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

2024
Views:

290

Pages:

5
Author:

Jabin F.

Section Factor Q_T^2 / (g A^3) = 1 Q^2 / g = A^3 / T Q / sqrt(g) = A sqrt(A/T) = Z_c Z_c = A sqrt(A/T) = Q / sqrt(g) MCQs Q.1 Fr = 0.8 y = 1.5m, y_c = ? Rectangular Q = A.V y_c = (q^2/g)^(1/3) Q = B.y.V q = y.V q = 3.060 * 1.5 q = 4.603 m^3/s/m Fr = V / sqrt(gD) 0.8 = V / sqrt(gy) 0.8 = V / sqrt(9.8 * 1.5) V = 3.060 m/s y_c = ((4.603)^2 / 9.81)^(1/3) y_c = 1.29 m MCQs Q.2 Rectangular B = 4.5m Q = 100m^3/s. Diagram of a rectangular channel with depth y_c and width 4.5. y_c = ? y_c = (q^2/g)^(1/3) q = Q/B = 100/4.5 y_c = (((100/4.5)^2) / 9.81)^(1/3) y_c = 3.69m MCQ Q.3 y1 = 0.1 m V = 6 m/s y2 = ? Alternate depth. q = Vy = 0.1 * 6 q = 0.6 E1 = E2 y1 + (V^2 / 2g) = y2 + (V^2 / 2g) q1 = q2 y1 + (q1^2 / (2g y1^2)) = y2 + (q2^2 / (2g y2^2)) 0.1 + (0.6^2 / (2 * 9.8 * 0.1^2)) = y2 + (0.6^2 / (2 * 9.8 * y2^2)) Conjugate depth -> y2 = 1.929, 0.1, -0.09507 y2 / y1 = 1/2 [-1 + sqrt(1 + 8 Fr^2)] Q.4 Rectangular channel Q = 100 m^3/s fr = 0.8 V / sqrt(gD) = 0.8 V / sqrt(gy) = 0.8 ← 5m → D = A/T = By/B D = y Q = AV 100 = 5 * y * V V = 20/y (20/y) / sqrt(gy) = 0.8 y = 4m