Question

Consider a field to be planted with corn. Corn requires 100 kg N per hectare and cannot
use N2 from the atmosphere (as soybeans can). The farmer is going to apply manure to
supply the nitrogen. The manure has the following composition:Total N: 12 g N per kg manure, of this
- Organic N: 6 g N per kg manure
- NH4: 6 g N per kg manure
Total phosphates (P2O5): 5 g P per kg manure
Total potassium (K2O): 4 g K per manure
But not all the N is biochemically available to the corn. For example, only 33% of the
organic nitrogen is available. 50% of the inorganic nitrogen (e.g., NH4) is available. Given
these efficiencies, how much manure should be applied? How much P is applied to the field
as a result of this manure load?

Answer 1

Let x kg of manure be applied

Now, mass of organic N = 6x g = 0.006*x kg

Mass of inorganic N = 6x g = 0.006*x Kg

Mass of organic N available to corn = 33% of 0.006*x = 0.00198*x kg

Mass of inorganic N available to corn = 50% of 0.006*x = 0.003*x kg

Now, total N required per hectare = 100 kg

Thus, (0.00198*x) + (0.003*x) = 100

or, 0.00498*x = 100

or, x = manure to be applied = 20080.32 kg.............(1)

Now, Total P applied = 5g per kg of manure = 0.005 kg per kg of manure = 0.005*20080.32 = 100.402 kg........(2)