--- title: "Francis design (90 m, 60 m³/s, 94%)" subject: "Hydropower Engineering" chapter: 7 key: "semester_7/hydropower_engineering:francis-design-90m-60cumec-94" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/francis-design-90m-60cumec-94 site: BCT Engineering --- # Francis design (90 m, 60 m³/s, 94%) Hydropower Engineering · Chapter 7: Hydro-electric Machines - Asked in: 2074 Ashwin, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/francis-design-90m-60cumec-94 - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7 ## Question A hydropower plant has a design discharge of 60 m³/s and net head of 90 m. Design a Francis turbine for this power plant (number of turbines, specific speed, diameter and setting of turbine). Take turbine efficiency 94%. ## Short answer Use 2 units of 24.9 MW; specific speed about 131, speed 230.8 rpm, runner diameter about 2.43 m; setting height is at most about 3.35 m above tailwater. ## Answer #### Step 1: Power and units $$ P = \eta\gamma Q H = 0.94 \times 9.81 \times 60 \times 90 = 49{,}796\ \text{kW} $$ Number of units: a single 50 MW unit would need a large runner (3.6 m) and shaft. Adopt **2 units**, each $Q = 30$ m³/s and $P_u = 24{,}898$ kW (gives easier transport, maintenance and part-load operation). #### Step 2: Specific speed and rated speed For Francis turbines (metric units, kW, m, rpm) the empirical relation of Schweiger and Gregory is $$ N_s = \frac{3763}{H^{0.854}} + 49.4 = 130.1 $$ $$ N = \frac{N_s H^{5/4}}{\sqrt{P_u}} = 228.5\ \text{rpm} $$ Synchronous speed at 50 Hz: $N = 6000/p$. Adopt **$p = 26$ poles, $N = 230.8$ rpm**. Actual specific speed: $$ N_s = \frac{N\sqrt{P_u}}{H^{5/4}} = 131.4 $$ #### Step 3: Runner diameter Peripheral speed $u = K_u\sqrt{2gH}$ with $K_u = 0.7$ (range 0.6 to 0.9): $$ u = 0.7\sqrt{2 \times 9.81 \times 90} = 29.41\ \text{m/s}, \qquad D = \frac{60u}{\pi N} = \frac{60 \times 29.41}{\pi \times 230.8} = 2.43\ \text{m} $$ #### Step 4: Setting (Thoma cavitation coefficient) $$ \sigma_c = 0.0432\left(\frac{N_s}{100}\right)^2 = 0.0745 $$ Sea-level site assumed: $H_{atm} = 10.3$ m, $H_v = 0.24$ m. $$ H_s = H_{atm} - H_v - \sigma_c H = 10.3 - 0.24 - 0.0745 \times 90 = 3.35\ \text{m} $$ This is the highest permissible level of the runner outlet above tailwater; in practice set lower (safety margin 1 to 2 m, and reduce $H_{atm}$ by about 1 m per 900 m of site elevation). **Answer: 2 units of 24.9 MW; $N_s \approx 131$; $N = 230.8$ rpm; $D \approx 2.43$ m; setting $H_s \le 3.35$ m above tailwater level.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are 2 units adopted instead of one 50 MW unit? A. To halve the head B. The specific speed would be negative C. A single unit needs a large runner and shaft; two units ease transport, maintenance and part-load operation D. A single unit gives higher efficiency Answer: C. Splitting the plant into 2 units of 24.9 MW gives easier transport, maintenance and part-load running. (Key point: Number of units) 2. Which power is used in N = Ns H^(5/4)/sqrt(P)? A. The power of one unit, 24,898 kW B. The water power at gross head C. The total plant power 49.8 MW D. The generator rating in MVA Answer: A. Specific speed applies to a single turbine, so Pu per unit is used. (Key point: Power per unit use) 3. Why is 230.8 rpm adopted rather than 228.5 rpm? A. It is the nearest synchronous speed 6000/26 for 50 Hz B. It sets sigma_c to zero C. It is the runaway speed D. It gives a smaller runner Answer: A. Synchronous speeds are 6000/p, and p = 26 poles gives 230.8 rpm. (Key point: Speed adopted) 4. What does Hs = 3.35 m represent? A. The highest permissible level of the runner outlet above tailwater, set lower in practice B. The draft tube length C. The depth of the runner below tailwater D. The net head loss Answer: A. Hs is the upper limit for cavitation-free operation, and a safety margin of 1 to 2 m is used in practice. 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