---
title: "Francis turbine design (45 m³/s, 150 m)"
subject: "Hydro Power"
chapter: 7
key: "semester_6/hydro_power:francis-design-45"
source: https://www.bctengineer.com/q/semester_6/hydro_power/francis-design-45
site: BCT Engineering
---

# Francis turbine design (45 m³/s, 150 m)

Hydro Power · Chapter 7: Hydro-Electric Machines

- Asked in: 2081 Asoj, 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-45
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-7

## Question

Design a Francis turbine which is to be installed at a hydroelectric station where atmospheric pressure is 10.3 m of water and vapour pressure is 0.2 m of water. The turbine should have a design discharge of 45 m³/s, a net head of 150 meters and an efficiency of 85%.

## Short answer

Turbine output 56.28 MW, speed 333.33 rpm (18 poles), specific speed 150.6. Runner outlet diameter 2.13 m, inlet diameter 2.19 m, inlet height 0.28 m. Required setting is 3.22 m below 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.

**Data:** $Q = 45$ m³/s, $H = 150$ m, $\eta = 0.85$, $H_{atm} = 10.3$ m, $H_v = 0.2$ m of water, frequency $f = 50$ Hz.

#### 1. Power output

$$
P = \eta \rho g Q H = 0.85 \times 9.81 \times 45 \times 150 = 56284.9\ \text{kW}
$$

#### 2. Trial specific speed and speed

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

$$
\begin{aligned}
N_s &= \frac{3470}{H^{0.625}} = \frac{3470}{150^{0.625}} = \frac{3470}{22.912} = 151.5 \\
N &= \frac{N_s H^{5/4}}{\sqrt{P}} = \frac{151.5 \times 524.95}{237.24} = 335.1\ \text{rpm}
\end{aligned}
$$

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

The generator must run at $N = 120f/p$. The nearest synchronous speed is with $p = 18$ poles:

$$
\begin{aligned}
N &= \frac{120 \times 50}{18} = 333.33\ \text{rpm} \\
N_s &= \frac{333.33\sqrt{56284.9}}{150^{1.25}} = 150.6
\end{aligned}
$$

This lies in the Francis range (about 50 to 450).

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

Peripheral velocity coefficient:

$$
\begin{aligned}
K_u &= 0.31 + 0.0025 N_s = 0.31 + 0.0025 \times 150.6 = 0.687 \\
D_3 &= \frac{84.5 K_u \sqrt{H}}{N} = \frac{84.5 \times 0.687 \times \sqrt{150}}{333.33} = 2.132\ \text{m}
\end{aligned}
$$

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

$$
\begin{aligned}
D_1 &= \left(0.4 + \frac{94.5}{N_s}\right) D_3 = 1.027 \times 2.132 = 2.190\ \text{m} \\
H_1 &= (0.094 + 0.00025 N_s) D_3 = 0.1317 \times 2.132 = 0.281\ \text{m}
\end{aligned}
$$

$H_1$ 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:

$$
\sigma_c = 7.54 \times 10^{-5} N_s^{1.41} = 7.54 \times 10^{-5} \times 150.6^{1.41} = 0.0888
$$

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

$$
\begin{aligned}
H_s &= H_{atm} - H_v - \sigma_c H \\
&= 10.3 - 0.2 - 0.0888 \times 150 \\
&= -3.22\ \text{m}
\end{aligned}
$$

The negative sign means the runner outlet must be placed **3.22 m below the minimum tailwater level** (submerged setting) to avoid cavitation.

#### Summary of design

| Item | Value |
|---|---|
| Power output | 56.28 MW |
| Speed | 333.33 rpm (18 poles) |
| Specific speed $N_s$ | 150.6 |
| Outlet diameter $D_3$ | 2.13 m |
| Inlet diameter $D_1$ | 2.19 m |
| Inlet / guide vane height $H_1$ | 0.28 m |
| Thoma coefficient $\sigma_c$ | 0.089 |
| Suction head $H_s$ | -3.22 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 = 56284.9 kW; trial Ns = 3470/H^0.625 = 151.5, then N = Ns H^(5/4)/sqrt(P) = 335.1 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 333.33 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 = 18 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 50/18 = 333.33 rpm, and Ns is recalculated as 150.6. (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 = 2.132 m, then D1 = (0.4 + 94.5/Ns) D3 = 2.190 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 0.281 m. (Key point: Runner diameters)

4. 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 = -3.22 m; the negative sign means the runner outlet must be below the tailwater level
   Answer: D. With sigma_c = 0.0888 from Ns, Hs = 10.3 - 0.2 - 0.0888 x 150 = -3.22 m, so the runner is submerged to avoid cavitation. (Key point: Setting against cavitation)

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