Given. No manure analysis → use Table 3 (Typical Manure Analysis by Livestock Type), Solid Manure, Poultry–broilers row: %Total N = 2.73, %P = 1.30, %K = 1.45. Application Oct 15 falls in the "Early Fall" window (Sept 21–Nov 9) per Table 4's own date bands; "incorporated within 24 hrs" selects the Incorporated (<24 hours) columns of both tables. Section C's own conversion factors: %available P₂O₅ = %P×0.92; %available K₂O = %K×1.08; and the "Convert to METRIC" block gives kg/tonne = %×10 for all three nutrients.
Find. Available N, P₂O₅ and K₂O, in kg/tonne of manure, for this application.
Approach. Available N uses Method 1 (Table 3 %Total N × Table 4's Available-N proportion for Solid Poultry–Broilers, Incorporated/Early Fall); available P₂O₅ and K₂O use Section C's oxide-conversion factors directly on Table 3's %P/%K (no incorporation/season adjustment applies to P or K).
Step 1 — Available N (Method 1). Table 4, Solid Poultry – Broilers row, Incorporated (<24 h) / Early Fall column: available-N factor = 0.39.
$$\%N_{\text{avail}} = 2.73\% \times 0.39 = 1.0647\%$$
$$\boxed{N_{\text{avail}} = 1.0647\% \times 10 = 10.65\ \text{kg/tonne}}$$
Step 2 — Available P₂O₅. Section C: %P × 0.92 = % available P₂O₅, then ×10 for kg/tonne.
$$\%P_2O_{5,\text{avail}} = 1.30\% \times 0.92 = 1.196\%$$
$$\boxed{P_2O_{5,\text{avail}} = 1.196\% \times 10 = 11.96\ \text{kg/tonne}}$$
Step 3 — Available K₂O. Section C: %K × 1.08 = % available K₂O, then ×10 for kg/tonne.
$$\%K_2O_{\text{avail}} = 1.45\% \times 1.08 = 1.566\%$$
$$\boxed{K_2O_{\text{avail}} = 1.566\% \times 10 = 15.66\ \text{kg/tonne}}$$
