University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

467

Pages:

7
Author:

Jabin F.

Consider a circular channel section of any Radius 'r' and depth of flow 'y'. As the depth of flow varies, the shape of flow area also varies due to convergence of channel boundaries at the top. Wetted area and wetted perimeter also changes, therefore area of the flow section can-not be consider as constant. => Case I -> cond^n of max. discharge. Case II -> cond^n of max mean Velocity. (I) cond^n of max. discharge through the circular channel -> Area of Sector = (1/2) * r^2 * θ Area of Δ is = 2 * (1/2 * r * sin(θ/2) * r * cos(θ/2)) = r^2 * (2 * sin(θ/2) * cos(θ/2)) / 2 = (r^2 / 2) * sin(θ) Total A = Area of Sector - Area of Δ A = (1/2) * r^2 * θ - (1/2) * r^2 * sin(θ) A = (r^2 / 2) * (θ - sinθ) wetted perimeter = θ * r By chezy's formula -> Q = A * C * sqrt(R * So) Q = A * C * sqrt((A/P) * So) Q = C * sqrt(A^3 / P * So) C and So are constant. => Qmax when (A^3 / P) max d(A^3 / P) / dθ = (P * d(A^3)/dθ - A^3 * dP/dθ) / P^2 0 = (P * 3A^2 * dA/dθ - A^3 * dP/dθ) / P^2 0 = 3P * (dA/dθ) - A * (dP/dθ) = 3(rθ) * (r^2 / 2) * (1 - cosθ) - (r^2 / 2) * (θ - sinθ) * (r) = 3r^3 * θ / 2 * (1 - cosθ) - r^3 / 2 * (θ - sinθ) Divide by r^3 / 2: 0 = 3θ(1 - cosθ) - (θ - sinθ) 3θ - 3θcosθ - θ + sinθ = 0 2θ - 3θcosθ + sinθ = 0 Solve θ = 308° Approximately By chezy's Method. y = r - r * cos(θ/2) = r(1 - cos(θ/2)) = r(1 - cos(308°/2)) = r(1 - cos(154°)) put r = D/2 y = 0.95 D R = A / P = (r^2 / 2) * (θ - sinθ) / (rθ) = (r / 2θ) * (θ - sinθ) = (D / 4θ) * (θ - sinθ) R = (r / 2θ) * (θ - sinθ) R = (r / (2 * 308 * π / 180)) * (308 * π / 180 - sin(308°)) R = (D / (4 * 308 * π / 180)) * (308 * π / 180 - sin(308°)) R = 0.29 D