Question 4 of 5: Passive Landfill Gas Control and MSW Composition Analysis

Question 4: Passive Landfill Gas Control and MSW Composition Analysis (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.

(a) Passive control of landfill gases. Passive control relies on the natural pressure and concentration gradients generated by the landfill itself (gas generation pressure, diffusion down a concentration gradient) to route landfill gas along a preferred path of least resistance, without mechanical blowers or vacuum extraction. Rather than actively pulling gas to a collection point, a passive system provides a permeable or impermeable engineered feature that channels migrating gas to atmosphere at a controlled, monitored location (typically vented gravel-filled trenches or vertical vent risers), or blocks its lateral migration outright, relying on the landfill's own internal gas pressure (which can reach several kPa above atmospheric) to drive flow through it.

Perimeter Interceptor Trenches are permeable barriers — a trench excavated around the landfill perimeter (or between the landfill and a structure needing protection), backfilled with clean, highly permeable gravel, and vented to atmosphere at intervals via vertical risers, sometimes with a synthetic membrane on the outward-facing wall to force flow up the trench rather than continuing to migrate laterally past it. Migrating gas encounters the trench's much lower flow resistance than the surrounding native soil, is preferentially channelled up through the gravel, and vents passively to atmosphere rather than continuing to migrate off-site through the native soil.

Slurry Walls take the opposite approach — an impermeable barrier, typically a soil-bentonite or cement-bentonite slurry trench keyed into an underlying low-permeability confining layer (clay or bedrock), that physically blocks lateral gas migration rather than channelling it. A slurry wall is used where no natural impermeable layer exists near the surface to key an interceptor trench's own barrier membrane into, or where a fully continuous, positive cutoff is required (e.g. immediately adjacent to an occupied structure) rather than a permeable pathway that still permits some residual lateral seepage past its ends. Slurry walls are frequently paired with a vented interceptor trench or a passive vent riser system on the protected side, since blocking migration alone still leaves the gas pressure that built up against the wall needing somewhere to vent.

Both measures are "passive" in the same sense: once installed, neither requires ongoing mechanical energy input to function, in contrast to an active extraction system's network of wells and blowers maintaining a vacuum (see Question 3(a)'s gas generation discussion for why sustained pressure exists to drive either passive mechanism in the first place). Passive systems are lower capital and operating cost than active extraction but provide less certain, less adjustable control, and are typically specified for smaller landfills or as a supplementary perimeter safeguard alongside an active interior gas collection system on larger sites.

(b) Percent moisture, dry-solids content and bulk density of the analyzed MSW.

Given. A 100 kg composite sample split across five components (table above), each with its own moisture fraction and as-discarded density.

Find. The overall percent moisture content, percent dry-solids content, and bulk (as-discarded) density of this MSW.

Approach. Moisture and dry-solids fractions are each component's own mass fraction summed and divided by the 100 kg total; bulk density requires converting each component's mass to its own volume (mass ÷ its own density), summing those volumes, and dividing the total sample mass by the total volume — component densities must never be averaged directly by mass.

1. Moisture mass per component and total moisture. Each component's moisture mass is its sample mass times its stated moisture fraction, e.g. paper $45\times0.07=3.15$ kg, organics $35\times0.70=24.50$ kg (the table above gives all five). Summing: $$m_{moisture} = 3.15+24.50+0.21+0.20+0.24 = 28.30\ \text{kg}$$
2. Overall percent moisture. $$\%\text{Moisture} = \dfrac{m_{moisture}}{m_{tot}}\times100\% = \dfrac{28.30}{100}\times100\% = \boxed{28.3\%}$$
3. Overall percent dry solids. By mass balance, dry solids $= m_{tot}-m_{moisture} = 100-28.30 = 71.70$ kg. $$\%\text{Dry solids} = \dfrac{71.70}{100}\times100\% = \boxed{71.7\%}$$
4. Volume of each component (mass ÷ density): paper $45/100=0.4500$ m³; organics $35/300=0.1167$ m³; metal $7/480=0.0146$ m³; glass $10/160=0.0625$ m³; ash $3/480=0.0063$ m³.
5. Total volume and overall bulk density. $$V_{tot} = 0.4500+0.1167+0.0146+0.0625+0.0063 = 0.6500\ \text{m}^3$$ $$\rho_{overall} = \dfrac{m_{tot}}{V_{tot}} = \dfrac{100\ \text{kg}}{0.6500\ \text{m}^3} = \boxed{153.8\ \text{kg/m}^3}$$