Quantity of Water Numerical

Quantity of Water Numerical | Geometrical, Arithmetical & More

In this article, we will discuss the quantity of water Numerical with formulas. numerical.

 

  1. Formulas  

 

  i. Arithmetical Increase Method  

 

Pn = Po + nC

 

Where,

Po = Number of the present population

Pn = Number of the population at time nth time unit

C = Average increase in population

 

  ii. Geometrical Increase Method  

 

Pn = Po(1+r/100)n

 

Where,

Po = Number of the present population

Pn = Number of the population at time nth time unit

r = Average of the % increase

 

  iii. Incremental Increase Method  

 

Pn = Po + nc + ( n(n+1)i )/2

 

Where,

Po = Number of the present population

Pn = Number of the population at time nth time unit

i = Incremental increase per unit time

 

  iv. Decreasing Rate Growth Method  

 

Pn = Po nj=1 [1+(Io-jd)/100]

 

where,

Po = Number of the present population

Pn = Number of the population at time nth time unit

Io= % increase in population in last known yr % time unit

j = Average decrease in % increment % time unit

 

Quantity of Water Numerical

 

  2. Quantity of Water Numerical  

Determine the population of the town in the year 2026 by a) Arithmetical, b) Geometrical, c) Incremental increase and d) Decreased rate of growth method.

Year AD196119711981199120012011
Population180002700038000510006600083000

 

Solution:

n = (2026-2011)/10 = 1.5

Present or last known population (Po)= 83000

Year ADPopulationIncrease in population% increase in populationIncremental increaseDecrease in % increase
196118000
197127000900050
1981380001100040.7420009.26
1991510001300034.2120006.53
2001660001500029.4120004.80
2011830001700025.7520003.66
Total65000/5180.11/58000/424.25/4
AverageC= 13000r= 36.022i=+2000d= +6.0625

 

i) Arithmetic Method:

Pn = Po +nC

P1.5 = 83,000 + 1.5 x 13000

P1.5 = 102500 nos

 

ii) Geometric Method:

Pn = Po(1+r/100)n

P1.5 = 83000(1+36.022/100)1.5

P1.5 = 1,31,672 nos

 

iii) Incremental Method

Pn = Po + nc + ( n(n+1)i )/2

P1.5 = 83,000+1.5 x 13000 +1.5 x 2000 x (1.5+1)/2

P1.5 = 1,06,250 nos

 

iv) Decreased rate growth method

 

Pn = Po nj=1 [1+(Io-jd)/100]

P1.5 = 83000 (1+25.75-6.0625/100)(1+25.75-1.5 x 6.0625/100)

P1.5 = 1,15,887 nos

 

2. Estimate the population of a town for the design year 2031 AD by any three methods and calculate the design quantity of water in lit/day.

The census data are as follows:

Year AD1971198119912001
Population40000450005500062000

 

Also, consider fire demand and losses and wastage.

Solution:

n = (2031-2001)/10 = 3

Present or last known population (Po)= 62000

Year ADPopulationIncrease in population%increase in populationIncremental increase
197140,000
198145,0005,00012.50
199155,00010,00022.225000
200162,0007,00012.73-3000
Total22000/347.45/32000/2
AverageC=7333r=15.82i=+1000

 

i) Arithmetic Method:

Pn = Po +nC

P3 = 62,000 + 3 x 7333

P3 =83999 nos

 

ii) Geometric Method:

Pn = Po(1+r/100)n

P3 = 62000(1+15.822/100)3

P3 = 96326 nos

 

iii) Incremental Method

Pn = Po + nc + ( n(n+1)i )/2

P3 = 62,000 + 3 x 7333+ 3((3+1)/2) x 1000

P3 = 89,999 nos

 

Taking design year (2031 AD) population as forecasted by geometrical method = 96326 nos

Assume per capita water demand for the town as 112 lpcd.

 

a). Domestic demand for town = Population x per capita demand

= 96326 x 112

= 10788512 lit/day

 

b). Assume 1 lpcd of water for fire fighting

= 96326 x 1 lpcd

= 96326 lit/day

 

c) Assume losses demand

Therefore, Total water demand

TD = Domestic demand + Fire demand + Losses demand.

TD = 10788512+96326+0.15TD

0.85 TD = 10884838

Total water demand= 12805692 lit/day

 

3. In rural villages, the survey is carried out in the year 2072 BS and the following data is obtained.

Solution,

Population = 5320 nos

Annual population growth rate = 1.7%

No. of cows = 4030 nos

No. of goats = 1520 nos

No. of chickens = 5500 nos

No. of students = 200 boarders & 1020 day scholars

No. of VDC offices = 5

No. of Tea shops = 3

If the base year is taken as 2075 BS and the design period is 20 years. Calculate the total water demand of the village for the service year.

Solution:

Survey year= 2072,

Total water demand in design year(2095)=?

For design year, n= base period + design peroid

=3+20

=23 years

Design year population by geometrical forecasting method;

Pn = Po(1+r/100)n

P23 = 5320 (1+1.7/100)23

P23 = 7840 nos

Calculation of water demand;

a. Domestic demand= 7840 @ 45 lpcd = 352800 lit/day

 

b. Livestock demand

For cows = 4030 nos @ 45 lit/animal/day = 181350 lit/day

For goats = 1520 nos @ 20 lit/goat/day = 30400 lit/day

For chickens = 5500 nos @ 0.2 lit/bird/day =1100 lit/day

Total livestock demand= 212850 lit/day

Comparing livestock demand with 20% of domestic demand;

20% of domestic demand = 70560 lit/day

Total livestock demand is greater than 20% of domestic demand so take livestock demand = 70560 lit/day

c. Commercial and Institutional demand

For day scholars students= 1020 nos @ 10 lpcd = 10200 lit/day

For boarders= 200nos @ 65 lpcd = 13000 lit/day

For VDC office = 5 @ 500 lit/office/day= 2500 lit/day

For Tea Sop = 3 @ 500 lit/teastall/day= 1500 lit/day

Total commercial & institutional demand= 27200 lit/day

Total water demand for design year= DD+LD+ID

= 352800+70560+27200

=4,50,560 lit/day

 

Read Also: Examination of Water

 

4. Safe yield of a proposed spring is 5 lps and per capita, water demand is 65 lpcd. calculate the current population that can be taken under the scheme if design period is 20 years and the population growth rate is 1.7% per annum

Solution:

Safe yield of proposed spring (Q) = 5 x 60 x 60 x24 = 432000 lit/day

Per capita demand = Q/P20

P20 = 432000/65 = 6647 nos

Let Po be the current population for the scheme;

P20 for the scheme = P0(1+1.7/100)20 = 4744 nos

The Current Population (Po) that can take under the scheme is 4744 nos.

 

Read Also: Impurities in Water

 

5. In a rural village, the survey is carried out in the year 2072 BS and the following data is obtained:

Population = 5000 nos

Annual population growth rate = 1.5 %

Annual growth rate for students = 1%

No. of cows = 50 nos

No. of goats = 250 nos

No. of chickens = 2000 nos

No. of schools 2 with overall 350-day scholars students

No of VDC offices=  2

No. of Tea shops = 3

If the base year is taken as 2075 BS and.the design period is 20 calculate the total water demand of the village for the service year.

Solution,
Survey year = 2072,

Total water demand in Design year (2095) = ?

For design year, n = Base period + design period = 3+20

= 23 years

Design year population by geometrical forecasting method;
Pn = Po(1+r/100)n

P23= 5000(1+1.5/100)23

=7042 nos. 1

Similarly, Students population (P23) = 350 (1+1/100)23

= 440 nos,

 

Calculation of water demand;

a. Domestic demand = 7042 @ 45 Iped = 316890 lit/day

 

b. Livestock demand

For Cows = 50 nos @ 45 lit/animal/day = 2250 lit/day

For goats = 250 nos @ 20 litigoat/day 5000 lit/day

For chickens 2000 nos @ 0.20 lit/bird/day = 400 lit/day

Total livestock demand = 7650 lit/day

Comparing livestock demand with 20% of domestic demand;

20% of domestic demand = 633 78 lit/day.

Total livestock demand is less than 20% of domestic demand so livestock demand = 7650 lit/day

c. Commercial and Institutional demand For day scholars students = 440 nos @ 10 Ipcd = 4400 lit/day

For VDC office 2 a 500 lit/office/day 1000 lit/day

For Tea Sop = 3 @ 500 lit/teastall/day= 1500 lit/day

Total commercial & institutional demand= 6900 lit/day

Total water demand for design year= DD+LD+ID

= 31689000+7650+6900

= 331440 lit/day