Environmental Engineering

Air pollution and waste treatment

Air pollutants are grouped first by physical state. Gaseous pollutants include sulphur dioxide, nitrogen oxides and carbon monoxide, while particulate pollutants are suspended solid or liquid matter. Dust, smoke, fly ash and fumes are all solid particles, whereas mist and spray are liquid droplets dispersed in air. The liquid forms of particulate pollutants are therefore mist and spray.

Thermal treatment of organic waste is distinguished by oxygen supply and energy balance. Pyrolysis is the thermal decomposition of organic material in the absence or near absence of oxygen. Because heat must be supplied to break chemical bonds, pyrolysis is endothermic. Combustion is the rapid oxidation of fuel in excess oxygen. It releases heat, so it is exothermic, and it cannot proceed without oxygen.

Each control device works by a specific mechanism, and the pairing of device and pollutant is tested directly.

• Cyclone separator — uses centrifugal force to throw particles outward. It suits coarse dust, not very fine fly ash. e.g., A cyclone is a common first-stage collector ahead of finer units.
• Electrostatic precipitator — charges particles, which migrate to collector plates in an electric field. It is efficient for fine fly ash.
• Wet lime scrubber — absorbs sulphur dioxide into lime slurry, forming calcium sulphite and sulphate.
• Catalytic converter — oxidises carbon monoxide to carbon dioxide over a catalyst.
• Flaring — burns combustible waste gas so that carbon monoxide and hydrocarbons are oxidised.
• Ammonia injection — reduces nitrogen dioxide and other nitrogen oxides to nitrogen and water in selective catalytic reduction.

When collectors are placed in series, their penetrations multiply. The outlet concentration equals the inlet concentration times the product of (1-\eta_i) over all units. For example, two units of 70 % and 60 % efficiency give an overall penetration of 0.3 \times 0.4 = 0.12, so an inlet of 20 mg/m^3 leaves at 2.4 mg/m^3 and the overall efficiency is 88 %.

Dispersion in the atmosphere is governed by lower-layer meteorology. The maximum mixing depth D_{max} is the height to which pollutants are mixed vertically. It is usually capped by a temperature inversion, so it need not equal the depth of the unstable layer. Vertical dispersion takes place only up to D_{max}. The ventilation coefficient is the product of D_{max} and the mean wind speed within the mixing layer, so it depends on D_{max}. A smaller mixing depth confines the same emissions to a smaller volume of air, which raises the air pollution potential.

The wind speed profile follows the power law u = u_1(z/z_1)^p, where the exponent p increases with atmospheric stability. With Turner's values, p \approx 0.15 for a very unstable atmosphere and p \approx 0.25 for a neutral one, so the very unstable exponent is the smaller. For a continuous elevated source, the ground-level centreline concentration is C = \frac{Q}{\pi\sigma_y\sigma_z u}\exp\left(-\frac{H^2}{2\sigma_z^2}\right). It varies inversely with wind speed u, so a higher wind speed lowers the downwind concentration.

Greenhouse gases are largely transparent to incoming short-wavelength solar radiation, which reaches and warms the earth surface. The surface re-emits this energy as long-wavelength terrestrial radiation. Greenhouse gases absorb that outgoing radiation and re-emit part of it downward.

Source reduction, meaning lower consumption of raw materials, is among the most effective ways to reduce solid waste because it removes waste at its origin. Solid waste is still generated when raw materials are converted into goods, through processing losses, rejected products and packaging.

When a biodegradable effluent enters a stream, the dissolved oxygen deficit follows the Streeter–Phelps relation, in which deoxygenation and reaeration act together. If deoxygenation exceeds reaeration, the minimum dissolved oxygen occurs downstream at the critical point, not at the outfall. Organic matter in an activated sludge process is degraded mainly by chemoheterotrophic bacteria, which obtain both carbon and energy from organic compounds. An oligotrophic lake is young, low in nutrients and low in productivity, whereas a eutrophic lake is nutrient-rich and highly productive.

Water and wastewater quality analysis

Quality analysis rests on a few standard calculations. Every concentration is mass per volume of the actual sample, so the mass found in the sample must be divided by the sample volume.

Total solids are the residue left after drying at about 103 to 105 °C. Volatile solids are the part lost on ignition at about 550 °C. The volatile solids mass equals the dried residue minus the fixed residue after ignition, both measured after subtracting the empty crucible mass. For example, 0.5 g of dried residue that leaves 0.3 g after ignition in 100 mL of sample contains 0.2 g of volatile solids, which is 2000 mg/L.

The most probable number (MPN) of coliform organisms is read from an index table giving the MPN per 100 mL for each combination of positive tubes at a standard set of portion sizes. If the portions actually used are smaller than the table portions by a factor, the MPN per 100 mL is the table value multiplied by that same factor. For example, portions one-thousandth of the table size mean the MPN per 100 mL is 1000 times the table index.

Ammonia nitrogen exists in equilibrium as ammonium ion and free ammonia: NH_4^+ \rightleftharpoons NH_3 + H^+, with pK_a \approx 9.25 at 25 °C. The ratio [NH_3]/[NH_4^+] = 10^{\,pH-9.25}. Below pH 9.25 the ammonium ion predominates, above it free ammonia predominates, and at neutral pH NH_4^+ is overwhelmingly predominant.

Hardness is measured by EDTA titration with Eriochrome Black T (EBT) as indicator in a buffer near pH 10. EBT forms a wine-red complex with Ca^{2+} and Mg^{2+}. EDTA forms a more stable complex with these ions and displaces EBT. At the end point the liberated free indicator is blue, so the colour changes from wine red to blue.

Concentrations are often expressed in milliequivalents per litre, \text{meq/L} = \text{mg/L} \div \text{equivalent weight}. For hardness reported as CaCO_3, the equivalent weight is 50 mg/meq, so 350 mg/L as CaCO_3 equals 7 meq/L.

Total nitrogen is the sum of all nitrogen species, each expressed as mass of N. Convert each species with its nitrogen fraction: for NH_3 it is 14/17 and for NO_3^- (as NO_3) it is 14/62. For example, 17 mg/L of NH_3 contains 14 mg N/L and 31 mg/L of NO_3^- contains 7 mg N/L, so the total is 21 mg N/L.

Wood pulping effluent is a good example of COD exceeding BOD_5. Lignin and related compounds are oxidised chemically but are only slowly biodegraded, so the COD is greater than the 5-day BOD and the two are not equal. In general the ultimate BOD is at least equal to BOD_5, so the ratio BOD_5/BOD_u cannot exceed 1.

The chemical oxygen demand of a compound follows from complete oxidation. For C_aH_bO_cN_d the reaction is C_aH_bO_cN_d + \frac{4a+b-2c-3d}{4}O_2 \rightarrow aCO_2 + \frac{b-3d}{2}H_2O + dNH_3. The COD is the moles of oxygen per mole times 32 g per mole, divided by the molar mass. For example, for C_4H_8O_2 (molar mass 88) the oxygen demand is (16+8-4)/4 = 5 mol O_2 per mol, so the COD is 5 \times 32/88 = 1.82 g O_2 per g of compound, and 100 mg/L of the compound gives about 182 mg/L COD. The TOC and TKN values give the carbon and nitrogen content, and the molar mass then fixes the remaining atoms.

Population equivalent is the total BOD load of a sewage flow divided by the standard BOD contributed per person per day. The load is the flow multiplied by the concentration. A flow of 40 million L/day at 250 mg/L BOD_5 carries 10 000 kg/day, and with a standard of 0.05 kg per person per day the population equivalent is 200 000 persons.

The BOD of a sample is the dissolved oxygen depletion multiplied by the dilution factor, BOD_5 = (D_1 - D_2) \times P, where P is the fraction of raw sewage in the bottle. For example, 5 mL of sewage in a 300 mL bottle gives P = 1/60. A depletion of 3 mg/L then gives BOD_5 = 3 \times 60 = 180 mg/L.

Activated sludge process design

Activated sludge design rests on three quantities: the hydraulic retention time \theta = V/Q, the mean cell residence time (sludge age) \theta_c, and the food-to-microorganism ratio F/M = QS_0/(XV). The sludge age is the biomass in the system divided by the biomass removed per day, \theta_c = \frac{VX}{Q_w X_r + Q_e X_e}. When the effluent solids X_e are negligible, this reduces to \theta_c = \frac{VX}{Q_w X_r}.

Waste sludge flow follows from this relation. For example, an aeration tank of 800 m^3 at 2.5 kg/m^3 of MLVSS, an underflow of 10 kg/m^3 and a sludge age of 8 days gives Q_w = \frac{800 \times 2.5}{8 \times 10} = 25 m^3/d. The BOD values do not enter this calculation.

The thickening area of a clarifier is set by the solids mass flow and the design solids flux G, so A = QX/G. For example, a flow of 4800 m^3/d is 200 m^3/h, and at 2 kg/m^3 the solids load is 400 kg/h. With a design flux of 2 kg/m^2·h the area is 200 m^2.

The aeration volume is found from the food-to-microorganism ratio, V = \frac{QS_0}{(F/M)X}. For example, 3000 m^3/d of wastewater with 200 mg/L of biodegradable organics carries QS_0 = 600 kg/d. At F/M = 0.3 d^{-1} and X = 2 kg/m^3, the volume is V = 600/(0.3 \times 2) = 1000 m^3.

Effluent solids add to the solids lost from the system. For example, a 400 m^3 tank at 2.5 kg/m^3 holds 1000 kg of biomass. Wasting 4 m^3/d at 8 kg/m^3 removes 32 kg/d, and an effluent flow of 1000 m^3/d carrying 8 mg/L of biomass removes 8 kg/d more. The total loss is 40 kg/d, so \theta_c = 1000/40 = 25 days.

For a settler with recycle, the solids balance is (Q+Q_r)X = (Q_r+Q_w)X_r when effluent solids are negligible. Combined with the wasted solids Q_wX_r = VX/\theta_c, this gives the wasted flow. For example, with Q = 4000 m^3/d, Q_r = 2000 m^3/d, X = 3 kg/m^3, V = 10\,000 m^3 and \theta_c = 5 d, the wasted solids are 6000 kg/d. At X_r = 6 kg/m^3 the wasted flow is Q_w = 1000 m^3/d, which satisfies the balance 18\,000 = 3000 \times 6 kg/d.

Water hardness calculation

Hardness is caused by polyvalent metal cations, mainly Ca^{2+} and Mg^{2+}, and is reported as an equivalent mass of CaCO_3. Each cation is converted with the factor 100/M, where M is its atomic mass, so 1 mmol of Ca^{2+} or Mg^{2+} equals 100 mg/L as CaCO_3. For example, 80 mg/L of Ca^{2+} gives 200 mg/L as CaCO_3, and 12 mg/L of Mg^{2+} gives 50 mg/L as CaCO_3. Total hardness is the sum of all such contributions.

Temporary, or carbonate, hardness is the CaCO_3 equivalent of bicarbonate, found with the factor 50/61. For example, 61 mg/L of HCO_3^- corresponds to 50 mg/L as CaCO_3. Temporary hardness cannot exceed total hardness, and permanent hardness is the total minus the temporary value.

Carbonate hardness equals the smaller of total hardness and total alkalinity, both as CaCO_3. Non-carbonate hardness is the excess of total hardness over this value. Fe^{2+} and Mn^{2+} also count toward hardness. For example, with total hardness 180 mg/L and alkalinity 120 mg/L, the carbonate hardness is 120 mg/L and the non-carbonate hardness is 60 mg/L.

When two salts are dissolved in water, their masses can be found from two conditions: the sum of the masses equals the total dissolved solids, and the hardness from the two salts equals the total hardness. MgCl_2 (molar mass 95 g/mol) and CaSO_4 (molar mass 136 g/mol) each supply one divalent cation per mole. For example, 95 mg/L of MgCl_2 and 136 mg/L of CaSO_4 give 1 mmol of each cation, a total hardness of 200 mg/L as CaCO_3 and TDS of 231 mg/L. For unknown masses x and y, the equations x+y = \text{TDS} and \frac{100x}{95}+\frac{100y}{136} = \text{TH} form a linear pair solved by substitution.

Carbonate speciation sets the pH of many natural waters. For the bicarbonate–carbonate pair, pH = pK_2 + \log\frac{[CO_3^{2-}]}{[HCO_3^-]}, with pK_2 \approx 10.3. For example, if the carbonate-to-bicarbonate ratio is 0.1, the pH is 10.3 - 1 = 9.3.

Hypochlorous acid dissociates as HOCl \rightleftharpoons H^+ + OCl^-, with pK_a \approx 7.5. Since [OCl^-]/[HOCl] = 10^{\,pH-pK_a}, the fraction of free chlorine present as HOCl is \frac{1}{1+10^{\,pH-pK_a}}. At pH 9.3 the ratio is about 63, so HOCl is only about 1.6 % of free chlorine. At pH 7.0 the same expression gives about 76 %.

Before a speciation result is trusted, check electroneutrality. The sum of cation equivalents in meq/L must equal the sum of anion equivalents. Strong acid anions such as Cl^- and SO_4^{2-} are fully dissociated, which is why their pK values are negative.

Water treatment processes and mechanisms

Each treatment process works through a physical or chemical mechanism, and exam statements test the pairing of process and mechanism.

• Reverse osmosis — pressure drives water through a semipermeable membrane. Solutes accumulate at the membrane surface, creating concentration polarization that reduces flux and salt rejection.
• Trickling filter — ponding occurs when excess biofilm clogs the media, so water collects on the surface instead of percolating.
• Coagulation — hydrolysed metal salts neutralise the surface charge of colloids, destabilising particles so that they aggregate. This is charge neutralization.
• Adsorption — the Freundlich isotherm q = KC^{1/n} describes equilibrium uptake of solute on a solid such as activated carbon.

Particulate matter can enclose bacteria and viruses and shield them from disinfectant, so removing turbidity improves disinfection. Ammonia reacts with free chlorine to form chloramines, which increases the chlorine demand rather than decreasing it. Phosphorus is a limiting nutrient in many fresh waters, and its addition stimulates algal and aquatic growth, leading to eutrophication. Calcium and magnesium cause hardness and add to the total dissolved solids.

Quicklime is used to supply alkalinity. The reactions are CaO + H_2O \rightarrow Ca(OH)_2 and Ca(OH)_2 + Ca(HCO_3)_2 \rightarrow 2CaCO_3 + 2H_2O. One mole of CaO is therefore needed per mole of Ca(HCO_3)_2, that is 56 g of CaO per 162 g of Ca(HCO_3)_2. For example, a deficit of 81 kg/d of Ca(HCO_3)_2 needs 81 \times 56/162 = 28 kg/d of CaO. Alkalinity already present in the raw water supplies part of the demand and must be credited before the lime dose is calculated.

Aeration oxidises dissolved ferrous iron (Fe^{2+}) and manganese (Mn^{2+}) to Fe(III) and Mn(IV). These form insoluble hydroxides or oxides that settle and are removed by filtration. Cadmium and zinc are not removed by aeration. They are removed by precipitation at higher pH, adsorption or ion exchange.

Water filtration bed design

Backwashing expands a granular filter bed by forcing water upward, and the bed expansion is described by the Richardson–Zaki relation V = v_t\,\varepsilon^{n}, where V is the superficial backwash velocity, v_t is the terminal settling velocity of a single particle, and \varepsilon is the expanded porosity. The exponent n depends on the particle Reynolds number: it is about 4.8 at very low Reynolds numbers and falls to about 2.4 at high ones, so it must be taken from the design data or chart for the given particle size.

The solid volume in the bed per unit plan area does not change during expansion, so the expanded depth follows from L_1(1-\varepsilon_1) = L_2(1-\varepsilon_2). Once the service and backwash porosities are known, this relation gives the expanded depth directly, and it does not depend on particle size.

After backwashing, a multimedia bed stratifies because particles settle in order of their terminal velocities. Under Stokes' law, v \propto (G-1)d^2, so for layering the lighter or finer media must settle more slowly than the denser media below them. For example, a coal particle with G = 1.4 must have a diameter below 1 mm to settle more slowly than a sand grain of G = 2.6 and diameter 0.5 mm, since 0.4\,d^2 < 1.6 \times 0.25.

Sedimentation basin removal efficiency

In an ideal rectangular sedimentation basin with plug flow and uniform inlet concentration, the performance is controlled by the surface overflow rate v_0 = Q/A, where Q is the flow rate and A is the plan area. A particle with settling velocity v_s \geq v_0 is removed completely, while a particle with v_s < v_0 is removed in the fraction v_s/v_0.

• Surface overflow rate — flow per unit plan area, the design parameter that governs removal. e.g. Q = 300 m^3/h on A = 120 m^2 gives v_0 = 2.5 m/h.
• Removal fraction — v_s/v_0 for particles that settle slower than the overflow rate. e.g. v_s = 0.5 m/h gives 0.2.
• Basin depth — does not change the ideal fraction removed, provided the flow remains plug-flow.

The settling velocity of a particle is obtained separately, usually from Stokes' law, d = \sqrt{\dfrac{18\nu v_s}{(G-1)g}}. For a particle with G = 2.5, v_s = 2\times10^{-4} m/s and \nu = 10^{-6} m^2/s, the minimum diameter is about 15.6 µm.

Municipal solid waste heating value

The higher heating value (HHV) of a fuel is the heat released when it burns completely and the water formed is condensed. Moisture in waste absorbs heat for evaporation, so the heating value on a wet basis is reduced, and the dry-basis value is obtained by dividing by the dry fraction: \text{HHV}_{dry} = \text{HHV}_{wet}/(1-w), where w is the moisture fraction. For example, waste with an HHV of 10 MJ/kg and 20 % moisture has a dry-basis value of 10/0.80 = 12.5 MJ/kg.

Refuse-derived fuel (RDF) is produced by shredding, drying, and removing inert and metallic fractions from MSW. Because the moisture and inert material are removed, RDF has a higher heating value per unit mass than the unprocessed waste. RDF can be produced in several physical forms, such as loose fluff, pellets, and fine powder, and it can be co-fired with other fuels.

The energy content of mixed MSW is the mass-weighted sum of the component energy contents, so the composition of paper, plastics and food waste must be known to estimate it.

Water disinfection agents

Disinfectants inactivate microorganisms by oxidation or by damage to cell structures. The common active agents in water treatment are chlorine species (free chlorine, hypochlorous acid, and hypochlorite), ozone, chlorine dioxide, and the hydroxyl radical formed in advanced oxidation. Chloride ion is the reduced product of chlorine and has no disinfecting action, so it is not an active disinfectant.

Free chlorine in water exists as hypochlorous acid, HOCl, and hypochlorite, OCl^-, in equilibrium: HOCl \rightleftharpoons H^+ + OCl^-. The fraction present as HOCl is found from \alpha = \dfrac{1}{1+10^{\,\text{pH}-\text{pK}}}. At pH 7.0 with pK = 7.5, \alpha = 1/(1+10^{-0.5}) = 0.76, so HOCl is the dominant form. Since HOCl is the more effective germicide, the disinfection efficiency falls as pH rises.

When chlorine is reported in mg/L as Cl_2, the molar mass of Cl_2 is 71 g/mol, so 1 mg/L corresponds to 1/71 mmol/L, or 14.08 µmol/L. This conversion is needed before the HOCl fraction can be expressed in µmol/L.

Sewage sludge composting

Composting is the aerobic decomposition of organic solids by thermophilic and mesophilic microorganisms. A bulking agent, such as wood chips, is mixed with the sludge to increase the porosity and air permeability of the solid mixture, which allows oxygen to reach the microbial population.

Bacteria, fungi and actinomycetes are all active in the decomposition, and actinomycetes are important in the later stages when the temperature falls. Excess moisture, particularly from rainfall on open windrows, can produce leachate, which must be collected and treated.

In-vessel systems may be built as horizontal or agitated reactors that operate in plug-flow mode, so they can be designed for a defined retention time.

Mixing sludge with organic municipal solid waste is controlled by the water balance. For example, mixing 1 kg of material with 25 % water and s kg of material with 75 % water to obtain a 50 % water content gives 0.25 + 0.75s = 0.5(1+s), so s = 1 kg. The bulk density of the mixture is the total mass divided by the total volume of the components.

Anaerobic decomposition of organic matter

Anaerobic decomposition of organic matter proceeds through hydrolysis, acidogenesis and methanogenesis, and the end products depend on the composition of the substrate. Carbohydrates such as glucose give mainly carbon dioxide and methane, together with hydrogen as an intermediate. The overall reaction is \mathrm{C_6H_{12}O_6 \rightarrow 3CH_4 + 3CO_2}, which is balanced in carbon, hydrogen and oxygen.

• Methane (CH_4) — the main combustible product of methanogenesis, the basis of biogas.
• Carbon dioxide (CO_2) — formed with methane from carbohydrates and organic acids.
• Hydrogen sulphide (H_2S) — formed only from sulphur-containing substrates, such as sulphur amino acids or sulphate reduced by sulphate-reducing bacteria.
• Ammonia (NH_3) — released from the degradation of nitrogen-containing organic matter.

Glucose contains no sulphur, so H_2S is not formed in its anaerobic decomposition.

The conversion of COD into biomass is found from the yield coefficient Y (g VSS per g COD) and the COD equivalent of the cell formula. For \mathrm{C_5H_7O_2N} the COD equivalent is about 1.42 g COD per g VSS, so a yield of Y = 0.05 converts about 0.05 \times 1.42 = 7.1 % of the COD into biomass.

Sedimentation basin design

For a circular primary settling tank, the plan area is found from the surface loading rate, A = Q/\text{SOR}, and the detention time gives the tank volume, V = Qt. The water depth is then H = V/A, which is equivalent to H = \text{SOR} \times t. For example, a flow of 5000 m^3/d with a surface loading of 30 m^3/m^2/d and a 2-hour detention time gives A = 166.7 m^2 and H = 30 \times 2/24 = 2.5 m.