Irodov 1.336: A Taper Multiplies the Reynolds Number

Problem Irodov 1.336

A pipeline carries a steady flow of fluid, and its cross-sectional radius shrinks gradually along its length according to $r(x) = r_0 e^{-\alpha x}$ , with $\alpha = 0.50$ m $^{-1}$ and $x$ measured from the pipeline’s inlet. For two cross-sections of the pipe separated by $\Delta x = 3.2$ m along the pipe’s axis, find the ratio of the Reynolds number at the downstream (narrower) section to the Reynolds number at the upstream (wider) section.

The pipe's radius, normalized to its inlet value, plotted against distance along the pipe. The two cross-sections in question, marked 1 (upstream, wider) and 2 (downstream, narrower), sit a fixed axial distance Δ x apart on this same exponential taper. — The pipe’s radius, normalized to its inlet value, plotted against distance along the pipe. The two cross-sections in question, marked $1$ (upstream, wider) and $2$ (downstream, narrower), sit a fixed axial distance $\Delta x$ apart on this same exponential taper.

Given

$r = r_0e^{-\alpha x}$
$\alpha = 0.50$ m $^{-1}$
$\Delta x = 3.2$ m

Find

Find

Re_2/Re_1

, the ratio of Reynolds numbers at the two sections

Solution

Step 1Velocity from continuity

The fluid is incompressible and the pipe carries one steady volume flow rate $Q$ past every cross-section, so the mean speed at a section of radius $r$ follows from dividing $Q$ by that section’s own area:

Q = v\,(\pi r^2) \quad\Longrightarrow\quad v = \frac{Q}{\pi r^2}. \tag{1}

Step 2Reynolds number as a function of radius

The Reynolds number is built from the mean speed and the section’s own diameter $d = 2r$ . Substituting the mean speed (1),

\begin{align*}Re &= \frac{\rho v d}{\eta} \\ &= \frac{\rho}{\eta}\left(\frac{Q}{\pi r^2}\right)(2r)\end{align*}

Canceling one power of $r$ ,

Re = \dfrac{2\rho Q}{\pi\eta}\cdot\dfrac{1}{r}

(2)

Step 3The ratio between two sections

Using (2) at the two sections $x_1$ and $x_2 = x_1+\Delta x$ , every factor except $r$ is common to both and cancels in the ratio, leaving only the radii:

\begin{align*}\frac{Re_2}{Re_1} &= \frac{r_1}{r_2} \\ &= \frac{r_0 e^{-\alpha x_1}}{r_0 e^{-\alpha(x_1+\Delta x)}} \\ &= e^{-\alpha x_1+\alpha x_1+\alpha\Delta x}.\end{align*}

The inlet radius $r_0$ cancels and the two exponents subtract, so the starting station $x_1$ drops out and only the separation is left in the exponent. Substituting the given values,

\alpha\Delta x = (0.50)(3.2) = 1.6,

\dfrac{Re_2}{Re_1} = e^{\alpha\Delta x} \approx 5.0

(3)

Check

The ratio is dimensionless, as a ratio of two like quantities must be, and $\alpha\Delta x=1.6$ is itself dimensionless since $\alpha$ carries units of inverse length. The narrower, downstream section carries the larger Reynolds number, and (2) shows why its two competing dependences do not cancel: squeezing the same flow rate through a smaller bore raises the speed as $1/r^2$ while the diameter falls only as $r$ , so the product of the two, and with it $Re$ , still rises as $1/r$ . That is exactly the trend that drives a pipeline toward turbulence as it tapers down.

Remark

Because $Re\propto 1/r\propto e^{\alpha x}$ , the ratio between any two sections depends only on their separation $\Delta x$ , never on where along the pipe that pair sits: moving both taps further downstream by the same amount would leave $Re_2/Re_1$ unchanged, while doubling $\Delta x$ would square it.