---
title: "Francis runner design (40 m head, 10 m³/s)"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:francis-turbine-velocity-triangle-40m"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/francis-turbine-velocity-triangle-40m
site: BCT Engineering
---

# Francis runner design (40 m head, 10 m³/s)

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2072 Chaitra, 2+2+1+2+1+4 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-turbine-velocity-triangle-40m
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

A Francis turbine works under a head of 40 m and discharge Q = 10 m³/s. The speed of the runner is 300 rpm. At the inlet tip of the runner vane, the speed ratio is $K_u = 0.85$ and flow ratio $K_f = 0.3$. If the overall efficiency and hydraulic efficiency of the turbine are 80% and 90% respectively, assume discharge at the outlet is radial and velocity of flow is constant. Determine: a) power developed in kW; b) diameter and width of runner at inlet; c) guide vane angle at inlet; d) specific speed of turbine; e) diameter of runner at outlet. Dimension suitably the powerhouse (length, breadth and height) with a sketch, if three such turbines were used in a power plant. Assume suitably any requirements for calculations.

## Short answer

Power = 0.8×9.81×10×40 = 3139 kW. Inlet diameter 1.52 m, width 0.25 m, guide vane angle 29.5°, specific speed 167.1, outlet diameter 1.23 m. For three units powerhouse dimensions approx. 25.5 m × 13 m × 19 m.

## Answer

#### Data
$H = 40$ m, $Q = 10\ \text{m}^3/\text{s}$, $N = 300$ rpm, $K_u = 0.85$, $K_f = 0.3$, $\eta_o = 0.80$, $\eta_h = 0.90$. Radial discharge at outlet: $V_{w2} = 0$, and $V_{f2} = V_{f1}$.

$\sqrt{2gH} = \sqrt{2 \times 9.81 \times 40} = 28.01$ m/s.

#### a) Power developed
$$
P = \eta_o\gamma Q H = 0.8 \times 9.81 \times 10 \times 40 = 3139\ \text{kW}
$$

#### b) Inlet diameter and width
$u_1 = K_u\sqrt{2gH} = 0.85 \times 28.01 = 23.81$ m/s

$$
D_1 = \frac{60u_1}{\pi N} = \frac{60 \times 23.81}{\pi \times 300} = 1.516\ \text{m}
$$

$V_{f1} = K_f\sqrt{2gH} = 0.3 \times 28.01 = 8.40$ m/s

$$
B_1 = \frac{Q}{\pi D_1 V_{f1}} = \frac{10}{\pi \times 1.516 \times 8.40} = 0.25\ \text{m}
$$

(vane thickness neglected)

#### c) Guide vane angle at inlet
From $\eta_h = \dfrac{V_{w1}u_1}{gH}$:

$$
V_{w1} = \frac{0.9 \times 9.81 \times 40}{23.81} = 14.83\ \text{m/s}
$$

$$
\tan\alpha = \frac{V_{f1}}{V_{w1}} = \frac{8.40}{14.83} \Rightarrow \alpha = 29.5^\circ
$$

Runner vane angle at inlet: since $V_{w1} < u_1$, $\tan\phi = \dfrac{V_{f1}}{u_1 - V_{w1}} = \dfrac{8.40}{8.98}$, so $\phi = 43.1^\circ$ with the backward tangent. Measured from the direction of motion of the wheel, the vane angle is $\theta = 180^\circ - 43.1^\circ = 136.9^\circ$ (an obtuse, backward-curved vane).

#### d) Specific speed
$$
N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{300\sqrt{3139}}{40^{1.25}} = 167.1
$$

#### e) Outlet diameter
The flow velocity is constant, $V_{f2} = 8.40$ m/s, and the water leaves without whirl in the axial direction, so the outlet flow area is a full circle:

$$
\frac{\pi}{4}D_2^2 = \frac{Q}{V_{f2}} \Rightarrow D_2 = \sqrt{\frac{4 \times 10}{\pi \times 8.40}} = 1.23\ \text{m}
$$

#### Powerhouse dimensions for three units (assumed thumb rules)
Each unit: 3.14 MW, $D_1 = 1.52$ m. Assume a vertical shaft and a spiral casing outer diameter of about $2.7D_1 \approx 4.1$ m.

| Item | Rule used | Value |
|---|---|---|
| Unit bay (centre to centre) | $\approx 3.6D_1$ | 5.5 m |
| Erection (service) bay | one unit bay + | 7 m |
| End clearance | | 2 m |
| **Length** | $3 \times 5.5 + 7 + 2$ | **about 25.5 m** |
| **Breadth** | generator + valve/gallery + control side | **about 13 m** |
| Draft tube depth | $\approx 2.5D_1$ | 3.8 m |
| Distributor to generator floor | | 4.5 m |
| Generator floor to crane rail | generator height + lifting clearance | 8.5 m |
| **Height** (floor to roof truss) | 3.8 + 4.5 + 8.5 + 2 (roof) | **about 19 m** |

```text
  PLAN (3 units + erection bay)        Section
 +--------+--------+--------+-----+      roof
 | unit 1 | unit 2 | unit 3 | EB  |    ==========  crane rail
 |  (G)   |  (G)   |  (G)   |     |     |  G  |  generator floor
 +--------+--------+--------+-----+     | TURB|  turbine floor
  25.5 m x 13 m                          \DT /  draft tube
```

**Answer: $P = 3139$ kW; $D_1 = 1.52$ m, $B_1 = 0.25$ m; $\alpha = 29.5^\circ$; $N_s = 167.1$; $D_2 = 1.23$ m; powerhouse about 25.5 m x 13 m x 19 m.**

## Check yourself

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

1. How is the power developed obtained?
   A. P = eta_o Q H
   B. P = gamma Q H
   C. P = eta_h gamma Q H
   D. P = eta_o gamma Q H = 3139 kW
   Answer: D. The overall efficiency applies to shaft power: 0.8 x 9.81 x 10 x 40 = 3139 kW. (Key point: Power developed)

2. How is the inlet width B1 found?
   A. B1 = Q/(pi D1 Vf1) = 0.25 m
   B. B1 = Q/(pi D1^2)
   C. B1 = Q/u1
   D. B1 = D1/6
   Answer: A. Discharge passes through the inlet cylinder area pi D1 B1 at flow velocity Vf1 = 8.40 m/s. (Key point: Inlet width)

3. How is the inlet whirl velocity Vw1 found?
   A. From hydraulic efficiency: Vw1 = eta_h g H/u1 = 14.83 m/s
   B. From Vw1 = u1
   C. From overall efficiency
   D. From Kf only
   Answer: A. With radial outlet, eta_h = Vw1 u1/(gH), so Vw1 = 0.9 x 9.81 x 40/23.81. (Key point: Whirl velocity)

4. How is the guide vane angle found?
   A. tan(alpha) = Vf1/Vw1, giving 29.5 degrees
   B. sin(alpha) = Kf
   C. tan(alpha) = u1/Vw1
   D. tan(alpha) = Vw1/Vf1
   Answer: A. Guide vane angle is between the absolute velocity and the tangent: tan(alpha) = 8.40/14.83. (Key point: Guide vane angle)

## Related in Hydropower Engineering

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- [Francis overall efficiency with draft tube](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-draft-tube-efficiency-32mw) — Hydropower Engineering, chapter 7
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