---
title: "Jet striking fixed and moving curved vanes"
subject: "Fluid Mechanics"
chapter: 4
key: "semester_4/fluid_mechanics_mechanical:jet-on-fixed-and-moving-vane"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/jet-on-fixed-and-moving-vane
site: BCT Engineering
---

# Jet striking fixed and moving curved vanes

Fluid Mechanics · Chapter 4: Basic Equations for Fluid Flow

- Asked in: year not given, 8 marks
- Video: [FORCE EXERTED BY TANGENTIAL JET ON A FIXED CURVED SYMMETRICAL AND UNSYMMETRICAL PLATE](https://www.youtube.com/watch?v=HAU8REIM2vg) — Mech PM
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/jet-on-fixed-and-moving-vane
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-4

## Question

A jet of water 50 mm in diameter and moving at 25 m/s strikes a smooth curved vane which deflects the jet through 165°. Find (a) the force exerted on the vane in the direction of the jet when the vane is fixed, and (b) the force, the work done per second and the efficiency when the vane (a single vane) moves at 10 m/s in the direction of the jet. Assume no friction, so that the relative velocity is the same at inlet and outlet.

## Short answer

For a fixed vane, the force is 2.41 kN. When the vane moves at 10 m/s, the force is 868.5 N, the work done per second is 8.69 kW, and the efficiency is 56.6%.

## Answer

#### Given data
$d = 0.05\ \text{m}$, $V = 25\ \text{m/s}$, deflection angle $\theta = 165^\circ$, $u = 10\ \text{m/s}$.

Jet area: $a = \dfrac{\pi}{4}(0.05)^2 = 1.9635\times10^{-3}\ \text{m}^2$

For a smooth vane with deflection $\theta$, the change of velocity component in the jet direction (relative to the vane) is $V_r(1 - \cos\theta)$.

$$
1 - \cos165^\circ = 1 + 0.9659 = 1.9659
$$

#### (a) Fixed vane
All the jet strikes the vane, $\dot m = \rho a V$ and $V_r = V$:

$$
\begin{aligned}
F &= \rho a V^2 (1 - \cos\theta) \\
&= 1000 \times 1.9635\times10^{-3} \times 625 \times 1.9659 \\
&= 2412.6\ \text{N}
\end{aligned}
$$

#### (b) Moving vane (single vane)
For a single moving vane the mass striking the vane per second is only $\rho a (V - u)$, because the vane is moving away from the nozzle. The relative velocity is $V - u = 15\ \text{m/s}$.

$$
\begin{aligned}
F &= \rho a (V - u)^2 (1 - \cos\theta) \\
&= 1000 \times 1.9635\times10^{-3} \times 15^2 \times 1.9659 \\
&= 868.5\ \text{N}
\end{aligned}
$$

Work done per second (power) = force $\times$ vane velocity:

$$
P = F u = 868.5 \times 10 = 8685\ \text{W} = 8.685\ \text{kW}
$$

Kinetic energy of the jet per second:

$$
\text{KE} = \tfrac12 \rho a V^3 = 0.5 \times 1000 \times 1.9635\times10^{-3} \times 15\,625 = 15\,340\ \text{W}
$$

$$
\eta = \frac{8685}{15\,340} = 0.566 = 56.6\%
$$

**Answer: (a) $F = 2.41\ \text{kN}$; (b) $F = 868.5\ \text{N}$, $P = 8.69\ \text{kW}$, $\eta = 56.6\%$**

## Check yourself

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

1. For a fixed smooth vane deflecting the jet by theta, the force in the jet direction is:
   A. rho a V^2 (1 - cos theta)
   B. rho a V^2 (1 + cos theta)
   C. rho a V (1 - cos theta)
   D. rho a V^2 sin theta
   Answer: A. The change in the jet-direction velocity is V(1 - cos theta), multiplied by mass flow rho a V. (Key point: Force on fixed vane)

2. With d = 50 mm, V = 25 m/s and theta = 165 degrees, the force on the fixed vane is about:
   A. 868 N
   B. 1227 N
   C. 2413 N
   D. 4825 N
   Answer: C. F = 1000 x 1.9635e-3 x 625 x 1.9659 = 2412.6 N. (Key point: Value for fixed vane)

3. For a single moving vane, the mass of water striking it per second is:
   A. rho a V
   B. rho a (V - u)
   C. rho a (V + u)
   D. rho a u
   Answer: B. The vane moves away from the nozzle, so it only intercepts jet water at relative rate (V - u). (Key point: Mass flow for a moving single vane)

4. The efficiency of the moving vane is the work done per second (F u = 8.69 kW) divided by:
   A. the force on the fixed vane
   B. the jet kinetic energy per second, 0.5 rho a V^3 = 15.3 kW
   C. the weight of the jet
   D. 0.5 rho a (V - u)^3
   Answer: B. Efficiency = 8685/15340 = 56.6%, comparing to the full jet kinetic energy rate. (Key point: Efficiency)

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## Related in other subjects

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