USAGE AND DEMAND OF WATER SUPPLY 1
2 TOPIC 2
USAGE AND DEMAND OF WATER O classification of water demand/water usage: a. Domestic, b. Commercial, c. Industrial, d. Agricultural, e. Public, f. Non revenue water (NRW) O factors effecting the population growth O Estimate future population using various methods: a.Arithmetic Increase Method, b. Geometric Increase Method, c. Incremental Increase Method, d. Decreasing Rate of Growth. O Calculate the water demand (WDn) using the water demand formula (service factor& design factor) additional demand to fulfil new development 3
Water usage O Domestic demand O Commercial /trade purpose O Industrial O Agriculture O Public or civic use O Non Revenue Water (NRW) 4
5 FACTORS INFLUENCE WATER DEMAND POPULATION TYPES OF INDUSTRY & COMMERCIAL CLIMATE WATER QUALITY AWARENESS CLO 1, PLO1
6 POPULATION BIG CITY SMALL TOWN
7 CLIMATE MORE IN SUMMER LESS IN WINTER
8 TYPES OF INDUSTRY
9 TYPES OF COMMERCIAL
10 WATER QUALITY
11 AWARENESS
Factors Effecting The Population Growth O Birth Rate O Mortality Rate O Migration Rate 12
13 CLO 1, PLO1
Basic Formula For Water Demand O WDn = Pn x q x F 1 x F2 …. + Dm O WDn = water demand at the end of year “n” O Pn = projected population at the end of year “n” O q = per capita consumption at the end of year “n” O F 1 = service factor at the end of year “n” O F2 = design factor at the end of year “n” O Dm = additional demand at the end of year “n” 14
15 O Pn = projected population at the end of year “n” Pn = Po x ( 1 + r ) n Pn = Population estimation at year ‘n’ Po = Total population for the year before r = Increased of population rate n = no of years
Example 1 O The following data obtained from Kampung Merah in 2007. calculate the water demand in 2012. O Total household = 6000 households O Average household member = 6 people O Per capita water consumption = 270 liters/day O Population growth = 2.65% per year O Industrial water needs = 1/3 of the population needs O Design factor = 2.4 O Percentage of NRW = 15% O Water supply coverage = 97% 16
Pn = Po x ( 1 + r )n Po = 6000 x 6 = 36,000 r= 2.65% a year = 2.65/100 = 0.0265 n= 2012- 2007 = 5 years Pn = Po x ( 1 + r )n P 2008 = 36000 ( 1 + 0.0265 )5 = 41,030 People 17
O WDn = Pn x q x F 1 x F2 …. + Dm q = 270 litre/capita/day F1 = 97% = 0.97 F2 = 2.4 Dm = 15%= 0.15 Industri = 1/3 of the population needs wD = ( 41,030 x 270 x 0.97 x 2.4) + [ 1/3 ( 41030 x 270 x 2.4)] + 0.15 [(41030 x 270) + ( 1/3 x 41030 x 270)] = 25.79 x 10˄6 + 8.86 x 10 ˄6 + 2.22 x 10 ˄6 = 36.87 x 10˄6 litres/ day 18
19 Problem 1 Based on the data for the 1994 below, calculate the population on 2000. a) No. of houses = 5000 unit b) average of person = 5 c) Percentage of population growth = 2.75% a year d) Per capita water consumption = 275 liters/day e) Industrial water needs : 1/3 of the population need f) Percentage of NRW = 10% g)Water supply coverage = 90% h) Design factor = 1.5
Problem 2 O The data given is collected from Taman Aman in year 2012. Estimate the daily water demand if water supply coverage 97%. O i. total population = 520,000 O ii. Water usage per capita = 270 liters/day O iii. Industry water demand = 1/3 from requirements of the population O iv. Design factor = 1.5 O v. NRW percentage = 15% 20
Problem 3 O Given the total daily water per capacity is 230 liter/ day (q), population is 32003, calculate the water demand estimation (WDn) if the services factor (F1) is 0.98 and the design factor (F2) is 2.5. assume that there is no additional demand. 21
Estimate Future Population OArithmetic Increase Method OGeometric Increase Method OIncremental Increase Method ODecreasing Rate of Growth Method 22
23 ARITMETIC METHOD Assumption used when population increase with constant rate.
24 Example - ARITMETIC METHOD Based on the data given below, estimate the population growth for the resident A for the year 2020 using the Aritmetic Method. Year 1970 1980 1990 2000 2010 Population 12,550 14,756 18,215 21,943 26,434 2,206 3,459 3,728 4,491
25 ARITMETIC METHOD 2,206 + 3,459 + 3,728 + 4,491 4 3,471 org / 10 year 347.1/1 year P2020 = 26,434 + 3,471 = 29,905 persons P2021_= 29905 + 347.1 = 30252 person.
PROBLEM 1 YEAR POPULATION 1970 12,550 1980 14,569 1990 17,770 2000 22,271 2010 28,112 26 The table shows data the population from 1970 to 2010. Estimate population for 2020 and 2030 using arithmetic method.
27 UNIFORM PERCENTAGE METHOD/GEOMETRIC INCREASE METHOD This method assumes the percentage increase in population from decade to decade as constant.
28 UNIFORM PERCENTAGE METHOD Based on the data given below, estimate the population growth for the resident B for the year 2010 using the Uniform Percentage Method. Year 1960 1970 1980 1990 2000 Population 86,300 101,900 130,000 141,500 147,750 15,600 28,100 11,500 6,250
29 UNIFORM PERCENTAGE METHOD Year 1960 1970 1980 1990 2000 Population 86,300 101,900 130,000 141,500 147,750 Population Growth Percentage of population growth 15,600 28,100 11,500 6,250 15,600 / 86,300 x 100% = 18.1% 28,100 / 101,900 x 100% = 27.6% 11,500 / 130,000 x 100% = 8.8% 6,250 / 141,500 x 100% = 4.4%
30 UNIFORM PERCENTAGE METHOD 15,600 / 86,300 x 100% = 18.1% 128,100 / 101,900 x 100% = 27.6% 11,500 / 130,000 x 100% = 8.8% 6,250 / 141,500 x 100% = 4.4% 18.1% + 27.6% + 8.8% + 4.4% 4 = 14.7% P2010 = 147,750 + ( 14.7/100 x 147,750 ) = 169,469 persons
PROBLEM Year Population 1940 1950 1960 1970 8000 12000 17000 22500 Based on the data given below, estimate the population growth for the resident A for the year 1980, 1990, 2000 using the Uniform Percentage Method.
INCREMENTAL INCREASE METHOD O This method is improvement over the above two methods. The average increase in the population is determined by the arithmetical method and to this is added the average of the net incremental increase once for each future decade.
The population after ‘n’ decade O where O P = population at present O Ia = Average Arithmetical increase O I c = Average incremental increase O n = decade
SOLUTION Year Population Increase in Population Incremental increase i.e Increment on the increase 1940 1950 1960 1970 8000 12000 17000 22500 - 4000 5000 5500 - - 1000 500 Total 14500 1500 Average 4833 750
Use formula Year Expected Population 1980 1990 2000 22500 + 1 x (4833 + 750) = 28083 28083 + 1 x (4833 + 750) = 33666 33666 + 1 x (4833 + 750) = 39249
PROBLEM 36 Based on the data given below, estimate the population growth for the resident A for the year 1980, 1990, 2000 using the INCREMENTAL INCREASE METHOD Year Population 1940 1950 1960 1970 9500 13000 18500 24500
DECREASING RATE OF GROWTH METHOD O In this method, the average decrease in the percentage increase is worked out, and is then subtracted from the latest percentage increase to get the percentage increase of next decade
Year Populatio n Increase in Population Percentage increase in population Decrease in the percentage increase 1940 1950 1960 1970 8000 12000 17000 22500 - 4000 5000 5500 - (4000/8000) x 100 = 50 (5000/12000) x 100 =41.7 (5500/17000) x 100 = 32.4 - - +8.3 +9.3 Total 14500 17.6 Average 4833 8.8
• Now the population at the end of various decades shall be as follows: Year Net percentage increase in population Expected Population 1980 1990 2000 32.4 – 8.8 = 23.6 23.6 – 8.8 = 14.8 14.8 – 8.8 = 6.0 22500 + 23.6/100 x 22500 = 27810 27810 + 14.8/100 x 27810 = 31926 31926 + 6/100 x 31926 = 33842
PROBLEM 40 Year Population 1940 1950 1960 1970 8250 13000 18500 24500 Based on the data given below, estimate the population growth for the resident A for the year 1980 and 1990 using the DECREASING RATE OF GROWTH METHOD
Activity (Jigsaw) 41 Students are placed into "home groups" and "expert groups" and are each assigned a different subtopic within the same general topic. Students work on researching their topics with others who have the same topic (their expert group) and then return back to their home group to teach them about their topic. All the information come together to form a complete design exercise
YEAR POPULATION 2012 55000 2013 58000 2014 65000 42 Calculate the expected population in 2020 by arithmetic increase method.