Dairy-Herd Composting Mixture: 22-Agric-B11 Principles of Waste Management, May 2015, Q3

Question 3: Dairy-Herd Composting Mixture (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.

Quantity	Value
Herd size (stated)	200 Holstein cows
Animal units, AU (as printed in source)	100 × 1,200/1,000 = 120 AU
Fraction of day in barn	6 h / 24 h = 0.25
Manure production rate	108 lb/d per 1,000 lb AU
Manure moisture content	87%
Straw addition	10 bales/d × 60 lb/bale = 600 lb/d
Straw moisture content	15%

Check: the source prints "Number of AU = 100 × 1,200/1,000 = 120," using a multiplier of 100 even though the herd is stated as 200 cows just above it — the printed AU figure is a genuine inconsistency in the exam's own data. Because the manure-production/nitrogen/volatile-solids rates that follow are all keyed to "AU" as printed, AU = 120 is used as given below rather than silently recomputed as 200×1,200/1,000 = 240 AU. Check also: target composting mixture moisture content = 55% (mid-point of the standard 50–60% range recommended for active composting, Rynk NRAES-54) — not stated numerically by the question.

Find. A suitable amendment strategy, and the mass of amendment needed per kilogram of manure to bring the mixture to the target moisture content.

Approach. Prorate the tabulated per-AU manure rates by both the animal-unit count and the barn-confinement fraction to get the actual daily scraped manure mass, then solve a two-component wet-mass/water-mass balance between manure and straw for the amendment ratio that lands the mixture at the target moisture content.

Daily scraped manure quantity. The tabulated rate (108 lb/d per 1,000 lb AU) is a full 24-hour production rate; only the fraction of the day the herd spends in the barn is actually scraped and collected, so $$\dot m_{manure} = 108\ \tfrac{\text{lb}}{\text{d}\cdot\text{AU-1000lb}} \times 120\ \text{AU} \times 0.25 = \boxed{3{,}240\ \text{lb/d}}$$ (the same barn-confinement proration applies to nitrogen production, $0.71\times120\times0.25=21.3$ lb N/d, and volatile solids, $11\times120\times0.25=330$ lb VS/d, though only the manure mass and moisture are needed for this part.)
Set up the moisture-balance for one unit mass of manure. Per kilogram (or pound) of wet manure, the water content is $MC_{manure}=0.87$ kg water. Let $x$ = kg of wet straw added per kg of wet manure; the straw itself carries $0.15x$ kg of water. The mixture's moisture content is then $$MC_{mix} = \frac{0.87(1) + 0.15\,x}{1 + x}$$
Solve for the amendment ratio at the target moisture content. Setting $MC_{mix}=0.55$ (target) and solving for $x$: $$0.87 + 0.15x = 0.55(1+x) \;\Rightarrow\; x = \frac{0.87-0.55}{0.55-0.15} = \boxed{0.80\ \text{kg straw per kg manure (wet basis)}}$$ Checking: mixture water $=0.87(1)+0.15(0.80)=0.99$ kg over mixture mass $1.80$ kg gives $MC_{mix}=0.99/1.80=55.0\%$ — consistent.
Compare the required rate to what is actually being added. At $3{,}240$ lb/d of scraped manure, the target ratio calls for $0.80\times3{,}240=\boxed{2{,}592\ \text{lb straw/d}}$, but only 10 bales (600 lb/d) are currently being added — roughly a 4.3× shortfall. The bedding straw alone therefore cannot bring the pile to a compostable moisture content.

Quantity	Result
Daily scraped manure	3,240 lb/d
Required amendment ratio (target MC 55%)	0.80 kg straw per kg manure (wet)
Required daily straw addition	2,592 lb/d
Actual current straw addition	600 lb/d (10 × 60 lb bales) — about 23% of what is needed

1) Suggested amendment(s): continue using wheat straw as the primary bulking/moisture-control amendment — it is already the cheapest, highest C:N source on the farm and requires no new supply chain — but the current bedding rate (10 bales/d) needs to increase roughly four-fold to reach the target 55% mixture moisture content, or a secondary bulking agent (dry wood chips, sawdust, or spent bedding) should be blended in alongside the straw if quadrupling straw purchases is not practical, since any of these bring both extra dry matter and high C:N capacity to a manure stream that is simultaneously too wet (87%) and too nitrogen-rich on its own.

Topic: Compost feedstock moisture balancing.

Key relations: $\dot m = (\text{rate per AU-1000lb}) \times \text{AU} \times f_{barn}$; two-component moisture mass balance $MC_{mix}=(m_{water,1}+m_{water,2})/(m_{wet,1}+m_{wet,2})$.

Why this works: Composting needs the pile's free water content low enough (typically 50–60%) that air can still move through the pore spaces between particles; a dry, absorbent, high-C:N bulking agent like straw simultaneously lowers the blended moisture content and raises the blended carbon:nitrogen ratio toward the 25–30:1 optimum, which is why the same amendment is the standard answer to both moisture and C:N problems in composting design.

Common pitfall: applying the tabulated per-AU manure/nitrogen/volatile-solids rates as if they were already daily totals, without prorating by both the animal-unit count and the barn-confinement fraction — the full 24-hour total is a distractor once a "6 hours in the barn" confinement schedule is stated.

Source: Rynk et al., On-Farm Composting Handbook (NRAES-54) — moisture and C:N targets, bulking-agent selection; MWPS-18 Livestock Waste Facilities Handbook — manure production/AU rate tables.

Question 3 of 5: Dairy-Herd Composting Mixture

Question 3: Dairy-Herd Composting Mixture (25 marks)