---
title: "Reservoir capacity with irrigated areas"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:reservoir-capacity-areas-79j"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/reservoir-capacity-areas-79j
site: BCT Engineering
---

# Reservoir capacity with irrigated areas

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2079 Jestha, 7 marks
- Video: [Water requirement of crop Numerical || Sum of Discharge Required and Required capacity of Reservoir](https://www.youtube.com/watch?v=E-VERQbA-dw) — Professor Nidhi Trivedi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/reservoir-capacity-areas-79j
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

The base period, duty of water, and area under irrigation for various crops under a canal system are given in the table below. If the losses in the reservoir and canals are respectively 15%, 25%, determine the reservoir capacity.

| Crop | Wheat | Sugarcane | Cotton | Rice | Vegetables |
|---|---|---|---|---|---|
| Base period B (Days) | 120 | 320 | 180 | 120 | 120 |
| Duty, D (ha/cumec) | 1800 | 1600 | 1500 | 800 | 700 |
| Area irrigated (ha) | 15000 | 10000 | 5000 | 7500 | 5000 |

## Short answer

Compute field volume for each crop as V = (A/D) × B × 86400. Total field volume 482.30 Mm³. Divide by (1–0.25) for canal loss → 643.06 Mm³, then by (1–0.15) for reservoir loss. Reservoir capacity ≈ 756.54 million m³.

## Answer

**Given:** reservoir loss = 15%, canal loss = 25%; areas, base periods and duties as in the table.

**Method.** For each crop, discharge needed at the field $Q = A/D$; volume for the base period $V = Q\times B\times 86400$ m$^3$. The losses are taken as a fraction of the water entering that part of the system, so field volume is divided by $(1-\text{loss})$.

| Crop | Area (ha) | Duty (ha/cumec) | Q = A/D (cumec) | B (days) | Volume = Q x B x 86400 (Mm$^3$) |
|---|---|---|---|---|---|
| Wheat | 15,000 | 1,800 | 8.333 | 120 | 86.400 |
| Sugarcane | 10,000 | 1,600 | 6.250 | 320 | 172.800 |
| Cotton | 5,000 | 1,500 | 3.333 | 180 | 51.840 |
| Rice | 7,500 | 800 | 9.375 | 120 | 97.200 |
| Vegetables | 5,000 | 700 | 7.143 | 120 | 74.057 |

**Total volume at the field:** $V_f = 482.297\times10^6$ m$^3$

**Allow canal loss (25%):**

$$
V_{canal\ head} = \frac{482.297}{1-0.25} = 643.063\times10^6\ \text{m}^3
$$

**Allow reservoir loss (15%):**

$$
V_{reservoir} = \frac{643.063}{1-0.15} = 756.545\times10^6\ \text{m}^3
$$

**Answer: reservoir capacity $\approx$ 756.54 million m$^3$ (0.7565 Gm$^3$).**

## Check yourself

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

1. Which method gives the water volume needed by each crop at the field?
   A. Q = A/D, then V = Q x B x 86400
   B. V = A x D x B
   C. V = D/A x B
   D. V = Q x B in hectares
   Answer: A. Discharge at the field times the base period in seconds gives the volume in m^3. (Key point: Volume per crop)

2. Which crop needs the largest volume and why?
   A. Wheat, because it has the largest area
   B. Sugarcane, 172.8 Mm^3, because its base period (320 days) is longest
   C. Rice, because its duty is the lowest
   D. Cotton, because its base period is 180 days
   Answer: B. Sugarcane: 10000/1600 = 6.25 cumec for 320 days gives 172.8 Mm^3, the largest. (Key point: Which crop dominates)

3. How are the canal and reservoir losses applied to the field volume?
   A. Multiply by (1 - loss) each time
   B. Add the loss percentages to the volume once
   C. Divide successively by (1 - canal loss) and (1 - reservoir loss)
   D. Subtract the losses from the volume
   Answer: C. Losses are fractions of water entering each part, so each stage needs volume/(1 - loss). (Key point: Loss correction)

4. What is the reservoir capacity?
   A. About 643.1 million m^3 (canal loss only)
   B. About 482.3 million m^3 (field volume)
   C. About 693.3 million m^3 (losses added)
   D. About 756.5 million m^3
   Answer: D. Divide 482.297 by 0.75 to get 643.06, then by 0.85 to get 756.54 Mm^3. (Key point: Final capacity)

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