A jet of water 60 mm in diameter with a velocity of 15m/s strikes a flat plate inclined at an angle of 250 to the axis of the jet. Calculate the normal force exerted on the plate (a) when the plate is stationary, (b) when the plate is moving at 4.5 m/s in the direction of jet and (c) the work done per sec and the efficiency for case b.

Fluid Mechanics Problem Solution

Problem Statement

A jet of water 60 mm in diameter with a velocity of 15 m/s strikes a flat plate inclined at an angle of 25° to the axis of the jet. Calculate:
a) The normal force exerted on the plate when it is stationary
b) The normal force exerted on the plate when it is moving at 4.5 m/s in the direction of the jet
c) The work done per sec and the efficiency for case (b)

Inclined Jet Impact on Flat Plate Diagram

Given Data

Diameter of jet (d) 60 mm = 0.06 m

Area of jet (A) π/4 × (0.06)² = 0.00283 m²

Velocity of jet (V) 15 m/s

Density of water (ρ) 1000 kg/m³

Angle of inclination (θ) 25°

Velocity of plate (u) 4.5 m/s

Solution Approach

To solve this problem, we’ll apply the principles of momentum conservation for a jet striking an inclined plate. The key difference from a normal impact is that only the component of velocity perpendicular to the plate causes a normal force on the plate.

We’ll analyze:

The normal force when the plate is stationary
The normal force when the plate is moving
The work done and efficiency when the plate is moving

Calculations

Part A: Normal Force on Stationary Plate

Step 1: For a jet striking an inclined stationary plate, the normal force is given by:

F = ρ × A × V² × Sin θ

This equation represents the rate of change of momentum in the direction normal to the plate.

Step 2: Substituting the given values:

F = 1000 × 0.00283 × 15² × Sin 25°

F = 1000 × 0.00283 × 225 × 0.4226

F = 269 N

Normal force on stationary plate (F) = 269 N

Part B: Normal Force on Moving Plate

Step 1: When the plate moves in the direction of the jet, the relative velocity is:

V_relative = V – u = 15 – 4.5 = 10.5 m/s

Step 2: The normal force on the moving plate is:

F_p = ρ × A × (V – u)² × Sin θ

F_p = 1000 × 0.00283 × 10.5² × Sin 25°

F_p = 1000 × 0.00283 × 110.25 × 0.4226

F_p = 132 N

Normal force on moving plate (F_p) = 132 N

Part C: Work Done and Efficiency

Step 1: Work done per second is the product of force and velocity:

W = F_p × u = 132 × 4.5 = 594 J/s = 594 W

Step 2: Kinetic energy of the jet per second:

KE/s = ½ × (ρ × A × V) × V² = ½ × ρ × A × V³

KE/s = ½ × 1000 × 0.00283 × 15³

KE/s = ½ × 1000 × 0.00283 × 3375

KE/s = 4776 J/s = 4776 W

Step 3: Efficiency is the ratio of work done to kinetic energy:

η = W / (KE/s) = 594 / 4776 = 0.12 = 12%

Work done per second (W) = 594 J/s
Efficiency (η) = 12%

Detailed Explanation

Physics of Inclined Jet Impact

When a water jet strikes an inclined plate, the force exerted has two components: normal to the plate and tangential along the plate. Only the normal component contributes to the force perpendicular to the plate surface. The sine of the angle of inclination (θ) determines what fraction of the jet momentum creates a normal force.

Effect of Plate Motion

When the plate moves in the direction of the jet, the relative velocity between the jet and the plate decreases. This reduces the momentum change rate and consequently the force exerted on the plate. The force is proportional to the square of the relative velocity, which explains the significant reduction from 269 N to 132 N when the plate moves at 4.5 m/s.

Efficiency Analysis

The efficiency of 12% is lower than what would be achieved with a normal impact (perpendicular to the jet). This lower efficiency occurs because:

Only a component of the jet momentum (Sin θ = Sin 25° = 0.4226) contributes to the normal force
A significant portion of the kinetic energy is directed tangentially along the plate surface
Energy is lost in friction and turbulence as the water flows along the inclined surface

Comparison with Perpendicular Impact

For the same jet striking a plate perpendicularly (θ = 90°):

The normal force would be greater by a factor of 1/Sin 25° (approximately 2.37 times larger)
The efficiency would be significantly higher

Optimal Conditions

For an inclined plate with angle θ, the theoretical maximum efficiency occurs when the plate velocity (u) is one-third of the jet velocity (V) multiplied by Sin θ. In this case:

u_optimal = V × Sin θ / 3 = 15 × Sin 25° / 3 = 15 × 0.4226 / 3 = 2.11 m/s

Since our plate velocity (4.5 m/s) exceeds this optimal value, we achieve a lower-than-maximum possible efficiency.

Practical Applications

This principle is applied in:

Design of turbine blades and vanes where the angle of incidence affects power generation efficiency
Hydraulic machines with inclined surfaces
Deflectors and diverters in fluid control systems
Splash plates and shields in industrial processes

Engineering Significance

Understanding the forces exerted by fluid jets on inclined surfaces is crucial in:

Optimizing energy extraction in hydropower systems
Designing spray nozzles and irrigation systems
Analyzing erosion patterns on inclined surfaces
Calculating loads on structures subjected to fluid jets at various angles

This problem demonstrates the importance of considering both the angle of impact and the relative motion when analyzing fluid-structure interactions in engineering applications.