Question

Kenny Kernel, a corn producer from Wildorado, TX, annually applies 250 lb/acre of anhydrous ammonia (NH3, 82-0-0). The producer decided to switch to liquid urea ammonium nitrate (UAN-28) fertilizer (28-0-0) for the upcoming year because of availability and lower price. If 28-0-0 weighs 10.7 lbs per gallon, how many gallons of 28-0-0/acre must be applied to equal the quantity of N previously added as 82-0-0?

Answer 1

Amount of anhydrous ammonia(82-0-0) applied/acre= 250lb

82-0-0 implies 82%(w/w) of nitrogen in the fertilizer

i.e., 100 gms of fertilizer will have 82gms of N

Thus, amount of nitrogen added previously= 0.82*250lb= 205lb

Thus, previously 205lb of nitrogen were added per acre.

On switching to UAN-28, (28-0-0 implies 28% w/w)

Weight of 1 gallon= 10.7lbs

Weight of nitrogen in 1 gallon= 0.28*10.7lbs= 2.996lbs

Thus, no. of gallons required to add 205lbs of nitrogen= 205/2.996=68.42gallons