Question

(b) The cross-section of a flow in a straight reach of river during a flood is shown in Figure 1. The main channel has a Manning's roughness of 0.022 whereas the flood banks are grassed and have a Manning's roughness of 0.035. The longitudinal slope of the channel So is 0.0002 n 0.035 1) 1.5 m 3 m 4 m 3 m n 0.022 6 m Figure 1 Using the Horton Equation for composite roughness, determine (i) The equivalent Manning's roughness coefficient of the cross-section. (15%) (ii) The total discharge in the channel assuming uniform flow conditions. (15%) (iii) The energy coefficient, a, for the cross-section. and (30%) (iv) Suggest a reason why your answers obtained from the Horton Equation may be naccurate (1 0%) (v) Suggest an alternative method, giving a reason why this method may provide improved estimates of resistance and dischar ge (10%)

Answer 1

hydralic radius R = A/p = 49/25.312 = 1.935 m theta = k squareroot s_0 s_0 = 0.0002 k = 1/n_e AR^2/3 n_e = 0.0556 A = 49 m^2 R = 7.935 total discharge of channel theta = 1/0.0556 times 49 times 1.935^2/3 squareroot 0.0002 17.985 times 49 times 1.5528 squareroot 0.0002 Q = 60.5222 m^3 /sec \r\n total discharge in the channel assuming uniform flow conditions Q = k squareroot S_0 where k = 1/n A R^2/3 area = A + B = [1.5 times 11] + (6 + (1 times 25) 25 = 16.5 + 21.25 37.75 m^2 Area = D + E Area D = -[(3 times 1.5) + 1/2 times 1.5 times 1.5] + [(3 times 1.5) + (1/2 times 1.5 times 15)] = [4.5 + 1.125] + [4.5 + 1.125] = 5.625 + 5.625 = 11.25 m^2 total area A = (A + B) + (C + D) = 37.75 + 11.25 = 49 m^2 neutral perimeter p = 6 + 3.535 = 3.535 + 2 + 2.121 = 2.121 25.312 m