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Francis turbine design (45 m³/s, 150 m)

Short answer

Turbine output 56.28 MW, speed 333.33 rpm (18 poles), specific speed 150.6. Runner outlet diameter 2.13 m, inlet diameter 2.19 m, inlet height 0.28 m. Required setting is 3.22 m below tailwater.

Design a Francis turbine which is to be installed at a hydroelectric station where atmospheric pressure is 10.3 m of water and vapour pressure is 0.2 m of water. The turbine should have a design discharge of 45 m³/s, a net head of 150 meters and an efficiency of 85%.

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=45Q = 45 m³/s, H=150H = 150 m, η=0.85\eta = 0.85, Hatm=10.3H_{atm} = 10.3 m, Hv=0.2H_v = 0.2 m of water, frequency f=50f = 50 Hz.

1. Power output

P=ηρgQH=0.85×9.81×45×150=56284.9 kWP = \eta \rho g Q H = 0.85 \times 9.81 \times 45 \times 150 = 56284.9\ \text{kW}

2. Trial specific speed and speed

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

Ns=3470H0.625=34701500.625=347022.912=151.5N=NsH5/4P=151.5×524.95237.24=335.1 rpm\begin{aligned} N_s &= \frac{3470}{H^{0.625}} = \frac{3470}{150^{0.625}} = \frac{3470}{22.912} = 151.5 \\ N &= \frac{N_s H^{5/4}}{\sqrt{P}} = \frac{151.5 \times 524.95}{237.24} = 335.1\ \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=18p = 18 poles:

N=120×5018=333.33 rpmNs=333.3356284.91501.25=150.6\begin{aligned} N &= \frac{120 \times 50}{18} = 333.33\ \text{rpm} \\ N_s &= \frac{333.33\sqrt{56284.9}}{150^{1.25}} = 150.6 \end{aligned}

This lies in the Francis range (about 50 to 450).

4. Runner outlet (throat) diameter D3D_3

Peripheral velocity coefficient:

Ku=0.31+0.0025Ns=0.31+0.0025×150.6=0.687D3=84.5KuHN=84.5×0.687×150333.33=2.132 m\begin{aligned} K_u &= 0.31 + 0.0025 N_s = 0.31 + 0.0025 \times 150.6 = 0.687 \\ D_3 &= \frac{84.5 K_u \sqrt{H}}{N} = \frac{84.5 \times 0.687 \times \sqrt{150}}{333.33} = 2.132\ \text{m} \end{aligned}

5. Runner inlet diameter D1D_1 and inlet height H1H_1

D1=(0.4+94.5Ns)D3=1.027×2.132=2.190 mH1=(0.094+0.00025Ns)D3=0.1317×2.132=0.281 m\begin{aligned} D_1 &= \left(0.4 + \frac{94.5}{N_s}\right) D_3 = 1.027 \times 2.132 = 2.190\ \text{m} \\ H_1 &= (0.094 + 0.00025 N_s) D_3 = 0.1317 \times 2.132 = 0.281\ \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=7.54×10−5×150.61.41=0.0888\sigma_c = 7.54 \times 10^{-5} N_s^{1.41} = 7.54 \times 10^{-5} \times 150.6^{1.41} = 0.0888

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

Hs=Hatm−Hv−σcH=10.3−0.2−0.0888×150=−3.22 m\begin{aligned} H_s &= H_{atm} - H_v - \sigma_c H \\ &= 10.3 - 0.2 - 0.0888 \times 150 \\ &= -3.22\ \text{m} \end{aligned}

The negative sign means the runner outlet must be placed 3.22 m below the minimum tailwater level (submerged setting) to avoid cavitation.

Summary of design

ItemValue
Power output56.28 MW
Speed333.33 rpm (18 poles)
Specific speed NsN_s150.6
Outlet diameter D3D_32.13 m
Inlet diameter D1D_12.19 m
Inlet / guide vane height H1H_10.28 m
Thoma coefficient σc\sigma_c0.089
Suction head HsH_s-3.22 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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