Flow regime calculation: A liquid flows at 11,400 L·h^-1 through a 4 cm-ID pipe. The liquid has density 1 g·mL^-1 and dynamic viscosity 0.001 kg·m^-1·s^-1. Identify the flow regime using the Reynolds number.

Introduction:
Determining the flow regime (laminar, transitional, turbulent) in process lines is fundamental for pressure drop, mixing, and mass-transfer correlations. The Reynolds number, Re = ρ * v * D / μ, is the standard criterion for Newtonian liquids in pipes.

Given Data / Assumptions:

• Volumetric flow rate Q = 11,400 L·h^-1 = 11.4 m^3·h^-1.
• Pipe inner diameter D = 0.04 m.
• Density ρ = 1 g·mL^-1 = 1000 kg·m^-3.
• Dynamic viscosity μ = 0.001 kg·m^-1·s^-1.
• Newtonian behavior; smooth pipe; single-phase liquid.

Concept / Approach:
Compute average velocity v = Q / A, where A = π * D^2 / 4. Then calculate Reynolds number Re = ρ * v * D / μ. Use the conventional thresholds: Re < 2100 (laminar), 2100–4000 (transitional), > 4000 (turbulent).

Step-by-Step Solution:
Convert flow rate: Q = 11.4 m^3·h^-1 = 11.4 / 3600 ≈ 0.003167 m^3·s^-1.Compute area: A = π * 0.04^2 / 4 ≈ 0.001257 m^2.Find velocity: v = Q / A ≈ 0.003167 / 0.001257 ≈ 2.52 m·s^-1.Compute Re: Re = 1000 * 2.52 * 0.04 / 0.001 ≈ 1.01 × 10^5.Classify regime: Re ≫ 4000 ⇒ turbulent.

Verification / Alternative check:
Even with modest roughness, a Reynolds number around 10^5 is strongly turbulent. Predictions of friction factor and pressure drop should therefore use turbulent correlations (e.g., Blasius for smooth pipes or Moody chart with appropriate relative roughness).

Why Other Options Are Wrong:
Laminar: requires Re < 2100; not satisfied.

Transitional: 2100–4000; much lower than the computed Re.

Any regime: data are sufficient; the regime is clearly turbulent.

Common Pitfalls:

• Failing to convert units (L·h^-1 to m^3·s^-1) correctly.
• Using kinematic viscosity instead of dynamic viscosity without dividing by density.

Final Answer:
Turbulent regime