22-Agric-B11 Principles of Waste Management · May 2014
Question 3 of 6: Composting Mixture Design
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2014 — 04-Agric-B11, Principles of Waste Management. Three-hour, open-book exam; any non-communicating calculator is permitted. Format: do Questions 1 and 2 plus any three of Questions 3–6 (five questions total). All six questions are solved below as a complete study resource.
Reference texts: Curtis, Environmental Management in Animal Agriculture (air emissions, odour generation); MWPS-18, Livestock Waste Facilities Handbook (manure storage tank sizing, manure pumping systems); Rynk et al., On-Farm Composting Handbook (NRAES-54) (composting C:N, moisture content, air requirement); Sommer & Christensen (eds.), Animal Manure Recycling: Treatment and Management (land application, nutrient/zinc-loading management); Metcalf & Eddy/Tchobanoglous, Wastewater Engineering: Treatment and Resource Recovery (aerated lagoon kinetics, sludge yield, oxygen requirements).
Given. 500-cow Holstein herd, in the barn 8 h/day; manure production 60 kg/day/cow (24 h basis), 85% moisture, 0.67% N (wet basis) $\equiv$ 0.40 kg N/day/cow; bedding = fifty 60-lb bales of wheat straw/day, 15% moisture, C:N = 80; waste is scraped from the barn (only the barn-time fraction of manure is collected).
Quantity
Value
Herd size / barn time
500 cows / 8 h of 24 h
Manure production (24 h basis)
60 kg/day/cow, 85% MC
Manure N
0.67% wet, 0.40 kg N/day/cow (24 h)
Straw bedding
50 bales × 60 lb/day, 15% MC, C:N = 80
Find. Moisture content and C:N ratio of the manure–straw mixture, the total air requirement for active composting, and a comment on suitability for optimum composting.
Figure 3. Mass balance for the barn-scraped composting mixture (only the 8-h barn fraction of daily manure is collected).
Approach. Prorate manure production to the 8-h barn-collection window, convert straw bale mass to SI, form a mass-weighted moisture content and a carbon/nitrogen mass balance for the mixture, then estimate active-phase oxygen and air demand from the carbon available for biological oxidation.
Manure actually collected (8-h barn fraction). Only manure deposited while the herd is in the barn is scraped up, so daily production is prorated by $8/24$:
$$m_{manure} = 500 \times 60 \times \tfrac{8}{24} = 10{,}000\ \text{kg/day}\ (\text{85\% MC})$$
$$N_{manure} = 500 \times 0.40 \times \tfrac{8}{24} = 66.67\ \text{kg N/day}$$
Dry manure mass $= 10{,}000 \times 0.15 = 1{,}500\ \text{kg/day}$; water $= 10{,}000\times0.85=8{,}500\ \text{kg/day}$.
Mixture moisture content. Total wet mass $= 10{,}000+1{,}360.8=11{,}360.8\ \text{kg/day}$; total water $=8{,}500+204.1=8{,}704.1\ \text{kg/day}$:
$$MC_{mix} = \dfrac{8{,}704.1}{11{,}360.8}\times100 = \boxed{76.6\%}$$
Mixture carbon and C:N.Check: neither material's carbon content is given directly. Dairy-manure dry matter is taken as ≈40% carbon (a typical value, Rynk/NRAES-54), which is self-consistent with the manure's own 4.44% dry-basis N (66.67/1,500) giving a manure C:N of 40/4.44 ≈ 9:1. Wheat-straw dry matter is taken as 0.5% N, which reproduces the given C:N = 80 at the standard 40% straw-carbon figure. Manure carbon $=1{,}500\times0.40=600\ \text{kg C/day}$. Straw N $=1{,}156.7\times0.005=5.78\ \text{kg N/day}$, straw carbon $= 5.78\times80=462.7\ \text{kg C/day}$. Totals: $N_{tot}=66.67+5.78=72.45\ \text{kg/day}$, $C_{tot}=600+462.7=1{,}062.7\ \text{kg/day}$:
$$\left(\dfrac{C}{N}\right)_{mix} = \dfrac{1{,}062.7}{72.45} = \boxed{14.7:1}$$
Air requirement. Assume (Check, typical active-phase value, Haug 1993) that 50% of the mixture's carbon is biologically oxidized during active composting: $C_{ox}=0.50\times1{,}062.7=531.3\ \text{kg C/day}$. Stoichiometric oxygen demand from $\text{C}+\text{O}_2\to\text{CO}_2$ (32 kg O&sub2; per 12 kg C):
$$O_{2,req} = 531.3 \times \dfrac{32}{12} = 1{,}417\ \text{kg O}_2/\text{day}$$
Air is 23.2% O&sub2; by mass, so $m_{air}=1{,}417/0.232=6{,}108\ \text{kg air/day}$; at $\rho_{air}\approx1.2\ \text{kg/m}^3$ this is $6{,}108/1.2=5{,}090\ \text{m}^3/\text{day}=3.53\ \text{m}^3/\text{min}$ on a pure stoichiometric basis. Applying a practical oversizing factor of 2 (Check, typical for non-ideal aeration/mixing efficiency):
$$\boxed{Q_{air,design}\approx 7.1\ \text{m}^3/\text{min}\ (\approx425\ \text{m}^3/\text{h})}$$
4) Suitability comment. The optimum composting window is roughly $C{:}N \approx 25$–$30{:}1$ and $MC \approx 50$–$60\%$. This mixture comes in at $C{:}N = 14.7{:}1$ (well below optimum — nitrogen-rich relative to available carbon) and $MC = 76.6\%$ (well above optimum — excessively wet). As mixed, it is not well suited to optimum composting: the low C:N risks excess ammonia volatilization/odour and nitrogen loss rather than microbial immobilization, while the high moisture content saturates pore space, restricts oxygen diffusion into the pile, and favours anaerobic (odorous) zones. Both problems point to the same fix — adding more dry, carbon-rich bulking material (additional straw or another bulking agent) beyond the fifty bales/day used here, which simultaneously raises the C:N ratio and lowers the moisture content toward their optimum ranges.
