7.1 Fluid Properties, Reynolds Number, and Flow Regimes

7.1 Fluid Properties, Reynolds Number, and Flow Regimes

Quick Answer: Density ρ and viscosity μ (with ν = μ/ρ) control inertia-to-viscous balance via Re = ρvD/μ. Pipe flow is laminar at low Re (≲ 2100), turbulent at high Re (≳ 4000). For incompressible steady flow, A₁v₁ = A₂v₂ keeps volumetric flow continuous.

Domain C of the UPDA/MMUP Chemical exam (transport phenomena — fluid, heat, mass — about 20% of this guide's planning allocation) opens with fluid mechanics. Heat and mass transfer later reuse the same dimensionless thinking (Re appears inside Nusselt and Sherwood correlations). Master density, viscosity, Reynolds number, and continuity here before Bernoulli, friction, and pumps in the next sections.

Why Fluid Properties Matter on UPDA Chemical

Qatar process plants move hydrocarbons, cooling water, seawater, amines, glycols, and utilities through long pipe networks. Exam stems rarely ask for research-level rheology; they ask whether flow is laminar or turbulent, how velocity changes with diameter, and which property drives pressure drop. Wrong units on μ or ρ are the fastest way to misclassify Re by orders of magnitude.

Density ρ

Density is mass per unit volume:

ρ = m / V

Liquids are nearly incompressible for most plant hydraulic estimates: ρ changes little with pressure and modestly with temperature. Water near ambient is often taken as ρ ≈ 1000 kg/m³ (more precisely ~997 kg/m³ at 25 °C). Light hydrocarbons may be 600–800 kg/m³; dense brines exceed 1000 kg/m³.

Gases have density that depends strongly on T and P (ideal gas: ρ = PM/(RT), or PM/(ZRT) when compressibility matters). Volumetric flow of gas at actual conditions is not the same as standard-condition flow—always ask “actual m³/h or standard m³/h?”

Hydrostatic pressure from a liquid column: ΔP = ρ g h. At ρ = 1000 kg/m³, g ≈ 9.81 m/s², a 10 m water column gives ΔP ≈ 98.1 kPa ≈ 0.97 bar — the classic “10 m of water ≈ 1 bar” rule of thumb.

Viscosity μ and Kinematic Viscosity ν

Dynamic (absolute) viscosity μ measures resistance to shear. In SI, μ is in Pa·s (equivalent to N·s/m² or kg/(m·s)). Older literature uses poise or centipoise (cP):

1 cP = 0.001 Pa·s = 10⁻³ kg/(m·s)

Water at ~20 °C has μ ≈ 1 cP = 0.001 Pa·s. Many organic liquids are 0.3–3 cP at ambient; heavy oils and syrups can be tens to thousands of cP.

Kinematic viscosity is:

ν = μ / ρ

Units: m²/s (or cSt: 1 cSt = 10⁻⁶ m²/s). Kinematic viscosity appears naturally when Re is written as vD/ν.

Temperature effects (exam intuition):

• Liquid viscosity falls sharply as T rises (hot oil pumps easier than cold oil).
• Gas viscosity rises slowly with T.
• Density of liquids falls modestly with T; gas density falls as 1/T at fixed P.

Newtonian Fluids (Exam Basics)

A Newtonian fluid has shear stress τ proportional to shear rate γ̇:

τ = μ γ̇

with μ independent of shear rate (though μ may still depend on T and composition). Water, light oils, most gases, and many clear process liquids are treated as Newtonian on licensing exams.

Non-Newtonian fluids (polymer solutions, slurries, some emulsions) have μ that depends on shear rate—power-law, Bingham plastic, etc. UPDA Chemical items that do not name a rheology model almost always imply constant μ Newtonian behavior.

Reynolds Number

The Reynolds number is the ratio of inertial forces to viscous forces:

Re = ρ v D / μ = v D / ν

Mean velocity in a full circular pipe:

v = Q / A , A = π D² / 4

so v = 4Q / (π D²). Mass flow ṁ = ρ Q = ρ A v.

Laminar, Transitional, and Turbulent Pipe Flow

Common circular pipe engineering criteria (smooth-entry, fully developed context):

Exact transition depends on entrance conditions, roughness, and disturbances—exam MCQs use the standard thresholds above, not research-edge nuance.

Why Re matters:

1. Friction factor correlations differ (Hagen–Poiseuille / f = 64/Re laminar Darcy form vs Moody chart turbulent).
2. Heat/mass transfer correlations change with regime.
3. Mixing and residence-time assumptions (plug flow vs laminar dispersion).

Worked Example: Reynolds Number for Cooling Water

Problem. Cooling water flows in a full circular pipe: D = 50 mm = 0.050 m, volumetric flow Q = 0.010 m³/s, ρ = 997 kg/m³, μ = 8.9×10⁻⁴ Pa·s (≈ 25 °C).

Step 1 — Area and velocity.

A = π (0.050)² / 4 = 1.963×10⁻³ m²
v = Q/A = 0.010 / 1.963×10⁻³ ≈ 5.09 m/s

Step 2 — Reynolds number.

Re = ρ v D / μ = (997)(5.09)(0.050) / (8.9×10⁻⁴)
≈ 253.7 / 8.9×10⁻⁴ ≈ 2.85×10⁵

Step 3 — Regime. Re ≫ 4000 → turbulent. Expect Moody-chart friction factors, not laminar f = 64/Re alone.

Worked Example: Same Pipe, Viscous Oil (Laminar)

Keep D = 0.050 m and v = 1.0 m/s, but μ = 0.50 Pa·s (heavy oil), ρ = 900 kg/m³.

Re = (900)(1.0)(0.050) / 0.50 = 45 / 0.50 = 90 → strongly laminar.

Exam lesson: High viscosity or low velocity/diameter drives laminar flow even in “large” pipes. Always compute Re; do not assume plant liquids are turbulent.

Continuity for Incompressible Flow

Mass continuity for steady flow in a single stream without accumulation:

ṁ = ρ₁ A₁ v₁ = ρ₂ A₂ v₂ = constant

For incompressible fluid (ρ₁ = ρ₂ = ρ), this collapses to volume continuity:

A₁ v₁ = A₂ v₂ = Q

Reducer example. Liquid at 2.0 m/s in a 100 mm ID line enters a 50 mm ID section. A ∝ D², so A₂/A₁ = (0.5)² = 0.25, thus v₂ = v₁ / 0.25 = 8.0 m/s. Kinetic energy and erosion risk rise; Bernoulli will convert some pressure into velocity head (Section 7.2).

Compressible gases: use ṁ = ρ A v with local ρ, or mass-flux form. Do not force A₁v₁ = A₂v₂ if density changes significantly between stations.

Consistent Units Checklist

Convert mm → m, cP → Pa·s, and L/min → m³/s before computing Re. A factor-of-1000 error on μ (forgetting cP vs Pa·s) moves Re by 1000× and flips the regime answer.

UPDA Chemical Exam Workflow

1. Identify fluid (liquid vs gas) and whether ρ is constant.
2. Get D (internal diameter), Q or ṁ, and properties ρ, μ at process T.
3. Compute v = Q/A (or v = ṁ/(ρ A)).
4. Compute Re = ρvD/μ and classify regime.
5. Only then choose laminar or turbulent friction / correlation paths (Section 7.3).
6. For geometry changes, apply A₁v₁ = A₂v₂ (incompressible) before energy balances.

Common Traps

• Using outer diameter instead of inner diameter for Re and velocity.
• Mixing mass flow and volumetric flow without density.
• Applying A₁v₁ = A₂v₂ to a gas that expands through a large pressure drop.
• Memorizing “water is always turbulent” without calculating Re for small tubes or low flow.
• Confusing μ with ν in the Re formula (missing or double-counting ρ).
• Forgetting that transition band exists—options may say “cannot be sure” near Re = 3000.

Section 7.2 places these velocities into the Bernoulli equation and mechanical energy balance; Section 7.3 turns regime knowledge into friction losses and pump power.