--- title: "Francis turbine design (140 m³/s, 120 m)" subject: "Hydro Power" chapter: 7 key: "semester_6/hydro_power:francis-design-140" source: https://www.bctengineer.com/q/semester_6/hydro_power/francis-design-140 site: BCT Engineering --- # Francis turbine design (140 m³/s, 120 m) Hydro Power · Chapter 7: Hydro-Electric Machines - Asked in: 2079 Chaitra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/francis-design-140 - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-7 ## Question 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%. ## 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. ## Answer 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:** $Q = 140$ m³/s, $H = 120$ m, $\eta = 0.94$, $H_{atm} = 10.3$ m, $H_v = 0.2$ m of water (assumed: sea-level atmosphere and water at about 20°C), frequency $f = 50$ Hz. #### 1. Power output $$ P = \eta \rho g Q H = 0.94 \times 9.81 \times 140 \times 120 = 154919.5\ \text{kW} $$ #### 2. Trial specific speed and speed Empirical relation for Francis turbines (metric, $P$ in kW): $$ \begin{aligned} N_s &= \frac{3470}{H^{0.625}} = \frac{3470}{120^{0.625}} = \frac{3470}{19.929} = 174.1 \\ N &= \frac{N_s H^{5/4}}{\sqrt{P}} = \frac{174.1 \times 397.17}{393.60} = 175.7\ \text{rpm} \end{aligned} $$ #### 3. Synchronous speed and actual specific speed The generator must run at $N = 120f/p$. The nearest synchronous speed is with $p = 34$ poles: $$ \begin{aligned} N &= \frac{120 \times 50}{34} = 176.47\ \text{rpm} \\ N_s &= \frac{176.47\sqrt{154919.5}}{120^{1.25}} = 174.9 \end{aligned} $$ 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 $D_3$ Peripheral velocity coefficient: $$ \begin{aligned} K_u &= 0.31 + 0.0025 N_s = 0.31 + 0.0025 \times 174.9 = 0.747 \\ D_3 &= \frac{84.5 K_u \sqrt{H}}{N} = \frac{84.5 \times 0.747 \times \sqrt{120}}{176.47} = 3.919\ \text{m} \end{aligned} $$ #### 5. Runner inlet diameter $D_1$ and inlet height $H_1$ $$ \begin{aligned} D_1 &= \left(0.4 + \frac{94.5}{N_s}\right) D_3 = 0.940 \times 3.919 = 3.686\ \text{m} \\ H_1 &= (0.094 + 0.00025 N_s) D_3 = 0.1377 \times 3.919 = 0.540\ \text{m} \end{aligned} $$ $H_1$ 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: $$ \sigma_c = 7.54 \times 10^{-5} N_s^{1.41} = 7.54 \times 10^{-5} \times 174.9^{1.41} = 0.1096 $$ Maximum suction head (height of runner outlet above tailwater): $$ \begin{aligned} H_s &= H_{atm} - H_v - \sigma_c H \\ &= 10.3 - 0.2 - 0.1096 \times 120 \\ &= -3.05\ \text{m} \end{aligned} $$ 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 | Item | Value | |---|---| | Power output | 154.92 MW | | Speed | 176.47 rpm (34 poles) | | Specific speed $N_s$ | 174.9 | | Outlet diameter $D_3$ | 3.92 m | | Inlet diameter $D_1$ | 3.69 m | | Inlet / guide vane height $H_1$ | 0.54 m | | Thoma coefficient $\sigma_c$ | 0.110 | | Suction head $H_s$ | -3.05 m | A spiral casing, 16 to 24 guide vanes and an elbow draft tube starting at diameter about $D_3$ complete the unit. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the power and a first trial speed obtained? A. P = eta rho g Q H = 154919.5 kW; trial Ns = 3470/H^0.625 = 174.1, then N = Ns H^(5/4)/sqrt(P) = 175.7 rpm B. P = rho g Q only; N is taken equal to the specific speed C. P = eta Q H with no g; N = 120 f D. P = rho g Q H with no efficiency; N = Ns x sqrt(P) Answer: A. Power includes efficiency, the empirical relation gives a trial Ns from the head, and N follows from the specific speed definition. (Key point: Power and trial speed) 2. Why is the trial speed changed to 176.47 rpm? A. The runner diameter must be a whole number of metres B. The generator must run at a synchronous speed N = 120f/p, so the nearest value with p = 34 poles is used C. The Thoma coefficient fixes the speed exactly D. Speed is always rounded to the nearest 100 rpm Answer: B. Generator speed is tied to frequency: N = 120 x 50/34 = 176.47 rpm, and Ns is recalculated as 174.9. (Key point: Synchronous speed) 3. Which relations give the outlet (throat) diameter D3 and the inlet diameter D1? A. D3 = Q/H and D1 = D3/2 B. D3 = 60u/(pi N) with Ku = 0.46 and D1 = 12 D3 C. Ku = 0.31 + 0.0025 Ns, D3 = 84.5 Ku sqrt(H)/N = 3.919 m, then D1 = (0.4 + 94.5/Ns) D3 = 3.686 m D. D3 = 84.5 sqrt(H) only, and D1 = D3 always Answer: C. Both diameters follow from empirical formulas depending on Ns, H and N; the guide vane height H1 is 0.540 m. (Key point: Runner diameters) 4. Atmospheric and vapour heads were not given. How does the design proceed, and what is Hs? A. Standard values (Hatm = 10.3 m, Hv = 0.2 m) are assumed; Hs = 10.3 - 0.2 - 0.1096 x 120 = -3.05 m, so the runner is submerged B. The cavitation check is skipped because the data is missing C. Hatm and Hv are taken as zero, giving Hs = -13.2 m D. Hatm = 10.3 m and Hv = 0.2 m are added to sigma_c H, giving Hs = +23.6 m Answer: A. Missing data is replaced by standard sea-level values, and the negative Hs means the runner outlet sits 3.05 m below the minimum tailwater. 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