Question

A vertical shaft Francis turbine rotates at 300 rpm. The available head at the inlet to the turbine is 200 m. The tip speed of the rotor is 40 m/s. Water leaves the runner of the turbine without whirl. Velocity at the exit of the draft tube is 3.5 m/s. The head losses in different components of the turbine are : 

(i) stator and guide vanes : 5.0 m,

(ii) rotor: 10 m, and 

(iii) draft tube: 2 m.

Flow rate through the turbine is 20 m³/s. Take g = 9.8 m/s². The hydraulic efficiency of the turbine is _____% (round off to one decimal place).

Answer 1

Concept:

H_inlet = H_turbine + H_{L, stator + Guide} + H_rotor + H_{draft tube} + \(\frac{{V_3^2}}{{2g}}\) ... (i)

\({η _H} = \frac{{{H_{turbine}}}}{{{H_{inlet}}}}\)

Where, η_H is the hydraulic efficiency, HL, stator + Guide = Head loss in stator and guide vanes, V₃ = Velocity at the exit of the draft tube, g = acceleration due to gravity

Calculation:

Given:

Hinlet = 200 m, HL, stator + Guide = 5 m, Hrotor = 10 m, Hdraft tube = 2 m, V₃ = 3.5 m/s, g = 9.8 m/s2

From equation (i)

200 = Hturbine + 5 + 10 + 2 + \(\frac{{3.5^2}}{{2 \times 9.8}}\)

Hturbine = 182.375 m

∴ \({η _H} = \frac{{{H_{turbine}}}}{{{H_{inlet}}}}\) = \(\frac {182.375}{200}\) = 0.9118 = 91.87 %