Example 1

Determine the value of the three types of manure for a corn-soybean rotation with the following soil test recommendations: For 150 bushels per acre corn: 195 pounds N, zero P₂O₅ and 60 pounds K₂O. For 40 bushels per acre soybeans: zero N, zero P₂O₅ and 75 pounds K₂O (soil test results were 116 [very high] for phosphorus, 280 for potassium, and 2.9 for organic matter).

The last crop was soybeans, so we can take credit for 30 pounds N from the soybeans and apply 165 pounds of N for corn.

Using all the slurry manure from 150 head (Table 20), the available N is 3,420 + 6,045 = 9,465 pounds of N; the available P₂O₅ is 8,250 pounds x 0.95 = 7,838 pounds; and the available K₂O is 11,850 pounds x 0.95 = 11,258 pounds. With 9,465 pounds of available N, we can apply 165 pounds of N to 57 acres. Using all the manure will also apply 8,250 pounds of P₂O₅ and 11,850 pounds of K₂O (Table 20). For the 57 acres of corn, the soil test calls for zero P₂O₅ and 3,420 pounds K₂O. No return should be expected from the 8,250 pounds of P₂O₅ . To have 3,420 pounds of K₂O available, we must apply 3,600 pounds at 0.95 percent availability. Adding the value of 9,465 pounds of N at $0.30 per pounds = $2,840 to the value of 3,420 pounds of K₂O at $0.15 per pounds = $513, the total value is $3,353. If all the P₂O₅ and K₂O applied had been needed, the value would have been $6,096 (Table 20).

For soybeans and applying the solid manure at a rate to supply 75 pounds of K₂O per acre, the 11,258 pounds of available K₂O will fertilize 11,258/75 = 150 acres of soybeans. Since the soil test did not call for any N or P₂O₅ , the only value to be credited to the manure is for 11,258 pounds of available K₂O at $0.15 = $1,689 (Table 20).

Caution
Since the soil test P is very high, any addition of manure increases the risk of downstream pollution from phosphorous and may not be allowed under developing regulations.

Table for example 1
Value of manure nutrients used by 150 bushels per acre corn (based on soil test)