Polymer Blends and Composites: Miscibility, Morphology, Alloys, Fibre and Particulate Reinforcement
2. Thermodynamics of mixing: Flory–Huggins
Mixing is spontaneous only if ΔG_mix = ΔH_mix − TΔS_mix < 0. In the Flory–Huggins lattice model, each lattice site has a reference volume V_r (about that of a repeat unit), a chain of component i occupies N_i sites, and φ_i is the volume fraction. Per mole of lattice sites:
Flory–Huggins free energy: ΔG_mix / RT = (φ₁/N₁) ln φ₁ + (φ₂/N₂) ln φ₂ + χ φ₁φ₂
The two logarithmic terms are the combinatorial entropy of mixing (always favourable); χφ₁φ₂ is the enthalpic term, with χ the interaction parameter (χ > 0 unfavourable). For dispersion-force-only pairs, χ = V_r(δ₁ − δ₂)²/RT, with δ the Hildebrand solubility parameter, so χ cannot be negative. Why ΔG_mix is small for polymers: the entropy terms carry 1/N. For two polymers of N = 1000 the entropy gain per site is about a thousand times smaller than for small molecules, so even a tiny positive χ defeats it. Two mixing conditions follow. The critical interaction parameter is χ_c = ½(1/√N₁ + 1/√N₂)², at the composition φ₁c = √N₂/(√N₁ + √N₂); for N₁ = N₂ = N, χ_c = 2/N and φ₁c = ½. Miscibility across all compositions needs χ < χ_c, which for long chains means χ nearly zero or negative.
Hence specific interactions (hydrogen bonding, dipole–dipole, ion–dipole, charge transfer) matter: they make mixing exothermic and χ negative. Miscible pairs rely on them, for example PVC with polycaprolactone or with high-acrylonitrile nitrile rubber, PMMA with PVDF, polyphenylene oxide with polystyrene. Pairs with only dispersion forces (polyethylene with polystyrene, polystyrene with PMMA at high molar mass) are immiscible.
3. Phase behaviour and how miscibility is recognised
χ depends on temperature, so a blend has a phase diagram. Where χ falls as T rises (enthalpy dominated, as for non-polar pairs and polymer solutions in poor solvents), there is an upper critical solution temperature (UCST): the blend is miscible above it. Blends held together by specific interactions, which weaken on heating, and by free-volume (equation-of-state) differences, typically show a lower critical solution temperature (LCST): miscible when cool, separating on heating, which is the usual trap in processing. The boundary is the binodal; inside it lies the spinodal, where ∂²ΔG/∂φ² < 0 (for the model: 1/(N₁φ₁) + 1/(N₂φ₂) − 2χ < 0). Between binodal and spinodal the blend is metastable and demixes by nucleation and growth (droplets); inside the spinodal it demixes spontaneously by spinodal decomposition (interconnected, co-continuous domains).
- Glass transition test. A miscible blend shows one Tg between those of the components, given for compatible pairs by the Fox equation 1/Tg = w₁/Tg₁ + w₂/Tg₂ (w mass fractions, T in kelvin). An immiscible blend shows two Tg, at the component values. Two Tg that have moved towards each other mean partial miscibility: two phases, each rich in one component. One Tg, measured by DSC or DMA, implies mixing on the scale of a few tens of nanometres (roughly 10–30 nm).
- Clarity (a single-phase blend is clear if the refractive indices match), cloud-point measurement on heating (LCST), and microscopy or scattering for domain size.
Polymer eutectics. When two crystallisable polymers are miscible in the melt but form separate crystals, the blend behaves like a simple eutectic system: each melting point is depressed by the other component (for a miscible blend, the Nishi–Wang analysis links the depression to a negative χ), the two liquidus lines meet at a eutectic composition with the lowest melting temperature, and that composition crystallises as a fine mixture of both. Melting-point depression of a crystalline component in a miscible blend is therefore another test of miscibility.
4. Phase morphology, compatibilisers and polymer alloys
Most commercial blends are immiscible, so their properties are set by morphology: droplets in a matrix, co-continuous or lamellar. The minor phase is dispersed when its volume fraction is small; near the middle of the composition range both phases are continuous. The lower-viscosity phase tends to be the matrix or to encapsulate the other. Droplet size comes from a balance of viscous stress, which stretches and breaks drops, against interfacial tension Γ, which holds them round; breakup occurs when the capillary number η·γ̇·R/Γ exceeds a critical value of order one, so smaller drops need either higher shear or lower Γ.
Because immiscible pairs have high Γ and weak adhesion across the interface, they give coarse, unstable morphology and brittle failure. A compatibiliser fixes this. It is usually a block or graft copolymer whose segments are miscible with, or identical to, each phase, so it sits at the interface. It lowers Γ (finer drops), stabilises them against coalescence and entangles with both phases to transmit stress. It can be added (SEBS for polystyrene–polyolefin blends) or formed in situ by reactive compatibilisation, for example a maleic-anhydride-grafted polypropylene whose anhydride reacts with the amine end groups of nylon during melt mixing, making a PP–nylon graft at the interface.
5. Plastic–plastic, rubber–plastic and rubber–rubber blends
| Family | Examples | What the blend achieves |
|---|---|---|
| Plastic–plastic | PPO/PS (miscible); PC/ABS and PC/PBT (alloys); nylon/PP with PP-g-MA; PVC/PCL (miscible); HDPE/LDPE | Heat resistance with processability (PPO/PS); impact plus heat resistance (PC/ABS); lower moisture uptake (nylon/PP); cost |
| Rubber–plastic | HIPS and ABS (rubber grafted into a glassy matrix); PP with EPDM or ethylene–propylene rubber; PP/EPDM thermoplastic vulcanisate; PVC with nitrile rubber; rubber-toughened nylon | Impact strength (rubber particles trigger crazing and shear yielding) or a thermoplastic elastomer; nitrile rubber acts as a permanent, non-migrating plasticiser for PVC |
| Rubber–rubber | NR/BR and SBR/BR (tyre treads and sidewalls); NR/SBR; NBR/EPDM or CR/NBR for oil and ozone balance | Wear, resilience, grip, ageing and cost, with the cure shared by both phases |
6. Composites: FRP, particulate, long and short fibre
In a fibre-reinforced plastic (FRP) the fibre carries the load and the matrix (unsaturated polyester, vinyl ester, epoxy, or thermoplastics such as nylon and PP) holds the fibres in place, transfers load into them by interfacial shear, protects them from damage and environment, and gives shape. In a particulate composite the filler (calcium carbonate, talc, silica, mica, carbon black) raises stiffness and lowers cost and shrinkage; strength rises only for fine particles bonded to the matrix.
Long (continuous) fibres, aligned. Along the fibres both phases strain equally (iso-strain, Voigt): E_c = E_f V_f + E_m V_m, with V the volume fraction. Across the fibres they carry equal stress (iso-stress, Reuss): 1/E_c = V_f/E_f + V_m/E_m, which is far lower and is a lower bound. Convert mass fractions with V_f = (w_f/ρ_f) / (w_f/ρ_f + w_m/ρ_m). The fibre’s share of the longitudinal load is E_f V_f / E_c. Strength follows the same form only when the fibre is the stronger phase and fails first.
Short fibres. Load enters a fibre only through the interface, so a fibre is stressed fully only beyond a minimum length. With fibre strength σ_f, diameter d and interfacial shear strength τ, the critical fibre length is l_c = σ_f d / (2τ). A fibre with l < l_c can never be loaded to fracture; it pulls out, and the composite is little stronger than the filled matrix. For l > l_c the Kelly–Tyson strength is σ_c = σ_f V_f (1 − l_c/2l) + σ_m* (1 − V_f), with σ_m* the matrix stress at the fibre-fracture strain. Aligned short fibres therefore approach continuous-fibre performance only when l ≥ 10 l_c or so. Randomly oriented fibres lose efficiency: the Krenchel factor multiplies the fibre term by 3/8 for random orientation in a plane and 1/5 for random in three dimensions.
7. Reinforcement and reinforcing fibres: natural and synthetic
Reinforcement means a rise in stiffness, strength, tear and abrasion resistance that exceeds simple dilution, and needs a large filler–polymer interface bonded to the matrix. In rubber, fine carbon black (10–100 nm) and precipitated silica reinforce: small, high-surface-area, structured particles immobilise rubber at their surface (bound rubber) and raise modulus by far more than the hydrodynamic prediction E = E₀(1 + 2.5φ), for which the Guth–Gold form E = E₀(1 + 2.5φ + 14.1φ²) adds a particle-contact term. Coarse, weakly bonded fillers such as calcium carbonate or clay act as extenders. A coupling agent, such as a silane on glass or a maleic-anhydride-grafted polyolefin on natural fibre, bonds the filler to the matrix.
| Fibre | Origin and make | Features |
|---|---|---|
| E-glass | Drawn from a borosilicate-type melt, sized with silane | Density about 2.5 g/cm³, E about 72 GPa, cheap, good strength; heavy, moderate stiffness; the workhorse of FRP |
| Carbon | From PAN precursor: stabilised in air at about 200–300 °C, then carbonised above 1000 °C in inert gas (pitch is the other precursor) | Density about 1.8 g/cm³; very high specific stiffness and strength, fatigue resistant, conductive; costly, brittle, galvanic with metals |
| Aramid | Poly(p-phenylene terephthalamide), liquid-crystalline spinning from sulfuric acid | Density about 1.4 g/cm³; high tensile strength and toughness, low compressive strength; ballistic armour, ropes, tyre cord |
| Natural: jute, flax, hemp, sisal, coir | Plant bast or leaf fibres of cellulose (crystalline) bound by hemicellulose and lignin | Low density (about 1.3–1.5 g/cm³), low cost, renewable, low abrasion of tools; hydrophilic, so poor adhesion to hydrophobic matrices, variable properties, processing limit near 200 °C |
8. Worked numericals
Critical interaction parameter. For N₁ = 400 and N₂ = 1600: χ_c = ½(1/20 + 1/40)² = ½(0.075)² = 0.0028. A pair with χ = 0.004 is immiscible over part of the composition range; one with χ = 0.002 is miscible at every composition.
Fox Tg. A 50 : 50 (mass) miscible blend of components with Tg = 373 K and 483 K: 1/Tg = 0.5/373 + 0.5/483 = 0.0023757 K⁻¹, so Tg = 421 K. Plain averaging of temperatures gives 428 K, which overstates the blend’s Tg; the Fox form weights the lower-Tg component more.
Rule of mixtures. An aligned E-glass/epoxy ply with V_f = 0.40, E_f = 72 GPa, E_m = 3.0 GPa: E₁ = 72(0.40) + 3.0(0.60) = 30.6 GPa along the fibres; across the fibres, 1/E₂ = 0.40/72 + 0.60/3.0 = 0.2056, so E₂ = 4.9 GPa, six times lower.
Critical length. σ_f = 2400 MPa, d = 12 µm, τ = 40 MPa: l_c = 2400 × 12 / (2 × 40) = 360 µm = 0.36 mm, so a fibre should be at least several millimetres long to be efficient. Mixing free energy in joules. For the case in section 2 at 300 K: ΔG = 0.000569 × 8.314 × 300 = 1.4 J per mole of lattice sites.
Key takeaways
- Blends and composites give a property balance and cost saving faster than new chemistry; partners need overlapping melt-processing windows, a workable viscosity ratio and a plan for the interface.
- Flory–Huggins: ΔG_mix/RT = (φ₁/N₁)lnφ₁ + (φ₂/N₂)lnφ₂ + χφ₁φ₂. The entropy terms carry 1/N, so long chains gain almost nothing; miscibility needs χ < χ_c = ½(1/√N₁ + 1/√N₂)² (2/N if equal), i.e. specific interactions.
- UCST blends mix on heating; LCST blends (specific interactions, free volume) separate on heating. One Tg (Fox) means miscible, two Tg immiscible, two shifted Tg partial miscibility; crystalline partners show melting-point depression and eutectic behaviour.
- Immiscible blends are controlled by morphology (droplet, co-continuous; low-viscosity phase tends to be the matrix). A block, graft or reactively formed copolymer compatibiliser lowers interfacial tension, refines drops and stabilises them; a compatibilised blend is an alloy.
- Continuous aligned fibres: E_c = E_fV_f + E_mV_m along the fibres (Voigt) and 1/E_c = V_f/E_f + V_m/E_m across (Reuss), with volume fractions. Short fibres need l > l_c = σ_f d/(2τ), and Kelly–Tyson with Krenchel factors gives strength.
- Glass is cheap and heavy, carbon stiff, strong and costly, aramid tough with poor compression; natural fibres (jute, flax, sisal) are light and renewable but hydrophilic and need alkali treatment or coupling agents; fine carbon black or silica reinforces rubber.
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
What is the main reason a manufacturer blends existing polymers instead of developing a new polymer?
Show answer
Answer: B — It is usually faster and cheaper, and can combine properties such as stiffness with toughness
Blending exploits existing, approved polymers, so a property balance is reached at lower cost and development time. Most blends are not single-phase, a blend has no covalent links between the partners (that is a copolymer), and blends are still melt-processed.Two polymers each have N = 1000 lattice sites per chain. Which statement best explains why mixing them is rarely spontaneous?
Show answer
Answer: D — The entropy gain per site scales as 1/N and is tiny, so even a small positive χ above 2/N makes ΔG_mix positive
The combinatorial entropy of mixing is always favourable but scaled by 1/N, so for long chains it is about a thousand times smaller per site than for small molecules. With χ_c = 2/N = 0.002, a typical positive χ of 0.01 is well above it. The enthalpy term per site is not larger for polymers, and the entropy of mixing is positive, not negative.A miscible polymer–polymer blend held together by hydrogen bonding is clear at 25 °C but turns cloudy when heated above 160 °C. What does this indicate, and what is the cause?
Show answer
Answer: A — A lower critical solution temperature: the specific interactions weaken on heating, so χ rises past χ_c and the blend demixes
Hydrogen bonds weaken as temperature rises, and free-volume differences grow, so χ increases and crosses χ_c: demixing on heating is an LCST, which is common for specifically interacting blends and is why processing temperature must be below the cloud point. UCST behaviour would be clear on heating and cloudy on cooling. Phase separation on heating is thermodynamically possible, and a melting point would not make a molten blend cloudy.A polystyrene–polyethylene blend has coarse polyethylene droplets and breaks in a brittle way. Adding a small amount of a polystyrene–polyethylene diblock copolymer improves it. What is the mechanism?
Show answer
Answer: D — Each block sits in its own phase, so the copolymer lies at the interface, lowers the tension and resists coalescence
A compatibiliser works at the interface: lower interfacial tension gives smaller drops at the same shear, the copolymer layer hinders coalescence, and chain entanglement into both phases transmits stress. It lowers, not raises, the interfacial tension, does not chemically crosslink, and does not act by dissolving in one phase.DSC of a 50 : 50 blend of two amorphous polymers (Tg = 100 °C and 200 °C) shows two glass transitions, at 112 °C and 185 °C. What is the correct interpretation?
Show answer
Answer: C — The blend is partially miscible: two phases, each rich in one component, with each Tg shifted towards the other
Two Tg mean two phases. If the phases were pure the Tg would remain at 100 and 200 °C; their movement towards each other shows that each phase contains some of the other polymer. A single Tg between the two would indicate full miscibility, and amorphous polymers have no crystalline phases.Short glass fibres of length 0.2 l_c (l_c is the critical fibre length) are moulded into a thermoplastic. What is the expected result?
Show answer
Answer: A — The fibres cannot be loaded to their fracture stress, pull out, and the composite is barely stronger than the matrix
With l < l_c the shear transferred through the interface cannot raise the fibre stress to σ_f, so failure is by pull-out and the fibres act almost as particulate filler. Fibres reach full strength only when l is well above l_c, and length, not just V_f, matters.Which fibre is made from a PAN precursor by stabilisation in air at about 200–300 °C followed by carbonisation above 1000 °C in an inert atmosphere?
Show answer
Answer: C — Carbon fibre, valued for very high specific stiffness
Polyacrylonitrile is oxidatively stabilised and then carbonised to give carbon fibre of high specific stiffness. Glass is drawn from a molten glass melt, aramid is spun from sulfuric acid solution, and jute is a plant fibre that is not heat-treated in this way.Untreated jute fibre bonds poorly to a polypropylene matrix. Which statement gives the cause and a suitable remedy?
Show answer
Answer: C — The fibre is hydrophilic and polar while PP is non-polar, so alkali treatment and a maleic-anhydride-grafted PP are used
Cellulose fibre is polar and absorbs water, while PP is non-polar, so the interface is weak. Alkali treatment cleans and roughens the surface, and a maleic-anhydride-grafted polyolefin reacts with the fibre –OH groups while entangling with the matrix. Jute is hydrophilic, not hydrophobic; it does not melt (it chars above about 200 °C); and it is lighter than glass, not too dense.Which of the following is a rubber–plastic blend in which the rubber is a dispersed phase?
Show answer
Answer: A — Polypropylene toughened with dispersed ethylene–propylene rubber
Impact polypropylene carries rubber particles in a plastic matrix, which makes it a rubber–plastic blend. Natural rubber with polybutadiene is rubber–rubber, and polyphenylene oxide with polystyrene and HDPE with LDPE are plastic–plastic.Which of the following statements about polymer blend miscibility are correct?
Show answer
Answer: A — Polyphenylene oxide is miscible with polystyrene in all proportions, and the blend shows a single Tg; B — Two polymers of N = 1000 each with χ = 0.01 fall above the critical value for full miscibility, which is 0.002
PPO/PS is the standard miscible pair, and χ_c = 2/N = 0.002 for N = 1000, so χ = 0.01 gives an immiscible region. The entropy term is small for long chains, so it does not guarantee miscibility, and many specifically interacting blends show an LCST rather than a UCST.Which of the following statements about the morphology and compatibilisation of immiscible blends are correct?
Show answer
Answer: A — Maleic-anhydride-grafted polypropylene can compatibilise PP with nylon by reacting with the nylon amine end groups to form a graft copolymer in situ; B — A compatibiliser lowers the interfacial tension, so the dispersed droplets become smaller and resist coalescence
Reactive compatibilisation and interfacial-tension reduction are the two standard mechanisms. A compatibiliser lowers, not raises, the tension, and the continuous phase depends on viscosity ratio as well as composition: the lower-viscosity phase tends to encapsulate the other, and co-continuity is common near the middle of the range.Which of the following statements about fibre-reinforced composites are correct?
Show answer
Answer: A — The matrix transfers load into the fibres by interfacial shear and protects them from damage and the environment; C — A fibre with length below l_c = σ_f d/(2τ) can never be stressed to its fracture strength by interfacial shear; D — A silane sizing on glass fibre improves adhesion between the fibre and the resin
The matrix roles, the critical-length condition and the use of silane coupling agents are all standard. Along the fibres the strain is equal, so the parallel (Voigt) model E = E_fV_f + E_mV_m applies; the series model describes the transverse direction.A blend of two polymers with N₁ = N₂ = 100 lattice sites per chain has χ = 0.03. Use ΔG_mix/RT = (φ₁/N₁) ln φ₁ + (φ₂/N₂) ln φ₂ + χ φ₁φ₂ per mole of lattice sites, with R = 8.314 J mol⁻¹ K⁻¹. At T = 300 K and φ₁ = φ₂ = 0.5, what is ΔG_mix in J per mole of lattice sites? Give the answer to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 1.4
Entropy term = 2 × (0.5/100) × ln 0.5 = 0.01 × (−0.693147) = −0.006931. Enthalpy term = 0.03 × 0.5 × 0.5 = 0.0075. ΔG/RT = 0.000569. RT = 8.314 × 300 = 2494.2 J/mol, so ΔG = 0.000569 × 2494.2 = 1.42, i.e. 1.4 J per mole of sites.A miscible 50 : 50 (by mass) blend consists of polystyrene (Tg = 373 K) and polyphenylene oxide (Tg = 483 K). Using the Fox equation 1/Tg = w₁/Tg₁ + w₂/Tg₂ with temperatures in kelvin, what is the blend Tg in kelvin? Give the answer to the nearest whole number.
Numerical answer — type the value.
Show answer
Answer: 421
1/Tg = 0.5/373 + 0.5/483 = 0.0013405 + 0.0010352 = 0.0023757 K⁻¹, so Tg = 1/0.0023757 = 420.9 K, i.e. 421 K. The arithmetic mean, 428 K, is not the Fox result.A unidirectional glass/epoxy composite contains 60 mass % E-glass fibre (density 2.5 g/cm³, modulus 72 GPa) in an epoxy matrix (density 1.2 g/cm³, modulus 3.0 GPa). Take the volume fraction from the mass fraction and use the rule of mixtures E = E_fV_f + E_mV_m along the fibres. What is the longitudinal modulus in GPa? Give the answer to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 31.9
Per 100 g: glass volume = 60/2.5 = 24.0 cm³, matrix volume = 40/1.2 = 33.33 cm³, so V_f = 24.0/57.33 = 0.4186 and V_m = 0.5814. E = 72 × 0.4186 + 3.0 × 0.5814 = 30.14 + 1.74 = 31.9 GPa. Using 0.6 as V_f would give 44.4 GPa.The critical fibre length is l_c = σ_f d / (2τ). For fibres of tensile strength σ_f = 2400 MPa and diameter d = 12 µm in a matrix with interfacial shear strength τ = 40 MPa, what is l_c in micrometres? Give the answer as a whole number.
Numerical answer — type the value.
Show answer
Answer: 360
l_c = 2400 × 12 / (2 × 40) = 28800/80 = 360 µm (MPa cancels). This is 0.36 mm, so practical short fibres of a few millimetres exceed l_c by roughly an order of magnitude.