---
title: "Parsons stage: work, power and efficiency"
subject: "Turbo Machines"
chapter: 2
key: "semester_7/turbo_machines:parsons-stage-numerical"
source: https://www.bctengineer.com/q/semester_7/turbo_machines/parsons-stage-numerical
site: BCT Engineering
---

# Parsons stage: work, power and efficiency

Turbo Machines · Chapter 2: Velocity Vector Diagram

- Asked in: year not given, 6 marks
- Video: [Problem 5 based on reaction turbine (parson's turbine) ,unit=2,TURBO](https://www.youtube.com/watch?v=i98HAcNbMbU) — EduGrown
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/turbo_machines/parsons-stage-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-2

## Question

A stage of a Parsons turbine has equal fixed and moving blade angles of 20 degrees at outlet. The mean blade speed is 200 m/s and the blade speed ratio $U/C_1$ is 0.7. For a steam flow of 12 kg/s, find (a) the inlet blade angle of the moving blade, (b) the work done per kg and the power developed, (c) the enthalpy drop supplied to the stage and (d) the blade efficiency.

## Short answer

From inlet triangle: C1=285.7 m/s, Cw1=268.5 m/s, Vw1=68.5 m/s, tanβ1=Ca/Vw1 gives β1=55.0°. Work w = U(2C1 cosα1 - U) = 67.4 kJ/kg, power 809 kW. Stage enthalpy drop Δh = w + C2^2/2 = 74.5 kJ/kg. Blade efficiency ηb = 90.4%.

## Answer

**Data:** $\alpha_1 = \beta_2 = 20^\circ$, $U = 200$ m/s, $\rho = U/C_1 = 0.7$, $\dot m = 12$ kg/s. For a Parsons stage: $V_{r2} = C_1$, $C_2 = V_{r1}$, $\beta_1 = \alpha_2$.

#### (a) Inlet velocity triangle
$$
\begin{aligned}
C_1 &= U/0.7 = 285.7\ \text{m/s} \\
C_{w1} &= C_1\cos 20^\circ = 268.5\ \text{m/s}, \quad C_{a} = C_1\sin 20^\circ = 97.7\ \text{m/s} \\
V_{w1} &= C_{w1} - U = 68.5\ \text{m/s} \\
\tan\beta_1 &= C_a/V_{w1} \Rightarrow \beta_1 = 55.0^\circ
\end{aligned}
$$
$$
V_{r1} = \sqrt{68.5^2 + 97.7^2} = 119.3\ \text{m/s} = C_2
$$

#### (b) Work and power
$$
w = U(2C_1\cos\alpha_1 - U) = 200 \times (2 \times 268.5 - 200) = 67.4\ \text{kJ/kg}
$$
$$
P = 12 \times 67.4 = 809\ \text{kW}
$$

#### (c) Energy supplied (entry velocity neglected)
$$
\frac{C_2^2}{2} = \frac{119.3^2}{2} = 7.12\ \text{kJ/kg}
$$
$$
\Delta h = w + \frac{C_2^2}{2} = 67.4 + 7.12 = 74.5\ \text{kJ/kg}
$$

#### (d) Blade efficiency
$$
\eta_b = \frac{w}{\Delta h} = \frac{67.4}{74.5} = 0.904
$$

Check with the formula $\eta_b = \dfrac{2\rho(2\cos\alpha_1 - \rho)}{1 + 2\rho\cos\alpha_1 - \rho^2} = \dfrac{1.4 \times 1.1794}{1.8256} = 0.904$.

**Answer:** $\beta_1 = 55.0^\circ$; $w = 67.4$ kJ/kg; $P = 809$ kW; $\Delta h = 74.5$ kJ/kg; $\eta_b = 90.4\%$.

## Check yourself

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

1. How is the steam velocity C1 found in this problem?
   A. C1 = U x 0.7
   B. C1 = U/0.7 = 285.7 m/s
   C. C1 = U cos 20
   D. C1 = U + 0.7
   Answer: B. The blade speed ratio U/C1 = 0.7 gives C1 = 200/0.7 = 285.7 m/s. (Key point: Finding C1)

2. Which formula gives the work per kg for the Parsons stage?
   A. U (C1 cos a1 - U)
   B. C1^2/2
   C. U (2 C1 cos a1 - U)
   D. U C1
   Answer: C. w = U(Cw1 + Vw2 - U) = U(2 C1 cos a1 - U) = 67.4 kJ/kg. (Key point: Work per kg)

3. What is the enthalpy drop supplied to the stage?
   A. 67.4 kJ/kg
   B. 7.12 kJ/kg
   C. 40.8 kJ/kg
   D. 74.5 kJ/kg
   Answer: D. Supply = work + exit kinetic energy = 67.4 + C2^2/2 = 67.4 + 7.12 = 74.5 kJ/kg. (Key point: Energy supplied)

4. Blade efficiency is work done divided by:
   A. Energy supplied (work plus exit kinetic energy)
   B. Inlet kinetic energy only
   C. Exit kinetic energy
   D. Rotor enthalpy drop only
   Answer: A. eta_b = 67.4/74.5 = 0.904. (Key point: Blade efficiency)

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