Replacing nutrients removed from cutting hay

The abundant pasture production last spring led to many paddocks being cut for hay or silage.

Removing this feed from a paddock also removes nutrients, particularly potassium but also phosphorus and sulfur. If these nutrients are not replaced, you run the risk of pastures not performing as well as they should, and this may lead to the appearance of low-fertility plants such as dandelions and sweet vernal grass.

Table 1 highlights how much phosphorus, potassium and sulfur is removed in every tonne of dry matter (DM).

Table 1. Nutrients removed with each tonne of dry matter harvested

Where a range is shown in Table 1 for the amount of nutrient removed, use the higher value if the hay is cut earlier in the season or has a large amount of green leafy material.

Two examples provide an illustration of the calculations required to estimate the amount of nutrients that need to be replaced.

Example 1: Hay

Thirty 5 m × 4 m round bales were baled off a three-hectare pasture paddock late in the season.

The harvested weight of these bales was approximately 350 kg each (wet weight), so 10,500 kg (30 × 350 kg) was harvested off the paddock. The harvested weight needs to be converted to a DM weight. Hay has a dry matter of roughly 85 per cent, so 10,500 kg × 0.85 = 8925 kgDM or 8.9 tDM.

So how much potassium, phosphorus and sulfur was removed? For late-cut hay, 15 kg of potassium, 2 kg of phosphorus and 2 kg of sulfur are removed for every tonne of DM harvested:

• 15 kg/tDM × 8.9 tDM = 133.5 kg potassium removed.
• 2 kg/tDM × 8.9 tDM = 17.8 kg phosphorus removed.
• 2 kg/tDM × 8.9 tDM = 17.8 kg sulfur removed.

To convert the amount of nutrient removed from the paddock in the hay to kilograms of nutrient per hectare, simply divide the amount of nutrient removed in the hay by the number of hectares it was cut from. In this case, it was a three-hectare paddock:

* Potassium—133.5 kg ÷ 3 ha = 44.5 kg/ha needs to be reapplied.

* Phosphorus—17.8 kg ÷ 3 ha = 6 kg/ha needs to be reapplied.

* Sulfur—17.8 kg ÷ 3 ha = 6 kg/ha needs to be reapplied.

Example 2: Silage

Fifteen 4 m×4 m round bales of silage were harvested off a three-hectare paddock. The weight of these bales was approximately 700 kg each, so a total of 10,500 kg (15 × 700 kg) was harvested. Baled silage has a DM of about 40% (but this can vary), so 10,500 kg × 0.4 = 4200 kgDM or 4.2 tDM.

So how much potassium, phosphorus and sulfur has been removed? For silage, 27 kg of potassium, 4.3 kg of phosphorus and 2 kg of sulfur were removed for every tonne of DM harvested:

• 27 kg/tDM × 4.2 tDM = 113 kg potassium removed.
• 4.3 kg/tDM × 4.2 tDM = 18 kg phosphorus removed.
• 2 kg/tDM × 4.2 tDM = 8.4 kg sulfur removed.

Removal per hectare:

• Potassium—113 kg ÷ 3 ha = 38 kg/ha needs to be reapplied.
• Phosphorus—18 kg ÷ 3 ha = 6 kg/ha needs to be reapplied.
• Sulfur—8.4 kg ÷ 3 ha = 2.8 kg/ha needs to be reapplied.

When contacting your fertiliser supplier, quote the figures for kilograms of nutrient per hectare you require to replace the nutrients. The supplier will either have a product that closely matches the nutrient content of what you require or be able to blend a product to match. For example, a super potash 1&1 fertiliser will have approximately 4.4 per cent phosphorus, 25 per cent potassium and 5.5 per cent sulfur, so applying 150 kg/ha of this product would supply 37.5 kg/ha of potassium, 6.6 kg/ha of phosphorus and 8.2 kg/ha of sulfur, which is close to requirements in both of these examples.

These calculations assume that none of the fodder cut is being fed back onto the paddock it is cut from. For example, if you think that about 30 per cent of the fodder will be fed back onto the paddock it was cut from, you can reduce the replacement nutrient requirements by 30 per cent.

Note that these example are just replacing the nutrients that have been removed from the paddock by harvesting the fodder and do not include a maintenance application—the annual maintenance fertiliser requirements that should still be applied.