3.6.1 fertilizer materials

8/13/2019 3.6.1 Fertilizer Materials

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Fertilizer Materialsand Calculation

Nutrient Management

Part 1: Understanding the PalayCheck Sy

stem


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Fertilizerany substance which is added to the soil

or applied to the plant to supply the elements

essential for plant nutrition

Fertilizer gradethe guaranteed percentage of total

nitrogen (N), available phosphorus (P2O5), and water

soluble potassium (K2O), in that order, in a fertilizer

material

FERTILIZER TERMINOLOGIES

16% N

20% P2O5

0% K2O16-20-0


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Total plant foodthe total percentage of nutrients

(usually of N,P,K and S) in a fertilizer material

17-0-17-12 = 46% total plant food14-14-14-12 = 54% total plant food

Carriera material that is necessary to hold the

fertilizer element in a form suitable for use

100%46% total plant food = 54% carrier



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Fillera material used to provide the fertilizer with the

desired amount of nutrient per unit of weight, to make

application easier, or to reduce the absorption of water

from the air

Fertilizer recommendationthe amount (kg) of N,P2O5,

and K2O recommended per hectare

90+30+60 = 90 kg N, 30 kg P2O5, and 60 kg K2O per

hectare application



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Classification of Fertilizer Materials

According to form

Natural organic

Chemical or inorganic

Natural chemical

Synthetic chemical

According to fertilizerelements present

Single

Incomplete

Complete

Mixed


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1. Natural organicany fertilizer of plant and/or animalorigin that has undergonedecomposition.

Examples:Commercial organic fertilizer (OF),

compost, chicken manure

Organic

Matter (%) C:N Ratio

Moisture

Content (%)

N+P2O5+K2

O (%)

Pure Organic 20 12:1 35 7

Fortified

Organic 20 12:1 35 > 7

According to form

Commercial OF may either be pure organic or fortified organic:


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2. Chemical or inorganicany fertilizer material which isnatural or synthetic product of chemical reaction of

certain raw materials.

Natural chemicalrock phosphate, Chilean nitrate

of soda, most potassium materials

Synthetic chemicalammonium sulfate, ordinary

superphosphate, ammonium phosphate, complete

fertilizer; Urea is synthetic but in organic form.

According to form


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1. Single fertilizerscontain only one of the 3 majorfertilizer elements (N,P, and K)

Examples:Urea, Ammonium sulfate, ordinary

superphosphate, muriate of potash, sulfate of potash

2. Incomplete fertilizerscontain 2 of the 3 major

fertilizer elements

Examples:ammonium phosphate, 17-0-17, potassium

nitrate, potassium phosphate, urea superphosphate

(20-10-0)

According to number of fertilizer elements present


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3. Complete fertilizerscontain all the 3 major fertilizerelements

Examples:14-14-14; 17-7-17; 10-15-15; 12-12-12

4. Mixed fertilizerscontain 2 or more of the major

fertilizer elements that are supplied by 2 or more

fertilizer materials

Examples:ammonium phosphate + muriate of potash;

urea superphosphate (20-10-0)

According to number of fertilizer elements present


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Why study fertilizer grades

and fertilizer computation?

Consider this:

If Shane needs a teaspoon of sugar to sweeten her

juice but there is no sugar available, can she use honey

instead?

If Aling Mena does not have enough soy sauce for the

adobe shes cooking, can she add salt? How much?

If Mang Ado needs to apply fertilizer at a recommended

rate of 90+30+60 but no complete fertilizer is available,

what other fertilizers and how much of it should he use

to meet the recommendation?


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50 kg

Complete fertilizer

14% nitrogen

14% phosphorus

14% potassium

50 kg sack:

7 kg nitrogen

7 kg phosphorus

7 kg potassium

Common fertilizer grades in the market

50 kg

Ammonium phosphate

16% nitrogen

20% phosphorus

0% potassium

50 kg sack:

8kg nitrogen

10kg phosphorus

0kg potassium


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50 kg

Urea

46% nitrogen0% phosphorus

0% potassium

50 kg sack:

23 kg nitrogen0 kg phosphorus

0 kg potassium

Other common fertilizers:

Ammonium sulfate (21-0-0)

Solophos (0-18-0)

Muriate of potash (0-0-60)Urea superphosphate (21-10-0)

46% x 50 kg

100


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Fertilizer Calculation

Amount of Fertilizer

Material (kg)

Recommended Rate (kg nutrient/ha) x Area (ha)=

Percent Nutrient in Fertilizer Material

If Mang Ado needs to apply fertilizerat a recommended rate of

90+30+60 but no complete fertilizer

is available, what other fertilizers

and how much of it should he use tomeet the recommendation?


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Use a combination of

incomplete and single


Use a combination of 3

single fertilizers


complete and single

fertilizer materialsOPTIONS


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Option 1. To calculate the amount of fertilizer

materials for 1 ha at the recommended

rate of 90 + 30 + 60 using acombination of single fertilizers:

21-0-0 (ammonium sulfate)

0-18-0 (ordinary superphosphate) 0-0-60 (muriate of potash or potassium chloride)


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Therefore, Mang Ado must apply 428.6 kg 21-

0-0, 166.7 kg 0-18-0, and 100.0 kg 0-0-60 to

satisfy the recommended rate of 90 + 30 + 60.

Amount of 21-0-0 (kg) 90 kg N/ha x 1 ha

0.21= = 428.6

Amount of 0-18-0 (kg) 30 kg P2O5/ha x 1 ha

0.18= = 166.7

Amount of 0-0-60 (kg)

60 kg K2O/ha x 1 ha

0.60= = 100.0


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16-20-0 (ammonium phosphate 46-0-0 (urea)

0-0-60 (muriate of potash)

Option 2. To calculate the amount of fertilizer

materials for 1 ha at the recommended

rate of 90 + 30 + 60 using a

combination of incomplete and single

fertilizers:


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Amount of 16-20-0 (kg) 30 kg P2O5/ha x 1 ha

0.20= = 150.0

Amount of 46-0-0 (kg) 90 kg Nkg N from 150 kg 16-20-0/ha x 1 ha0.46

=

90(150 x 0.16)

0.46= 9024

0.46= 66

0.46=

= 143.5

Amount of 0-0-60 (kg)60 kg K2O/ha x 1 ha

0.60= = 100.0

Therefore, he must apply 150.0 kg 16-20-0, 143.5

kg 46-0-0, and 100.0 kg 0-0-60 to satisfy the

recommended rate of 90 + 30 + 60


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Amount of 14-14-14 (kg)30 kg P2O5/ha x 1 ha

0.14= = 214.3

Therefore, he must apply 214.3 kg 14-14-14,

130.4 kg 46-0-0, and 50.0 kg 0-0-60 to satisfy the

recommended rate of 90 + 30 + 60

Amount of 46-0-0 (kg) 90 kg N

kg N from 214.3 kg 14-14-14/ha x 1 ha0.46

=

90(214.3 x 0.14)

0.46= 9030

0.46= 60

0.46= 130.4=

Amount of 0-0-60 (kg) 60 kg K2Okg K2O from 214.3 kg 14-14-14/ha x 1 ha

0.60=

60(214.3 x 0.14)

0.60

=6030

0.60

=30

0.60

= 50.0=


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Determine the least

expensive fertilizer

combination that can

satisfy the recommended

rate (90 + 30 + 60).


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21-0-0428.6 kg

50 kg/bag= = P 4,714.60x P 550/bag

0-18-0166.7 kg

50 kg/bag= = P 1,833.70x P 550/bag

0-0-60

100.0 kg

50 kg/bag= = P 1,650.00x P 825/bag

TOTAL COST = P 8,198.30

Cost of option 1. combination of single fertilizers:


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16-20-0150.0 kg

50 kg/bag= = P 2,460.00x P 820/bag

46-0-0143.5 kg

50 kg/bag= = P 2,554.30x P 890/bag

0-0-60100.0 kg

50 kg/bag= = P 1,650.00x P 825/bag


Cost of option 2. combination of

incomplete and

single fertilizers:


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14-14-14214.3 kg

50 kg/bag= = P 3,471.66x P 810/bag

46-0-0130.4 kg

50 kg/bag= = P 2,321.12x P 890/bag

0-0-60

50.0 kg

50 kg/bag= = P 825.00x P 825/bag


Cost of option 3. combination of complete

and single fertilizers:


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incomplete and single


Use a combination of 3

single fertilizers


complete and single



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CREDITS

Instructional presentation designer:Ms. Ev Parac

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Produced in 2011.Text: 0920-911-1398

3.6.1 fertilizer materials

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