Question

So question. We are covering fertilizer calculations in class. we did not go over examples and i know its easy but i still get confused because one little mistake can ruin the problem. Here is the question: If the K requirement of bent grass is 1lb K/1000 ft2 per year and all 18 greens of a golf course is equal to the acreage given in Q9 ( green acres 1.8), how many lb of 0-0-60 must you purchase to meet the K needs of all greens?

Answer 1

Solution –

Given data –

K requirement of bend grass = 1 lb k/1000 ft² per year

Total acreage = 1.8 green acre = 78408 ft²

Fertilizer = 0-0-60

Here 0-0-60 means the percentage by weight of N - P as P₂O₅ - K as K₂O.

So, in our given problem we have 60 % by weight K₂O.

Also, lb K = lb K₂O × 0.83 (Learn)

Total requirement of K = 1 * 78408 / 1000

                                             = 78.408 lb

Let the amount of fertilizer we need by x

The amount of K₂O in fertilizer = 0.6 * x

The amount of K = 0.6 * x * 0.83 = 0.498 * x

The amount of K in fertilizer = the requirement of K

0.498 * x = 78.408

x = 157.446 lb

The amount of 0-0-60 needed is 157.446 lb or 158 lb (approx.)