D is the diameter of a circular sewer and $$\alpha $$ is the side of a square section sewer. If both are hydraulically equivalent, the relationship which holds good, is A. $$\pi {{\text{D}}^{\frac{8}{3}}} = 4{{\text{b}}^{\frac{8}{3}}}$$ B. $$\pi {{\text{D}}^{\frac{3}{8}}} = 4{{\text{b}}^{\frac{3}{8}}}$$ C. $$\pi {{\text{D}}^{\frac{2}{3}}} = 4{{\text{b}}^{\frac{2}{3}}}$$ D. $$\pi {{\text{D}}^{\frac{3}{2}}} = 4{{\text{b}}^{\frac{3}{3}}}$$

The correct answer is $\pi D^{2/3} = 4b^{2/3}$.

Hydraulically equivalent sewers have the same discharge capacity. The discharge capacity of a sewer is given by the Manning equation:

$$Q = \frac{1}{n} \frac{A}{R^2} S$$

where $Q$ is the discharge, $n$ is the Manning roughness coefficient, $A$ is the cross-sectional area, $R$ is the hydraulic radius, and $S$ is the slope of the sewer.

For a circular sewer, the cross-sectional area is $A = \pi D^2/4$ and the hydraulic radius is $R = D/2$. For a square sewer, the cross-sectional area is $A = b^2$ and the hydraulic radius is $R = b/2$.

Substituting these expressions into the Manning equation, we get:

$$Q = \frac{1}{n} \frac{\pi D^2/4}{(D/2)^2} S = \frac{2}{n} \pi D^2 S$$

and

$$Q = \frac{1}{n} \frac{b^2}{(b/2)^2} S = 2n b^2 S$$

Equating these two expressions, we get:

$$\frac{2}{n} \pi D^2 S = 2n b^2 S$$

or

$$\pi D^2 = 4nb^2$$

or

$$\pi D^{2/3} = 4b^{2/3}$$

Therefore, the relationship which holds good is $\pi D^{2/3} = 4b^{2/3}$.