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Oil Dynamic Viscosity Example

Easy DifficultyFE Fluid Mechanics

Calculate dynamic viscosity of oil from shear stress and velocity data.

[Diagram: Flat plate moving over oil layer - to be added]

Concept

Dynamic viscosity (μ) is a fluid property that measures resistance to flow and shear deformation. Newton's law of viscosity states that shear stress is proportional to the velocity gradient: τ = μ \frac{d u}{d y} τ = shear stress (N/m² or Pa) μ = dynamic viscosity (Pa\cdots or N\cdots/m²) d u / d y = velocity gradient (s⁻¹) For a thin layer of fluid between parallel plates with one moving and one stationary, the velocity profile is linear.

A flat plate is pulled at constant velocity V = 0.5 m/s over a layer of oil with thickness t = 2 mm. The force required per unit area (shear stress) is τ = 150 N/m². Find: Dynamic viscosity μ of the oil (Pa\cdots) Assume linear velocity distribution in the oil layer.

V = 0.5 m/s t = 2 mm = 0.002 m τ = 150 N/m^{2}

For parallel flow between plates, the shear stress relates to viscosity and velocity gradient: τ = μ \frac{d u}{d y} With linear velocity distribution, the velocity profile is u (y) = V \frac{y}{t}

For linear velocity distribution from 0 at the bottom to V at the top: \frac{d u}{d y} = \frac{V}{t} \frac{d u}{d y} = \frac{0.5}{0.002} = 250 s^{- 1}

Rearrange Newton's law to solve for μ : μ = \frac{τ}{d u / d y} μ = \frac{150}{250} μ = 0.6 N \cdot s/m^{2} = 0.6 Pa \cdot s

Viscosity is often expressed in centipoise (cP) where 1 Pa\cdots = 1000 cP: μ = 600 cP This value is consistent with medium-weight oils (SAE 40-50 grade oils have viscosities of 400-900 cP at room temperature).

μ = 0.6 Pa \cdot s = 0.6 N \cdot s/m^{2} Alternative: μ = 600 cP

τ = μ \frac{d u}{d y} Notation: τ = shear stress, μ = dynamic viscosity, d u / d y = velocity gradient. For linear profile between plates: d u / d y = V / t .