--- title: "Dynamic equation of GVF, wide rectangular form" subject: "Hydraulics" chapter: 8 key: "semester_4/hydraulics:gvf-dynamic-equation-wide-rectangular" source: https://www.bctengineer.com/q/semester_4/hydraulics/gvf-dynamic-equation-wide-rectangular site: BCT Engineering --- # Dynamic equation of GVF, wide rectangular form Hydraulics · Chapter 8: Non-uniform gradually varied flow (GVF) - Asked in: 2082 Kartik, 8 marks; 2069 Bhadra, 8 marks (asked 2 times) - Video: [Differential Equation of GVF | Gradually Varied Flow | Hydraulics and Fluid Mechanics](https://www.youtube.com/watch?v=DUSh2cq8iSg) — The Content Paradox - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/gvf-dynamic-equation-wide-rectangular - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8 ## Question Derive the dynamic equation of Gradually Varied Flow (GVF) and convert the derived equation for the case of wide rectangular channel, using Manning's equation, into following form: $\frac{dy}{dx} = \frac{S_o\left[1 - \left(\frac{y_n}{y}\right)^{10/3}\right]}{1 - \left(\frac{y_c}{y}\right)^3}$ where $S_o$ = bed slope, $y_n$ = normal depth, $y_c$ = critical depth. ## Short answer From dH/dx = -Sf and dH = dz + dy + d(V²/2g), derive dy/dx = (S₀ - Sf)/(1 - Fr²). For a wide rectangle, using Manning’s equation, Sf/S₀ = (yₙ/y)^(10/3) and Fr² = (yc/y)³, giving dy/dx = S₀[1 - (yₙ/y)^(10/3)] / [1 - (yc/y)³]. ## Answer #### Dynamic equation of GVF **Assumptions:** steady flow; hydrostatic pressure distribution (small slope); the channel is prismatic; the energy loss is given by the uniform-flow formula using the local depth and velocity; $\alpha = 1$. Take a short reach $dx$ of an open channel. The total head above a datum is $$ H = z + y + \frac{V^2}{2g} $$ where $z$ is the bed level, $y$ the depth and $V = Q/A$. The bed slope is $S_0 = -dz/dx$ and the energy slope is $S_f = -dH/dx$. Differentiating with respect to $x$: $$ \frac{dH}{dx} = \frac{dz}{dx} + \frac{dy}{dx} + \frac{d}{dx}\left(\frac{V^2}{2g}\right) $$ $$ -S_f = -S_0 + \frac{dy}{dx} + \frac{d}{dy}\left(\frac{Q^2}{2gA^2}\right)\frac{dy}{dx} $$ Since $dA = T\,dy$: $$ \frac{d}{dy}\left(\frac{Q^2}{2gA^2}\right) = -\frac{Q^2}{gA^3}\frac{dA}{dy} = -\frac{Q^2 T}{gA^3} = -Fr^2 $$ Therefore $$ -S_f = -S_0 + \frac{dy}{dx}\left(1 - Fr^2\right) $$ $$ \boxed{\frac{dy}{dx} = \frac{S_0 - S_f}{1 - Fr^2}}\qquad Fr^2 = \frac{Q^2 T}{gA^3} $$ **Physical meaning:** the numerator is the difference between the bed slope and the slope of the energy line. If $S_0 > S_f$, depth tends to increase in the flow direction (for subcritical flow); the denominator decides the sign by flow regime ($1 - Fr^2 > 0$ subcritical, $< 0$ supercritical). When $y = y_n$, $S_f = S_0$ and $dy/dx = 0$ (uniform flow). When $y = y_c$, $Fr = 1$ and $dy/dx \to \infty$ (the surface meets the critical depth line steeply). #### Wide rectangular form using Manning's equation For a wide rectangular channel, $R \approx y$, $A = by$ and $q = Q/b$. **Friction slope** from Manning's equation $V = \frac{1}{n}R^{2/3}S_f^{1/2}$: $$ S_f = \frac{n^2 V^2}{R^{4/3}} = \frac{n^2 q^2}{y^{10/3}} $$ At normal depth $S_f = S_0$ and $y = y_n$: $$ S_0 = \frac{n^2 q^2}{y_n^{10/3}} \;\Rightarrow\; \frac{S_f}{S_0} = \left(\frac{y_n}{y}\right)^{10/3} $$ **Froude number:** at critical depth $q^2 = g\,y_c^3$, so $$ Fr^2 = \frac{q^2}{g y^3} = \left(\frac{y_c}{y}\right)^3 $$ Substituting into the dynamic equation: $$ \boxed{\frac{dy}{dx} = \frac{S_0\left[1 - \left(\dfrac{y_n}{y}\right)^{10/3}\right]}{1 - \left(\dfrac{y_c}{y}\right)^3}} $$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The dynamic equation of GVF is: A. dy/dx = (S0 - Sf)/(1 - Fr^2) B. dy/dx = (S0 + Sf)/(1 - Fr^2) C. dy/dx = (S0 - Sf)/(1 + Fr^2) D. dy/dx = (1 - Fr^2)/(S0 - Sf) Answer: A. Differentiating total head H = z + y + V^2/2g gives dy/dx = (S0 - Sf)/(1 - Fr^2). (Key point: Basic GVF equation) 2. Why does d/dy (Q^2/2gA^2) equal -Fr^2? A. Because A is constant B. Because dA = T dy, giving -Q^2 T/(g A^3) C. Because V = sqrt(g y) D. Because Sf = S0 Answer: B. d/dy(Q^2/2gA^2) = -(Q^2/gA^3) dA/dy = -Q^2 T/(gA^3) = -Fr^2. (Key point: Velocity head derivative) 3. For a wide rectangular channel with R = y, Manning gives Sf as: A. n q / y^(2/3) B. n^2 q^2 / y^(10/3), so Sf/S0 = (yn/y)^(10/3) C. n^2 q^2 / y^3 D. g y^3 / q^2 Answer: B. Sf = n^2V^2/R^(4/3) = n^2 q^2 / y^(10/3), and at yn it equals S0. (Key point: Friction slope for wide channel) 4. How does Fr^2 become (yc/y)^3? A. Fr^2 = q^2/(g y^3) and q^2 = g yc^3 at critical depth B. Fr = 1 at every depth C. Because yc = yn D. Because T = y Answer: A. Substituting q^2 = g yc^3 into q^2/(g y^3) gives (yc/y)^3. (Key point: Froude number substitution) 5. What happens to dy/dx when y approaches yn and when y approaches yc? A. Both give dy/dx = 0 B. dy/dx = 0 at yn (uniform flow); dy/dx tends to infinity at yc C. dy/dx tends to infinity at yn; 0 at yc D. dy/dx = S0 at both Answer: B. At yn, Sf = S0 so numerator is 0; at yc, Fr = 1 so the denominator is 0. (Key point: Limiting behaviour) ## Related in Hydraulics - [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 - [GVF in 5 m channel, depth 0.25 m to 0.40 m](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Agvf-5m-channel-1-85-m3s-direct-step) — Hydraulics, chapter 8, asked 2× - [GVF in 12 m wide channel, 2 steps](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Agvf-30m3s-12m-wide-two-steps) — Hydraulics, chapter 8 - [GVF in 4 m channel, graphical integration](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Agvf-4m-channel-2-m3s-graphical-integration) — Hydraulics, chapter 8, asked 2× - [Define gradually varied flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Adefine-gradually-varied-flow) — Hydraulics, chapter 5, asked 2× - [Normal depth in compound channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acompound-channel-normal-depth-berms) — Hydraulics, chapter 6 ## Related in other subjects - [Economical rectangular condition and trapezoidal bed slope](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atrapezoidal-channel-bed-slope) — Hydro Power, 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× - [Efficient rectangular section and triangular channel depth](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-normal-depth-rect) — Hydro Power, chapter 1, asked 2× - [Uniform flow in rectangular channel (Chezy)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Arectangular-channel-chezy-flow) — Hydro Power, chapter 1