--- title: "Draft tube functions and inlet pressure" subject: "Hydro Power" chapter: 7 key: "semester_6/hydro_power:draft-tube-pressure-inlet" source: https://www.bctengineer.com/q/semester_6/hydro_power/draft-tube-pressure-inlet site: BCT Engineering --- # Draft tube functions and inlet pressure Hydro Power · Chapter 7: Hydro-Electric Machines - Asked in: 2082 Chaitra, 1+2+3 marks; 2082 Kartik, 1+2+3 marks (asked 2 times) - Video: [Draft Tube](https://www.youtube.com/watch?v=X3aypwGhMME) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/draft-tube-pressure-inlet - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-7 ## Question What is a draft tube? Write down its functions. Show that the pressure at inlet of draft tube is less than atmospheric pressure. ## Short answer A draft tube is a diverging passage from reaction turbine runner to tailrace. It recovers exit kinetic energy, creates suction allowing turbine above tailwater without head loss. Bernoulli gives p1/ρg = pa/ρg - Hs - (V1²-V2²)/(2g) + hf; thus p1 < atmospheric. ## Answer A **draft tube** is a gradually expanding pipe or passage that connects the runner exit of a **reaction turbine** to the tailrace, with its outlet always submerged below the tailwater level. #### Functions 1. **Recovers kinetic energy:** the increasing area reduces the high exit velocity, converting kinetic energy into pressure energy instead of wasting it. 2. **Creates a negative head (suction) at the runner exit**, so the turbine can be installed above the tailwater level without losing head, and the full head $H$ is used. 3. **Allows turbine to be set above tailwater** for easy inspection and maintenance. 4. **Increases turbine output and efficiency** (net head on the runner increases). 5. Discharges water safely to the tailrace with low exit velocity. #### Pressure at inlet of the draft tube is below atmospheric ```text runner ===[ ]=== (1) inlet, p1, V1 \ / | \/ | Hs (above TWL) ~~~~~~~~|~~~~~~~~~|~~~~ TWL | | | y |__| (2) outlet, V2 ``` Let $H_s$ = height of draft tube inlet above the tailwater, $y$ = depth of outlet below tailwater, $V_1, V_2$ = velocities at inlet and outlet, $h_f$ = loss in the draft tube, $p_a$ = atmospheric pressure. Take the datum at section 2. Bernoulli between 1 and 2: $$ \frac{p_1}{\rho g} + \frac{V_1^2}{2g} + (H_s + y) = \frac{p_2}{\rho g} + \frac{V_2^2}{2g} + 0 + h_f $$ At the outlet, $\dfrac{p_2}{\rho g} = \dfrac{p_a}{\rho g} + y$. Substituting: $$ \begin{aligned} \frac{p_1}{\rho g} + \frac{V_1^2}{2g} + H_s + y &= \frac{p_a}{\rho g} + y + \frac{V_2^2}{2g} + h_f \\ \frac{p_1}{\rho g} &= \frac{p_a}{\rho g} - H_s - \left(\frac{V_1^2 - V_2^2}{2g} - h_f\right) \end{aligned} $$ Since $H_s > 0$ and $V_1 > V_2$ with $\dfrac{V_1^2 - V_2^2}{2g} > h_f$ (the draft tube recovers more energy than it loses), the bracketed term is positive. Hence $$ \frac{p_1}{\rho g} < \frac{p_a}{\rho g} $$ i.e. the pressure at the draft tube inlet (runner exit) is **less than atmospheric** (a vacuum). This suction is the gain in head given by the draft tube. It must not fall below the vapour pressure, otherwise cavitation occurs; this limits $H_s$ (Thoma's criterion). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is a draft tube? A. A converging pipe before the runner B. A gradually expanding passage from the reaction turbine runner exit to the tailrace, with its outlet submerged C. A pipe that carries water from the reservoir D. A tube used only in Pelton turbines Answer: B. It is a diverging tube from the runner exit to the tailrace, with the outlet below tailwater. (Key point: Definition of draft tube) 2. How does the draft tube let the turbine be set above tailwater without losing head? A. By increasing flow velocity at the exit B. By creating a negative pressure head (suction) at the runner exit C. By raising the tailwater level D. By reducing the net head Answer: B. The suction at the runner exit means the height Hs is not lost and the full head is used. (Key point: Function: suction) 3. From the Bernoulli derivation, p1/(rho g) equals: A. pa/(rho g) + Hs + (V1^2 - V2^2)/2g B. pa/(rho g) - Hs - [(V1^2 - V2^2)/2g - hf] C. pa/(rho g) - y D. pa/(rho g) + hf Answer: B. Subtracting Hs and the positive recovered-energy term from atmospheric head gives a pressure below atmospheric. (Key point: Result of Bernoulli) 4. Why is the bracketed term in the derivation positive? A. Because V2 > V1 B. Because the draft tube recovers more kinetic energy ((V1^2-V2^2)/2g) than the friction loss hf C. Because hf is zero D. Because Hs is negative Answer: B. V1 > V2 and the energy recovered exceeds the loss, so p1 is less than atmospheric. (Key point: Why below atmospheric) ## Related in Hydro Power - [Conical draft tube inlet pressure and efficiency](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aconical-draft-tube-numerical) — Hydro Power, chapter 7 - [Draft tube function and turbine selection criteria](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Adraft-tube-turbine-selection) — Hydro Power, chapter 7 - [Why draft tube is used in Francis turbine](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Adraft-tube-francis) — Hydro Power, chapter 7 - [Draft tube objectives and governors](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Adraft-tube-objectives-governors) — Hydro Power, chapter 7 ## Related in other subjects - [Functions of draft tube](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Adraft-tube-functions) — Hydropower Engineering, chapter 7 - [Max height of straight draft tube (Francis)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-draft-tube-max-height-11760kw) — Hydropower Engineering, chapter 7 - [Francis overall efficiency with draft tube](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-draft-tube-efficiency-32mw) — Hydropower Engineering, chapter 7 - [3-phase IM principle; speed less than sync](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Athree-phase-im-principle-slip) — Electrical Machine, chapter 5