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The maintenance P and K requirements of a paddock in any given year will depend on the yield of the paddock and weather the paddock was primarily grazed or if there were, surplus bales cut on the paddock. Figure 1 highlights the amount of P and K removed from a cover of 1,500 grass DM/ha. Up to 10 times more K can be removed from a paddock when it is cut for bales v’s grazing. When grazed cows will recycle 90% of the K in the grass back on to the paddock in dung and urine. If baled 100% of the K in the grass is removed from the field. A typical bale of silage contains 10 units of K and 1.7 units of P.
For example if a paddock was grazed 6 times so far this year and will be grazed another three times before the end of the year the maintenance requirements can be worked out roughly as follows:
9 grazing x 3 units K/grazing = 27 units K/acre
9 grazing x 1.7 units P/grazing = 15.3 units P/acre
This changes significantly, where surplus bales are taken. For example if the same paddock was grazed 8 times in the year and cut for surplus bales once at a yield of 4 bales/acre the maintenance requirements for K are as follows:
8 grazing x 3 units K/grazing = 24 units K/acre
4 bales/acre x 10 units K/bale = 40 units K/acre
Total = 64 units K/acre
These figures can be used to work out the K and P maintenance requirements of your individual paddocks. An application of 2,000 gallons of slurry will provide anywhere from 40 to 65 units K/acre and 8 to 12 units P/acre. Using slurry is the cheapest ways to meet these P and K maintenance requirements but often there is not enough slurry to go around the entire farm. Now is a good time of the year to assess what has been applied P and K wise year to date and determine if sufficient P and K has been applied via slurry or fertiliser to meet P and K maintenance requirements.