# Fluid Mechanics Problem: Reducing Bend ## Problem Summary - **Water flow rate, \( Q \) = .25 m³/s** - **Inlet pressure at section 1, \( P_1 \) = 150 kPa (gauge)** - **Weight acting, \( W \) = 142 N** (20 cm right of point A) - **Water density, \( \rho \) = 100 kg/m³** - **Section 1 diameter, \( D_1 \) = 30 \) cm = .3 m** - **Section 2 diameter, \( D_2 \) = 20 \) cm = .2 m** - **The bend is in the vertical plane** We are asked to: a) Find velocity 1, velocity 2, and gauge pressure 2. b) Find the forces required to hold the bend in space. c) Find the moment the support system must resist. --- ## Step 1: Find Velocity at Section 1 and Section 2 ### **A. Calculate Velocities** **Area of a circle:** \[ A = \frac{\pi D^2}{4} \] #### **Section 1:** \[ A_1 = \frac{\pi (.3)^2}{4} = .0707 \ m^2 \] \[ v_1 = \frac{Q}{A_1} = \frac{.25}{.0707} = 3.54 \ \text{m/s} \] #### **Section 2:** \[ A_2 = \frac{\ (.2)^2}{4} = .0314 \ m^2 \] \[ v_2 = \frac{Q}{A_2} = \frac{.25}{.0314} = 7.96 \ \text{m/s} \] --- ## Step 2: Find Gauge Pressure at Section 2 Using Bernoulli’s Equation Assume both sections are at the same elevation (vertical height cancels). \[ \frac{P_1}{\rho g} + \frac{v_1^2}{2g} = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} \] \[ P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2) \] Plug in values: - \(P_1 = 150,000\) Pa - \(\rho = 100 \ \text{kg/m}^3\) - \(v_1 = 3.54 \ \text{m/s}\) - \(v_2 = 7.96 \ \text{m/s}\) \[ P_2 = 150,000 + \frac{1}{2} \times 100 \times (3.54^2 - 7.96^2) \] \[ = 150,000 + 500 \times (12.52 - 63.36) \] \[ = 150,000 + 500 \times (-50.84) \] \[ = 150,000 - 25,420 \] \[ = 124,580\ \text{Pa} = 124.6 \ \text{kPa (gauge)} \] --- ## **Summary Table** | Section | Area (m²) | Velocity (m/s) | Gauge Pressure (kPa) | |---------|-----------|----------------|----------------------| | 1 | .0707 | 3.54 | 150 | | 2 | .0314 | 7.96 | 124.6 | --- ## **Graph: Velocity and Pressure vs. Section** ```plaintext Velocity (m/s) 8 | * 7 | * 6 | 5 | 4 | * 3 | * 2 | 1 | +------------------------- Section 1 Section 2 Pressure (kPa) 160 | * 150 | * 140 | 130 | 120 | * 110 | * 100 +------------------------- Section 1 Section 2 ``` *Graph shows how velocity increases and pressure decreases from section 1 to section 2.* --- ## **Final Answers (Step-by-Step):** ### **a) Velocities and Pressure** - **Velocity at Section 1 (\( v_1 \)):** \( 3.54 \) m/s - **Velocity at Section 2 (\( v_2 \)):** \( 7.96 \) m/s - **Gauge Pressure at Section 2 (\( P_2 \)):** \( 124.6 \) kPa --- ## **Next Steps** If you need solutions for (b) and (c), let me know!

# Fluid Mechanics Problem: Reducing Bend ## Problem Summary - **Water flow rate, \( Q \) = 0.25 m³/s** - **Inlet pressure at section 1, \( P_1 \) = 150 kPa (gauge)** - **Weight acting, \( W \) = 142 N** (20 cm right of point A) - **Water density, \( \rho \) = 1000 kg/m³** - **Section 1 diameter, \( D_1 \) = 30 cm = 0.3 m** - **Section 2 diameter, \( D_2 \) = 20 cm = 0.2 m** - **The bend is in the vertical plane** We are tasked with solving the following: 1. Find velocity at section 1, velocity at section 2, and gauge pressure at section 2. 2. Find the forces required to hold the bend in space. 3. Find the moment the support system must resist. --- ## Step 1: Find Velocity at Section 1 and Section 2 ### **A. Calculate Velocities** **Area of a circle:** \[ A = \frac{\pi D^2}{4} \] #### **Section 1:** \[ A_1 = \frac{\pi (0.3)^2}{4} = 0.0707 \ m^2 \] \[ v_1 = \frac{Q}{A_1} = \frac{0.25}{0.0707} \approx 3.54 \ \text{m/s} \] #### **Section 2:** \[ A_2 = \frac{\pi (0.2)^2}{4} = 0.0314 \ m^2 \] \[ v_2 = \frac{Q}{A_2} = \frac{0.25}{0.0314} \approx 7.96 \ \text{m/s} \] --- ## Step 2: Find Gauge Pressure at Section 2 Using Bernoulli’s Equation Assuming both sections are at the same elevation: \[ \frac{P_1}{\rho g} + \frac{v_1^2}{2g} = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} \] Rearranging gives: \[ P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2) \] ### **Plug in values**: - \(P_1 = 150,000\) Pa - \(\rho = 1000 \ \text{kg/m}^3\) - \(v_1 = 3.54 \ \text{m/s}\) - \(v_2 = 7.96 \ \text{m/s}\) \[ P_2 = 150,000 + \frac{1}{2} \times 1000 \times (3.54^2 - 7.96^2) \] \[ = 150,000 + 500 \times (12.52 - 63.36) \] \[ = 150,000 + 500 \times (-50.84) \] \[ = 150,000 - 25,420 \] \[ = 124,580\ \text{Pa} = 124.6 \ \text{kPa (gauge)} \] --- ## **Summary Table** | Section | Area (m²) | Velocity (m/s) | Gauge Pressure (kPa) | |---------|-----------|----------------|----------------------| | 1 | 0.0707 | 3.54 | 150 | | 2 | 0.0314 | 7.96 | 124.6 | --- ## Step 3: Find Forces Required to Hold the Bend ### **A. Force Calculation** Using the pressure acting on the surface area of the bend: 1. **Calculate Force at Section 1:** \[ F_1 = P_1 \times A_1 = 150,000 \times 0.0707 \approx 10,605 \ \text{N} \] 2. **Calculate Force at Section 2:** \[ F_2 = P_2 \times A_2 = 124,580 \times 0.0314 \approx 3,911 \ \text{N} \] ### **B. Net Force on the Bend:** \[ F_{net} = F_1 - F_2 = 10,605 - 3,911 \approx 6,694 \ \text{N} \] --- ## Step 4: Find the Moment the Support System Must Resist ### **A. Moment Calculation** Given that the weight \( W \) acts 20 cm from point A, the moment \( M \) is: \[ M = W \times d = 142 \times 0.2 = 28.4 \ \text{Nm} \] --- ## **Final Answers** ### **a) Velocities and Pressure** - **Velocity at Section 1 (\( v_1 \)):** \( 3.54 \) m/s - **Velocity at Section 2 (\( v_2 \)):** \( 7.96 \) m/s - **Gauge Pressure at Section 2 (\( P_2 \)):** \( 124.6 \) kPa ### **b) Forces Required to Hold the Bend** - **Net Force on the Bend:** \( 6,694 \) N (upward). ### **c) Moment the Support System Must Resist** - **Moment:** \( 28.4 \) Nm. --- If you need further assistance or additional calculations, please let me know!