---
title: "Reservoir capacity: rice, wheat, sugarcane"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:reservoir-capacity-areas-71bh"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/reservoir-capacity-areas-71bh
site: BCT Engineering
---

# Reservoir capacity: rice, wheat, sugarcane

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2071 Bhadra, 8 marks
- Video: [Irrigation Reservoir Capacity 🔥|Numerical Solution|Irrigation Efficiency |Duty, Delta & Base Period🔥](https://www.youtube.com/watch?v=dFl4bY-ab6A) — DIPAK  MANDAL
- 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-71bh
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

The base period, intensity of irrigation and duty of various crops under a canal irrigation system are given in the table below. Find the reservoir capacity if the canal losses are 18% and reservoir losses are 14%.

| Crop | Base periods (days) | Duty at the field (Ha/Cumecs) | Area under the crop (hectares) |
|---|---|---|---|
| Rice | 120 | 850 | 3000 |
| Wheat | 120 | 1700 | 4500 |
| Sugarcane | 360 | 750 | 5400 |
| Vegetables | 120 | 650 | 1200 |
| Cotton | 200 | 1300 | 2200 |

## Short answer

For each crop, Q = A/D, volume = Q × B × 86400. Total field volume 336.37 Mm³. After 18% canal loss: 410.21 Mm³; after 14% reservoir loss: 476.99 Mm³. Reservoir capacity ≈ 476.99 million m³.

## Answer

**Given:** canal loss = 18%, reservoir loss = 14%; area, base period and duty of each crop as in the table (the duty is at the field).

**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$^3$) |
|---|---|---|---|---|---|
| Rice | 3,000 | 850 | 3.529 | 120 | 36.593 |
| Wheat | 4,500 | 1,700 | 2.647 | 120 | 27.445 |
| Sugarcane | 5,400 | 750 | 7.200 | 360 | 223.949 |
| Vegetables | 1,200 | 650 | 1.846 | 120 | 19.141 |
| Cotton | 2,200 | 1,300 | 1.692 | 200 | 29.243 |

**Total volume at the field:** $V_f = 336.370\times10^6$ m$^3$

**Allow canal loss (18%):**

$$
V_{canal\ head} = \frac{336.370}{1-0.18} = 410.208\times10^6\ \text{m}^3
$$

**Allow reservoir loss (14%):**

$$
V_{reservoir} = \frac{410.208}{1-0.14} = 476.986\times10^6\ \text{m}^3
$$

**Answer: reservoir capacity $\approx$ 476.99 million m$^3$ (0.4770 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. Rice, because it has the lowest duty
   B. Sugarcane, 223.9 Mm^3, because of its 360-day base period and 7.2 cumec demand
   C. Wheat, because its base period is 120 days
   D. Cotton, because it has the highest duty
   Answer: B. Sugarcane: 5400/750 = 7.2 cumec for 360 days gives 223.9 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 410.2 million m^3 (canal loss only)
   B. About 336.4 million m^3 (field volume)
   C. About 452.5 million m^3 (losses multiplied)
   D. About 477.0 million m^3
   Answer: D. 336.370/0.82 = 410.21, then /0.86 = 476.99 Mm^3. (Key point: Final capacity)

## Related in Irrigation and Drainage Engineering

- [Reservoir capacity with intensity table](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Areservoir-capacity-40000ha-table-78p) — Irrigation and Drainage Engineering, chapter 2
- [Reservoir storage for wheat, rice, maize](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Areservoir-supply-40000ha-wheat-rice-maize) — Irrigation and Drainage Engineering, chapter 2
- [Reservoir capacity with irrigated areas](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Areservoir-capacity-areas-79j) — Irrigation and Drainage Engineering, chapter 2
- [Duty, delta, base period, crop period](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Adefine-duty-delta-base-crop-period) — Irrigation and Drainage Engineering, chapter 2
- [Outlet discharge from kor table (rice, wheat)](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Aoutlet-discharge-rice-wheat-kor) — Irrigation and Drainage Engineering, chapter 2
- [Duty and minor discharge, wheat 400 ha](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Aminor-wheat-400ha-duty-discharge) — Irrigation and Drainage Engineering, chapter 2