Transcript Slide 1

Part 2: Answers to practical
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
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
Problem 1
How many lb of N are in one ton of 34-0-0?
1 ton = 2000 lb 34-0-0
2000 lb x 100 or 2000 lb
34
0.34 (my preference)
= 680 lb N
INCORRECT!! REDO!!
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: 200 lb x 0.21 = 42 lb N/ac
lb S/ac: 200 x 0.24 = 48 lb S/ac (a high rate!)
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: 5 tons x 2000 lb/ton = 10000 lb
Total lb hay DM: 10000 lb x 0.85 DM concentration = 8500 lb DM
Total lb K removed: 8500 lb DM x 0.025 K = 212 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: 6 tons DM x 2000 lb/ton = 12000 lb DM
Total N, P, and K removals, lb/ac:
12000 x 0.04 = 480 lb N
12000 x 0.003 = 36 lb P
12000 x 0.03 K = 360 lb K
Total P2O5 and K2O removals, lb/ac: 36 lb P x 2.29 = 82 lb P2O5
360 lb K x 1.2 = 432 lb K2O
Does N need to be replaced? Not by fertilizer; rely instead on N2
fixation by properly-nodulated alfalfa at correct soil pH (> 6.5).
Problem 5
If a fertilizer spreader applies 10 lb of material to a 300square foot area, approximately how many tons would
it apply over an acre (43,560 ft2/ac)?
300 = 0.0069
43560
10 lb material = 1452 lb/ac
0.0069 ac
1452 lb
2000 lb/ton
= 0.73 ton material/ac
Problem 6
If a fertilizer dealer mixes 1000 lb each of 45-0-0, 0-45-0, and 0-060, approximately what analysis of fertilizer has the dealer made?
1000 lb 45-0-0 = 450 lb N
450 lb N in 3000 lb blend = 15% N
1000 lb 0-45-0 = 450 lb P2O5 450 lb P2O5 “
“ = 15% P2O5
1000 lb 0-0-60 = 600 lb K2O 600 lb K2O
“
“ = 20% K2O
(15-15-20)
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?
Same approach as above, but 1000 lb 11-52-0 contains 110 lb N as
well as 520 lb P2O5, so total N in 3000 lb blend = 450 lb from 45-0-0
+ 110 lb N from MAP = 560 lb = 18.7% N in blend (19-17-20).
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 (45-0-0),
diammonium phosphate (DAP, 16-48-0), and muriate of potash
(KCl, 0-0-60).
A. To meet these nutrient requirements, the amount of DAP
(tons) to be applied to the whole field is:
Total field P2O5 requirement: 150 lb x 40 ac = 6000 lb P2O5
DAP requirement: 6000 lb = 12500 lb = 6.25 tons DAP.
0.48
*Note DAP also provides 12500 lb x 0.16 N = 2000 lb N
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: 120 lb x 40 ac = 4800 lb N
Less N being provided by DAP: 4800 – 2000 lb from DAP = 2800 lb
Urea requirement: 2800 lb N = 6222 lb urea = 3.11 tons
0.45
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: 180 lb/ac x 40 ac = 7200 lb K2O
KCl requirement: 7200 = 12000 lb KCl = 6.0 tons
0.60
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: 6.25 tons (= 2000 lb N + 6000 lb P2O5)
Urea: 3.11 tons (= 2800 lb N)
KCl: 6.0 tons (= 7200 lb K2O)
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
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?
Using previous table, assign available N credit of 70 lb/ac, so
urea requirement will be to meet 150-70 = 80 lb N/ac
(= 178 lb 45-0-0)
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?
I have no simple answer to this, other than listing the information
that we need to provide to answer this, some or all of which you no
doubt covered in your workshop:
P fixation by soil:
Available P released from parent material:
Available P from mineralization of organic matter including manure:
Other?
Problem 7
Urease activity is greatest:
[ ] a. Below 50o F
[x] b. Between 50o F and 100o F
[ ] c. Under dry soil conditions
[ ] d. Below pH 6.5
Urease is the enzyme excreted by soil microbes that
converts urea to ammonium ion (NH4+). Its activity
level is temperature-, pH-, and moisture-dependent
(see McCauley et al. 2009).
(McCauley et al. 2009)