A jet of water of diameter 150 mm strikes a flat plate normally with a velocity of 12 m/s. The plate is moving with a velocity of 6 m/s in the direction of the jet and away from the jet. Find : (i) the force exerted by the jet on the plate, (ii) work done by the jet on the plate per second, (iii) power of the jet, and (iv) efficiency of the jet.

Force of a Jet on a Moving Plate

Problem Statement

Given Data & Constants

Diameter of jet, $d = 150 \, \text{mm} = 0.15 \, \text{m}$
Velocity of jet, $V = 12 \, \text{m/s}$
Velocity of plate, $u = 6 \, \text{m/s}$
Density of water, $\rho = 1000 \, \text{kg/m}^3$

Solution

(i) The Force Exerted by the Jet on the Plate ($F$)

The force on a moving plate is determined by the mass of water striking the plate per second and the change in its velocity. The mass striking is based on the relative velocity ($V-u$).

$$ \text{Area of jet, } A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.15)^2 \approx 0.01767 \, \text{m}^2 $$ $$ F = \rho A (V - u)^2 $$ $$ F = 1000 \times 0.01767 \times (12 - 6)^2 $$ $$ F = 1000 \times 0.01767 \times (6)^2 = 1000 \times 0.01767 \times 36 $$ $$ F \approx 636.17 \, \text{N} $$

(ii) Work Done by the Jet on the Plate per Second

Work done per second is the force exerted on the plate multiplied by the velocity of the plate.

$$ \text{Work Done per second} = F \times u $$ $$ \text{Work Done} = 636.17 \, \text{N} \times 6 \, \text{m/s} \approx 3817 \, \text{J/s} \text{ or } \text{W} $$

(iii) Power of the Jet

The power of the jet is its initial kinetic energy per second.

$$ \text{Power of Jet} = \frac{1}{2} (\text{mass flow rate}) V^2 = \frac{1}{2} (\rho A V) V^2 = \frac{1}{2} \rho A V^3 $$ $$ \text{Power} = \frac{1}{2} \times 1000 \times 0.01767 \times (12)^3 $$ $$ \text{Power} = 500 \times 0.01767 \times 1728 \approx 15268 \, \text{W} $$

(iv) Efficiency of the Jet

Efficiency is the ratio of the useful work done on the plate to the initial power available in the jet.

$$ \eta = \frac{\text{Work Done per second}}{\text{Power of Jet}} $$ $$ \eta = \frac{3817}{15268} \approx 0.25 $$

Final Results:

(i) Force exerted by the jet: $ \approx 636.2 \, \text{N} $

(ii) Work done per second: $ \approx 3817 \, \text{W} $ (or 3.82 kW)

(iii) Power of the jet: $ \approx 15268 \, \text{W} $ (or 15.27 kW)

(iv) Efficiency of the jet: $ \approx 25\% $

Explanation of Key Concepts

Force on a Moving Plate: Unlike a stationary plate, the mass of water striking a moving plate depends on the *relative velocity* between the jet and the plate ($V-u$). The change in velocity is also this same relative velocity. This is why the force formula becomes $F = \rho A (V - u)^2$.
Work Done: This is the useful output of the system. It's the force that the jet successfully applies to the plate multiplied by the distance the plate moves per second (its velocity, $u$).
Power of the Jet: This is the total energy available in the jet per second. It represents the maximum possible work that could be extracted under ideal conditions.
Efficiency: This measures how effectively the jet's kinetic energy is converted into useful work on the plate. The efficiency is not 100% because a significant amount of kinetic energy remains in the water as it splashes away from the moving plate.

Problem Statement

Given Data & Constants

Solution

(i) The Force Exerted by the Jet on the Plate (\(F\))

(ii) Work Done by the Jet on the Plate per Second

(iii) Power of the Jet

(iv) Efficiency of the Jet

Explanation of Key Concepts

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Problem Statement

Given Data & Constants

Solution

(i) The Force Exerted by the Jet on the Plate (\(F\))

(ii) Work Done by the Jet on the Plate per Second

(iii) Power of the Jet

(iv) Efficiency of the Jet

Explanation of Key Concepts

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