Question 4 of 5: In-Vessel Composting of Cattle-Farm Waste

Question 4: In-Vessel Composting of Cattle-Farm Waste (25 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.

Find. Ash-free empirical chemical formula; a suitable amendment strategy; total (theoretical) air requirement; maximum volume reduction.

Approach. Convert the elemental mass data to moles for the empirical formula, use the raw mass ratio of carbon to nitrogen (a separate, standard composting parameter) together with the waste's moisture content to diagnose and quantify an amendment need, then scale the elemental data to the actual ash-free organic mass and apply an aerobic stoichiometric oxidation balance to estimate air demand and the maximum theoretical volume reduction.

1. Dry solids, ash, and ash-free organic mass. $m_{dry} = 20{,}000\times0.25=5{,}000\ \text{kg}$; ash $=5{,}000\times0.05=250\ \text{kg}$; ash-free organic fraction $$m_{organic} = 5{,}000-250 = \boxed{4{,}750\ \text{kg}}$$
2. Ash-free empirical chemical formula. Converting each relative mass to moles ($n_i=m_i/MW_i$): C $=120/12=10$, H $=17/1=17$, O $=100/16=6.25$, N $=8/14=0.571$, P $=2/31=0.065$. Normalizing to N = 1 (the standard composting-formula convention), $$\boxed{\text{C}_{17.5}\text{H}_{29.8}\text{O}_{10.9}\text{N}\ \text{P}_{0.11}}$$
3. Raw feedstock C:N ratio and moisture. The composting C:N ratio is a MASS ratio (distinct from the mole-basis empirical formula above); since the table's relative units are already mass units, $$C{:}N = 120/8 = \boxed{15{:}1}$$ well below the ideal composting range of 25–30:1, so the feedstock is nitrogen-rich. Moisture content follows from the dry-solids fraction: $$MC = 1-0.25 = \boxed{75\%}$$ far above the ideal 50–60% active-composting range, so the feedstock is also too wet.
4. Amendment quantity to correct both parameters. Both defects point to the same fix: a high-carbon, low-moisture bulking agent. Using typical wheat-straw properties (%N = 0.67%, C:N = 80 ⇒ %C = 80×0.67% = 53.6%) and the waste's own %C = 2,308/5,000 = 46.2%, %N = 153.8/5,000 = 3.08% of total dry solids (scaling the table to the actual 4,750 kg organic mass), solving the dry-mass mixing balance for a target blended C:N = 25 $$\frac{0.462+0.536\,x}{0.0308+0.0067\,x}=25 \;\Rightarrow\; x=\boxed{0.83\ \text{kg dry straw per kg dry waste}}$$ Scaled to the whole batch, $5{,}000\times0.83\approx\boxed{4{,}170\ \text{kg dry straw}}$ ($\approx$4,910 kg wet straw at 15% MC) is needed. Checking the resulting moisture (4 kg wet waste per kg dry waste at 75% MC, blended with wet straw at 15% MC) gives $MC_{mix}\approx63\%$ — a substantial improvement from 75%, though still slightly above the 50–60% ideal, so continued mechanical turning and pile porosity remain the practical levers for the remaining gap.
5. Total air requirement. Scaling the table (basis mass 120+17+100+8+2=247) to the actual 4,750 kg organic mass gives a scale factor $4{,}750/247=19.23$, so actual masses are C = 2,308 kg, H = 327 kg, O = 1,923 kg, N = 154 kg. The aerobic stoichiometric oxidation of $C_aH_bO_cN_d$ is $$C_aH_bO_cN_d+\left(a+\tfrac{b}{4}-\tfrac{c}{2}-\tfrac{3d}{4}\right)O_2 \rightarrow a\,CO_2+\tfrac{b-3d}{2}H_2O+d\,NH_3$$ Converting the scaled masses to moles and substituting gives $x_{O_2}\approx205{,}700\ \text{mol} = 6{,}583\ \text{kg O}_2$. At air's 23.2% O₂ mass fraction and an air density of 1.2 kg/m³, $$m_{air}=6{,}583/0.232=28{,}373\ \text{kg} \;\Rightarrow\; V_{air}=28{,}373/1.2=\boxed{23{,}640\ \text{m}^3}$$ (a mass-balance check confirms the CO₂+NH₃+H₂O product mass equals the organic input plus this O₂ mass, within rounding.)
6. Maximum volume reduction. The theoretical maximum occurs at 100% oxidation of the ash-free organic fraction, i.e. all 4,750 kg of volatile solids converts to gas, leaving only the 250 kg ash behind — a maximum mass loss of $4{,}750/5{,}000=\boxed{95\%}$ of total dry solids. Assuming roughly constant compost bulk density (no other data given), this corresponds to a maximum theoretical volume reduction on the same order (≈95%); real composting typically destroys only 40–60% of volatile solids, so a realistic field volume reduction is considerably lower, on the order of 40–60% of the 95% volatile fraction (≈38–57%).