---
title: "Jet on flat plate, force and efficiency"
subject: "Fluid Mechanics"
chapter: 7
key: "semester_3/fluid_mechanics:jet-on-flat-plate-efficiency"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/jet-on-flat-plate-efficiency
site: BCT Engineering
---

# Jet on flat plate, force and efficiency

Fluid Mechanics · Chapter 7: Momentum principle and flow analysis

- Asked in: 2075 Chaitra, 3+3+4 marks
- Video: [Impact of Jet on  horizontal flat plate](https://www.youtube.com/watch?v=K8C3BSB5XPE) — NPTEL - Special Lecture Series
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/jet-on-flat-plate-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-7

## Question

A jet of water with a velocity U and jet area A strikes a flat plate normal to it. Determine the force of impingement, power developed and efficiency (i) when the plate is at rest. (ii) when the plate is permitted to move along the direction of a velocity u. Also determine condition of maximum possible efficiency. (iii) what would be the possible maximum efficiency if series of plates were to face the jet in quick succession?

## Short answer

Stationary plate: force ρAU², zero power, efficiency 0. Moving plate: force ρA(U-u)², efficiency 2u(U-u)²/U³, max 8/27 (29.6%) at u=U/3. Series of plates gives maximum efficiency 50% at u=U/2.

## Answer

**Notation:** jet velocity $U$, area $A$, density $\rho$. The kinetic power available in the jet is $\dfrac{1}{2}\rho AU^3$.

#### (i) Plate at rest

$$
\begin{aligned}
F &= \rho AU^2\\
\text{Power developed} &= F\times 0 = 0\\
\eta &= 0
\end{aligned}
$$

A stationary plate receives a force but does no work, so its efficiency is zero (the jet's energy is carried away as kinetic energy of the deflected sheet).

#### (ii) Plate moving with velocity u in the direction of the jet

Only the relative velocity $(U - u)$ hits the plate, and the mass flow striking the plate is $\rho A(U - u)$:

$$
\begin{aligned}
F &= \rho A(U - u)^2\\
P &= Fu = \rho A(U - u)^2u\\
\eta &= \frac{P}{\frac{1}{2}\rho AU^3} = \frac{2u(U - u)^2}{U^3}
\end{aligned}
$$

**Condition for maximum efficiency:**

$$
\frac{d\eta}{du} = 0\;\Rightarrow\; (U - u)^2 - 2u(U - u) = 0\;\Rightarrow\; U - 3u = 0\;\Rightarrow\; u = \frac{U}{3}
$$

$$
\eta_{max} = \frac{2(U/3)(2U/3)^2}{U^3} = \frac{8}{27} = 29.6\ \%
$$

#### (iii) Series of plates (e.g. wheel with many vanes)

With plates arriving in quick succession, the whole jet discharge $\rho AU$ is used, not just $\rho A(U - u)$:

$$
\begin{aligned}
F &= \rho AU(U - u)\\
P &= Fu = \rho AU(U - u)u\\
\eta &= \frac{2u(U - u)}{U^2}
\end{aligned}
$$

$$
\frac{d\eta}{du} = 0\;\Rightarrow\; U - 2u = 0\;\Rightarrow\; u = \frac{U}{2},\qquad \eta_{max} = \frac{2(U/2)(U/2)}{U^2} = 0.5 = 50\ \%
$$

| Case | Force | Max efficiency | At |
|---|---|---|---|
| Fixed plate | $\rho AU^2$ | 0 | all |
| Single moving plate | $\rho A(U-u)^2$ | 29.6 % | $u = U/3$ |
| Series of plates | $\rho AU(U-u)$ | 50 % | $u = U/2$ |

**Answer: (i) $F = \rho AU^2$, $P = 0$, $\eta = 0$; (ii) $F = \rho A(U-u)^2$, $\eta = 2u(U-u)^2/U^3$, maximum $8/27 = 29.6\%$ at $u = U/3$; (iii) $\eta_{max} = 50\%$ at $u = U/2$.**

## Check yourself

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

1. Why is the efficiency zero for a fixed plate?
   A. The force is zero
   B. Force exists but the plate does no work as it does not move
   C. Velocity is zero
   D. The jet is absorbed
   Answer: B. Power = F x 0 = 0. (Key point: Fixed plate)

2. What is the force on a single moving plate?
   A. rho A U^2
   B. rho A U (U - u)
   C. rho A (U - u)^2
   D. rho A u^2
   Answer: C. Only the relative velocity (U - u) hits the plate and the mass flow is rho A (U - u). (Key point: Moving single plate force)

3. At what plate speed and efficiency is the single-plate maximum?
   A. u = U/3, 29.6%
   B. u = U/2, 50%
   C. u = U/2, 29.6%
   D. u = U/3, 50%
   Answer: A. Setting d(eta)/du = 0 gives u = U/3 and eta = 8/27. (Key point: Max efficiency single plate)

4. Why does a series of plates reach 50%?
   A. The plate speed is zero
   B. The jet is reflected
   C. Friction is removed
   D. The entire jet discharge rho A U is used, giving eta = 2u(U-u)/U^2 with max at u = U/2
   Answer: D. With quick succession, mass flow striking the plates is rho A U not rho A (U - u). (Key point: Series of plates)

## Related in Fluid Mechanics

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## Related in other subjects

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