Discharge Calculation of an Open Channel Using Manning's Formula

Problem: A trapezoidal channel shown below has a Manning's coefficient of n=0.013 and is laid at a slope of 0.0028. The depth of flow is 2.0ft. What is the flow rate at ft ³?
Exam Name:Sylhet Gas Field 2017, Easteni Refinery Ltd 2017
 

Solution: In English unit, Manning's formula is given by the following equation,

\[V=\frac{1.486}{n}\:R^{\frac{2}{3}}S^{\frac{1}{2}}\]

Here, S= 0.0028
          n=0.013
          R=A/P
         A= {(8+14)/2}×2
            = 22 ft ²
         P= 8+ {2×√(2² + 3² )}
           = 15.21


R= A/P
   =22/15.21
   =1.45

ஃ V= (1.485/0.013) ✕ 1.45^(2/3) ✕ 0.0028^(1/2)
       =7.74 ft/s


We know Discharge, Q=AV
                                     = (22 ✕ 7.74) ft³/s
                                     = 17.18 ft³/s

Answer: 17.18 ft³/s

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