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Four Steps in Fertilizer Calculations. Determine the area Determine the rate per area Determine the amount of nutrient (area X rate) Convert amount of nutrient to amount of fertilizer. Two Basic Types of Calculations (nutrient vs. fertilizer).

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four steps in fertilizer calculations
Four Steps in Fertilizer Calculations
  • Determine the area
  • Determine the rate per area
  • Determine the amount of nutrient (area X rate)
  • Convert amount of nutrient to amount of fertilizer
two basic types of calculations nutrient vs fertilizer
Two Basic Types of Calculations(nutrient vs. fertilizer)
  • You have a given amount of fertilizer and you need to calculate how much nutrient is in it
  • You have a given amount of a nutrient to apply and you need to calculate how much fertilizer to use
memorize and understand this
Memorize and Understand This!
  • Weight of fertilizer X % of Nutrient = Weight of nutrient
  • # of urea X 46% N = # of N
  • Weight of fertilizer = (Weight of nutrient)/(% of nutrient)
  • # of urea = (# of N)/(46% N)
converting from fertilizer to nutrient
Converting From Fertilizer to Nutrient :
  • How many pounds of N are in a 50# bag of ammonium nitrate (35-0-0)?
  • 50# fert X 0.35 = 17.5 #N
  • 50# fert X (0.35#N/#fert) = 17.5 #N
converting from nutrient to fertilizer
Converting From Nutrient to Fertilizer :
  • How many pounds of ammonium nitrate (35-0-0) do you need to give you 5# of N?
  • x# fert X 0.35 = 5 #N
  • x# fert X (0.35#N/#fert) = 5 #N
  • x# fert = (5 #N)/(0.35#N/#fert) = 14.3 # fertilizer
example 1
Example 1
  • The area of a home lawn is 15,000 ft2
  • You want to apply 1#N per 1000 ft2 using ammonium sulfate (21-0-0)
  • Total amount of N = (1#N/K) X (15K) = 15#N
  • Total amount of fertilizer = 15#N/0.21 = 71# ammonium sulfate
example 1 more difficult
Example 1, more difficult
  • You want to apply 1#N per 1000 ft2 to the same lawn, using 75% IBDU and 25% urea
  • 15#N X 75% = 11.25#N from IBDU (31%N)
  • x# IBDU = (11.25#N)/(0.31#N/#IBDU) = 36.3# IBDU
  • 15#N X 25% = 3.75#N from urea (46%N)
  • x# urea = (3.75#N)/(0.46#N/#urea) = 8.2# urea
example 2 acres
Example 2, Acres
  • 1 acre = approx. 43,000 ft2 = 43K
  • 1#N/K X (43K/acre) = 43#N/acre
  • How much ammonium nitrate (33-0-0) would you buy for 75 acres if you want to apply a total of 4#N/K during the next year, in 4 separate applications? It comes in 50 # bags. How many bags would you order?
continued
Continued
  • 1#N/K = 43#N/acre
  • 4#N/K = ?#N/acre = 172#N/acre
  • For 75 acres, need: 172#N/acre X 75 acres = 12900#N
  • x # AN = (12900#N)/(0.33#N/#fert) = 39091 #AN
  • 39091 #AN/(50 #AN/bag) = 782 bags
example 3
Example 3
  • You are growing 25 acres of tall fescue and plan on applying a total of 4 #N on the following schedule: 2# N as MU (39-0-0) on May 1 1# N as 26-4-8 on Mar. 1 1# N as 26-4-8 on Oct. 1
  • Question 1: How much of each fertilizer do you need to order?
question 1 calculations
Question 1 Calculations:
  • Will apply 2#/K of both fertilizers
  • (2#N/K)(43K/acre)(25 acres) = 2150#N
  • (2150 #N)/(0.39 #N/# MU) = 5513 # MU
  • (2150 #N)/(0.26 #N/# 26-4-8) = 8269 # fert
example 3 continued
Example 3 continued
  • You also want to apply 1.5 # actual P/1000 and 3 # actual K/1000 ft2.
  • Question 2: How much actual P and K are applied in the 26-4-8 applications?
question 2 calculations
Question 2 calculations
  • There are a couple ways to approach this.
  • Figure out how much fertilizer is applied per 1000 ft2:
  • (8269 # fert)/(25 acres) = 331 # fert/acre (331 # fert/acre)/(43K ft2/acre) = 7.7 #fert/K
  • Now calculate how much actual P and K are applied in 7.7 # fert/1000 ft2.
continued1
Continued...
  • 7.7 # fert/1000 ft2 (26-4-8)
  • For P: (7.7 #fert)(.04 #P2O5/#fert) = 0.3# P2O5/1000 ft2
  • (0.3# P2O5)(0.44# P/#P2O5) = 0.14 # P/1000
  • To apply a total of 1.5 actual P/1000, need 1.36# P/1000 (1.5-0.14) from some other source. Let’s use super phosphate (0-20-0).
continued2
Continued...
  • Need 1.36 #P/1000 from 0-20-0 for 25 acres
  • How much total P? (1.36#P/K)(43K/acre)(25 acres) = 1462 #P
  • How much 0-20-0 is needed? First calculate the amount of P2O5 equal to 1462 #P: (1462#P)/(0.44#P2O5/#P) = 3323 # P2O5
  • How much 0-20-0 is needed?: (3323 # P2O5)/(0.20# P2O5 /# fert) = 16614 # 0-20-0
continued3
Continued...
  • Now figure out how much K must be applied:
  • Remember we applied 7.7 #26-4-8/1000
  • How much actual K is applied in 7.7 # fert/1000 ft2? (7.7#fert/1000)(0.08#K2O/#fert) = 0.62# K2O
  • (0.62# K2O)(0.83#P/# K2O) = 0.51# K/1000
continued4
Continued...
  • To apply a total of 3 actual K/1000, need 2.5# K/1000 (3-0.5) from some other source. Let’s use KCl (0-0-60)
  • How much extra K is needed? (2.5#K/1000)(43K/acre)(25 acres) = 2688 #K
  • Convert from actual K to K2O: (2688#K)/(0.83#K2O/#K) = 3238 # K2O
continued5
Continued...
  • How much 0-0-60 is needed?: (3238#K2O)/(0.60#K2O/#fert) = 5397 # of KCl