18-Env-A6 Solid Waste Engineering and Management · May 2013

Question 15 of 17: Moisture Content and Density of MSW

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examination, May 2013 — 04-Env-A6 / 18-Env-A6, Solid Waste Engineering and Management. 3 hours duration, closed book, one letter-sized aid sheet permitted. All 17 questions constitute a complete paper (100 marks total).

Reference texts: Tchobanoglous, Theisen & Vigil, Integrated Solid Waste Management: Engineering Principles and Management Issues; Vesilind, Worrell & Reinhart, Solid Waste Engineering; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.); CCME, Guidance Document on Landfill Gas Management.

Check: several questions on this paper (Q1, Q13, Q14, Q17) supply an incomplete data set and explicitly invite the candidate to "make and state" assumptions. Every assumed value below is called out where it is introduced and is chosen from standard solid-waste-engineering practice; the governing METHOD, not the specific assumed number, is what the exam is testing.

Question 15: Moisture Content and Density of MSW (7 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. A 100 kg composite sample split into six components (Table 1 above), each with its own moisture split and as-discarded density.

Find. The overall (as-discarded) moisture content and bulk density of the MSW.

Approach. Moisture content is the ratio of total moisture mass to total sample mass; bulk density requires converting each component's mass to volume (mass ÷ its own density) and dividing total mass by total volume.

Overall moisture content. Sum each row's moisture mass: $3.2+14.0+0.2+0.2+0.2+3.0 = 20.8$ kg. $$MC = \dfrac{20.8}{100}\times100\% = \boxed{20.8\%}$$
Volume of each component (mass ÷ density): paper $45/80=0.563$ m³; organics $20/300=0.067$ m³; metal $7/480=0.015$ m³; glass $10/160=0.063$ m³; ash $3/480=0.006$ m³; misc. $15/160=0.094$ m³.
Total volume and overall density. $$V_{tot} = 0.563+0.067+0.015+0.063+0.006+0.094 = 0.806\ \text{m}^3$$ $$\rho_{overall} = \dfrac{100\ \text{kg}}{0.806\ \text{m}^3} = \boxed{124\ \text{kg/m}^3}$$

Quantity	Value
Overall moisture content (as-discarded)	20.8%
Overall (as-discarded) density	124 kg/m³

Topic: Composite moisture content and density of MSW.

Key relations: $MC = \sum m_{moisture,i} / \sum m_i$; $\rho_{overall} = \sum m_i / \sum(m_i/\rho_i)$ (volumes add, not densities).

Why this works: Mass is conserved additively across components, but density is an intensive property — the only correct way to combine per-component densities into a bulk value is via each component's own volume, then dividing total mass by total volume.

Common pitfall: mass-averaging the individual densities directly ($\sum f_i \rho_i$) instead of volume-averaging — this overweights dense, low-volume components (metal, glass) and gives a bulk density far too high; here that shortcut would give ≈254 kg/m³ instead of the correct 124 kg/m³.

Source: Tchobanoglous, Theisen & Vigil, Integrated Solid Waste Management, Ch. 3 (physical properties: moisture content and density of MSW).

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