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Draft tube: purpose, setting and efficiency

Short answer

A draft tube recovers runner-exit kinetic energy, allows the turbine above tailwater, and raises effective head. Its efficiency is ηd = [(V1²−V2²)/2g − hf]/(V1²/2g), and maximum setting is Hs,max = Hatm − Hv − ηd V1²/2g, equivalently Hatm − Hv − σcH.

Why is a draft tube provided at the outlet of a runner of a reaction turbine? Derive the equation for the maximum permissible turbine setting and efficiency of the draft tube.

Purpose of a draft tube

A draft tube is a gradually expanding pipe connecting the runner outlet to the tailrace. It is provided to:

  • allow the turbine to be set above tailwater level (for easy access) without losing the head between the runner and the tailwater;
  • regain kinetic energy at the runner outlet by reducing the velocity from V1V_1 to V2V_2 and converting it into pressure, thus creating a suction (low pressure) at the runner exit and increasing the effective head;
  • discharge the water safely into the tailrace.
   runner outlet (1)
      |  |            ^
      |  |            | H_s
 -----+  +----------  | 
       \  /  TWL ~~~~~|~~~~
        \/   (2)  exit under water

Maximum permissible setting

Let section 1 be the draft tube inlet (runner outlet), at height HsH_s above tailwater, with velocity V1V_1 and pressure p1p_1. Section 2 is the draft tube outlet submerged in the tailwater, with velocity V2V_2 and pressure ≈patm\approx p_{atm}. Bernoulli's equation between 1 and 2 (datum at tailwater), with loss hfh_f in the tube:

p1γ+V122g+Hs=patmγ+V222g+hf\frac{p_1}{\gamma} + \frac{V_1^2}{2g} + H_s = \frac{p_{atm}}{\gamma} + \frac{V_2^2}{2g} + h_f p1γ=Hatm−Hs−[V12−V222g−hf]\frac{p_1}{\gamma} = H_{atm} - H_s - \left[\frac{V_1^2 - V_2^2}{2g} - h_f\right]

Draft tube efficiency

It is the ratio of the kinetic energy actually regained to the kinetic energy at the inlet:

ηd=V12−V222g−hfV122g\eta_d = \frac{\dfrac{V_1^2 - V_2^2}{2g} - h_f}{\dfrac{V_1^2}{2g}}

Substituting in the pressure equation,

p1γ=Hatm−Hs−ηdV122g\frac{p_1}{\gamma} = H_{atm} - H_s - \eta_d\frac{V_1^2}{2g}

To avoid cavitation, p1≥pvp_1 \ge p_v (vapour pressure), so

Hs,max=Hatm−Hv−ηdV122gH_{s,max} = H_{atm} - H_v - \eta_d\frac{V_1^2}{2g}

In terms of the Thoma coefficient σ=Hatm−Hv−HsH\sigma = \dfrac{H_{atm} - H_v - H_s}{H}:

Hs,max=Hatm−Hv−σcHH_{s,max} = H_{atm} - H_v - \sigma_c H

A smaller σc\sigma_c allows a higher setting. Because the exit velocity head V22/2gV_2^2/2g and the friction loss hfh_f are not recovered, ηd\eta_d is less than 1, typically 0.75 to 0.90.

✓ Answer checked by Claude Opus 5.5 (high effort)

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