Soils 205-90

Lecture 18- Nutrient Management

Videos                                                             Pages in text.

Note:  Homework set #3 due by Friday Apr. 13
 29,30		  

669-736

A. Goals of Nutrient Management

    - cost-effective production of high-quality plants

    - efficient use and conservation of nutrient resources

    - maintenance or enhancement of soil quality

    - protection of the environment beyond the soil (best management practices)

 

B. Organic Nutrient Sources

1. Animal manure

  • Disposal vs utilization

- 50,000-head beef feedlot » 90,000 Mg manure/year

- at 15 Mg/hectares (7 tons/acre) » 150 kg N/ha

» 6000 hectares (1500 acres)

  • Composition

- highly variable

- examples:

Manure % water % N % P % K ppm Fe ppm Zn
Beef 80 1.8 1.1 2.5 5,000 8
Dairy 75 2.3 0.6 2.7 1,800 165
Horse 63 2.1 0.3 1.0 - 125
Swine 72 3.9 0.6 1.2 1,100 390
Poultry 35 4.1 2.5 2.6 1,000 480
Green 85 2.5 0.2 2.1 100 40
Alfalfa hay 21 2.4 0.2 1.2 - -
Wheat straw 6 0.5 0.1 0.5 - -
Leaf waste - 1.0 0.1 0.4 733 67
  • Storage and Management

- minimize nutrient loss and water pollution

- liquid lagoons

- drying

- composting

- anaerobic digestion (biogas production)

2. Sewage effluent and sludge

  • Effluent

- liquid following sewage treatment

-irrigation or overland flow

nutrient use by plants

adsorption by soil colloids

degradation of organic compounds

  • Sludge

- biosolids

suspension, partially dried cake, or dried

  • Composition of sludge (highly variable)
% water % N % P % K ppm Fe ppm Zn
80 3.3 2.3 0.3 16,000 1,740

- heavy metals determined by industrial input

3. Industrial by-products

  • municipal garbage

-mostly, incinerated or landfilled

limited landfill space and air quality concerns

- composting of organics

with nutrient-rich manures, sewage sludge, ...

  • food-processing wastes

- aquaculture, potato processing, cheese, ...

tomato pomace: 1.8 % N, 0.3 % P, 0.1 % K

fish meal: 11 % N, 4 % P, 3 % K

  • wood products

- sawdust, logyard waste, woodash, bark, ..

- high C:N ratio = need N

sawdust: 0.02 % N, 0.05 % P, 0.1 % K

- slow decomposition = mulch value

4. Utilization of organic nutrient sources

  • governed by N content and application
  • N mineralization rate varies with source
Organic Material % of N mineralized
1st year 2nd year 3rd year
Feed lot-aged 35 10 5
Dairy manure-fresh 50 15 5
Dairy Lagoon liquid 45 15 6
Poultry 70 8 5
Sewage sludge 25 10 5
Composted sludge 10 5 5
Activated sludge 40 15 5

amount needed to supply N required by plants

Year 1

Crop: winter rapeseed in northern Idaho
Yield potential: 3,000 lb/A
Estimated total N needed: 255 lb/A (FG)
N source: composted sewage sludge (35 % water)
Sludge analysis: 0.15 % mineral N and 3.8 % total N
year 1 mineralization: 10 % (above)
organic N = total - mineral = 3.8 - 0.15 = 3.65 %
available N:
inorganic = 0.15 % Þ 0.15 lb N/100 lb dry sludge
organic = 3.65 % Þ 3.65 lb organic-N/100 lb sludge
10 % of 3.65 = 0.365 lb N/100 lb dry sludge
0.15 + 0.365 = 0.515 lb total available N/100 lb dry sludge
(255 lb N/A)(100 lb dry/0.515 lb N)(T dry/2000 lb dry) = 25 T dry/A
at 35 % moisture content - (25 T dry/A)(100 T waste/65 T dry)
= 38 T/A wet composted sludge
(» 171 lb/100 ft2)

Year 2

Crop: spring wheat in northern Idaho
Yield potential: 80 bu/A
Estimated total N needed: 184 lb/A (FG)
N source: activated sewage sludge (47 % water)
Sludge analysis: 0.2 % mineral N and 4.4 % total N
year 1 mineralization: 40 % (above)
mineralizable N from composted sludge in year 2 (5 % from above):
(25 T /A)(2000 lb /T)(3.65 lb org-N/100 lb) x 5 % = 91 lb/A
N needed from activated sludge = 184 - 91 = 93 lb/A
organic N = total - mineral = 4.4 - 0.2 = 4.2 %
available N:
inorganic = 0.2 % Þ 0.2 lb N/100 lb dry sludge
organic = 4.2 % Þ 4.2 lb organic-N/100 lb sludge
40 % of 4.2 = 1.68 lb N/100 lb dry sludge
0.2 + 1.68 = 1.88 lb total available N/100 lb dry sludge
(93 lb N/A)(100 lb dry/1.88 lb N)(T dry/2000 lb dry) = 2.5 T dry/A
at 47 % moisture content - (2.5 T dry/A)(100 T waste/53 T dry)
= 4.7 T/A wet activated sludge
(» 21 lb/100 ft2)

 

5. Potential for compost use (Battelle Institute for Composting Council)

Compost
source		 Potential
supply
106 yd3		 Current
supply
106 yd3
Municipal 60 2
Biosolids 6 4
Horticultural 30 10
Agricultural 6 <0.6
TOTAL 102 <17
Application Potential
demand
106 yd3		 Currently
met
%
Landscaping 2.0 <20
Topsoil 3.7 <5
Bagged 8.0 80
Landfill cover 0.6 <5
Mine reclamation 0.2 <5
Container nurseries 0.9 50
Field nurseries 4.0 <1
Sod production 20.0 <1
Silviculture 104.0 <1
Agriculture 895 0 <2
TOTAL 1,038.4 <2
  • Conclusion:

ü only 2 % of current demand is met by compost

ü potential supply » 10 % of potential demand

ü "challenge is to develop market, not create them"

 

C. Inorganic Fertilizer Materials

1. N carriers

N2 (atmos) ¾( + natural gas)® NH3

  • compressed to liquid ¾® anhydrous ammonia

- 82 % N

  • + CO2 ¾® urea

- CO(NH2)2

- 45 % N

  • + O2 ¾® HNO3 ¾(+ NH3)® NH4NO3 (ammonium nitrate)

- 33 % N

  • + H3PO4 ¾® Ammonium Phosphates

- (NH4)2HPO4 and NH4H2PO4

- 11 - 21 % N

  • + H2SO4 ¾® (NH4)2SO4 (ammonium sulfate)

- 21 % N

  • N solutions

- NH4NO3 and urea dissolved in water

- 27 - 32 % N

2. P materials

Œ Water-soluble ü

ý Þ available

Citrate-soluble þ

(15 % ammonium citrate)

Ž Insoluble Þ Þ unavailable

(rock P)

142 g P2O5/2 mole P
¾¾¾¾¾¾¾¾¾ = 2.3 g P2O5/g P or 2.3 P2O5/1 P
31 g P/mole P

25 % P2O5 x 1/2.3 (P/P2O5) = 11 % P

and

13 % P x 2.3 (P2O5/P) = 30 % P2O5

  • rock-P + H2SO4 ¾® superphosphate

- 16 - 21 % available P2O5

+ additional H2SO4 ¾® H3PO4

  • H3PO4 + P rock ¾® triple superphosphate

- 40 - 47 % available P2O5

  • H3PO4 + NH3 ¾® ammonium phosphates

- 20 - 50 % P2O5 + N

  • Rock P

- apatites [Ca3(PO4)2.CaX]

X = F, OH, CO3,...

- highly insoluble

- some available in acid soils

3. K materials

94 g K2O/2 mole K
¾¾¾¾¾¾¾¾¾ = 1.2 g K2O/g K
39 g K/mole K

5 % K x 1.2 g K2O/g K = 6% K2O

8 % K2O x 1 g K/1.2 g K2O = 6.7 % K

- KCl (60 % K2O)

- K2SO4 (50 % K2O)

- KNO3(44 % K2O + 13 % N)

 

D. Fertilizer Label

· Fertilizer laws (state)

· N-P-K

ü Total N

ü Available (citrate soluble) P2O5

ü Water soluble K2O

for example a 12-16-5 fertilizer material contains:

- 12 % total N

- 16 % available P2O5 (7 % P)

- 5 % water-soluble K2O (4 % K)

(12-7-4 by element)

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Problems (based on the above fertilizer material):

1. ?? lbs/A to add 150 lbs/A of N?

(150 lbs N/A)(100 lbs fert./12 lb N) = 1,250 lbs fert/A

2. ?? for 40 lbs P/A?

(40 lb P/A)(2.3 P2O5 /P)(100 lb fert/16 lbs P2O5) = 575 lbs fert/A

******************************************************************

 

E. Limiting Factor Concept - "Liebig’s Barrel"

 

F. Application methods

1. Broadcast - with or without tillage

  • uniformly spread - larger areas
  • raise the general fertility level of soil
  • subject to reaction with soil

2. Banding

  • closer to seed
  • less reaction with soil
  • use lower amounts than broadcast
  • supplement broadcast

3. Irrigation water - "fertigation"

  • limited amounts and care

4. Foliar Sprays

  • esp. micronutrients
  • care for "burning"

G. Response curve

1. Yield vs profit

 

2. low fertility status = high return from fertilizer

high fertility status = lower return from fertilizer

- "squeezing out" maximum yield may be costly

 

H. Soil testing

1. Sample collection - representative sample

2. Quick chemical test for soil nutrient status

3. Commonly - pH, O.M., available P and K, NO3-N, (lime requirement)

4. Different tests in various locations

5. Requires calibration of soil test values to crop response

- need field calibration data for each crop and area

6. FG (fertilizer guide) for each crop and area

 

I. Plant tissue analysis

1. Plant part

potato petioles

sugar beet petioles

onion bulb

grain stalk

2. Depends on plant, environment

and age or time of year

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