CHANNEL section is called most efficient ifCost is the MINIMUM for a given discharge.It can pass maximum discharge for a given cross-sectional area, resistance coefficient, and bottom slope.Discharge is maximum when R is becoming maximum and Perimeter BECOMES minimum as \(Q = AV = A\frac{1}{n}{R^{2/3}}{S^{1/2}}\) and \(R = \frac{A}{P}\)where,Q = dischargeV = velocity of flowS = bottom slopen = resistance coefficientA = Area of the flow, P = Wetted Perimeter, R = hydraulic radiusCalculation:Rectangular channel sectionArea of the flow, A = ByWetted Perimeter, P = B + 2yFrom the area, put value of B in TERMS of A and y in Wetted perimeter equation:\(B = \frac{A}{y}\)\(P = \frac{A}{y} + 2y\)Since, P is minimum for most efficient section, then\(\frac{{dP}}{{dy}} = 0\)\(\frac{{dP}}{{dy}} = - \frac{A}{{{y^2}}} + 2 = 0\)PUTTING A = By, we getB = 2yHydraulic Radius, \(R = \frac{A}{P} = \frac{{By}}{{B + 2y}} = \frac{{2{y^2}}}{{4y}} = \frac{y}{2}\)Rectangular channel sectionTrapezoidal channel sectionR = y / 2A = 2y2 T = 2yP = 4yD = yR = y / 2\(A = \sqrt 3 {y^2}\)\(T = \frac{{4y}}{{\sqrt 3 }}\)\(P = \frac{{2y}}{{\sqrt 3 }}\) D = 3y / 4