A reducing bend with an angle of 60° is installed in a pipeline carrying water. The initial diameter is 300 mm and the final diameter is 150 mm. Water flows at a rate of 330 lps and the pressure at the commencement of the bend is 3.1 bar. The friction loss in the pipe is assumed as 10% of the kinetic energy at the exit of the bend. Determine the force exerted by the reducing bend.

Reducing Bend Problem Solution

Problem Statement

Given Data

Diameter at Inlet (d₁) 300 mm = 0.3 m

Diameter at Outlet (d₂) 150 mm = 0.15 m

Flow Rate (Q) 330 lps = 0.33 m³/s

Inlet Pressure (P₁) 3.1 bar = 3.1×10⁵ N/m²

Friction Loss 10% of kinetic energy at exit

Bend Angle (θ) 60°

Solution Approach

To determine the force exerted by the reducing bend, we first compute the cross-sectional areas and velocities at both sections. Next, using Bernoulli’s equation (with a head loss equal to 10% of the exit kinetic energy) we determine the pressure at the exit. Finally, we apply the conservation of momentum in both horizontal and vertical directions to obtain the resultant force and its direction.

Calculations

Basic Parameters Calculation

Step 1: Calculate the cross-sectional areas at the inlet and outlet.

A₁ = π/4 × (0.3)² ≈ 0.07068 m²

A₂ = π/4 × (0.15)² ≈ 0.01767 m²

Step 2: Determine velocities using Q = A×V.

V₁ = Q / A₁ = 0.33 / 0.07068 ≈ 4.67 m/s

V₂ = Q / A₂ = 0.33 / 0.01767 ≈ 18.67 m/s

Step 3: Compute the loss head as 10% of the exit kinetic energy.

h_L = 0.1 × (V₂²)/(2g) = 0.1 × (18.67²)/(2×9.81) ≈ 1.77 m

Step 4: Use Bernoulli’s equation (with Z₁ = Z₂) between sections 1 and 2.

P₁/(ρg) + V₁²/(2g) = P₂/(ρg) + V₂²/(2g) + h_L

With ρ = 1000 kg/m³ and g = 9.81 m/s²:

P₂ ≈ 129256 N/m²

Force Analysis

Step 5: Compute the horizontal force (F_x) using momentum conservation.

(P₁A₁ – P₂A₂Cosθ) – F_x = ρQ (V₂Cosθ – V₁)

F_x = (P₁A₁ – P₂A₂Cos60) + ρQ (V₁ – V₂Cos60)

F_x ≈ (310000×0.07068 – 129256×0.01767×Cos60) + 1000×0.33 (4.67 – 18.67×Cos60)

F_x ≈ 19229 N

Step 6: Compute the vertical force (F_y).

F_y – P₂A₂Sinθ = ρQ (V₂Sinθ – 0)

F_y = P₂A₂Sin60 + ρQ V₂Sin60

F_y ≈ 129256×0.01767×Sin60 + 1000×0.33×18.67×Sin60

F_y ≈ 7314 N

Step 7: Determine the resultant force and its direction.

F_R = √(F_x² + F_y²) ≈ √(19229² + 7314²) ≈ 20573 N

Direction, θ_R = Tan^-1(F_y/F_x) ≈ Tan^-1(7314/19229) ≈ 20.80°

Resultant Force: 20573 N at 20.80° (to the right and downward)

Detailed Explanation

Key Concepts

This problem involves applying Bernoulli’s equation with a correction for friction loss (10% of the exit kinetic energy) to obtain the outlet pressure. Then, conservation of momentum is used to calculate the forces acting on the reducing bend.

Force Components

Horizontal Force (F_x): Arises from the difference in pressure forces and the change in momentum along the flow direction.

Vertical Force (F_y): Results from the vertical component of the momentum change and the pressure force acting vertically.

Engineering Implications

The calculated force of approximately 20.6 kN is crucial for designing support structures and anchorage systems to safely accommodate the forces imposed by the fluid flow in reducing bends.

Flow Effects

Variations in the flow rate or friction loss would significantly affect the exit velocity and pressure, thereby changing the magnitude and direction of the force. Accurate estimation is vital for reliable hydraulic system design.