Free nitrogen from irrigation!

The nitrogen (N) in your irrigation water can be credited to your total nitrogen application. How much it contributes depends on its nitrate-nitrogen (NO3-N) concentration. This can vary between irrigation water sources. For example: Lake Ontario < 1ppm, surface pond 8 ppm, well 17 ppm.

As a general rule of thumb: if the NO3-N concentration is greater than 10 ppm, it should be credited towards the crop's total nitrogen requirement. In fact, at some points during the crop's development, this might be enough to meet its weekly N requirement. That's free nitrogen every time you irrigate.

To determine how much free nitrogen one irrigation event contributes, you need to know the following:

  1. NO3-N concentration in the irrigation water, in parts per million (ppm)
  2. Irrigation system flow rate (gpm, US)
  3. How long the irrigation event is in hours
  4. Area being irrigated (ac)

How to calculate:

Step 1

  • Determine NO3-N concentration (ppm) by completing an irrigation water analysis

Step 2

  • Determine lbs of N in an acre-inch water by multiplying irrigation NO3-N ppm by 0.23 Lbs of N per acre-inch = NO3-N ppm x 0.23

Step 3

  • Determine the inches of water applied during an irrigation event using the following equation: D = (Q x T) ÷ (449 x A)
    • D = water applied, inches
    • Q = irrigation system flow rate, gpm
    • T = irrigation duration, hours
    • A = area irrigated, acres

    Step 4

    • Determine how many lbs N per acre are applied in one irrigation event by multiplying the answer from Step 2 by answer from Step 3.
    • Lbs N/acre applied in one irrigation event = lbs N/acre-inch (Step 2) x inches of water applied in one irrigation event (Step 3)
    • Multiply this answer by 1.12 to get kg N/ha


    Step 1

    • From irrigation water analysis: 15 ppm NO3-N

    Step 2

    • Determine lbs of N in an acre-inch water: Lbs N per acre-inch water = 15 ppm x 0.23 = 3.5

    Step 3

    • Determine inches of water applied during an irrigation event using: D = (Q x T) ÷ (449 x A)
    • Where:
      • Q, irrigation system flow rate = 1000 gpm
      • T, irrigation duration = 54 hours
      • A, acres irrigated = 40 acres
    • D = (1000 x 54) ÷ (449 x 40) = 3.0 inches

    Step 4

    • Lbs N/acre in one irrigation event = Step 2 x Step 3 = 3.5 x 3.0 = 10.5
    • Multiplying this answer by 1.12 gives kg N/ha = 10.5 x 1.12 = 11.8

    In this example one irrigation event supplies almost 12 kg N/ha. This provides about half of the weekly N requirement for processing tomatoes for weeks 4-9 (OMAF and MRA Publication 363 Vegetable Production Recommendations 2010-11, Table 9-103).

    By crediting the nitrogen provided from irrigation water, the nitrogen fertilizer added for fertigation can be reduced. Free nitrogen.

    Ref: Drip Irrigation of Processing Tomatoes, Publication 3506. 2008. University of California Agriculture and Natural Resources.

For more information:
Toll Free: 1-877-424-1300