--- title: "Program for normal depth in trapezoidal channel" subject: "Hydraulics" chapter: 6 key: "semester_4/hydraulics:algorithm-normal-depth-trapezoidal" source: https://www.bctengineer.com/q/semester_4/hydraulics/algorithm-normal-depth-trapezoidal site: BCT Engineering --- # Program for normal depth in trapezoidal channel Hydraulics · Chapter 6: Uniform flow in open channel - Asked in: 2076 Baisakh, 8 marks - Video: [Find Normal Depth In Open Channel: Newton-Raphson & Excel Solver Methods for All Channel Shapes](https://www.youtube.com/watch?v=F1Z_5P2JmTk) — Give Value for Water - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/algorithm-normal-depth-trapezoidal - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6 ## Question Write an algorithm, flow chart and computer program in any high level language to determine normal depth in a trapezoidal channel. ## Short answer Use bisection on f(y)=AR^{2/3} - nQ/√S₀ =0, with A=(b+zy)y, P=b+2y√(1+z²). Set lo=0.0001, hi=1; while f(hi)<0, hi*=2. Repeat mid=(lo+hi)/2: if f(mid)>0, hi=mid else lo=mid, until hi-lo<1e-6. Output mid as normal depth. ## Answer The normal depth $y_n$ satisfies Manning's equation, which cannot be solved directly for $y$. It is found numerically by the bisection method on $$ f(y)=AR^{2/3}-\frac{nQ}{\sqrt{S_0}}=0,\qquad A=(b+zy)y,\quad P=b+2y\sqrt{1+z^2},\quad R=\frac AP $$ $f(y)$ increases with $y$, so the root is unique. #### Algorithm 1. Read $b$, $z$, $n$, $Q$, $S_0$. 2. Compute the target value $K_0=nQ/\sqrt{S_0}$. 3. Set lower limit $y_{lo}=0.0001$ and upper limit $y_{hi}=1$; while $f(y_{hi})<0$, double $y_{hi}$. 4. Repeat: $y_{mid}=(y_{lo}+y_{hi})/2$. If $f(y_{mid})>0$ set $y_{hi}=y_{mid}$, else $y_{lo}=y_{mid}$. 5. Stop when $y_{hi}-y_{lo}<10^{-6}$ (or 100 iterations). 6. Print $y_n=(y_{lo}+y_{hi})/2$. #### Flow chart ```text [Start] | [Read b,z,n,Q,S0] | [ylo=0.0001, yhi=1] | /-----------\ yes < f(yhi) < 0 ? >-----> [yhi = 2*yhi] --+ \-----------/ | | no <-------------------------+ v [ymid = (ylo+yhi)/2]<---------+ | | /-----------\ yes | < f(ymid) > 0 >--> [yhi=ymid] | \-----------/ | | no | [ylo=ymid] | | | /---------------\ no | < yhi-ylo < 1e-6 ? >----------+ \---------------/ | yes [Print yn = (ylo+yhi)/2] | [Stop] ``` #### C program ```c #include #include /* f(y) = A*R^(2/3) - n*Q/sqrt(S): zero at the normal depth */ double f(double y, double b, double z, double n, double Q, double S) { double A = (b + z * y) * y; /* flow area */ double P = b + 2.0 * y * sqrt(1.0 + z * z); /* wetted perimeter */ double R = A / P; /* hydraulic radius */ return A * pow(R, 2.0 / 3.0) - n * Q / sqrt(S); } int main(void) { double b, z, n, Q, S, lo, hi, mid; int i; printf("Enter b (m), z (H:1V), n, Q (m3/s), S0: "); scanf("%lf %lf %lf %lf %lf", &b, &z, &n, &Q, &S); lo = 0.0001; hi = 1.0; while (f(hi, b, z, n, Q, S) < 0.0) /* bracket the root */ hi *= 2.0; for (i = 0; i < 100; i++) { /* bisection */ mid = 0.5 * (lo + hi); if (f(mid, b, z, n, Q, S) > 0.0) hi = mid; else lo = mid; if (hi - lo < 1e-6) break; } printf("Normal depth yn = %.4f m\n", 0.5 * (lo + hi)); return 0; } ``` #### Sample run Input: `b = 5, z = 1.5, n = 0.015, Q = 20, S0 = 0.0004` Output: `Normal depth yn = 1.7563 m` (The result was checked independently: $A=13.409$ m², $R=1.183$ m, $AR^{2/3}=15.000$ and $nQ/\sqrt{S_0}=15.000$.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the normal depth found numerically? A. Manning's equation cannot be solved directly for y in a trapezoid B. Depth is not defined in a trapezoid C. Manning's equation is invalid D. Q is unknown Answer: A. A and R depend on y in a way that cannot be rearranged for y. (Key point: Why numerical) 2. What function is set to zero in the bisection? A. f(y) = A R^(2/3) - nQ/sqrt(S0) B. f(y) = A - Q C. f(y) = R - y/2 D. f(y) = Fr - 1 Answer: A. Manning's equation rearranged gives A R^(2/3) = nQ/sqrt(S0). (Key point: Function) 3. If f(ymid) > 0, what is done? A. Set ylo = ymid B. Stop C. Set yhi = ymid D. Double yhi Answer: C. f increases with y, so f > 0 means ymid is too large and the root lies below it. (Key point: Bisection step) 4. What is the purpose of doubling yhi while f(yhi) < 0? A. To reduce iterations B. To make sure the root lies between ylo and yhi C. To compute critical depth D. To avoid division by zero Answer: B. A negative f at yhi means the root is above it, so the upper limit is raised until it brackets the root. (Key point: Bracketing) ## Related in Hydraulics - [Program for critical depth in trapezoidal channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aalgorithm-critical-depth-trapezoidal) — Hydraulics, chapter 7 - [Normal depth in compound channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acompound-channel-normal-depth-berms) — Hydraulics, chapter 6 - [Program for uniform depth in rectangular channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aalgorithm-uniform-depth-rectangular) — Hydraulics, chapter 6 - [Program for alternate depths in rectangular channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aalgorithm-alternate-depths-rectangular) — Hydraulics, chapter 7 - [Bed slope for critical normal depth](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Abed-slope-for-critical-flow-at-normal-depth) — Hydraulics, chapter 7 - [Distance to within 5% of normal depth](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Awide-channel-5m3s-depth-3-5m-5-percent-normal) — Hydraulics, chapter 8 ## Related in other subjects - [Efficient trapezoidal section and triangular channel depth](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-normal-depth-trap) — Hydro Power, chapter 1, asked 2× - [High level languages and natural language](https://www.bctengineer.com/question.md?key=semester_1%2Fcomputer_programming%3Ahll-similar-natural-language) — Computer Programming, chapter 1 - [Normal depth in triangular channel (1:1)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-normal-depth-0-0004) — Hydro Power, chapter 1, asked 2× - [Remez exchange algorithm with flow chart](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Aremez-flowchart-explain) — Digital Signal Analysis and Processing, chapter 5, asked 4×