New December 1997

Download a free PDF of this publication (129KB). PDF help

Order printed copies
G9330, Calculating the Value of Manure as a Fertilizer Source

Price: $0.50
Availability: 1000+

Contents

Work sheet to calculate fertilizer value per load of manure
Example input form
Comments
Additional information

Related publications

Use our feedback form for questions or comments about G9330.

Search publications

Search only the MU Extension publications.

Calculating the Value of Manure as a Fertilizer Source

John Lory
Department of Agronomy and Commercial Agriculture Program

Ray Massey
Department of Agricultural Economics and Commercial Agriculture Program

Manure has value on a field only if it offsets the need to purchase other nutrients or soil amendments. The following work sheets allow you to calculate the fertilizer value of the manure for a specific field. Assemble the information specified in the following input form before completing the work sheet. An example work sheet is included at the end of this guide.

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 nutrients in 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 (pound per ton or pound per 1,000 galllon)
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.
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)
Need for phosphorus and potassium should be based on soil testing results, and nitrogen need depends on the crop grown and yield goal.
Step 3 Fertilizer costs
1. Fertilizer N (cost per pound)
2. Fertilizer P₂O₅ (cost per pound)
3. Fertilizer K₂O (cost per pound)
Step 4 Other information
1. Spreader capacity (tons or 1,000 gallons)

Work sheet to calculate fertilizer value per load of manure

	P₂O₅	K₂O
1. Manure nutrient content (pound per ton or pound per 1,000 galllons) from input form, step 1
2. Availability**	1.0	1.0
3. Available nutrients*** (pound per ton or pound per 1,000 galllons) multiply line 1 x line 2
4. Spreader capacity (tons or 1,000 gallons) from input form, step 4
5. Nutrients per load (pound per ton or pound per 1,000 galllons) multiply line 3 x line 4
6. Crop need (pound per acre) from input form, step 2
7. Loads of manure needed (loads per acre) divide line 6 by line 5
8. Loads applied per acre**** select one value from line 7
9. Effective manure fertilizer value (pounds per load) If line 7 is greater than or equal to line 8, then line 9 = line 5 If line 7 is less than line 8, then line 9 = line 6 ÷ line 8
10. Fertilizer cost (cost per pound) from input form, step 3
11. Manure value (cost per nutrient per load) multiply line 9 x line 10
12. Value per load (cost per load) sum of values in line 11

Organic N = total N 2 ammonia N
Ammonia N availability can range from 0.20 for surface-applied manure to 1.0 for incorporated manure. Organic N availability ranges from 0.25 to 0.6 in the year of application, depending on manure type. Missouri Department of Natural Resources regulations may stipulate that certain availability coefficients be used for MDNR-permitted waste facilities.
Add available ammonia N and organic N in line 3 to determine available total N in line 3.
****Select from line 7 the number of loads per acre 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 7.

Example

This example is based on a surface application of poultry manure on corn. See example work sheet (Figure 2).

Example input form

Step 2 Crop nutrient need
Step 1 Total nutrients in manure (pounds per ton or pounds per 1,000 gallons)
1. Total nitrogen (TN or TKN):	54 pounds per ton
2. Inorganic nitrogen (IN):	10 pounds per ton
3. Phosphorus (P₂O₅): (P₂O₅ = P x 2.29)	60 pounds per ton
4. Potassium (K₂O): (K₂O = K x 1.20)	39 pounds per ton
1. Nitrogen (N):	150 pounds per acre
2. Phosphorus (P₂O₅): (P₂O₅ = P x 2.29)	40 pounds per acre
3. Potassium (K₂O): (K₂O = K x 1.20)	175 pounds per acre
Step 3: Fertilizer costs
1. Fertilizer N (cost per pound):	0.33 per pounds
2. Fertilizer P₂O₅ (cost per pound):	0.30 per pounds
3. Fertilizer K₂O (cost per pound):	0.12 per pounds
Step 4: Other information
1. Spreader capacity (tons or 1,000 gallons):	5 tons

Example work sheet

	Total Nitrogen	Ammonia Nitrogen	Organic Nitrogen*	P₂O₅	K₂O
1. Manure nutrient content (pound per ton or pound per 1,000 galllons) from input form, step 1	54	10	44	60	39
2. Availability		0.4	0.5	1.0	1.0
3. Available nutrients (pound per ton or pound per 1,000 galllons) multiply line 1 x line 2	26	4	22	60	39
4. Spreader capacity (tons or 1,000 gallons) from input form, step 4	5			5	5
5. Nutrients per load (pound per ton or pound per 1,000 galllons) multiply line 3 x line 4	130			300	195
6. Crop need (pounds per acre) from input form, step 2	150			40	175
7. Loads of manure needed (loads per acre) divide line 6 by line 5	1.15			0.13	0.90
8. Loads applied per acre select one value from line 7	0.90			0.90	0.90
9. Effective manure fertilizer value (pounds per load) per If line 7 is greater than or equal to line 8, then line 9 = line 5 If line 7 is less than line 8, then line 9 = line 6 ÷ line 8	130			44	195
10. Fertilizer cost (cost per pound) from input form, step 3	0.33			0.30	0.12
11. Manure value (cost per nutrient per load) multiply line 9 x line 10	42.90			13.20	23.40
12. Value per load (cost per load) sum of values in line 11	42.90 + 13.20 + 23.40 = $79.50 per load

Comments

This manure would have a value of $79.50 per load if applied at 0.90 load per acre (4.5 tons per acre). The value per acre would be $71.55 ($79.50 per load x 0.90 load 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 N per acre; we will apply 125 pounds N per load x 0.9 load per acre = 112 pounds N per acre. The crop will need 38 pounds per acre (150 - 112 = 38) additional fertilizer N.

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.

Additional information

Download manrvalu.exe, an EXCEL® spreadsheet program of this worksheet, from File Sharing on Agricultural Electronic Bulletin Board (AgEBB.)

G9330, new December 1997