Transport Phenomena in Polymers: Problem Set

Problem 1: The Power of Molecular Weight (Momentum Transfer)

Problem Statement

Two grades of polystyrene (PS) are used in injection molding:

For entangled polymers, viscosity follows: \(\eta \propto M^{3.4}\)

a) Calculate the viscosity of PS-300
b) Which grade requires more injection pressure? By what factor?
c) Why might you still choose the high-MW grade despite processing difficulty?

Solution

Part (a): Viscosity of PS-300

Using the scaling law:

\[\frac{\eta_2}{\eta_1} = \left(\frac{M_2}{M_1}\right)^{3.4}\]

Substituting values:

\[\frac{\eta_2}{1000} = \left(\frac{300,000}{100,000}\right)^{3.4} = 3^{3.4} = 46.6\]
\[\boxed{\eta_2 = 46,600 \text{ Pa·s}}\]

Part (b): Injection Pressure Comparison

Injection pressure is proportional to viscosity (for same flow rate):

\[\boxed{\text{PS-300 requires 46.6× more pressure!}}\]

Part (c): Why Choose High-MW Grade?

Higher MW gives better mechanical properties (strength, toughness, impact resistance) — worth the processing challenge for structural parts.

Key Insight: Small changes in molecular structure cause HUGE changes in transport properties

Problem 2: Glass Transition and Three Transport Properties

Problem Statement

Poly(methyl methacrylate) (PMMA, aka Plexiglass) has Tg = 105°C. Below are properties at two temperatures:

Property At 25°C At 150°C Units
Viscosity η 1012 104 Pa·s
Thermal diffusivity α 0.10 0.12 mm²/s
O₂ diffusivity D 10-14 10-9 m²/s

a) Calculate the ratio of properties (high T / low T) for each transport property
b) Which property is most sensitive to temperature crossing Tg?
c) Explain physically why mass transfer shows the largest change

Solution

Part (a): Property Ratios

Part (b): Most Sensitive Property

\[\boxed{\text{Viscosity is most sensitive (10}^8\text{ change), followed by mass diffusivity (10}^5\text{)}}\]

Part (c): Physical Explanation

Key Insight: Tg is a "transport property cliff" — small temperature change, massive property change

Problem 3: Polymer Membrane for Food Packaging (Mass Transfer)

Problem Statement

You're designing a chip bag. Oxygen causes staleness. Two polymer options:

Polymer Thickness (mm) O₂ Permeability P (cm³·mm/m²·day·atm)
LDPE 0.05 8,000
EVOH 0.01 5

Outside air: 21% O₂ (0.21 atm), Inside bag: 0% O₂ initially
Bag area: 0.1 m²

a) Calculate oxygen flux (cm³/day) through each material
b) If 10 cm³ of O₂ makes chips taste stale, how long until staleness for each?
c) Why is EVOH barrier layer so thin compared to LDPE?

Solution

Part (a): Oxygen Flux

Using Fick's Law through membranes:

\[J = P \cdot A \cdot \frac{\Delta p}{L}\]

LDPE:

Area conversion: 0.1 m² = 1000 cm²

\[J_{LDPE} = 8000 \times 0.1 \times \frac{0.21}{0.05} = 800 \times 4.2\] \[\boxed{J_{LDPE} = 3,360 \text{ cm}^3/\text{day}}\]

EVOH:

\[J_{EVOH} = 5 \times 0.1 \times \frac{0.21}{0.01} = 0.5 \times 21\] \[\boxed{J_{EVOH} = 10.5 \text{ cm}^3/\text{day}}\]

Part (b): Time to Staleness

\[t = \frac{\text{Critical O}_2}{\text{Flux}}\]
\[\boxed{t_{LDPE} = \frac{10}{3360} = 0.003 \text{ days} = 4 \text{ minutes (useless!)}}\]
\[\boxed{t_{EVOH} = \frac{10}{10.5} = 0.95 \text{ days} = 23 \text{ hours}}\]

Part (c): Why Thin EVOH Layer?

EVOH has 1600× lower permeability than LDPE! Even at 1/5 the thickness, it's still 320× better barrier. EVOH is expensive, so we use thin layers in multilayer films (LDPE/EVOH/LDPE structure).

Key Insight: Material selection trumps geometry for mass transfer applications

Problem 4: Viscous Heating in Extrusion (Coupled Heat + Momentum)

Problem Statement

Molten polyethylene (PE) is extruded through a die:

a) Calculate the viscous heat generation rate per unit volume
b) If no heat is removed, how fast does temperature rise?
c) Why is this a problem in polymer processing?

Solution

Part (a): Viscous Heat Generation Rate

Viscous dissipation rate:

\[\dot{q}''' = \eta \dot{\gamma}^2\] \[\dot{q}''' = 500 \times (100)^2 = 500 \times 10,000\] \[\boxed{\dot{q}''' = 5,000,000 \text{ W/m}^3 = 5 \text{ MW/m}^3}\]

Part (b): Temperature Rise Rate

Temperature rise rate (adiabatic):

\[\frac{dT}{dt} = \frac{\dot{q}'''}{\rho c_p}\] \[\frac{dT}{dt} = \frac{5,000,000}{800 \times 2000} = \frac{5,000,000}{1,600,000}\] \[\boxed{\frac{dT}{dt} = 3.1 \text{ °C/s} = 186 \text{ °C/min}}\]

Part (c): Processing Problems

Solution: Die cooling channels to remove heat!

Key Insight: Momentum transfer (shear) GENERATES heat → must solve coupled transport equations

Problem 5: Drug Delivery Patch (Time-Dependent Mass Transfer)

Problem Statement

A transdermal nicotine patch releases drug through a polymer membrane:

a) Calculate the diffusion time constant through the membrane
b) What is the steady-state flux?
c) After how long does release rate drop to 1% of initial?

Solution

Part (a): Diffusion Time Constant

\[\tau = \frac{L^2}{D}\] \[\tau = \frac{(10^{-4})^2}{5 \times 10^{-12}} = \frac{10^{-8}}{5 \times 10^{-12}}\] \[\boxed{\tau = 2000 \text{ s} = 33 \text{ minutes}}\]

This is how long to reach steady-state diffusion!

Part (b): Steady-State Flux

Steady-state flux (Fick's Law):

\[J = D \frac{C_0 - 0}{L} = 5 \times 10^{-12} \times \frac{100}{10^{-4}}\] \[J = 5 \times 10^{-12} \times 10^6\] \[\boxed{J = 5 \times 10^{-6} \text{ kg/(m}^2\cdot\text{s)} = 5 \text{ mg/(m}^2\cdot\text{s)}}\]

Part (c): Long-Term Release

For depletion from a finite reservoir (patch runs out), the time depends on patch volume and area. Patches are designed to maintain steady release for hours to days by using thick reservoirs or rate-controlling membranes.

Key Insight: Polymer selection controls drug release kinetics → "programmable" mass transfer

Problem 6: Reptation and Polymer Processing (Advanced)

Problem Statement

For entangled polymer chains, the reptation time (time for chain to escape its "tube") is:

\[\tau_{rep} = \frac{L^4}{D} \propto M^3\]

where L is chain length and D is self-diffusion coefficient.

Polystyrene samples:

a) Calculate reptation time for PS-200
b) In extrusion at 1000 s⁻¹ shear rate, which sample "sees" steady flow?
c) What happens if processing time < reptation time?

Solution

Part (a): Reptation Time for PS-200

Using scaling:

\[\frac{\tau_2}{\tau_1} = \left(\frac{M_2}{M_1}\right)^3\] \[\frac{\tau_2}{0.1} = \left(\frac{200,000}{50,000}\right)^3 = 4^3 = 64\] \[\boxed{\tau_{rep} = 6.4 \text{ s}}\]

Part (b): Steady vs. Elastic Flow

Processing time scale = \(1/\dot{\gamma}\) = 1/1000 = 0.001 s

Deborah Number (compares relaxation to process time):

\[De = \frac{\tau_{rep}}{t_{process}}\]
\[\boxed{\text{Neither sample is in steady viscous flow! Both behave elastically.}}\]

Part (c): Consequences of Fast Processing

When process time < relaxation time:

Key Insight: Molecular relaxation time determines processing behavior → connects molecular dynamics to macro transport

Summary: The Big Picture

Transport Molecular Origin Polymer Example Key Equation
Momentum Chain entanglement Viscosity ∝ M3.4 τ = η·γ̇
Heat Phonons + chain wiggling k changes at Tg q = -k∇T
Mass Free volume hopping Permeability in films J = -D∇C
All three are controlled by the same thing: molecular mobility