A 450 reducing bend is connected in a pipeline carrying water. The diameter at the inlet and outlet of the bend is 400 mm and 200 mm respectively. Find the force exerted by water on the bend if the intensity of pressure at the inlet is 215.8 kN/m², the rate of flow is 0.5 m³/s, and the loss of head in the bend is 1.25 m of oil (specific gravity 0.85).

Reducing Bend Problem Solution

Problem Statement

A 45° reducing bend is connected in a pipeline carrying water. The diameter at the inlet and outlet of the bend is 400 mm and 200 mm respectively. Find the force exerted by water on the bend if the intensity of pressure at the inlet is 215.8 kN/m², the rate of flow is 0.5 m³/s, and the loss of head in the bend is 1.25 m of oil (specific gravity 0.85).

Given Data

Diameter at Inlet (d₁) 400 mm = 0.4 m

Diameter at Outlet (d₂) 200 mm = 0.2 m

Flow Rate (Q) 0.5 m³/s

Inlet Pressure (P₁) 215.8 kN/m² = 215800 N/m²

Loss of Head 1.25 m of oil (sp.gr. 0.85)

Bend Angle (θ) 45°

Solution Approach

To determine the force exerted by the water on the reducing bend, we apply Bernoulli’s equation (with head loss) and the conservation of momentum. We begin by calculating the cross-sectional areas, velocities, and converting the loss of head into equivalent water head. Then, we resolve the momentum changes in both the horizontal and vertical directions.

Calculations

Basic Parameters Calculation

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

A₁ = π/4 × d₁² = π/4 × (0.4)² ≈ 0.1256 m²

A₂ = π/4 × d₂² = π/4 × (0.2)² ≈ 0.0314 m²

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

V₁ = Q / A₁ = 0.5 / 0.1256 ≈ 3.98 m/s

V₂ = Q / A₂ = 0.5 / 0.0314 ≈ 15.92 m/s

Step 3: Convert the loss of head from oil to water terms.

Head loss in oil = 1.25 m, sp.gr. = 0.85

h_L = 0.85 × 1.25 ≈ 1.0625 m

Step 4: Apply Bernoulli’s equation (assuming 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, solving gives:

P₂ ≈ 86574 N/m²

Force Analysis

Step 5: Calculate the force in the X direction using momentum conservation.

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

F_x = (P₁A₁ – P₂A₂Cos45) + ρQ (V₁ – V₂Cos45)

F_x ≈ (215800×0.1256 – 86574×0.0314×Cos45) + 1000×0.5 (3.98 – 15.92×Cos45)

F_x ≈ 21544 N

Step 6: Calculate the force in the Y direction.

F_y – P₂A₂Sinθ = ρQ (V₂Sinθ – 0)

F_y = P₂A₂Sin45 + ρQ V₂Sin45

F_y ≈ 86574×0.0314×Sin45 + 1000×0.5×15.92×Sin45

F_y ≈ 7551 N

Step 7: Determine the resultant force and its direction.

F_R = √(F_x² + F_y²) ≈ √(21544² + 7551²) ≈ 22829 N

Direction, θ_R = Tan^-1(F_y/F_x) ≈ Tan^-1(7551/21544) ≈ 19.30°

Resultant Force: 22829 N at 19.30° (to the right and downward)

Detailed Explanation

Key Principles

This problem uses Bernoulli’s equation (with a head loss correction) to determine the outlet pressure and the conservation of momentum to evaluate the forces in the horizontal and vertical directions.

Force Components

Horizontal Component (F_x): Determined by the difference in pressure forces and the momentum change in the horizontal direction.

Vertical Component (F_y): Accounts for the vertical momentum of the flow and the pressure force acting in the vertical direction.

Engineering Relevance

The calculated resultant force of approximately 22.8 kN is critical in designing the support and anchorage for the reducing bend. It ensures that the structure can safely resist both the pressure-induced forces and the momentum changes due to the abrupt change in flow area.

Impact of Flow Conditions

Any variation in the flow rate or the head loss will significantly affect the velocities and pressures at the sections, thereby altering the net force on the bend. It highlights the importance of accurate parameter estimation in hydraulic design.