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3. (10 pts) You grow a corn crop that yields 150 bushels ac-1, with the following...

3. (10 pts) You grow a corn crop that yields 150 bushels ac-1, with the following nutrient concentrations: 2% N and 0.25% P.

a. (7 pts) Compute the yield in kg ha-1 in addition to the crop removal rate, in kg ha-1, for N and P. Assume dry weights (i.e., 0% moisture) for yields measured in bushels.

b. (3 pts) How much urea (46-0-0) do you need to add (in lb ac-1) to meet the N removal computed in 3.a?


1 P = P2O5 x 0.436
1 K = K2O x 0.83
lb. ac-1 = multiply by 1.12 to get kg ha-1
kg ha-1 = multiply by 0.89 to get lb. ac-1
1 bushel of corn = 56 lb. bu-1
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Answer #1

3. Given that corn crop that yields 150 bushels ac-1 with 2% N and 0.25% P nutrient concentrations.

a. Also given that 1 bushel of corn = 56 lb. bu-1

Then 150 bushels ac-1 of corn = 56 x 150 = 8400 lb. ac-1

In order to get the corn yields in kg ha-1, we have to multiply lb. ac-1 by 1.12

8400 lb. ac-1 = 8400 x 1.12 = 9408 kg ha-1.

Therefore, corn yield =  9408 kg ha-1.

Nutrients removal: The standard nutrient removal rate obtained from the International Plant Nutrition Institute (IPNI) is,

0.67 lb/yield for N and 0.35 lb/yield for P2O5.

N removal rate at 8400 lb. ac-1 is 8400 x 0.67 = 5628 lb. ac-1 = 5628 lb. ac-1 x 1.12 = 6303 kg ha-1.

Therefore, N removal rate = 6303 kg ha-1.

Similarly for P2O5 removal rate at 8400 lb. ac-1 is 8400 x 0.35 = 2940 lb. ac-1 = 2940 lb. ac-1 x 1.12 = 3293 kg ha-1.

Given that 1 P = P2O5 x 0.436, by using this we can calculate P removal rate.

P = 3293 of P2O5 x 0.436 = 1436 kg ha-1.

Therefore, P removal rate = 1436 kg ha-1.

b. Now calculate the amount of urea 46-0-0 (N-P-K) need to be added (in lb ac-1) to meet the N removal rate (5628 lb. ac-1)

We don’t know the information about the soil, so we can consider it as typical mineral soil, its hectare–furrow slice of surface soil will weigh 2.2 million kilograms.

46 Kg N in 2200000 Kg of soil or 46 g N in 2200 Kg of soil

Ammonium nitrate contains 33 % of N, that means 33 mg of N present in 100 mg ammonium nitrate.

(46 x 100)/33 = 139 mg ammonium nitrate in 2.2 kg of soil.

139 x 5628 = 7823 lb. ac-1

So approximately 7823 lb. ac-1 of urea (46-0-0) to meet the N removal of 5628 lb. ac-1.

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