--- title: "Program for critical depth in trapezoidal channel" subject: "Hydraulics" chapter: 7 key: "semester_4/hydraulics:algorithm-critical-depth-trapezoidal" source: https://www.bctengineer.com/q/semester_4/hydraulics/algorithm-critical-depth-trapezoidal site: BCT Engineering --- # Program for critical depth in trapezoidal channel Hydraulics · Chapter 7: Energy and Momentum Principles in Open channel flow - Asked in: 2068 Magh, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/algorithm-critical-depth-trapezoidal - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7 ## Question Write algorithm and programme coding in any high level language (C or Fortran) for determination of critical depth in trapezoidal channel section. ## Short answer The algorithm solves f(y) = Q^2 T − g A^3 = 0 for trapezoidal section with A = (b+zy)y, T = b+2zy by bisection. Bracketing the root by doubling y_high until f changes sign, then iterating to tolerance 1e-6. Example: for Q=6, b=2.5, z=2, yc = 0.691 m. The C program implements this. ## Answer **Principle.** Critical flow requires $\dfrac{Q^2 T}{g A^3} = 1$. For a trapezoidal section with bed width $b$ and side slope $z$ (z H : 1 V): $$ A = (b + zy)y,\quad T = b + 2zy $$ Define $f(y) = Q^2 T - gA^3$. The critical depth is the root of $f(y) = 0$. $f > 0$ for $y < y_c$ and $f < 0$ for $y > y_c$, so the bisection method always converges. **Algorithm** 1. Start. 2. Read $Q$, $b$, $z$. Set $g = 9.81$ and tolerance $\varepsilon = 10^{-6}$. 3. Set $y_{low} = 10^{-6}$ and $y_{high} = 1$. 4. While $f(y_{high}) > 0$, set $y_{high} = 2\,y_{high}$ (bracket the root). 5. Repeat: $y_{mid} = (y_{low} + y_{high})/2$. - If $f(y_{mid}) > 0$, set $y_{low} = y_{mid}$; otherwise set $y_{high} = y_{mid}$. 6. Stop repeating when $y_{high} - y_{low} < \varepsilon$. 7. Print $y_c = (y_{low} + y_{high})/2$. 8. Stop. **Program in C** ```c #include #include #define G 9.81 /* f(y) = Q^2*T - g*A^3 ; zero at critical depth */ double f(double y, double Q, double b, double z) { double A = (b + z * y) * y; double T = b + 2.0 * z * y; return Q * Q * T - G * A * A * A; } int main(void) { double Q, b, z, lo, hi, mid, tol = 1e-6; int i; printf("Enter discharge Q (m3/s): "); scanf("%lf", &Q); printf("Enter bed width b (m): "); scanf("%lf", &b); printf("Enter side slope z (z H : 1 V): "); scanf("%lf", &z); lo = 1e-6; hi = 1.0; while (f(hi, Q, b, z) > 0.0) /* expand until sign changes */ hi *= 2.0; for (i = 0; i < 100 && (hi - lo) > tol; i++) { mid = 0.5 * (lo + hi); if (f(mid, Q, b, z) > 0.0) lo = mid; /* root is deeper */ else hi = mid; /* root is shallower */ } mid = 0.5 * (lo + hi); printf("Critical depth yc = %.4f m\n", mid); return 0; } ``` **Sample run:** $Q = 6$, $b = 2.5$, $z = 2$ gives **$y_c = 0.6912\ \text{m}$**. Setting $z = 0$ gives a rectangular channel and $b = 0$ gives a triangular channel, so the same program handles all three shapes. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What function is driven to zero to find the critical depth? A. f(y) = Q^2 T - g A^3 B. f(y) = Q - A V C. f(y) = A - T y D. f(y) = y - V^2/2g Answer: A. Critical flow needs Q^2 T/(g A^3) = 1, i.e. Q^2 T - g A^3 = 0. (Key point: Root function) 2. In the bisection step, if f(ymid) > 0 what is done? A. Set yhigh = ymid B. Stop the loop C. Set ylow = ymid, since the root is deeper D. Double ymid Answer: C. f > 0 means y < yc, so the root lies above ymid and ylow is moved up. (Key point: Bisection logic) 3. Why is yhigh doubled while f(yhigh) > 0? A. To increase accuracy B. To make sure yhigh is deeper than yc so the root is bracketed C. To avoid dividing by zero in A D. To convert units Answer: B. The root must lie between ylow and yhigh; doubling continues until f changes sign. (Key point: Bracketing) 4. How can the same program handle rectangular and triangular channels? A. By changing g B. By changing the tolerance C. It cannot D. Set z = 0 for rectangular and b = 0 for triangular Answer: D. A = (b + zy)y and T = b + 2zy reduce to the rectangular and triangular forms for those values. 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