Determine the discharge through the following sections for a normal depth of 4.6 ft: n=0.012, and blttom slope S = 0.20% Solution: a. A rectangular section 18.4 ft wide. 1). Section area A = ft; 2). Wetted perimeter Pw = ft; 3). Hydraulic radius Rh = ft: 4). Discharge/flow rate Q = b. A circular section 18.4 ft in diameter. 1). Inner angle = rad= = cfs: 2). Section area A = ft; 3). Hydraulic radius Rh = ft cfs: 4). Discharge/flow rate Q = c. A right angled triangular section. 1). Side slope m = 2). Section area A = ft; 3). Hydraulic radius Rh = ft: cfs:. 4). Discharge/flow rate Q = d. A trapezoidal section with a bottom width of 18.4 ft and side slope of V:H=1:2. 1). Section area A = ft: 2). Wetted perimeter Pw = 3). Hydraulic radius Rh = ft: 4). Discharge/flow rate Q = ft: cfs: