--- title: "Bed, hydraulic and energy slopes" subject: "Hydraulics" chapter: 5 key: "semester_4/hydraulics:bed-hydraulic-energy-slopes-not-parallel" source: https://www.bctengineer.com/q/semester_4/hydraulics/bed-hydraulic-energy-slopes-not-parallel site: BCT Engineering --- # Bed, hydraulic and energy slopes Hydraulics · Chapter 5: Basics of Open channel flow - Asked in: 2071 Bhadra, 4 marks - Video: [LGCIV2051 OC17 - Changes in bed slope](https://www.youtube.com/watch?v=mdzpkzqthwA) — Hydraulics and beyond - Sandra Soares-Frazao - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/bed-hydraulic-energy-slopes-not-parallel - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-5 ## Question Define bed slope, hydraulic slope and energy slope. Why for non-uniform flow, these slopes are not parallel to each other, explain with neat sketch. ## Short answer Bed slope S0 = -dz/dx, hydraulic slope Sw = -d(z+y)/dx, energy slope Sf = -dH/dx with H = z+y+αV^2/2g. In non-uniform flow, depth and velocity change, so S0, Sw, and Sf are not equal and lines are not parallel. Only in uniform flow with constant depth and velocity do all three slopes coincide and lines become parallel. ## Answer #### Definitions - **Bed slope** ($S_0$): slope of the channel bottom, $S_0=-\dfrac{dz}{dx}=\sin\theta\approx\tan\theta$. - **Hydraulic slope** ($S_w$): slope of the hydraulic grade line, which is the free water surface: $S_w=-\dfrac{d(z+y)}{dx}$. - **Energy slope** ($S_f$): slope of the energy grade line (EGL), the rate at which total head $H=z+y+\alpha V^2/2g$ is lost: $S_f=-\dfrac{dH}{dx}$. ```text EGL -----\ ____ V1^2/2g \ -----\ WSL ----------\ \------ \ y1 y2 Bed ----------------\--------\--- datum ...................... ``` #### Why they are not parallel in non-uniform flow 1. In uniform flow depth $y$ and velocity $V$ are constant, so $\dfrac{dy}{dx}=0$ and $\dfrac{d}{dx}\left(\dfrac{V^2}{2g}\right)=0$. Then the fall of the water surface equals the fall of the bed, and the fall of the EGL equals both: $S_0=S_w=S_f$. All three lines are parallel. 2. In non-uniform flow, depth and velocity change along the channel. Therefore $z$, $z+y$ and $z+y+V^2/2g$ fall at different rates. 3. The relations are $S_w=S_0-\dfrac{dy}{dx}$ and $S_f=S_w-\dfrac{d}{dx}\left(\dfrac{\alpha V^2}{2g}\right)$. 4. For a rising depth (backwater) $dy/dx>0$ so $S_wS_0$. Velocity head also changes, so $S_f\neq S_w$. 5. Friction loss $S_f$ is always positive, so the EGL always falls in the flow direction. Only when $dy/dx=0$ and $dV/dx=0$ do the three slopes become equal. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the energy slope Sf represent? A. The slope of the bed B. The slope of the water surface C. The rate at which total head z + y + aV^2/2g is lost per unit length D. The velocity gradient Answer: C. Sf = -dH/dx is the slope of the energy grade line. (Key point: Energy slope) 2. Why are the three slopes equal in uniform flow? A. Because depth and velocity are constant, so z, z+y and total head fall at the same rate B. Because friction is zero C. Because the bed is horizontal D. Because discharge is zero Answer: A. Constant y and V mean dy/dx = 0 and the velocity head is constant, so S0 = Sw = Sf. (Key point: Uniform flow equality) 3. For a backwater curve with depth increasing downstream (dy/dx > 0), how are Sw and S0 related? A. Sw > S0 B. Sw = S0 C. Sw < S0, since Sw = S0 - dy/dx D. Sw is negative always Answer: C. Sw = S0 - dy/dx, so rising depth makes the water surface fall less steeply than the bed. (Key point: Relation Sw and S0) 4. Why does the EGL always fall in the flow direction? A. Because bed slope is positive B. Because depth always decreases C. Because velocity head is always decreasing D. Because friction loss Sf is always positive Answer: D. Energy is lost to friction, so Sf is positive and the EGL falls downstream. 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