Example Prescription for Swine (hog, pig) Barn and Lagoon (pond, cell) Manure Treatment

For the purpose of explaining our treatment protocol, we will describe a preferred system configuration and assume certain waste and system parameters based upon the aggregate data typical of the swine industry. Listed prices are based upon a July startup. Products and prices will require adjustment for a fall or winter startup. This treatment does not require the presence of a plenum, but will not interfere with its operation, if present.

Our approach for each client includes thorough analysis of actual test results from the clients waste, combined with a complete process analysis for the in-place system. These investigations lead to our recommendation of the most cost-effective products and/or design modifications to produce maximum performance from that system.

Parameters of a typical swine producing operation
1,000 adult pigs per barn

10 barns per location, elevated 7 feet above lagoon
1) animals are on slotted floor above waste collected in waste pits that drain to a manifold, which empties into a lagoon.
2) the barn has adjustable side curtains

Anaerobic lagoon - (9 MG volume)
1) anaerobic
2) waste is drained by 10 inch PVC pipes, 400 to 500 ft., from central sump pump in barns

Flow: .05 to .08 MGD
COD: 20,000 ppm (1/1.4 VS)
BOD: 8,000 ppm (40% COD)
TS: 3.5% volume = 35,000 ppm
VS: 80% TS = 28,000 ppm
TKN: 1200 ppm
NH3-N: 650 ppm
oPO4:
3% TS (1,050 ppm)
Lignin: 8% TS = 2,800 ppm
Cellulose: 30% TS
Sulfur: 3200 ppm
(0.011 lbs. sulfur/head per day per Dr. Alan Sutton/Purdue University)
pH: 6.7 to 7.7
Calcium: 1032 ppm
Magnesium: 348 ppm
C/N ratio: 6 to 18 (ideal is 20 to 30)

 

Aerobic lagoon (5 MG volume)
1) aerated between 2 & 3 ppm oxygen
2) recharge system from lagoon 2 to the waste pits
Flow: .05 to .08 MGD
Retention time is 63 days. Retention time measures the amount of time bacteria are actually present in the lagoons to be treated, per cycle.



The refractory fraction of the volatile solids = 1 - 0.83 - (0.028 x lignin% volatile solids)
2.8g/L lignin / 28 g/L VS= 10% lignin(VS basis)
R = 1-biodegradable fraction = 1-(0.83-(0.028 x 10(% VS))
Refractory fraction = 1-0.55=0.45



Org. N = TKN-NH3-N = 550 mg/L / 35,000 mg/L = 0.016% of TS
Protein = Org N x 6.25 = 0.1% of TS = 0.1 g/L = 100 mg/L

Biodegradable VS = 0.55 x 28 g/L = 15.4 g/L

for Double Lagoon Schematic

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