---
title: "Field channel discharge for gram and wheat"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:field-channel-gram-wheat-2000ha"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/field-channel-gram-wheat-2000ha
site: BCT Engineering
---

# Field channel discharge for gram and wheat

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2068 Chaitra (old course), 6 marks
- 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/field-channel-gram-wheat-2000ha
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

A field channel has a culturable command area of 2000 hectares. The intensity of irrigation for gram is 30% and for wheat is 50%. Gram has a kor period of 18 days and kor depth of 12 cm, while wheat has a kor period of 15 days & kor depth of 15 cm. Calculate the discharge of the field channel.

## Short answer

Gram area 600 ha, Q = (600×10⁴×0.12)/(18×86400) = 0.463 cumec. Wheat area 1000 ha, Q = (1000×10⁴×0.15)/(15×86400) = 1.157 cumec. Both are rabi crops, so total design discharge = 0.463 + 1.157 = 1.62 cumec.

## Answer

**Formula:** $Q = \dfrac{A\,d}{T\times 86400}$ ($A$ in m$^2$, $d$ in m, $T$ = kor period in days).

**Areas:** gram $= 0.30\times 2000 = 600$ ha; wheat $= 0.50\times 2000 = 1000$ ha.

**Gram** ($T = 18$ days, $d = 0.12$ m):

$$
Q_g = \frac{6000000\times 0.12}{18\times 86400} = 0.463\ \text{m}^3/\text{s}
$$

**Wheat** ($T = 15$ days, $d = 0.15$ m):

$$
Q_w = \frac{10000000\times 0.15}{15\times 86400} = 1.157\ \text{m}^3/\text{s}
$$

Gram and wheat are both rabi crops, grown at the same time on different parts of the command area (30% + 50% = 80% of the CCA). Their kor waterings can therefore coincide, so the field channel must be able to carry both demands together:

$$
Q = Q_g + Q_w = 0.463 + 1.157 = 1.620\ \text{m}^3/\text{s}
$$

**Answer: design discharge of the field channel = 1.62 cumec** (gram 0.463 cumec + wheat 1.157 cumec).

## Check yourself

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

1. What areas of gram and wheat are watered?
   A. 600 ha gram and 1000 ha wheat
   B. 300 ha and 500 ha
   C. 1200 ha and 800 ha
   D. 600 ha and 1400 ha
   Answer: A. Intensity x CCA: 0.30 x 2000 and 0.50 x 2000. (Key point: Crop areas)

2. What is the wheat kor discharge (1000 ha, 15 days, 15 cm)?
   A. 0.463 cumec
   B. 1.157 cumec
   C. 1.62 cumec
   D. 0.116 cumec
   Answer: B. Q = 1e7 x 0.15/(15 x 86400) = 1.157 cumec. (Key point: Wheat discharge)

3. Which discharge is adopted and why?
   A. 1.62 cumec, since both are always irrigated together
   B. 0.46 cumec, the smaller one
   C. 1.16 cumec, since kor waterings fall at different times and wheat governs
   D. 0.81 cumec, the average
   Answer: C. The two kor waterings do not coincide, so the larger value is used. (Key point: Design discharge)

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