Problem Statement
A jet of water of diameter 100 mm moving with a velocity of 30 m/s strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate.
Given Data & Constants
- Diameter of jet, \(d = 100 \, \text{mm} = 0.1 \, \text{m}\)
- Velocity of jet, \(V = 30 \, \text{m/s}\)
- Deflection angle = 120°
- Angle of the jet at outlet, \(\theta = 180^\circ - 120^\circ = 60^\circ\)
- Density of water, \(\rho = 1000 \, \text{kg/m}^3\)
Solution
1. Calculate the Area of the Jet (A)
2. Calculate the Force Exerted by the Jet (\(F_x\))
The force exerted by the jet in its original direction is the rate of change of momentum in that direction.
The force exerted by the jet in the direction of the jet is approximately \(10603 \, \text{N}\) or \(10.6 \, \text{kN}\).
Explanation of the Force Calculation
When a jet strikes a curved plate, the plate changes the direction of the water's momentum. According to Newton's Second Law, this change of momentum requires a force. The plate exerts a force on the water to deflect it, and the water exerts an equal and opposite force on the plate.
- Initial Momentum: The water approaches with a velocity \(V\) in the x-direction.
- Final Momentum: Assuming no friction, the water leaves with the same speed \(V\), but at an angle \(\theta\) to the original direction. The component of this velocity in the original x-direction is \(-V \cos\theta\). The negative sign is crucial because this velocity component is now pointing in the opposite direction.
- Change in Velocity: The total change in velocity in the x-direction is \(V - (-V \cos\theta) = V(1 + \cos\theta)\).
- Force: The force is the mass flow rate (\(\rho A V\)) multiplied by this change in velocity, giving the formula \(F_x = \rho A V^2 (1 + \cos\theta)\). This principle is fundamental to how impulse turbines, like the Pelton wheel, generate power.
