## How to Calculate and Solve for Capacity of the Sprinkler System | Irrigation Water Requirement

The image above represents capacity of the sprinkler system.

To compute for the capacity of the sprinkler system, four essential parameters are needed and these parameters are Area to be Irrigated (A), Net Depth of Water Application (d), Number of Drop allowed for the Completion of One Irrigation (F), Number of Actual Operating Hours per Day (H) and Water Application Efficiency (E).

The formula for calculating capacity of the sprinkler system:

Q = 27800 . A . d / F . H . E

Where:

Q = Capacity of the Sprinkler System
A = Area to be Irrigated
d = Net Depth of Water Application
F = Number of Drop allowed for the Completion of One Irrigation
H = Number of Actual Operating Hours per Day
E = Water Application Efficiency

Let’s solve an example;
Find the capacity of the sprinkler system when the area to be irrigated is 2, the net depth of water application is 8, the number of drop allowed for the completion of one irrigation is 10, the number of actual operating hours per day is 6 and the water application efficiency is 4.

This implies that;

A = Area to be Irrigated = 2
d = Net Depth of Water Application = 8
F = Number of Drop allowed for the Completion of One Irrigation = 10
H = Number of Actual Operating Hours per Day = 6
E = Water Application Efficiency = 4

Q = 27800 . A . d / F . H . E
Q = 27800 . (2) . (8) / (10) . (6) . (4)
Q = 444800 / 240
Q = 1853.3

Therefore, the capacity of the sprinkler system is 1853.3.

Calculating for the Area to be Irrigated when the Capacity of the Sprinkler System, the Net Depth of Water Application, the Number of drop Allowed for the Completion of One Irrgiation, the Number of Actual Operating Hours per Day and the Water Application Efficiency is Given.

A = Q x F . H . E / 27800 . d

Where;

A = Area to be Irrigated
Q = Capacity of the Sprinkler System
d = Net Depth of Water Application
F = Number of Drop allowed for the Completion of One Irrigation
H = Number of Actual Operating Hours per Day
E = Water Application Efficiency

Let’s Solve an example;
Find the area to be irrigated when the capacity of the sprinkler system is 4, the net depth of water application is 6, number of drop allowed for the competion of one irrigation is 2, number of actual operating hours per day is 3 and the water application efficiency is 8.

This implies that;

Q = Capacity of the Sprinkler System = 4
d = Net Depth of Water Application = 6
F = Number of Drop allowed for the Completion of One Irrigation = 2
H = Number of Actual Operating Hours per Day = 3
E = Water Application Efficiency = 8

A = Q x F . H . E / 27800 . d
A = 4 x 2 . 3 . 8 / 27800 . 6
A = 192 / 166800
A = 0.00115

Therefore, the area to be irrigated is 0.00115.

## How to Calculate and Solve for Water Application Efficiency | Methods of Application of Water

The image above represents water application efficiency.

To compute for water application efficiency, two essential parameters are needed and these parameters are Water stored in the Root Zone of the Plant (wsand Water delivered to the Irrigated Plot (wf).

The formula for calculating water application efficiency:

Ea = ws / wf

Where:

Ea = Water Application Efficiency
ws = Water stored in the Root Zone of the Plant
wf = Water delivered to the Irrigated Plot

Let’s solve an example;
Find the water application efficiency when the water stored in the root zone of the plant is 45 and the water delivered to the irrigated plot is 9.

This implies that;

ws = Water stored in the Root Zone of the Plant = 45
wf = Water delivered to the Irrigated Plot = 9

Ea = ws / wf
Ea = 45 / 9
Ea = 5

Therefore, the water application efficiency is 5.

Calculating the Water Stored in the Root Zone of the Plant when the Water Application Efficiency and the Water Delivered to the Irrigated Plot is Given.

ws = Ea x wf

Where;

ws = Water stored in the Root Zone of the Plant
Ea = Water Application Efficiency
wf = Water delivered to the Irrigated Plot

Let’s solve an example;
Find the water stored in the root zone of the plant when the water application efficiency is 15 and the water delivered to the irrigated plot is 4.

This implies that;

Ea = Water Application Efficiency = 15
wf = Water delivered to the Irrigated Plot = 4

ws = Ea x wf
ws = 15 x 4
ws = 60

Therefore, the water stored in the root zone of the plant is 60.