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-
Q
2
gA3
C
B

where g = 9.81 m
s
2
, Ac = the cross-sectional area (m2

), and B = 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

B = 3 + y
AC = 3y +
y
2
2

Solve for the critical depth use initial guesses of xl = 0.5 and xu = 2.5, and
iterate until the approximate error falls below 1% or the number of
iterations exceeds 10.

To solve for the critical depth \( y \) of the trapezoidal channel, we are tasked with finding the root of the given equation using the bisection method. The equation for the critical depth \( y \) can be written as:

\[
0 = 1 - \frac{Q^2}{gA_c^3B}
\]

Where:
- \( Q = 20 \, \text{m}^3/\text{s} \) (flow rate),
- \( g = 9.81 \, \text{m}/\text{s}^2 \) (gravitational constant),
- \( A_c = 3y + \frac{y^2}{2} \) (cross-sectional area),
- \( B = 3 + y \) (surface width).

We are asked to perform the bisection method with the initial guesses \( x_l = 0.5 \) and \( x_u = 2.5 \), iterating until either the approximate error is less than 1%, or the number of iterations exceeds 10.

### Steps to Follow:

1. **Define the function**: We need to express the equation in terms of \( y \), where:
   \[
   f(y) = 1 - \frac{Q^2}{gA_c^3B}
   \]
   with the given expressions for \( A_c \) and \( B \).

2. **Bisection Method Setup**:
   - Initial lower bound \( x_l = 0.5 \),
   - Initial upper bound \( x_u = 2.5 \),
   - Stopping criteria: relative error less than 1% or 10 iterations.

3. **Iteration Process**:
   - Compute the midpoint \( x_r = \frac{x_l + x_u}{2} \),
   - Evaluate \( f(x_r) \),
   - Depending on the sign of \( f(x_r) \), adjust the bounds and calculate the approximate error \( \varepsilon \).

Let me now proceed to solve this iteratively using Python to implement the bisection method.The bisection method converged to a solution after 10 iterations. The critical depth \( y \) is approximately \( 0.6699 \, \text{m} \), and the approximate error after the final iteration was \( 0.29\% \), which is below the 1% tolerance threshold.

Here’s a summary of the results:

- Final critical depth: \( y \approx 0.6699 \, \text{m} \)
- Approximate error: \( 0.29\% \)
- Number of iterations: 10

Would you like further details on the iteration process or have any specific questions?

Here are 5 related questions to explore further:
1. How does changing the initial guesses affect the convergence of the bisection method?
2. What would happen if the flow rate \( Q \) were increased?
3. How does the bisection method compare to other root-finding methods like Newton's method?
4. Can the critical depth equation be solved symbolically for specific cases?
5. What are the physical implications of a critical depth in open-channel flow?

**Tip:** The bisection method is simple but requires many iterations for high precision. Faster convergence methods like Newton-Raphson can be more efficient but require good initial guesses.

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 - (Q^2/gA_c^3B), where g = 9.81 m/s^2, Ac = the cross-sectional area (m^2), and B = 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 B = 3 + y and A_c = 3y + y^2/2. Solve for the critical depth using initial guesses of xl = 0.5 and xu = 2.5, and iterate until the approximate error falls below 1% or the number of iterations exceeds 10.

This solution explores the critical depth calculation for a trapezoidal channel with a flow rate of 20 m³/s. Using the bisection method, the problem iterates with initial guesses of y between 0.5 and 2.5 until the error falls below 1% or 10 iterations are completed. Suitable for engineering students studying fluid mechanics.

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