---
title: "Maximum efficiency on series of semicircular vanes"
subject: "Fluid Mechanics"
chapter: 7
key: "semester_3/fluid_mechanics:semicircular-vanes-max-efficiency"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/semicircular-vanes-max-efficiency
site: BCT Engineering
---

# Maximum efficiency on series of semicircular vanes

Fluid Mechanics · Chapter 7: Momentum principle and flow analysis

- Asked in: 2075 Asoj, 5 marks
- Video: [Force Exerted by a Jet of Water on a Series of Vanes](https://www.youtube.com/watch?v=6msfWq-ZUfY) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/semicircular-vanes-max-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-7

## Question

When a jet of fluid strikes series of semicircular vanes, show that the maximum efficiency of the system is 1.

## Short answer

For a series of vanes, mass flow is ρaV. Force is 2ρaV(V-u), work done 2ρaV(V-u)u. Efficiency = 4u(V-u)/V², maximum of 1 (100%) occurs at u=V/2, where exit absolute velocity becomes zero.

## Answer

When a jet strikes a **series** of vanes, the whole jet discharge ($\rho aV$) is used on the moving vanes, because new vanes keep replacing the old ones. The efficiency is the work done on the vanes divided by the kinetic energy supplied by the jet. For semicircular vanes it reaches a maximum of 1 (100%) when the vane speed is $u = V/2$.

#### Setup

Let $V$ = jet velocity, $u$ = vane velocity (same direction), $a$ = jet area. Take the vanes as smooth, so the relative speed is unchanged.

```text
   jet V -->   ____
 ============>/    \   vane moves
              \____/ -> u
```

- Relative velocity at inlet: $V_{r1} = V - u$.
- The semicircular vane turns the jet through $180^\circ$, so the relative velocity at exit is $V_{r2} = V - u$ in the opposite direction.
- Absolute exit velocity along the jet: $V_{2x} = u - (V-u) = 2u - V$.

#### Force and work

Mass flow striking the series of vanes = $\rho a V$.

$$
\begin{aligned} F &= \rho a V\,[V - (2u - V)] = 2\rho a V (V-u) \\ \text{Work done per second} &= F u = 2\rho a V (V-u)\,u \end{aligned}
$$

#### Efficiency

Input power (kinetic energy of the jet per second) $= \tfrac12 \rho a V \cdot V^2$.

$$
\eta = \frac{2\rho a V (V-u)\,u}{\tfrac12 \rho a V^3} = \frac{4u(V-u)}{V^2}
$$

#### Condition for maximum

$$
\frac{d\eta}{du} = \frac{4}{V^2}(V - 2u) = 0 \;\Rightarrow\; u = \frac V2
$$

The second derivative is $-8/V^2 < 0$, so this is a maximum.

$$
\eta_{max} = \frac{4\,(V/2)(V/2)}{V^2} = 1
$$

**Hence the maximum efficiency is 1 (100%)** at $u = V/2$. At this speed the exit absolute velocity is $2u - V = 0$, so the jet leaves with no kinetic energy and all of its energy is given to the vanes.

## Check yourself

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

1. Why is the full jet discharge (rho a V) used for a series of moving vanes?
   A. The vanes move faster than the jet
   B. New vanes keep replacing old ones, so all the jet fluid is intercepted
   C. The jet area increases as the vanes move
   D. The vanes are stationary
   Answer: B. With a series of vanes, no fluid is wasted, so the whole discharge rho a V strikes the vanes. (Key point: Series of vanes: mass flow)

2. For a smooth semicircular vane moving at u in the jet direction, what is the absolute exit velocity along the jet?
   A. 2u - V
   B. V - 2u
   C. V - u
   D. u - V/2
   Answer: A. Relative exit velocity is V - u reversed, so V2x = u - (V - u) = 2u - V. (Key point: Exit velocity)

3. What is the efficiency of the semicircular vane series in terms of u and V?
   A. 2u(V - u)/V^2
   B. u(V - u)/V^2
   C. 4u(V + u)/V^2
   D. 4u(V - u)/V^2
   Answer: D. Work 2 rho a V (V-u) u divided by input 1/2 rho a V^3 gives 4u(V-u)/V^2. (Key point: Efficiency expression)

4. At what vane speed is the efficiency maximum, and what is its value?
   A. u = V, efficiency 1
   B. u = V/2, efficiency 0.5
   C. u = V/2, efficiency 1
   D. u = V/3, efficiency 0.59
   Answer: C. d(eta)/du = 0 gives u = V/2, and eta = 4(V/2)(V/2)/V^2 = 1. (Key point: Condition for maximum)

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