Solid and Hazardous Waste

7% of exam

Solid waste management, collection, and disposal, landfill leachate and gas, mass and energy balances, hazardous waste compatibility, site characterization, and waste treatment and disposal.

4 concepts

Exam areas

B. Solid waste management

Solid Waste Management and Collection

Estimate MSW generation and composition, apply the waste hierarchy, size collection routing and transfer, and compute material recovery and diversion/recycling rates.

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C. Solid waste disposal

Landfills: Airspace, Leachate, and Gas

Size landfill airspace from compacted density and compaction ratio, estimate leachate by water balance, predict landfill gas from anaerobic stoichiometry, and value MSW as fuel.

A modern sanitary landfill is an engineered bioreactor, not a hole in the ground: waste is placed in cells, compacted, and covered daily, while a liner and leachate-collection system capture the contaminated water that drains through, and a gas system manages the methane that anaerobic decomposition generates for decades. The FE Environmental exam tests the arithmetic that ties these together — how many cubic yards of airspace a community's waste consumes, how much leachate a cover lets percolate, and how much gas a mass of organics yields. Almost every error here is a density or a unit error, because the same waste has a loose, a compacted, and an in-place density. The governing relations live in the FE Reference Handbook — Environmental Engineering (waste density, the soil-cover water balance, the clay-liner breakthrough time, and the anaerobic stabilization stoichiometry that sits in Chemistry and Biology).

Airspace and compaction

Airspace is the landfill's product: the permitted volume that waste and cover soil fill. You convert an incoming mass rate to a volume rate by dividing by the in-place compacted density — never the loose, as-collected density. Waste is set out loose at the curb (roughly

100

–

300\,\text{lb/yd}^3

); a packer truck compacts it to about

500

–

800\,\text{lb/yd}^3

for haul, and the landfill compactor crushes it further to an in-place density of about

1{,}000

–

1{,}400\,\text{lb/yd}^3

. The compaction ratio is the loose volume divided by the in-place compacted volume, equal to the ratio of in-place to loose density — about

4

–

8

from curbside-loose to in-place; a ratio of

6

means each loose cubic yard occupies one-sixth of the airspace it would uncompacted. Because daily and intermediate cover soil also consume airspace, divide the waste volume by an airspace utilization factor (or add a cover allowance, typically

10

–

25\%

) to get total airspace consumed.

V = \dfrac{M}{\rho_c}, \qquad CR = \dfrac{V_{loose}}{V_{compacted}} = \dfrac{\rho_c}{\rho_{loose}}

Cover, settlement, and overburden density

In-place density is not constant with depth. As lifts accumulate, the overburden pressure compresses lower waste, raising its specific weight along a saturating curve. The handbook models this with two empirical constants

a

and

b

so that the specific weight climbs from an initial compacted value toward an asymptote as pressure grows. For exam sizing you usually use a single representative compacted density, but recognize that a deeper landfill stores more mass per cubic yard, and that long-term settlement (often

10

–

25\%

of waste depth) reclaims airspace that can be re-permitted.

SW_p = SW_i + \dfrac{p}{a + b\,p}

Leachate by water balance

Leachate is precipitation that infiltrates the cover, exceeds the soil's storage, and percolates into the waste. A landfill cover is a one-dimensional water budget: precipitation in, minus runoff, minus evapotranspiration, minus the change in soil moisture storage, equals percolation into the waste. Once the cover soil reaches field capacity the storage term goes to zero and percolation tracks net infiltration directly. That percolation, multiplied by the landfill footprint, is the leachate the collection system must handle — convert inches over acres to gallons with

1\,\text{acre-in} = 27{,}154\,\text{gal}

PER_{sw} = P - R - ET - \Delta S_{LC}

The liner and breakthrough time

A compacted-clay liner does not stop leachate; it slows it. The handbook's breakthrough-time relation estimates how many years leachate needs to migrate through a clay liner of thickness

d

under hydraulic head

h

, given porosity

\eta

and hydraulic conductivity

K

. Thicker, lower-conductivity, higher-porosity liners buy more time; standing leachate head (a clogged collection system) shortens it. Modern subtitle-D liners pair this clay with a geomembrane and a leachate-collection blanket precisely so the head

h

stays near zero.

t = \dfrac{d^{2}\,\eta}{K\,(d + h)}

Landfill gas: stoichiometry

Buried organics decompose anaerobically to a roughly half-methane, half-carbon-dioxide gas. For complete stabilization the handbook balances a generic organic

\mathrm{C_aH_bO_cN_d}

with water to methane, carbon dioxide, and ammonia, fixing the molar yields from the elemental formula. The methane and carbon-dioxide coefficients are set entirely by

a,b,c,d

— once you write a representative formula for the waste, the gas split follows. Multiply the moles of methane by the molar gas volume (

22.4\,\text{L/mol}

at STP) to get a theoretical gas yield per unit mass of organics.

\mathrm{C_aH_bO_cN_d} + r\,\mathrm{H_2O} \rightarrow m\,\mathrm{CH_4} + s\,\mathrm{CO_2} + d\,\mathrm{NH_3}, \quad m = \dfrac{4a + b - 2c - 3d}{8}

Landfill gas: first-order generation rate

Stoichiometry gives the total gas a mass of waste can ever yield; it says nothing about when. Field gas-recovery design uses a first-order decay model (the basis of EPA's LandGEM): the methane generation rate is proportional to the mass of degradable carbon remaining, so each year's deposited waste contributes a decaying exponential. The methane potential

L_0

(

\text{m}^3/\text{Mg}

of waste) and decay constant

k

(

\text{yr}^{-1}

) are the two parameters; arid sites have small

k

(slow), wet sites large

k

. The total recoverable gas integrates to

L_0

per unit mass, tying the kinetic model back to the stoichiometric ceiling. Note this LandGEM-style model is an EPA design tool, not a formula printed on the FE reference sheet — expect it only conceptually.

Q_{CH_4} = k\,L_0\,M\,e^{-k\,t}

MSW as fuel

Waste-to-energy and landfill-gas recovery both turn waste into energy, so the FE expects you to estimate a heating value. Given an ultimate (elemental) analysis, the modified Dulong formula estimates the higher heating value from carbon, hydrogen, oxygen, sulfur, and nitrogen mass percentages (percent by weight); the oxygen term penalizes already-oxidized fuel (

\mathrm{H} - \mathrm{O}/8

is the available hydrogen). As-discarded MSW runs about

4{,}500

–

6{,}500\,\text{Btu/lb}

; dry paper and plastics are far higher, wet food waste far lower. Subtract the latent heat of the water formed and the moisture present to get the lower heating value that a combustor actually delivers. The FE reference sheet supplies only a tabulated energy content by waste component, not the Dulong formula, so treat this textbook correlation as background you carry in, not a provided equation.

HHV\;(\text{Btu/lb}) = 145\,C + 610\left(H - \tfrac{O}{8}\right) + 40\,S + 10\,N

Exam strategy

Lead every landfill problem by writing down which density you have — loose, compacted/in-place, or baled — and convert mass to airspace with the compacted value only. Keep cover soil in the budget: add

10

–

25\%

or divide by an airspace utilization factor. For leachate, the water balance is just bookkeeping in inches; reach field capacity and

\Delta S \to 0

, then scale by area with

27{,}154\,\text{gal/acre-in}

. For gas, decide whether the question wants a theoretical maximum (stoichiometry, moles

\times\,22.4\,\text{L/mol}

) or a rate over time (first-order,

k

and

L_0

). Watch ton (mass,

2{,}000\,\text{lb}

) versus yd

^3

(volume), and Mg/tonne (

1{,}000\,\text{kg}

) versus US ton.

Key equations

Airspace from compacted density

V = \dfrac{M}{\rho_c}

Volume of airspace a waste mass $M$ consumes; $\rho_c$ = in-place compacted density ( $\approx 1{,}000$ – $1{,}400\,\text{lb/yd}^3$ ). Use compacted, never loose, density.

Compaction ratio

CR = \dfrac{V_{loose}}{V_{compacted}} = \dfrac{\rho_c}{\rho_{loose}}

How many loose volumes collapse into one compacted volume; about $4$ – $8$ from curbside-loose to in-place landfill density. Dimensionless.

Volume reduction

VR = \dfrac{V_i - V_f}{V_i}\times 100\%

Percent volume saved by compaction or baling; $V_i$ initial (loose), $V_f$ final (compacted). Related to $CR$ by $VR = 1 - 1/CR$ .

Overburden specific weight

SW_p = SW_i + \dfrac{p}{a + b\,p}

In-place specific weight ( $\text{lb/yd}^3$ ) at overburden pressure $p$ ( $\text{lb/in}^2$ ); $SW_i$ initial compacted ( $\approx 1{,}000$ ), $a,b$ empirical constants.

Site life

L = \dfrac{V_{airspace}}{\dot V_{waste} + \dot V_{cover}}

Years of capacity = permitted airspace divided by annual airspace consumed by compacted waste plus cover soil.

Cover water balance (percolation)

PER_{sw} = P - R - ET - \Delta S_{LC}

Leachate percolation (in.) = precipitation $-$ runoff $-$ evapotranspiration $-$ change in cover-soil storage. At field capacity $\Delta S_{LC}\to 0$ .

Clay-liner breakthrough time

t = \dfrac{d^{2}\,\eta}{K\,(d + h)}

Years for leachate to penetrate a clay liner; $d$ thickness (ft), $\eta$ porosity, $K$ hydraulic conductivity (ft/yr), $h$ leachate head (ft).

Anaerobic stabilization (Buswell)

m = \dfrac{4a + b - 2c - 3d}{8}, \quad s = \dfrac{4a - b + 2c + 3d}{8}

Methane ( $m$ ) and carbon-dioxide ( $s$ ) moles from complete stabilization of $\mathrm{C_aH_bO_cN_d}$ . Ammonia coefficient $= d$ ; water $r = (4a - b - 2c + 3d)/4$ .

First-order gas generation

Q_{CH_4} = k\,L_0\,M\,e^{-k\,t}

Methane rate from a waste mass $M$ ; $L_0$ = methane potential ( $\text{m}^3/\text{Mg}$ ), $k$ = decay constant ( $\text{yr}^{-1}$ ), $t$ = age. Basis of EPA LandGEM.

Gas flux through cover (Fick)

N_A = D\,\eta_{gas}^{4/3}\,\dfrac{C_{A,fill} - C_{A,atm}}{L}

Diffusive (outward) flux of gas $A$ through cover of depth $L$ ; $D$ diffusion coefficient, $\eta_{gas}$ gas-filled porosity (the $\eta_{gas}^{4/3}$ factor is the porosity (tortuosity) correction that converts the free-gas diffusion coefficient $D$ into the effective diffusivity $D\,\eta_{gas}^{4/3}$ ). Writing the driving force as $C_{fill}-C_{atm}$ makes outward flux positive.

Modified Dulong heating value

HHV = 145\,C + 610\left(H - \tfrac{O}{8}\right) + 40\,S + 10\,N

Higher heating value (Btu/lb) from ultimate-analysis mass percents $C,H,O,S,N$ . The $O/8$ term removes hydrogen already bound as water.

Worked examples

Airspace consumption and site life

Problem. A community delivers $500\,\text{tons/day}$ of MSW to a landfill, operating $6\,\text{days/week}$ ( $312\,\text{day/yr}$ ). The compactor achieves an in-place density of $1{,}200\,\text{lb/yd}^3$ , and daily cover soil adds $20\%$ to the waste volume. If the permitted airspace is $5.0\times 10^{6}\,\text{yd}^3$ , estimate the site life.

Solution. Annual mass:

500\,\tfrac{\text{ton}}{\text{day}}\times 2{,}000\,\tfrac{\text{lb}}{\text{ton}}\times 312\,\tfrac{\text{day}}{\text{yr}} = 3.12\times 10^{8}\,\text{lb/yr}

. Compacted waste volume:

V = M/\rho_c = 3.12\times 10^{8} / 1{,}200 = 2.60\times 10^{5}\,\text{yd}^3/\text{yr}

. Add cover:

2.60\times 10^{5}\times 1.20 = 3.12\times 10^{5}\,\text{yd}^3/\text{yr}

. Life:

L = 5.0\times 10^{6} / 3.12\times 10^{5} = 16.0\,\text{yr}

. Sanity check:

260{,}000\,\text{yd}^3

from

\sim 156{,}000\,\text{ton/yr}

\approx 0.6\,\text{ton/yd}^3 = 1{,}200\,\text{lb/yd}^3

, matching the input density. Final answer:

\approx 16.0\,\text{years}

L = \dfrac{5.0\times 10^{6}\,\text{yd}^3}{(3.12\times 10^{5})(1.20)\,\text{yd}^3/\text{yr}} \approx 16.0\ \text{yr}

Leachate from a water balance

Problem. A $20$ -acre landfill cap receives $40\,\text{in/yr}$ of precipitation. Runoff is $15\%$ of precipitation, evapotranspiration is $18\,\text{in/yr}$ , and the cover soil has reached field capacity ( $\Delta S \approx 0$ ). Estimate the leachate generation rate in gal/day.

Solution. Runoff:

R = 0.15\times 40 = 6\,\text{in/yr}

. Percolation:

PER_{sw} = P - R - ET - \Delta S = 40 - 6 - 18 - 0 = 16\,\text{in/yr}

. Volume over the footprint:

16\,\text{in}\times 20\,\text{acre}\times 27{,}154\,\tfrac{\text{gal}}{\text{acre-in}} = 8.69\times 10^{6}\,\text{gal/yr}

. Rate:

8.69\times 10^{6}/365 = 2.38\times 10^{4}\,\text{gal/day}

. Sanity check:

16\,\text{in/yr}

40\%

of the

40\,\text{in}

that falls — a reasonable net infiltration for a modest cap with no barrier layer; a geomembrane cap would cut this by an order of magnitude. Final answer:

\approx 2.38\times 10^{4}\,\text{gal/day}

(

\approx 23{,}800\,\text{gpd}

Q_{leach} = \dfrac{16\times 20\times 27{,}154}{365} \approx 2.38\times 10^{4}\ \text{gal/day}

Theoretical methane from cellulosic waste

Problem. Approximate the rapidly biodegradable fraction of MSW as cellulose, $\mathrm{C_6H_{10}O_5}$ (molar mass $162\,\text{g/mol}$ ). Using complete anaerobic stabilization, find the moles of methane per mole of cellulose and the theoretical methane volume per kilogram at STP ( $22.4\,\text{L/mol}$ ).

Solution. Here

a=6,\ b=10,\ c=5,\ d=0

m = \dfrac{4(6)+10-2(5)-3(0)}{8} = \dfrac{24}{8} = 3\ \text{mol CH}_4

s = \dfrac{4(6)-10+2(5)+0}{8} = \dfrac{24}{8} = 3\ \text{mol CO}_2

; the reaction is

\mathrm{C_6H_{10}O_5 + H_2O \rightarrow 3\,CH_4 + 3\,CO_2}

. Mass check: LHS

162 + 18 = 180\,\text{g}

; RHS

3(16) + 3(44) = 48 + 132 = 180\,\text{g}

. Balanced. Per kilogram:

\dfrac{1000}{162}\times 3 = 18.5\,\text{mol CH}_4

; volume

= 18.5\times 22.4 = 415\,\text{L CH}_4/\text{kg}

. Sanity check: a

50/50

\mathrm{CH_4/CO_2}

split is exactly what landfill gas exhibits, and

\sim 0.4\,\text{m}^3\,\mathrm{CH_4}

per kg of pure cellulose is the right order (

L_0

for whole MSW is lower because much of MSW is non-degradable). Final answer:

3\,\text{mol CH}_4

/mol,

\approx 415\,\text{L CH}_4/\text{kg}

V_{CH_4} = \dfrac{1000\,\text{g}}{162\,\text{g/mol}}\times 3 \times 22.4\,\tfrac{\text{L}}{\text{mol}} \approx 415\ \text{L/kg}

Heating value from ultimate analysis

Problem. A paper-rich waste stream has an ultimate analysis (dry, mass %): $C = 48\%$ , $H = 6.4\%$ , $O = 37.6\%$ , $N = 1.2\%$ , $S = 0.4\%$ . Estimate the higher heating value with the modified Dulong formula.

Solution. Available hydrogen:

H - O/8 = 6.4 - 37.6/8 = 6.4 - 4.70 = 1.70

HHV = 145(48) + 610(1.70) + 40(0.4) + 10(1.2)

= 6{,}960 + 1{,}037 + 16 + 12 = 8{,}025\,\text{Btu/lb}

. Sanity check: paper and cardboard run

7{,}000

–

8{,}000\,\text{Btu/lb}

dry, so

\sim 8{,}000\,\text{Btu/lb}

is right for a paper-dominated stream; mixing in wet food waste would pull the as-received value down toward

5{,}000\,\text{Btu/lb}

. Final answer:

\approx 8{,}025\,\text{Btu/lb}

(HHV, dry basis).

HHV = 145(48) + 610\left(6.4 - \tfrac{37.6}{8}\right) + 40(0.4) + 10(1.2) \approx 8{,}025\ \text{Btu/lb}

Common pitfalls

•Using loose or baled density to size airspace. Mass-to-volume must use the in-place compacted density ( $\sim 1{,}200\,\text{lb/yd}^3$ ); loose curbside (as-collected) density ( $\sim 200\,\text{lb/yd}^3$ ) overstates airspace by the compaction ratio.
•Forgetting cover soil. Daily and intermediate cover consume $10$ – $25\%$ of airspace; ignoring it overstates site life. Add the allowance or divide by an airspace utilization factor.
•Confusing ton, tonne, and yd $^3$ . A US ton is $2{,}000\,\text{lb}$ (mass), a metric tonne/Mg is $1{,}000\,\text{kg}$ , and yd $^3$ is volume — never equate them without a density.
•Dropping the storage term wrongly in the water balance. $\Delta S \to 0$ only after the cover reaches field capacity; during wet-up some precipitation is stored and leachate lags rainfall.
•Treating stoichiometric gas yield as the field rate. Buswell gives the lifetime maximum ( $L_0$ ); the year-by-year rate follows the first-order $k\,L_0\,M\,e^{-kt}$ decay, and only the degradable fraction of MSW counts.
•Mixing HHV and LHV. The modified Dulong gives the higher heating value; a combustor delivers the lower heating value after subtracting the latent heat of formed and inherent water — significant for wet MSW.
•Assuming the clay liner stops leachate. It only delays breakthrough; design pairs it with a geomembrane and keeps the head $h$ near zero so $t = d^2\eta /[K(d+h)]$ stays large.

References

NCEES FE Reference Handbook — Environmental Engineering
NCEES FE Reference Handbook — Chemistry and Biology — anaerobic stabilization stoichiometry
Tchobanoglous, Theisen & Vigil, Integrated Solid Waste Management — density, water balance, energy content
U.S. EPA LandGEM, Landfill Gas Emissions Model — first-order gas generation

D. Hazardous waste compatibility

Hazardous Waste: RCRA, CERCLA, and Compatibility

Distinguish RCRA cradle-to-grave management from CERCLA cleanup, identify hazardous waste by listing or the four characteristics, and apply compatibility, incineration DRE, and stabilization.

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E. Site characterization

Site Characterization and Soil Sampling

Run Phase I/II assessments, design EPA soil and groundwater sampling with QA/QC and data-quality objectives, delineate contamination, and convert soil concentrations to remediation mass.

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