Continuity in cylindrical coordinates, radial flow

(a) Continuity equation in cylindrical coordinates

With velocity components $v_r$ (radial), $v_\theta$ (tangential) and $v_z$ (axial):

$$\frac{1}{r}\frac{\partial (r v_r)}{\partial r} + \frac{1}{r}\frac{\partial v_\theta}{\partial \theta} + \frac{\partial v_z}{\partial z} = 0$$

• First term: net radial outflow per unit volume. The factor $r$ appears because the flow area $r\,d\theta\,dz$ grows with $r$.
• Second term: net tangential outflow per unit volume. The factor $1/r$ appears because the arc length of the element is $r\,d\theta$.
• Third term: net axial outflow, the same as in the Cartesian form.

For compressible unsteady flow the term $\partial\rho/\partial t$ is added and $\rho$ is included inside the derivatives. For axisymmetric flow, $\partial/\partial\theta = 0$.

(b) Radial flow between discs

Given $b = 0.02\ \text{m}$, $Q = 0.04\ \text{m}^3/\text{s}$. The flow is radial and symmetric, so $v_\theta = 0$, $v_z = 0$, and the flow area at radius $r$ is $2\pi r b$.

From continuity (a volume balance), $v_r = \dfrac{Q}{2\pi r b}$, so $r v_r = \dfrac{Q}{2\pi b} = \text{constant}$.

Hence

$$\frac{1}{r}\frac{\partial (r v_r)}{\partial r} = \frac{1}{r}\frac{\partial}{\partial r}\left(\frac{Q}{2\pi b}\right) = 0$$

and the cylindrical continuity equation is satisfied.

Velocities:

$$\begin{aligned} v_r(0.10) &= \frac{0.04}{2\pi \times 0.10 \times 0.02} = 3.18\ \text{m/s} \\ v_r(0.25) &= \frac{0.04}{2\pi \times 0.25 \times 0.02} = 1.27\ \text{m/s} \end{aligned}$$

The velocity falls inversely with radius, as the flow area increases.

Answer: $v_r = 3.18\ \text{m/s}$ at $r = 0.10\ \text{m}$ and $1.27\ \text{m/s}$ at $r = 0.25\ \text{m}$