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

2024
Views:

230

Pages:

5
Author:

Jabin F.

Most Economical Rectangular Section -> The most Economical Section Signify That the construction cost is minimum. Q = (A/n) R^(2/3) S^(1/2) A, n, s -> constant. Q ∝ R^(2/3) Q ∝ (A/P)^(2/3) ↑ Q ∝ (1/P)^(2/3) ✓ Q ↑ ⇒ P ↓ The most Efficient or Economical channel is one which can pass Maxi. discharge for a given cross/s Area, resistance coeff. and Slope. for Economical channel section perimeter should be minimum. Why perimeter should be mini.? The cost of lining in lined canals is 2 - 2.5 times more than that of unlined canal, hence if the perimeter is less/min than cost of linning will be less and it will be most efficient Trapezoidal section B = Slant length And this is equal to the half of Regular hexagon. P = B + 2y * sqrt(1+m^2) = (2y / sqrt(3)) + (4y / sqrt(3)) = (6y / sqrt(3)) P = (6y / sqrt(3)) A = ((2y / sqrt(3)) + (1/sqrt(3))y) y = (3y / sqrt(3)) * y A = sqrt(3) y^2 T = B + 2my = (2y / sqrt(3)) + 2 * (y / sqrt(3)) T = (4y / sqrt(3)) ⇒ R = A/P = (sqrt(3) y^2 / 6y) * sqrt(3) R = y/2 same as Rectangular for Most Economic Rectangular and Trapezoidal section hydraulic Radius is depth/2 in Both case. D = A/T = (sqrt(3)y^2 * sqrt(3)) / 4y D = (3/4)y