Manure Problem Solving December 17, 2009 CraigW. Yohn WVU Extension Agent and Certified Crop Advisor

WVU Soil Testing Lab Recommendations: Clover\Tall Grass Mix Greater than 30% Legume Soil Test ResultsPlant Food Recommendations PKNP2O5P2O5 K2OK2O Low Med High Very High 0250

WVU Soil Testing Lab Recommendations: Clover\Tall Grass Mix Less than 30% Legume Soil Test ResultsPlant Food Recommendations PKNP2O5P2O5 K2OK2O Low Med High Very High

WVU Soil Testing Lab Recommendations: Corn Silage Soil Test ResultsPlant Food Recommendations PKNP2O5P2O5 K2OK2O Low Med High Very High 15040

WVU Soil Testing Lab Recommendations: Corn Grain Soil Test ResultsPlant Food Recommendations PKNP2O5P2O5 K2OK2O Low Med High Very High 15040

Plant Available Nutrients  P = P  K = K  N = sum of Plant Available Nitrogen

Plant Available Nitrogen Step 1: Determine the amount of organic and inorganic N in the manure Total N (TKN) – Inorganic N (NH 4 + -N) = Organic N Step 2: Estimate the amount of organic N that will mineralize during the year Organic N * Mineralization Factor = Organic N Available First Year

Plant Available Nitrogen Step 1: Determine the amount of organic and inorganic N in the manure Total N (TKN) – Inorganic N (NH 4 + -N) = Organic N Step 2: Estimate the amount of organic N that will mineralize during the year Organic N * Mineralization Factor = Organic N Available First Year

Plant Available Nitrogen Step 1: Determine the amount of organic and inorganic N in the manure Total N (TKN) – Inorganic N (NH 4 + -N) = Organic N Step 2: Estimate the amount of organic N that will mineralize during the year Organic N * Mineralization Factor = Organic N Available First Year

Plant Available Nitrogen Step 1: Determine the amount of organic and inorganic N in the manure Total N (TKN) – Inorganic N (NH 4 + -N) = Organic N Step 2: Estimate the amount of organic N that will mineralize during the year Organic N * Mineralization Factor = Organic N Available First Year

Plant Available Nitrogen Step 3: Estimate the amount of Inorganic N (NH 4 + -N) that will be available following land application Inorganic N (NH 4 + -N) * Volatilization Factor = Available Inorganic N (NH 4 + -N) Step 4: To calculate PAN = Organic N Available First Year (Step 2) + Available Inorganic N (NH 4 + -N) (lbs./ton or lbs. per 1,000 gallons)

Plant Available Nitrogen Step 3: Estimate the amount of Inorganic N (NH 4 + -N) that will be available following land application Inorganic N (NH 4 + -N)* Volatilization Factor = Available Inorganic N (NH 4 + -N) Step 4: To calculate PAN = Organic N Available First Year (Step 2) + Available Inorganic N (NH 4 + -N) (lbs./ton or lbs. per 1,000 gallons)

Plant Available Nitrogen Step 3: Estimate the amount of Inorganic N (NH 4 + -N) that will be available following land application Inorganic N (NH 4 + -N)* Volatilization Factor = Available Inorganic N (NH 4 + -N) Step 4: To calculate PAN = Organic N Available First Year (Step 2) + Available Inorganic N (NH 4 + -N) (lbs./ton or lbs. per 1,000 gallons)

Plant Available Nitrogen Step 3: Estimate the amount of Inorganic N (NH 4 + -N) that will be available following land application Inorganic N (NH 4 + -N)* Volatilization Factor = Available Inorganic N (NH 4 + -N) Step 4: To calculate PAN = Organic N Available First Year (Step 2) + Available Inorganic N (NH 4 + -N) (lbs./ton or lbs. per 1,000 gallons)

Source for Information

Table 9.1- Manure Production

Table 9.8- Forms of Nitrogen

Table 9.9- Nitrogen Mineralization

Table Ammonia – N Availability

Geometry Lesson Rectangle measured in feet –(L +L)*(W+W)*H = Cubic Feet Square measured in feet –L*W*H = Cubic Cylinder measured in feet – * L (in feet) = cubic feet – D = circumference

Common Conversion Factors 7.5 gallons = 1 cubic foot 1.25 cubic feet = 1 bushel 27,154 gallons = 1 acre inch 1 acre = 43, 500 square feet 1 gallon liquid manure = 8 pounds K 2 O = K * 1.2 P 2 O 5 = P * to 62 lbs. of dry manure = 1 cubic foot

Problem 1 A producer has a 150 dairy cattle. The average weight of each cow is 1400 pounds. The cattle are kept year round in a feedlot. The manure is scraped into a manure pit which is emptied as a liquid. What is the estimated annual volume of manure collected from the feedlot? a. 13,030 cubic feet b. 93,075 cubic feet c. 130,305 cubic feet d. None of the above

Problem 1 What is the estimated annual volume of manure collected from the feedlot? a. 13,030 cubic feet b. 93,075 cubic feet c. 130,305 cubic feet d. None of the above 1400 lbs. /1000 = 1.4 AU 1.4 AU * 1.7 ft 3 of manure production/ day = 2.38 ft 3 per day per cow 2.38 ft 3 per day * 150 cows = 357 ft 3 per day per herd 357 ft 3 per day per herd * 365 days = 130,305 ft 3 of manure

Problem 2 A soil test reveals that to produce 125 bushels of corn and maintain present soil fertility, the producer needs to apply 125 pounds of N, 50 pounds of P 2 O 5 and 40 pounds of K 2 O. The producer has poultry litter available with an analysis of TKN : pounds per ton, Ammonium –N – 21.12; P 2 O5 – 56.34; K 2 O – What is the application rate to meet P requirements of the a. 1 ton per acre b..75 ton per acre c..89 ton per acre d. None of the above

Problem 2 What is the application rate to meet P requirements of the a. 1 ton per acre b..75 ton per acre c..89 ton per acre d. None of the above 50 pounds of P 2 O 5 required lbs. P 2 O 5 per ton of litter.89 tons per acre

Problem 2 Continued A soil test reveals that to produce 125 bushels of corn and maintain present soil fertility, the producer needs to apply 125 pounds of N, 50 pounds of P 2 O 5 and 40 pounds of K 2 O. The producer has poultry litter available with an analysis of TKN : pounds per ton, Ammonium –N – 21.12; P2O5 – 56.34; K2O – How much nitrogen is available to the plant if the litter was not incorporated? a pounds b pounds c pounds d pounds

Problem 2 Continued How much nitrogen is available to the plant if the litter was not incorporated? a pounds b pounds c pounds d pounds (39.75 lbs. of N – 21.12) = (18.63 *.60) + ( 0 * 21.12) = = 11.2 pounds of Nitrogen

Problem 2 Continued How much nitrogen is available to the plant if the litter was not incorporated? a pounds b pounds c pounds d pounds (39.75 lbs. of N – 21.12) = (18.63 *.60) + ( 0 * 21.12) = = 11.2 pounds of Nitrogen

Problem 3 A producer is applying poultry litter to a field. He is driving his tractor at a speed of 5 miles/hour. He is trying to calibrate his spreader. To calibrate the spreader, the producer laid out a 20 feet long by 10 feet wide tarp on which he collected 15 pounds of litter in one pass. What is the producer’s approximate application rate per acre? a. 2.5 tons b. 1 ton c. 1.5 tons d. 2 tons

Problem 3 What is the producer’s approximate application rate per acre? a. 2.5 tons b. 1 ton c. 1.5 tons d. 2 tons 10 * 20 = /43,560 = acres 15/ = x/1; 15 =.00459x 15/ = 3,268 pounds/ 2000 pounds per ton X = 1.63 tons per acre

Problem 4 A dairyman had applied 2,000 gallons of liquid dairy manure per acre to a 50 acre field the previous fall with no incorporation after taking off 40 bushels of soybeans. He is rotating to corn for silage in this field and the manure pit is full again. The nutrient requirement is 145 – N; 40 – P 2 O 5 and 40 – K 2 O. The manure analysis for fall and spring is TKN – 38.05; Ammonium –N – 29.86; P 2 O 5 – 17.03; K 2 O What is the application rate per acre to meet phosphorous requirements. a. 2,000 gallons b. 1,000 gallons c. 1,500 gallons d. None

Problem 4 What is the application rate per acre to meet phosphorous requirements. a. 2,000 gallons b. 1,000 gallons c. 1,500 gallons d. None 2 * = (Fall application) = 5.94 lbs. of P 2 O 5 required – 5.94 = excess

Problem 4 continued A dairyman had applied 2,000 gallons of liquid dairy manure per acre to a 50 acre field the previous fall with no incorporation after taking off 40 bushels of soybeans. He is rotating to corn for silage in this field and the manure pit is full again. The nutrient requirement is 145 – N; 40 – P 2 O 5 and 40 – K 2 O. The manure analysis is TKN – 38.05; Ammonium –N – 29.86; P 2 O 5 – 17.03; K 2 O Would the 1,000 gallons per acre meet the K 2 O needs of the crop? a. Yes b. No

Problem 4 continued A dairyman had applied 2,000 gallons of liquid dairy manure per acre to a 50 acre field the previous fall with no incorporation after taking off 40 bushels of soybeans. He is rotating to corn for silage in this field and the manure pit is full again. The nutrient requirement is 145 – N; 40 – P 2 O 5 and 40 – K 2 O. The manure analysis is TKN – 38.05; Ammonium –N – 29.86; P 2 O 5 – 17.03; K 2 O Would the 1,000 gallons per acre meet the K 2 O needs of the crop? a. Yes b. No

Problem 4 continued Would the 1,000 gallons per acre meet the K 2 O needs of the crop? a. Yes b. No 2 * = – 40 = = excess pounds

Problem 4 continued How much additional nitrogen will be required for the crop of corn? a. 120 pounds b. 100 pounds c. 135 pounds d. 60 pounds

Problem 4 continued How much additional nitrogen will be required for the crop of corn? a. 120 pounds b. 100 pounds c. 135 pounds d. 60 pounds Soybeans – 40 * 1lb/ bushel = 40 lbs of N ((38.05 – 29.86)*.35)+ (29.86 * 0) = 2.87 lbs. per 1, * 2 = 5.7 lbs of N from fall application of manure 2.87 pounds from spring application of manure 145 lbs. required – ~50 pounds = 100 pounds

Questions