Question 4 of 6: Land Application Nutrient Management

Question 4: Land Application Nutrient Management (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Surface-applied liquid swine manure, 5 wt% dry solids; total production 3,000 m³; starter N credit 20 kg/ha; corn N requirement 160 kg/ha/yr; Zn accumulation limit 50 kg/ha; manure (per kg dry solids): organic N 0.02, $\text{NH}_4$-N 0.12, $\text{NO}_3^-$-N 0; ammonia volatilization 30% of applied $\text{NH}_4$-N; organic-N mineralization 40% in the year applied, 20% in the following year (at steady annual application); manure density 1000 kg/m³; Zn content 500 mg/kg dry solids.

Find. The annual (2nd-year) manure application rate per hectare, the total land area required, and the useful life of the site before the cumulative zinc limit is reached.

Approach. Build the plant-available nitrogen supplied per kg of dry solids applied at steady annual application (current year's $\text{NH}_4$-N after volatilization, plus current year's organic-N mineralization, plus the residual mineralization credit from the previous year's application); size the application rate to meet the crop's manure-supplied N need (crop requirement less the starter credit); scale to land area from total manure volume; then find the useful life from the same steady annual dry-solids rate against the Zn limit.

1. Available nitrogen per kg dry solids, steady 2nd-year application. $\text{NH}_4$-N is applied and (largely) taken up or lost within the year it is applied, after a 30% volatilization loss: $0.12\times(1-0.30)=0.084$. Organic N mineralizes 40% in its own application year and, at steady annual application, the previous year's organic N contributes a further 20% in year 2: $0.02\times0.40+0.02\times0.20=0.008+0.004=0.012$. Total: $$N_{avail} = 0.084+0.012 = \boxed{0.096\ \text{kg available N/kg DS}}$$
2. Application rate. The starter application already supplies 20 kg N/ha directly, so the manure need only supply $160-20=140\ \text{kg N/ha}$: $$\text{DS rate} = \dfrac{140}{0.096} = 1{,}458.3\ \text{kg DS/ha}$$ At 5% solids content, the wet manure mass rate is $1{,}458.3/0.05=29{,}167\ \text{kg/ha}$, and at $\rho=1000\ \text{kg/m}^3$ this is $$\boxed{29.17\ \text{m}^3/\text{ha (2nd-year rate)}}$$
3. Land area requirement. $$A = \dfrac{3{,}000\ \text{m}^3}{29.17\ \text{m}^3/\text{ha}} = \boxed{102.9\ \text{ha}}$$
4. Useful life of the site (zinc limit). At the steady 2nd-year dry-solids rate of 1,458.3 kg DS/ha/yr and 500 mg Zn/kg DS $=5\times10^{-4}\ \text{kg Zn/kg DS}$: $$m_{Zn} = 1{,}458.3\times5\times10^{-4} = 0.729\ \text{kg Zn/ha/yr}$$ $$\text{Life} = \dfrac{50}{0.729} = \boxed{68.6\ \text{years}}$$