Question

3. Assume that (because of environmental regulations) the maximum rate of poultry litter that a pasture can receive is that which supplies an amount of P equal to that taken up by the forage. The forage absorbs 250 kg N/ha and 40 kg P/ha.

a) What would be the poultry litter rate (lbs/A) you could apply if the litter contains 3.5% N and 1.7% P?

b) If only 50% of the N in the poultry litter is available to the forage, how much ammonium nitrate ((kg/ha) would you have to add to make sure the amount of available N applied with the litter and ammonium nitrate is equal to that taken up by the forage (250 kg N/ha)?

Answer 1

3) Ans:

Important conversions

1 ha = 2.4711 Acre   1 Kg =2.20462 lbs 1 lbs = 0.4536 Kg

R(max) = Maximum rate of Poultry litter that Pasture receive

= Amount of 'P' taken up by forage

= 40 Kg 'P' / ha = 40 Kg 'P' / 2.4711 Acre = 16.1875 Kg P /A

= 16.1875 x 2.20462 lbs P /A = 35.6873 lbs P/A   

Amount of 'N' can be taken up by forage

= 250 Kg 'N' / ha = 250 Kg 'N' / 2.4711 Acre = 101.17 Kg N /A

= 101.17 x 2.20462 lbs P /A = 223.04 lbs N /A

a) What would be the poultry litter rate (lbs/A) you could apply if the litter contains 3.5% N and 1.7% P? Ans: Poultry litter rate (lbs/A) =

R(max) = Maximum rate of Poultry litter that Pasture receive

= Amount of 'P' taken up by forage

= 40 Kg 'P' / ha = 40 Kg 'P' / 2.4711 Acre = 16.1875 Kg P /A

= 16.1875 x 2.20462 lbs P /A = 35.6873 lbs P/A   

So, Poultry litter rate should provide 35.6873 lbs P /A . But litter contains only 1.7 % P

therefore,  Wt. of Litter/100% = Wt. of P/1.7%

Wt. of Litter/ 100% = 35.6873 lbs 'P' / 1.7%

Therefore, we could apply Poultry Litter at the rate = 2099.253 lbs LITTER / A

b) If only 50% of the N in the poultry litter is available to the forage, how much ammonium nitrate ((kg/ha) would you have to add to make sure the amount of available N applied with the litter and ammonium nitrate is equal to that taken up by the forage (250 kg N/ha)?

Ans: Now, Maximum Poultry Litter rate = 2099.253 lbs/A

But, Litter contains only 3.5% N

therefore wt. of N in 2099.253 lbs litter can be given as,

But only 50% i.e. 1/2 of total Nitrogen from litter is available for Forage

So, Available N = 73.474 lbs / 2 = 36.737 lbs

Actually N absorption rate = Amount of 'N' can be taken up by forage

= 250 Kg 'N' / ha = 250 Kg 'N' / 2.4711 Acre = 101.17 Kg N /A

= 101.17 x 2.20462 lbs P /A = 223.04 lbs N /A

Now, Available N = 73.474 lbs / 2 = 36.737 lbs N /A

= 36.737 /2.20462 Kg N /A = 16.664 Kg N/A

= 16.664 x 2.4711 Kg N /ha = 41.18 Kg N/ha

N defficiency = Maximum N absorption (Kg/ha) - Available N (Kg/ha)

= 250 Kg 'N' / ha -  41.18 Kg N/ha

= 208.82 Kg N / ha Should provided in the form of Ammonium Nitrate.

Calculate % N in NH₄NO₃ (Ammonium Nitrate) :

MW of NH₄NO₃ = 80 units and wt. of N in NH₄NO₃ = 28 units

therefore, % N in NH₄NO₃ = (28 x 100 ) / 80 = 35 %

WEIGHT of NH₄NO₃TO PROVIDE 208.82 Kg N which is 35% in NH₄NO₃ can be given as,

WEIGHT of NH₄NO₃ = (208.82 x 100) / 35 = 596.63 Kg NH₄NO₃ / ha

Answer is 596.63 Kg NH₄NO₃ / ha

Ammonium nitrate ((kg/ha) would you have to add to make sure the amount of available N applied with the litter and ammonium nitrate is equal to that taken up by the forage (250 kg N/ha) = 596.63 Kg NH₄NO₃ / ha

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