Author:

Blaine R. Hanson
Irrigation and Drainage Specialist
University of California, Davis

Drought Tip 92-29 is a publication series developed as a cooperative effort by the following organizations:

California Department of Water Resources - Water Conservation Office
Department of Land, Air and Water Resources University of California
USDA Drought Response Office
USDA Soil Conservation Service If you have comments or suggestions, please email lawrweb@ucdavis.edu.

Last reviewed December 19, 2002

Drought Tip 92-29
Irrigation Water Management Made Simple

Effective water management requires knowing the relationships between flowrate into the field, acres irrigated, irrigation times, and the amount of applied water. The following equation can be used to provide this information:

Q x T = 449 x A x D

where Q = flow rate of water being applied to the field in gallons per minute; T = actual hours used to irrigate the field; A = acres irrigated; D = inches of water applied.

The following questions can be answered using this equation:

How many inches of water are applied during an irrigation?

D = (Q x T)/(449 x A)

Example 1: How many inches are applied to 40 acres where the pump flow rate is 300 gpm and irrigation time is 22 hours per day for 10 days?

Q = 300 gpm;
A = 40 acres;
T = 22 hours per day x 10 days = 220 hours;
D = (300 gpm x 220 hours)/(449 x 40 acres) = 3.7 inches.

What flow rate is needed to irrigate the field?

Q = (449 x A x D)/T

Example 2: What flow rate is needed to apply 3 inches of water over 100 acres with a total irrigation time of 180 hours?

Q = (449 x 100 acres x 3 inches)/180 hours = 748 gallons per minute

How many acres can I irrigate with my water supply?

A = (Q x T)/(449 x D)

Example 3: How many acres can I irrigate with a flow rate of 200 gpm? The desired depth to be applied is 3 inches per set, each set is 12 hours long, and the number of sets is 10.

T = 12 hours per set x 10 sets = 120 hours;

A = (200 gpm x 120 hours)/(449 x 3 inches) = 17.8 acres or 18 acres.

How long should I irrigate?

T = (449 x A x D)/Q

Example 4: How long should I operate my irrigation system to apply 4 inches over 80 acres? The flow rate into the field is 600 gpm.

T = (449 x 80 acres x 4 inches)/600 gpm = 239 hours of operation.