Question

What is the irrotationally condition for a conical flow?

(a) Vθ = \(\frac {∂(V_r )}{∂θ}\)

(b) Vϕ = \(\frac {∂(V_r )}{∂ϕ}\)

(c) Vθ = \(\frac {1}{r} \frac {∂(V_θ )}{∂θ}\)

(d) Vθ = \(\frac {∂(V_r )}{∂θ}\)Vr

I have been asked this question in an interview.

This interesting question is from Quantitative Formulation in division Linearized and Conical Flows of Aerodynamics

Answer 1

Right choice is (a) Vθ = \(\frac {∂(V_r )}{∂θ}\)

The explanation is: If we apply Crocco’s theorem in spherical coordinates we get,

∇ × V = \(\frac {1}{r^2 sinθ} \begin {vmatrix} e_r & re_θ & (rsinθ) e_ϕ \\

\frac {∂}{∂r} & \frac {∂}{∂θ} & \frac {∂}{∂ϕ} \\

V_r & rV_θ & (rsinθ) V_ϕ \\

\end {vmatrix}\) = 0

On expanding this we get,

∇ × V = \(\frac {1}{r^2 sinθ} \bigg [ \)er (\(\frac {∂}{∂θ}\)(rsinθ)Vϕ – \(\frac {∂(rV_θ)}{∂ϕ}\)) – reθ(\(\frac {∂}{∂r}\)(rsinθ)Vϕ – \(\frac {∂(V_r)}{∂ϕ}\)) + (rsinθ)eϕ\( \bigg( \frac {∂(rV_θ)}{∂r} – \frac {∂(V_r)}{∂θ} \bigg )  \bigg ] \) = 0

For this equation to be valid, the terms inside the bracket are zero. Taking the last bracket term,

\( \frac {∂(rV_θ)}{∂r} – \frac {∂(V_r)}{∂θ}\) = 0

Using chain rule to expand this, we get

r\( \frac {∂(V_θ )}{∂r}\) + Vθ\( \frac {∂(r)}{∂r} – \frac {∂(V_r )}{∂θ}\) = 0

Based on the conical flow assumptions, \( \frac {∂}{∂r}\) = 0 and \( \frac {∂}{∂ϕ}\) = 0. Applying this the equation reduced to

\(\frac {∂(V_r )}{∂θ}\) = 0

Which results in the irrotationally condition for a conical flow as Vθ = \(\frac {∂(V_r )}{∂θ}\).