JNTU B.Tech II HYDRAULICS AND HYDRAULIC MACHINERY Previous Question Paper (August/September - 2022 - R15)

For a trapezoidal channel with area A, depth y, bed width b, and side slope 1:n (horizontal to vertical), the area is A = (b + ny)y, which gives b = (A/y) - ny. The wetted perimeter is P = b + 2y√(1 + n²). Substituting b, we get P = (A/y) - ny + 2y√(1 + n²). For maximum discharge, the wetted perimeter P must be minimum for a constant area A. Differentiating P with respect to y and setting it to zero: dP/dy = -A/y² - n + 2√(1 + n²) = 0. Substituting A = (b + ny)y: -(b + ny)y/y² - n + 2√(1 + n²) = 0, which simplifies to b + 2ny = 2y√(1 + n²). Dividing by 2, we get (b + 2ny)/2 = y√(1 + n²). The hydraulic mean depth R = A/P. For the most efficient section, the top width T = b + 2ny, so R = A / (b + 2ny) = (b + ny)y / (b + 2ny). Substituting the condition (b + 2ny) = 2y√(1 + n²), we find R = (b + ny)y / (2y√(1 + n²)). Further geometric analysis shows that for a trapezoidal section, the condition of maximum efficiency is R = y/2.

Given discharge Q = 11.25 m³/s and velocity v = 0.75 m/s, the area A = Q/v = 11.25 / 0.75 = 15 m². For a trapezoidal section, A = (b + ny)y. With side slope 1:2 (n=2), A = (b + 2y)y = 15. For the most efficient section, the hydraulic mean depth R = y/2 and the side slope condition is b + 2ny = 2y√(1 + n²). Substituting n=2, b + 4y = 2y√(1 + 2²) = 2y√5, so b = 2y(√5 - 2). Substituting b into the area equation: [2y(√5 - 2) + 2y]y = 15, which simplifies to 2y²(√5 - 1) = 15. Solving for y: y² = 15 / (2 * 1.236) = 6.068, so y = 2.46 m. Then b = 2 * 2.46 * (2.236 - 2) = 1.16 m. The wetted perimeter P = b + 2y√(1 + n²) = 1.16 + 2 * 2.46 * √5 = 12.16 m. R = A/P = 15 / 12.16 = 1.23 m. Using Manning's formula v = (1/n) * R^(2/3) * S^(1/2), we have 0.75 = (1/0.025) * (1.23)^(2/3) * S^(1/2). Calculating S: S^(1/2) = (0.75 * 0.025) / (1.23)^(2/3) = 0.01875 / 1.148 = 0.0163. Thus, S = 0.000266.