A jet of water of diameter 100 mm moving with a velocity of 15 m/s strikes a curved plate at its centre with a velocity of 7 m/s in the direction of the jet. The jet is deflected through an angle of 150°. Assuming the plate smooth find : (i) force exerted on the plate in the direction of the jet, (ii) power of the jet, and (iii) efficiency.

Force of a Jet on a Moving Curved Plate

Problem Statement

A jet of water of diameter 100 mm moving with a velocity of 15 m/s strikes a curved plate at its centre with a velocity of 7 m/s in the direction of the jet. The jet is deflected through an angle of 150°. Assuming the plate smooth find : (i) force exerted on the plate in the direction of the jet, (ii) power of the jet, and (iii) efficiency.

Given Data & Constants

  • Diameter of jet, \(d = 100 \, \text{mm} = 0.1 \, \text{m}\)
  • Velocity of jet, \(V = 15 \, \text{m/s}\)
  • Velocity of plate, \(u = 7 \, \text{m/s}\)
  • Deflection angle = 150°
  • Angle of the jet at outlet, \(\theta = 180^\circ - 150^\circ = 30^\circ\)
  • Density of water, \(\rho = 1000 \, \text{kg/m}^3\)

Solution

(i) Force Exerted on the Plate (\(F_x\))

The force on a moving curved plate is determined by the mass of water striking the plate per second (based on relative velocity) and the change in velocity of the jet.

$$ \text{Area of jet, } A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.1)^2 \approx 0.007854 \, \text{m}^2 $$ $$ F_x = \rho A (V - u) [ (V-u) - (-(V-u)\cos\theta) ] $$ $$ F_x = \rho A (V - u)^2 (1 + \cos\theta) $$ $$ F_x = 1000 \times 0.007854 \times (15 - 7)^2 \times (1 + \cos(30^\circ)) $$ $$ F_x = 1000 \times 0.007854 \times (8)^2 \times (1 + 0.866) $$ $$ F_x = 1000 \times 0.007854 \times 64 \times 1.866 \approx 937.8 \, \text{N} $$

(ii) Power of the Jet (Work Done on Plate)

The power delivered to the plate is the work done per second, which is the force in the direction of motion multiplied by the plate's velocity.

$$ \text{Power} = F_x \times u $$ $$ \text{Power} = 937.8 \, \text{N} \times 7 \, \text{m/s} \approx 6564.6 \, \text{W} $$

(iii) Efficiency

Efficiency is the ratio of the useful power delivered to the plate to the initial kinetic energy of the jet.

$$ \text{Initial Kinetic Energy of Jet per second} = \frac{1}{2} (\rho A V) V^2 = \frac{1}{2} \rho A V^3 $$ $$ KE_{jet} = \frac{1}{2} \times 1000 \times 0.007854 \times (15)^3 \approx 13240 \, \text{W} $$ $$ \eta = \frac{\text{Power}}{\text{Initial KE of Jet}} = \frac{6564.6}{13240} \approx 0.4958 $$
Final Results:

(i) Force exerted on the plate: \( \approx 937.8 \, \text{N} \)

(ii) Power (Work done on the plate): \( \approx 6.56 \, \text{kW} \)

(iii) Efficiency: \( \approx 49.6\% \)

Explanation of Key Concepts

  • Force on a Moving Curved Plate: The force calculation is based on the impulse-momentum principle. The mass of water striking the plate is determined by the relative velocity (\(V-u\)). The change in velocity is also based on this relative velocity being deflected through the angle \(\theta\).
  • Power: This is the useful work done by the jet on the plate. It's the component of force acting in the direction the plate is moving, multiplied by the plate's speed.
  • Efficiency: This measures how effectively the initial kinetic energy of the water jet is converted into useful mechanical work on the plate. The efficiency is less than 100% because some kinetic energy is always lost; the water still has velocity when it leaves the plate.

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