---
title: "Francis design (90 m, 40 m³/s, 86%)"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:francis-design-90m-40cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/francis-design-90m-40cumec
site: BCT Engineering
---

# Francis design (90 m, 40 m³/s, 86%)

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2079 Bhadra, 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-40cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

A proposed hydropower development having a net head of 90 m and design discharge of 40 m³/s uses a Francis turbine. Taking turbine efficiency 0.86, calculate specific speed, turbine diameter and setting of the turbine.

## Short answer

Power is 30.4 MW; specific speed about 126, synchronous speed 200 rpm, runner diameter about 2.81 m; setting height is at most about 3.9 m above tailwater.

## Answer

#### Step 1: Power and units
$$
P = \eta\gamma Q H = 0.86 \times 9.81 \times 40 \times 90 = 30{,}372\ \text{kW}
$$

One unit is adopted ($P_u = 30{,}372$ kW).

#### Step 2: Specific speed and rated speed
For Francis turbines (metric units, kW, m, rpm) the empirical relation of Schweiger and Gregory is

$$
N_s = \frac{3763}{H^{0.854}} + 49.4 = 130.1
$$

$$
N = \frac{N_s H^{5/4}}{\sqrt{P_u}} = 206.9\ \text{rpm}
$$

Synchronous speed at 50 Hz: $N = 6000/p$. Adopt **$p = 30$ poles, $N = 200$ rpm**. Actual specific speed:

$$
N_s = \frac{N\sqrt{P_u}}{H^{5/4}} = 125.7
$$

#### Step 3: Runner diameter
Peripheral speed $u = K_u\sqrt{2gH}$ with $K_u = 0.7$ (range 0.6 to 0.9):

$$
u = 0.7\sqrt{2 \times 9.81 \times 90} = 29.41\ \text{m/s}, \qquad D = \frac{60u}{\pi N} = \frac{60 \times 29.41}{\pi \times 200} = 2.81\ \text{m}
$$

#### Step 4: Setting (Thoma cavitation coefficient)
$$
\sigma_c = 0.0432\left(\frac{N_s}{100}\right)^2 = 0.0683
$$

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

$$
H_s = H_{atm} - H_v - \sigma_c H = 10.3 - 0.24 - 0.0683 \times 90 = 3.91\ \text{m}
$$

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: $P = 30.4$ MW; specific speed $N_s \approx 126$ (design value 130); $N = 200$ rpm; runner diameter $D \approx 2.81$ m; setting $H_s \le 3.9$ m above tailwater level.**

## Check yourself

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

1. How is the design specific speed of the Francis turbine estimated?
   A. From Ns = 0.7 sqrt(2gH)
   B. From the empirical Schweiger and Gregory relation Ns = 3763/H^0.854 + 49.4
   C. From Ns = 85.49/H^0.243
   D. From Ns = N sqrt(Q)/H^(3/4) with N unknown
   Answer: B. The Francis empirical relation gives Ns = 130.1 for H = 90 m. (Key point: Specific speed relation)

2. How is the rated speed fixed after computing 206.9 rpm?
   A. Taken as synchronous speed 6000/p, using 30 poles for 200 rpm
   B. Taken as 0.7 sqrt(2gH)
   C. Rounded to the nearest 100 rpm
   D. Increased by 10% for runaway
   Answer: A. For 50 Hz the speed must be 6000/p, so p = 30 gives N = 200 rpm. (Key point: Rated speed)

3. How is the runner diameter computed?
   A. u = Ku sqrt(2gH) with Ku = 0.7, D = 60u/(pi N) = 2.81 m
   B. D = Q/(pi N)
   C. D = sqrt(4Q/(pi V))
   D. D = d x 10
   Answer: A. Peripheral speed u = 29.41 m/s and D = 60u/(pi N). (Key point: Runner diameter)

4. How does the setting Hs depend on the Thoma coefficient?
   A. Hs = Hatm - Hv - sigma_c H
   B. Hs = sigma_c H - Hatm
   C. Hs = Hatm - sigma_c
   D. Hs = Hatm + Hv - sigma_c H
   Answer: A. Hs = 10.3 - 0.24 - 0.0683 x 90 = 3.91 m is the highest permissible runner outlet level above tailwater. (Key point: Setting)

## Related in Hydropower Engineering

- [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, 60 m³/s, 94%)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-design-90m-60cumec-94) — Hydropower Engineering, chapter 7
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- [Francis design (150 m, 25 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-specific-speed-diameter-setting-150m-25cumec) — Hydropower Engineering, chapter 7, asked 2×
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- [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

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