Let's break down the **fluid analysis of air flow through the S-pipe** step by step. --- ## **Step 1: Problem Recap and Assumptions** - **Pipe shape**: S-pipe, smooth bends. - **Air at 25°C** (room temp). - **Pipe diameter**: Inner radius = 3 mm ⟹ diameter = 6 mm. - **Inlet velocity**: \( V_{inlet} = 5 \) mm/s \( = .005 \) m/s. - **Outlet**: Atmospheric pressure. - **Boundary condition**: No-slip at wall. - **All dimensions**: mm (convert to meters if needed). --- ## **Step 2: Properties of Air at 25°C** - **Density (\(\rho\))**: ~1.184 kg/m³ - **Dynamic Viscosity (\(\mu\))**: \(1.849 \times 10^{-5}\) kg/(m·s) --- ## **Step 3: Area and Flow Rate Calculation** ### **Pipe Cross-sectional Area (A):** \[ A = \pi r^2 = \pi \times (.003)^2 = 2.827 \times 10^{-5} \text{ m}^2 \] ### **Volumetric Flow Rate (Q):** \[ Q = V_{inlet} \times A = .005 \times 2.827 \times 10^{-5} = 1.4135 \times 10^{-7} \text{ m}^3/\text{s} \] --- ## **Step 4: Reynolds Number Calculation** \[ Re = \frac{\rho V D}{\mu} \] \[ D = 2r = .006 \text{ m} \] \[ Re = \frac{1.184 \times .005 \times .006}{1.849 \times 10^{-5}} = \frac{3.552 \times 10^{-5}}{1.849 \times 10^{-5}} \approx 1.92 \] - **Flow Regime:** \(Re < 230\) ⟹ **Laminar flow** --- ## **Step 5: Velocity Distribution** For a **laminar flow in a pipe**, the velocity profile is **parabolic**: \[ u(r) = u_{max} \left[1 - \left(\frac{r}{R}\right)^2 \right] \] Where: - \(u_{max} = 2 \times u_{avg}\) - \(u_{avg} = V_{inlet} = .005 \) m/s - \(R = .003\) m So, \[ u_{max} = 2 \times .005 = .01 \text{ m/s} \] \[ u(r) = .01 \left[1 - \left(\frac{r}{.003}\right)^2 \right] \] --- ## **Step 6: Pressure Drop Calculation** For **laminar flow**, pressure drop (\(\Delta P\)) in a straight pipe: \[ \Delta P = \frac{8 \mu L Q}{\pi r^4} \] ### **Total Length (L):** - From the diagram, let's estimate: - Horizontal: \(20+40+20 = 80\) mm - 2 bends, each \(\approx \pi \times 10\) mm = \(31.42\) mm per bend \(= 62.84\) mm - Total \(L \approx 80 + 62.84 = 142.84\) mm \(= .14284\) m ### **Plug values:** \[ \Delta P = \frac{8 \times 1.849 \times 10^{-5} \times .14284 \times 1.4135 \times 10^{-7}}{\pi \times (.003)^4} \] Calculate denominator: \[ \pi \times (.003)^4 = \pi \times 8.1 \times 10^{-11} = 2.545 \times 10^{-10} \] Calculate numerator: \[ 8 \times 1.849 \times 10^{-5} \times .14284 \times 1.4135 \times 10^{-7} \approx 2.981 \times 10^{-12} \] So, \[ \Delta P = \frac{2.981 \times 10^{-12}}{2.545 \times 10^{-10}} \approx .0117 \text{ Pa} \] **This is a very small pressure drop due to the very low velocity and small pipe size.** --- ## **Step 7: Effects of Bends** - For very **low Reynolds number**, losses due to bends are negligible compared to viscous (frictional) losses. - At higher velocities, you’d add **minor losses** using loss coefficients, but for \(Re \sim 2\), you can ignore. --- ## **Step 8: Final Answers** ### **Velocity Distribution** - **Parabolic profile**, with centerline (max) velocity: - \(u_{max} = .01\) m/s at center - \(u = \) m/s at wall (no-slip) - **Formula:** \(u(r) = .01 \left[1 - \left(\frac{r}{.003}\right)^2 \right]\) (for \(r\) from to .003 m) ### **Pressure Distribution** - **Pressure drops by \(.012\) Pa** (rounded) from inlet to outlet. - If outlet is atmospheric, **inlet pressure = atmospheric + .012 Pa** --- ## **Summary Table** | Parameter | Value | |------------------------|----------------------| | Inlet Velocity | .005 m/s | | Max Velocity (Center) | .01 m/s | | Reynolds Number | 1.92 (laminar flow) | | Pressure Drop (\(\Delta P\)) | .012 Pa | | Velocity Profile | Parabolic | --- ### **Interpretation** - The flow is extremely slow and dominated by viscous effects. - The pressure drop is negligible. - The velocity profile is parabolic, as expected for laminar pipe flow. --- **If you need velocity/pressure plots or more details, please specify!**