Chapter 13

Question 1

Why does potential flow exist in inviscid, compressible, adiabatic, reversible flows when vorticity is initially zero?

Answer 1

Because taking the curl of the momentum equation shows that vorticity remains zero if it starts zero, so the velocity field can be written as the gradient of a potential.

Question 2

What is the form of the velocity field in compressible potential flow with zero vorticity upstream?

Answer 2

V = ∇φ, where φ is the velocity potential.

Question 3

How is the momentum equation integrated for compressible potential flow?

Answer 3

It gives ∂φ/∂t + V²/2 + h = h₀, where h₀ is constant in space for irrotational flow.

Question 4

For a perfect gas in steady compressible potential flow, how is temperature related to velocity?

Answer 4

T/T₀ = 1 − V²/(2h₀).

Question 5

What steady conservation equation determines the velocity potential?

Answer 5

∇·(ρ∇φ) = 0.

Question 6

How is density expressed using isentropic relations and the energy equation?

Answer 6

ρ/ρ₀ = (1 − V²/(2h₀))^{1/(γ−1)}.

Question 7

What nonlinear equation results when substituting ρ(V) into ∇·(ρ∇φ)=0?

Answer 7

∇·[∇φ (1−V²/(2h₀))^{1/(γ−1)}] = 0.

Question 8

What assumption leads to a simplified linear equation for compressible potential flow?

Answer 8

Moderate compressibility corrections, letting u and V ≈ upstream value U and dropping small higher-order terms.

velocity and pressure changes are small compared to the main flow

Question 9

What linearized equation is obtained before introducing Mach number?

Answer 9

φₓₓ + φᵧᵧ = 0, with a compressibility correction factor applied.

Question 10

What is the Prandtl–Glauert equation for subsonic compressible flow?

Answer 10

(1 − M²) φₓₓ + φᵧᵧ = 0.

Question 11

How does coordinate scaling reduce the Prandtl–Glauert equation to Laplace’s equation?

Answer 11

Use x’ = x / sqrt(1−M²), making the equation φₓ’ₓ’ + φᵧᵧ = 0.

Question 12

How does the Prandtl–Glauert factor affect lift coefficient predictions?

Answer 12

Lift is increased by 1 / sqrt(1−M²).

Question 13

Why does the Prandtl–Glauert approximation fail near M ≈ 1?

Answer 13

It predicts singular forces at M = 1 (Mach gets infinitely large at 1), but real forces remain finite; nonlinear transonic effects dominate.

Question 14

Why is potential flow theory often invalid in supersonic flows with strong or curved shocks?

Answer 14

Shocks are non-isentropic and curved shocks generate vorticity, violating potential flow assumptions.

Question 15

Why is the Prandtl–Glauert approximation not valid near stagnation points?

Answer 15

Because the linearized assumptions used break down there; however, the stagnation region contributes no lift.

because it relies on small disturbance theory, assuming flow velocity changes are small and linear, which is violated where flow dramatically slows to zero at the stagnation point.

Question 16

What is Prandtl-Glauert?

Answer 16

The Prandtl-Glauert (P-G) method linearizes compressible flow by assuming small perturbations (tiny changes in velocity/pressure) relative to a uniform inviscid flow, treating flow as steady, irrotational, and isentropic, and strictly keeping the free-stream Mach number well below 1, thereby simplifying nonlinear equations into a linear form for moderate subsonic speeds (up to ~0.8), but it breaks down near Mach 1 due to the Prandtl-Glauert singularity where nonlinearity becomes dominant.