--- title: "Flow through partially and fully submerged orifice" subject: "Fluid Mechanics" chapter: 6 key: "semester_3/fluid_mechanics:submerged-orifice-flow-expression" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/submerged-orifice-flow-expression site: BCT Engineering --- # Flow through partially and fully submerged orifice Fluid Mechanics · Chapter 6: Flow measurement - Asked in: 2078 Kartik, 2+2 marks - Video: [Discharge through Full & partially Submerged orifice | Derivation | Fluid Mechanics](https://www.youtube.com/watch?v=rhQ0pROAaH8) — Sudip Dhakal - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/submerged-orifice-flow-expression - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-6 ## Question Derive an expression for flow through partially and fully submerged orifice. ## Short answer For fully submerged orifice, Q = Cd a √(2g(H1-H2)) where H1, H2 are upstream and downstream heads. Partially submerged flow sums free part Q1 = (2/3)Cd b √(2g)[H^(3/2) - H1^(3/2)] and submerged part Q2 = Cd b (H2 - H) √(2gH), where H is the level difference, H1 top edge, H2 bottom edge. ## Answer A **submerged orifice** discharges below the surface of the liquid on the downstream side. It may be fully or partially submerged. #### Fully submerged orifice Let the upstream and downstream liquid surfaces be at heights $H_1$ and $H_2$ above the centre of the orifice, and $H = H_1 - H_2$. Apply Bernoulli between a point in the upstream tank (negligible velocity) and the vena contracta, where the pressure is the downstream hydrostatic pressure $\gamma H_2$: $$ H_1 + 0 = H_2 + \frac{V_c^2}{2g}\;\Rightarrow\; V_c = \sqrt{2g\,(H_1 - H_2)} = \sqrt{2gH} $$ With orifice area $a$ and coefficient of contraction $C_c$, $Q = C_c a V_c$. Including the velocity coefficient $C_v$ gives $C_d = C_cC_v$: $$ Q = C_d\,a\sqrt{2g\,(H_1 - H_2)} $$ The discharge depends only on the head difference. #### Partially submerged orifice Here the downstream level lies between the top and bottom edges of the orifice, and the flow is the sum of two parts. Let $b$ be the width of the (rectangular) orifice, $H_1$ and $H_2$ the depths from the upstream surface to the top and bottom edges, and $H$ the difference between the upstream and downstream levels ($H_1 < H < H_2$). - **Free part (top edge to the downstream level):** the pressure outside is atmospheric, so it behaves like a free rectangular orifice with the head varying from $H_1$ to $H$: $$ Q_1 = \frac{2}{3}C_d b\sqrt{2g}\left[H^{3/2} - H_1^{3/2}\right] $$ - **Submerged part (downstream level to bottom edge):** height $(H_2 - H)$ under the constant head difference $H$: $$ Q_2 = C_d\,b\,(H_2 - H)\sqrt{2gH} $$ Total discharge: $$ Q = \frac{2}{3}C_d b\sqrt{2g}\left[H^{3/2} - H_1^{3/2}\right] + C_d b\,(H_2 - H)\sqrt{2gH} $$ ```text ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ upstream level | | | | | H1 | | v ____ top edge | H | | free part H2 | | | | v ~~~~~~~~~|~~~~|~~~~~~~ downstream level | | | submerged part v |____| bottom edge ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a fully submerged orifice, what head drives the discharge? A. The upstream head H1 above the orifice centre B. The downstream head H2 above the orifice centre C. The sum of the upstream and downstream heads D. The difference between upstream and downstream levels, H1 - H2 Answer: D. Bernoulli from the upstream tank to the vena contracta, where pressure is gamma*H2, gives V = sqrt(2g(H1 - H2)). Discharge depends only on the head difference. (Key point: Head for fully submerged flow) 2. In deriving the fully submerged orifice velocity, what pressure is taken at the vena contracta? A. The upstream hydrostatic pressure gamma*H1 B. The downstream hydrostatic pressure gamma*H2 C. Zero absolute pressure D. Atmospheric pressure Answer: B. The jet discharges below the downstream surface, so the pressure at the vena contracta equals the downstream hydrostatic pressure gamma*H2. (Key point: Pressure at vena contracta) 3. How is the discharge of a partially submerged orifice obtained? A. Add a free-orifice part above the downstream level and a submerged part below it B. Subtract the submerged discharge from the free discharge C. Use the downstream head only D. Treat it as fully submerged using the full orifice height Answer: A. The part above the downstream level behaves as a free rectangular orifice, and the part below discharges under constant head H, so Q = Q1 + Q2. (Key point: Partial submergence split) 4. For the submerged part of a partially submerged orifice, which expression gives its discharge? A. (2/3)*Cd*b*(H2 - H)*sqrt(2g*H1) B. Cd*b*(H2 - H)*sqrt(2gH) C. (2/3)*Cd*b*sqrt(2g)*H^(3/2) D. Cd*b*H*sqrt(2g*H2) Answer: B. The submerged strip has height (H2 - H) and acts under the constant head difference H, so Q2 = Cd*b*(H2 - H)*sqrt(2gH). 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