Chemical, fertilization through pivot system and rate example

Chemical, fertilization through pivot system and rate example

One of the most daunting tasks in using a center pivot for chemigation or fertigation is calculating the injection rate of the fertilizer or chemical.

During times where we need high inputs on irrigated land but field conditions are not suitable for ground equipment, you may work with putting chemicals and fertilizer through the pivot.  Here is information on how to use the pivot irrigation and an example of how to calculate rates.

According to University of Georgia Extension irrigation specialist Dr. Wes Porter:

“Due to excessive rainfall during the growing season and in some cases excessive plant growth and height, it becomes difficult and sometimes impossible to enter a field to apply the proper chemicals and/or fertilizers. 

“In this case the addition of an injection pump for chemigation and fertigation can be very advantageous. A center pivot can typically walk around the field when the moisture level is much higher than can a ground based sprayer.  Thus, one main advantage is the ability to apply nutrients at critical periods of crop demand.”

One of the most daunting tasks in using a center pivot for chemigation or fertigation is calculating the injection rate of the fertilizer or chemical, Porter says.

Steps for calculating fertilizer injection rate:

  • Determine the irrigated area (acres)
  • Determine the required application rate of product (in gallons per acre)
  • Determine the amount required
  • Determine the injection rate

For a practical example, Porter says, let’s assume that you want to apply 30 lbs N/ac of UAN-32 through a 1,500 ft long center pivot at a rate of 0.3 inches in 12 hours (one complete circle).

  1. Irrigated area = = 3.14 * 1,5002= 7,065,000 ft2
  2. Divide ft2 by 43,560 to get acres = 7,065,000 ÷ 43,560 = 162.2 acres
  3. Determine application rate: = 30 lbs N/ac ÷ 3.5 lb N/gal = 8.6 gal/ac
  4. Determine required amount:  = 8.6 gal/ac * 162.2 acres = 1390.3 gallons
  5. Injection Rate:  = 1390.3 gal ÷ 12 h = 115.9 gal/h

More information on this practice can be found by the University of Tennessee Extension at “Fertigation of Row Crops Using Overhead Irrigation.”

(More blog posts by Sawyer can be found at Thomas County Ag.)

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