# Problem Restatement A fluid with the following properties: - **Density \(\rho = 100~\mathrm{kg/m^3}\)** - **Kinematic viscosity \(v = 1.01 \times 10^{-6}~\mathrm{m^2/s}\)** is contained in a tank of heighth = 2~\mathrmm}\). The fluid flows to the atmosphere through a horizontal pipe of: - **Length \(L = 10~\mathrm{m}\)** - **Diameter \(D = 25~\mathrm{mm} = .025~\mathrm{m}\)** - **Pipe roughness \(\epsilon = 1.5~\mathrm{mm} = .0015~\mathrm{m}\)** Find the **exit velocity \(v\)** of the fluid. --- # Step 1: Calculate the Ideal Velocity (Neglecting Losses) By Bernoulli’s principle (no losses): \[ \frac{v^2}{2g} = h \implies v = \sqrt{2gh} \] Using \(g = 9.81~\mathrm{m/s^2}\): \[ v_{ideal} = \sqrt{2 \times 9.81 \times 2} = \sqrt{39.24} \approx 6.266~\mathrm{m/s} \] --- # Step 2: Head Losses Due to Pipe Friction The actual velocity will be lower due to pipe friction (major loss). Use the **Darcy-Weisbach equation**: \[ h_f = f \frac{L}{D} \frac{v^2}{2g} \] The energy equation (from tank surface to outlet, both open to atmosphere): \[ h = h_f + \frac{v^2}{2g} \] \[ h = f \frac{L}{D} \frac{v^2}{2g} + \frac{v^2}{2g} \] \[ h = \left( f \frac{L}{D} + 1 \right) \frac{v^2}{2g} \] \[ v = \sqrt{ \frac{2gh}{ 1 + f \frac{L}{D} } } \] --- # Step 3: Estimate the Darcy Friction Factor (\(f\)) First, estimate the Reynolds number: \[ \text{Dynamic viscosity: } \mu = \rho \nu = 100 \times 1.01 \times 10^{-6} = 1.01 \times 10^{-3}~\mathrm{Pa \cdot s} \] \[ Re = \frac{v D}{\nu} \] Assume \(v \approx 2~\mathrm{m/s}\) (just to estimate \(Re\)): \[ Re = \frac{2 \times .025}{1.01 \times 10^{-6}} \approx \frac{.05}{1.01 \times 10^{-6}} \approx 4.95 \times 10^4 \] This is turbulent flow (\(Re > 400\)). Use the **Colebrook-White equation** or the **Swamee-Jain approximation**: \[ f = .25 \left[ \log_{10} \left( \frac{\epsilon}{3.7D} + \frac{5.74}{Re^{.9}} \right) \right]^{-2} \] Calculate \(\frac{\epsilon}{D} = \frac{.0015}{.025} = .06\) Plug in the numbers: - \(\frac{\epsilon}{3.7D} = \frac{.0015}{3.7 \times .025} = \frac{.0015}{.0925} \approx .0162\) - \(\frac{5.74}{Re^{.9}} = \frac{5.74}{(4.95 \times 10^4)^{.9}} \approx \frac{5.74}{1.98 \times 10^4} \approx .00029\) - So, inside the log: \(.0162 + .00029 \approx .0165\) - \(\log_{10}(.0165) \approx -1.782\) - \([-1.782]^{-2} = (1/1.782)^2 \approx .315\) - \(f = .25 \times .315 \approx .079\) --- # Step 4: Solve for Actual Velocity Recall: \[ v = \sqrt{ \frac{2gh}{ 1 + f \frac{L}{D} } } \] Plug in: - \(f = .079\) - \(L = 10~\mathrm{m}\) - \(D = .025~\mathrm{m}\) - \(g = 9.81~\mathrm{m/s^2}\) - \(h = 2~\mathrm{m}\) Calculate \(f \frac{L}{D} = .079 \times \frac{10}{.025} = .079 \times 400 = 31.6\) So denominator: \(1 + 31.6 = 32.6\) \[ v = \sqrt{ \frac{2 \times 9.81 \times 2}{32.6} } = \sqrt{ \frac{39.24}{32.6} } = \sqrt{1.203} \approx 1.10~\mathrm{m/s} \] --- # Step 5: Check and Refine (Iterate if Needed) Calculate actual \(Re\): \[ Re = \frac{1.10 \times .025}{1.01 \times 10^{-6}} \approx \frac{.0275}{1.01 \times 10^{-6}} \approx 2.72 \times 10^4 \] Update \(f\): - \(\frac{5.74}{(2.72 \times 10^4)^{.9}} \approx \frac{5.74}{8.33 \times 10^3} \approx .00069\) - \(\frac{\epsilon}{3.7D} \approx .0162\) (same) - Total inside log: \(.0162 + .00069 = .0169\) - \(\log_{10}(.0169) \approx -1.772\) - \([-1.772]^{-2} \approx .318\) - \(f = .25 \times .318 \approx .08\) So the friction factor is pretty stable. --- # Step 6: Final Answer ## **Exit Velocity** \[ \boxed{v \approx 1.1~\mathrm{m/s}} \] --- ## **Summary Table** | Parameter | Value | Units | |-----------|-------|-------| | Fluid density (\(\rho\)) | 100 | kg/m³ | | Kinematic viscosity (\(v\)) | \(1.01 \times 10^{-6}\) | m²/s | | Pipe length (L) | 10 | m | | Pipe diameter (D) | .025 | m | | Surface roughness (\(\epsilon\)) | .0015 | m | | Tank height (h) | 2 | m | | Exit velocity (v) | 1.1 | m/s | --- ## **Notes** - The calculation assumes the only significant loss is due to friction in the pipe (no minor losses). - If minor losses (entry/exit, bends, etc.) are significant, the actual velocity will be slightly lower. - For pipes with very high roughness, the flow may be in the fully rough regime; here, we are close to that, so the friction factor is dominated by roughness. --- ## **References** - Moody, L. F. (1944). Friction factors for pipe flow. Transactions of the ASME. - White, F. M. (2011). Fluid Mechanics (7th Edition). --- **No image is necessary for this calculation.**