---
title: "Field capacity from applied water data"
subject: "Irrigation and Drainage Engineering"
chapter: 2
key: "semester_6/irrigation_and_drainage_engineering:field-capacity-from-applied-water"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/field-capacity-from-applied-water
site: BCT Engineering
---

# Field capacity from applied water data

Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements

- Asked in: 2062 Kartik (old course), 8 marks
- Video: [Find the field capacity of soil | Numerical | Irrigation Engineering | SBK Concept](https://www.youtube.com/watch?v=3iJN3B0_ONE) — SBK Concept
- 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-capacity-from-applied-water
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2

## Question

Determine the field capacity of the soil from following data: depth of root zone = 1.5 m; present water content = 5%; dry density of soil = 1600 kg/m³; water applied to the soil = 700 m³; water loss due to evaporation = 10%; area of the plot = 1 hectare.

## Short answer

Effective water stored in soil is 630 m³, equivalent to a depth of 63 mm. The increase in moisture content is 2.625%, so field capacity is 7.62%.

## Answer

**Given:** root zone depth $D = 1.5$ m, present moisture $= 5\%$ (by dry weight), $\gamma_d = 1600$ kg/m$^3$ so $\gamma_d/\gamma_w=1.6$, water applied $= 700$ m$^3$, evaporation loss $= 10\%$, area $= 1$ ha $= 10^4$ m$^2$. The applied water brings the soil exactly to field capacity.

**Step 1: Water actually stored in the soil**

$$
V = 700\times(1-0.10) = 630\ \text{m}^3
$$

**Step 2: Depth of water stored**

$$
d = \frac{630}{10^4} = 0.063\ \text{m} = 63\ \text{mm}
$$

**Step 3: Increase in moisture content.** Using $d = \Delta w\cdot\dfrac{\gamma_d}{\gamma_w}\cdot D$:

$$
\Delta w = \frac{d}{(\gamma_d/\gamma_w)\,D} = \frac{0.063}{1.6\times1.5} = 0.02625 = 2.625\%
$$

**Step 4: Field capacity**

$$
FC = 5 + 2.625 = 7.625\%
$$

**Answer: field capacity $\approx$ 7.62%.**

## Check yourself

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

1. Of the 700 m^3 applied, how much is taken as stored in the soil?
   A. 630 m^3, after removing the 10% evaporation loss
   B. 700 m^3, since evaporation is ignored
   C. 770 m^3, since losses are added
   D. 70 m^3, the evaporation loss itself
   Answer: A. Only 90% of the applied water reaches the soil: 700 x 0.9 = 630 m^3. (Key point: Stored water)

2. What depth of water is stored over the 1 ha plot?
   A. 630 mm
   B. 63 mm
   C. 7 mm
   D. 0.63 mm
   Answer: B. Depth = volume/area = 630/10000 = 0.063 m = 63 mm. (Key point: Depth of water)

3. Which relation gives the increase in moisture content from the stored depth?
   A. d = delta w x gamma_w x D / gamma_d
   B. d = delta w x D only
   C. d = delta w x (gamma_d/gamma_w) x D
   D. d = delta w x (gamma_d/gamma_w) / D
   Answer: C. Depth of water equals moisture fraction times relative dry density times root zone depth, so delta w = 0.063/(1.6 x 1.5) = 2.625%. (Key point: Moisture increase formula)

4. How is field capacity obtained once the increase is known?
   A. Subtract the increase from present moisture: 2.375%
   B. Use only the increase: 2.625%
   C. Multiply: 5 x 2.625 = 13.1%
   D. Add the increase to the present moisture: 5 + 2.625 = 7.625%
   Answer: D. The applied water brings the soil exactly to field capacity, so FC = present moisture + increase. (Key point: Field capacity)

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