Water is flowing in a trapezoidal channel at a rate of Q Squareroot 37 m3/s. The

Question

Water is flowing in a trapezoidal channel at a rate of Q Squareroot 37 m3/s. The critical depth y for such a channel must satisfy the equation 0 = 1 - Q^2/g A^3_c B where g = 9.81 m/s^2, A_C is the cross sectional area (in m^2) and B is the width of the channel at the surface (in m). For this case, the width and cross-sectional area can be related to the depth 'y' by B = 3 + y and A_C = 3y + y^2/2. Solve for the critical depth correct to three decimal digits by (i) bisection method (ii) fixed point iteration method.

Explanation / Answer

1.) Bisection Method-

Replacing the values of Q, B and A in the above formula we arrive at the function

Answer - Critical Depth upto three Decimal digits is 0.718m

1.) Bisection Method-

Replacing the values of Q, B and A in the above formula we arrive at the function

f(y) =1.225* y^6 + 22.05*y^5 + 132.3*y^4 + 264.6*y^3 -37y -111 Assuming the depth is between 0 and 1 meters we try to solve this equation using Bisection method in an excel spreadsheet. ** look for videos online how to use a spreadsheet for bisection method Guess A Guess B Midpoint F(A) F(B) F(Midpoint) Error 0 1 0.5 -111 272.175 -87.4480469 1 0.5 1 0.75 -87.448 272.175 20.18926392 0.5 0.5 0.75 0.625 -87.448 20.18926392 -47.1621451 0.25 0.625 0.75 0.6875 -47.1621 20.18926392 -17.3830642 0.125 0.6875 0.75 0.71875 -17.3831 20.18926392 0.360507024 0.0625 0.6875 0.71875 0.703125 -17.3831 0.360507024 -8.76317721 0.03125 0.703125 0.71875 0.710938 -8.76318 0.360507024 -4.26539113 0.015625 0.710938 0.71875 0.714844 -4.26539 0.360507024 -1.96859231 0.007813 0.714844 0.71875 0.716797 -1.96859 0.360507024 -0.80809731 0.003906 0.716797 0.71875 0.717773 -0.8081 0.360507024 -0.22481095 0.001953 0.717773 0.71875 0.718262 -0.22481 0.360507024 0.067593819 0.000977

Answer - Critical Depth upto three Decimal digits is 0.718m

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