22-Mec-A6 Fluid Machinery · December 2019

Question 2 of 6: Ideal Flow from a Radial Velocity Potential

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: Ideal Flow from a Radial Velocity Potential (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A planar potential flow described in polar coordinates $(r,\theta)$ by $\phi=-\dfrac{A}{2\pi}\ln r$ with $A\gt 0$; the flow is irrotational and incompressible, so both a potential $\phi$ and a stream function $\psi$ exist.

Find. $\psi(r,\theta)$; the shape of the $\phi=$ const and $\psi=$ const families; $V_r$ and the identification of the flow; and the physical interpretation of $A$.

1. (c & setup) Velocity components. In polar coordinates $V_r=\dfrac{\partial\phi}{\partial r}$ and $V_\theta=\dfrac{1}{r}\dfrac{\partial\phi}{\partial\theta}$. With $\phi=-\dfrac{A}{2\pi}\ln r$, $$\boxed{V_r=\frac{\partial\phi}{\partial r}=-\frac{A}{2\pi r},\qquad V_\theta=0 .}$$ Since $A\gt 0$, $V_r\lt 0$ everywhere (flow directed inward) and $V_r\propto 1/r$ — this is a two-dimensional line sink at the origin.
2. (a) Stream function. For an incompressible plane flow $V_r=\dfrac{1}{r}\dfrac{\partial\psi}{\partial\theta}$ and $V_\theta=-\dfrac{\partial\psi}{\partial r}$. Matching $V_r=-\dfrac{A}{2\pi r}$ gives $\dfrac{\partial\psi}{\partial\theta}=-\dfrac{A}{2\pi}$, and $V_\theta=0$ gives $\dfrac{\partial\psi}{\partial r}=0$. Hence $$\boxed{\psi=-\frac{A}{2\pi}\,\theta .}$$
3. (b) Equipotentials and streamlines. $\phi=$ const requires $\ln r=$ const, i.e. $r=$ const — the equipotentials are concentric circles centred on the origin. $\psi=$ const requires $\theta=$ const — the streamlines are radial rays through the origin. The two families intersect at right angles, as they must for a potential flow (Figure 2).
4. (d) Physical meaning of $A$. The volume flow rate per unit depth crossing any circle of radius $r$ is $$Q=\oint V_r\,r\,d\theta=\int_0^{2\pi}\!\Big(-\frac{A}{2\pi r}\Big)r\,d\theta=-A .$$ So $A$ is the strength of the sink: the magnitude of the volumetric flow rate (per unit length normal to the plane) that the sink withdraws from the field. The negative sign confirms that fluid is being absorbed at the origin (a drain/well); had $\phi$ carried a $+$ sign, the same $A$ would be the discharge of a source.