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Francis turbine design (55 m³/s, 25 m)

Short answer

Turbine output 11.47 MW, speed 250 rpm (24 poles), specific speed 478.9. Runner outlet diameter 2.55 m, inlet diameter 1.52 m, inlet height 0.54 m. Runner must be set 1.53 m below tailwater to avoid cavitation.

Design a Francis turbine for a hydroelectric station with a design discharge of 55 m³/s, a net head of 25 m, and an efficiency of 85%. The turbine will be installed at a location where the atmospheric pressure is equivalent to 10 m of water, and the vapor pressure is 0.2 m of water.

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.

Data: Q=55Q = 55 m³/s, H=25H = 25 m, η=0.85\eta = 0.85, Hatm=10H_{atm} = 10 m, Hv=0.2H_v = 0.2 m of water, frequency f=50f = 50 Hz.

1. Power output

P=ηρgQH=0.85×9.81×55×25=11465.4 kWP = \eta \rho g Q H = 0.85 \times 9.81 \times 55 \times 25 = 11465.4\ \text{kW}

2. Trial specific speed and speed

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

Ns=3470H0.625=3470250.625=34707.477=464.1N=NsH5/4P=464.1×55.90107.08=242.3 rpm\begin{aligned} N_s &= \frac{3470}{H^{0.625}} = \frac{3470}{25^{0.625}} = \frac{3470}{7.477} = 464.1 \\ N &= \frac{N_s H^{5/4}}{\sqrt{P}} = \frac{464.1 \times 55.90}{107.08} = 242.3\ \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=24p = 24 poles:

N=120×5024=250 rpmNs=25011465.4251.25=478.9\begin{aligned} N &= \frac{120 \times 50}{24} = 250\ \text{rpm} \\ N_s &= \frac{250\sqrt{11465.4}}{25^{1.25}} = 478.9 \end{aligned}

This is at the upper end of the Francis range (about 50 to 450, fast runners up to about 500). It is a fast (high specific speed) Francis runner; at such a low head a Kaplan turbine could also be considered, but the design is done for a Francis turbine as asked.

4. Runner outlet (throat) diameter D3D_3

Peripheral velocity coefficient:

Ku=0.31+0.0025Ns=0.31+0.0025×478.9=1.507D3=84.5KuHN=84.5×1.507×25250=2.547 m\begin{aligned} K_u &= 0.31 + 0.0025 N_s = 0.31 + 0.0025 \times 478.9 = 1.507 \\ D_3 &= \frac{84.5 K_u \sqrt{H}}{N} = \frac{84.5 \times 1.507 \times \sqrt{25}}{250} = 2.547\ \text{m} \end{aligned}

5. Runner inlet diameter D1D_1 and inlet height H1H_1

D1=(0.4+94.5Ns)D3=0.597×2.547=1.521 mH1=(0.094+0.00025Ns)D3=0.2137×2.547=0.544 m\begin{aligned} D_1 &= \left(0.4 + \frac{94.5}{N_s}\right) D_3 = 0.597 \times 2.547 = 1.521\ \text{m} \\ H_1 &= (0.094 + 0.00025 N_s) D_3 = 0.2137 \times 2.547 = 0.544\ \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=7.54×10−5×478.91.41=0.4534\sigma_c = 7.54 \times 10^{-5} N_s^{1.41} = 7.54 \times 10^{-5} \times 478.9^{1.41} = 0.4534

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

Hs=Hatm−Hv−σcH=10−0.2−0.4534×25=−1.53 m\begin{aligned} H_s &= H_{atm} - H_v - \sigma_c H \\ &= 10 - 0.2 - 0.4534 \times 25 \\ &= -1.53\ \text{m} \end{aligned}

The negative sign means the runner outlet must be placed 1.53 m below the minimum tailwater level (submerged setting) to avoid cavitation.

Summary of design

ItemValue
Power output11.47 MW
Speed250 rpm (24 poles)
Specific speed NsN_s478.9
Outlet diameter D3D_32.55 m
Inlet diameter D1D_11.52 m
Inlet / guide vane height H1H_10.54 m
Thoma coefficient σc\sigma_c0.453
Suction head HsH_s-1.53 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)

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