Notes on this paper
National Exams, December 2016 — 04-Env-B6, Agricultural Waste Management (3 hours, open book, all questions to be attempted, 100 marks total).
Reference texts: Rynk et al., On-Farm Composting Handbook (NRAES-54); OMAFRA, Nutrient Management Act, 2002 and O. Reg. 267/03 / Nutrient Management Protocol (NMAN); Metcalf & Eddy, Wastewater Engineering (anaerobic digestion chapter); ASABE Standards (manure storage, land application equipment).
Question 12: Available N, P₂O₅, K₂O — Solid Broiler Manure (10 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. No manure analysis is available, so Table 3 (typical values) applies; application is Oct 15, incorporated within 24 h.
Table 3 — solid poultry (broilers), typical analysis
%Total N %P %K 2.73 1.30 1.45
Find. Available N, P₂O₅ and K₂O, each in kg/tonne, for solid broiler manure applied Oct 15 and incorporated within 24 hours.
Approach. Method 1 (no lab analysis, fall-applied): %Total N (Table 3) × the available-N factor (Table 4) gives %available N; P₂O₅ and K₂O use their own fixed unit-conversion factors from Section C (no timing/incorporation adjustment applies to P or K).
Step 1 — classify the application timing. Oct 15 falls in Table 4's "Early Fall" window (Sept 21–Nov 9), and the manure is incorporated within 24 hours, so Table 4's "Incorporated (<24 hours) / Early Fall" column applies. For "Solid Poultry — Broilers" that factor is 0.39.Step 2 — available nitrogen.
$$\%N_{\text{avail}} = 2.73\% \times 0.39 = 1.065\%$$
$$\boxed{N_{\text{avail}} = 1.065\% \times 10 = 10.6\ \text{kg/tonne}}$$Step 3 — available P₂O₅. Section C's fixed conversion (%P × 0.92 = %available P₂O₅) applies with no seasonal factor:
$$\%P_2O_{5,\text{avail}} = 1.30\% \times 0.92 = 1.196\%$$
$$\boxed{P_2O_{5,\text{avail}} = 1.196\% \times 10 = 12.0\ \text{kg/tonne}}$$Step 4 — available K₂O. Section C's fixed conversion (%K × 1.08 = %available K₂O):
$$\%K_2O_{\text{avail}} = 1.45\% \times 1.08 = 1.566\%$$
$$\boxed{K_2O_{\text{avail}} = 1.566\% \times 10 = 15.7\ \text{kg/tonne}}$$
Final Results — Question 12
Nutrient Available content N 10.6 kg/tonne (1.065%) P₂O₅ 12.0 kg/tonne (1.196%) K₂O 15.7 kg/tonne (1.566%)
Topic: Available-nutrient calculation for land-applied manure (Method 1, no lab analysis, fall application).
Key relations: %available N = %Total N × available-N factor (Table 4, timing- and incorporation-dependent); %available P₂O₅ = %P × 0.92; %available K₂O = %K × 1.08 (fixed molecular-weight-style conversions, timing-independent); kg/tonne = %available × 10.
Why this works: Nitrogen is volatile (ammonia loss to the atmosphere) and mineralizes over time, so its plant-available fraction genuinely depends on WHEN and HOW manure is applied; phosphorus and potassium are not volatile, so their "available" figures are simply unit conversions to the oxide-equivalent form used for fertilizer recommendations, unaffected by application timing.
Common pitfall: Applying Table 4's timing/incorporation factor to P₂O₅/K₂O as well as N — only nitrogen availability is timing-sensitive in this framework.
Source: OMAFRA Nutrient Management Protocol, Available Nitrogen Method 1 and Section C conversion factors (adapted from Barry, Beauchamp et al., University of Guelph, 2000).
