Calculation of Soil Moisture (With Example)

After reading this article you will learn about the calculation of soil moisture.

The soil moisture calculations in percentages on a weight basis have been commonly used, but this does not give a true picture of soil-moisture relationships. Two soils may have similar moisture content on a weight basis but not on a volume basis.

Calculations on a volume basis are more meaningful and practical because water is retained in the soil within a given volume and plant roots also absorb moisture from a volume of soil.

To change percentage soil moisture on a weight basis to percentage soil moisture on a volume basis, the following calculation is given:

% moisture by volume = % moisture by weight × bulk density of the soil.

When calculated for a depth of 12 inches, this value indicates water in inches per foot depth. On an acre or hectare basis, this value is written as acre or hectare inches per acre or hectare foot of soil.

Problem 1:

Calculate

(a) The total water presently contained in the top 30 cm,

(b) The depth to which 27.5 mm (1.1 inch) of irrigation would wet this uniform soil and

(c) The available water the soil contains in the top 30 cm when the soil is at field capacity. The measurement of the soil is as follows:

Present water content – 18%

Water content at field capacity – 23%

Permanent wilting percentage – 9%

Bulk density of 0-30 cm depth surface soil – 1.3 g/c.c.

Solution:

(a) Depth of water (d_w)

= 1.3 ×18/100 × 30 cm

= 7.0 cm

So, the depth of total water present in the top 30 cm of soil is 7.0 cm.

(b) To calculate the depth of wetting by a 27.5 mm (1.1 inch) irrigation, the following equation is substituted.

or, 27.5 mm = 1.3 ×23 – 18/100 × d_s

or, 27.5 mm = 1.3 ×5/100 × d_s

or, d_s = 27.5mm × 100/1.3 × 5

= 42.3 mm

So 42.3 mm depth of soil will be wetted.

(c) To calculate the total possible plant available water in the top 30 cm, when the soil is wetted equals field capacity minus permanent wilting percentage. So the plant available water (d_w) is

d_w= 1.3 ×23 – 9/100 × 30 cm

= 1.3 × 14/100× 30 cm

= 1.3 × 0.14 × 30 cm

= 5.46 cm.

So the top 30 cm of soil contains 5.46 cm of available water.

Problem 2:

A soil sample taken from a field is placed in the aluminium box, weighed, dried in an oven at 105°C (221°F) and reweighed.

The measurements are as follows:

Weight of moist soil plus aluminium box = 159 g

Weight of oven dried soil plus aluminium box = 134 g

Weight of empty aluminium box = 41 g

Calculate the moisture content of the soil.

Solution:

Weight of moist soil only = (159 — 41) g

= 118g

Weight of oven dried soil only = (134 – 41) g

= 93 g

= 118 – 93/93 × 100

= 26.9