Fluid properties – relation between kinematic and dynamic viscosity Is the following statement correct? “Kinematic viscosity equals the product of dynamic viscosity and the fluid density.”

Introduction / Context:Fluid mechanics uses both dynamic viscosity (μ) and kinematic viscosity (ν). Mixing up their relationship leads to wrong Reynolds numbers and pressure-drop calculations.

Given Data / Assumptions:

• Dynamic viscosity μ measures resistance to shear (units Pa·s).
• Kinematic viscosity ν measures momentum diffusivity (units m^2/s).
• Mass density ρ has units kg/m^3.

Concept / Approach:The correct relationship is ν = μ / ρ, not μ * ρ. Dividing by density converts force-based resistance into a diffusivity that governs how momentum spreads through a fluid, analogous to thermal diffusivity in heat transfer.

Step-by-Step Solution:Step 1: Write ν = μ / ρ.Step 2: Check units: (Pa·s) / (kg/m^3) = (N·s/m^2) / (kg/m^3) = (kg·m/s^2 · s / m^2) / (kg/m^3) = m^2/s.Step 3: If you multiplied μ * ρ, the units would be Pa·s·kg/m^3, which is dimensionally incorrect for ν.Step 4: Conclude the statement is false.

Verification / Alternative check:Common conversions: water at 20 °C has μ ≈ 1.0×10^-3 Pa·s and ρ ≈ 998 kg/m^3, giving ν ≈ 1.0×10^-6 m^2/s, consistent with data tables.

Why Other Options Are Wrong:

• Any “true” variant contradicts ν = μ / ρ.
• Temperature clauses do not fix the incorrect form; the functional relation is independent of temperature (though values of μ and ρ change with T).

Common Pitfalls:Confusing ν with dynamic viscosity units (Pa·s) or with the C.G.S. unit “stoke,” where 1 cSt = 10^-6 m^2/s.

Final Answer:No