A jet of 20mm in diameter moving with a velocity of 5m/s strikes a smooth plate, which is inclined at an angle of 200 to the horizontal. Compute the amount of flow on each side of the plate and the force exerted on the plate.

A water jet of 20mm diameter moving with a velocity of 5m/s strikes a fixed plate at an angle of 20° to the horizontal. Compute the amount of flow on each side of the plate and the force exerted on the plate.

Fluid Mechanics Problem Solution

Problem Statement

A water jet of 20mm diameter moving with a velocity of 5m/s strikes a fixed plate at an angle of 20° to the horizontal. Compute the amount of flow on each side of the plate and the force exerted on the plate.

Water Jet Striking an Inclined Plate

Given Data

Diameter of jet (d) 20 mm = 0.02 m
Velocity of jet (V) 5 m/s
Angle of inclination of jet with horizontal (θ) 20°
Area of jet (A) π/4 × (0.02)² = 0.000314 m²
Density of water (ρ) 1000 kg/m³

Solution Approach

When a water jet strikes a fixed plate, it divides into two flows: one going up and one going down along the plate. We’ll first calculate the total flow rate, then determine how it divides on each side of the plate using momentum principles, and finally calculate the force exerted on the plate.

Calculations

Total Flow Rate

Step 1: Calculate the total flow rate of the water jet:

Q = A × V = 0.000314 × 5 = 0.00157 m³/s

Flow Division on Each Side of the Plate

Step 2: Using the momentum equation in the direction of the jet:

ρA₁V² – ρA₂V² – ρAV²cosθ = 0

With Q = AV, Q₁ = A₁V, and Q₂ = A₂V, this equation becomes:

Q₁ – Q₂ = Qcosθ

Step 3: From the continuity equation, we know:

Q₁ + Q₂ = Q

Step 4: Solving these two equations:

Q₁ = Q/2(1 + cosθ)
Q₂ = Q/2(1 – cosθ)

Step 5: Substituting values:

Q₁ = 0.00157/2(1 + cos20°) = 0.00157/2(1 + 0.9397) = 0.00152 m³/s
Q₂ = 0.00157/2(1 – cos20°) = 0.00157/2(1 – 0.9397) = 0.00005 m³/s

Flow on Upper Side (Q₁) = 0.00152 m³/s

Flow on Lower Side (Q₂) = 0.00005 m³/s

Force Exerted on the Plate

Step 6: The normal force on the plate:

Fn = ρAV²sinθ = 1000 × 0.000314 × 5² × sin20° = 2.7 N

Step 7: The force in the horizontal direction:

Fx = Fncos(90° – 20°) = 2.7 × cos70° = 2.7 × 0.342 = 0.92 N

Normal Force on Plate (Fn) = 2.7 N

Horizontal Force on Plate (Fx) = 0.92 N

Detailed Explanation

Physics of Flow Division

When a water jet strikes a fixed plate at an angle, the flow divides into two parts: one flowing upward and one flowing downward along the plate. This division is not equal because of the initial momentum of the jet in the horizontal direction. The portion of the jet that flows upward (Q₁) is significantly larger than the portion flowing downward (Q₂) because the horizontal momentum favors the upward direction.

Momentum Conservation Principle

The calculation of flow division is based on the principle of momentum conservation. The incoming jet has momentum in both horizontal and vertical directions. When the jet strikes the plate, this momentum must be conserved in the flow along the plate, resulting in unequal division.

Force Analysis

The force exerted on the plate is due to the change in momentum of the water as it strikes and is deflected by the plate. This force has two components:

  • Normal force (Fn): Perpendicular to the plate surface
  • Horizontal force (Fx): In the direction parallel to the horizontal

Practical Implications

This problem has several practical applications:

  • Design of hydraulic machinery such as Pelton wheels and jet pumps
  • Analysis of erosion patterns in hydraulic structures
  • Planning of water diversion structures in irrigation systems
  • Understanding spray patterns in various industrial processes

Analysis of Results

The results show that:

  • Approximately 96.8% of the flow (0.00152 m³/s) moves along the upper side of the plate
  • Only about 3.2% of the flow (0.00005 m³/s) moves along the lower side
  • The normal force on the plate (2.7 N) is relatively small due to the modest jet size and velocity
  • The horizontal component of force (0.92 N) is only about 34% of the normal force due to the angle of incidence

This demonstrates how significantly the angle of incidence affects both flow distribution and force components in fluid jet impacts on surfaces.

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