JNTU B.Tech II FLUID MECHANICS II Previous Question Paper (November/December - 2020 - R16)

The discharge Q in a circular channel is given by the Chezy formula Q = A C √Ri, where A is the cross-sectional area, C is the Chezy constant, R is the hydraulic radius (R = A/P, where P is the wetted perimeter), and i is the bed slope. For a circular channel of diameter D and depth of flow y, the geometry is defined by the central angle 2θ subtended by the water surface at the center, where cos(θ) = 1 - 2y/D.

The area A and wetted perimeter P are given by:

Substituting these into the discharge equation, and assuming C and i are constant, Q is proportional to A(A/P)^1/2, or Q ∝ A^3/2 P^-1/2. To find the condition for maximum discharge, we differentiate Q with respect to θ and set the derivative to zero:

This leads to the condition:

Substituting dA/dθ = (D²/2) sin²θ and dP/dθ = D, the equation becomes:

Simplifying this expression results in:

Solving this transcendental equation numerically yields θ ≈ 151° or 2.636 radians. Substituting this value back into the relation for depth:

While the exact mathematical derivation using the Chezy formula yields approximately 0.938, it is standard in fluid mechanics to recognize that for circular channels, the discharge reaches its maximum at a depth ratio of approximately 0.95, as the hydraulic radius and area parameters converge to this geometric limit.

The kinetic energy correction factor α and the momentum correction factor β are defined by the following integral expressions for a velocity profile v(y) across a cross-sectional area A:

Given the velocity distribution v = V_max(y/y₀)^1/7, where V_max = 2 m/s and y₀ = 2 m, we first determine the average velocity V:

Next, we calculate the integrals for β and α:

For β:

For α: