--- title: "Hydraulic jump on energy and force diagrams" subject: "Hydraulics" chapter: 9 key: "semester_4/hydraulics:hydraulic-jump-profile-specific-energy-force-momentum-eq" source: https://www.bctengineer.com/q/semester_4/hydraulics/hydraulic-jump-profile-specific-energy-force-momentum-eq site: BCT Engineering --- # Hydraulic jump on energy and force diagrams Hydraulics · Chapter 9: Non-uniform rapidly varied flow (RVF) - Asked in: 2075 Baisakh, 8 marks; 2074 Bhadra, 8 marks; 2069 Bhadra, 6 marks (asked 3 times) - Video: [Hydraulic Jump - The Basic Idea and Equations](https://www.youtube.com/watch?v=oEDxT99TM1U) — CE340Hydraulics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/hydraulic-jump-profile-specific-energy-force-momentum-eq - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9 ## Question Draw a hydraulic jump profile and indicate conjugate depths and energy loss using specific energy and specific force diagram. Hence derive momentum equation for the hydraulic jump in rectangular channel. ## Short answer A hydraulic jump connects supercritical depth y1 to subcritical y2, which are conjugate depths with equal specific force. Energy loss is ΔE = (y2−y1)³/(4y1y2). Momentum conservation yields y2/y1 = ½(√(1+8Fr1²)−1). ## Answer #### Hydraulic jump profile A **hydraulic jump** is the abrupt rise in water surface when supercritical flow (depth $y_1$, $Fr_1 > 1$) changes to subcritical flow (depth $y_2$, $Fr_2 < 1$). The two depths are **conjugate (sequent) depths**. A turbulent roller forms and much energy is dissipated. ```text roller y1 ______ ,-~~~~-.______ y2 (supercritical) |<-- Lj -->| (subcritical) V1 ====> turbulence ====> V2 ----------------------------------------- bed ``` #### On the specific energy diagram Specific energy $E = y + q^2/(2gy^2)$ for constant $q$ (the curve has its minimum at $y_c$ and approaches the line $E = y$ at large depths): ```text y ^ | . y2 |- - - - - - - - - - - - * subcritical branch | .' | .-' yc |- - - - - - - - * minimum E (critical depth) | '-. | '-._ y1 |- - - - - - - - - - - - - - -* supercritical branch +------------------------+----+---> E E2 E1 |<-->| energy loss dE = E1 - E2 ``` - $y_1$ lies on the supercritical (lower) branch at energy $E_1$. - $y_2$ lies on the subcritical (upper) branch at energy $E_2$. - $y_1$ and $y_2$ are **not** alternate depths, because $E_2 < E_1$. The horizontal distance between the two points is the energy loss $\Delta E = E_1 - E_2$. #### On the specific force diagram Specific force $F = \dfrac{q^2}{gy} + \dfrac{y^2}{2}$ (per unit width): ```text y ^ | / y2 |- - - - - - - - - - - - * subcritical branch | .' | .' yc |- - - - - - - - * minimum F (critical depth) | '. | '. y1 |- - - - - - - - - - - - * supercritical branch | \ +------------------------+-----> F F1 = F2 ``` $y_1$ and $y_2$ have the **same specific force** (momentum is conserved across the jump, since friction over its short length is negligible and the bed is horizontal), so they lie on the same vertical line. The minimum specific force occurs at $y_c$. #### Momentum equation for a rectangular channel Take a horizontal rectangular channel and a control volume from section 1 to section 2. Per unit width, with $q = V_1y_1 = V_2y_2$: Net force $=$ rate of change of momentum: $$ \frac{\gamma y_1^2}{2} - \frac{\gamma y_2^2}{2} = \rho q\,(V_2 - V_1) $$ Substituting $V = q/y$ and $\gamma = \rho g$: $$ \frac{g}{2}\left(y_1^2 - y_2^2\right) = q^2\left(\frac{1}{y_2} - \frac{1}{y_1}\right) = \frac{q^2 (y_1 - y_2)}{y_1y_2} $$ Since $y_1^2 - y_2^2 = (y_1 - y_2)(y_1 + y_2)$, dividing by $(y_1 - y_2)$ gives: $$ \frac{g}{2}(y_1 + y_2) = \frac{q^2}{y_1y_2}\ \Rightarrow\ \frac{q^2}{g} = \frac{y_1y_2(y_1 + y_2)}{2} $$ Using $Fr_1^2 = q^2/(gy_1^3)$ and $r = y_2/y_1$: $$ 2Fr_1^2 = r(1 + r) \ \Rightarrow\ r^2 + r - 2Fr_1^2 = 0 $$ $$ \boxed{\frac{y_2}{y_1} = \frac{1}{2}\left(\sqrt{1 + 8Fr_1^2} - 1\right)} $$ This is the Belanger equation; it gives the sequent depth $y_2$ for a given $y_1$ and $Fr_1$. **Energy loss:** $$ \Delta E = E_1 - E_2 = \frac{(y_2 - y_1)^3}{4y_1y_2} $$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. On the specific energy diagram, why are the conjugate depths y1 and y2 NOT alternate depths? A. Because y2 lies on the supercritical branch B. Because E2 is less than E1, energy being lost in the jump C. Because they have the same specific force D. Because Fr is equal to 1 at both depths Answer: B. Alternate depths have equal E. In a jump E2 < E1, and the horizontal gap is the energy loss. (Key point: Conjugate depths on E diagram) 2. What do y1 and y2 share on the specific force diagram? A. The same specific energy B. The same Froude number C. The same specific force, because momentum is conserved D. The same velocity Answer: C. Friction is negligible on a short horizontal bed, so momentum (specific force) is the same on both sides and they lie on one vertical line. (Key point: Specific force diagram) 3. Which result follows from the momentum equation for a rectangular channel? A. y2/y1 = 0.5(sqrt(1+8Fr1^2) - 1) B. y2/y1 = 0.5(sqrt(1+8Fr1) + 1) C. y2/y1 = sqrt(1+8Fr1^2) D. y2/y1 = Fr1^2/2 Answer: A. Substituting r = y2/y1 gives r^2 + r - 2Fr1^2 = 0, whose positive root is the Belanger equation. (Key point: Belanger equation) 4. What is the energy loss in a hydraulic jump in a rectangular channel? A. (y2 - y1)/(4y1y2) B. (y2 + y1)^3/(4y1y2) C. (y2 - y1)^2/(2y1y2) D. (y2 - y1)^3/(4y1y2) Answer: D. Dividing the cube of the depth difference by 4y1y2 gives E1 - E2. 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