Immobilisation of soil nitrogen in heavy stubble loads

Page last updated: Friday, 3 November 2017 - 11:32am

Please note: This content may be out of date and is currently under review.

For growers who are able and/or willing to retain stubble the key question is:

How much fertiliser nitrogen at seeding will be used to breakdown the stubble instead of supplying the crop?

An air seeder seeding on stubble

The answer depends on your background soil N, how the stubble is treated (incorporated or left standing), the quality of the stubble and the timeframe between the break of the season and the sowing date.

Precisely determining the amount of soil N that may be immobilised is difficult, but a working example can provide some guidance regarding the likely impacts of different stubble loads on an N budget (see the scenario calculations below). It is important to remember that organic residues can take months or years to break down and therefore any N required to enable this is used over the same timeframe – so not all N will be immobilised at once.

Deep soil testing to determine background soil N prior to seeding is needed to determine fertiliser strategies. Additional N may also be made available if significant rainfall occurs prior to sowing due to the mineralisation of soil organic matter. Under optimal conditions (warm temperatures) in summer, N can be mineralised at 10-15 kg N/ha per day, though in autumn and winter this is more likely in the region of 1-5 kg N/ha per day.

Scenario (example only)                                           

A grower retains 5t/ha of wheat stubble (3.3t/ha grain yield at a harvest index of 0.4) and wants to estimate the likely impact of the stubble on soil N levels in the paddock. Assume a C:N (carbon:nitrogen) ratio of 120:1 in the stubble.

Step 1 Calculate the amount of C in stubble that is added to the soil:

Stubble (kg/ha) x (typical C% of stubble) = 5000 x 0.45 = 2250 kg C/ha.

Step 2 Calculate N present in the crop residues added to the soil, remembering that in this example stubble contains 2250kg C and wheat has a C:N ratio of about 120:1.

(Kg C/ha) ÷ (C:N) = 2250 ÷ 120 = 18.75 kg N/ha in the stubble.

Step 3 Allow for 30% of the C to be used by microbes for their growth and the remaining 70% to be respired as carbon dioxide. Microbes need N for growth but not for respiration. The amount of C used by the microbes is:

(Kg C/ha) x (0.3) = 2250 x 0.3 = 675 kg C/ha used by the microbes for their growth.

Step 4 Assume microbes have a C:N ratio of 12:1, as in they require 1kg of N for every 12kg of C to grow. The nitrogen required to decompose the stubble residue is:

(C used by microbes) ÷ (12) = 675 ÷ 12 = 56.25 kg N/ha is going to be used by the microbes for their growth.

Step 4 Calculate how much N has been immobilised. The organic matter from the stubble contained 18.75 kg N/ha and the microbes require 56.25 kg N/ha to grow. Thus the N balance = 18.75 - 56.25

= -37.5 kg N ha

This is a NEGATIVE N balance. Thus the nitrogen deficit (equivalent to 37.5 kg N/ha) will be sourced from the existing N reserves in the soil, N supply from soil organic matter turnover and subsequent fertiliser applications.

Note: Since stubble does not normally all break down in a year, it is more likely that this is the N required by microbes over a 12-36 month period and therefore it is likely that a lower amount of N would be required at the break of season than the total N deficit that this exmaple indicates. The highest risk period will be during the first 6-8 weeks after sowing.

Contact information

Craig Scanlan
+61 (0)8 9690 2174