A 150mm diameter pipe on the horizontal plane carries water under the head of 16m of water with the velocity of 3.5 m/s. Find the direction and magnitude of the pipe bend, if the axis of the bend was turned with angle 75°. Assume no loss of energy at the pipe bend.

Pipe Bend Force Calculation

Fluid Mechanics Problem Solution

Problem Statement

Given Data

Pipe diameter (d₁ = d₂)150 mm = 0.15 mFlow velocity (V₁ = V₂)3.5 m/sPressure head16 m of waterFluidWater (density ρ = 1000 kg/m³)Bend angle (θ)75°

Solution Approach

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

Calculate the cross-sectional area and discharge
Convert pressure head to pressure
Apply the momentum equation to determine the forces in both X and Y directions
Calculate the resultant force and its direction

Preliminary Calculations

Step 1: Calculate the cross-sectional area:

A = A₁ = A₂ = π/4 × d² = π/4 × 0.15² = 0.01767 m²

Step 2: Calculate the discharge:

Q = A × V = 0.01767 × 3.5 = 0.0618 m³/s

Step 3: Convert pressure head to pressure:

P = P₁ = P₂ = ρ × g × h = 1000 × 9.81 × 16 = 156,960 N/m²

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 = 3.5 m/s). At section 2, after the bend, the X-component of velocity is (V₂_x = V₂Cosθ = 3.5×Cos75° = 0.906 m/s).

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

Step 3: Solve for F_x:

F_x = (P₁A₁ - P₂A₂Cosθ) + ρQ(V₁ - V₂Cosθ) F_x = (156960 × 0.01767 - 156960 × 0.01767 × Cos75°) + 1000 × 0.0618 × (3.5 - 3.5 × Cos75°) F_x = 156960 × 0.01767 × (1 - Cos75°) + 1000 × 0.0618 × 3.5 × (1 - Cos75°) F_x = 2216 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 = 3.5×Sin75° = 3.38 m/s).

F_y - P₂A₂Sinθ = ρQ(V₂Sinθ - 0)

Step 3: Solve for F_y:

F_y = P₂A₂Sinθ + ρQV₂Sinθ F_y = 156960 × 0.01767 × Sin75° + 1000 × 0.0618 × 3.5 × Sin75° F_y = 2888 N

Resultant Force Calculation

Step 1: Calculate the magnitude of the resultant force:

F_R = √(F_x² + F_y²) F_R = √(2216² + 2888²) F_R = √(4,910,656 + 8,340,544) F_R = √13,251,200 = 3,640 N

Step 2: Calculate the direction of the resultant force:

θ = tan⁻¹(F_y/F_x) = tan⁻¹(2888/2216) = 52.5°

The resultant force exerted by the water on the bend is 3,640 N at an angle of 52.5° (to the right and downward).

Summary

The key parameters of the pipe system:
- Pipe diameter = 0.15 m
- Flow velocity = 3.5 m/s
- Cross-sectional area = 0.01767 m²
- Discharge (Q) = 0.0618 m³/s
- Pressure = 156,960 N/m²
The force components were calculated using the momentum equation:
- X-direction force: F_x = 2,216 N
- Y-direction force: F_y = 2,888 N
The resultant force on the bend:
- Magnitude: 3,640 N
- Direction: 52.5° from the X-axis (to the right and downward)

This problem demonstrates the application of 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 75° bend. Because there is no change in pipe diameter, the only factors affecting the force are the pressure and the change in direction of the flow.

A 150mm diameter pipe on the horizontal plane carries water under the head of 16m of water with the velocity of 3.5 m/s. Find the direction and magnitude of the pipe bend, if the axis of the bend was turned with angle 75°. Assume no loss of energy at the pipe bend.

Pipe Bend Force Calculation

Problem Statement

Given Data

Solution Approach

Preliminary Calculations

Force in X-Direction

Force in Y-Direction

Resultant Force Calculation

Summary

Keep learning