---
title: "Population from spring yield of 5 L/s"
subject: "Water Supply Engineering"
chapter: 3
key: "semester_5/water_supply_engineering:pop-s71"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pop-s71
site: BCT Engineering
---

# Population from spring yield of 5 L/s

Water Supply Engineering · Chapter 3: Quantity of Water

- Asked in: 2071 Shrawan, 8 marks
- Video: [Gravity Water Supply System | Unit: 8 | Water Supply Engineering | Prashant YT |CTEVT | 4th sem |](https://www.youtube.com/watch?v=GlKGf6Tma2U) — Prashant YT
- 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-s71
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-3

## Question

Safe yield of a proposed spring is 5 liter per second 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 population growth rate is 1.7% per annum.

## Short answer

Spring yield 5 l/s = 432000 l/d; at 65 lpcd it serves design population 6646. Current population = 6646/(1.017)^20 = 4744.

## Answer

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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The spring gives 5 l/s. What is the daily yield?
   A. 5,000 l/day
   B. 18,000 l/day
   C. 432,000 l/day
   D. 43,200 l/day
   Answer: C. 5 x 86,400 s = 432,000 l/day. (Key point: Daily yield)

2. The yield 432,000 l/day divided by 65 lpcd gives which population?
   A. The design-year population (end of 20 years)
   B. The current population
   C. The population at the start of base period
   D. The peak hour population
   Answer: A. The source must meet demand at the end of design period, so 6,646 is the design population. (Key point: Which population the yield serves)

3. How is the current population found from 6,646?
   A. Multiply by (1.017)^20
   B. Divide by 1.017 x 20
   C. Subtract 1.7% of 6,646
   D. Divide by (1.017)^20
   Answer: D. P_d = P_0(1+r)^n, so P_0 = P_d/(1.017)^20 = 4,744. (Key point: Current population)

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