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Natural Resources Extension Program
Estimating the volume of liquid manure wastewater
and run-off produced
Use this worksheet to calculate the amount of liquid
manure and run-off produced.
Example
3000 head at an average weight of 750 lbs
Critical
storage period = 150 days per year.
750 lb beef cattle = 5.3
gallons per day at 88.4 % moisture
25yr/24hr storm = 2.2 inches precipitation (see
map)
Lot area = 4.36 acres or 189,921 sq. ft. (1 acre
= 43,560 sq. ft.)
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1. 5.3 gallons x 88.4% = 39.78 gallons manure
2. 39.78 gallons x 150 days = 5967 gallons manure/head/year
3. 5967 gallons x 3000 head = 17,901,000 gallons
manure wastewater/year
4. Runoff = [(P - 0.5) ÷ 12 in./ft] x
LA |
| Variables: |
RO |
= lot runoff, ac-ft |
|
P |
= Precipitation, inches |
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LA |
= lot area, acres |
| RO = [(2.2 – 0.5) ÷
12] x 4.36 ac = 0.62 ac-ft |
| There are 350,850 gallons in
1 ac-ft. Therefore, 0.62 ac-ft equals 217,527
gallons. |
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| 5. Add lines 3 and 4: 17,901,000 + 217,527 = 18,118,527
gallons |
- # head ____________ and average weight _______________
- critical storage period = ________________
- # gallons/per day ___________
and % of wet material __________ (see
table)
- 25yr/24hr storm = __________inches precipitation
(see
map)
- Lot area = _______ acres or ______ sq. ft.
(1 acre = 43,560 sq. ft.)
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1. _____ gallons x _____ % moisture = 39.78
gallons manure
2. _____ gallons x ______ critical storage days
= ______ gallons manure/head/year
3. ____ gallons x ______head = _________ gallons
manure wastewater/year
4. Runoff = [(P - 0.5) ÷ 12 in./ft] x
LA
- [(____ inches precip.- 0.5) ÷ 12 in/ft.]
x _____ lot area acres = _____ ac-ft
- There are 350,850 gallons in 1 ac-ft. Therefore
= ____________ gallons runoff.
5. Total gallons of wastewater and runoff (add
lines 3 and 4) = _________ gallons.
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