Skip to main content

Draft tube functions and inlet 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.

What is a draft tube? Write down its functions. Show that the pressure at inlet of draft tube is less than atmospheric pressure.

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 HH 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

     runner
   ===[  ]===   (1) inlet, p1, V1
       \  /        |
        \/         | Hs (above TWL)
 ~~~~~~~~|~~~~~~~~~|~~~~ TWL
        |  |       | y
        |__| (2) outlet, V2

Let HsH_s = height of draft tube inlet above the tailwater, yy = depth of outlet below tailwater, V1,V2V_1, V_2 = velocities at inlet and outlet, hfh_f = loss in the draft tube, pap_a = atmospheric pressure. Take the datum at section 2. Bernoulli between 1 and 2:

p1ρg+V122g+(Hs+y)=p2ρg+V222g+0+hf\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, p2ρg=paρg+y\dfrac{p_2}{\rho g} = \dfrac{p_a}{\rho g} + y. Substituting:

p1ρg+V122g+Hs+y=paρg+y+V222g+hfp1ρg=paρg−Hs−(V12−V222g−hf)\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 Hs>0H_s > 0 and V1>V2V_1 > V_2 with V12−V222g>hf\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

p1ρg<paρg\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 HsH_s (Thoma's criterion).

✓ Answer checked by Claude Opus 5.5 (high effort)

All Hydro Power chapter 7 questions →