Skip to main content

Francis turbine design (140 m³/s, 120 m)

Short answer

Turbine output 154.92 MW, speed 176.47 rpm (34 poles), specific speed 174.9. Runner outlet diameter 3.92 m, inlet diameter 3.69 m, inlet height 0.54 m. Runner must be placed 3.05 m below minimum tailwater.

Design a Francis turbine for a site where the net head is 120 m and discharge is 140 m³/s. Take efficiency of the turbine is 94%.

A Francis turbine is designed by fixing its power, a synchronous speed, the specific speed and then the main runner dimensions and the safe setting (suction head) against cavitation. The empirical relations used are the standard ones of de Siervo and de Leva given in the ESHA small hydro guide. Atmospheric and vapour pressure heads are not given, so standard values are assumed.

Data: Q=140Q = 140 m³/s, H=120H = 120 m, η=0.94\eta = 0.94, Hatm=10.3H_{atm} = 10.3 m, Hv=0.2H_v = 0.2 m of water (assumed: sea-level atmosphere and water at about 20°C), frequency f=50f = 50 Hz.

1. Power output

P=ηρgQH=0.94×9.81×140×120=154919.5 kWP = \eta \rho g Q H = 0.94 \times 9.81 \times 140 \times 120 = 154919.5\ \text{kW}

2. Trial specific speed and speed

Empirical relation for Francis turbines (metric, PP in kW):

Ns=3470H0.625=34701200.625=347019.929=174.1N=NsH5/4P=174.1×397.17393.60=175.7 rpm\begin{aligned} N_s &= \frac{3470}{H^{0.625}} = \frac{3470}{120^{0.625}} = \frac{3470}{19.929} = 174.1 \\ N &= \frac{N_s H^{5/4}}{\sqrt{P}} = \frac{174.1 \times 397.17}{393.60} = 175.7\ \text{rpm} \end{aligned}

3. Synchronous speed and actual specific speed

The generator must run at N=120f/pN = 120f/p. The nearest synchronous speed is with p=34p = 34 poles:

N=120×5034=176.47 rpmNs=176.47154919.51201.25=174.9\begin{aligned} N &= \frac{120 \times 50}{34} = 176.47\ \text{rpm} \\ N_s &= \frac{176.47\sqrt{154919.5}}{120^{1.25}} = 174.9 \end{aligned}

This lies in the Francis range (about 50 to 450). One unit of about 155 MW is designed here; in practice the flow may be shared between two or more units of the same design.

4. Runner outlet (throat) diameter D3D_3

Peripheral velocity coefficient:

Ku=0.31+0.0025Ns=0.31+0.0025×174.9=0.747D3=84.5KuHN=84.5×0.747×120176.47=3.919 m\begin{aligned} K_u &= 0.31 + 0.0025 N_s = 0.31 + 0.0025 \times 174.9 = 0.747 \\ D_3 &= \frac{84.5 K_u \sqrt{H}}{N} = \frac{84.5 \times 0.747 \times \sqrt{120}}{176.47} = 3.919\ \text{m} \end{aligned}

5. Runner inlet diameter D1D_1 and inlet height H1H_1

D1=(0.4+94.5Ns)D3=0.940×3.919=3.686 mH1=(0.094+0.00025Ns)D3=0.1377×3.919=0.540 m\begin{aligned} D_1 &= \left(0.4 + \frac{94.5}{N_s}\right) D_3 = 0.940 \times 3.919 = 3.686\ \text{m} \\ H_1 &= (0.094 + 0.00025 N_s) D_3 = 0.1377 \times 3.919 = 0.540\ \text{m} \end{aligned}

H1H_1 is the height of the runner inlet, which is also the height of the guide vanes.

6. Cavitation check and setting of the turbine

Critical Thoma cavitation coefficient for Francis turbines:

σc=7.54×10−5Ns1.41=7.54×10−5×174.91.41=0.1096\sigma_c = 7.54 \times 10^{-5} N_s^{1.41} = 7.54 \times 10^{-5} \times 174.9^{1.41} = 0.1096

Maximum suction head (height of runner outlet above tailwater):

Hs=Hatm−Hv−σcH=10.3−0.2−0.1096×120=−3.05 m\begin{aligned} H_s &= H_{atm} - H_v - \sigma_c H \\ &= 10.3 - 0.2 - 0.1096 \times 120 \\ &= -3.05\ \text{m} \end{aligned}

The negative sign means the runner outlet must be placed 3.05 m below the minimum tailwater level (submerged setting) to avoid cavitation.

Summary of design

ItemValue
Power output154.92 MW
Speed176.47 rpm (34 poles)
Specific speed NsN_s174.9
Outlet diameter D3D_33.92 m
Inlet diameter D1D_13.69 m
Inlet / guide vane height H1H_10.54 m
Thoma coefficient σc\sigma_c0.110
Suction head HsH_s-3.05 m

A spiral casing, 16 to 24 guide vanes and an elbow draft tube starting at diameter about D3D_3 complete the unit.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Hydro Power chapter 7 questions →