A 60cm pipe is connected to a 30cm pipe by a standard reducer fitting. For the same flow of 0.9 m3/s of water and a pressure of 200Kpa, what force is exerted by the water on the reducer, neglecting any lost head?

Pipe Reducer Force Calculation – Fluid Mechanics Solution

Pipe Reducer Force Calculation

Fluid Mechanics Problem Solution

Problem Statement

A 60cm pipe is connected to a 30cm pipe by a standard reducer fitting. For the same flow of 0.9 m³/s of water and a pressure of 200kPa, what force is exerted by the water on the reducer, neglecting any lost head?

Given Data

Diameter of larger pipe (d₁) 60 cm = 0.6 m

Diameter of smaller pipe (d₂) 30 cm = 0.3 m

Flow rate (Q) 0.9 m³/s

Pressure at section 1 (P₁) 200 kPa = 200,000 N/m²

Fluid Water (ρ = 1000 kg/m³)

Gravitational acceleration (g) 9.81 m/s²

Head loss Neglected

Solution Approach

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

Calculate the cross-sectional areas of both pipes
Determine the velocities in each section
Apply Bernoulli’s equation to find the pressure at section 2
Use the momentum equation to calculate the force on the reducer

Preliminary Calculations

Step 1: Calculate the cross-sectional areas:

A₁ = π/4 × d₁² = π/4 × 0.6² = 0.2827 m²

A₂ = π/4 × d₂² = π/4 × 0.3² = 0.07068 m²

Step 2: Calculate the velocities:

V₁ = Q/A₁ = 0.9/0.2827 = 3.18 m/s

V₂ = Q/A₂ = 0.9/0.07068 = 12.73 m/s

Applying Bernoulli’s Equation

Step 1: Apply Bernoulli’s equation between sections 1 and 2 (assuming Z₁ = Z₂):

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

Step 2: Substitute the values:

200,000/(1000×9.81) + 3.18²/(2×9.81) = P₂/(1000×9.81) + 12.73²/(2×9.81)

20.39 + 0.51 = P₂/(1000×9.81) + 8.26

P₂/(1000×9.81) = 20.39 + 0.51 – 8.26 = 12.64

P₂ = 12.64 × 1000 × 9.81 = 124,030 N/m²

Force Calculation Using Momentum Equation

Step 1: Apply the momentum equation:

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

(P₁A₁ – P₂A₂) – F_x = ρQ(V₂ – V₁)

Step 2: Rearrange to solve for F_x:

F_x = (P₁A₁ – P₂A₂) – ρQ(V₂ – V₁)

F_x = (P₁A₁ – P₂A₂) + ρQ(V₁ – V₂)

Step 3: Substitute the values:

F_x = (200,000 × 0.2827 – 124,030 × 0.07068) + 1000 × 0.9 × (3.18 – 12.73)

F_x = (56,540 – 8,764) + 1000 × 0.9 × (-9.55)

F_x = 47,776 – 8,595 = 39,181 N

The force exerted by water on the reducer is 39,178 N to the right (in the direction of flow).

Summary

We determined the velocities in both pipe sections:
- V₁ = 3.18 m/s (60 cm pipe)
- V₂ = 12.73 m/s (30 cm pipe)
Using Bernoulli’s equation, we calculated the pressure at section 2:
- P₁ = 200,000 N/m² (given)
- P₂ = 124,030 N/m² (calculated)
The momentum equation was applied to find the force:
- Pressure force component: 47,776 N
- Momentum change component: -8,595 N
- Net force: 39,178 N to the right
The forces in the Y-direction balance each other, so F_y = 0.

This problem demonstrates the application of Bernoulli’s principle and the momentum equation in fluid mechanics to determine forces in pipe systems with changing cross-sectional areas. The reducer experiences a significant force due to both the pressure difference and the momentum change of the fluid.