§8.1 Characterization of Industrial Effluents: Textile, Tannery & Pharmaceutical
Industrial manufacturing generates wastewater streams exhibiting extreme variations in chemical composition, pH, temperature, toxicity, and recalcitrance. Effective treatment facility design requires precise characterization of sector-specific contaminants.
Sectoral Effluent Profiles
1. Textile Dyeing and Finishing Effluents:
- Characterized by deep chromophoric coloration (synthetic reactive, azo, disperse, and vat dyes), elevated temperatures ($40 - 60^\circ\text{C}$), alkaline pH ($9 - 11$), high chemical oxygen demand (COD: $1,500 - 4,000\ \text{mg}/\text{L}$), and high salinity (total dissolved solids, TDS: $5,000 - 15,000\ \text{mg}/\text{L}$) driven by sodium chloride ($NaCl$) and sodium sulfate ($Na_2SO_4$) used as dye exhaustion exhausting electrolytes.
- Low biodegradability ratio ($ ext{BOD}_5 / \text{COD} \approx 0.15 - 0.25$), rendering conventional single-stage biological treatment insufficient.
2. Tannery Effluents:
- Beamhouse operations (liming, dehairing, fleshing) release enormous concentrations of dissolved sulfides ($S^{2-}$), proteins, ammonium, and total suspended solids (TSS: $3,000 - 8,000\ \text{mg}/\text{L}$).
- Tanyard operations discharge acidic effluents laden with toxic basic chromium(III) sulfate ($Cr(OH)SO_4$), requiring segregated physicochemical treatment.
3. Pharmaceutical & Specialty Chemical Effluents:
- Contain recalcitrant Active Pharmaceutical Ingredients (APIs), antibiotics, heterocyclic solvents, and high organohalogen content, frequently displaying acute microbial biocidal activity.
§8.2 Primary Treatment Engineering: Equalization, DAF & Coagulation
Primary wastewater operations mitigate hydraulic shock loads, remove coarse suspended solids, and destabilize colloidal organic dispersion before biological processing.
Flow Equalization Basins
Industrial batch operations cause erratic volumetric flow rates and pollutant surges. Equalization basins operate as completely stirred tanks providing adequate hydraulic retention time ($HRT = 8 - 24\ \text{hours}$) to dampen peak loadings:
Dissolved Air Flotation (DAF)
DAF separates low-density suspended solids, emulsified oils, and grease ($O\&G$):
- A portion of clarified effluent is pressurized to $4 - 6\ \text{bar}$ in a retention tank saturated with dissolved air.
- Pressure release across an injection nozzle nucleates micro-bubbles ($20 - 50\ \mu\text{m}$ diameter) into the contact chamber.
- Micro-bubbles adhere to hydrophobic flocs, reducing apparent particle density below that of water:
The aggregate rises rapidly to the surface according to Stokes' Law, where mechanical scrapers skim off the concentrated float sludge.
Chemical Coagulation-Flocculation Mechanisms
Colloidal wastewater particles remain suspended due to mutually repulsive electrostatic zeta potentials (typically $-15$ to $-35\ \text{mV}$). Chemical coagulant dosing destabilizes colloids via four primary mechanisms:
1. Double Layer Compression: High ionic strength suppresses the electrostatic Debye length.
2. Charge Neutralization: Cationic multivalent hydrolyzing metal complexes (e.g. $[Al_{13}O_4(OH)_{24}(H_2O)_{12}]^{7+}$ in polyaluminum chloride, PAC) adsorb specifically onto negatively charged colloids, neutralizing net charge.
3. Enmeshment (Sweep Flocculation): Massive hydroxide precipitation ($Fe(OH)_3, Al(OH)_3$) traps non-reactive particles within a dense settling floc blanket.
4. Inter-particle Bridging: Long-chain anionic or cationic polyacrylamide polymers anchor simultaneously to multiple flocs, building large macro-flocs.
§8.3 Secondary Biological Treatment: Lawrence-McCarty Activated Sludge Design
Secondary treatment relies on a dynamic suspension of heterotrophic microorganisms (activated sludge) in an aerobic bioreactor to oxidize dissolved organic carbon into carbon dioxide and microbial biomass.
Lawrence-McCarty Steady-State Kinetic Formulation
The system couples Monod microbial growth kinetics with mass balance conservation across a completely mixed aeration basin and secondary clarifier:
1. Substrate Degradation (Effluent Soluble Substrate, $S$):
where:
- $\theta_c$: Mean cell residence time / sludge age ($\text{days}$).
- $Y$: Biomass synthesis yield coefficient ($\text{g VSS} / \text{g substrate}$).
- $k$: Maximum specific substrate utilization rate ($\text{day}^{-1}$).
- $K_s$: Half-velocity constant ($\text{mg}/\text{L}$).
- $b$: Endogenous biomass decay coefficient ($\text{day}^{-1}$).
Fundamental Principle: Effluent soluble substrate concentration $S$ is strictly governed by sludge age $\theta_c$, completely independent of influent substrate concentration $S_0$ or hydraulic retention time $\theta$.
2. Bioreactor Biomass Concentration ($X$, Mixed Liquor Volatile Suspended Solids, MLVSS):
where $\theta = V / Q$ is hydraulic retention time ($HRT$).
3. Food-to-Microorganism ($F/M$) Ratio:
Typical conventional activated sludge operations maintain $F/M$ between $0.2$ and $0.5\ \text{day}^{-1}$ to ensure dense bio-flocculation and prevent filamentous sludge bulking.
§8.4 Anaerobic Digestion: UASB Reactors, Methanogenesis & Biogas
Anaerobic biological treatment is ideally suited for high-strength industrial wastewaters ($ ext{COD} > 2,000 - 30,000\ \text{mg}/\text{L}$), converting organic pollution directly into combustible methane without requiring expensive electrical aeration energy.
Multi-Stage Anaerobic Microbial Trophic Cascade
Organic macromolecules undergo sequential bioconversion across four phylogenetically distinct bacterial and archaeal guilds:
1. Hydrolysis: Extracellular enzymes (cellulases, proteases, lipases) hydrolyze insoluble polymers into soluble monomers (glucose, amino acids, fatty acids).
2. Acidogenesis: Fermentative bacteria convert monomers into volatile fatty acids (VFAs: propionate, butyrate, valerate), lactic acid, alcohols, $H_2$, and $CO_2$.
3. Acetogenesis: Syntrophic obligate hydrogen-producing acetogenic bacteria oxidize VFAs into acetate, carbon dioxide, and hydrogen:
Thermodynamic constraint: This endergonic reaction proceeds only when hydrogen-consuming methanogens maintain extremely low hydrogen partial pressures ($P_{H2} < 10^{-4}\ \text{atm}$).
4. Methanogenesis: Strictly anaerobic Euryarchaeota produce methane via two parallel metabolic pathways:
- Acetoclastic Methanogenesis (Methanosarcina, Methanosaeta): Cleaves acetate ($70\%$ of global biogas):
- Hydrogenotrophic Methanogenesis (Methanobacterium): Reduces $CO_2$ with $H_2$ ($30\%$ of biogas):
Upflow Anaerobic Sludge Blanket (UASB) Architecture
UASB reactors introduce influent through bottom distribution manifolds, flowing upward through a dense blanket of self-immobilized anaerobic granular sludge (settling velocity $> 30\ \text{m}/\text{h}$). A top three-phase separator (gas-liquid-solid / GLS separator) disengages biogas bubbles, allows clarified effluent discharge, and returns settled granules directly to the sludge bed without requiring mechanical recycling.
§8.5 Advanced Oxidation Processes (AOPs): Fenton, Photo-Fenton & Photocatalysis
Advanced Oxidation Processes (AOPs) destroy recalcitrant, non-biodegradable, or toxic xenobiotics by generating the non-selective hydroxyl radical ($\cdot OH$), one of the strongest chemical oxidants known ($E^\circ = +2.80\ \text{V}$ vs SHE).
The Classical Fenton Reaction
Discovered by H.J.H. Fenton in 1894, the reaction relies on ferrous iron catalyzing hydrogen peroxide decomposition under acidic conditions:
Regeneration of ferrous iron occurs slowly via reduction by hydrogen peroxide:
Because $k_1 \gg k_2$, ferric iron accumulates. The system operates optimally in the strict range of pH 2.8 to 3.2. At $\text{pH} > 3.5$, ferric iron hydrolyzes and precipitates as inactive $\text{Fe(OH)}_3$ sludge; at $\text{pH} < 2.5$, proton scavenging of hydroxyl radicals ($\cdot\text{OH} + \text{H}^+ + e^- \rightarrow \text{H}_2\text{O}$) and formation of stable $[\text{Fe(H}_2\text{O})_6]^{3+}$ complexes suppresses radical generation.
Photo-Fenton Catalysis
Irradiation with near-UV / visible light ($\lambda < 580\ \text{nm}$) dramatically accelerates oxidation by inducing photochemical reduction of aqueous ferric complexes:
This photo-reduction regenerates active $\text{Fe}^{2+}$ while producing an additional mole of hydroxyl radical.
Heterogeneous Photocatalysis ($TiO_2/UV$)
When titanium dioxide semiconductor particles are irradiated with UV light exceeding their bandgap energy ($E_g = 3.2\ \text{eV}$ for anatase, $\lambda \le 387\ \text{nm}$), electrons are promoted from the valence band ($VB$) to the conduction band ($CB$):
The valence band holes ($h^+_{VB}$) possess extreme oxidizing potential ($+2.7\ \text{V}$), directly oxidizing surface-adsorbed water and hydroxide ions into hydroxyl radicals:
§8.6 Membrane Separation Technologies: MF, UF, NF & Reverse Osmosis
Pressure-driven membrane processes separate solute species across engineered semi-permeable polymeric or ceramic barriers based on molecular size exclusion, steric hindrance, and electrostatic Donnan exclusion.
Membrane Classification Spectrum
1. Microfiltration (MF):
- Pore size: $0.1 - 10\ \mu\text{m}$, Operating pressure: $0.1 - 2.0\ \text{bar}$.
- Removes suspended solids, protozoa, and large bacteria; zero salt or color rejection.
2. Ultrafiltration (UF):
- Pore size: $0.005 - 0.1\ \mu\text{m}$ (Molecular Weight Cut-Off, MWCO: $1,000 - 100,000\ \text{Da}$), Pressure: $1 - 5\ \text{bar}$.
- Retains colloidal matter, viruses, proteins, and macro-polymers; serves as critical pretreatment for RO.
3. Nanofiltration (NF):
- Pore size: $0.001 - 0.005\ \mu\text{m}$ (MWCO: $200 - 1,000\ \text{Da}$), Pressure: $3 - 15\ \text{bar}$.
- Negatively charged surface; preferentially rejects multivalent divalent ions ($SO_4^{2-}, Ca^{2+}, Mg^{2+}$) via Donnan exclusion while passing monovalent salts ($NaCl$).
4. Reverse Osmosis (RO):
- Dense non-porous polyamide thin-film composite (TFC) skin layer, Operating pressure: $15 - 80\ \text{bar}$.
- Rejects $> 99.5\%$ of all dissolved inorganic ionic species and organic micropollutants.
Transport Models: Solution-Diffusion
Solvent water flux ($J_w$) across an RO membrane is governed by net driving pressure:
where $A$ is the pure water permeability coefficient ($\text{L}/(\text{m}^2\cdot\text{h}\cdot\text{bar})$), $\Delta P$ is the applied transmembrane hydrostatic pressure, and $\Delta \pi = \pi_{feed} - \pi_{permeate}$ is osmotic pressure differential, computed via the van 't Hoff equation:
Solute flux ($J_s$) is driven purely by concentration gradient, completely independent of hydraulic pressure:
where $B$ is the solute permeability coefficient.
§8.7 Chromium Geochemistry in Tannery Effluents: Cr(VI) vs Cr(III)
Tanning chemistry utilizes chromium for cross-linking collagen triple helices, generating effluent streams with high chromium concentrations.
Oxidation States & Toxicological Divergence
Chromium exists predominantly in two oxidation states exhibiting diametrically opposing environmental properties:
- Trivalent Chromium [Cr(III)]:
- Exists as the hard Lewis acid aqua-cation $[Cr(H_2O)_6]^{3+}$.
- Readily forms insoluble amorphous hydroxide precipitates ($Cr(OH)_3$) above pH 6.0 ($K_{sp} \approx 6.3 \times 10^{-31}$).
- Low cellular membrane permeability; essential trace nutrient in micro-quantities.
- Hexavalent Chromium [Cr(VI)]:
- Exists as soluble oxyanions: chromate ($CrO_4^{2-}$) and dichromate ($Cr_2O_7^{2-}$):
- Structural analogue to sulfate ($SO_4^{2-}$); actively transported through cellular sulfate permease channels.
- Potent mutagen, teratogen, and Class 1 human carcinogen, inducing intracellular DNA cross-linking.
Chemical Reduction and Precipitation Engineering
To remediate chromium-laden tannery wastewater, any incidental or deliberate $\text{Cr(VI)}$ must be chemically reduced to $\text{Cr(III)}$ prior to alkaline neutralization:
1. Acidification: Effluent is acidified to $\text{pH } 2.0 - 2.5$ using sulfuric acid ($H_2SO_4$).
2. Chemical Reduction: Ferrous sulfate ($\text{FeSO}_4$) or sodium metabisulfite ($\text{Na}_2\text{S}_2\text{O}_5$) is added:
3. Alkaline Neutralization and Precipitation: Lime ($\text{Ca(OH)}_2$) or caustic soda ($\text{NaOH}$) is dosed to adjust pH to $8.5 - 9.0$:
The settled hydroxide cake is dewatered via filter presses and stabilized in hazardous landfill facilities or recycled back into chromium tanning liquor via acid digestion.
§8.8 Zero Liquid Discharge (ZLD) Systems & Industrial Water Reclamation
Zero Liquid Discharge (ZLD) represents the pinnacle of industrial wastewater sustainability, eliminating all liquid effluent discharge while recovering $> 95 - 98\%$ of purified water for manufacturing reuse.
Comprehensive ZLD Process Flowsheet Architecture
A state-of-the-art textile or chemical ZLD facility integrates physical, biological, membrane, and thermal operations:
1. Biological Pre-treatment: Membrane Bioreactor (MBR) combining activated sludge aeration with submerged microfiltration membranes, achieving complete TSS removal and COD reduction down to $< 50\ \text{mg}/\text{L}$.
2. Polishing & Color Removal: Fixed-bed granular activated carbon (GAC) or ozone contactors eliminate trace residual refractory organics, protecting downstream reverse osmosis membranes from organic fouling.
3. Multi-Stage High-Recovery RO System:
- Primary and Secondary Brackish Water RO (BWRO) units operating at recoveries of $75 - 85\%$.
- High-Pressure RO (HPRO / Disc-Tube RO) operating up to $120\ \text{bar}$ concentrating the brine up to $100,000 - 140,000\ \text{mg}/\text{L}$ TDS.
4. Thermal Brine Concentration:
- Mechanical Vapor Recompression (MVR) Brine Falling Film Evaporators: Compresses evaporated steam mechanically to elevate its saturation temperature, reusing latent heat of vaporization to concentrate brine to near saturation ($250,000 - 300,000\ \text{mg}/\text{L}$ TDS).
5. Crystallization & Solid Salt Recovery:
- Forced circulation crystallizers precipitate sodium sulfate ($Na_2SO_4$) and sodium chloride ($NaCl$) crystals.
- Centrifuges and drying ovens produce high-purity dry industrial salt cakes suitable for reuse in textile dyeing baths, closing the material loop.
Worked Practice Problems (9 Challenge Exercises)
Multi-step solved problems covering barometric scale height, Leighton photostationary ozone equilibria, VOC-hydroxyl radical degradation lifetimes, aqueous SO2 bisulfite oxidation, acid rain carbonate dissolution, three-way catalytic converter stoichiometry, aerosol Stokes terminal settling, and atmospheric radon-222 secular equilibrium with line-by-line mathematical proofs.
A textile dyeing mill operates over an 8-hour shift producing fluctuating wastewater flows and COD loadings recorded hourly: Hour 1: $60\ \text{m}^3/\text{h}$, $2500\ \text{mg}/\text{L}$; Hour 2: $90\ \text{m}^3/\text{h}$, $3200\ \text{mg}/\text{L}$; Hour 3: $120\ \text{m}^3/\text{h}$, $3800\ \text{mg}/\text{L}$; Hour 4: $140\ \text{m}^3/\text{h}$, $4000\ \text{mg}/\text{L}$; Hour 5: $110\ \text{m}^3/\text{h}$, $3000\ \text{mg}/\text{L}$; Hour 6: $80\ \text{m}^3/\text{h}$, $2200\ \text{mg}/\text{L}$; Hour 7: $50\ \text{m}^3/\text{h}$, $1800\ \text{mg}/\text{L}$; Hour 8: $70\ \text{m}^3/\text{h}$, $2100\ \text{mg}/\text{L}$. The downstream biological treatment plant requires a constant average flow rate $\bar{Q}$. (a) Calculate the total 8-hour volume $V_{tot}$, the constant pumped outflow rate $\bar{Q}$ (in $\text{m}^3/\text{h}$), and the flow-weighted average influent COD concentration $\overline{\text{COD}}$. (b) Determine the minimum active equalization basin storage capacity $V_{eq}$ (in $\text{m}^3$) required to balance the hourly flow variations using cumulative volume mass analysis.
Step 1: Calculate total volume and constant outflow rate $\bar{Q}$ Hourly flow rates: $[60, 90, 120, 140, 110, 80, 50, 70]\ \text{m}^3/\text{h}$.
The constant pumping rate over the 8-hour period:
Step 2: Calculate flow-weighted average COD concentration Total mass of COD entering:
- Hour 1: $60 \times 2500 = 150,000\ \text{g}$
- Hour 2: $90 \times 3200 = 288,000\ \text{g}$
- Hour 3: $120 \times 3800 = 456,000\ \text{g}$
- Hour 4: $140 \times 4000 = 560,000\ \text{g}$
- Hour 5: $110 \times 3000 = 330,000\ \text{g}$
- Hour 6: $80 \times 2200 = 176,000\ \text{g}$
- Hour 7: $50 \times 1800 = 90,000\ \text{g}$
- Hour 8: $70 \times 2100 = 147,000\ \text{g}$
The flow-weighted concentration:
Step 3: Determine required equalization volume via mass curve analysis Calculate cumulative deviation from constant discharge ($Q_i - \bar{Q}$):
- Hour 1: $60 - 90 = -30\ \text{m}^3 \implies \text{Cum} = -30\ \text{m}^3$
- Hour 2: $90 - 90 = 0\ \text{m}^3 \implies \text{Cum} = -30\ \text{m}^3$
- Hour 3: $120 - 90 = +30\ \text{m}^3 \implies \text{Cum} = 0\ \text{m}^3$
- Hour 4: $140 - 90 = +50\ \text{m}^3 \implies \text{Cum} = +50\ \text{m}^3$
- Hour 5: $110 - 90 = +20\ \text{m}^3 \implies \text{Cum} = +70\ \text{m}^3$ (Maximum surplus)
- Hour 6: $80 - 90 = -10\ \text{m}^3 \implies \text{Cum} = +60\ \text{m}^3$
- Hour 7: $50 - 90 = -40\ \text{m}^3 \implies \text{Cum} = +20\ \text{m}^3$
- Hour 8: $70 - 90 = -20\ \text{m}^3 \implies \text{Cum} = 0\ \text{m}^3$
The maximum cumulative excess: $(\Delta V)_{max} = +70\ \text{m}^3$. The maximum cumulative deficit: $(\Delta V)_{min} = -30\ \text{m}^3$. The minimum active equalization storage volume required:
Conclusion: The average outflow is $90.0\ \text{m}^3/\text{h}$, flow-weighted COD is $3,051\ \text{mg}/\text{L}$, and minimum active basin capacity is $100\ \text{m}^3$.
An industrial wastewater stream of $Q = 2400\ \text{m}^3/\text{day}$ is treated with commercial alum (hydrated aluminum sulfate, $\text{Al}_2(\text{SO}_4)_3 \cdot 14\text{H}_2\text{O}$, molar mass $594.36\ \text{g}/\text{mol}$) at an optimal coagulation dose of $80.0\ \text{mg}/\text{L}$. Alum reacts quantitatively with natural bicarbonate alkalinity according to: $\text{Al}_2(\text{SO}_4)_3 \cdot 14\text{H}_2\text{O} + 6\text{HCO}_3^- \rightarrow 2\text{Al(OH)}_3\text{(s)} + 3\text{SO}_4^{2-} + 6\text{CO}_2 + 14\text{H}_2\text{O}$. (a) Calculate the daily mass of alum consumed in kilograms per day. (b) Calculate the mass of alkalinity consumed per milligram of alum dosed, expressed as $\text{mg } \text{CaCO}_3 / \text{mg alum}$ (molar mass $\text{CaCO}_3 = 100.09\ \text{g}/\text{mol}$). (c) If the raw wastewater has an initial alkalinity of $45.0\ \text{mg}/\text{L}$ as $\text{CaCO}_3$, calculate the residual alkalinity, and verify if hydrated lime ($\text{Ca(OH)}_2$) addition is needed to maintain a minimum buffering reserve of $20.0\ \text{mg}/\text{L}$ as $\text{CaCO}_3$.
Step 1: Calculate daily mass of commercial alum Given $Q = 2400\ \text{m}^3/\text{day}$ and $\text{Dose} = 80.0\ \text{mg}/\text{L} = 80.0\ \text{g}/\text{m}^3$:
Step 2: Calculate stoichiometric alkalinity consumption From the balanced chemical reaction:
$1\ \text{mol of alum}$ ($594.36\ \text{g}$) consumes $6\ \text{mol of } \text{HCO}_3^-$. In standard environmental water quality units, alkalinity is reported as equivalent $\text{CaCO}_3$. Since $1\ \text{mol of } \text{CaCO}_3$ ($100.09\ \text{g}$) neutralizes $2\ \text{mol of } H^+$ (or accepts $2\ \text{moles of charge}$):
The mass ratio of alkalinity consumption:
Step 3: Calculate alkalinity depletion and supplemental lime requirement Alum dose is $80.0\ \text{mg}/\text{L}$:
Residual alkalinity:
Because the residual alkalinity ($4.58\ \text{mg}/\text{L}$) falls far below the minimum safe buffering limit of $20.0\ \text{mg}/\text{L}$, pH will drop sharply into the acidic range, impairing floc formation. Alkalinity deficit to be made up by hydrated lime:
Conclusion: The mill consumes $192.0\ \text{kg/day}$ of alum. Each mg of alum destroys $0.505\ \text{mg } \text{CaCO}_3$, leaving a meager $4.58\ \text{mg}/\text{L}$ residual alkalinity. Supplemental lime dosing of at least $15.4\ \text{mg}/\text{L as } \text{CaCO}_3$ is mandatory.
Spent tannery chrome bath liquor has a volume of $V = 15.0\ \text{m}^3$ and contains hexavalent chromium at $[ ext{Cr(VI)}] = 420.0\ \text{mg}/\text{L}$. Reduction to trivalent chromium is carried out using commercial sodium metabisulfite ($\text{Na}_2\text{S}_2\text{O}_5$, $96.0\%$ purity, molar mass $190.11\ \text{g}/\text{mol}$) under acidic conditions according to: $2\text{Cr}_2\text{O}_7^{2-} + 3\text{S}_2\text{O}_5^{2-} + 10\text{H}^+ \rightarrow 4\text{Cr}^{3+} + 6\text{SO}_4^{2-} + 5\text{H}_2\text{O}$ (noting that $1\ \text{mol } \text{Cr}_2\text{O}_7^{2-}$ contains $2\ \text{mol of Cr}$). (a) Calculate the total mass of $\text{Cr(VI)}$ present in the batch in kilograms. (b) Determine the stoichiometric mass ratio of pure $\text{Na}_2\text{S}_2\text{O}_5$ to $\text{Cr(VI)}$ (in $\text{g } \text{Na}_2\text{S}_2\text{O}_5 / \text{g Cr}$). (c) Calculate the actual required mass of commercial $96.0\%$ metabisulfite including a $15\%$ excess to drive the reaction to completion.
Step 1: Calculate total mass of $\text{Cr(VI)}$ Given $V = 15.0\ \text{m}^3 = 15,000\ \text{L}$ and $[ ext{Cr(VI)}] = 420.0\ \text{mg}/\text{L} = 0.420\ \text{g}/\text{L}$:
Step 2: Determine stoichiometric mass ratio From the balanced redox equation:
$4\ \text{moles of Cr atoms}$ ($4 \times 51.996\ \text{g}/\text{mol} = 207.984\ \text{g}$) react with $3\ \text{moles of } \text{Na}_2\text{S}_2\text{O}_5$ ($3 \times 190.107\ \text{g}/\text{mol} = 570.321\ \text{g}$). The theoretical stoichiometric mass ratio:
Step 3: Calculate actual required commercial reagent mass Theoretical pure mass:
Applying $15\%$ excess ($1.15$) and accounting for $96.0\%$ purity ($0.960$):
Conclusion: The batch contains $6.30\ \text{kg of Cr(VI)}$, requiring $20.7\ \text{kg}$ of commercial sodium metabisulfite for complete chemical reduction to harmless Cr(III).
An activated sludge ETP treats textile effluent with flow rate $Q = 3600\ \text{m}^3/\text{day}$ and influent soluble $\text{BOD}_5$ $S_0 = 850\ \text{mg}/\text{L}$. Environmental discharge regulations mandate effluent soluble BOD $S \le 25\ \text{mg}/\text{L}$. Biological kinetic parameters are: synthesis yield $Y = 0.50\ \text{g VSS}/\text{g BOD}$, maximum utilization rate $k = 4.0\ \text{day}^{-1}$, half-velocity constant $K_s = 60\ \text{mg}/\text{L}$, and endogenous decay coefficient $b = 0.060\ \text{day}^{-1}$. The target MLVSS concentration in the aeration tank is maintained at $X = 3000\ \text{mg}/\text{L}$ ($3.0\ \text{kg}/\text{m}^3$). (a) Calculate the minimum mean cell residence time (sludge age, $\theta_c$) required to achieve $S = 25\ \text{mg}/\text{L}$. (b) Calculate the required aeration basin volume $V$ (in $\text{m}^3$) and hydraulic retention time $\theta$ (in hours). (c) Calculate the food-to-microorganism ($F/M$) ratio and daily excess biological sludge production $P_x$ (in $\text{kg VSS}/\text{day}$).
Step 1: Calculate required mean cell residence time $\theta_c$ From the Lawrence-McCarty substrate equation:
Given:
- $S = 25\ \text{mg}/\text{L}$
- $K_s = 60\ \text{mg}/\text{L}$
- $Y = 0.50$
- $k = 4.0\ \text{day}^{-1} \implies Y k = (0.50)(4.0) = 2.0\ \text{day}^{-1}$
- $b = 0.060\ \text{day}^{-1} \implies Y k - b = 2.0 - 0.060 = 1.940\ \text{day}^{-1}$
Rearranging to solve for $\theta_c$:
To ensure a robust safety factor against shock loads and provide excellent bio-flocculation, engineering practice applies a design factor of $3.5 \times$:
At $\theta_c = 6.63\ \text{days}$, effluent soluble BOD drops to:
Step 2: Calculate aeration basin volume and hydraulic retention time Using the design sludge age $\theta_c = 6.63\ \text{days}$ with $X = 3000\ \text{mg}/\text{L}$, $S_0 = 850\ \text{mg}/\text{L}$, and $S = 7.1\ \text{mg}/\text{L}$:
In hours:
Aeration basin volume:
Step 3: Calculate $F/M$ ratio and daily sludge production $P_x$
Daily biological excess waste sludge ($P_x$):
Conclusion: The system requires an aeration basin of $2,400\ \text{m}^3$ ($HRT = 16.0\ \text{h}$), operating at an $F/M$ of $0.425\ \text{day}^{-1}$, producing $1,086\ \text{kg VSS/day}$ of excess secondary sludge.
An Upflow Anaerobic Sludge Blanket (UASB) reactor processes high-strength distillery wash: wastewater flow $Q = 1200\ \text{m}^3/\text{day}$ with soluble COD $S_0 = 12,000\ \text{mg}/\text{L}$ ($12.0\ \text{kg}/\text{m}^3$). Design volumetric loading rate is $VLR = 8.0\ \text{kg COD}/(\text{m}^3\cdot\text{day})$ with an achieved COD removal efficiency of $\eta = 82.0\%$. (a) Calculate the required active liquid volume of the UASB reactor $V$ and the hydraulic retention time $\theta$ (in hours). (b) At standard conditions ($0^\circ\text{C}, 1\ \text{atm}$), $1.0\ \text{kg of COD}$ converted yields stoichiometrically $0.350\ \text{m}^3$ of pure methane gas ($CH_4$). Correcting to the mesophilic operating temperature of $35^\circ\text{C}$ ($308.15\ \text{K}$), calculate the daily volume of pure $CH_4$ produced in $\text{m}^3/\text{day}$. (c) If the biogas contains $65.0\%\ CH_4$ by volume and the lower heating value of methane is $35.8\ \text{MJ}/\text{m}^3$ at $35^\circ\text{C}$, calculate the daily thermal energy recovery potential in gigajoules (GJ/day).
Step 1: Calculate required UASB reactor volume and HRT Total daily COD mass loading:
With design volumetric loading rate $VLR = 8.0\ \text{kg COD}/(\text{m}^3\cdot\text{day})$:
Hydraulic retention time:
Step 2: Calculate daily methane generation at $35^\circ\text{C}$ Total COD destroyed:
At STP ($0^\circ\text{C}$, $1\ \text{atm}$), theoretical methane yield is $0.350\ \text{m}^3/\text{kg COD}$:
Adjusting for thermal expansion at $35^\circ\text{C}$ ($308.15\ \text{K}$):
Step 3: Calculate energy generation potential Total biogas volume ($65.0\%\ CH_4$):
Daily thermal energy produced from methane ($LHV = 35.8\ \text{MJ}/\text{m}^3$):
In continuous thermal power equivalent ($1\ \text{day} = 86,400\ \text{s}$):
Conclusion: The UASB volume is $1,800\ \text{m}^3$ ($HRT = 36\ \text{h}$), generating $4,662\ \text{m}^3/\text{day}$ of pure methane, yielding $166.9\ \text{GJ/day}$ ($1.93\ \text{MW}$ thermal power).
The destruction of a recalcitrant textile azo dye (Reactive Red 120, $D$) in a Photo-Fenton batch reactor follows second-order kinetics with hydroxyl radicals: $-\frac{d[D]}{dt} = k_{OH} [\cdot OH]_{ss} [D]$, where the second-order rate constant is $k_{OH} = 4.80 \times 10^9\ \text{M}^{-1}\text{s}^{-1}$. In the illuminated reactor, steady-state hydroxyl radical concentration is maintained at $[\cdot OH]_{ss} = 2.50 \times 10^{-12}\ \text{M}$. (a) Calculate the pseudo-first-order rate constant $k_{obs}$ in $\text{s}^{-1}$ and $\text{min}^{-1}$. (b) Calculate the dye half-life $t_{1/2}$ in minutes. (c) If natural chloride ions present in the dyeing bath ($[Cl^-] = 0.050\ \text{M}$) scavenge hydroxyl radicals via $\cdot OH + Cl^- \rightarrow ClOH^{\cdot-}$ ($k_{scav} = 4.30 \times 10^9\ \text{M}^{-1}\text{s}^{-1}$), depressing $[\cdot OH]_{ss}$ by $70.0\%$, calculate the new reaction time required to achieve $99.0\%$ color removal.
Step 1: Calculate pseudo-first-order rate constant $k_{obs}$ Given:
- $k_{OH} = 4.80 \times 10^9\ \text{M}^{-1}\text{s}^{-1}$
- $[\cdot OH]_{ss} = 2.50 \times 10^{-12}\ \text{M}$
Converting to per minute:
Step 2: Calculate dye half-life
Step 3: Calculate required reaction time under chloride scavenging When chloride depression reduces steady-state radical concentration by $70.0\%$, the remaining radical concentration is:
The reduced observed rate constant:
For $99.0\%$ color elimination, remaining dye concentration is $1.0\%$ ($[D]/[D]_0 = 0.010$):
Conclusion: In pure water, dye half-life is 57.8 seconds. High salinity scavenger interference extends the $99\%$ destruction time from $6.4\ \text{min}$ to 21.3 minutes.
A spiral-wound polyamide reverse osmosis (RO) membrane element treats tertiary textile effluent containing $C_b = 3500\ \text{mg}/\text{L}$ $NaCl$ at $25^\circ\text{C}$ ($T = 298.15\ \text{K}$, $R = 0.08314\ \text{L}\cdot\text{bar}/(\text{mol}\cdot\text{K})$). The membrane has pure water permeability $A = 3.50\ \text{L}/(\text{m}^2\cdot\text{h}\cdot\text{bar})$ and salt permeability $B = 0.850\ \text{L}/(\text{m}^2\cdot\text{h})$. Applied net hydraulic pressure is $\Delta P = 24.0\ \text{bar}$. Taking complete dissociation of $NaCl$ ($i = 2$, molar mass $58.44\ \text{g}/\text{mol}$): (a) Assuming zero permeate salt concentration as a first approximation, calculate the feed osmotic pressure $\pi_f$ (in bar) and initial net driving pressure $\Delta P - \pi_f$. (b) Using the coupled solution-diffusion equations $J_w = A (\Delta P - \Delta \pi)$ and $R_{obs} = 1 - \frac{C_p}{C_m} = \frac{J_w}{J_w + B}$, and assuming negligible concentration polarization ($C_m \approx C_b$), solve iteratively for the steady-state permeate flux $J_w$ (in $\text{L}/(\text{m}^2\cdot\text{h})$) and true observed salt rejection $R_{obs}$. (c) Calculate the permeate salt concentration $C_p$ in $\text{mg}/\text{L}$.
Step 1: Calculate feed osmotic pressure $\pi_f$ Feed salt concentration in molarity:
With van 't Hoff factor $i = 2$:
Step 2: Set up and solve the coupled quadratic equation for $J_w$ The osmotic pressure differential across the membrane:
From the solution-diffusion model:
Substituting into the water flux equation:
Multiplying both sides by $(J_w + B)$:
Rearranging into standard quadratic form:
Evaluate coefficients with given numerical values:
- $A = 3.50\ \text{L}/(\text{m}^2\cdot\text{h}\cdot\text{bar})$
- $B = 0.850\ \text{L}/(\text{m}^2\cdot\text{h})$
- $\Delta P = 24.0\ \text{bar}$
- $\pi_f = 2.969\ \text{bar}$
- $\Delta P - \pi_f = 24.0 - 2.969 = 21.031\ \text{bar}$
- $A (\Delta P - \pi_f) = 3.50 \times 21.031 = 73.6085$
- Linear coefficient: $b_{quad} = B - A(\Delta P - \pi_f) = 0.850 - 73.6085 = -72.7585$
- Constant coefficient: $c_{quad} = -A B \Delta P = -(3.50)(0.850)(24.0) = -71.40$
The quadratic equation:
Applying the quadratic formula:
Step 3: Calculate rejection $R_{obs}$ and permeate concentration $C_p$
Permeate salt concentration:
Conclusion: The system delivers a permeate flux of $73.7\ \text{L}/(\text{m}^2\cdot\text{h})$ at an observed rejection of $98.86\%$, purifying the feed from $3500\ \text{mg}/\text{L}$ down to $39.9\ \text{mg}/\text{L}$ TDS.
An advanced tertiary denitrification filter removes nitrate from an industrial wastewater effluent with flow rate $Q = 4800\ \text{m}^3/\text{day}$. Influent to the anoxic filter contains nitrate-nitrogen $[NO_3^--N] = 38.0\ \text{mg}/\text{L}$, nitrite-nitrogen $[NO_2^--N] = 2.0\ \text{mg}/\text{L}$, and dissolved oxygen $[O_2] = 4.5\ \text{mg}/\text{L}$. Methanol ($CH_3OH$, molar mass $32.04\ \text{g}/\text{mol}$) is dosed as the supplemental electron donor. The overall empirical stoichiometry including biomass synthesis is: $NO_3^- + 1.08\ CH_3OH + 0.24\ H_2CO_3 \rightarrow 0.056\ C_5H_7O_2N\text{(biomass)} + 0.47\ N_2 + 1.68\ H_2O + HCO_3^-$. In addition, methanol is consumed by $NO_2^-$ ($NO_2^- + 0.67\ CH_3OH \rightarrow \text{biomass} + 0.5\ N_2$) and by residual dissolved oxygen ($O_2 + 0.93\ CH_3OH \rightarrow \text{biomass} + CO_2 + H_2O$). Using the standard EPA empirical methanol requirement formula: $C_m = 2.47 [NO_3^--N] + 1.53 [NO_2^--N] + 0.87 [O_2]$. (a) Calculate the required methanol concentration $C_m$ in $\text{mg}/\text{L}$. (b) Determine the daily consumption of commercial methanol ($99.0\%$ purity, density $\rho = 0.792\ \text{kg}/\text{L}$) in liters per day. (c) Calculate the daily biological denitrification sludge production in $\text{kg dry solids}/\text{day}$.
Step 1: Calculate required methanol concentration $C_m$ Applying the EPA empirical relationship:
Substituting parameter values:
- Term 1 (Nitrate): $2.47 \times 38.0 = 93.86\ \text{mg}/\text{L}$
- Term 2 (Nitrite): $1.53 \times 2.0 = 3.06\ \text{mg}/\text{L}$
- Term 3 (Dissolved Oxygen): $0.87 \times 4.5 = 3.915\ \text{mg}/\text{L}$
Step 2: Calculate daily volume of commercial methanol Total mass of pure methanol per day:
Accounting for $99.0\%$ chemical purity and density $\rho = 0.792\ \text{kg}/\text{L}$:
Step 3: Calculate daily excess biological sludge production From the balanced stoichiometry for nitrate: $1\ \text{mol } NO_3^--N$ ($14.01\ \text{g N}$) produces $0.056\ \text{mol of biomass}$ ($C_5H_7O_2N$, molar mass $113.12\ \text{g}/\text{mol}$):
Nitrate nitrogen removed daily:
Sludge produced from nitrate reduction:
Adding cellular synthesis from dissolved oxygen respiration ($0.20\ \text{g VSS}/\text{g } O_2$):
Total biological sludge produced:
Conclusion: The system requires a dosing rate of $100.8\ \text{mg}/\text{L}$ of methanol ($617\ \text{L/day}$ of commercial methanol), generating $86.8\ \text{kg/day}$ of dry biological sludge.
A Zero Liquid Discharge (ZLD) plant concentrates High-Pressure Reverse Osmosis (HPRO) reject brine at $Q_f = 25.0\ \text{m}^3/\text{h}$ with TDS $C_f = 80,000\ \text{mg}/\text{L}$ ($8.0\ \text{wt}\%$, density $\rho_f = 1060\ \text{kg}/\text{m}^3$). The brine enters a Mechanical Vapor Recompression (MVR) falling film evaporator producing pure condensed distillate and concentrated liquor at $C_L = 280,000\ \text{mg}/\text{L}$ ($28.0\ \text{wt}\%$, density $\rho_L = 1200\ \text{kg}/\text{m}^3$). This concentrated liquor feeds a draft-tube baffle crystallizer yielding dry salt cake ($98.0\ \text{wt}\%$ dry solids) and mother liquor recycled to extinction. (a) Perform a solid and liquid mass balance on the MVR evaporator to calculate the mass flow rate of feed $m_f$ (in $\text{kg}/\text{h}$), concentrated liquor $m_L$ (in $\text{kg}/\text{h}$), and pure evaporated distillate $m_D$ (in $\text{kg}/\text{h}$ and $\text{m}^3/\text{h}$). (b) Calculate the daily mass of dry solid salt cake recovered (in metric tons/day). (c) The MVR compressor consumes $28.0\ \text{kWh}$ of electrical power per metric ton of water evaporated. Calculate the continuous operating electrical power demand of the MVR compressor in kilowatts (kW).
Step 1: Perform mass balance on MVR falling film evaporator Feed mass flow rate:
Mass fraction of dissolved solids:
- Feed: $w_f = 0.080$ ($8.0\ \text{wt}\%$)
- Concentrated liquor: $w_L = 0.280$ ($28.0\ \text{wt}\%$)
Salt mass conservation across the evaporator (pure vapor contains $0\%$ salt):
Total water evaporated and condensed as pure distillate:
Volumetric distillate water recovery ($ ho_w \approx 1000\ \text{kg}/\text{m}^3$):
Water recovery percentage in the evaporator:
Step 2: Calculate daily solid salt recovery from crystallizer Total salt entering the system per hour:
With crystallization producing salt cake at $98.0\ \text{wt}\%$ solids:
Daily salt cake production ($24\ \text{hours}$):
Step 3: Calculate MVR electrical compressor power demand Specific electrical energy consumption $= 28.0\ \text{kWh} / \text{metric ton evaporated}$. Evaporation rate:
Continuous electrical power consumption:
Conclusion: The MVR evaporator recovers $18.93\ \text{m}^3/\text{h}$ of pure distilled water ($454\ \text{m}^3/\text{day}$), producing $51.92\ \text{tons/day}$ of dry salt cake, drawing a steady compressor load of $530\ \text{kW}$.