Appendix D: Calculating Available Nutrients
from a Manure Analysis

Excerpt from Agronomy Guide for Field Crops (Appendices)

Order OMAFRA Publication 811: Agronomy Guide for Field Crops

Table of Contents

1. Introduction
2. Available Nitrogen
3. Available Phosphate
4. Available Potash
5. For example
6. Updates on Appendix D: Calculating Available Nutrients from a Manure Analysis
7. Related links...

Introduction

Information from a manure analysis, plus knowledge of the conditions at the time the manure was applied, can be used to provide a more precise estimate of the available nutrients in the manure applied to your fields.

Available nitrogen is determined by subtracting the ammonia losses to the air from the ammonium-N applied, and adding the mineralization from the organic N portion of the manure.

| Top of Page

Reductions in fertilizer phosphate and potash can be calculated by determining the available portion of the total P and K in the manure (40% for phosphorus and 90% for potassium) and multiplying by a factor to convert from the elemental form to the oxide form (fertilizer nutrients are expressed in the oxide form). Read the report carefully, since some labs make these conversions and report the nutrients available to replace fertilizer, while others report the elemental P and K in the manure.

The required calculations are summarized in the above tables. The calculations will be the same whether the units are in percentage, lb/1,000 gal, or kg/tonne. Be sure to keep the same units throughout the calculation. Some reports will provide ammonium-N contents in ppm (mg/kg, mg/L), while the other numbers are in percentages.

| Top of Page |

To convert ppm to percentage, divide by 10,000.

For example, a manure analysis is received for a liquid dairy manure, with dry matter content of 7%, total N of 0.65%, ammonium-N of 3,500 mg/L, P of 0.6%, and total K of 0.3%. All the nutrient contents are expressed as percentage of wet weight.

The manure is applied in spring immediately before planting under warm, dry conditions and is incorporated 3 days after application.

The ammonium-N content can be converted to percentage by dividing by 10,000, giving an ammonium-N content of 0.35%. The organic N content can then be calculated by subtracting the ammonium-N from the total N (0.65 - 0.35 = 0.3% organic N).

Ammonium losses in this situation will be calculated as ammonium-N (0.35%) times the factor from Table 2-11. Lime Requirements to Correct Soil Acidity Based on Soil pH and Soil Buffer pH, for the application conditions (0.65), giving an ammonium loss of 0.2275 (0.35 x 0.65 = 0.2275). The ammonium-N available to the crop is the initial ammonium-N minus the loss (0.35 - 0.2275 = 0.1225).

The available nitrogen from the organic portion is calculated as the organic-N times the availability factor in Table 2-12. Example Calculation of the Fineness Rating of a Limestone. For liquid dairy manure, this is 0.2, so the available organic N is 0.06% (0.3% x 0.2 = 0.06%).

Total available N is the ammonium-N plus organic-N, giving a total of 0.1825% (0.1225 + 0.06 = 0.1825). This percentage of available N times the manure application rate provides the amount of nitrogen available to the crop from the manure application. To convert the percentage of nutrient to kg/tonne, multiply times 10 (to lb/1,000 gal, multiply times 100).

Available phosphate is the phosphorus content (0.6%) times the availability factor (0.4), times the conversion to phosphate (2.29). This gives an available phosphate from this manure of 0.275% (0.6% x 0.4 x 2.29 = 0.275%).

Available potash is the potassium content (0.3%) times the availability factor (0.9) times the conversion to potash (1.2). This gives an available potash from this manure of 0.324% (0.3% x 0.9 x 1.2 = 0.324%).

| Top of Page |

Updates on Appendix D: Calculating Available Nutrients from a Manure Analysis

No updates available at this time.

| Top of Page |

Related links...

• Soil Fertility

| Top of Page |