---
title: "Francis turbine design (55 m³/s, 25 m)"
subject: "Hydro Power"
chapter: 7
key: "semester_6/hydro_power:francis-design-55"
source: https://www.bctengineer.com/q/semester_6/hydro_power/francis-design-55
site: BCT Engineering
---

# Francis turbine design (55 m³/s, 25 m)

Hydro Power · Chapter 7: Hydro-Electric Machines

- Asked in: 2082 Kartik, 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-55
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-7

## Question

Design a Francis turbine for a hydroelectric station with a design discharge of 55 m³/s, a net head of 25 m, and an efficiency of 85%. The turbine will be installed at a location where the atmospheric pressure is equivalent to 10 m of water, and the vapor pressure is 0.2 m of water.

## Short answer

Turbine output 11.47 MW, speed 250 rpm (24 poles), specific speed 478.9. Runner outlet diameter 2.55 m, inlet diameter 1.52 m, inlet height 0.54 m. Runner must be set 1.53 m below tailwater to avoid cavitation.

## 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.

**Data:** $Q = 55$ m³/s, $H = 25$ m, $\eta = 0.85$, $H_{atm} = 10$ m, $H_v = 0.2$ m of water, frequency $f = 50$ Hz.

#### 1. Power output

$$
P = \eta \rho g Q H = 0.85 \times 9.81 \times 55 \times 25 = 11465.4\ \text{kW}
$$

#### 2. Trial specific speed and speed

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

$$
\begin{aligned}
N_s &= \frac{3470}{H^{0.625}} = \frac{3470}{25^{0.625}} = \frac{3470}{7.477} = 464.1 \\
N &= \frac{N_s H^{5/4}}{\sqrt{P}} = \frac{464.1 \times 55.90}{107.08} = 242.3\ \text{rpm}
\end{aligned}
$$

#### 3. Synchronous speed and actual specific speed

The generator must run at $N = 120f/p$. The nearest synchronous speed is with $p = 24$ poles:

$$
\begin{aligned}
N &= \frac{120 \times 50}{24} = 250\ \text{rpm} \\
N_s &= \frac{250\sqrt{11465.4}}{25^{1.25}} = 478.9
\end{aligned}
$$

This is at the upper end of the Francis range (about 50 to 450, fast runners up to about 500). It is a fast (high specific speed) Francis runner; at such a low head a Kaplan turbine could also be considered, but the design is done for a Francis turbine as asked.

#### 4. Runner outlet (throat) diameter $D_3$

Peripheral velocity coefficient:

$$
\begin{aligned}
K_u &= 0.31 + 0.0025 N_s = 0.31 + 0.0025 \times 478.9 = 1.507 \\
D_3 &= \frac{84.5 K_u \sqrt{H}}{N} = \frac{84.5 \times 1.507 \times \sqrt{25}}{250} = 2.547\ \text{m}
\end{aligned}
$$

#### 5. Runner inlet diameter $D_1$ and inlet height $H_1$

$$
\begin{aligned}
D_1 &= \left(0.4 + \frac{94.5}{N_s}\right) D_3 = 0.597 \times 2.547 = 1.521\ \text{m} \\
H_1 &= (0.094 + 0.00025 N_s) D_3 = 0.2137 \times 2.547 = 0.544\ \text{m}
\end{aligned}
$$

$H_1$ 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:

$$
\sigma_c = 7.54 \times 10^{-5} N_s^{1.41} = 7.54 \times 10^{-5} \times 478.9^{1.41} = 0.4534
$$

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

$$
\begin{aligned}
H_s &= H_{atm} - H_v - \sigma_c H \\
&= 10 - 0.2 - 0.4534 \times 25 \\
&= -1.53\ \text{m}
\end{aligned}
$$

The negative sign means the runner outlet must be placed **1.53 m below the minimum tailwater level** (submerged setting) to avoid cavitation.

#### Summary of design

| Item | Value |
|---|---|
| Power output | 11.47 MW |
| Speed | 250 rpm (24 poles) |
| Specific speed $N_s$ | 478.9 |
| Outlet diameter $D_3$ | 2.55 m |
| Inlet diameter $D_1$ | 1.52 m |
| Inlet / guide vane height $H_1$ | 0.54 m |
| Thoma coefficient $\sigma_c$ | 0.453 |
| Suction head $H_s$ | -1.53 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 = 11465.4 kW; trial Ns = 3470/H^0.625 = 464.1, then N = Ns H^(5/4)/sqrt(P) = 242.3 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. 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 = 2.547 m, then D1 = (0.4 + 94.5/Ns) D3 = 1.521 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 0.544 m. (Key point: Runner diameters)

3. How is the safe turbine setting found, and what does the sign of Hs mean?
   A. Hs = Hatm + Hv + sigma_c H, so the runner is placed above the tailwater
   B. Hs = sigma_c x H only, and any sign is fine
   C. Hs = Hatm - Hv; sigma_c is not needed
   D. Hs = Hatm - Hv - sigma_c H = -1.53 m; the negative sign means the runner outlet must be below the tailwater level
   Answer: D. With sigma_c = 0.4534 from Ns, Hs = 10 - 0.2 - 0.4534 x 25 = -1.53 m, so the runner is submerged to avoid cavitation. (Key point: Setting against cavitation)

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