Interactive Lecture
Fluid Resistance to Deformation and the Rotation of Fluid Elements
by Dr Arnaldo Delli Carri
Two fundamental properties govern the internal mechanics of real fluid flows: viscosity, which describes a fluid's resistance to deformation by shear stress, and vorticity, which measures the local rotation of fluid elements. Together they explain why fluids resist flow, how energy is dissipated, and why rotating structures such as tornadoes, wing-tip vortices, and pipe eddies form and persist.
Viscosity links shear stress to the velocity gradient in a fluid — faster-moving layers drag slower ones. Vorticity is twice the angular velocity of a fluid element about its own axis. A flow with zero vorticity everywhere is called irrotational, and its analysis simplifies enormously; real flows near solid surfaces are always rotational because viscosity creates vorticity at walls.
The dynamic (absolute) viscosity \(\mu\) characterises how strongly adjacent fluid layers resist sliding past each other. For a Newtonian fluid — water, air, most oils — the shear stress \(\tau\) is directly proportional to the velocity gradient perpendicular to the flow direction:
Newton's Law of Viscosity:
\[\tau = \mu \frac{du}{dy}\]
Where:
The kinematic viscosity \(\nu\) normalises dynamic viscosity by density, combining both fluid resistance and inertia into a single parameter that appears naturally in the Reynolds number:
Kinematic Viscosity:
\[\nu = \frac{\mu}{\rho}\]
Where:
Common values at 20 °C: water \(\nu \approx 1.0 \times 10^{-6}\) m²/s; air \(\nu \approx 1.5 \times 10^{-5}\) m²/s.
Newton's law of viscosity applies to fluids where \(\mu\) is constant regardless of shear rate — these are Newtonian fluids. Many engineering and biological fluids deviate from this behaviour:
The vorticity vector \(\boldsymbol{\omega}\) is defined as the curl of the velocity field \(\mathbf{v}\):
Vorticity Vector:
\[\boldsymbol{\omega} = \nabla \times \mathbf{v}\]
In 2D planar flow (\(u, v\) in the \(x\)-\(y\) plane), only the \(z\)-component is non-zero:
\[\omega_z = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\]
Where:
Vorticity equals twice the angular velocity of a fluid element: \(\boldsymbol{\omega} = 2\,\boldsymbol{\Omega}\).
The circulation \(\Gamma\) around a closed contour \(C\) is the line integral of velocity, and by Stokes' theorem equals the flux of vorticity through the enclosed surface:
Circulation:
\[\Gamma = \oint_C \mathbf{v} \cdot d\mathbf{l} = \iint_S \boldsymbol{\omega} \cdot d\mathbf{S}\]
Where \(\Gamma\) = circulation (m²/s).
For a free (irrotational) vortex, \(\omega_z = 0\) everywhere except at the core, yet \(\Gamma \neq 0\) on any contour enclosing the core singularity.
A flow is irrotational if \(\boldsymbol{\omega} = 0\) everywhere. For a free vortex with tangential velocity \(v_\theta = \Gamma / (2\pi r)\), fluid elements translate along circular paths but do not spin about their own axes — the flow is irrotational outside the core. In contrast, a solid-body rotation (forced vortex) with \(v_\theta = \Omega\,r\) has uniform vorticity \(\omega_z = 2\Omega\) everywhere and is fully rotational.
Viscous flow between two infinite parallel plates (Couette flow). The bottom plate is stationary; the top plate moves at velocity \(U\). Adjust the dynamic viscosity and plate velocity to explore the linear velocity profile and the shear stress on the plates. The fluid element shows how layers rotate due to the velocity gradient.
Compare a free vortex (irrotational, \(\omega_z = 0\) outside the core) with a forced vortex / solid-body rotation (\(\omega_z = 2\Omega\) everywhere). The colour map shows vorticity magnitude at each grid point; velocity vectors show the flow direction. Toggle between modes and adjust the vortex strength.
| Concept | Equation | Description |
|---|---|---|
| Newton's Law of Viscosity | \(\tau = \mu \dfrac{du}{dy}\) | Shear stress proportional to velocity gradient (Newtonian fluids) |
| Kinematic Viscosity | \(\nu = \dfrac{\mu}{\rho}\) | Dynamic viscosity normalised by density (m²/s) |
| Couette Flow Profile | \(u(y) = U\dfrac{y}{h}\) | Linear velocity profile between stationary and moving plates |
| Vorticity Vector | \(\boldsymbol{\omega} = \nabla \times \mathbf{v}\) | Curl of velocity field; equals twice the angular velocity of a fluid element |
| 2D Vorticity | \(\omega_z = \dfrac{\partial v}{\partial x} - \dfrac{\partial u}{\partial y}\) | Out-of-plane vorticity component for 2D planar flows |
| Circulation | \(\Gamma = \oint_C \mathbf{v} \cdot d\mathbf{l}\) | Line integral of velocity around a closed contour (m²/s) |
| Free Vortex Velocity | \(v_\theta = \dfrac{\Gamma}{2\pi r}\) | Irrotational vortex; \(\omega_z = 0\) outside the core |
| Forced Vortex Velocity | \(v_\theta = \Omega\,r,\quad \omega_z = 2\Omega\) | Solid-body rotation; uniform vorticity throughout |
A machine component slides over a flat surface separated by a uniform oil film of thickness \(h = 0.4\) mm. The sliding velocity of the component is \(U = 2.5\) m/s and the surface beneath is stationary. The oil has dynamic viscosity \(\mu = 0.12\) Pa·s. Assuming Couette flow (linear velocity profile) across the film:
Show Solution
Solution:
Part (a) — Shear Stress
Given:
For Couette flow the velocity profile is linear: \(u(y) = U\,y/h\), so the velocity gradient is constant:
\[\frac{du}{dy} = \frac{U}{h} = \frac{2.5}{4 \times 10^{-4}} = 6250 \text{ s}^{-1}\]
Applying Newton's law of viscosity:
\[\tau = \mu\frac{du}{dy} = 0.12 \times 6250 = 750 \text{ Pa}\]
Answer: \(\tau = 750\) Pa.
Part (b) — Drag Force
Given: \(A = 80\) cm² \(= 80 \times 10^{-4}\) m² \(= 8 \times 10^{-3}\) m²
\[F = \tau \cdot A = 750 \times 8 \times 10^{-3} = 6.0 \text{ N}\]
Answer: \(F = 6.0\) N.
A 2D incompressible flow has the velocity field:
\[u = 3x^2y - y^3, \qquad v = -3xy^2 + x^3\]
Calculate the vorticity \(\omega_z\) at the point \((x, y) = (1, 2)\) m.
Show Solution
Solution:
Given:
The 2D vorticity is:
\[\omega_z = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\]
Compute each partial derivative:
\[\frac{\partial v}{\partial x} = \frac{\partial}{\partial x}(-3xy^2 + x^3) = -3y^2 + 3x^2\]
\[\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(3x^2 y - y^3) = 3x^2 - 3y^2\]
Therefore:
\[\omega_z = (-3y^2 + 3x^2) - (3x^2 - 3y^2) = -3y^2 + 3x^2 - 3x^2 + 3y^2 = 0\]
Answer: \(\omega_z = 0\) at every point in the domain — the flow is irrotational (it can be shown to be the real and imaginary parts of a complex potential, confirming the result).
Engine oil at 40 °C has dynamic viscosity \(\mu = 0.113\) Pa·s and density \(\rho = 876\) kg/m³.
Show Solution
Solution:
Part (a) — Kinematic Viscosity
\[\nu = \frac{\mu}{\rho} = \frac{0.113}{876} = 1.290 \times 10^{-4} \text{ m}^2/\text{s}\]
Converting to cSt:
\[\nu = \frac{1.290 \times 10^{-4}}{1 \times 10^{-6}} = 129.0 \text{ cSt}\]
Answer: \(\nu = 1.290 \times 10^{-4}\) m²/s = 129.0 cSt.
Part (b) — Reynolds Number
Given: \(D = 25\) mm \(= 0.025\) m, \(\bar{u} = 0.8\) m/s
\[Re = \frac{\bar{u}\,D}{\nu} = \frac{0.8 \times 0.025}{1.290 \times 10^{-4}} = \frac{0.020}{1.290 \times 10^{-4}} \approx 155\]
Answer: \(Re \approx 155\), which is \(\ll 2300\), so the flow is laminar. This is typical for viscous oils even at moderate velocities.