--- title: "Population 2088 by three methods, design year demand" subject: "Water Supply Engineering" chapter: 3 key: "semester_5/water_supply_engineering:pop-k72" source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pop-k72 site: BCT Engineering --- # Population 2088 by three methods, design year demand Water Supply Engineering · Chapter 3: Quantity of Water - Asked in: 2072 Kartik, 8 marks - Video: [water demand || calculation of total water demand || water supply engineering](https://www.youtube.com/watch?v=jGZDjVw_IHA) — Rabi 100 - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pop-k72 - Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-3 ## Question Determine the population of the town in the year 2088 by (a) Arithmetical increase method, (b) Geometrical increase method and (c) Decreased rate of growth method. The census population of the city is as follows: | Year | 2068 | 2058 | 2048 | 2038 | 2028 | |---|---|---|---|---|---| | Population | 65,500 | 57,000 | 47,000 | 37,000 | 29,000 | Calculate the design year and total water demand for a Nepalese town assuming the per capita demand of 120 lpcd. ## Short answer Forecasts for 2088: arithmetic 83750, geometric 98438, decreasing rate 77186; adopt mean 86458. At 120 lpcd, total demand is 10375 m3/d (120.1 l/s). ## Answer 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** | Decade | Increase | Growth rate (%) | Decrease in rate (%) | |---|---|---|---| | 2028-2038 | 8,000 | 27.59 | - | | 2038-2048 | 10,000 | 27.03 | 0.56 | | 2048-2058 | 10,000 | 21.28 | 5.75 | | 2058-2068 | 8,500 | 14.91 | 6.36 | Average decrease in rate $= 4.225\%$ per decade. | Year | Growth rate (%) | Population | |---|---|---| | 2078 | 10.69 | 72,500 | | 2088 | 6.46 | 77,186 | $P_{2088} = 77{,}186$ | Method | Population (2088) | |---|---| | (a) Arithmetical increase | 83,750 | | (b) Geometrical increase | 98,438 | | (c) Decreased rate of growth | 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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Data are given in descending year. What is done and what is n for 2088? A. Use as given; n = 20 decades B. Arrange ascending (2028 to 2068); n = 2 decades C. Arrange ascending; n = 4 decades D. Use as given; n = 2 decades but with negative increase Answer: B. Increases must be taken in time order; 2088 is 20 years, i.e. 2 decades, after the last census 2068. (Key point: Order and n) 2. What is the average geometric growth rate per decade? A. (65,500/29,000)^(1/5) - 1 B. (65,500-29,000)/29,000/4 C. (65,500/29,000)^(1/4) - 1 = 22.592% D. (65,500/29,000) - 1 Answer: C. There are 4 decade intervals in the 5 census records. (Key point: Geometric rate) 3. How is the 2078 growth rate 10.69% obtained? A. Last decade rate 14.91% minus average decrease 4.225% B. 14.91% plus 4.225% C. Average of all growth rates D. 27.59% minus 4.225% Answer: A. Each future decade's rate = previous rate - average decrease in rate. (Key point: Decreasing rate step) 4. How are the design population and total demand found? A. Geometrical value x 120 lpcd B. Mean of three methods (86,458) x 120 lpcd = 10,375 m^3/day C. Mean of three methods x 120 x 1.15 D. Arithmetical value x 120 lpcd Answer: B. The mean is adopted and Q = P x per capita demand. 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