Question

Which of these is the continuity equation for an axisymmetric flow?

(a) ρVθcot⁡θ + ρ\(\frac {∂(V_θ)}{∂θ}\) + Vθ\(\frac {∂(ρ)}{∂θ}\) = 0

(b) 2ρVr + ρVθcot⁡θ + ρ\(\frac {∂(V_θ)}{∂θ}\) + Vθ\(\frac {∂(ρ)}{∂θ}\) = 0

(c) 2ρVr + ρVθcot⁡θ = 0

(d) \(\frac {1}{r{^2}} \frac {∂}{∂r}\) (r^2ρVr) +  \(\frac {1}{r sinθ} \frac {∂}{∂θ}\)(ρVθsin⁡θ) + \(\frac {1}{r sinθ} \frac {∂(ρV_ϕ)}{∂ϕ}\) = 0

I have been asked this question in examination.

Question is from Quantitative Formulation topic in portion Linearized and Conical Flows of Aerodynamics

Answer 1

The correct choice is (b) 2ρVr + ρVθcot⁡θ + ρ\(\frac {∂(V_θ)}{∂θ}\) + Vθ\(\frac {∂(ρ)}{∂θ}\) = 0

To elaborate: The general continuity equation is given by

\(\frac {∂}{∂t}\) + ∇.(ρV) = 0.

Since the flow is assumbed to be steady, \(\frac {∂}{∂t}\) = 0.

For spherical coordinated of a cone, the del operator is expanded as

∇.(ρV) = \(\frac {1}{r{^2}} \frac {∂}{∂r}\)(r^2ρVr) +  \(\frac {1}{r sinθ} \frac {∂}{∂θ}\) (ρVθsin⁡θ) + \(\frac {1}{r sinθ} \frac {∂(ρV_ϕ)}{∂ϕ}\) = 0

Solving the partial derivatives, we get

\(\frac {1}{r{^2}}\bigg [ \)r^2\(\frac {∂}{∂r}\)(ρVr) + ρVr\(\frac {∂(r^2)}{∂r} \bigg ] + \frac {1}{r sinθ} \bigg [ \)ρVθ\(\frac {∂}{∂θ}\)(sin⁡θ ) + sin⁡θ\(\frac {∂(ρV_θ)}{∂θ} \bigg ] + \frac {1}{r sinθ} \frac {∂(ρV_ϕ)}{∂ϕ}\) = 0

This is equal to

\(\frac {1}{r{^2}}\bigg [ \)r^2\(\frac {∂}{∂r}\)(ρVr) + ρVr(2r)\( \bigg ] + \frac {1}{r sinθ} \bigg [ \)ρVθ(cos⁡θ) + sin⁡θ\(\frac {∂(ρV_θ)}{∂θ} \bigg ] + \frac {1}{r sinθ} \frac {∂(ρV_ϕ)}{∂ϕ}\) = 0

Since the flow properties are constant along a ray, \(\frac {∂}{∂r}\)(ρVr) = 0 and \( \frac {∂(ρV_ϕ)}{∂ϕ}\) = 0

The equations becomes \(\frac {1}{r{^2}}\)[ρVr(2r)] + \(\frac {1}{r sinθ} \bigg [ \)ρVθ(cos⁡θ) + sin⁡θ \(\frac {∂(ρV_θ)}{∂θ} \bigg ] \) + 0

Multiplying the final equation with r: 2ρVr + ρVθcot⁡θ + ρ\(\frac {∂(ρV_θ)}{∂θ}\) + Vθ\(\frac {∂(ρ)}{∂θ}\) = 0