---
title: "Storage tank capacity and water level, 20000 L/day"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:reservoir-level-river-20000"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/reservoir-level-river-20000
site: BCT Engineering
---

# Storage tank capacity and water level, 20000 L/day

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2078 Kartik, 8 marks; 2080 Baisakh, 8 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/reservoir-level-river-20000
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

A rural area has a design year demand of water 20000 liters per day. The demand is met by a continuous system of supply from a river source with measured safe yield of 0.25 lps. The consumption pattern is as follows: estimate the capacity of the storage (balancing) tank. What will be the water level in the tank at a given time (10 / 6 PM) if the cross section area of the tank is "A" ("X")?

| Time (Hour) | 5-7 | 7-12 | 12-17 | 17-19 | 19-5 |
|---|---|---|---|---|---|
| Consumption (%) | 25 | 35 | 20 | 20 | 0 |

## Short answer

Mass curve analysis with continuous inflow of 0.9 m³/h gives a balancing storage of 9 m³. At 10:00 the water depth is 4.3/A m and at 6 PM it is 2.7/A m, where A is the tank cross-sectional area in m².

## Answer

#### Given data
- Design-year demand = 20,000 L/day = 20 m³/day
- Safe yield of river = 0.25 lps = 0.25 × 86,400 = 21,600 L/day = 21.6 m³/day, i.e. 0.9 m³/h continuous inflow
- Consumption: 5–7 h 25%, 7–12 h 35%, 12–17 h 20%, 17–19 h 20%, 19–5 h 0%

#### Check of source and need of storage
The source yield (21.6 m³/day) exceeds the demand (20 m³/day), so the source is adequate for the day. But the demand rate in 5–7 h is 5/2 = 2.5 m³/h and in 7–12 h it is 1.4 m³/h, both more than the inflow of 0.9 m³/h. A balancing tank is therefore necessary.


#### Mass-curve table (analytical method)
Inflow: 0.9 m³/h for 24 h. Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

| Time | Demand (m³) | Inflow (m³) | Inflow − demand | Cumulative |
|---|---|---|---|---|
| 05:00–07:00 | 5.00 | 1.80 | -3.20 | -3.20 |
| 07:00–12:00 | 7.00 | 4.50 | -2.50 | -5.70 |
| 12:00–17:00 | 4.00 | 4.50 | +0.50 | -5.20 |
| 17:00–19:00 | 4.00 | 1.80 | -2.20 | -7.40 |
| 19:00–05:00 | 0.00 | 9.00 | +9.00 | +1.60 |

Total demand = 20.00 m³; total inflow = 21.60 m³.

#### Capacity
- Maximum cumulative surplus = 1.60 m³
- Maximum cumulative deficit = 7.40 m³
- Balancing capacity = max. surplus + max. deficit = 1.60 + 7.40 = **9.00 m³**
- Because the inflow in a day (21.60 m³) is more than the demand, the extra water simply overflows once the tank is full. The minimum storage that still meets every demand is the largest draw-down, 7.40 m³. The textbook value of 9.00 m³ is on the safe side.

#### Water level in the tank
With the 9.00 m³ tank, the night inflow (9.00 m³) fills it to the top by 05:00, and any extra overflows. So storage at any time = 9.00 m³ + cumulative value (counted from 05:00). The lowest storage is 9.00 − 7.40 = 1.60 m³, at 19:00.

- At 10:00: cumulative = −3.20 + (3/5)(−2.50) = −4.70, storage = 9.00 − 4.70 = 4.30 m³
- At 18:00 (6 PM): cumulative = −5.20 + (1/2)(−2.20) = −6.30, storage = 9.00 − 6.30 = 2.70 m³

For a tank of plan area $A$ (m²): level at 10:00 $= 4.30/A$ m and level at 6 PM $= 2.70/A$ m. For example, with a 3 m × 3 m tank ($A = 9$ m²), these are 0.48 m and 0.30 m above the floor. (If only the minimum 7.40 m³ tank were built, it would be full at 05:00 and empty at 19:00, giving $2.70/A$ and $1.10/A$.)

**Answer: Balancing tank capacity ≈ 9 m³ (practical 10 m³). With a 9 m³ tank, the water depth is 4.3/A m at 10:00 and 2.7/A m at 6 PM.**

## Check yourself

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

1. The yield (21.6 m3/day) exceeds the demand (20 m3/day). Why is a tank still needed?
   A. Because the source is unreliable
   B. Because demand rates in 5-7 h and 7-12 h exceed the inflow of 0.9 m3/h
   C. Because the yield is below demand
   D. Because the demand is 0% at night
   Answer: B. Peak-period demand rates (2.5 and 1.4 m3/h) are greater than the steady inflow, so storage is required. (Key point: Need of storage)

2. What is the continuous inflow of the source in m3/h?
   A. 0.25
   B. 0.9
   C. 21.6
   D. 2.5
   Answer: B. 0.25 lps x 3600 = 900 L/h = 0.9 m3/h. (Key point: Inflow rate)

3. How is the capacity 9.00 m3 obtained?
   A. 1.60 + 7.40
   B. 7.40 - 1.60
   C. 20 - 9
   D. 21.6 - 12.6
   Answer: A. Maximum surplus (1.60) plus maximum deficit (7.40) gives 9.00 m3. (Key point: Capacity)

4. At 10:00 the storage is 2.70 m3. For a tank of plan area A, what is the water level?
   A. 2.70 x A
   B. 2.70 / A
   C. 9 / A
   D. 7.40 / A
   Answer: B. Depth = volume / area = 2.70/A m. (Key point: Water level)

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