Interactive Lecture
Understanding Fluid Flow Through Conservation Laws
by Dr Arnaldo Delli Carri
The Navier-Stokes equations are the fundamental governing equations of fluid mechanics, describing the motion of viscous fluids. They express Newton's second law for fluid elements, accounting for pressure gradients, viscous forces, and body forces.
The Navier-Stokes equations represent conservation of momentum for a fluid continuum. Combined with mass conservation (continuity equation) and energy conservation, they form a complete system for predicting fluid behaviour — from aircraft aerodynamics to blood flow.
For an incompressible fluid, the continuity equation simplifies to:
\[\nabla \cdot \vec{u} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0\]
This states that the divergence of the velocity field must be zero for incompressible flow.
The momentum equation for an incompressible Newtonian fluid is:
Vector Form:
\[\rho\left(\frac{\partial \vec{u}}{\partial t} + \vec{u} \cdot \nabla \vec{u}\right) = -\nabla p + \mu \nabla^2 \vec{u} + \rho \vec{g}\]
where:
Component Form (2D):
\[\rho\left(\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y}\right) = -\frac{\partial p}{\partial x} + \mu\left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right)\]
\[\rho\left(\frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y}\right) = -\frac{\partial p}{\partial y} + \mu\left(\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2}\right)\]
The Reynolds number characterises the flow regime and indicates the relative importance of inertial forces to viscous forces:
\[Re = \frac{\rho U L}{\mu} = \frac{UL}{\nu}\]
where:
Flow Regimes:
Vorticity measures the local rotation of fluid elements and is defined as the curl of the velocity field:
\[\vec{\omega} = \nabla \times \vec{u}\]
In 2D: \(\omega_z = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\)
The vorticity transport equation is:
\[\frac{\partial \omega}{\partial t} + \vec{u} \cdot \nabla \omega = \nu \nabla^2 \omega\]
For certain flow conditions, the Navier-Stokes equations can be simplified:
Common Simplifications:
What This Simulation Shows:
This simulation demonstrates vortex shedding (von Kármán vortex street) — a fundamental phenomenon predicted by the Navier-Stokes equations. When fluid flows past an obstacle at moderate Reynolds numbers, vortices alternately form and detach from either side, creating an oscillating wake pattern.
The coloured particles are passive tracers that move with the fluid, visualising the velocity field. Red particles show clockwise rotation, blue shows counterclockwise rotation.
The Navier-Stokes equations underpin whole branches of engineering:
| Concept | Equation | Description |
|---|---|---|
| Continuity (Incompressible) | \(\nabla \cdot \vec{u} = 0\) | Mass conservation |
| Navier-Stokes (Vector) | \(\rho\left(\frac{\partial \vec{u}}{\partial t} + \vec{u} \cdot \nabla \vec{u}\right) = -\nabla p + \mu \nabla^2 \vec{u} + \rho \vec{g}\) | Momentum conservation |
| Reynolds Number | \(Re = \frac{\rho U L}{\mu} = \frac{UL}{\nu}\) | Ratio of inertial to viscous forces |
| Vorticity | \(\vec{\omega} = \nabla \times \vec{u}\) | Local rotation of fluid elements |
| Vorticity Transport | \(\frac{\partial \omega}{\partial t} + \vec{u} \cdot \nabla \omega = \nu \nabla^2 \omega\) | Evolution of vorticity |
| Strouhal Number | \(St = \frac{fD}{U}\) | Vortex shedding frequency (≈0.2 for cylinders) |
Water (ν = 1.0 × 10⁻⁶ m²/s) flows through a pipe of diameter D = 0.05 m at a velocity of U = 2 m/s. Calculate the Reynolds number and determine if the flow is laminar or turbulent. If the pipe diameter is doubled while maintaining the same flow rate, how does the Reynolds number change?
Show Solution
Part 1: Calculate Reynolds number
\[Re = \frac{UD}{\nu} = \frac{2 \times 0.05}{1.0 \times 10^{-6}} = 100,000\]
Since Re = 100,000 >> 4,000, the flow is turbulent.
Part 2: Effect of doubling diameter
If diameter doubles: \(D' = 0.1\) m
To maintain same flow rate Q: \(Q = U \cdot A = U' \cdot A'\)
\(U \cdot \pi (D/2)^2 = U' \cdot \pi (D'/2)^2\)
\(U' = U \cdot (D/D')^2 = 2 \times (0.05/0.1)^2 = 0.5\) m/s
New Reynolds number:
\[Re' = \frac{U' D'}{\nu} = \frac{0.5 \times 0.1}{1.0 \times 10^{-6}} = 50,000\]
The Reynolds number is halved when the diameter is doubled at constant flow rate.
Answer: Re = 100,000 (turbulent); doubling D at constant Q gives U' = 0.5 m/s and Re' = 50,000.
For laminar flow in a circular pipe, the Hagen-Poiseuille equation gives the pressure drop: \(\Delta p = \frac{32 \mu L U}{D^2}\). Calculate the pressure drop for oil (μ = 0.1 Pa·s) flowing at U = 0.5 m/s through a 10 m long pipe with D = 0.02 m. Verify that the flow is indeed laminar (assume ρ = 900 kg/m³).
Show Solution
Step 1: Check Reynolds number
\[\nu = \frac{\mu}{\rho} = \frac{0.1}{900} = 1.11 \times 10^{-4} \text{ m}^2/\text{s}\]
\[Re = \frac{UD}{\nu} = \frac{0.5 \times 0.02}{1.11 \times 10^{-4}} = 90\]
Since Re = 90 < 2300, the flow is laminar. ✓
Step 2: Calculate pressure drop
\[\Delta p = \frac{32 \mu L U}{D^2} = \frac{32 \times 0.1 \times 10 \times 0.5}{(0.02)^2}\]
\[\Delta p = \frac{16}{0.0004} = 40,000 \text{ Pa} = 40 \text{ kPa}\]
Physical interpretation: The high pressure drop is due to the high viscosity of oil and small pipe diameter. This relationship shows that pressure drop scales with \(1/D^4\) for constant flow rate, making diameter the most critical design parameter.
Answer: Δp = 40 kPa; Re = 90, confirming laminar flow.
A chimney with diameter D = 2 m is exposed to wind at U = 15 m/s. Using the Strouhal number relationship \(St = fD/U \approx 0.2\) for cylinders, calculate the vortex shedding frequency. If the chimney has a natural frequency of 0.8 Hz, is there a risk of resonance?
Show Solution
Step 1: Calculate shedding frequency
\[f = \frac{St \cdot U}{D} = \frac{0.2 \times 15}{2} = 1.5 \text{ Hz}\]
Step 2: Check for resonance
Natural frequency: \(f_n = 0.8\) Hz
Shedding frequency: \(f = 1.5\) Hz
Ratio: \(f/f_n = 1.5/0.8 = 1.875\)
Conclusion: While not exactly at resonance, these frequencies are close enough that vortex-induced vibrations could still be significant. The chimney should be designed with vortex suppression features (helical strakes, shrouds, or spoilers) or the natural frequency should be modified to avoid this wind speed range.
Real-world note: Vortex-induced vibration analysis is routinely used to design safe chimneys, offshore risers, and heat exchanger tubes. Helical strakes are a common vortex-suppression solution.
Answer: f = 1.5 Hz; f/f_n = 1.875 — not at resonance, but close enough to warrant vortex-suppression measures.