---
title: "Storage capacity, depth and water at outlet"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:storage-capacity-wheat-field-efficiency"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/storage-capacity-wheat-field-efficiency
site: BCT Engineering
---

# Storage capacity, depth and water at outlet

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2079 Chaitra, 6 marks
- Video: [Find Water Capacity, Requirement Water Depth, and,, Amount of the water needed at the canal outlet](https://www.youtube.com/watch?v=yWTLc9USyCc) — E.T.E. Suman
- 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/storage-capacity-wheat-field-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

Wheat is to be grown in a field having field capacity of 27% and permanent wilting point of 13%. Find the storage capacity in 80 cm depth of soil, if dry unit wt. of soil is 14.72 kN/m³. If irrigation water is to be supplied when soil moisture falls to 18%, find the water depth required to be supplied to the field if the field application efficiency is 80%. What is the amount of water needed at the canal outlet if the water lost in the water courses and the field channel is 15% of the outlet discharge?

## Short answer

Available moisture storage capacity in 80 cm is 168.06 mm. When moisture falls to 18%, net depth to restore FC is 108.04 mm; field supply is 135.05 mm (80% efficiency) and outlet requirement is 158.88 mm (15% conveyance loss).

## Answer

**Given:** FC = 27%, PWP = 13%, depth = 80 cm = 800 mm, $\gamma_d = 14.72$ kN/m$^3$, $\gamma_w=9.81$ kN/m$^3$, $\eta_a = 80\%$, watercourse and field channel loss = 15% of outlet discharge.

$$
\frac{\gamma_d}{\gamma_w} = \frac{14.72}{9.81} = 1.501
$$

**1. Storage capacity (available moisture) in 80 cm of soil**

$$
d_{AM} = \frac{27-13}{100}\times 1.501\times 800 = 168.06\ \text{mm}
$$

(The total water held at field capacity is $0.27\times1.501\times800 = 324.11$ mm, of which 168.06 mm is available to the crop.)

**2. Depth of water to be supplied when moisture falls to 18%** (net depth, to bring the soil back to FC)

$$
d_{net} = \frac{27-18}{100}\times 1.501\times 800 = 108.04\ \text{mm}
$$

With $\eta_a = 80\%$, the depth at the field:

$$
d_{field} = \frac{108.04}{0.80} = 135.05\ \text{mm}
$$

**3. Water needed at the canal outlet.** 15% of the outlet water is lost, so 85% reaches the field:

$$
d_{outlet} = \frac{135.05}{0.85} = 158.88\ \text{mm}\ (1589\ \text{m}^3/\text{ha})
$$

**Answer: storage capacity = 168.06 mm; depth to be supplied = 135.05 mm at the field; water at the outlet = 158.88 mm.**

## Check yourself

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

1. gamma_d/gamma_w is
   A. 14.72 x 9.81
   B. 1.0
   C. 14.72/9.81 = 1.501
   D. 0.67
   Answer: C. Dimensionless ratio. (Key point: Unit weight ratio)

2. Storage capacity in 80 cm is
   A. 324.11 mm
   B. 108.04 mm
   C. 135.05 mm
   D. 168.06 mm
   Answer: D. 0.14 x 1.501 x 800. (Key point: Storage capacity)

3. Net depth when moisture falls to 18% is
   A. 108.04 mm
   B. 168.06 mm
   C. 135.05 mm
   D. 158.88 mm
   Answer: A. (27-18)% to bring back to FC. (Key point: Net depth)

4. Outlet depth with 15% loss after field depth 135.05 is
   A. 155.3 mm
   B. 158.88 mm
   C. 116 mm
   D. 135.05 mm
   Answer: B. 135.05/0.85. (Key point: Outlet water)

## Related in Irrigation and Drainage Engineering

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