A jet of water of the diameter 100 mm moving with a velocity of 20 m/s strikes a curved fixed plate tangentially at one end at an angle of 30° to the horizontal. The jet leaves the plate at an angle of 20° to the horizontal. Find the force exerted by the jet on the plate in the horizontal and vertical directions.

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Force of a Jet on an Un-symmetrical Curved Plate

Problem Statement

A jet of water of the diameter 100 mm moving with a velocity of 20 m/s strikes a curved fixed plate tangentially at one end at an angle of 30° to the horizontal. The jet leaves the plate at an angle of 20° to the horizontal. Find the force exerted by the jet on the plate in the horizontal and vertical directions.

Given Data & Constants

  • Diameter of jet, \(d = 100 \, \text{mm} = 0.1 \, \text{m}\)
  • Velocity of jet, \(V = 20 \, \text{m/s}\)
  • Inlet angle, \(\theta = 30^\circ\)
  • Outlet angle, \(\phi = 20^\circ\)
  • Density of water, \(\rho = 1000 \, \text{kg/m}^3\)

Solution

1. Calculate Area and Mass Flow Rate

$$ \text{Area of jet, } A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.1)^2 \approx 0.007854 \, \text{m}^2 $$ $$ \text{Mass flow rate, } \dot{m} = \rho A V = 1000 \times 0.007854 \times 20 = 157.08 \, \text{kg/s} $$

2. Resolve Velocities into Components

We resolve the initial and final velocities into horizontal (x) and vertical (y) components. The curved plate reverses the direction of the horizontal velocity component.

$$ \text{Initial x-velocity, } V_{1x} = V \cos(\theta) = 20 \times \cos(30^\circ) \approx 17.32 \, \text{m/s} $$ $$ \text{Initial y-velocity, } V_{1y} = V \sin(\theta) = 20 \times \sin(30^\circ) = 10.0 \, \text{m/s} $$ $$ \text{Final x-velocity, } V_{2x} = -V \cos(\phi) = -20 \times \cos(20^\circ) \approx -18.79 \, \text{m/s} $$ $$ \text{Final y-velocity, } V_{2y} = V \sin(\phi) = 20 \times \sin(20^\circ) \approx 6.84 \, \text{m/s} $$

3. Calculate Force in Horizontal Direction (\(F_x\))

The force on the plate is the rate of change of momentum of the jet.

$$ F_x = \dot{m} (V_{1x} - V_{2x}) $$ $$ F_x = 157.08 \times (17.32 - (-18.79)) = 157.08 \times (17.32 + 18.79) $$ $$ F_x = 157.08 \times (36.11) \approx 5672.3 \, \text{N} $$

4. Calculate Force in Vertical Direction (\(F_y\))

$$ F_y = \dot{m} (V_{1y} - V_{2y}) $$ $$ F_y = 157.08 \times (10.0 - 6.84) = 157.08 \times (3.16) $$ $$ F_y \approx 496.3 \, \text{N} $$
Final Results:

Force in the horizontal direction: \( F_x \approx 5672.3 \, \text{N} \)

Force in the vertical direction: \( F_y \approx 496.3 \, \text{N} \)

Explanation of the Force Components

The forces are calculated based on the change in the jet's momentum in the horizontal (x) and vertical (y) directions.

  • Horizontal Force (\(F_x\)): The result is positive (5672.3 N). This is because the plate causes a large reversal in the horizontal momentum of the water (from +17.32 m/s to -18.79 m/s). To cause this change, the plate must exert a large force on the water to the left. By Newton's third law, the water exerts an equal and opposite force on the plate to the right.
  • Vertical Force (\(F_y\)): The result is positive (496.3 N), which means the force exerted by the jet on the plate is in the positive y-direction (upwards). This makes sense because the plate reduces the upward velocity of the water (from 10 m/s to 6.84 m/s), meaning the plate must exert a downward force on the water. By Newton's third law, the water exerts an equal and opposite (upward) force on the plate.

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