 2001. For 2011, $n = 1$ decade; for 2016, $n = 1.5$ decades.

Population in 2011 (n = 1)

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{83{,}000-18{,}000}{5} = 13{,}000.0$

$$P_{2011} = P_n + n\bar x = 83{,}000 + 1\times 13{,}000.0 = 96{,}000$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{83{,}000}{18{,}000}\right)^{1/5}-1 = 35.757\%$

$$P_{2011} = P_n(1+r)^n = 83{,}000\times(1+0.35757)^{1} = 112{,}678$$

Decreasing rate of growth method

Average decrease in rate $= 6.061\%$ per decade.

$P_{2011} = 99{,}348$

Population in 2016 (n = 1.5)

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{83{,}000-18{,}000}{5} = 13{,}000.0$

$$P_{2016} = P_n + n\bar x = 83{,}000 + 1.5\times 13{,}000.0 = 102{,}500$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{83{,}000}{18{,}000}\right)^{1/5}-1 = 35.757\%$

$$P_{2016} = P_n(1+r)^n = 83{,}000\times(1+0.35757)^{1.5} = 131{,}286$$

Decreasing rate of growth method

Average decrease in rate $= 6.061\%$ per decade.

Population at 2016 is obtained by linear interpolation between the decade values: $P_{2016} = 106{,}122$.

Summary

Answer: 2011: arithmetical 96,000, geometrical 112,678, decreased rate 99,348. 2016: arithmetical 102,500, geometrical 131,286, decreased rate 106,122.

Design period

The number of years for which a water supply scheme is designed to serve the growing population before it must be extended or replaced. It is counted from the base year to the design year and depends on the life of the component, cost of capital, ease of expansion and the growth rate. Typical values: intake and pipes 20 to 30 years, treatment plant 15 to 20 years, pumps 10 to 15 years, reservoirs 30 to 50 years. For rural schemes in Nepal it is commonly 15 to 20 years.

Base period

The time between the survey (data collection) year and the start of operation of the scheme (it covers design, approval and construction). Typical value: 2 to 5 years. The base year = survey year + base period, and the design year = base year + design period.

Peak hour demand

The maximum rate of demand in any hour of the day, which is the rate used for design of distribution system. Peak hour demand $\approx 1.5\times$ maximum daily demand $\approx 2.7\times$ average daily demand.

Types of water demand

Factors affecting the demand

Climate, size and nature of the town, standard of living, industry and commerce, quality and cost of water (tariff), metering, supply pressure, continuous or intermittent supply, leakage control, and habits and culture of the people.

Base year 2024 and design period 20 years give the design year $2024+20 = 2044$. The last census is 2021, so $n = (2044-2021)/10 = 2.3$ decades.

Population forecast

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{43{,}000-16{,}000}{4} = 6{,}750.0$

$$P_{2044} = P_n + n\bar x = 43{,}000 + 2.3\times 6{,}750.0 = 58{,}525$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{43{,}000}{16{,}000}\right)^{1/4}-1 = 28.037\%$

$$P_{2044} = P_n(1+r)^n = 43{,}000\times(1+0.28037)^{2.3} = 75{,}918$$

The two methods give 58,525 and 75,918. The data show an almost constant increase of 6000 to 8000 per decade, so the geometric result is too high. I adopt the mean of the two methods as design population.

$$P_{2044} = \frac{58{,}525+75{,}918}{2} = 67{,}221$$

Water demand

Domestic demand $= 67{,}221\times110/1000 = 7{,}394.4\ \mathrm{m^3/day}$.

Industrial demand is 25% and losses 15% of the total demand $Q$, so the domestic demand is the remaining 60%.

$$\begin{aligned} Q &= \frac{7{,}394.4}{1-0.25-0.15} = \frac{7{,}394.4}{0.60} = 12{,}323.9\ \mathrm{m^3/day}\\ \text{Industrial} &= 0.25Q = 3{,}081.0\ \mathrm{m^3/day}\\ \text{Losses} &= 0.15Q = 1{,}848.6\ \mathrm{m^3/day} \end{aligned}$$

Answer: Design population (2044) = 67,221; total water demand = 12,324 m$^3$/day (142.6 l/s).

Last census is 2021; for 2040, $n = (2040-2021)/10 = 1.9$ decades.

Population by geometrical increase method

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{42{,}000}{15{,}000}\right)^{1/4}-1 = 29.357\%$

$$P_{2040} = P_n(1+r)^n = 42{,}000\times(1+0.29357)^{1.9} = 68{,}493$$

Water demand in 2040

Domestic demand $= 68{,}493\times120/1000 = 8{,}219.2\ \mathrm{m^3/day}$.

Industrial demand (20%) and losses (15%) are fractions of the total demand $Q$, so domestic demand is 65% of $Q$.

$$\begin{aligned} Q &= \frac{8{,}219.2}{1-0.20-0.15} = \frac{8{,}219.2}{0.65} = 12{,}645.0\ \mathrm{m^3/day}\\ \text{Industrial} &= 0.20Q = 2{,}529.0\ \mathrm{m^3/day},\quad \text{Losses} = 0.15Q = 1{,}896.7\ \mathrm{m^3/day} \end{aligned}$$

Answer: Population in 2040 = 68,493; total water demand = 12,645 m$^3$/day (146.4 l/s).

The census interval is 10 years and the last census is 1991. For 2021, $n = 3.0$ decades; for 2028, $n = 3.7$ decades. "Changing rate of increase" is taken as the decreasing (changing) rate of growth method.

Population in 2021 (n = 3.0)

Incremental increase method

Increase per decade: 2,000, 3,000, 2,500, 6,600, 7,400, 5,550, 7,450, 6,500; incremental increase: 1,000, -500, 4,100, 800, -1,850, 1,900, -950. $\bar x = 5{,}125.00$, $\bar y = 642.86$

$$P_{2021} = P_n + n\bar x + \frac{n(n+1)}{2}\bar y = 61{,}000 + 3(5{,}125.00) + \frac{3(4)}{2}(642.86) = 80{,}232$$

Decreasing rate of growth method

Average decrease in rate $= -0.275\%$ per decade.

$P_{2021} = 86{,}800$

Population in 2028 (n = 3.7)

Incremental increase method

Increase per decade: 2,000, 3,000, 2,500, 6,600, 7,400, 5,550, 7,450, 6,500; incremental increase: 1,000, -500, 4,100, 800, -1,850, 1,900, -950. $\bar x = 5{,}125.00$, $\bar y = 642.86$

$$P_{2028} = P_n + n\bar x + \frac{n(n+1)}{2}\bar y = 61{,}000 + 3.7(5{,}125.00) + \frac{3.7(4.7)}{2}(642.86) = 85{,}552$$

Decreasing rate of growth method

Average decrease in rate $= -0.275\%$ per decade.

Population at 2028 is obtained by linear interpolation between the decade values: $P_{2028} = 94{,}715$.

Summary

Answer: Incremental increase: 2021 = 80,232, 2028 = 85,552. Changing rate of increase: 2021 = 86,800, 2028 = 94,715.

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.

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.

The design year is 2092 BS. The last census is 2068, so $n = (2092-2068)/10 = 2.4$ decades.

Population by incremental increase method

Incremental increase method

Increase per decade: 900, 1,200, 1,700, 2,000, 3,000; incremental increase: 300, 500, 300, 1,000. $\bar x = 1{,}760.00$, $\bar y = 525.00$

$$P_{2092} = P_n + n\bar x + \frac{n(n+1)}{2}\bar y = 16{,}000 + 2.4(1{,}760.00) + \frac{2.4(3.4)}{2}(525.00) = 22{,}366$$

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. Health post beds are taken at 100 l/bed/day. The 3 schools have 350 day scholars and 50 boarders in total, and the 2 health posts have 10 beds in total. Fire demand is neglected.

Losses and wastage $= 15\%$ of net demand $= 154{,}076$ l/day

$$Q_{avg} = 1{,}027{,}170 + 154{,}076 = 1{,}181{,}246\ \text{l/day} = 1{,}181.25\ \mathrm{m^3/day} = 13.67\ \text{l/s}$$

Answer: Design population (2092 BS) = 22,366; total water requirement = 1,181 m$^3$/day (13.7 l/s).

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

Design population

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

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.

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

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

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

Survey year 2074 BS; base period 5 years and design period 30 years, so the design year is $2074+5+30 = 2109$ BS. The last census is 2074, so $n = 35/10 = 3.5$ decades.

Population forecast

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{59{,}500}{45{,}500}\right)^{1/4}-1 = 6.937\%$

$$P_{2109} = P_n(1+r)^n = 59{,}500\times(1+0.06937)^{3.5} = 75{,}242$$

Decreasing rate of growth method

Average decrease in rate $= 1.102\%$ per decade.

Population at 2109 is obtained by linear interpolation between the decade values: $P_{2109} = 63{,}465$.

Water demand ($q = 110$ lpcd)

$$Q = P\times110\ \text{l/day}$$

The per capita allowance of 110 lpcd is taken to cover all domestic and public demands, so no separate losses or institutional demands are added.

Answer: (a) Geometrical: P = 75,242, Q = 8,277 m$^3$/day. (b) Decreasing rate: P = 63,465, Q = 6,981 m$^3$/day.

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

Design population

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

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.

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

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

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

Design year 2030; the last census is 2011, so $n = 1.9$ decades.

Population forecast

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{42{,}000-15{,}000}{4} = 6{,}750.0$

$$P_{2030} = P_n + n\bar x = 42{,}000 + 1.9\times 6{,}750.0 = 54{,}825$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{42{,}000}{15{,}000}\right)^{1/4}-1 = 29.357\%$

$$P_{2030} = P_n(1+r)^n = 42{,}000\times(1+0.29357)^{1.9} = 68{,}493$$

Incremental increase method

Increase per decade: 6,000, 6,000, 7,000, 8,000; incremental increase: 0, 1,000, 1,000. $\bar x = 6{,}750.00$, $\bar y = 666.67$

$$P_{2030} = P_n + n\bar x + \frac{n(n+1)}{2}\bar y = 42{,}000 + 1.9(6{,}750.00) + \frac{1.9(2.9)}{2}(666.67) = 56{,}662$$

The mean of the three methods is adopted: $P_{2030} = 59{,}993$.

Water demand

For a fully plumbed town the domestic demand is taken as 135 lpcd (IS 1172).

$$Q_{dom} = 59{,}993\times135 = 8{,}099{,}107\ \text{l/day} = 8{,}099\ \mathrm{m^3/day}$$

Industrial demand (20%) and losses (15%) are fractions of the total demand $Q$, so domestic demand is 65% of $Q$.

$$\begin{aligned} Q &= \frac{8{,}099}{0.65} = 12{,}460\ \mathrm{m^3/day}\\ \text{Industrial} &= 0.20Q = 2{,}492\ \mathrm{m^3/day},\quad \text{Losses} = 0.15Q = 1{,}869\ \mathrm{m^3/day} \end{aligned}$$

Answer: Population in 2030 = 59,993; total water demand = 12,460 m$^3$/day (144.2 l/s).

The census data are arranged in ascending order of year (2028 to 2068). Interval 10 years; for 2088, $n = 2$ decades.

Population forecast

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{65{,}500-29{,}000}{4} = 9{,}125.0$

$$P_{2088} = P_n + n\bar x = 65{,}500 + 2\times 9{,}125.0 = 83{,}750$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{65{,}500}{29{,}000}\right)^{1/4}-1 = 22.592\%$

$$P_{2088} = P_n(1+r)^n = 65{,}500\times(1+0.22592)^{2} = 98{,}438$$

Decreasing rate of growth method

Average decrease in rate $= 4.225\%$ per decade.

$P_{2088} = 77{,}186$

Design population and water demand

The design year is 2088 BS. The mean of the three methods is adopted as design population:

$$P_{2088} = \frac{83{,}750+98{,}438+77{,}186}{3} = 86{,}458$$ $$Q = 86{,}458\times120 = 10{,}374{,}966\ \text{l/day} = 10{,}375\ \mathrm{m^3/day}$$

Answer: Design year = 2088 BS; design population = 86,458; total water demand = 10,375 m$^3$/day (120.1 l/s).

The spring yield must supply the demand of the population at the end of the design period, so work backwards from the design year population.

Step 1: daily yield

$$Q = 5\ \text{l/s}\times86400 = 432{,}000\ \text{l/day}$$

Step 2: design population that can be served

$$P_{d} = \frac{Q}{q} = \frac{432{,}000}{65} = 6{,}646$$

Step 3: population today

$$P_{d} = P_0(1+r)^n \Rightarrow P_0 = \frac{6{,}646}{(1.017)^{20}} = 4{,}744$$

Answer: Current population that can be served = 4,744 (about 4,740 persons); design population after 20 years = 6,646.

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

Design population

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

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.

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

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

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

The last census is 2011, so for 2055, $n = (2055-2011)/10 = 4.4$ decades. Four methods are used.

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{57{,}000-32{,}000}{4} = 6{,}250.0$

$$P_{2055} = P_n + n\bar x = 57{,}000 + 4.4\times 6{,}250.0 = 84{,}500$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{57{,}000}{32{,}000}\right)^{1/4}-1 = 15.526\%$

$$P_{2055} = P_n(1+r)^n = 57{,}000\times(1+0.15526)^{4.4} = 107{,}565$$

Incremental increase method

Increase per decade: 4,000, 8,000, 8,000, 5,000; incremental increase: 4,000, 0, -3,000. $\bar x = 6{,}250.00$, $\bar y = 333.33$

$$P_{2055} = P_n + n\bar x + \frac{n(n+1)}{2}\bar y = 57{,}000 + 4.4(6{,}250.00) + \frac{4.4(5.4)}{2}(333.33) = 88{,}460$$

Decreasing rate of growth method

Average decrease in rate $= 0.962\%$ per decade.

Population at 2055 is obtained by linear interpolation between the decade values: $P_{2055} = 76{,}741$.

Summary

Design population and justification

The census increases are 4000, 8000, 8000 and 5000 per decade, so the growth is irregular and slowing in the last decade. The geometrical method gives an extreme value (it assumes the growth rate stays constant) and the decreasing rate method gives the lowest value. A mean of the four methods balances them, so the mean is adopted:

$$P_{2055} = \frac{84{,}500+107{,}565+88{,}460+76{,}741}{4} = 89{,}317$$

Daily water requirement

For a rural area, assume 45 lpcd and 15% losses and wastage.

$$Q = 89{,}317\times45\times1.15 = 4{,}622{,}130\ \text{l/day} = 4{,}622\ \mathrm{m^3/day}$$

Answer: Population in 2055 = 89,317 (adopted mean); daily water requirement = 4,622 m$^3$/day (53.5 l/s).

Survey year 2070 BS, base year 2073 BS and design period 20 years: the service (design) year is $2073+20 = 2093$ BS, which is 23 years after the survey.

Population and students in the design year

$$\begin{aligned} P_{2093} &= 5000\times(1.015)^{23} = 7{,}042\\ S_{2093} &= 350\times(1.01)^{23} = 440 \end{aligned}$$

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 numbers are taken as constant.

Losses and wastage $= 15\%$ of net demand $= 48{,}808$ l/day

$$Q_{avg} = 325{,}385 + 48{,}808 = 374{,}193\ \text{l/day} = 374.19\ \mathrm{m^3/day} = 4.33\ \text{l/s}$$

Answer: Total water demand of the village in 2093 BS = 374 m$^3$/day (4.33 l/s), for a design population of 7,042.

Census interval 10 years; last census 2011. For 2021, $n = 1$ decade; for 2026, $n = 1.5$ decades.

Population in 2021 (n = 1)

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{83{,}000-18{,}000}{5} = 13{,}000.0$

$$P_{2021} = P_n + n\bar x = 83{,}000 + 1\times 13{,}000.0 = 96{,}000$$

Decreasing rate of growth method

Average decrease in rate $= 6.061\%$ per decade.

$P_{2021} = 99{,}348$

Population in 2026 (n = 1.5)

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{83{,}000-18{,}000}{5} = 13{,}000.0$

$$P_{2026} = P_n + n\bar x = 83{,}000 + 1.5\times 13{,}000.0 = 102{,}500$$

Decreasing rate of growth method

Average decrease in rate $= 6.061\%$ per decade.

Population at 2026 is obtained by linear interpolation between the decade values: $P_{2026} = 106{,}122$.

Summary

Answer: Arithmetic increase: 2021 = 96,000, 2026 = 102,500. Decreased rate of growth: 2021 = 99,348, 2026 = 106,122.

The last census is 2000, so for 2025, $n = 2.5$ decades.

Population by incremental increase method

Incremental increase method

Increase per decade: 530, 745, 840, 2,515, 2,559; incremental increase: 215, 95, 1,675, 44. $\bar x = 1{,}437.80$, $\bar y = 507.25$

$$P_{2025} = P_n + n\bar x + \frac{n(n+1)}{2}\bar y = 14{,}339 + 2.5(1{,}437.80) + \frac{2.5(3.5)}{2}(507.25) = 20{,}153$$

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. Health post beds are taken at 100 l/bed/d; each of the 2 health posts is taken to have 5 beds. School staff are counted with the 125 day scholars of each school (4 schools, 500 persons at 5 l/d).

Losses and wastage $= 15\%$ of net demand $= 140{,}138$ l/day

$$Q_{avg} = 934{,}251 + 140{,}138 = 1{,}074{,}389\ \text{l/day} = 1{,}074.39\ \mathrm{m^3/day} = 12.44\ \text{l/s}$$

Answer: Population in 2025 = 20,153; total water requirement = 1,074 m$^3$/day (12.4 l/s).

The last census is 2000, so for 2030, $n = 3$ decades. Three methods are used.

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{62{,}000-40{,}000}{3} = 7{,}333.3$

$$P_{2030} = P_n + n\bar x = 62{,}000 + 3\times 7{,}333.3 = 84{,}000$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{62{,}000}{40{,}000}\right)^{1/3}-1 = 15.729\%$

$$P_{2030} = P_n(1+r)^n = 62{,}000\times(1+0.15729)^{3} = 96{,}100$$

Incremental increase method

Increase per decade: 5,000, 10,000, 7,000; incremental increase: 5,000, -3,000. $\bar x = 7{,}333.33$, $\bar y = 1{,}000.00$

$$P_{2030} = P_n + n\bar x + \frac{n(n+1)}{2}\bar y = 62{,}000 + 3(7{,}333.33) + \frac{3(4)}{2}(1{,}000.00) = 90{,}000$$

The mean of the three is adopted as the design population: $P_{2030} = 90{,}033$.

Design quantity of water

Assumptions: domestic and public demand 135 lpcd (IS 1172), losses and wastage 15% of the domestic demand, fire demand by the National Board of Fire Underwriters formula for a duration of 2 hours.

$$\begin{aligned} \text{Domestic} &= 90{,}033\times135 = 12{,}154{,}500\ \text{l/day}\\ \text{Losses} &= 0.15\times12{,}154{,}500 = 1{,}823{,}175\ \text{l/day}\\ Q_f &= 4637\sqrt{P}\,(1-0.01\sqrt{P}),\ P = 90.03\ \text{(thousand)}\\ &= 39{,}824\ \text{l/min}\\ \text{Fire volume (2 h)} &= 39{,}824\times120 = 4{,}778{,}849\ \text{l} \end{aligned}$$ $$Q_{total} = 12{,}154{,}500+1{,}823{,}175+4{,}778{,}849 = 18{,}756{,}524\ \text{l/day}$$

Answer: Design population (2030) = 90,033; design quantity of water = 18,756,524 l/day (18.76 million l/day), including losses and a 2-hour fire demand.