---
title: "Francis turbine velocity triangles and power"
subject: "Fluid Machines"
chapter: 3
key: "semester_5/fluid_machines:francis-turbine-numerical"
source: https://www.bctengineer.com/q/semester_5/fluid_machines/francis-turbine-numerical
site: BCT Engineering
---

# Francis turbine velocity triangles and power

Fluid Machines · Chapter 3: Water Turbine

- Asked in: year not given, 8 marks
- Video: [Francis turbine-velocity triangle(animation),work done, efficiency, design procedure](https://www.youtube.com/watch?v=IDW3bHq0UTo) — BEST MECHANICAL ENGINEERING
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/fluid_machines/francis-turbine-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-3

## Question

An inward flow Francis turbine works under a net head of 45 m at 400 rpm. The outer and inner diameters of the runner are 1.0 m and 0.5 m, and the width at inlet is 0.12 m. The hydraulic efficiency is 93 % and the water leaves the runner without whirl. The flow velocity is constant at 7 m/s through the runner and the thickness of blades may be neglected. Determine (a) the guide vane angle, (b) the runner vane angles at inlet and outlet, (c) the discharge, (d) the width of the runner at outlet and (e) the power developed by the runner.

## Short answer

u₁ = 20.94 m/s, from η_h = V_w₁ u₁/(gH) → V_w₁ = 19.60 m/s. Guide vane angle α = 19.7°. Runner inlet angle θ = 100.9°, outlet φ = 33.8°. Discharge Q = π D₁ B₁ V_f = 2.639 m³/s; width B₂ = 0.24 m; runner power = ρ Q V_w₁ u₁ = 1083.4 kW.

## Answer

**Given:** $H=45$ m, $N=400$ rpm, $D_1=1.0$ m, $D_2=0.5$ m, $B_1=0.12$ m, $\eta_h=0.93$, $V_{w2}=0$, $V_{f1}=V_{f2}=7$ m/s.

#### Blade speeds

$$
u_1=\frac{\pi D_1N}{60}=\frac{\pi\times1.0\times400}{60}=20.94\ \text{m/s},\qquad
u_2=\frac{\pi\times0.5\times400}{60}=10.47\ \text{m/s}
$$

#### Inlet whirl velocity

For $V_{w2}=0$: $\eta_h=\dfrac{V_{w1}u_1}{gH}$

$$
V_{w1}=\frac{0.93\times9.81\times45}{20.94}=19.60\ \text{m/s}
$$

#### (a) Guide vane angle

$$
\tan\alpha=\frac{V_{f1}}{V_{w1}}=\frac{7}{19.60}\ \Rightarrow\ \alpha=19.7^\circ
$$

#### (b) Runner vane angles

Since $V_{w1}<u_1$, the whirl component of the relative velocity ($V_{w1}-u_1$) is opposite to $u_1$, so the blade angle $\theta$ (measured from the direction of $u_1$) is obtuse, greater than 90 degrees.

$$
\tan(180^\circ-\theta)=\frac{V_{f1}}{u_1-V_{w1}}=\frac{7}{1.34}\ \Rightarrow\ 180^\circ-\theta=79.1^\circ,\ \ \theta=100.9^\circ
$$

At outlet ($V_{w2}=0$):

$$
\tan\phi=\frac{V_{f2}}{u_2}=\frac{7}{10.47}\ \Rightarrow\ \phi=33.8^\circ
$$

#### (c) Discharge

$$
Q=\pi D_1B_1V_{f1}=\pi\times1.0\times0.12\times7=2.639\ \text{m}^3/\text{s}
$$

#### (d) Width at outlet

Continuity: $\pi D_1B_1V_{f1}=\pi D_2B_2V_{f2}$, with $V_{f1}=V_{f2}$:

$$
B_2=\frac{D_1B_1}{D_2}=\frac{1.0\times0.12}{0.5}=0.24\ \text{m}=240\ \text{mm}
$$

#### (e) Power developed by runner

$$
P=\rho Q\,V_{w1}u_1=1000\times2.639\times19.60\times20.94=1083.4\ \text{kW}
$$

Check: water power $\rho gQH=1165.0$ kW, and $1083.4/1165.0=93.0\%=\eta_h$ (satisfied).

**Answer:** (a) $\alpha=19.7^\circ$; (b) $\theta=100.9^\circ$, $\phi=33.8^\circ$; (c) $2.639$ m³/s; (d) 0.24 m; (e) 1083.4 kW.

## Check yourself

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

1. With Vw2 = 0, the inlet whirl is from:
   A. eta_h = Vw1 u1/(gH)
   B. eta_h = Vf1/Vw1
   C. Q = Vw1 A
   D. eta_h = u1/V1
   Answer: A. Vw1 = 0.93 x 9.81 x 45 / 20.94 = 19.60 m/s. (Key point: Whirl velocity)

2. The guide vane angle satisfies:
   A. sin(alpha) = Vw1/Vf1
   B. tan(alpha) = u1/Vw1
   C. tan(alpha) = Vf1/Vw1, giving 19.7 degrees
   D. alpha = 90 degrees
   Answer: C. tan alpha = 7/19.60, so alpha = 19.7 degrees. (Key point: Guide vane angle)

3. The inlet blade angle is obtuse (100.9 degrees) because:
   A. Vw2 = 0
   B. Vf1 = 0
   C. Vw1 < u1
   D. Vw1 > u1
   Answer: C. Vw1 (19.6) < u1 (20.94), so the relative whirl is opposite to u1. (Key point: Inlet blade angle)

4. Outlet width B2 follows from continuity with equal Vf:
   A. B2 = B1
   B. B2 = 0.06 m
   C. B2 = D1 B1/D2 = 0.24 m
   D. B2 = B1 D2/D1
   Answer: C. pi D1 B1 Vf = pi D2 B2 Vf, so B2 = 1.0 x 0.12/0.5 = 0.24 m. (Key point: Outlet width)

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