Problem Statement: A stream on a plain has a reach that is described as clean and winding with some pools and some weeds. This reach has a reasonably constant slope of 0.0002. The stream cross-section of flow can be approximated as a trapezoid with bottom width equal to 5 feet and side slopes having horiz:vert = 3:1. Use the Manning equation with estimated maximum and minimum values of n for open channel flow in this stream to find the range of river discharge and water flow velocity to be expected for a 3 ft water depth.
Solution: As given above: bottom width, b = 5 ft; channel bottom slope, S = 0.0003; side slope, z = 3; and flow depth, y = 3 ft. From the table in the previous section, for a 'stream on plain, winding with some pools or shoals and some weeds or rocks', the maximum expected value of n is 0.050 and the minimum is 0.35. The second article in this series, 'Calculation of Hydraulic Radius for Open Channel Flow' gives an equation for hydraulic radius of a trapezoidal channel as follows:
RH = (by + zy2)/[ b + 2y(1 + z2)1/2 ].
Substituting known values gives: RH = [ (5)(3) + 3(3)2]/[5 + (2)(3)(1 + 32)1/2] = 1.75 ft.
Also, the trapezoid area is A = by + zy2 = (5)(3) + 3(32) = 42 ft2.
Values can now be substituted into the Manning equation [Q = (1.49/n)A(RH2/3)S1/2] to give:
For minimum n (0.035): Qmax = (1.49/0.035)(42)(1.752/3)(0.00031/2) = 45.0 cfs
For maximum n (0.050): Qmin = (1.49/0.050)(69)(2.882/3)(0.00031/2) = 31.5 cfs
Using V = Q/A: Vmax = 45.0/42 = 1.08 ft/sec; Vmin = 31.5/42 = 0.750 ft/sec