Let's solve the problem step by step: --- ## **Step 1: Understanding the Problem** - **Fluid:** Water (assume incompressible and steady flow) - **Pipe Geometry:** Diffuser (inlet diameter = 3 cm, outlet diameter = 6 cm, length = 10 cm) - **Boundary Conditions:** - **Inlet:** Uniform velocity \( V_1 = 10 \) mm/s \( = .01 \) m/s - **Outlet:** Atmospheric pressure (\( P_2 = P_{atm} \)) - **Walls:** No-slip boundary (velocity at wall = ) --- ## **Step 2: Calculate Cross-sectional Areas** \[ A = \pi r^2 \] - **Inlet:** - \( d_1 = 3 \) cm \( = .03 \) m → \( r_1 = .015 \) m - \( A_1 = \pi \times (.015)^2 = 7.07 \times 10^{-4} \) m² - **Outlet:** - \( d_2 = 6 \) cm \( = .06 \) m → \( r_2 = .03 \) m - \( A_2 = \pi \times (.03)^2 = 2.827 \times 10^{-3} \) m² --- ## **Step 3: Apply Continuity Equation (Incompressible Flow)** \[ Q = V_1 A_1 = V_2 A_2 \] Given \( V_1 = .01 \) m/s, \[ Q = .01 \times 7.07 \times 10^{-4} = 7.07 \times 10^{-6} \ \mathrm{m}^3/\mathrm{s} \] \[ V_2 = \frac{Q}{A_2} = \frac{7.07 \times 10^{-6}}{2.827 \times 10^{-3}} = 2.5 \times 10^{-3} \ \mathrm{m/s} = 2.5 \ \mathrm{mm/s} \] --- ## **Step 4: Apply Bernoulli’s Equation (Neglecting Elevation Change and Viscous Losses)** \[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \] Given: - \( P_2 = P_{atm} \) - \( V_1 = .01 \) m/s - \( V_2 = .0025 \) m/s - \( \rho \) (water) \( = 100 \) kg/m³ Solve for \( P_1 \): \[ P_1 = P_{atm} + \frac{1}{2} \rho (V_2^2 - V_1^2) \] \[ P_1 = P_{atm} + \frac{1}{2} \times 100 \times [(.0025)^2 - (.01)^2] \] \[ P_1 = P_{atm} + 500 \times (6.25 \times 10^{-6} - 1 \times 10^{-4}) \] \[ P_1 = P_{atm} + 500 \times (-9.375 \times 10^{-5}) = P_{atm} - .0469 \ \text{Pa} \] --- ## **Step 5: Pressure and Velocity Distributions** ### **Velocity Distribution** - **Inlet (centerline):** \( 10 \) mm/s - **Outlet (centerline):** \( 2.5 \) mm/s - **At walls:** Zero (no-slip) - **Between inlet and outlet:** Decreases gradually from 10 mm/s to 2.5 mm/s along the centerline. ### **Pressure Distribution** - **Inlet:** \( P_{atm} - .0469 \) Pa - **Outlet:** \( P_{atm} \) - **Pressure increases linearly along the length of the diffuser (for inviscid, incompressible flow).** --- ## **Step 6: Final Answers & Summary Table** | Location | Velocity (mm/s) | Pressure (relative to \(P_{atm}\)) (Pa) | |----------|-----------------|-----------------------------------------| | Inlet | 10 | -.0469 | | Outlet | 2.5 | | --- ## **Summary** - **Velocity decreases** from 10 mm/s to 2.5 mm/s along the pipe centerline. - **Pressure increases** from \( P_{atm} - .0469 \) Pa at the inlet to atmospheric at the outlet. - **At the walls, velocity is zero** (no-slip condition). --- ### **Assumptions** - Steady, incompressible, laminar flow. - Neglect viscous losses (ideal scenario). - No elevation change. If you need a more detailed velocity profile (e.g., considering boundary layer effects), let me know!

# Fluid Analysis of Water Flow through a Diffuser Pipe ## Step 1: Understanding the Problem - **Fluid**: Water (assumed incompressible and steady flow) - **Pipe Geometry**: Diffuser - Inlet diameter = 3 cm - Outlet diameter = 6 cm - Length = 10 cm - **Boundary Conditions**: - **Inlet**: Uniform velocity \( V_1 = 10 \) mm/s \( = 0.01 \) m/s - **Outlet**: Atmospheric pressure (\( P_2 = P_{atm} \)) - **Walls**: No-slip boundary condition ## Step 2: Calculate Cross-sectional Areas Using the formula for the area of a circle: \[ A = \pi r^2 \] - **Inlet**: - Diameter \( d_1 = 3 \) cm \( = 0.03 \) m → Radius \( r_1 = 0.015 \) m - Area \( A_1 = \pi \times (0.015)^2 = 7.07 \times 10^{-4} \) m² - **Outlet**: - Diameter \( d_2 = 6 \) cm \( = 0.06 \) m → Radius \( r_2 = 0.03 \) m - Area \( A_2 = \pi \times (0.03)^2 = 2.827 \times 10^{-3} \) m² ## Step 3: Apply Continuity Equation (Incompressible Flow) The continuity equation states that: \[ Q = V_1 A_1 = V_2 A_2 \] Calculating the flow rate at the inlet: \[ Q = 0.01 \times 7.07 \times 10^{-4} = 7.07 \times 10^{-6} \ \mathrm{m}^3/\mathrm{s} \] Solving for \( V_2 \): \[ V_2 = \frac{Q}{A_2} = \frac{7.07 \times 10^{-6}}{2.827 \times 10^{-3}} = 2.5 \times 10^{-3} \ \mathrm{m/s} = 2.5 \ \mathrm{mm/s} \] ## Step 4: Apply Bernoulli’s Equation Using Bernoulli's equation: \[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \] Given that \( P_2 = P_{atm} \) and substituting values: \[ P_1 = P_{atm} + \frac{1}{2} \rho (V_2^2 - V_1^2) \] Substituting \( \rho = 1000 \) kg/m³ (density of water): \[ P_1 = P_{atm} + \frac{1}{2} \times 1000 \times \left[(2.5 \times 10^{-3})^2 - (0.01)^2\right] \] Calculating: \[ P_1 = P_{atm} + 500 \times \left(6.25 \times 10^{-6} - 1 \times 10^{-4}\right) \] \[ P_1 = P_{atm} + 500 \times (-9.375 \times 10^{-5}) = P_{atm} - 0.0469 \ \text{Pa} \] ## Step 5: Pressure and Velocity Distributions ### Velocity Distribution - **At Inlet (centerline)**: \( 10 \) mm/s - **At Outlet (centerline)**: \( 2.5 \) mm/s - **At Walls**: Zero (no-slip) - **Between Inlet and Outlet**: Gradual decrease from 10 mm/s to 2.5 mm/s along the centerline. ### Pressure Distribution - **At Inlet**: \( P_{atm} - 0.0469 \) Pa - **At Outlet**: \( P_{atm} \) - **Pressure** increases linearly along the length of the diffuser. ## Step 6: Final Answers & Summary Table | Location | Velocity (mm/s) | Pressure (relative to \(P_{atm}\)) (Pa) | |----------|-----------------|-----------------------------------------| | Inlet | 10 | -0.0469 | | Outlet | 2.5 | | ## Summary - **Velocity** decreases from 10 mm/s to 2.5 mm/s along the pipe centerline. - **Pressure** increases from \( P_{atm} - 0.0469 \) Pa at the inlet to atmospheric at the outlet. - **At the walls**, the velocity is zero due to the no-slip condition. ### Assumptions - Steady, incompressible, laminar flow. - Neglect viscous losses (ideal scenario). - No elevation change. If further details or specific analyses are needed, please let me know!

# Fluid Analysis of Water Flow through a Diffuser Pipe ## Step 1: Understanding the Problem - **Fluid**: Water (assumed incompressible and steady flow) - **Pipe Geometry**: Diffuser - Inlet diameter = 3 cm - Outlet diameter = 6 cm - Length = 10 cm - **Boundary Conditions**: - **Inlet**: Uniform velocity \( V_1 = 10 \) mm/s \( = 0.01 \) m/s - **Outlet**: Atmospheric pressure (\( P_2 = P_{atm} \)) - **Walls**: No-slip boundary condition ## Step 2: Calculate Cross-sectional Areas Using the formula for the area of a circle: \[ A = \pi r^2 \] - **Inlet**: - Diameter \( d_1 = 3 \) cm \( = 0.03 \) m → Radius \( r_1 = 0.015 \) m - Area \( A_1 = \pi \times (0.015)^2 = 7.07 \times 10^{-4} \) m² - **Outlet**: - Diameter \( d_2 = 6 \) cm \( = 0.06 \) m → Radius \( r_2 = 0.03 \) m - Area \( A_2 = \pi \times (0.03)^2 = 2.827 \times 10^{-3} \) m² ## Step 3: Apply Continuity Equation (Incompressible Flow) The continuity equation states that: \[ Q = V_1 A_1 = V_2 A_2 \] Calculating the flow rate at the inlet: \[ Q = 0.01 \times 7.07 \times 10^{-4} = 7.07 \times 10^{-6} \ \mathrm{m}^3/\mathrm{s} \] Solving for \( V_2 \): \[ V_2 = \frac{Q}{A_2} = \frac{7.07 \times 10^{-6}}{2.827 \times 10^{-3}} = 2.5 \times 10^{-3} \ \mathrm{m/s} = 2.5 \ \mathrm{mm/s} \] ## Step 4: Apply Bernoulli’s Equation Using Bernoulli's equation: \[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \] Given that \( P_2 = P_{atm} \) and substituting values: \[ P_1 = P_{atm} + \frac{1}{2} \rho (V_2^2 - V_1^2) \] Substituting \( \rho = 1000 \) kg/m³ (density of water): \[ P_1 = P_{atm} + \frac{1}{2} \times 1000 \times \left[(2.5 \times 10^{-3})^2 - (0.01)^2\right] \] Calculating: \[ P_1 = P_{atm} + 500 \times \left(6.25 \times 10^{-6} - 1 \times 10^{-4}\right) \] \[ P_1 = P_{atm} + 500 \times (-9.375 \times 10^{-5}) = P_{atm} - 0.0469 \ \text{Pa} \] ## Step 5: Pressure and Velocity Distributions ### Velocity Distribution - **At Inlet (centerline)**: \( 10 \) mm/s - **At Outlet (centerline)**: \( 2.5 \) mm/s - **At Walls**: Zero (no-slip) - **Between Inlet and Outlet**: Gradual decrease from 10 mm/s to 2.5 mm/s along the centerline. ### Pressure Distribution - **At Inlet**: \( P_{atm} - 0.0469 \) Pa - **At Outlet**: \( P_{atm} \) - **Pressure** increases linearly along the length of the diffuser. ## Step 6: Final Answers & Summary Table | Location | Velocity (mm/s) | Pressure (relative to \(P_{atm}\)) (Pa) | |----------|-----------------|-----------------------------------------| | Inlet | 10 | -0.0469 | | Outlet | 2.5 | | ## Summary - **Velocity** decreases from 10 mm/s to 2.5 mm/s along the pipe centerline. - **Pressure** increases from \( P_{atm} - 0.0469 \) Pa at the inlet to atmospheric at the outlet. - **At the walls**, the velocity is zero due to the no-slip condition. ### Assumptions - Steady, incompressible, laminar flow. - Neglect viscous losses (ideal scenario). - No elevation change.