How to Determine the Frequency of Irrigation – Problem & Solution

7th June 202118th October 2021 Malay Sautya No Comments

How to Determine the Frequency of Irrigation

Mathematical Problem

Following data are given, determine the frequency of irrigation

Field capacity of soil = 45 %
Permanent wilting point = 19 %
Density of soil = 1.5 g/cm³
Depth of root zone = 65 cm
Daily consumptive use of water = 18 mm

Solution:

Available moisture = Field Capacity – Permanent wilting point = 45 – 19 = 26

Let readily available moisture = is 75 % of the available moisture,

So, readily available moisture is = 26 × 0.75 = 19.5 %

Optimum moisture content = 45 – 19.5 = 25.5 %

Now, by applying irrigation water the moisture content is to be raised from 25.5 % to 45 %.

Calculate the depth of water to be applied in each watering, by using the following equation:

$D_{w} = \frac{W_{s}\times d}{W_{w}}\times [F_{c}-M_{0}]$

Where,

W_s = Unit Weight of soil or density of soil = 1.5 g/cm³
W_w = Density of water = 1g/cm³
d = Depth of root zone = 65 cm
F_c = Field capacity = 45 %
M₀ = Optimum moisture content = 25.5 %

After putting the all value,

$D_{w} = \frac{1.5\times 65}{1}\times\left [ 0.45 - 0.255 \right ]$

$Or, D_{w} = 97.5 \times 0.195 = 19.01$ cm

Now, we can calculate the frequency of irrigation by using the following equation

$f_{w} = \frac{D_{w}}{C_{u}}$

Where,

f_w = Frequency of watering or frequency of irrigation
C_u = Daily consumptive use of water =
D_w = Depth of water to be applied in each watering

After puting the all value, the frequency of irrigation

$f_{w} = \frac{19.01}{1.8} = 10.56 = 11 days (say)$

Related

Irrigation Engineering, Mathematical Example

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