If the straight sides of a triangular section of a lined canal with circular bottom of radius D, make an angle $$\theta $$ with horizontal, the hydraulic mean depth is A. $${\text{D}}$$ B. $$\frac{{\text{D}}}{2}$$ C. $$\frac{{\text{D}}}{3}$$ D. $$\frac{{\text{D}}}{4}$$ E. $$\frac{{\text{D}}}{5}$$

The correct answer is $\frac{D}{2}$.

The hydraulic mean depth is defined as the ratio of the cross-sectional area of the flow to the wetted perimeter. In this case, the cross-sectional area is a right triangle with base $D$ and height $2D\sin\theta$. The wetted perimeter is the sum of the length of the base and the length of the two sides, which is $2D+2D\cos\theta$. Therefore, the hydraulic mean depth is

$$\frac{A}{P} = \frac{\frac{1}{2}D^2\sin^2\theta + D^2\cos^2\theta}{2D+2D\cos\theta} = \frac{D}{2}.$$

Option A is incorrect because it is the radius of the circular bottom of the canal, which is not the hydraulic mean depth. Option B is incorrect because it is half of the hydraulic mean depth. Option C is incorrect because it is one-third of the hydraulic mean depth. Option D is incorrect because it is one-fourth of the hydraulic mean depth. Option E is incorrect because it is one-fifth of the hydraulic mean depth.