18-Env-A6 Solid Waste Engineering and Management · May 2017
Question 6 of 16: Moisture Content and Density of MSW
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination, May 2017 — 04-Env-A6 / 18-Env-A6, Solid Waste Engineering and Management. 3 hours duration, closed book, non-communicating calculator permitted. All 16 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; ISO 14040/14044, Environmental Management — Life Cycle Assessment.
Q6 below is solved from the six printed per-component rows, which are unambiguous exam-given data — see the callout at Q6 for the arithmetic. Table 2's "5.800 kJ/kg" organics value (period instead of comma) is read as 5,800 kJ/kg.
Question 6: Moisture Content and Density of MSW (8 marks)
Given. A 100 kg composite MSW sample broken into 6 components (Table above): each component's mass, moisture fraction, dry-solids fraction, and typical density.
Find. The overall moisture content (%) and overall bulk density (kg/m³) of this MSW.
Approach. Moisture content sums the six components' printed moisture masses over the 100 kg total; density is found from total mass over total VOLUME (each component's mass divided by its own density), not by averaging the six densities directly.
The per-component rows are the actual exam-given data, so they — not the extraction's totals row — are used below.
Overall moisture content: sum the printed component moisture masses over the total sample mass.
$$MC = \frac{3.2+14.0+0.2+0.2+0.2+3.0}{100}\times 100\% = \frac{20.8}{100}\times100\%$$
$$MC = \boxed{20.8\%}$$
Bulk density: total mass divided by total volume (volume-weighted, not mass-weighted). Each component's volume is its mass divided by its own density, $V_i = m_i/\rho_i$:
$$V = \frac{45}{80}+\frac{20}{300}+\frac{7}{480}+\frac{10}{160}+\frac{3}{480}+\frac{15}{160}$$
$$V = 0.5625+0.0667+0.0146+0.0625+0.0063+0.0938 = 0.8063\ \text{m}^3$$
Substituting into $\rho = m_{total}/V_{total}$:
$$\rho = \frac{100\ \text{kg}}{0.8063\ \text{m}^3} = \boxed{124.0\ \text{kg/m}^3}$$
A naive mass-weighted average of the six component densities ($\sum f_i\rho_i$) gives 184 kg/m³ — over 40% higher than the correct volume-weighted result — because it overweights the small-mass, high-density components (metal, ashes) that in fact occupy very little of the sample's actual volume. Density must always be computed from total mass over total volume, never averaged directly.
