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Practical review of fertilizer
management problems
Fundamentals of Nutrient Management
December 16-17, 2009
West Virginia University Extension Service
T.C. Griggs
Division of Plant & Soil Sciences, WVU
Review of N2 fixation
• Plant nitrogen (N) requirement can be met via uptake
of NH4+ or NO3- from soil.
• Atmospheric N2 fixed into NH3 (then amino acids )
via symbiotic relationship with rhizobia:
Rhizobium spp.
Mesorhizobium spp.
Bradyrhizobium spp.
Ensifer (prev. Sinorhizobium) spp.
• Nitrogenase, the N2-fixing enzyme synthesized in
nodules, reduces N2 that diffuses in to NH3, which is
converted to amino acids or amides for transport.
Legume
Alfalfa
Birdsfoot trefoil
Crownvetch
Cicer milkvetch
Crimson clover
Hairy vetch
Kura clover
Lentil
Red clover
Soybean
Subterranean clover
Sweetclover
White clover
N2 fixed (lb N/ac-yr)
70-200
44-150
98
140
57
99
17-158
149-168
60-200
20-200
52-163
120
115-180
Legume nodules
Effective nodules
have a pinkish,
rather than
whitish, interior
Nodules are easily
stripped away from
roots during
careless sampling
Birdsfoot trefoil
Red clover
Match legume species with correct inoculum species
(e.g., types ‘A’, ‘B’, ‘K’, etc. from EMD Crop BioScience,
formerly Nitragin, Inc.)
N2 fixation
• Nitrogenase is inactivated by O2.
• To keep O2 away from nitrogenase, the O2carrier leghemoglobin synthesized in nodules
captures O2 before it reaches nitrogenase, then
releases it for respiration to drive N2-fixation.
• Leghemoglobin with captured O2 imparts pink
color to healthy nodules - a good check of
nodule effectiveness (effective vs. ineffective
nodules).
N2 fixation and transfer to grass depend on:
• N transfer to grass is primarily from nodule
sloughing, legume leaf litter and plant death,
and grazing animal urine and manure.
• Site conditions including adequate P, S, B,
and pH > 6.0-6.5.
• Legume density and health.
• Leaf area for growth (defoliation management).
Residual nitrogen contributions (‘credits’)
from legumes
Species and density
Lb available N/ac*
Alfalfa, 25-49% of stand
80
Red clover, 25-49% of stand
70
Soybeans harvested for grain
1 per bu/ac harvested
*Lb/ac = pound(s)/acre (43,560 ft2)
WVDA, 2009
Example annual nutrient uptakes by forage crops. If crops are fed
on-farm, some nutrients recycle back to soil in manure and
urine. If crops are exported off-farm, nutrients must be replaced
by mineralization of soil organic matter, nutrient release from
soil parent material, or fertilizer and manure applications.
Species
DM yield
N
P
K
lb/ac
Alfalfa
16,000
450
35
400
Orchardgrass
12,000
300
45
310
Silage corn
24,000
240
45
250
Note higher N-use efficiency (lb DM production/lb N) for silage
corn (C4 grass) than for orchardgrass (C3 grass).
Feed values expressed as elements and fertilizer values expressed
as oxides. Crude protein (CP) = N x 6.25 (17% CP = 2.7% N)
E. Rayburn (ed.). 2006. Forage production for pasture-based livestock production. NRAES-172.
Natural Resource, Agriculture, and Engineering Service, Cooperative Extension, Ithaca, NY.
Fertilizer mixtures
• Simple (‘straight’) fertilizer: single nutrient,
e.g. urea, ammonium nitrate, or triple
superphosphate.
• Compound or mixed fertilizer: contains at least
2 of the 3 primary macronutrients N, P, and K,
e.g., diammonium phosphate (DAP).
• Complete fertilizer: contains the 3 primary
macronutrients N, P, and K, e.g., 10-20-20.
Definition of a ‘unit’ of fertilizer
• Variable interpretations; check with dealer/
applicator/grower to be sure what is meant.
• Typically, 1 unit is equivalent to 1 lb of plant nutrient,
e.g., 1 unit of N = 1 lb of N.
• Example: spreading urea (46-0-0) at 300 lb bulk
material/ac delivers 138 lb N/ac or 138 ‘units’ of N/ac
(300 x 0.46 = 138).
• Sticking to lb of nutrient applied is probably safer
terminology.
Fertilizer grade and P and K interconversions
• Soil test results are usually in elemental form for N, P,
and K (ppm or lb/ac of available element).
• Fertilizer grading and application recommendations are
usually in terms of oxide forms for P and K.
• P2O5 and K2O are remnants from when contents of
minerals were expressed as the oxides formed upon
heating.
• Grade: Percentages by weight of
total N - available P2O5 - soluble K2O (and S, if a fourth).
• Nutrient budgets and calculations often concern
elemental, rather than oxide forms, e.g., N, P, and K
rather than N, P2O5, and K2O. Why do we do this to
ourselves?
P and K interconversions
• (No interconversions necessary for N)
• Phosphorus (element vs oxide):
P x 2.29 = P2O5
P2O5 x 0.44 = P
• Potassium (element vs oxide):
K x 1.2 = K2O
K2O x 0.83 = K
Common fertilizer sources, grades, and prices
(09/25/09), delivered locally
Grade*
(% N-available P2O5-soluble K2O)
Price**
($/ton material)
Urea
46-0-0
449
NH4NO3 (ammonium nitrate)
34-0-0
Not available
(NH4)2SO4 (ammonium sulfate)
21-0-0 (+ 24 S)
339
DAP (diammonium phosphate)
18-46-0
519
Triple super phosphate
0-46-0
529
KCl (muriate of potash)
0-0-60
950
19-19-19
19-19-19
609
20-10-10
20-10-10
449
Material
*Nutrient concentrations can vary slightly among sources of a material. See label.
**Add approx. $8.50/ac to spread
Note differing costs of P from DAP vs triple superphosphate (what are they?)
Fertilizer application rates - 1
• Fertilizer rate recommendations are typically given in
lb/ac of N, P2O5, K2O, and S.
• To convert recommendations to lb/ac of fertilizer
material:
lb nutrient recommended/ac x 100
% nutrient in fertilizer material
= lb fertilizer material needed/ac
Example 1: To supply 120 lb N/ac using ammonium
nitrate (34-0-0):
120 x 100 = 353 lb/ac of 34-0-0
34
Mahler, 2002
Fertilizer application rates - 2
Example 2: To supply 120 lb N/ac using urea (45-0-0)
and 40 lb P2O5/ac using triple superphosphate (0-44-0):
For N, 120 x 100 = 267 lb/ac of 45-0-0
45
For P2O5, 40 x 100 = 91 lb/ac of 0-44-0
44
Mahler, 2002
Fertilizer application rates - 3
Example 3: To supply 120 lb N/ac using urea (45-0-0)
and DAP (16-48-0), and 40 lb P2O5/ac using DAP:
Start with the amount of DAP needed to meet the
P2O5 requirement: 40 x 100 = 83 lb/ac of 16-48-0.
48
Then calculate N supplied by rate of DAP application:
83 lb x 0.16 = 13 lb N/ac
Lastly, calculate remaining N requirement to be met
by urea: 120 – 13 = 107 lb N/ac from urea
107 x 100 = 238 lb/ac of 45-0-0
45
Example problems - 1
Q. How much 34-0-0 is needed to apply 30 lb of N?
A. 88 lb (30 ÷ 0.34) of 34-0-0.
Q. If 15-8-10 is used to apply 45 lb of N, how much P2O5
and K2O is also be applied?
A. 300 lb (45 ÷ 0.15) of 15-8-10 is used to apply 45 lb of
N. Therefore, 24 lb of P2O5 (300 X 0.08) and 30 lb of
K2O (300 X 0.10) are also applied.
(McCauley et al. 2009)
Example problems - 2
Q. How much urea (46-0-0) is needed to apply 135 lb
N/ac to 30 ac?
A. First, calculate how much urea is needed to provide
135 lb of N/ac. This is 293.5 lb (135 ÷ 0.46). The total
urea needed for 30 ac is then 8,804 lb (293.5 X 30) or
4.4 tons (2,000 lb/ton).
(McCauley et al. 2009)
Example problems - 3
Q. How much N, P, and K are in a 50 lb bag of 16-6-12?
Recall that the numbers in the fertilizer grade are
percentages by weight and can be expressed as
decimal fractions (i.e., 6% = 0.06).
A. Since a conversion factor is not required for N,
N content = 0.16 x 50 lb = 8 lb of N
The conversion factor for P2O5 is 0.44, so
P content = 0.06 x 0.44 x 50 lb = 1.3 lb of P
The conversion factor for K2O is 0.83, so
K content = 0.12 x 0.83 x 50 lb = 5 lb of K
(McCauley et al. 2009)
Problem 1
How many lb of N are in one ton of 34-0-0?
Problem 2
A producer has applied 200 lb/ac of ammonium
sulfate (21-0-0-24). How much N and S were
applied/ac?
lb N/ac:
lb S/ac:
Problem 3
How many lb of K are removed from the soil by 5 tons
of alfalfa hay, assuming the hay has 2.5% K in the dry
matter (DM)? Assume that air-dry hay is approx. 85%
DM, i.e., 15% H2O (‘moisture’)
Total lb air-dry hay:
Total lb hay DM:
Total lb K removed:
Problem 4
How much fertilizer N, P2O5, and K2O will be needed to
replace the N, P, and K removed in alfalfa hay yielding 6
tons dry matter (DM)/ac annually (from 4 harvests)? The
DM contains 4% N, 0.3% P, and 3% K
Total DM yield, lb/ac:
Total N, P, and K removals, lb/ac:
Total P2O5 and K2O removals, lb/ac:
Does N need to be replaced?
P2O5 and K2O replacement required, lb/ac:
Problem 5
If a fertilizer spreader applies 10 lb of material to
a 300-square foot area, approximately how
many tons would it apply over an acre (43,560
ft2/ac)?
Problem 6
If a fertilizer dealer mixes 1000 lb each of 45-0-0,
0-45-0, and 0-0-60, approximately what analysis of
fertilizer has the dealer made?
If a fertilizer dealer mixes 1000 lb each of 45-0-0,
11-52-0 (MAP), and 0-0-60, approximately what
analysis of fertilizer has the dealer made?
Fertilizing a 40-ac field for corn - 1
A farmer will fertilize a 40-acre field for corn. Nutrient
requirements for the crop are: 120 lb N/ac, 150 lb
P2O5/ac, and 180 lb K2O/ac.
Commercial fertilizers that are available are urea,
diammonium phosphate (DAP), and muriate of potash
(KCl).
A. To meet these nutrient requirements, the amount of
DAP (tons) to be applied to the whole field is:
Total field P2O5 requirement:
DAP requirement:
Fertilizing a 40-ac field for corn - 2
B. The urea (tons) that will be added to the blend to
meet N requirements for the entire field is:
Total field N requirement:
Less N being provided by DAP:
Urea requirement:
C. The muriate of potash (tons) that will be added to
the blend to meet K requirements for the entire field
is:
Total field K2O requirement:
KCl requirement:
Fertilizing a 40-ac field for corn - 3
D. What rate of blended material (complete fertilizer)
will be applied/ac to meet crop requirements?
DAP:
Urea:
KCl:
Fertilizing a 40-ac corn crop following
grass-legume hay
Based on soil test results, recommended nutrient
application rates for a corn crop are: 150 lb N/ac, 60 lb
P2O5 /ac, and 140 lb K2O/ac.
However, this corn is following a hay crop that was a
grass and red clover mixture. Red clover constituted
about 40% of the crop stand.
How much urea will be needed to meet the crop N
requirement?
Nutrient availability in a field
Soil test results show that available P in a field was 180
lb/acre. A corn crop that was harvested from this field
removed 20 lb P/acre. How much plant-available P was
left in the field after the crop harvest?
What information do we need to answer this?
P fixation by soil:
Available P released from parent material:
Available P from mineralization of organic matter
including manure:
Problem 7
Urease activity is greatest:
[ ] a. Below 50o F
[ ] b. Between 50o F and 100o F
[ ] c. Under dry soil conditions
[ ] d. Below pH 6.5
(McCauley et al. 2009)
Additional resources
• D.B. Beegle. Undated. Comparing fertilizer materials. Agronomy
Facts 6. Penn State Univ. Dept. of Crop and Soil Sciences Cooperative Extension,
http://cropsoil.psu.edu/extension/facts/agfacts6.cfm
• G.D. Binford. 2006. Commercial fertilizers. Chap. 8 in The midAtlantic nutrient management handbook. MAWP 06-02. MidAtlantic Regional Water Program,
http://www.mawaterquality.org/publications/pubs/manhcompl
ete.pdf
• R.L. Mahler. 2002. Fertilizer primer. Terminology, calculations,
and application. CIS 863. Univ. of Idaho Extension,
http://www.cals.uidaho.edu/edComm//pdf/CIS/CIS0863.pdf
• A. McCauley, C. Jones, and J. Jacobsen. 2009. Commercial
fertilizers and soil amendments. Nutrient Management Module
No. 10. 4449-10. Montana State Univ. Extension,
http://msuextension.org/publications/AgandNaturalResources/
4449/4449_10.pdf
Add next time
• Pre-sidedress N (NO3?) test.
• Fall stalk N test (for lux. consumption NO3?);
D. Beegle?
• Cap soybean N credit at 40 lb? (Craig Yohn
may have been the person who mentioned
this).
Also pH-dependent:
•Al (toxic, like Fe)
•bacteria
•fungi
•mineralization of
OM
•earthworms
(optimum for all of
above is pH 6-7)
•pH scale is
logarithmic; pH 5
and 7 are each 10x
different from pH 6
Grass response to N rates
Typical cool-season forage grass N
rates range up to 150-200 lb
N/ac-yr. Forage production
responds to higher N rates, but
above 180 lb N/ac there is
increased risk of NO3 leaching to
groundwater
Growth increment =
10-50 lb forage DM/lb N:
Lower efficiency for C3 and at
higher application rates
Higher efficiency for C4 and at
lower application rates
For WV cool-season grasses,
assume 20-30 lb forage DM
production/lb N applied (how
does the value of 25 lb forage DM
compare to cost of 1 lb N?)
Interaction of N x S; alleviating soil S
deficiency improves efficiency of N
response; same concept holds for
other nutrients
Tall fescue response to inorganic and organic N sources
E. Rayburn (ed.). 2006. Forage production for pasture-based livestock production. NRAES-172.
Natural Resource, Agriculture, and Engineering Service, Cooperative Extension, Ithaca, NY.
Example of timothy response to N rates