Midterm Prep - Ch4

Question 1

What is the governing equation for incompressible, inviscid, irrotational flow?

Answer 1

The flow reduces to solving Laplace’s equation: ∇²φ = 0
Solutions of this equation are called harmonic functions.

Question 2

What boundary conditions are required to solve Laplace’s equation in aerodynamic problems?

Answer 2

At infinity: velocity → free-stream condition.

On a stationary surface: (u · n) = 0.

On a moving surface: (u - u_body) · n = 0.
Tangential velocity is unconstrained in inviscid flow.

Question 3

What is the significance of solving Laplace’s equation in aerodynamics?

Answer 3

Once φ is known, velocity is u = ∇φ. Pressure is found using Bernoulli’s equation, and aerodynamic forces are obtained from pressure integration over surfaces.

Question 4

Why can solutions to Laplace’s equation be linearly combined?

Answer 4

Because ∇² is linear: if ∇²φ = 0 and ∇²ψ = 0, then C1φ + C2ψ is also a solution. [SUPERPOSITION PRINCIPLE]

Question 5

Give an example of using superposition to construct a new flow solution.

Answer 5

Combining uniform horizontal flow Ux and uniform vertical flow Vy gives the solution Ux + Vy, representing a uniform flow inclined to the x-axis.

Question 6

What flow is obtained by superposing Ux and a 2D source (Q / 2π) ln r?

Answer 6

The Rankine ogive flow, representing a uniform stream disturbed by a source (or sink), forming a streamlined body shape.

Question 7

Why is the derivative of a harmonic function also harmonic?

Answer 7

Because derivatives commute with the Laplacian:
∇²(∂φ/∂A) = ∂(∇²φ)/∂A = 0
if φ is harmonic.

A derivative is the limit of the distance between two functions. If these functions are harmonic, so will the derivative be.

Question 8

What physical structure results from taking the derivative of a source potential with respect to its position?

Answer 8

A dipole (doublet) representing a source and a nearby sink.

2D: φ ≈ (A cosθ) / r

3D: φ ≈ (A cosθ) / r²

Question 9

How can the flow around a cylinder or sphere be constructed?

Answer 9

By superposing a uniform flow potential with a dipole potential aligned with the free-stream direction.

Question 10

How do complex functions help generate harmonic solutions in 2D?

Answer 10

If F(z) is analytic, where z = x + i y, then both Re[F(z)] and Im[F(z)] satisfy Laplace’s equation.
Example: F(z) = z² = x² − y² + 2i xy → x² − y² and 2xy are harmonic

Question 11

What is the advantage of using conformal mapping?

Answer 11

If z = z(w) is a mapping and F(z) is analytic, then G(w) = F[z(w)] is analytic too. This generates new harmonic solutions in transformed geometries.

Question 12

hat boundary conditions apply for flow around a stationary sphere?

Answer 12

At infinity: φ → Ux

At the surface (r = R): radial velocity ur = 0

Question 13

What is the general form of the potential for uniform flow plus a dipole?

Answer 13

φ = U ( r cosθ + (a cosθ) / r² )

Question 14

How is the dipole strength a determined?

Answer 14

Apply ∂φ/∂r = 0 at r = R:
cosθ (1 - 2a / R³) = 0 → a = R³ / 2

Question 15

What is the final potential for flow around a stationary sphere?

Answer 15

φ = U cosθ ( r + R³ / (2 r²) )

Question 16

What is the physical meaning of the dipole term?

Answer 16

It represents the induced velocity from the sphere’s presence, required to satisfy the no-penetration condition on the surface.

Question 17

How does the boundary condition change for a breathing sphere?

Answer 17

Instead of ∂φ/∂r = 0, the boundary condition is:
(∂φ/∂r) at r = R = dR/dt

Question 18

What additional potential term is needed for a breathing sphere?

Answer 18

Because the radius of the sphere changes with time, d(phi)/d(r) doesn’t = 0 but it = dR/dt. This can’t be implemented by changing the strength of the dipole but it can be by introducing A source term:
φ = -C(t) / r
with C / R² = dR/dt

Question 19

What is ∂φ/∂t for a breathing sphere?

Answer 19

∂φ/∂t = (3 U C(t) cosθ) / (2 r²) - (1 / r) dC/dt

Question 20

What is the general velocity vector u from ∇φ?

Answer 20

u = U î + A e_r + B e_θ
A = C / r² - U cosθ (R³ / r³)
B = U sinθ (R³ / (2 r³))

Question 21

What is the key property of the stream function in 2D incompressible flow?

Answer 21

A stream function satisfies automatically the incompressibility condition.
Because of that (doing some math) ew can see It automatically satisfies continuity. If:
u = ∂ψ / ∂y
v = -∂ψ / ∂x
then ∂u/∂x + ∂v/∂y = 0.

Question 22

What are the stream function relations in 2D polar coordinates?

Answer 22

r u_r = ∂ψ / ∂θ
u_θ = -∂ψ / ∂r

Question 23

What are the relations in axisymmetric coordinates (r, z)?

Answer 23

r u_z = ∂ψ / ∂r
r u_r = -∂ψ / ∂z

Question 24

What are the relations in spherical coordinates (r, θ) with no azimuthal dependence?

Answer 24

r² sinθ u_r = ∂ψ / ∂θ
r sinθ u_θ = -∂ψ / ∂r

Question 25

What condition must hold for a stream function to exist?

Answer 25

The problem must involve only two independent variables (e.g., planar 2D, axisymmetric, or flows with one ignorable coordinate).

Question 26

What equation does ψ satisfy in a 2D irrotational flow?

Answer 26

∇²ψ = 0 since vorticity ω = -∇²ψ = 0 in potential flow.

Question 27

What does a constant ψ line represent?

Answer 27

A streamline. Since u · ∇ψ = 0, ψ is constant along the flow direction.

Question 28

What does ψ₁ - ψ₂ represent between two streamlines?

Answer 28

The volume flow rate per unit depth (in 2D) or per unit azimuth (in axisymmetric flow) between the two streamlines.

Question 29

How does the order of the equation for ψ differ from that for velocity?

Answer 29

Writing velocity in terms of ψ adds one derivative order, and taking the curl to eliminate pressure adds another. The ψ equation is thus two orders higher than the Navier–Stokes equations.

Question 30

Why is the order of the ψ equation reduced by three in 2D irrotational flows?

Answer 30

1. Neglect viscous terms → order drops by 1. 2. Zero vorticity simplifies the equation → order drops by 2 more. Final result: a second-order Laplace equation for ψ.