Hydraulic radius of a complex shape I'm working on a project involving thermoacoustics, and one of the important parameters is known as the hydraulic radius. If you have a pipe with some odd geometry, the hydraulic radius is its cross-sectional area of flow (i.e.: of the area not occupied by the shape) divided by its wetted perimeter.
The project I'm working on uses a "stack" for a certain component of the piping system. This comprises a series of plates with some thickness t and spacing x that fit within a tube. One of the plates is bisected widthwise by a diameter of the tube. If the thickness, number of plates, and tube diameter are known, what is the hydraulic radius of this shape?

The white area inside the red circle is the cross-sectional flow area
The perimeter of that white area is the wetted perimeter
The ratio of these two values is the value in question
 A: If a given region has a low point of y, then the angle formed between the lower-left and upper-left of the region and the downward line are $\theta_1$ and $\theta_2$ respectively.

$Perimeter(ABCD)$ is $|\overline{AD}| + |\overline{BC}| + Len(AB) + Len(CD)$
$$ \begin{align}
|\overline{AD}| = & 2R \sin(\theta_1) \\
|\overline{BC}| = & 2R \sin(\theta_2) \\
Len(AB) = & \frac{\theta_2 - \theta_1}{2\pi} \cdot 2\pi R = R(\theta_2 - \theta_1)\\
Len(BC) = &  R(\theta_2 - \theta_1)\\
\end{align}
$$
Hence, $Perimeter(ABCD) = 2R[\sin(\theta_1)+\sin(\theta_2)+\theta_2-\theta_1 ]$
$Area(ABCD)$ can be computed as $Area(OBMC) - Area(OAMD)$.
Given $y_1=|\overline{ME}|$ for arbitrary $y_1$, and your $t = |\overline{EF}|, y_2=y_1+t$:
$$ \begin{align}
& Area(OAMD) = \pi R^2 \frac{\theta_1}{\pi} = \theta_1 R^2 \\
& Area(OAED) = R^2\sin(\theta_1)\cos(\theta_1)\\
& \implies Area(AEDM) = R^2[ \theta_1 - \sin(\theta_1)\cos(\theta_1)] \\
\end{align}
$$
So the area of the region is the difference of the two:
$$ Area(ABCD) = Area(BCM) - Area(ADM) = \\
R^2[( \theta_2 - \sin(\theta_2)\cos(\theta_2) ) - 
( \theta_1 - \sin(\theta_1)\cos(\theta_1) ) ]
$$
$\theta_1$ can be calculated as: $$\theta_1 = \cos^{-1}\left(\frac{R-y}{R}\right), y=|\overline{ME}| $$
Note: this is just the area and perimeter for one of the white areas.  I don't know if it's valid to combine all of the regions for hydraulic radius.
A: One reasonable approximation would be assuming that every white rectangle has some width $w_i$ and height $x$. So your hydraulic radius, according to your description, should be computed as
$$\frac{\text{area}}{\text{perimeter}} =
\frac{\sum_i w_i x}{\sum_i 2w_i + 2x} \approx
\frac{x\sum_i w_i}{2\sum_i w_i} =
\frac x2$$
This approximation works if


*

*The number of plates is sufficiently high, so that the rectangle approximation is sound

*The special cases at the ends of the sequence, where shapes differ even further from the rectangular case, can be ignored as well

*You have $x\ll w_i$ so that you can ignore the short sides of the rectangles


If this is not enough, then I fear you'll have to do actual computations for all the dimensions involved. I doubt you'll end up with a simple decription without such approximations.
