Polymer Flow, Modulus–Temperature Behaviour and Rheometry

This chapter answers the first half of Section XE5.6, Polymer Rheology, of XE5 Polymer Science and Engineering, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam. The syllabus reads: “Flow of Newtonian and non-Newtonian fluids, different flow equations, dependence of shear modulus on temperature, molecular/segmental deformations at different zones and transitions; Measurements of rheological parameters by capillary, parallel plate, cone-plate rheometer.” The chapter teaches Newtonian and non-Newtonian flow and the power-law, Bingham, Cross and Carreau equations, the shear thinning of melts, the temperature dependence of modulus and the molecular motions in each zone, the WLF and Arrhenius shifts, and the three rheometers with their working formulae. The numericals are power-law flow rate, capillary wall shear stress and rate with the Bagley and Rabinowitsch corrections, cone-plate viscosity and the WLF shift factor.

1. Newtonian and non-Newtonian flow

Rheology is the study of deformation and flow. In simple shear a layer of fluid moves past another, the shear stress is τ = F/A (Pa) and the shear rate is γ̇ = dv/dy (s⁻¹), the velocity gradient. A Newtonian fluid obeys τ = μ γ̇ with a viscosity μ that does not change with γ̇: water, oils, dilute solutions. A polymer melt is non-Newtonian: its apparent viscosity η = τ/γ̇ depends on the shear rate (and on temperature, pressure and time), so the flow curve of τ against γ̇ is not a straight line through the origin. The ratio is still called a viscosity, but it is a point on a curve, not a constant.

Types of time-independent flow
BehaviourViscosity trendExample
NewtonianConstantWater, mineral oil
Shear-thinning (pseudoplastic)Falls with γ̇Polymer melts and solutions, most rubbers
Shear-thickening (dilatant)Rises with γ̇Concentrated particle suspensions
Bingham plasticNo flow below a yield stress, then linearPastes, highly filled compounds

Why melts thin. At low rates the entangled coils are disturbed little and the viscosity sits on a zero-shear plateau η₀. At higher rates the chains disentangle faster than they re-entangle and orient along the flow, so the viscosity falls; at very high rates a second, lower plateau η∞ is approached. Because η₀ grows steeply with molar mass (in proportion to M below the critical entanglement mass and to about M^3.4 above it), a broad distribution starts to thin at a lower rate. Processing happens in the thinning region, 10 to 10⁴ s⁻¹ for extrusion and injection moulding.

2. The flow equations

Power law (Ostwald–de Waele): τ = K γ̇ⁿ, so the apparent viscosity is η = K γ̇^(n−1). K is the consistency index (Pa·sⁿ) and n the flow index. n = 1 is Newtonian (K = μ); n < 1 is shear thinning (polymer melts typically 0.2 to 0.6); n > 1 is shear thickening. On a log–log plot η against γ̇ is a straight line of slope n − 1. Worked example: K = 1.0 × 10⁴ Pa·sⁿ and n = 0.4. At γ̇ = 100 s⁻¹, η = 10⁴ × 100^(−0.6) = 631 Pa·s; at 10 s⁻¹, η = 2512 Pa·s. A tenfold rise in rate lowers η by 10^0.6 = 3.98. The law fits only the middle of the curve: it predicts infinite η as γ̇ → 0 and zero η as γ̇ → ∞.

Bingham: τ = τ_y + μ_p γ̇ for τ > τ_y, and no flow for τ ≤ τ_y; τ_y is the yield stress and μ_p the plastic viscosity. The Herschel–Bulkley form τ = τ_y + K γ̇ⁿ combines both. For a full melt curve, the Cross model η = η∞ + (η₀ − η∞)/[1 + (λ γ̇)^m] and the Carreau model, which has a similar shape with a different functional form, both give a zero-shear plateau η₀, a power-law decline and, with η∞, a high-shear plateau; λ is a characteristic time, and 1/λ marks where thinning begins. Qualitatively, a larger λ means thinning starts at a lower rate, which is what a higher molar mass or a broader distribution does.

Flow in a tube. For steady laminar flow in a tube of radius R and length L, a force balance gives the wall shear stress τ_w = R ΔP/(2L) for any fluid. For a Newtonian fluid, Q = π R⁴ ΔP/(8 μ L) (Hagen–Poiseuille) and the wall shear rate is 4Q/(π R³). For a power-law fluid, Q = [π n/(3n + 1)] R³ (τ_w/K)^(1/n) and the wall shear rate is [(3n + 1)/(4n)] × 4Q/(π R³), larger than the Newtonian value 4Q/(π R³) because the plug-like profile has a steeper wall gradient. Worked example: K = 1.0 × 10⁴ Pa·sⁿ, n = 0.4, R = 2 mm, L = 0.1 m and ΔP = 1.5 MPa. τ_w = 0.002 × 1.5 × 10⁶/(2 × 0.1) = 15 000 Pa; τ_w/K = 1.5, and 1.5^(1/0.4) = 1.5^2.5 = 2.756; π n/(3n + 1) = π × 0.4/2.2 = 0.5712; Q = 0.5712 × (2 × 10⁻³)³ × 2.756 = 1.259 × 10⁻⁸ m³/s = 12.6 mm³/s.

⚠️ The wall stress formula holds for every fluid; the rate formula does not
τ_w = R ΔP/(2L) comes from a force balance and needs no constitutive law. The rate 4Q/(π R³) is exact only for a Newtonian fluid, so for a melt it is called the apparent shear rate and must be corrected.

3. Modulus versus temperature: the five zones and their molecular motions

Measure the shear (relaxation) modulus of an amorphous, uncrosslinked polymer at a fixed time of about 10 s while raising the temperature. It falls by many decades, giving five zones:
(1) glassy, G near 10⁹ Pa, the chains frozen and only bond bending, stretching and local side-group motion possible, so the response is hard and elastic;
(2) glass transition (leathery), where G falls by about three decades within about 20 K as cooperative segmental motion of some 10 to 50 backbone atoms switches on, which defines Tg;
(3) rubbery plateau, G near 10⁵ to 10⁶ Pa, where entanglements act as temporary crosslinks and only segments between entanglements move;
(4) rubbery flow, where the chains begin to slip through entanglements and the modulus falls;
(5) liquid flow, where whole chains reptate and disentangle, so the polymer flows with a viscosity that falls steadily with temperature.

Zones, motions and what governs the length of each
ZoneMolecular motionEffect of the structure
GlassyLocal vibration, side-group and bond motionModulus almost the same for all polymers
TransitionCooperative segmental motion startsTg rises with stiffness, falls with plasticiser
Rubbery plateauSegments move, entanglements holdPlateau lengthens with molar mass; crosslinking extends it to decomposition
FlowReptation and disentanglement of whole chainsAbsent for a crosslinked network, which stays rubbery

Two corollaries are asked often. A crosslinked rubber has no flow zone: its plateau modulus G ≈ ν R T (ν is the number of moles of network chains per unit volume) stays flat until decomposition and rises with crosslink density and with temperature. A semicrystalline polymer holds a higher modulus above Tg, because crystallites act as crosslinks, until melting at Tm.

4. Time–temperature superposition: WLF and Arrhenius shifts

Warming a polymer has the same effect on its response as waiting longer or measuring more slowly: every relaxation process speeds up with temperature by the same shift factor a_T = τ(T)/τ(T_ref) (a_T < 1 above T_ref). Curves measured at several temperatures can be slid along the log-time or log-frequency axis onto one master curve at T_ref that covers many decades. This holds for thermorheologically simple materials only, and not across Tg of a blend or across a melting point.

WLF equation (Williams–Landel–Ferry), valid from Tg to about Tg + 100 K: log₁₀ a_T = − C₁ (T − T_ref)/(C₂ + T − T_ref). With T_ref = Tg the approximately universal constants are C₁ = 17.44 and C₂ = 51.6 K. Worked example: at T = Tg + 20 K, log₁₀ a_T = −17.44 × 20/(51.6 + 20) = −348.8/71.6 = −4.87, so a_T = 1.3 × 10⁻⁵: relaxation is about 75 000 times faster than at Tg. Far above Tg, beyond about Tg + 100 K, free-volume effects fade and the Arrhenius form is used, ln a_T = (Ea/R)(1/T − 1/T_ref) with T in kelvin. Example: Ea = 50 kJ/mol, T_ref = 200 °C and T = 230 °C: ln a_T = (50 000/8.314)(1/503.15 − 1/473.15) = 6014 × (−1.2601 × 10⁻⁴) = −0.758, so a_T = 0.47.

⚠️ Which shift to use, and the sign
Use WLF between Tg and Tg + 100 K, and Arrhenius well above that. For T > T_ref both give a_T < 1: the process is faster, so the master curve is shifted to shorter times. The WLF constants 17.44 and 51.6 K apply with T_ref = Tg; with another T_ref they change.

5. The capillary rheometer, Bagley and Rabinowitsch corrections

A capillary rheometer pushes melt from a barrel through a die of radius R and length L with a piston at a set speed (so Q is known) and records the pressure ΔP. It covers shear rates of about 10 to 10⁵ s⁻¹, closer to processing than any rotational instrument. The working formulae are τ_w = ΔP R/(2L) (wall shear stress) and γ̇_a = 4Q/(π R³) (apparent wall shear rate), from which the apparent viscosity η_a = τ_w/γ̇_a follows. Two corrections make these true values.

  • Bagley correction (entrance pressure). The measured ΔP also contains the pressure drop ΔP_ent that drives the converging flow into the die. At a constant apparent rate, ΔP is plotted against L/R for dies of several lengths: ΔP = ΔP_ent + 2 τ_w (L/R). The slope is 2τ_w and the intercept at L/R = 0 is ΔP_ent. A long die (L/R above about 40) or a correction from a zero-length die can be used.
  • Rabinowitsch correction (non-Newtonian profile). The true wall shear rate is γ̇_w = [(3n′ + 1)/(4n′)] γ̇_a, where n′ = d ln τ_w/d ln γ̇_a is the local slope of the log–log flow curve. For a Newtonian fluid n′ = 1 and the factor is 1; for a melt with n′ = 0.5 it is 1.25.
  • Other errors. Viscous heating raises the melt temperature at high rates; pressure raises viscosity; wall slip and melt fracture (a distorted extrudate above a critical τ_w) limit the usable range.

Worked example (Bagley). At one apparent rate, two dies give ΔP = 3.0 MPa at L/R = 20 and ΔP = 4.6 MPa at L/R = 40. The slope is (4.6 − 3.0)/(40 − 20) = 0.08 MPa per unit L/R, which equals 2τ_w, so τ_w = 0.04 MPa = 40 kPa, and the intercept is 3.0 − 20 × 0.08 = 1.4 MPa. (Rabinowitsch) For R = 0.5 mm and Q = 1.0 × 10⁻⁸ m³/s, γ̇_a = 4Q/(π R³) = 4 × 10⁻⁸/(π × 1.25 × 10⁻¹⁰) = 101.9 s⁻¹; with n′ = 0.5, γ̇_w = 1.25 × 101.9 = 127.3 s⁻¹.

6. Parallel-plate and cone-plate rheometers

In rotational rheometers a sample is sheared between a rotating and a fixed surface, and the torque T at angular speed Ω is measured. They reach low rates (about 10⁻³ to 10² s⁻¹), allow oscillatory tests, and measure normal forces, but they are limited at high rates by edge fracture, sample ejection and inertia.

Working formulae (Newtonian, small gap or cone angle)
GeometryShear rateShear stressRemarks
Parallel plates, radius R, gap hAt the rim γ̇_R = Ω R/h; varies linearly with radiusτ_R = 2T/(π R³) (Newtonian)Gap adjustable: good for stiff melts and filled compounds; non-uniform rate
Cone and plate, radius R, cone angle θ₀ (small, in radians)γ̇ = Ω/θ₀, uniform over the sampleτ = 3T/(2π R³)Viscosity η = 3 T θ₀/(2π R³ Ω); N₁ = 2F/(π R²) from the axial force F

Worked example. A cone-plate measurement with R = 12.5 mm, θ₀ = 0.04 rad, Ω = 0.5 rad/s and torque T = 0.020 N·m gives γ̇ = 0.5/0.04 = 12.5 s⁻¹ and τ = 3 × 0.020/(2π × (0.0125)³) = 0.06/(1.227 × 10⁻² × 10⁻³) ≈ 4889 Pa, so η = 4889/12.5 = 391 Pa·s. For parallel plates at Ω = 5 rad/s, R = 12.5 mm and h = 1 mm, the rim rate is 5 × 12.5/1 = 62.5 s⁻¹. For a shear-thinning melt the parallel-plate stress needs a correction, τ_R = [T/(2π R³)] (3 + d ln T/d ln γ̇_R), which reduces to 2T/(π R³) when d ln T/d ln γ̇_R = 1.

🎯 Why cone-plate is preferred for a flow curve
The gap grows in proportion to the radius, so the local rate Ωr/(r θ₀) = Ω/θ₀ is the same everywhere: one stress and one rate describe the whole sample, with no correction for the shape of the curve. The price is a small gap at the cone tip, so it is unsuitable for filled compounds with large particles, and the sample volume must be exact.

Key takeaways

  • A Newtonian fluid has τ = μ γ̇ with constant μ; polymer melts shear-thin because chains disentangle and orient. Power law τ = K γ̇ⁿ gives η = K γ̇^(n−1) (n < 1 thinning); Bingham has a yield stress τ_y; Cross and Carreau give η₀ and η∞ plateaus.
  • For tube flow τ_w = R ΔP/(2L) always; for a power-law fluid Q = [π n/(3n + 1)] R³ (τ_w/K)^(1/n), and the wall rate is [(3n + 1)/(4n)] times 4Q/(π R³).
  • The modulus of an uncrosslinked amorphous polymer falls through glassy, transition, rubbery plateau, rubbery flow and liquid flow zones; segmental motion starts at Tg, entanglements hold the plateau, reptation gives flow, and crosslinking or crystallinity removes the flow zone.
  • Time–temperature superposition shifts curves by a_T: WLF from Tg to Tg + 100 K, log a_T = −17.44 (T − Tg)/(51.6 + T − Tg), and Arrhenius ln a_T = (Ea/R)(1/T − 1/T_ref) well above; a_T < 1 above the reference.
  • Capillary rheometer: τ_w = ΔP R/(2L), γ̇_a = 4Q/(π R³); the Bagley plot (ΔP against L/R) removes the entrance loss, and the Rabinowitsch factor (3n′ + 1)/(4n′) gives the true wall rate and is 1 for a Newtonian fluid.
  • Parallel plates: γ̇_R = Ω R/h at the rim (non-uniform), τ_R = 2T/(π R³) for Newtonian; cone-plate: γ̇ = Ω/θ₀ (uniform), τ = 3T/(2π R³), η = 3Tθ₀/(2π R³ Ω).

Practice questions (17)

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.

  1. Which statement defines a Newtonian fluid?

    1. Stress is proportional to shear rate, with viscosity independent of rate at a fixed temperature
    2. Its viscosity falls steadily as the shear rate is raised, because the chains disentangle and orient
    3. It does not flow until the applied stress exceeds a yield stress, above which stress rises linearly
    4. Its viscosity is the same at every temperature and pressure, and at every shear rate and time
    Show answer

    Answer: A — Stress is proportional to shear rate, with viscosity independent of rate at a fixed temperature

    A Newtonian fluid has τ = μ γ̇ with μ independent of γ̇ at a given temperature; μ itself still depends on temperature and pressure, so the last claim is wrong. A falling viscosity is shear thinning, and a yield stress marks a Bingham fluid.

  2. Why does the apparent viscosity of a polymer melt fall at high shear rates?

    1. Frictional heating always melts the crystalline regions of the melt completely
    2. The molar mass of the chains is halved by the shear in every flow situation
    3. The chains disentangle faster than they re-entangle and orient with the flow
    4. The melt becomes a glass because the shear raises the Tg
    Show answer

    Answer: C — The chains disentangle faster than they re-entangle and orient with the flow

    Disentanglement and chain orientation lower the resistance to flow; this is the molecular origin of shear thinning, seen even in an amorphous melt with no crystals. Chain scission and viscous heating can occur, but they are side effects, not the cause, and the shear does not turn the melt into a glass.

  3. A power-law melt has a flow index n = 0.5. If the shear rate is doubled, by what factor does the apparent viscosity change?

    1. It is multiplied by 2, as if the fluid were shear thickening
    2. It is multiplied by 1/√2, about 0.71
    3. It is multiplied by 0.5, as for a fluid with n = 0
    4. It is unchanged, as for a Newtonian fluid
    Show answer

    Answer: B — It is multiplied by 1/√2, about 0.71

    η = K γ̇^(n−1), so η₂/η₁ = 2^(n−1) = 2^(−0.5) = 0.71. A factor of 0.5 would need n = 0, and no change would need n = 1, a Newtonian fluid.

  4. A highly filled paste does not flow under its own weight, and above a threshold stress it flows with a stress that rises linearly with the shear rate. Which model describes it best?

    1. The Newtonian model, τ = μ γ̇, with a constant μ
    2. The power-law model, τ = K γ̇ⁿ with n < 1
    3. The Cross model with a zero-shear plateau
    4. The Bingham model, τ = τ_y + μ_p γ̇
    Show answer

    Answer: D — The Bingham model, τ = τ_y + μ_p γ̇

    A yield stress followed by linear flow is the definition of a Bingham plastic. A power-law or Cross fluid flows at any stress, however small (the Cross model even has a finite zero-shear viscosity), so neither can describe a material with a threshold.

  5. Two uncrosslinked, amorphous polystyrene samples have the same chemistry, one of molar mass 100 000 g/mol and one of 400 000 g/mol. Compared at the same temperature range, which is expected for the modulus–temperature curves?

    1. The glassy modulus and the plateau modulus are both about four times higher for the larger sample, in proportion to molar mass
    2. The smaller sample has the longer plateau because shorter chains entangle more
    3. The curves coincide to the end of the plateau, and the larger sample has a much longer plateau
    4. Neither shows a plateau, since only crosslinked polymers have one
    Show answer

    Answer: C — The curves coincide to the end of the plateau, and the larger sample has a much longer plateau

    The glassy modulus and the entanglement plateau modulus depend on chemistry, not on chain length, so the curves coincide through the glass transition and the start of the plateau. Longer chains stay entangled for longer, so the plateau extends to a higher temperature before flow. Uncrosslinked, high-molar-mass polymers do show a plateau, held by entanglements.

  6. Which molecular process switches on at the glass transition of an amorphous polymer?

    1. Breaking of the covalent bonds of the backbone
    2. Melting of crystallites into a disordered liquid
    3. Reptation of entire chains out of their entanglements
    4. Cooperative motion of several backbone atoms
    Show answer

    Answer: D — Cooperative motion of several backbone atoms

    At Tg the cooperative micro-Brownian motion of chain segments (of the order of tens of atoms) becomes possible within the experimental time. Whole-chain reptation occurs only much higher, in the flow zone, melting is a first-order transition of crystals, and covalent bond breaking is degradation.

  7. A relaxation time follows an Arrhenius shift, ln a_T = (Ea/R)(1/T − 1/T_ref), with Ea = 50 kJ/mol, R = 8.314 J mol⁻¹ K⁻¹, T in kelvin and T_ref = 200 °C. What is a_T at 230 °C (take 0 °C = 273.15 K)?

    1. 0.47
    2. 2.13
    3. 0.22
    4. 0.69
    Show answer

    Answer: A — 0.47

    T_ref = 473.15 K and T = 503.15 K. 1/T − 1/T_ref = −1.2601 × 10⁻⁴ K⁻¹ and Ea/R = 6014 K, so ln a_T = −0.758 and a_T = 0.47. The value 2.13 is its reciprocal, which comes from reversing the sign: above T_ref the relaxation is faster, so a_T < 1.

  8. In capillary rheometry, a Bagley plot of ΔP against L/R at a constant apparent shear rate is made to correct for which effect?

    1. The non-parabolic velocity profile of a shear-thinning melt in the die
    2. The extra pressure drop as melt converges into the die entrance
    3. The rise of melt temperature through viscous heating
    4. Slip of the melt along the wall of the barrel and die
    Show answer

    Answer: B — The extra pressure drop as melt converges into the die entrance

    The measured pressure includes the entrance loss, and the intercept at L/R = 0 gives it. The velocity-profile effect is corrected by the Rabinowitsch factor, which is the tempting confusion, not by the Bagley plot.

  9. A parallel-plate rheometer of radius 12.5 mm and gap 1.0 mm is rotated at 5.0 rad/s. What is the shear rate at the rim, and how does it vary over the plate?

    1. 12.5 s⁻¹ at the rim, since this equals R/h, and the same at every radius
    2. 62.5 s⁻¹ at the rim, and the same value at every radius of the plate
    3. 62.5 s⁻¹ at the rim, falling linearly to zero at the axis
    4. 5.0 s⁻¹ everywhere, since the rate equals the angular speed
    Show answer

    Answer: C — 62.5 s⁻¹ at the rim, falling linearly to zero at the axis

    γ̇(r) = Ω r/h, so at r = R, γ̇ = 5.0 × 12.5/1.0 = 62.5 s⁻¹, and it falls linearly to zero at the centre. A uniform rate is the property of the cone-plate geometry, not of parallel plates.

  10. Which statements about flow equations are correct?

    1. A power-law fluid with n < 1 is shear thinning, and with n = 1 it is Newtonian
    2. A shear-thickening fluid corresponds to a power-law index n < 1
    3. The Cross model gives a zero-shear viscosity plateau and a high-shear plateau
    4. A Bingham fluid flows at any stress, however small
    Show answer

    Answer: A — A power-law fluid with n < 1 is shear thinning, and with n = 1 it is Newtonian; C — The Cross model gives a zero-shear viscosity plateau and a high-shear plateau

    n < 1 means η falls with γ̇ (thinning) and n = 1 means constant η, so the first claim holds; the Cross model includes both η₀ and η∞. Thickening needs n > 1, and a Bingham fluid needs τ above τ_y before it flows.

  11. Which statements about the modulus–temperature behaviour of polymers are correct?

    1. The glassy modulus of an organic polymer is of the order of 10⁹ Pa
    2. In the rubbery plateau the entanglements act as temporary crosslinks
    3. A crosslinked rubber has no flow zone and its plateau persists up to decomposition
    4. In the glassy state the backbone segments move freely over long distances
    Show answer

    Answer: A — The glassy modulus of an organic polymer is of the order of 10⁹ Pa; B — In the rubbery plateau the entanglements act as temporary crosslinks; C — A crosslinked rubber has no flow zone and its plateau persists up to decomposition

    These three claims are the standard picture: glassy modulus near 1 GPa, entanglements as physical crosslinks, and permanent crosslinks removing the flow zone. In the glass the long-range segmental motion is frozen, so the last claim is the opposite of the truth.

  12. Which statements about the capillary rheometer are correct?

    1. The apparent wall shear rate is 4Q/(π R³)
    2. In the Bagley plot the intercept at L/R = 0 is the entrance pressure drop
    3. The Rabinowitsch factor (3n′ + 1)/(4n′) differs from 1 even for a Newtonian fluid
    4. The wall shear stress is ΔP R/(2L), provided the pressure drop is that of fully developed flow
    Show answer

    Answer: A — The apparent wall shear rate is 4Q/(π R³); B — In the Bagley plot the intercept at L/R = 0 is the entrance pressure drop; D — The wall shear stress is ΔP R/(2L), provided the pressure drop is that of fully developed flow

    The statements on the apparent rate, the Bagley intercept and the wall stress are the working formulae and the Bagley logic. For a Newtonian fluid n′ = 1, so the Rabinowitsch factor is exactly (3 + 1)/4 = 1; it matters only for non-Newtonian melts.

  13. A polymer melt follows the power law τ = K γ̇ⁿ with K = 1.0 × 10⁴ Pa·sⁿ and n = 0.4. It flows steadily through a circular tube of radius R = 2.0 mm and length L = 0.10 m under a pressure drop ΔP = 1.5 MPa. Use τ_w = R ΔP/(2L) and Q = [π n/(3n + 1)] R³ (τ_w/K)^(1/n) and find the volumetric flow rate in mm³/s, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 12.6

    τ_w = 0.002 × 1.5 × 10⁶/(2 × 0.10) = 15 000 Pa, so τ_w/K = 1.5 and 1.5^2.5 = 2.756. π × 0.4/2.2 = 0.5712 and R³ = 8 × 10⁻⁹ m³, so Q = 0.5712 × 8 × 10⁻⁹ × 2.756 = 1.259 × 10⁻⁸ m³/s = 12.6 mm³/s. Using n = 0.4 as the exponent instead of 1/n = 2.5 is the common slip.

  14. For a melt flowing through capillary dies at one fixed apparent shear rate, the pressure drop is 3.0 MPa for a die with L/R = 20 and 4.6 MPa for a die with L/R = 40. Use the Bagley relation ΔP = ΔP_ent + 2 τ_w (L/R) and find the wall shear stress τ_w in kPa.

    Numerical answer — type the value.

    Show answer

    Answer: 40

    The two points give a slope of (4.6 − 3.0)/(40 − 20) = 0.08 MPa per unit L/R, and this slope equals 2τ_w. So τ_w = 0.04 MPa = 40 kPa, and the entrance drop is 3.0 − 20 × 0.08 = 1.4 MPa. Using ΔP R/(2L) directly with the raw 3.0 MPa would give 75 kPa, which includes the entrance loss.

  15. A capillary of radius 0.50 mm carries a melt at Q = 1.0 × 10⁻⁸ m³/s. The apparent wall shear rate is γ̇_a = 4Q/(π R³). The local slope of the log–log flow curve is n′ = 0.5, and the Rabinowitsch correction is γ̇_w = [(3n′ + 1)/(4n′)] γ̇_a. Find the true wall shear rate in s⁻¹, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 127.3

    R³ = (0.5 × 10⁻³)³ = 1.25 × 10⁻¹⁰ m³, so γ̇_a = 4 × 10⁻⁸/(π × 1.25 × 10⁻¹⁰) = 101.86 s⁻¹. The factor is (1.5 + 1)/(2) = 1.25, so γ̇_w = 127.3 s⁻¹.

  16. A cone-plate rheometer has plate radius R = 12.5 mm and cone angle θ₀ = 0.040 rad. At Ω = 0.50 rad/s the torque is T = 0.020 N·m. For a Newtonian-type reading use τ = 3T/(2π R³) and γ̇ = Ω/θ₀, and find the viscosity η = τ/γ̇ in Pa·s, to the nearest whole number.

    Numerical answer — type the value.

    Show answer

    Answer: 391

    R³ = (0.0125)³ = 1.953 × 10⁻⁶ m³, so τ = 3 × 0.020/(2π × 1.953 × 10⁻⁶) = 4889 Pa. γ̇ = 0.50/0.040 = 12.5 s⁻¹, so η = 4889/12.5 = 391 Pa·s. Using the plate-plate stress 2T/(π R³) would give a wrong value.

  17. Use the WLF equation log₁₀ a_T = − C₁ (T − T_ref)/(C₂ + T − T_ref) with T_ref = Tg, C₁ = 17.44 and C₂ = 51.6 K. Find log₁₀ a_T at a temperature 20 K above Tg, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: -4.87

    T − T_ref = 20 K, so log₁₀ a_T = −17.44 × 20/(51.6 + 20) = −348.8/71.6 = −4.87. The negative sign is correct above T_ref: the response is faster.