Let's break down and solve the problem **step by step**. Here’s a structured approach: --- ## **Given Data** - **Clay-water slurry properties:** - Yield stress (\(\tau_y\)) = **.7 N/m²** - Density (\(\rho\)) = **1.2 kg/L = 120 kg/m³** - Viscosity (\(\mu\)) = **.045 Pa·s** - **Flow rate required:** \(.004 \mathrm{~m}^3/\mathrm{s}\) - **Tube:** - Length (\(L\)) = **3 m** - Diameter (\(D\)) = **3/8 inch ID = .375 \text{ in} = .009525 \text{ m}** - **Height from tank to pipe** = **.4 m** - **Other details:** - Ball and globe valve - **Pressurized air** is applied to tank (pressure to be found) - Assume **smooth copper tube** - **Account for fittings and elevation changes** --- ## **Step 1: List Required Head Losses** ### **1. Elevation Head (\(h_z\)):** \[ h_z = .4 \text{ m} \] ### **2. Pipe Friction Loss (\(h_f\)):** \[ h_f = f \frac{L}{D} \frac{v^2}{2g} \] ### **3. Fittings Loss (\(h_{\text{fittings}}\)):** \[ h_{\text{fittings}} = K_{\text{total}} \frac{v^2}{2g} \] - Assume \(K_{\text{ball valve}} \approx .05\) - Assume \(K_{\text{globe valve}} \approx 10\) - Total \(K = .05 + 10 = 10.05\) ### **4. Slurry (Bingham plastic) Correction:** --- ## **Step 2: Calculate Flow Velocity** \[ Q = v \cdot A \implies v = \frac{Q}{A} \] \[ A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (.009525)^2 = 7.13 \times 10^{-5} \text{ m}^2 \] \[ v = \frac{.004}{7.13 \times 10^{-5}} = 56.1 \text{ m/s} \] > **This velocity is extremely high for such a small pipe!** > Let's **double-check the diameter**: - **3/8 in** = .375 in × .0254 m/in ≈ **.00953 m** (correct) --- ## **Step 3: Calculate Head Losses** ### **Pipe Friction Loss for Bingham Plastic (Slurry)** The head loss for **Bingham plastics** is more complicated than for Newtonian fluids. #### **Bingham Plastic Flow (Simplified)** The total pressure drop is given by: \[ \Delta P = \frac{4L}{D} \left(\tau_y + \frac{\mu v}{D/2}\right) \] or, for head loss, \[ h_f = \frac{\Delta P}{\rho g} \] #### **Calculate Shear Rate:** For Bingham plastics, wall shear stress: \[ \tau_w = \tau_y + \mu \left(\frac{8v}{D}\right) \] --- ## **Step 4: Calculate Required Pressure at Tank Top** The total head to be supplied by the pressurized air (ignoring atmospheric pressure, since it acts at both ends): \[ \frac{P_{\text{air}}}{\rho g} = h_z + h_{\text{pipe}} + h_{\text{fittings}} \] --- ### **Let's Plug in the Numbers** #### **1. Area and Velocity** \[ A = 7.13 \times 10^{-5} \text{ m}^2 \] \[ v = \frac{.004}{7.13 \times 10^{-5}} = 56.1 \text{ m/s} \] #### **2. Friction Loss (Estimate)** \[ \tau_y = .7 \text{ N/m}^2, \quad \mu = .045 \text{ Pa}\cdot\text{s} \] \[ \tau_{\text{wall}} = \tau_y + \mu \frac{8v}{D} \] \[ \tau_{\text{wall}} = .7 + .045 \times \frac{8 \times 56.1}{.009525} \] \[ = .7 + .045 \times 47132 \] \[ = .7 + 212.9 = 2121.6 \text{ N/m}^2 \] #### **Total Pressure Drop Across Pipe:** \[ \Delta P = \tau_{\text{wall}} \frac{4L}{D} \] \[ = 2121.6 \cdot \frac{4 \times 3}{.009525} \] \[ = 2121.6 \cdot 126.4 = 2.677 \times 10^{6} \text{ Pa} \] #### **Convert to Head Loss:** \[ h_f = \frac{\Delta P}{\rho g} = \frac{2.677 \times 10^6}{120 \times 9.81} \approx 227.5 \text{ m} \] #### **Fittings Loss:** \[ h_{\text{fittings}} = K \frac{v^2}{2g} = 10.05 \cdot \frac{(56.1)^2}{2 \times 9.81} \] \[ = 10.05 \cdot \frac{3148.2}{19.62} \] \[ = 10.05 \cdot 160.5 = 1613. \text{ m} \] #### **Total Head:** \[ h_{\text{total}} = h_z + h_f + h_{\text{fittings}} \] \[ = .4 + 227.5 + 1613. \approx 1841 \text{ m} \] #### **Required Pressure at Tank:** \[ P_{\text{air}} = \rho g h_{\text{total}} = 120 \times 9.81 \times 1841 \] \[ = 21,650,292 \text{ Pa} \] \[ = 21.65 \text{ MPa} \] --- ## **Step 5: Final Answer** ### **Required Gauge Pressure at Tank Top** **The required gauge pressure to maintain a flow rate of .004 m³/s through the 3 m long, 3/8 in ID copper pipe, accounting for the elevation and all losses, is:** \[ \boxed{21.7 \text{ MPa}} \] --- ## **Summary Table** | Head Loss Component | Value (m) | |---------------------|-----------| | Elevation | .4 | | Pipe friction (Bingham) | 227.5 | | Fittings | 1613. | | **Total** | **1841** | --- ### **Notes** - This **extremely high pressure** (over 200 atm) is mainly due to the high velocity through a very small pipe (3/8 in) and the high viscosity of the slurry. **Check if flow rate and pipe size are practical!** - If a more reasonable pressure is needed, **increase the pipe diameter** or **reduce the flow rate**. - All calculations use standard fluid mechanics and Bingham plastic formulas. --- If you have more specific data for the loss coefficients or need to verify assumptions, let me know!