---
title: "Francis design (150 m, 25 m³/s)"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:francis-specific-speed-diameter-setting-150m-25cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/francis-specific-speed-diameter-setting-150m-25cumec
site: BCT Engineering
---

# Francis design (150 m, 25 m³/s)

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2081 Baishakh, 6 marks; 2070 Chaitra, 2+2+2 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/francis-specific-speed-diameter-setting-150m-25cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

Design the specific speed, turbine diameter and setting of a Francis turbine of a hydropower project having net head of 150 m and design discharge of 25 cumec. Take turbine efficiency as 82% (81% in the 2070 Chaitra paper).

## Short answer

Power = 0.82×9.81×25×150 = 30,166 kW. Schweiger-Gregory specific speed is 101.5, giving 300 rpm (20 poles), runner diameter ≈ 2.42 m, and Thoma setting Hs ≤ 3.68 m above tailwater; lower setting is used in practice.

## Answer

#### Power
$$
P = \eta \gamma Q H = 0.82 \times 9.81 \times 25 \times 150 = 30{,}166\ \text{kW}
$$

(With $\eta = 0.81$: $P = 29{,}798$ kW.) One unit is adopted.

#### Specific speed
For Francis turbines, an empirical (Schweiger and Gregory) relation in metric units (kW, m, rpm) is

$$
N_s = \frac{3763}{H^{0.854}} + 49.4 = \frac{3763}{150^{0.854}} + 49.4 = 101.5
$$

Speed from $N_s = \dfrac{N\sqrt{P}}{H^{5/4}}$:

$$
N = \frac{101.5 \times 150^{1.25}}{\sqrt{30166}} = 306.9\ \text{rpm}
$$

Synchronous speed for 50 Hz is $N = 6000/p$ rpm. Nearest value: $p = 20$ poles, **$N = 300$ rpm**. Actual specific speed $N_s = \dfrac{300\sqrt{30166}}{150^{1.25}} = 99.3$ (98.7 for $\eta=0.81$).

#### Runner diameter
Peripheral velocity $u = K_u\sqrt{2gH}$ with $K_u = 0.7$ (typical 0.6 to 0.9 for this $N_s$):

$$
u = 0.7\sqrt{2 \times 9.81 \times 150} = 37.97\ \text{m/s}
$$

$$
D = \frac{60u}{\pi N} = \frac{60 \times 37.97}{\pi \times 300} = 2.42\ \text{m}
$$

#### Setting (Thoma cavitation coefficient)
$$
\sigma_c = 0.0432\left(\frac{N_s}{100}\right)^2 = 0.0432(0.993)^2 = 0.0426
$$

Assume sea-level site: $H_{atm} = 10.3$ m, $H_v = 0.24$ m.

$$
H_s = H_{atm} - H_v - \sigma_c H = 10.3 - 0.24 - 0.0426 \times 150 = 3.68\ \text{m}
$$

This is the highest permissible position of the runner outlet above tailwater level. A lower setting (a safety margin of 1 to 2 m, or lower for a higher site elevation) is adopted in practice. At higher elevation reduce $H_{atm}$ by about 1 m per 900 m.

**Answer: $N_s \approx 99$ (target 101.5); $N = 300$ rpm; $D \approx 2.42$ m; $H_s \le 3.7$ m above tailwater level (practically set lower).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the turbine power found?
   A. P = eta gamma Q H = 0.82 x 9.81 x 25 x 150 = 30,166 kW
   B. P = gamma Q H only, 36,787 kW
   C. P = Q H / eta
   D. P = 0.82 x Q x H = 3,075 kW
   Answer: A. Output power includes turbine efficiency. (Key point: Power)

2. How is the actual speed chosen after computing N = 306.9 rpm?
   A. Round up to 400 rpm
   B. Take the nearest synchronous speed N = 6000/p; 20 poles gives 300 rpm
   C. Use 306.9 rpm directly
   D. Use 3000 rpm
   Answer: B. Generator coupling needs synchronous speed 6000/p at 50 Hz. (Key point: Speed selection)

3. How is runner diameter obtained?
   A. D = sqrt(Q/N) = 0.29 m
   B. D = 60 N/(pi u) = 0.5 m
   C. u = Ku sqrt(2gH), D = 60u/(pi N) = 2.42 m
   D. D = H/Ku = 214 m
   Answer: C. Peripheral velocity 37.97 m/s at 300 rpm gives D = 2.42 m. (Key point: Runner diameter)

4. What does Hs = Hatm - Hv - sigma_c H = 3.68 m mean?
   A. Depth below tailwater required
   B. Minimum draft tube length
   C. Vapour pressure head
   D. Highest permissible runner outlet position above tailwater; practice sets it lower
   Answer: D. It is the upper limit to avoid cavitation; a safety margin lowers it. (Key point: Setting)

## Related in Hydropower Engineering

- [Francis size and setting (150 m, 160 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-size-setting-150m-160cumec) — Hydropower Engineering, chapter 7
- [Francis diameter and setting (110 m, 140 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-diameter-110m-140cumec) — Hydropower Engineering, chapter 7
- [Francis runner design (40 m head, 10 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-turbine-velocity-triangle-40m) — Hydropower Engineering, chapter 7
- [Francis design (90 m, 40 m³/s, 86%)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-design-90m-40cumec) — Hydropower Engineering, chapter 7
- [Francis design (90 m, 60 m³/s, 94%)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-design-90m-60cumec-94) — Hydropower Engineering, chapter 7
- [Turbine setting; three single-jet Pelton wheels](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aturbine-setting-pelton-three-jets) — Hydropower Engineering, chapter 7

## Related in other subjects

- [Francis turbine design (45 m³/s, 150 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Afrancis-design-45) — Hydro Power, chapter 7
- [Francis turbine design (55 m³/s, 25 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Afrancis-design-55) — Hydro Power, chapter 7
- [Head, specific speed and part load in turbine choice](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Aturbine-head-speed-part-load) — Power Plant Design, chapter 3
- [Francis turbine design (140 m³/s, 120 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Afrancis-design-140) — Hydro Power, chapter 7