The discharge of water through a 130° bend is 30 litres/s. The bend is lying in the horizontal plane and the diameters at the entrance and exit are 200mm and 100mm respectively. The pressure measured at the entrance is 100 kN/m². What is the magnitude and direction of the force exerted by the water on the bend?

Pipe Bend Force Calculation – Fluid Mechanics Solution

Pipe Bend Force Calculation

Fluid Mechanics Problem Solution

Problem Statement

The discharge of water through a 130° bend is 30 litres/s. The bend is lying in the horizontal plane and the diameters at the entrance and exit are 200mm and 100mm respectively. The pressure measured at the entrance is 100 kN/m². What is the magnitude and direction of the force exerted by the water on the bend?

Given Data

Pipe diameter at entrance (d₁) 200 mm = 0.2 m

Pipe diameter at exit (d₂) 100 mm = 0.1 m

Flow rate (Q) 30 litres/s = 0.03 m³/s

Fluid Water (density ρ = 1000 kg/m³)

Pressure at entrance (P₁) 100 kN/m² = 100,000 Pa

Bend angle 130° (in horizontal plane)

Deflection angle (θ) 180° – 130° = 50°

Solution Approach

To find the resultant force exerted by the water on the bend, we need to:

Calculate the cross-sectional areas and velocities at entrance and exit
Apply Bernoulli’s equation to find the pressure at the exit (section 2)
Determine the forces in both X and Y directions using the momentum equation
Calculate the resultant force and its direction

Preliminary Calculations

Step 1: Calculate the cross-sectional areas:

A₁ = π/4 × d₁² = π/4 × 0.2² = 0.0314 m²

A₂ = π/4 × d₂² = π/4 × 0.1² = 0.00785 m²

Step 2: Calculate the velocities:

V₁ = Q/A₁ = 0.03/0.0314 = 0.95 m/s

V₂ = Q/A₂ = 0.03/0.00785 = 3.82 m/s

Applying Bernoulli’s Equation

Step 1: Apply Bernoulli’s equation between sections 1 and 2 (in a horizontal plane, Z₁ = Z₂):

P₁/ρg + V₁²/2g + Z₁ = P₂/ρg + V₂²/2g + Z₂

Step 2: Substitute the values:

100,000/(1000×9.81) + 0.95²/(2×9.81) = P₂/(1000×9.81) + 3.82²/(2×9.81)

10.19 + 0.046 = P₂/(1000×9.81) + 0.743

P₂/(1000×9.81) = 10.19 + 0.046 – 0.743 = 9.493

P₂ = 9.493 × 1000 × 9.81 = 93,155 Pa

Force in X-Direction

Step 1: Apply the momentum equation in the X-direction:

∑Forces in X direction = Rate of change of momentum in X direction

(P₁A₁ + P₂Cosθ A₂) – F_x = ρQ(V₂_x – V₁_x)

Step 2: At section 1, the velocity is entirely in the X-direction (V₁_x = V₁ = 0.95 m/s). At section 2, after the bend, the X-component of velocity is (V₂_x = -V₂Cosθ = -3.82×Cos50° = -2.46 m/s).

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

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

Step 3: Solve for F_x:

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

F_x = (100,000 × 0.0314 + 93,155 × 0.00785 × Cos50°) + 1000 × 0.03 × (0.95 + 3.82 × Cos50°)

F_x = (3,140 + 472) + 1000 × 0.03 × (0.95 + 2.46)

F_x = 3,612 + 1000 × 0.03 × 3.41 = 3,612 + 100 = 3,712 N

Force in Y-Direction

Step 1: Apply the momentum equation in the Y-direction:

∑Forces in Y direction = Rate of change of momentum in Y direction

F_y – P₂Sinθ A₂ = ρQ(V₂_y – V₁_y)

Step 2: At section 1, there is no Y-component of velocity (V₁_y = 0). At section 2, after the bend, the Y-component of velocity is (V₂_y = V₂Sinθ = 3.82×Sin50° = 2.93 m/s).

F_y – P₂A₂Sinθ = ρQ(V₂Sinθ – 0)

F_y – P₂A₂Sinθ = ρQV₂Sinθ

Step 3: Solve for F_y:

F_y = P₂A₂Sinθ + ρQV₂Sinθ

F_y = 93,155 × 0.00785 × Sin50° + 1000 × 0.03 × 3.82 × Sin50°

F_y = 560 + 1000 × 0.03 × 2.93 = 560 + 88 = 648 N

Resultant Force Calculation

Step 1: Calculate the magnitude of the resultant force:

F_R = √(F_x² + F_y²)

F_R = √(3,712² + 648²)

F_R = √(13,778,944 + 419,904)

F_R = √14,198,848 = 3,768 N

Step 2: Calculate the direction of the resultant force:

θ = tan⁻¹(F_y/F_x) = tan⁻¹(648/3,712) = 10°

The resultant force exerted by the water on the bend is 3,768 N at an angle of 10° (to the right and downward).

Summary

The fluid velocities in the pipe were calculated:
- Entrance velocity (V₁) = 0.95 m/s
- Exit velocity (V₂) = 3.82 m/s
Using Bernoulli’s equation, we determined the pressure at the exit:
- P₁ = 100,000 Pa (given)
- P₂ = 93,155 Pa (calculated)
The force components were calculated using the momentum equation:
- X-direction force: F_x = 3,712 N
- Y-direction force: F_y = 648 N
The resultant force on the bend:
- Magnitude: 3,768 N
- Direction: 10° from the X-axis (to the right and downward)

This problem demonstrates the application of Bernoulli’s principle and the momentum equation in fluid mechanics to determine forces on pipe bends. The resultant force is significant due to both the pressure forces and the momentum change of the fluid as it changes direction through the 130° bend. The reduction in pipe diameter also contributes to the change in fluid velocity and pressure, affecting the overall force on the bend.