Seminole Crop E News

Agricultural News for Farmers and Agribusiness in SW Georgia

Injecting through the Pivot

Posted by romeethredge on May 7, 2015

Dr. Wes Porter, UGA/AU Extension Irrigation Specialist gives us this irrigation injection information.

Traditionally we typically like to apply fertilizers and chemicals using either ground based (both fertilizers and chemicals) or aerial application methods.  However, we often times are under a time and suitable field working day constraint when it comes to applying many products.  Farmers also typically put more production inputs into irrigation crop land.  It’s known that irrigation has a higher yield potential and helps to protect that yield potential especially in dry years.  Thus, we want to do a more thorough job and provide the crops under irrigation with a more robust production plan to ensure we capture the full yield potential. 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.

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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.  The following process and example can be found at along with more information about specifics on Fertigation of Row-crops.

Steps for calculating fertilizer injection rate:

  1. Determine the irrigated area (acres)

  2. Determine the required application rate of product (in gallons per acre)

  3. Determine the amount required

  4. Determine the injection rate

For a practical example:

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

    1. Divide ft2 by 43,560 to get acres = 7,065,000 ÷ 43,560 = 162.2 acres

  2. Determine application rate: = 30 lbs N/ac ÷ 3.5 lb N/gal = 8.6 gal/ac

  3. Determine required amount:  = 8.6 gal/ac * 162.2 acres = 1390.3 gallons

  4. Injection Rate:  = 1390.3 gal ÷ 12 h = 115.9 gal/h

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