Q1: D is the diameter of a circular sewer and α is the side of a square section sewer. If both are hydraulically equivalent, the relationship which holds good, is

ANS:A - πD^8/3 = 4 b^8/3

When two sewer sections are hydraulically equivalent, it means that they have the same flow capacity for a given flow rate and slope. In this case, we're comparing a circular sewer with diameter D to a square sewer with side α. The relationship between their dimensions can be determined using the hydraulic radius (R) of each section. For a circular sewer, the hydraulic radius (RD​) is given by half the diameter: RD​=D/2​ For a square sewer, the hydraulic radius (Rb​) is equal to half the side length: Rb​=α/2​ Since the two sewer sections are hydraulically equivalent, their hydraulic radii are equal: RD​=Rb​ D/2​=α​/2 D=α So, the diameter of the circular sewer is equal to the side of the square sewer. Given that D=α, we can substitute α for D in the options provided:

1. πD8/3=4⋅b8/3 becomes 8/3=4⋅8/3πα8/3=4⋅b8/3
2. πD3/8=4⋅b3/8 becomes  πα3/8=4⋅b^3/8
3. πD2/3=4⋅b2/3 becomes πα2/3=4⋅b^2/3
4. πD3/2=4⋅b3/2 becomes  πα3/2=4⋅b^3/2

Since all options have πα and 4b on both sides, none of these relationships hold true