---
title: "Kaplan turbine discharge and blade angles"
subject: "Fluid Machines"
chapter: 3
key: "semester_5/fluid_machines:kaplan-turbine-numerical"
source: https://www.bctengineer.com/q/semester_5/fluid_machines/kaplan-turbine-numerical
site: BCT Engineering
---

# Kaplan turbine discharge and blade angles

Fluid Machines · Chapter 3: Water Turbine

- Asked in: year not given, 8 marks
- Video: [Problem on Kaplan turbine  / Fluid mechanics](https://www.youtube.com/watch?v=IdbLi3iA1XE) — Edutech Guru Engineering Learning-By Priyanka ma'am
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/fluid_machines/kaplan-turbine-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-3

## Question

A Kaplan turbine develops 9000 kW of shaft power under a net head of 20 m. The overall efficiency is 88 % and the hydraulic efficiency is 93 %. The runner has an outer diameter of 4.0 m and a hub diameter of 1.8 m, and runs at 120 rpm. Assuming that the flow velocity is uniform over the annulus and the water leaves without whirl, calculate the discharge, the flow velocity, and the guide vane angle and runner blade angles at inlet and outlet at the mean diameter of the runner.

## Short answer

Discharge Q = P/(ρgHηo) = 52.13 m³/s. Flow velocity Vf = Q/A = 5.201 m/s. Guide vane angle α = arctan(Vf/Vw1) = 27.4°. Runner blade inlet angle θ = 147.6°, outlet angle φ = 15.9°.

## Answer

**Given:** $P=9000$ kW, $H=20$ m, $\eta_o=0.88$, $\eta_h=0.93$, $D_o=4.0$ m, $D_b=1.8$ m, $N=120$ rpm, $V_{w2}=0$, $u_1=u_2$ (axial flow).

#### Discharge

$$
Q=\frac{P}{\rho g H\eta_o}=\frac{9\times10^{6}}{1000\times9.81\times20\times0.88}=52.13\ \text{m}^3/\text{s}
$$

#### Flow velocity

Annular flow area:

$$
A=\frac{\pi}{4}(D_o^2-D_b^2)=\frac{\pi}{4}(4.0^2-1.8^2)=10.022\ \text{m}^2
$$

$$
V_{f1}=V_{f2}=\frac{Q}{A}=\frac{52.13}{10.022}=5.201\ \text{m/s}
$$

#### Blade speed at mean diameter

$D_m=\dfrac{4.0+1.8}{2}=2.90$ m

$$
u_1=u_2=\frac{\pi D_mN}{60}=\frac{\pi\times2.90\times120}{60}=18.22\ \text{m/s}
$$

#### Whirl velocity at inlet

$$
\eta_h=\frac{V_{w1}u_1}{gH}\ \Rightarrow\ V_{w1}=\frac{0.93\times9.81\times20}{18.22}=10.014\ \text{m/s}
$$

#### Angles

Guide vane angle:

$$
\tan\alpha=\frac{V_{f1}}{V_{w1}}=\frac{5.201}{10.014}\ \Rightarrow\ \alpha=27.4^\circ
$$

Runner blade angle at inlet. Here $V_{w1}<u_1$, so the blade angle is obtuse:

$$
\tan(180^\circ-\theta)=\frac{V_{f1}}{u_1-V_{w1}}=\frac{5.201}{8.21}\ \Rightarrow\ 180^\circ-\theta=32.4^\circ,\ \ \theta=147.6^\circ
$$

Runner blade angle at outlet:

$$
\tan\phi=\frac{V_{f2}}{u_2}=\frac{5.201}{18.22}\ \Rightarrow\ \phi=15.9^\circ
$$

(Angles of blades measured from the direction of blade motion. Because $u$ changes from hub to tip, the blade is twisted; the angles at the other radii are found in the same way with the local $u$.)

**Answer:** $Q=52.13$ m³/s; $V_f=5.201$ m/s; guide vane angle $27.4^\circ$; blade angle $147.6^\circ$ at inlet and $15.9^\circ$ at outlet.

## Check yourself

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

1. Discharge is from:
   A. Q = A u
   B. Q = P/(rho g H eta_o) = 52.13 m3/s
   C. Q = P/(rho H)
   D. Q = P/(rho g H eta_h)
   Answer: B. Overall efficiency is used since P is shaft power. (Key point: Discharge)

2. Flow velocity uses annular area:
   A. pi Dm
   B. pi/4 Do^2
   C. pi/4 (Do^2 - Db^2) = 10.02 m2
   D. pi/4 Db^2
   Answer: C. Vf = 52.13/10.022 = 5.20 m/s. (Key point: Flow area)

3. Blade speed is calculated at:
   A. Hub diameter
   B. Tip diameter
   C. Draft tube
   D. Mean diameter 2.9 m, u = 18.22 m/s
   Answer: D. u = pi x 2.9 x 120/60 = 18.22 m/s. (Key point: Blade speed)

4. Inlet whirl uses which efficiency?
   A. Mechanical
   B. Hydraulic: Vw1 = eta_h g H/u1 = 10.01 m/s
   C. None
   D. Overall
   Answer: B. With Vw2 = 0, eta_h = Vw1 u1/(gH). (Key point: Whirl)

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