---
title: "Jet on series of curved vanes: power and efficiency"
subject: "Fluid Machines"
chapter: 1
key: "semester_5/fluid_machines:series-of-moving-vanes-numerical"
source: https://www.bctengineer.com/q/semester_5/fluid_machines/series-of-moving-vanes-numerical
site: BCT Engineering
---

# Jet on series of curved vanes: power and efficiency

Fluid Machines · Chapter 1: Introduction of Turbomachime and Dynamic Action of Fluid

- Asked in: year not given, 8 marks
- Video: [Problem-1 on force exerted by jet on series of curved vanes](https://www.youtube.com/watch?v=ARLYFOK5nXw) — mona yadav
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/fluid_machines/series-of-moving-vanes-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-1

## Question

A jet of water of 50 mm diameter and velocity 40 m/s strikes a series of curved vanes mounted on a wheel and moving at 15 m/s in the direction of the jet. The jet enters tangentially at the inlet tip of the vanes, and the vane tip angle at outlet is 20 degrees to the direction of motion. Assuming no friction loss, calculate the force on the vanes, the power developed and the efficiency of the wheel.

## Short answer

Inlet: Vw1=40 m/s, Vr1=25 m/s. Outlet: Vr2=25 m/s, Vw2 = u - Vr2 cos20° = -8.49 m/s. Force F = ρaV(Vw1 - Vw2) = 3809 N. Power = Fu = 57.13 kW. Efficiency = 90.9%.

## Answer

**Given:** $d=50$ mm, $V=V_1=40$ m/s, $u=15$ m/s (same at inlet and outlet), outlet vane angle $\phi=20^\circ$, no friction so $V_{r2}=V_{r1}$.

Jet area $a=\frac{\pi}{4}(0.05)^2=1.9635\times10^{-3}$ m². Because the vanes form a series on a wheel, the **entire** jet flow is used: $m=\rho a V=78.54$ kg/s.

#### Inlet triangle

The jet enters tangentially, so $V_{w1}=V=40$ m/s and $V_{r1}=V-u=25$ m/s.

#### Outlet triangle

```text
 Outlet (vane moves right, water leaves backwards)

   Vr2 = 25 m/s
        \
         \ 20 deg
   <------*-------------> u = 15 m/s
   Vw2 = u - Vr2 cos20  (negative = backwards)
```

$V_{r2}=V_{r1}=25$ m/s. Component of $V_{r2}$ opposite to motion: $V_{r2}\cos20^\circ=23.49$ m/s.

$$
V_{w2} = u - V_{r2}\cos\phi = 15 - 23.49 = -8.49\ \text{m/s}
$$

The negative sign means the exit whirl is opposite to the direction of the vane. The change in whirl velocity is $V_{w1} - V_{w2} = 40 + 8.49 = 48.49$ m/s.

#### Force, power, efficiency

$$
\begin{aligned}
F &= \rho a V\,(V_{w1}-V_{w2}) = 78.54\times48.49 = 3809\ \text{N}\\
P &= F u = 3809\times15 = 57129\ \text{W} = 57.13\ \text{kW}\\
\text{Jet power} &= \tfrac12\rho a V^3 = 62.83\ \text{kW}\\
\eta &= \frac{P}{\text{jet power}} = \frac{57.13}{62.83} = 90.9\%
\end{aligned}
$$

Check with the formula $\eta=\dfrac{2u(V-u)(1+\cos\phi)}{V^2}=\dfrac{2\times15\times25\times(1+0.9397)}{1600}=90.9\%$, which agrees.

The remaining energy is lost as the kinetic energy of the water leaving the vanes, with $V_2=\sqrt{V_{w2}^2+V_{f2}^2}=12.05$ m/s.

**Answer:** Force $=3809$ N, power $=57.13$ kW, efficiency $=90.9\%$.

## Check yourself

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

1. For a series of vanes on a wheel, which mass flow is used?
   A. rho a (V+u)
   B. rho a (V-u)
   C. rho a u
   D. rho a V, the entire jet
   Answer: D. The whole jet is used: m = rho a V = 78.54 kg/s. (Key point: Mass flow)

2. The relative velocity at inlet is:
   A. V - u = 25 m/s
   B. V = 40 m/s
   C. u = 15 m/s
   D. V + u = 55 m/s
   Answer: A. The jet enters tangentially: Vr1 = 40 - 15 = 25 m/s, and Vr2 = Vr1 without friction. (Key point: Inlet relative velocity)

3. Exit whirl Vw2 = u - Vr2 cos(20) = -8.49 m/s. The negative sign means:
   A. Exit whirl is opposite to the vane motion
   B. Water leaves with zero speed
   C. Efficiency is negative
   D. Flow reverses at inlet
   Answer: A. Negative means the exit whirl is opposite to the direction of blade motion, which increases (Vw1 - Vw2). (Key point: Exit whirl)

4. The force and efficiency are:
   A. 3809 N and 90.9%
   B. 1227 N and 29.6%
   C. 2194 N and 43.5%
   D. 3809 N and 50%
   Answer: A. F = 78.54 x 48.49 = 3809 N; P = 57.13 kW; jet power 62.83 kW; eta = 90.9%. (Key point: Force and efficiency)

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