### University of Missouri Extension

G9330, Revised June 2013

# Calculating the Value of Manure as a Fertilizer Source

##### John Lory Commercial Agriculture Program

Manure has value on a field only if it offsets the need to purchase other nutrients or soil amendments. The work sheet below allows you to calculate the fertilizer value of the manure for a specific field. Assemble the information specified in the input form (Figure 1) before completing the work sheet. An example work sheet is included in the following section.

Manure has characteristics that may reduce its value relative to that of commercial chemical fertilizers. Manure can be a less dependable nitrogen source and is perceived as a source of weed seeds. Because manure is an unbalanced fertilizer source, meeting crop needs for one nutrient may result in application of too much or too little of other crop nutrients. Low nutrient concentration in manure increases handling and application costs. Manure has positive attributes as well. It is a slow-release fertilizer, and the organic material can improve soil quality.

The work sheet in this guide determines the value of manure only as a nutrient source, similar to commercial chemical fertilizers, regardless of its other positive or negative attributes. Actual economic value of manure must be negotiated between the manure seller and the buyer.

Input form
Information needed to complete the work sheet.

Step 1
Total nutrients in manure (pounds per ton or pounds per 1,000 gallon)
1. Total nitrogen (TN or TKN)
2. Inorganic nitrogen (IN)
3. Organic nitrogen (ON)
(ON = TN - IN)
4. Phosphorus (P2O5)
(P2O5 = P x 2.29)
5. Potassium (K2O)
(K2O = K x 1.20)
You should get these numbers from a laboratory analysis of manure from your farm. Without a lab test, you can use a table of typical nutrient values for manure, but you should realize that the nutrient value of your manure could easily be half or double the tabular value. See MU Extension publication EQ215, Laboratory Analysis of Manure, for more information on manure testing. See MU Extension publication EQ201, Reduce Environmental Problems With Proper Land Application of Animal Manure, for average nutrient values of several types of livestock manure.
Step 2
Crop nutrient need (pounds per acre)
1. Nitrogen (N)
2. Phosphorus (P2O5)
(P2O5 = P x 2.29)
3. Potassium (K2O)
(K2O = K x 1.20)
The need for phosphorus and potassium should be based on soil testing results, and nitrogen need depends on the crop grown and yield goal. For more information, see MU Extension publications G9217, Soil Sampling Hayfields and Row Crops, and G9112, Interpreting Missouri Soil Test Reports.
Step 3
Fertilizer costs
1. Fertilizer N (dollars per pound)
2. Fertilizer P2O5 (dollars per pound)
3. Fertilizer K2O (dollars per pound)
Step 4
Other information
1. Spreader capacity (tons or 1,000 gallons)

Figure 1. Information on this input form will be needed to calculate the fertilizer value per load of manure.

## Work sheet to calculate fertilizer value per load of manure

Total N Inorganic N1 Organic N P2O5 K2O
1. Manure nutrient content (pounds per ton or pounds per 1,000 gallons) from input form, step 1
2. Nutrient availability2       1.0 1.0
3. Available nutrients3 in manure (pounds per ton or pounds per 1,000 gallons)
Multiply line 1 by line 2.
0 0 0 0 0
4. Crop need (pounds per acre) from input form, step 2
5.Quantity of manure needed (tons per acre or 1,000 gallons per acre)
Divide line 4 by line 3.

6. Quantity of manure to apply4 (tons per acre or 1,000 gallons per acre)
Select one value from line 5.

7. Available nutrients that will be applied (pounds per acre)
Multiply line 6 by line 3.

8.Nutrients with value (pounds per acre)
Select the value of line 4 or line 7, whichever is lower.
0 0 0
9. Fertilizer cost (dollars per pound) from input form, step 3 0 0 0
10. Manure value by nutrient (dollars per acre)
Multiply line 8 by line 9.
\$0 \$0 \$0
11. Manure value per acre (dollars per acre)
Sum the values in line 10.
\$0
12. Manure value per unit of manure (dollars per ton or dollars per 1,000 gallons)
Divide line 11 by line 6.

## Example

This example is based on a surface application of poultry manure on corn. See example work sheet below.

#### Example inputs

Step 1. Total nutrients in manure (pounds per ton or pounds per 1,000 gallons)

1. Total nitrogen (TN or TKN): 54 lb/ton
2. Inorganic nitrogen (IN): 10 lb/ton
3. Organic nitrogen (ON) 44 lb/ton
(ON = TN - IN)
4. Phosphorus (P2O5): 50 lb/ton
(P2O5 = P × 2.29)
5. Potassium (K2O): 39 lb/ton
(K2O = K × 1.20)

Step 2. Crop nutrient need (pounds per acre)

1. Nitrogen (N): 150 lb/acre
2. Phosphorus (P2O5): 40 lb/acre
(P2O5 = P × 2.29)
3. Potassium (K2O): 175 lb/acre
(K2O = K × 1.20)

Step 3. Fertilizer costs

1. Fertilizer N (\$/lb): 0.50/lb
2. Fertilizer P2O5 (\$/lb): 0.55/lb
3. Fertilizer K2O (\$/lb): 0.52/lb

Step 4. Other information

1. Spreader capacity (tons or 1,000 gallons): 5 tons

#### Example work sheet

 Total N Ammonia N1 Organic N P2O5 K2O 1. Manure nutrient content (pounds per ton or pounds per 1,000 gallons) from input form, step 1 54 10 44 60 39 2. Nutrient availability2 0.4 0.5 1.0 1.0 3. Available nutrients3 (pounds per ton or pounds per 1,000 gallons) Multiply line 1 by line 2. 26 4 22 50 39 4.Crop need (pounds per acre) from input form, step 2 150 40 175 5.Quantity of manure needed (tons per acre or 1,000 gallons per acre) Divide line 4 by line 3. 5.8 0.8 45 6. Quantity of manure to apply4 (tons per acre or 1,000 gallons per acre) Select one value from line 5. 4.5 4.5 4.5 7. Available nutrients that will be applied (pounds per acre) Multiply line 6 by line 3. 117 225 176 8. Nutrients with value (pounds per acre) Select the value of line 4 or line 7, whichever is lower. 117 40 175 9. Fertilizer cost (dollars per pound) from input form, step 3 \$0.50 \$0.55 \$0.52 10. Manure value by nutrient (dollars per acre) Multiply line 8 by line 9. \$58.50 \$22.00 \$91.00 11. Manure value per acre (dollars per acre) Sum the values in line 10. \$58.50 + \$22.00 + \$91.00 = \$171.50 12. Manure value per unit of manure (dollars per ton or dollars per 1,000 gallons) Divide line 11 by line 6. \$171.50 ÷ 4.5 = \$38.11
##### 4Select from line 5 the quantity of manure to be applied per acre based on whether you want to meet the N, P or K needs of the crop. If you want to meet or exceed all crop nutrient needs, select the highest value in line 5.

This manure would have a value of \$171.50 per acre if applied at 4.5 tons per acre. The value per ton would be \$38.11 (\$171.50 per acre ÷ 4.5 tons per acre).

In this example, manure is applied to meet K2O requirements of the crop. Consequently, N is underapplied and additional fertilizer N will be needed. The crop requires 150 pounds of N per acre; we will apply 125 pounds N per acre. The crop will need 33 pounds additional fertilizer N per acre (150 – 117 = 33).

In this example, applying manure to meet the K2O requirements of the crop results in applying substantially more P2O5 than is needed as indicated by soil test. Excess P2O5 in the manure does not contribute to the value of the manure because the nutrients are not required for crop growth.