T4 · Rheology / FlowPolymer RheologyGATE 2027 Syllabus (XE5)Shop-Floor Standard

Ostwald-de Waele Power-Law Shear Thinning Model

Models pseudoplastic shear-thinning behavior of polymer melts where apparent melt viscosity (eta) decreases non-linearly with increasing shear rate (gamma_dot).

Mathematical Canonical Relation

\tau = K \cdot \dot{\gamma}^n \quad \Rightarrow \quad \eta(\dot{\gamma}) = K \cdot \dot{\gamma}^{n-1}

Plain Formula: tau = K * gamma_dot^n => eta = K * gamma_dot^(n-1)

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Consistency Index K (Pa·s^n) (Pa·s^n)

Range: 500 – 40000 Pa·s^n

Flow Index n (n < 1) (dimensionless)

Range: 0.15 – 0.85 dimensionless

Shear Rate (s⁻¹) (s⁻¹)

gamma_dot

Range: 1 – 50000 s⁻¹

Calculated Engineering Result:

52.6 Pa·s (Stress = 131 kPa)

Engineering Verification:Standard molding flow range.

When & Where to Apply

Use when modeling polymer flow through extrusion dies, injection mold gates, or capillary viscometers.

Common Engineering Pitfalls

Confusing consistency index K (Pa·s^n) with dynamic viscosity at low shear
Assuming power law holds at zero shear rate (does not predict zero-shear viscosity plateau eta_0)

Physical Assumptions & Validity Limits

Steady-state shear flow
Isothermal melt conditions
Pseudoplastic flow behavior index n < 1.0

Variable Symbols, SI Units & Dimensional Specifications

Symbol	Variable Meaning	SI Unit	Typical Industrial Range
\tau	Shear stress	Pa (N/m²)	—
\eta	Apparent viscosity	Pa·s	—
K	Consistency index	Pa·s^n	8500
\dot{\gamma}	Shear rate	s⁻¹	2500
n	Flow behavior index (n < 1)	dimensionless	0.35

Worked Industrial Numerical Example: Polypropylene High Shear Gate Viscosity Drop

Problem Statement:

A PP injection molding grade resin has consistency index K = 8,500 Pa·s^n and power law index n = 0.35 at 230°C. Calculate apparent viscosity at gate shear rate 10,000 s⁻¹.

Step-by-Step Numerical Solution:

Apply Power-Law viscosity form: eta = K * (gamma_dot)^(n - 1)
Compute exponent: n - 1 = 0.35 - 1.0 = -0.65
Calculate shear factor: (10,000)^(-0.65) = 0.0025119
Multiply by K: eta = 8500 * 0.0025119 = 21.35 Pa·s

Final Answer: eta = 21.4 Pa·s

💡 Engineering Interpretation: Viscosity drops by 89x under high injection shear rate compared to extrusion shear rates (10 s⁻¹), allowing high-velocity cavity filling.

Authoritative Technical Reference & Source Provenance

Polymer Rheology: Fundamentals and Applications by R. S. Lenk (Elsevier, 1978), 1st Edition, Page 88-95.

Connected Curriculum Lesson

Non-Newtonian Flow Behavior & Power-Law Model

Study Full Lesson