Lecture 14

Question 1

In the high Reynolds number limit for incompressible flows, why is it valid to drop the viscous term from the Navier-Stokes equations? What assumption must hold?

Answer 1

The viscous term becomes negligible because inertial forces dominate. Mathematically, 1/𝑅𝑒 ∇^2𝑢≪ ∂𝑢/∂𝑡 +𝑢⋅∇𝑢. This holds when 𝑅𝑒≫1, so viscosity plays a role only in thin boundary layers or wakes.

Question 2

What paradox arises when we neglect viscosity in steady high-Re flow past a body, and what does it imply?

Answer 2

The D’Alembert paradox: inviscid, incompressible, irrotational flow yields zero drag (𝐹_drag=0). This contradicts experimental results and implies the necessity of viscous boundary layers to resolve the discrepancy.

Question 3

Derive Bernoulli’s equation from the Navier-Stokes equations for steady, inviscid, irrotational flow.

Answer 3

Starting with:
∂𝑢/∂𝑡+𝑢⋅∇𝑢=−1/𝜌 ∇𝑝+𝑔

In steady, irrotational flow:
𝜔=∇×𝑢=0⇒𝑢⋅∇𝑢=∇(∣𝑢∣^2).
So:
∇(𝑝+1/2 𝜌∣𝑢∣^2+𝜌𝑔𝑧)=0⇒𝑝+1/2𝜌𝑢^2+𝜌𝑔𝑧=const
along streamlines.

Question 4

In steady inviscid flow, how does pressure vary across streamlines? Derive it.

Answer 4

Perpendicular to a streamline (direction 𝑛),
∂𝑝_𝑑/∂𝑛=𝜌𝑢^2𝜅
where 𝜅 is the local curvature of the streamline. This shows pressure increases on the outer side of curvature (centrifugal balance).

Question 5

What velocity field is associated with an infinitely long, straight vortex filament of strength Γ?

Answer 5

From Biot–Savart law:
𝑢_𝜃=Γ/2𝜋𝑟
The flow is purely tangential and decays as 1/𝑟, where 𝑟 is the radial distance from the vortex core.

Question 6

Why does a vortex filament velocity profile become parabolic near the core in viscous fluids?

Answer 6

Near the core, viscosity dominates and smooths out the infinite gradient from the inviscid ∼1/𝑟 profile. The result is a solid-body rotation in the core and a parabolic pressure profile due to viscous effects.

Question 7

Describe the velocity interaction of two identical point vortices placed symmetrically in an inviscid fluid.

Answer 7

Each vortex induces a velocity at the location of the other. The symmetry leads to induced downward motion at the midpoint, forming a jet-like flow. Streamlines are deflected downward, and the vortex pair translates together.

Question 8

What does the method of images accomplish in inviscid flow? How does it relate to solid boundaries?

Answer 8

It imposes a boundary condition 𝑢⋅𝑛=0 (no penetration) by placing an image vortex of opposite sign across the wall. The real and image vortex together create streamlines that are tangent to the wall, mimicking a solid boundary.

Question 9

Why can’t you impose both no-slip and no-flow-through conditions in inviscid flow?

Answer 9

The governing equation is first-order in space, so it only allows one boundary condition per wall. For inviscid flow, only the no-penetration condition 𝑢⋅𝑛=0 is enforced—no-slip requires viscosity and boundary layers.

Question 10

What happens to the pressure gradient at 𝑟→0 near a vortex filament in an inviscid flow?

Answer 10

Since 𝑢_𝜃∼1/𝑟, the pressure gradient ∂_𝑟𝑝∼𝜌𝑢_𝜃^2/𝑟∼1/𝑟^3, which diverges. In reality, viscosity regularizes this and creates a finite core.