Water is flowing in a trapezoidal channel at a rate of Q 5 20 m3/s. The critical depth y for such a channel must satisfy the equ
ation 2 0 5 1 2 gA3 B c where g 5 9.81 m/s2, Ac 5 the cross-sectional area (m2), and B 5 the width of the channel at the surface (m). For this case, the width and the cross-sectional area can be related to depth y by
1 answer:
Answer / Explanation:
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Complete Question
Water is flowing in a trapezoidal channel at a rate of Q=20m³/s . The critical depth y for such a channel must satisfy the equation:
0 = 1 − Q² / gA³c . B Where g= 9.81m /s² and Ac = the cross-section area can be related to depth y by B = 3+y and Ac = 3y+y²/2. Solve for the critical depth using (a). the graphical method,
Answer:
Given the equation,
0 = 1 − Q² / gA³c. B
Now substituting the given value g= 9.81m /s² , Q =20m³/s, B = 3+y, and Ac = 3y+y²/2,
We get:
0 = 1 - 20² / (9.81) ( 3y + y²/2)³ (3+y)
Hence we choose f(y) = 1 - 40.7747 / (3y + y²/2)³ . (3 + y) and solve for f(y) = 0
Therefore,
To solve using a graph, we take twelve sample points ( starting at y = 0.25 in step of 0.25m and plot a graph using MS- Excel. Kindly find the graph below.
2) As evident from the sample point and the graph function f(y) gets close to zero at y = 1.5, hence the root of f(y) = 0 is Xr = 1.5
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