Chemistry / Industrial Chemistry Chemical Processes, Transport Dynamics & Quality Assurance 100% Free Open Access
Chapter 3 β€’ Theory & Derivations

Unit 3: Sugar and Starch Industries: Extraction, Refining & Bioproducts

Comprehensive industrial chemical engineering treatise on carbohydrate technologies: sugarcane milling, continuous liming/carbonatation/sulphitation clarification, multiple-effect evaporation thermodynamics, vacuum pan massecuite boiling and crystallization, raw sugar refining, by-product valorization (bagasse, molasses fermentation into ethanol), and corn/cassava wet milling into native starches, glucose syrups, and dextrins.

Β§3.1 Raw Carbohydrate Feedstocks: Sugarcane vs Sugar Beet Biochemistry

Industrial sucrose ($\text{C}_{12}\text{H}_{22}\text{O}_{11}$, $\alpha\text{-D-glucopyranosyl-(1}\to\text{2)-}\beta\text{-D-fructofuranoside}$, $M = 342.30\text{ g/mol}$) is extracted from two major commercial agricultural crops:

1. Sugarcane (*Saccharum officinarum*): Tropical C4 perennial grass. Mature cane stalks contain $11 - 16\text{ wt}\%$ sucrose, $0.5 - 1.5\%$ reducing sugars (glucose and fructose), $11 - 16\%$ fiber (cellulose, hemicellulose, lignin), and $68 - 75\%$ water.

2. Sugar Beet (*Beta vulgaris*): Temperate biennial root crop. Contains $15 - 20\text{ wt}\%$ sucrose, negligible reducing sugars ($< 0.1\%$), $4 - 5\%$ pulp, and $75 - 80\%$ water.

Sugar Inversion and Optical Rotation

Sucrose is a non-reducing disaccharide without an anomeric free hemiacetal group. In acidic aqueous media or via yeast $\beta$-fructofuranosidase (invertase), sucrose undergoes hydrolysis into equimolar amounts of D-glucose and D-fructose:

$$\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{H}^+ / \text{Invertase}} \text{C}_6\text{H}_{12}\text{O}_6\text{ (D-Glucose)} + \text{C}_6\text{H}_{12}\text{O}_6\text{ (D-Fructose)}$$

This hydrolysis is termed inversion because of the reversal of optical polarization:

  • Pure sucrose is dextrorotatory: $[\alpha]_D^{20} = +66.5^\circ$
  • D-Glucose is dextrorotatory: $[\alpha]_D^{20} = +52.7^\circ$
  • D-Fructose is strongly levorotatory: $[\alpha]_D^{20} = -92.4^\circ$

The equimolar equiconcentration mixture (invert sugar) has a net levorotatory optical rotation:

$$[\alpha]_{D, \text{net}}^{20} = \frac{+52.7^\circ + (-92.4^\circ)}{2} = -19.85^\circ$$

Because invert sugars are hygroscopic and hinder sucrose crystallization, industrial extraction operations must maintain strictly alkaline to neutral pH conditions ($\text{pH } 7.2 - 8.2$) to minimize sucrose inversion losses.

Β§3.2 Cane Milling, Juice Extraction & Shredding Mechanics

Juice extraction from sugarcane stalks involves physical cell rupturing followed by mechanical squeezing or countercurrent liquid diffusion.

1. Cane Preparation

Raw cane stalks arriving at the factory are dumped onto feeding tables and pass through:

  • Cane Knives: High-speed rotating blades ($500 - 600\text{ rpm}$) that level and chop stalks into small billets.
  • Heavy-Duty Shredders: Hammer-mill rotors with pivoting hammers that shatter rind cells without expressing juice, achieving an Open Cell Index ($\text{OCI}$) exceeding $88 - 92\%$.

2. Multi-Mill Tandems & Compound Imbibition

The prepared cane passes through a tandem of 4 to 6 three-roller mills. Each mill comprises a Top roller, Feed roller, and Discharge roller arranged in a triangular configuration with a central trash plate:

  • Compound Imbibition: To extract residual sugar from the fiber matrix without excessive dilution, fresh imbibition water ($65 - 75^\circ\text{C}$, $20 - 30\text{ wt}\%$ on cane) is applied exclusively to the bagasse entering the final mill.
  • The expressed thin juice from the final mill is recycled backward to spray bagasse entering the penultimate mill, moving countercurrently against the advancing fiber mat.
  • Modern milling tandems achieve total sucrose extraction efficiencies ($\text{Pol Extraction}$) of $95.5 - 97.0\%$, discharging moist bagasse ($48 - 50\text{ wt}\%$ moisture, $1.5 - 2.5\%\text{ Pol}$) directly to factory steam boilers.

Multiple-Effect Evaporator Heat Transfer & Boiling Point Elevation Table

In sugarcane and beet sugar refining, thin juice is concentrated from $14 - 16^\circ\text{Brix}$ to $65 - 70^\circ\text{Brix}$ syrup across a quintuple-effect evaporator train:

| Evaporator Effect | Operating Pressure ($\text{bar abs}$) | Vapor Temp ($^\circ\text{C}$) | Juice Brix ($^\circ\text{Bx}$) | BPE ($\text{K}$) | Boiling Temp ($^\circ\text{C}$) | Overall $U$ ($\text{W/(m}^2\cdot\text{K)}$) | |---|---|---|---|---|---|---| | Effect 1 | $2.05\text{ bar}$ | $121.0^\circ\text{C}$ | $18.5^\circ\text{Bx}$ | $0.6\text{ K}$ | $121.6^\circ\text{C}$ | $2,600\text{ W/(m}^2\cdot\text{K)}$ | | Effect 2 | $1.45\text{ bar}$ | $110.3^\circ\text{C}$ | $24.0^\circ\text{Bx}$ | $1.0\text{ K}$ | $111.3^\circ\text{C}$ | $2,100\text{ W/(m}^2\cdot\text{K)}$ | | Effect 3 | $0.98\text{ bar}$ | $99.1^\circ\text{C}$ | $32.5^\circ\text{Bx}$ | $1.8\text{ K}$ | $100.9^\circ\text{C}$ | $1,600\text{ W/(m}^2\cdot\text{K)}$ | | Effect 4 | $0.55\text{ bar}$ | $83.7^\circ\text{C}$ | $45.0^\circ\text{Bx}$ | $3.5\text{ K}$ | $87.2^\circ\text{C}$ | $1,100\text{ W/(m}^2\cdot\text{K)}$ | | Effect 5 | $0.20\text{ bar}$ | $60.1^\circ\text{C}$ | $68.0^\circ\text{Bx}$ | $8.4\text{ K}$ | $68.5^\circ\text{C}$ | $650\text{ W/(m}^2\cdot\text{K)}$ |

Boiling Point Elevation ($\text{BPE}$) of sugar solutions follows the empirical DΓΌhring line relationship:

$$\text{BPE} = \frac{0.070 \cdot B}{100 - B} \cdot \left(\frac{T_{\text{sat}} + 273.15}{100}\right)^2 \quad (\text{K})$$

where $B$ is sucrose concentration in $^\circ\text{Brix}$ and $T_{\text{sat}}$ is saturation temperature of pure water in $^\circ\text{C}$.

Β§3.3 Juice Clarification: Defecation, Sulphitation & Carbonatation Technologies

Raw sugarcane juice is an opaque, turbid, acidic liquid ($\text{pH } 5.0 - 5.5$) containing suspended bagasse particles, colloidal proteins, polysaccharides, organic acids, polyphenols, and coloring matter.

Clarification Chemical Processes

``` SUGAR JUICE CLARIFICATION PATHS β”‚ β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”Όβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β–Ό β–Ό β–Ό SIMPLE DEFECATION SULPHITATION PROCESS CARBONATATION PROCESS Milk of Lime [Ca(OH)β‚‚] Milk of Lime + SOβ‚‚ gas Excess Ca(OH)β‚‚ + COβ‚‚ gas Target pH: 7.2 - 7.6 Target pH: 6.8 - 7.0 Double Carbonatation (pH 10.5 ──► 8.5) Calcium Phosphate Floc Calcium Sulfite [CaSO₃] Floc Bulk CaCO₃ Crystal Surface Adsorption β”‚ β”‚ β”‚ β–Ό β–Ό β–Ό Raw Sugar Production Direct Plantation White Sugar Refined White Sugar (> 99.8% Pol) ```

1. Simple Defecation (Raw Sugar Manufacture)

Raw juice is heated to $70 - 75^\circ\text{C}$ and treated with milk of lime ($\text{Ca(OH)}_2$, $0.05 - 0.10\text{ wt}\%\text{ CaO}$ on juice) to neutralize organic acids and raise pH to $7.4 - 7.8$. Soluble natural phosphate ($\text{PO}_4^{3-}$, native concentration adjusted to $> 300\text{ ppm}$) reacts with calcium ions to form a flocculent tricalcium phosphate precipitate:

$$3\text{Ca}^{2+} + 2\text{PO}_4^{3-} \to \text{Ca}_3(\text{PO}_4)_2\downarrow$$

The juice is heated to boiling ($103 - 105^\circ\text{C}$) to flash off entrained air bubbles and fed to continuous multideck clarifiers (Dorr-Oliver clarifiers) with synthetic polyacrylamide flocculants. Clear juice overflows the top, while settled muds are filtered on rotary vacuum drum filters.

2. Sulphitation (Plantation White Sugar)

Simultaneous addition of milk of lime and sulfur dioxide gas ($\text{SO}_2$, generated by burning sulfur in air) precipitates insoluble calcium sulfite:

$$\text{SO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{SO}_3$$
$$\text{Ca(OH)}_2 + \text{H}_2\text{SO}_3 \to \text{CaSO}_3\downarrow + 2\text{H}_2\text{O}$$

The $\text{CaSO}_3$ precipitate adsorbs gums, waxes, and colored impurities. Simultaneously, the reducing sulfite ion ($\text{SO}_3^{2-}$) bleaches melanoidin and polyphenolic pigments by chemical reduction.

3. Continuous Carbonatation (Refined Sugar Standard)

Juice is treated with a large excess of lime ($1.5 - 2.5\text{ wt}\%\text{ CaO}$) followed by sparging with flue gas carbon dioxide ($28 - 32\%\text{ CO}_2$ from lime kilns):

$$\text{Ca(OH)}_2 + \text{CO}_2 \to \text{CaCO}_3\downarrow + \text{H}_2\text{O}$$

The massive precipitation of microcrystalline calcite ($\text{CaCO}_3$) creates an extensive surface area that occludes colloidal impurities, delivering clarified liquor suitable for white refined sugar.

Β§3.4 Multiple-Effect Evaporator Stations & Calandria Steam Economy

Clarified juice enters the evaporator station at $13 - 16^\circ\text{Bx}$ (percent dissolved solids) and must be concentrated to thick syrup at $60 - 68^\circ\text{Bx}$, requiring the evaporation of approximately $75 - 80\text{ wt}\%$ of the initial water content.

The Physics of Multiple-Effect Evaporation (Norbert Rillieux, 1843)

In a multiple-effect evaporator, live boiler exhaust steam ($1.8 - 2.4\text{ bar}$, $115 - 125^\circ\text{C}$) is supplied exclusively to the heating calandria of the First Effect. The water vapor boiled off from the first effect is used as the heating medium in the calandria of the Second Effect, which is maintained at a lower pressure and boiling point. This cascading sequence continues through 4 or 5 effects in series, ending in a barometric condenser operating under deep vacuum ($0.15 - 0.25\text{ bar}$, boiling point $55 - 65^\circ\text{C}$).

Steam Economy Principle

For an ideal $N$-effect evaporator with negligible heat loss:

$$\text{Steam Economy} = \frac{\text{Total kg of Water Evaporated across train}}{\text{kg of Live Steam supplied to 1st Effect}} \approx 0.85 \times N$$

For a standard Quadruple-Effect station ($N = 4$), $1\text{ kg}$ of live exhaust steam evaporates $3.2 - 3.5\text{ kg}$ of water.

Boiling Point Elevation (BPE)

As juice concentrates across the train, the boiling point of the solution exceeds the boiling point of pure water at the same pressure, governed by the DΓΌhring rule and Raoult's law:

$$\Delta T_b = K_b \cdot m \cdot i$$

In the 4th effect ($65^\circ\text{Bx}$ syrup), BPE reaches $5.5 - 7.5^\circ\text{C}$, reducing the effective temperature driving force ($\Delta T_{\text{eff}} = T_{\text{steam}} - T_{\text{boil}} - \text{BPE}$) available for heat transfer.

Β§3.5 Vacuum Pan Massecuite Boiling & Crystallization Kinetics

Syrup from the evaporators ($65^\circ\text{Bx}$) is converted into crystalline sugar within single-effect vacuum pans operating under controlled vacuum ($0.15 - 0.20\text{ bar}$, boiling at $60 - 68^\circ\text{C}$ to avoid carmelization and thermal color development).

Supersaturation and Crystallization Regimes

The crystallization driving force is governed by the supersaturation coefficient ($SS$):

$$SS = \frac{(S / W)_{T}}{(S / W)_{T, \text{sat}}}$$

where $(S/W)$ is the mass ratio of dissolved sucrose to water:

1. Metastable Zone ($1.00 < SS < 1.20$): Existing crystal seed faces grow by mass-transfer diffusion without spontaneous nucleation.

2. Intermediate Zone ($1.20 < SS < 1.40$): Existing crystals grow, and false grain (uncontrolled secondary nucleation) occurs if agitated.

3. Labile Zone ($SS > 1.40$): Spontaneous homogeneous nucleation occurs, generating irregular fine crystals.

The Three-Boiling Scheme (A, B, C Massecuites)

To maximize sucrose recovery from mother liquor, sugar factories employ a multi-strike boiling sequence:

  • A-Strike (High Purity, $\sim 85 - 90\%$ Purity): Boiled from virgin syrup and seeded with fine slurry. Discharged as A-Massecuite into centrifugals; yields commercial raw sugar and A-Molasses.
  • B-Strike ($\sim 75 - 78\%$ Purity): Boiled from A-Molasses and syrup; yields B-Sugar and B-Molasses.
  • C-Strike ($\sim 58 - 62\%$ Purity, Low Grade): Boiled from B-Molasses; cooled slowly in continuous water-cooled crystallizers over $24 - 48\text{ hours}$ to exhaust mother liquor down to $SS \to 1.0$. Centrifugation separates low-grade C-Sugar (recycled as seed magma) from Final Blackstrap Molasses ($30 - 35\%$ sucrose, $15 - 20\%$ reducing sugars, purity $< 35\%$), from which no further crystallization is economically viable.

Sucrose Hydrolysis (Inversion) Kinetics & Optical Mutarotation

During heating at acidic $\text{pH}$, sucrose hydrolyzes into equimolar D-glucose and D-fructose:

$$\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \overset{\text{H}^+ / \text{Invertase}}{\longrightarrow} \text{C}_6\text{H}_{12}\text{O}_6\text{ (D-Glucose)} + \text{C}_6\text{H}_{12}\text{O}_6\text{ (D-Fructose)}$$
  • Optical Inversion Phenomenon:
  • Pure sucrose is dextrorotatory: $[\alpha]_D^{20} = +66.5^\circ$.
  • Inverted hydrolysate contains D-glucose ($[\alpha]_D^{20} = +52.7^\circ$) and strongly levorotatory D-fructose ($[\alpha]_D^{20} = -92.4^\circ$).
  • Net specific optical rotation shifts from positive to negative:
$$[\alpha]_{D, \text{invert}}^{20} = \frac{+52.7^\circ + (-92.4^\circ)}{2} = -19.85^\circ$$

This reversal of polarimetric sign from $+66.5^\circ \to -19.85^\circ$ designates the reaction as sugar inversion.

Β§3.6 Raw Sugar Refining: Affination, Decolorization & Granulation

Raw sugar crystals ($97.0 - 98.5\%$ sucrose) are covered with a thin film of dark, sticky mother liquor containing ash, invert sugars, and colorants. Refining removes this molasses film and purifies sucrose to $> 99.9\%$ purity.

Industrial Refining Sequence

1. Affination: Raw sugar crystals are mixed with hot, concentrated affination syrup ($70 - 75^\circ\text{Bx}$, $65 - 75^\circ\text{C}$) to soften the outer molasses film without dissolving the crystalline sucrose core. The magma is spun in high-speed basket centrifugals, washing off $85\%$ of surface impurities to produce Affined Sugar ($99.0 - 99.3\%\text{ Pol}$).

2. Melting & Clarification: Affined sugar is dissolved in hot demineralized water to produce raw liquor ($60 - 65^\circ\text{Bx}$) and clarified by phosphatation or carbonatation.

3. Decolorization: Clarified liquor passes through deep beds of granular activated carbon (GAC), bone char, or macroporous styrenic/acrylic strong-base anion exchange resins:

  • Polymeric colorants (melanoidins from Maillard reactions, caramels, and alkaline degradation products of fructose) carry negative charges and bind to quaternary ammonium resin sites ($\text{Resin-N}^+\text{Me}_3\cdots\text{Color}^-$).
  • Effluent liquor reaches $< 20\text{ ICUMSA}$ color units (sparkling water-white).

4. Refined Sugar Boiling & Conditioned Granulation: White liquor is boiled in vacuum pans, centrifuged, and dried in rotary drum granulators using dehumidified air to moisture $< 0.03\text{ wt}\%$.

Β§3.7 Industrial Starch Processing: Wet Milling, Enzymatic Hydrolysis & Dextrins

Starch is an endosperm energy-storage polysaccharide comprising two $\alpha$-glucan polymers:

  • Amylose ($20 - 30\text{ wt}\%$): Linear chain of $\alpha\text{-(1}\to\text{4)-linked}$ D-glucopyranose units ($M \approx 10^5 - 10^6\text{ g/mol}$) forming helical coils.
  • Amylopectin ($70 - 80\text{ wt}\%$): Branched polymer containing $\alpha\text{-(1}\to\text{4)}$ linear links with $\alpha\text{-(1}\to\text{6)}$ branch points every $24 - 30$ glucose residues ($M \approx 10^7 - 10^8\text{ g/mol}$).

1. Corn Wet Milling Extraction

Shelled corn kernels ($70\%$ starch, $10\%$ protein, $4.5\%$ oil, $2\%$ fiber) undergo a 4-step wet separation:

1. Steeping: Kernels soak in warm water ($50 - 52^\circ\text{C}$) containing $0.10 - 0.20\text{ wt}\%\text{ SO}_2$ and lactic acid for $24 - 48\text{ hours}$. $\text{SO}_2$ reduces disulfide crosslinks in the glutelin protein matrix, freeing starch granules.

2. Germ Separation: Coarsely ground corn passes through hydrocyclones; low-density germ ($50\%$ corn oil) floats and is recovered for corn oil pressing.

3. Fiber Washing: Finely milled slurry passes through curved DSM screens to remove coarse fiber (bran/pericarp).

4. Starch-Gluten Separation: The starch-gluten slurry passes to high-speed disk-nozzle centrifuges. Insoluble corn gluten protein (zein, density $\rho = 1.1\text{ g/cm}^3$) exits in the overflow, while dense starch granules ($\rho = 1.5\text{ g/cm}^3$) discharge in the underflow ($> 99.5\%$ pure starch).

2. Enzymatic Conversion into Sweeteners

Starch slurry ($30 - 35\text{ wt}\%$ solids) is converted into commercial syrups via two-stage enzymatic digestion:

1. Liquefaction: Slurry is adjusted to $\text{pH } 5.8 - 6.2$, stabilized with $\text{Ca}^{2+}$ ($50\text{ ppm}$), and injected with thermostable endo-$\alpha$-amylase (*Bacillus licheniformis*) in a steam jet cooker ($105^\circ\text{C}$ for $5\text{ min}$, then $95^\circ\text{C}$ for $90\text{ min}$):

$$\text{Starch} \xrightarrow{\alpha\text{-Amylase}} \text{Maltodextrins} \quad (\text{DE } 10 - 18)$$

The enzyme cleaves internal $\alpha\text{-(1}\to\text{4)}$ bonds, reducing viscosity.

2. Saccharification: Liquefied maltodextrin is cooled to $60^\circ\text{C}$, adjusted to $\text{pH } 4.2 - 4.5$, and incubated with fungal glucoamylase (*Aspergillus niger*):

$$\text{Maltodextrins} \xrightarrow{\text{Glucoamylase}} \text{D-Glucose (Dextrose)} \quad (\text{DE } 95 - 98)$$

Glucoamylase hydrolyzes both $\alpha\text{-(1}\to\text{4)}$ and $\alpha\text{-(1}\to\text{6)}$ linkages from non-reducing chain ends.

3. Isomerization to High Fructose Corn Syrup (HFCS-42): Purified dextrose liquor ($95\text{ DE}$) passes through immobilized glucose isomerase columns (*Streptomyces*), isomerizing glucose into an equilibrium mix of $42\%\text{ fructose}$ and $50 - 52\%\text{ glucose}$.

Β§3.8 Simulated Moving Bed (SMB) Chromatography, HFCS-55 & Starch Bioplastics

Starch processing extends beyond basic syrups into high-fructose corn syrups (HFCS) and renewable bioplastics that displace petroleum-derived polymers.

1. Simulated Moving Bed (SMB) Chromatographic Fructose Enrichment

Enzymatic glucose isomerase converts glucose into an equilibrium mixture containing at most $42\text{ wt}\%$ fructose (HFCS-42). To formulate beverage-grade HFCS-55 ($55\text{ wt}\%$ fructose, matching cane sucrose sweetness), the syrup must be chromatographically enriched:

  • Separation Principle: Resins functionalized with calcium sulfonate ($\text{Ca}^{2+}$) form weak coordination complexes with fructose oxygen atoms (fructose contains a flexible furanose ring that coordinates strongly with divalent calcium), while glucose passes through unretarded.
  • Simulated Moving Bed Technology: Continuous counter-current solid-liquid adsorption is achieved without physically moving the solid resin by cyclically shifting the fluid feed, eluent (pure water), extract (pure fructose), and raffinate (pure glucose) ports across $8 - 24$ interconnected packed columns:
$$u_{\text{port}} = \frac{L_{\text{zone}}}{\Delta t_{\text{switch}}}$$

The extracted stream yields $90\text{ wt}\%$ pure fructose, which is blended with HFCS-42 to formulate high-purity HFCS-55.

2. Starch Biopolymer Engineering: Polylactic Acid (PLA)

Starch is an industrial agricultural feedstock for biodegradable polyesters:

1. Lactic Acid Fermentation: Starch hydrolysate (pure glucose) is fermented anaerobically by *Lactobacillus delbrueckii* at $45 - 50^\circ\text{C}$ and $\text{pH } 5.5 - 6.5$:

$$\text{C}_6\text{H}_{12}\text{O}_6 \longrightarrow 2\text{CH}_3\text{-CH(OH)-COOH} \quad (\Delta G^\circ < 0)$$

2. Lactide Ring Formation: Lactic acid is pre-polymerized into low-molecular-weight oligomers, then catalytically depolymerized at $200^\circ\text{C}$ under vacuum with tin(II) octoate catalyst to yield cyclic dimer Lactide:

$$2\text{CH}_3\text{-CH(OH)-COOH} \longrightarrow \text{Lactide} + 2\text{H}_2\text{O}$$

3. Ring-Opening Polymerization (ROP): Purified lactide undergoes coordinate-insertion ring-opening polymerization:

$$n\,(\text{Lactide}) \overset{\text{Sn(Oct)}_2}{\longrightarrow} [-\text{CH(CH}_3)\text{-CO-O-}]_n \quad (\text{PLA}, \bar{M}_w > 100,000\text{ g/mol})$$

PLA exhibits tensile modulus comparable to polystyrene ($E \approx 3.5\text{ GPa}$) and degrades completely into water and $\text{CO}_2$ in industrial composting facilities.

University Honors Industrial Case Study: Dextran Contamination & Enzymatic Polysaccharide Mitigation

When sugarcane experiences post-harvest delays in tropical climates, epiphytic bacteria (Leuconostoc mesenteroides) metabolize sucrose into extracellular high-molecular-weight Dextran ($\alpha\text{-(1}\to\text{6)}$-linked D-glucan with $\alpha\text{-(1}\to\text{3)}$ branch points, $M_w > 10^6\text{ g/mol}$):

$$\text{Sucrose} \overset{\text{Dextransucrase}}{\longrightarrow} \text{Dextran} + \text{D-Fructose}$$
  • Process Havoc: Dextran drastically elevates juice and syrup viscosity, distorting sucrose crystal growth into elongated needle-shaped "cigar" crystals that cannot be separated in centrifugal baskets.
  • Biotechnological Remedy: Industrial cane mills inject thermo-tolerant fungal dextranase enzymes ($50 - 65^\circ\text{C}$) into mixed juice clarifiers, cleaving internal $\alpha\text{-(1}\to\text{6)}$ linkages into low-viscosity isomalto-oligosaccharides, restoring normal cubic crystal habit and sucrose crystallization velocity.
Easy Example 3.1: Cane Juice Defecation & Lime Consumption Mass Balance

A sugar mill processes $150.0\text{ metric tons/h}$ of raw sugarcane juice containing $15.0\text{ wt}\%$ dissolved solids ($15.0^\circ\text{Bx}$) and $12.5\text{ wt}\%$ sucrose. The raw juice has an initial natural acidity requiring $0.80\text{ kg of available CaO}$ per metric ton of raw juice for neutralization and calcium phosphate defecation. Commercial quicklime available at the factory has an active purity of $85.0\text{ wt}\%\text{ CaO}$.

  1. Calculate the hourly consumption of commercial quicklime in metric tons per hour.
  2. The factory slakes this quicklime with water to prepare an $8.0^\circ\text{Be}$ milk of lime suspension ($75.0\text{ g/L available CaO}$, suspension density $\rho = 1.06\text{ kg/L}$). Calculate the required volumetric flow rate of milk of lime in Liters per hour.
  3. If defecation removes $80.0\%$ of the initial non-sugar dissolved impurities into filter mud, and $99.5\%$ of the initial sucrose is preserved in clarified juice, calculate the sucrose purity ($\% \text{ Pol / Brix}$) of the clarified juice.

Step 1: Commercial Quicklime Mass

Hourly raw juice flow rate: $\dot{m}_{\text{juice}} = 150.0\text{ t/h}$. Available $\text{CaO}$ required:

$$\dot{m}_{\text{CaO, pure}} = 150.0\text{ t/h} \times 0.80\text{ kg/t} = 120.0\text{ kg/h}$$

Accounting for $85.0\%$ purity:

$$\dot{m}_{\text{quicklime}} = \frac{120.0\text{ kg/h}}{0.850} \approx 141.18\text{ kg/h} \approx 0.1412\text{ t/h}$$

The mill consumes $141.2\text{ kg/h}$ of commercial quicklime.

Step 2: Milk of Lime Volumetric Rate

With milk of lime containing $75.0\text{ g/L}$ active $\text{CaO}$:

$$\dot{V}_{\text{lime}} = \frac{120.0 \times 10^3\text{ g/h}}{75.0\text{ g/L}} = 1,600.0\text{ L/h}$$

The dosing pump delivers $1,600\text{ Liters/hour}$ ($1.60\text{ m}^3/\text{h}$) of milk of lime.

Step 3: Clarified Juice Purity

Initial juice composition per hour ($150.0\text{ t/h}$):

  • Total Brix (solids): $150.0 \times 0.150 = 22.50\text{ t/h}$
  • Sucrose: $150.0 \times 0.125 = 18.75\text{ t/h}$
  • Non-sugar impurities: $22.50 - 18.75 = 3.75\text{ t/h}$
  • Initial Purity: $\frac{18.75}{22.50} \times 100\% = 83.33\%$

In clarified juice:

  • Sucrose retained: $18.75 \times 0.995 = 18.656\text{ t/h}$
  • Non-sugars remaining ($80\%$ removed, $20\%$ remain): $3.75 \times 0.20 = 0.750\text{ t/h}$
  • Total Brix in clarified juice: $18.656 + 0.750 = 19.406\text{ t/h}$

Clarified juice purity:

$$\text{Purity} = \frac{18.656\text{ t/h}}{19.406\text{ t/h}} \times 100\% \approx 96.14\%$$

Clarification elevates the juice purity from $83.3\%$ to $96.1\%$.

Intermediate Example 3.2: Quadruple-Effect Evaporator Steam Economy & Water Evaporation

A sugar factory quadruple-effect evaporator train concentrates $120.0\text{ metric tons/h}$ of clarified juice from $15.0^\circ\text{Bx}$ to thick syrup at $65.0^\circ\text{Bx}$. Live exhaust steam at $2.20\text{ bar}$ ($123.3^\circ\text{C}$, latent heat $\lambda_s = 2,193\text{ kJ/kg}$) is supplied to the 1st effect calandria at a rate of $27.0\text{ metric tons/h}$.

  1. Calculate the total mass of water evaporated across the quadruple-effect train in metric tons per hour.
  2. Determine the production rate of $65.0^\circ\text{Bx}$ thick syrup in metric tons per hour.
  3. Calculate the global Steam Economy of the evaporator train ($\text{kg of water evaporated per kg of live steam}$).

Step 1: Total Water Evaporation Rate

Let feed juice rate be $F = 120.0\text{ t/h}$ at $x_F = 15.0^\circ\text{Bx} = 0.150$. Let syrup product rate be $S$ at $x_S = 65.0^\circ\text{Bx} = 0.650$. By conservation of dissolved solids:

$$F \cdot x_F = S \cdot x_S$$
$$120.0\text{ t/h} \times 0.150 = S \times 0.650$$
$$18.00\text{ t/h (solids)} = S \times 0.650 \implies S = \frac{18.00}{0.650} \approx 27.692\text{ metric tons/h}$$

Total water evaporated:

$$W_{\text{evap}} = F - S = 120.0 - 27.692 = 92.308\text{ metric tons/h}$$

The evaporator train evaporates $92.31\text{ metric tons/h}$ of water.

Step 2: Thick Syrup Production Rate

The station yields $27.69\text{ metric tons/h}$ of $65.0^\circ\text{Bx}$ syrup for pan boiling.

Step 3: Steam Economy

Given live exhaust steam consumption $\dot{m}_{\text{steam}} = 27.0\text{ t/h}$:

$$\text{Steam Economy} = \frac{W_{\text{evap}}}{\dot{m}_{\text{steam}}} = \frac{92.308\text{ t/h}}{27.0\text{ t/h}} \approx 3.419\text{ kg water / kg steam}$$

The quadruple-effect station achieves a high industrial steam economy of $3.42$.

Intermediate Example 3.3: A-Massecuite Vacuum Pan Crystallization Yield

A vacuum pan boils a batch of A-Massecuite from $40.0\text{ metric tons}$ of syrup and A-seed magma. The dropped massecuite weighs $40.0\text{ metric tons}$ and has a dry substance content of $DS = 90.0\text{ wt}\%$ ($90.0^\circ\text{Bx}$) and a sucrose purity of $P_M = 85.0\%$. Centrifugal purging separates this massecuite into raw sugar crystals (purity $P_S = 98.5\%$, moisture $0.50\%$, so $DS_S = 99.5\%$) and mother-liquor A-Molasses (purity $P_L = 70.0\%$, $DS_L = 75.0\%$). Using the S-J-M (Cobenze) recovery formula:

$$\text{Sugar Yield Fraction } R = \frac{S(M - L)}{M(S - L)}$$

where $M, S, L$ are the decimal purities of Massecuite, Sugar, and Molasses:

  1. Calculate the theoretical percentage recovery of sucrose into raw sugar crystals.
  2. Determine the mass of commercial raw sugar crystals recovered from the $40.0\text{ ton}$ strike in metric tons.
  3. Determine the mass of A-Molasses discharged.

Step 1: S-J-M Sucrose Recovery Percentage

Given:

  • Massecuite purity: $M = 0.850$
  • Sugar purity: $S = 0.985$
  • Molasses purity: $L = 0.700$
$$R = \frac{S(M - L)}{M(S - L)} = \frac{0.985 \times (0.850 - 0.700)}{0.850 \times (0.985 - 0.700)}$$
$$R = \frac{0.985 \times 0.150}{0.850 \times 0.285} = \frac{0.14775}{0.24225} \approx 0.609907$$

The recovery of sucrose into crystal raw sugar is $61.0\%$ of total sucrose.

Step 2: Mass of Raw Sugar Crystals Produced

Total dry substance in $40.0\text{ tons}$ of massecuite:

$$m_{\text{DS, massecuite}} = 40.0\text{ t} \times 0.900 = 36.00\text{ metric tons}$$

Total sucrose in massecuite:

$$m_{\text{sucrose, total}} = 36.00\text{ t} \times 0.850 = 30.60\text{ metric tons}$$

Sucrose recovered in raw sugar:

$$m_{\text{sucrose, sugar}} = 30.60\text{ t} \times 0.609907 \approx 18.663\text{ metric tons}$$

Mass of raw sugar crystals ($98.5\%\text{ sucrose on dry basis}$, $99.5\%\text{ DS}$):

$$m_{\text{raw sugar}} = \frac{18.663\text{ t}}{0.985 \times 0.995} = \frac{18.663}{0.980075} \approx 19.043\text{ metric tons}$$

The batch produces $19.04\text{ metric tons}$ of commercial raw sugar.

Step 3: Mass of A-Molasses

By overall mass balance:

$$m_{\text{molasses}} = 40.00 - 19.043 = 20.957\text{ metric tons}$$

The centrifugals discharge $20.96\text{ metric tons}$ of heavy A-molasses for the B-strike.

Easy Example 3.4: Molasses Ethanol Fermentation Theoretical Yield

A distillery valorizes blackstrap molasses by continuous anaerobic yeast fermentation (Saccharomyces cerevisiae). The plant feeds $50.0\text{ metric tons/day}$ of molasses containing $48.0\text{ wt}\%$ total fermentable sugars (calculated as invert sugar equivalent, $M = 180.16\text{ g/mol}$). Under Gay-Lussac stoichiometry:

$$\text{C}_6\text{H}_{12}\text{O}_6 \to 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$$
  1. Calculate the maximum theoretical mass of anhydrous ethanol ($\text{C}_2\text{H}_5\text{OH}$, $M = 46.07\text{ g/mol}$) obtainable per day in metric tons.
  2. If real fermentation reaches Pasteur efficiency of $90.5\%$ (due to biomass growth and by-product glycerol/succinate formation), and distillation recovery is $98.0\%$, calculate the daily production of anhydrous ethanol in Liters (density $\rho_{\text{EtOH}} = 0.789\text{ kg/L}$).

Step 1: Theoretical Ethanol Production

Mass of fermentable sugars fed per day:

$$m_{\text{sugar}} = 50.0\text{ t/day} \times 0.480 = 24.00\text{ metric tons/day}$$

Moles of hexose sugar:

$$n_{\text{sugar}} = \frac{24.00 \times 10^6\text{ g}}{180.16\text{ g/mol}} \approx 133,215\text{ mol/day}$$

Theoretical moles of ethanol formed ($2:1$ stoichiometry):

$$n_{\text{EtOH}} = 2 \times 133,215 = 266,430\text{ mol/day}$$

Theoretical mass of ethanol:

$$m_{\text{EtOH, theo}} = 266,430\text{ mol} \times 46.07\text{ g/mol} \approx 12,274,430\text{ g/day} \approx 12.274\text{ metric tons/day}$$

(Note: $0.5114\text{ kg ethanol per kg hexose}$).

Step 2: Actual Ethanol Output

Accounting for $90.5\%$ fermentation yield and $98.0\%$ distillation recovery:

$$m_{\text{EtOH, actual}} = 12.274\text{ t/day} \times 0.905 \times 0.980 \approx 10.886\text{ metric tons/day}$$

Volume of ethanol:

$$V_{\text{EtOH}} = \frac{10,886\text{ kg/day}}{0.789\text{ kg/L}} \approx 13,797\text{ Liters/day}$$

The distillery produces $13,800\text{ Liters/day}$ of fuel-grade anhydrous bioethanol.

Intermediate Example 3.5: Bagasse Higher Heating Value (HHV) & Boiler Energy Balance

A sugar factory boiler burns bagasse directly from the final milling tandem. The bagasse analysis is:

  • Moisture ($W$): $50.0\text{ wt}\%$
  • Sucrose/Brix ($S$): $2.0\text{ wt}\%$
  • Fiber ($F$): $46.0\text{ wt}\%$
  • Ash ($A$): $2.0\text{ wt}\%$

The Lower Heating Value (LHV) of moist bagasse in $\text{kJ/kg}$ is estimated by the Hugot formula:

$$\text{LHV} = 18,260 - 20,940 \, W - 18,260 \, S - 31.4 \, A$$

(where $W, S, A$ are decimal mass fractions).

  1. Calculate the LHV of this bagasse in $\text{kJ/kg}$.
  2. If the factory generates $60.0\text{ metric tons/h}$ of bagasse, calculate the total thermal heat input rate in Megawatts ($\text{MW}_{\text{th}}$).
  3. If the boiler thermal efficiency is $68.0\%$ producing steam at $45.0\text{ bar}$ and $440^\circ\text{C}$ ($\Delta h = 2,850\text{ kJ/kg steam}$ from feedwater at $105^\circ\text{C}$), calculate the steam generation rate in metric tons per hour.

Step 1: Bagasse LHV Calculation

Given:

  • $W = 0.500$
  • $S = 0.020$
  • $A = 0.020$
$$\text{LHV} = 18,260 - (20,940 \times 0.500) - (18,260 \times 0.020) - (31.4 \times 0.020)$$
$$\text{LHV} = 18,260 - 10,470 - 365.2 - 0.63 = 7,424.17\text{ kJ/kg}$$

The lower heating value is $7,424\text{ kJ/kg}$ ($7.424\text{ MJ/kg}$).

Step 2: Thermal Energy Input Rate

Bagasse feed rate:

$$\dot{m}_{\text{bagasse}} = 60.0\text{ t/h} = \frac{60,000\text{ kg}}{3600\text{ s}} \approx 16.667\text{ kg/s}$$

Thermal input power:

$$\dot{Q}_{\text{in}} = (16.667\text{ kg/s}) \times (7,424.17\text{ kJ/kg}) = 123,736\text{ kW} \approx 123.74\text{ MW}_{\text{th}}$$

The bagasse fuel generates $123.7\text{ MW}_{\text{th}}$ of raw combustion power.

Step 3: Steam Generation Rate

Useful thermal energy transferred to steam ($68.0\%$ efficiency):

$$\dot{Q}_{\text{steam}} = 123,736\text{ kW} \times 0.680 \approx 84,140.5\text{ kW} = 84,140.5\text{ kJ/s}$$

Steam production rate:

$$\dot{m}_{\text{steam}} = \frac{84,140.5\text{ kJ/s}}{2,850\text{ kJ/kg}} \approx 29.523\text{ kg/s}$$

In metric tons per hour:

$$\dot{m}_{\text{steam}} = 29.523\text{ kg/s} \times 3.6 = 106.28\text{ metric tons/h}$$

The boilers generate $106.3\text{ metric tons/h}$ of high-pressure steam, powering turbo-generators and providing exhaust steam for the entire factory.

Intermediate Example 3.6: Corn Wet Milling Starch-Gluten Centrifugal Separation

A corn wet milling plant feeds $40.0\text{ metric tons/h}$ of a starch-gluten slurry to a high-speed disk-stack nozzle centrifuge. The feed contains $30.0\text{ wt}\%$ dry solids, of which $88.0\text{ wt}\%$ is pure starch and $12.0\text{ wt}\%$ is corn gluten protein. Centrifugal separation yields:

  • An underflow starch stream containing $40.0\text{ wt}\%$ dry solids, with a starch purity of $99.2\%$ (on dry basis).
  • An overflow gluten stream containing $8.0\text{ wt}\%$ dry solids.

Assuming $98.5\%$ of the initial starch is recovered into the underflow product:

  1. Calculate the mass rate of dry starch recovered in the underflow in metric tons per hour.
  2. Determine the total mass flow rate of the underflow slurry stream in metric tons per hour.
  3. Calculate the protein content ($\%$) of the dry solids in the overflow gluten stream.

Step 1: Dry Starch in Underflow

Total solids fed per hour:

$$\dot{m}_{\text{solids, in}} = 40.0\text{ t/h} \times 0.300 = 12.00\text{ metric tons/h}$$

Composition of feed solids:

  • Starch: $12.00\text{ t/h} \times 0.880 = 10.56\text{ metric tons/h}$
  • Gluten: $12.00\text{ t/h} \times 0.120 = 1.44\text{ metric tons/h}$

Starch recovered in underflow ($98.5\%$ recovery):

$$\dot{m}_{\text{starch, underflow}} = 10.56\text{ t/h} \times 0.985 = 10.4016\text{ metric tons/h}$$

The plant recovers $10.40\text{ metric tons/h}$ of dry starch.

Step 2: Underflow Slurry Flow Rate

Total dry solids in underflow (at $99.2\%$ starch purity):

$$\dot{m}_{\text{solids, underflow}} = \frac{10.4016\text{ t/h}}{0.992} \approx 10.4855\text{ metric tons/h}$$

At $40.0\text{ wt}\%$ dry solids concentration:

$$\dot{m}_{\text{underflow slurry}} = \frac{10.4855\text{ t/h}}{0.400} \approx 26.214\text{ metric tons/h}$$

The underflow discharges $26.21\text{ metric tons/h}$ of starch slurry.

Step 3: Protein Content of Overflow Gluten Stream

Gluten remaining in overflow: Total gluten fed ($1.44\text{ t/h}$) minus gluten lost in underflow ($10.4855 - 10.4016 = 0.0839\text{ t/h}$):

$$\dot{m}_{\text{gluten, overflow}} = 1.44 - 0.0839 = 1.3561\text{ metric tons/h}$$

Starch escaping into overflow ($1.5\%$ of $10.56\text{ t/h}$):

$$\dot{m}_{\text{starch, overflow}} = 10.56 \times 0.015 = 0.1584\text{ metric tons/h}$$

Total solids in overflow:

$$\dot{m}_{\text{solids, overflow}} = 1.3561 + 0.1584 = 1.5145\text{ metric tons/h}$$

Protein concentration in dried corn gluten meal:

$$\% \text{ Protein} = \frac{1.3561\text{ t/h}}{1.5145\text{ t/h}} \times 100\% \approx 89.54\%$$

The dried gluten meal contains $89.5\%\text{ protein}$, sold as a high-protein animal feed.

Easy Example 3.7: Dextrose Equivalent (DE) & Starch Liquefaction Kinetics

A starch processing plant produces maltodextrin and glucose syrups. Dextrose Equivalent (DE) is defined as the percentage of reducing sugars calculated as D-glucose on a dry substance basis:

$$\text{DE} = \frac{\text{Mass of reducing sugars as D-glucose}}{\text{Total dry substance mass}} \times 100\%$$

For a linear polymer of degree of polymerization $\overline{DP}_n$:

$$\text{DE} \approx \frac{100}{\overline{DP}_n}$$
  1. Compute the theoretical $\overline{DP}_n$ of a low-DE maltodextrin possessing $\text{DE} = 12.5$.
  2. An enzymatic jet cooker liquefies $5,000\text{ kg}$ of pure starch ($M_0 = 162.14\text{ g/mol}$) with $\alpha$-amylase to $\text{DE} = 16.0$. During hydrolysis, each glucosidic cleavage adds one molecule of water ($18.02\text{ g/mol}$). Calculate the mass of water chemically incorporated into the hydrolysate.
  3. Determine the final dry mass of the resulting maltodextrin.

Step 1: Degree of Polymerization ($\overline{DP}_n$)

Using the relation:

$$\overline{DP}_n = \frac{100}{\text{DE}} = \frac{100}{12.5} = 8.0$$

The maltodextrin has an average chain length of $8.0$ glucose units.

Step 2: Chemically Incorporated Water

Initial moles of anhydroglucose units in $5,000\text{ kg}$ starch:

$$n_{\text{AGU}} = \frac{5,000 \times 10^3\text{ g}}{162.14\text{ g/mol}} \approx 30,837.5\text{ mol}$$

At $\text{DE} = 16.0$, the final average degree of polymerization is:

$$\overline{DP}_n = \frac{100}{16.0} = 6.25$$

Number of polymer chains after liquefaction:

$$N_{\text{chains}} = \frac{n_{\text{AGU}}}{\overline{DP}_n} = \frac{30,837.5}{6.25} \approx 4,934\text{ mol}$$

Assuming the initial starch chains were very long ($\overline{DP} > 1,000$), each created chain represents the cleavage of one glucosidic bond and the consumption of one molecule of water:

$$n_{\text{water consumed}} \approx N_{\text{chains}} = 4,934\text{ mol}$$

Mass of water chemically incorporated:

$$m_{\text{water}} = 4,934\text{ mol} \times 18.015\text{ g/mol} \approx 88,886\text{ g} \approx 88.89\text{ kg}$$

The reaction consumes $88.89\text{ kg}$ of water.

Step 3: Final Maltodextrin Dry Mass

$$m_{\text{final}} = m_{\text{starch}} + m_{\text{water}} = 5,000.0 + 88.89 = 5,088.89\text{ kg}$$

The process yields $5,088.9\text{ kg}$ of dry maltodextrin solids.

Hard Example 3.8: Simulated Moving Bed (SMB) Chromatographic HFCS-55 Formulation

A corn sweetener refinery operates a continuous Simulated Moving Bed (SMB) chromatographic separation unit to manufacture beverage-grade High-Fructose Corn Syrup (HFCS-55, $55.0\text{ wt}\%\text{ fructose}$, $41.0\text{ wt}\%\text{ glucose}$, $4.0\text{ wt}\%\text{ oligosaccharides}$ on dry solids).

  • The feed entering the SMB unit is enzymatically isomerized syrup (HFCS-42) containing $42.0\text{ wt}\%\text{ fructose}$ and $58.0\text{ wt}\%\text{ non-fructose sugars}$ at a flow rate of $\dot{m}_{\text{feed}} = 20.0\text{ metric tons/h}$ (dry solids).
  • The SMB unit separates the feed into:
  • Extract stream: Enriched fructose containing $90.0\text{ wt}\%\text{ fructose}$ and $10.0\text{ wt}\%\text{ glucose}$.
  • Raffinate stream: Enriched glucose containing $10.0\text{ wt}\%\text{ fructose}$ and $90.0\text{ wt}\%\text{ glucose}$ (recycled back to the isomerization reactors).
  • A portion of the $90.0\text{ wt}\%$ fructose extract is blended with bypass $42.0\text{ wt}\%$ syrup to formulate the commercial HFCS-55 product.
  1. Apply component mass balances across the SMB unit to calculate the dry solids mass flow rates of the Extract and Raffinate streams in metric tons per hour.
  2. Determine the fructose recovery efficiency in the Extract stream.
  3. Calculate the blending ratio (mass of $90\%$ Extract to mass of bypass $42\%$ syrup) required to produce the $55.0\text{ wt}\%\text{ fructose}$ product.

Step 1: SMB Material Balance

Let $E$ be the Extract flow rate and $R$ be the Raffinate flow rate (dry tons/h). Total mass balance:

$$E + R = 20.0\text{ metric tons/h} \implies R = 20.0 - E$$

Fructose mass balance:

$$0.90 \cdot E + 0.10 \cdot R = 0.42 \times 20.0 = 8.40\text{ metric tons/h}$$

Substitute $R = 20.0 - E$:

$$0.90 \cdot E + 0.10(20.0 - E) = 8.40$$
$$0.90 \cdot E + 2.00 - 0.10 \cdot E = 8.40$$
$$0.80 \cdot E = 8.40 - 2.00 = 6.40$$
$$E = \frac{6.40}{0.80} = 8.00\text{ metric tons/h}$$
$$R = 20.00 - 8.00 = 12.00\text{ metric tons/h}$$

The SMB produces $8.00\text{ metric tons/h}$ Extract and $12.00\text{ metric tons/h}$ Raffinate.

Step 2: Fructose Recovery Efficiency

Fructose in Extract:

$$m_{\text{fructose, extract}} = 0.90 \times 8.00 = 7.20\text{ metric tons/h}$$

Total fructose entering in feed:

$$m_{\text{fructose, feed}} = 8.40\text{ metric tons/h}$$
$$\% \text{ Recovery} = \frac{7.20\text{ t}}{8.40\text{ t}} \times 100\% = 85.71\%$$

Step 3: Blending for HFCS-55

Let $m_E$ be mass of $90\%$ extract blended with $m_B$ of $42\%$ bypass syrup to yield $55\%$ product:

$$0.90 \cdot m_E + 0.42 \cdot m_B = 0.55 \cdot (m_E + m_B)$$
$$0.90 \cdot m_E - 0.55 \cdot m_E = 0.55 \cdot m_B - 0.42 \cdot m_B$$
$$0.35 \cdot m_E = 0.13 \cdot m_B$$
$$\frac{m_E}{m_B} = \frac{0.13}{0.35} \approx 0.3714$$

Or in percentage:

$$\% \text{ Extract in blend} = \frac{0.13}{0.13 + 0.35} \times 100\% = \frac{0.13}{0.48} \times 100\% = 27.08\%$$

Blending requires $0.371\text{ parts}$ of $90\%$ Extract per $1.0\text{ part}$ of $42\%$ syrup ($27.1\text{ wt}\%$ extract).

Easy Example 3.9: Polylactic Acid (PLA) Ring-Opening Polymerization Molar Mass

A polymer reactor synthesizes bio-based polylactic acid (PLA) via ring-opening polymerization of L,L-lactide ($M_{\text{monomer}} = 144.13\text{ g/mol}$) initiated by 1-dodecanol ($\text{C}_{12}\text{H}_{25}\text{OH}$, $M_{\text{init}} = 186.34\text{ g/mol}$) with tin(II) 2-ethylhexanoate catalyst.

  • Monomer feed: $m_{\text{lactide}} = 500.0\text{ kg}$ ($3,469.1\text{ mol}$).
  • Initiator added: $m_{\text{init}} = 1.295\text{ kg}$ ($6.950\text{ mol}$).
  • The reaction runs at $180^\circ\text{C}$ to a monomer conversion of $p = 96.0\%$.

Each initiator molecule grows exactly one linear polymer chain.

  1. Calculate the initial monomer-to-initiator molar ratio ($[M]_0 / [I]_0$).
  2. Determine the theoretical number-average degree of polymerization ($\overline{DP}_n$) of the resulting PLA polymer.
  3. Calculate the number-average molecular weight ($\bar{M}_n$) of the finished PLA resin in $\text{g/mol}$.

Step 1: Monomer-to-Initiator Ratio

$$[M]_0 / [I]_0 = \frac{3,469.1\text{ mol}}{6.950\text{ mol}} \approx 499.15 \approx 500$$

Step 2: Number-Average Degree of Polymerization ($\overline{DP}_n$)

Each lactide dimer molecule contains two lactic acid repeat units. In ring-opening polymerization:

$$\overline{DP}_n = \left(\frac{[M]_0}{[I]_0}\right) \times p = 499.15 \times 0.960 = 479.18\text{ lactide units} \ (958.4\text{ lactic acid units})$$

Step 3: Number-Average Molecular Weight ($\bar{M}_n$)

$$\bar{M}_n = M_{\text{init}} + (\overline{DP}_n \times M_{\text{monomer}})$$
$$\bar{M}_n = 186.34 + (479.18 \times 144.13) = 186.34 + 69,064.2 = 69,250.5\text{ g/mol}$$

The resulting PLA bioplastic has a number-average molar mass of $\bar{M}_n = 69,250\text{ g/mol}$, suitable for extrusion of compostable packaging films.

Solved Honors Problems & Derivations

Step-by-step rigorous solutions with full physical, thermodynamic, and process engineering validation.