---
title: "Force of jet on symmetrical curved vane"
subject: "Fluid Machines"
chapter: 1
key: "semester_5/fluid_machines:jet-on-curved-vane-numerical"
source: https://www.bctengineer.com/q/semester_5/fluid_machines/jet-on-curved-vane-numerical
site: BCT Engineering
---

# Force of jet on symmetrical curved vane

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

- Asked in: year not given, 6 marks
- Video: [FLUID MECHANICS: FORCE EXERTED BY JET ON A FIXED CURVED SYMMETRICAL PLATE](https://www.youtube.com/watch?v=y46dSDVodbU) — Mech PM
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/fluid_machines/jet-on-curved-vane-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-1

## Question

A jet of water 60 mm in diameter and moving at 30 m/s strikes at the centre of a symmetrical curved vane which deflects the jet through 160 degrees (each half leaves at 20 degrees to the original jet direction, back towards the source). Neglecting friction, find (a) the force exerted on the vane in the direction of the jet when it is fixed, and (b) the force and power developed when the vane moves at 10 m/s in the direction of the jet.

## Short answer

(a) Fixed vane: F = ρaV²(1+cos20°) = 4936 N. (b) Moving vane: relative velocity (V-u)=20 m/s, mass flow ρa(V-u), F = ρa(V-u)²(1+cos20°) = 2194 N, power = Fu = 21.94 kW.

## Answer

**Given:** $d=0.06$ m, $V=30$ m/s, deflection $\theta=160^\circ$, so the exit angle with the original jet direction is $180^\circ-20^\circ$ and the exit velocity component along the jet is $-V\cos20^\circ$ (opposite to the jet).

Jet area $a=\frac{\pi}{4}(0.06)^2=2.827\times10^{-3}$ m².

#### (a) Fixed vane

The vane is symmetrical and the jet splits into two equal halves, which leave with velocity $V$ (no friction, no change in speed). Change of velocity along the jet per kg: $V - (-V\cos20^\circ) = V(1+\cos20^\circ)$.

$$
\begin{aligned}
F &= \rho a V^2 (1+\cos 20^\circ)\\
&= 1000\times2.827\times10^{-3}\times 30^2\times(1+0.9397)\\
&= 4936\ \text{N}
\end{aligned}
$$

#### (b) Vane moving at 10 m/s in the jet direction

A single vane moving away receives mass flow $\rho a (V-u)$ per second, and the relative speed $(V-u)=20$ m/s is unchanged along the vane (no friction). Relative to the vane the jet is turned in the same way, so

$$
\begin{aligned}
F &= \rho a (V-u)^2 (1+\cos20^\circ)\\
&= 1000\times2.827\times10^{-3}\times 20^2\times(1+0.9397)\\
&= 2194\ \text{N}\\
P &= F\,u = 2194\times10 = 21937\ \text{W} = 21.94\ \text{kW}
\end{aligned}
$$

**Answer:** (a) 4936 N (fixed vane); (b) 2194 N and 21.94 kW (moving vane).

## Check yourself

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

1. For the fixed symmetrical vane, the change of velocity along the jet per kg is:
   A. V cos20
   B. V(1 + cos20)
   C. V(1 - cos20)
   D. 2V
   Answer: B. Exit component is -V cos20 (backwards), so change = V + V cos20 = V(1+cos20). (Key point: Velocity change)

2. For the moving vane (single), which velocity is used in the force?
   A. Relative velocity V-u = 20 m/s
   B. V = 30 m/s
   C. V+u = 40 m/s
   D. u = 10 m/s
   Answer: A. The mass flow and relative speed are both based on (V-u) = 20 m/s: F = rho a (V-u)^2 (1+cos20) = 2194 N. (Key point: Moving vane flow)

3. The force on the fixed vane in the answer is about:
   A. 2194 N
   B. 4936 N
   C. 21.9 kN
   D. 1227 N
   Answer: B. F = 1000 x 2.827e-3 x 900 x 1.9397 = 4936 N. (Key point: Fixed vane force)

4. The power developed on the moving vane is:
   A. Force times (V-u)
   B. Half of jet kinetic energy
   C. Force times vane velocity u
   D. Force times jet velocity
   Answer: C. P = F u = 2194 x 10 = 21.94 kW. (Key point: Power)

## Related in Fluid Machines

- [Jet on series of curved vanes: power and efficiency](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Aseries-of-moving-vanes-numerical) — Fluid Machines, chapter 1
- [Force of jet on flat plate, stationary and moving](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Ajet-on-flat-plate-numerical) — Fluid Machines, chapter 1

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