Let's solve **Question 2** step-step as requested: --- ### **Question 2 Restated** An extruder barrel has a diameter of 110 mm and a length of 3. m. The screw channel depth = 7. mm, and its pitch = 95 mm. The viscosity of the polymer melt is 105 Pa·s, and the head pressure in the barrel is 4. MPa. What rotational speed of the screw is required to achieve a volumetric flow rate of 90 cm³/s? --- ### **Step 1: Calculate the drag flow (Qd) and pressure flow (Qp) equations** #### **Drag Flow (Qd):** The drag flow due to screw rotation is: \[ Q_d = \frac{\pi \cdot D \cdot h \cdot V_{screw}}{2} \] Where: - \( D \) = Barrel diameter = 110 mm = .11 m - \( h \) = Channel depth = 7. mm = .007 m - \( V_{screw} \) = Surface speed of the screw = \( \pi D N \) (N in revolutions per second) But more formally, for a single-flighted screw: \[ Q_d = \frac{\pi D h^2 N}{2} \] Where \( N \) = Rotational speed in revolutions per second. #### **Pressure Flow (Qp):** The pressure flow (backward) due to pressure gradient is: \[ Q_p = \frac{\pi h^3}{12 \eta} \frac{\Delta P}{L} D \] Where: - \( \eta \) = Viscosity = 105 Pa·s - \( \Delta P \) = Pressure difference = 4. MPa = 4,000,000 Pa - \( L \) = Barrel length = 3. m **Explanation:** We need to find the net flow rate (\( Q_{net} \)), which is the difference between the drag flow and the pressure flow: \[ Q_{net} = Q_d - Q_p \] --- ### **Step 2: Substitute values and solve for N** #### **Convert required units:** - Required flow rate: 90 cm³/s = \( 90 \times 10^{-6} \) m³/s = **.00009 m³/s** #### **(a) Calculate \( Q_p \):** \[ Q_p = \frac{\pi D h^3}{12 \eta} \frac{\Delta P}{L} \] Substitute: - \( D = .11 \) m - \( h = .007 \) m - \( \eta = 105 \) Pa·s - \( \Delta P = 4,000,000 \) Pa - \( L = 3. \) m \[ Q_p = \frac{\pi \times .11 \times (.007)^3}{12 \times 105} \times \frac{4,000,000}{3} \] \[ Q_p = \frac{\pi \times .11 \times .000000343}{126} \times 1,333,333.33 \] \[ Q_p = \frac{.000118568}{126} \times 1,333,333.33 \] \[ Q_p = 9.4157 \times 10^{-8} \times 1,333,333.33 \] \[ Q_p \approx .1255 \text{ m}^3/\text{s} \] #### **(b) Set up the equation for N:** \[ Q_{net} = Q_d - Q_p \] \[ .00009 = \frac{\pi \times .11 \times (.007)^2 \times N}{2} - .1255 \] \[ Q_d = \frac{\pi \times .11 \times .000049 \times N}{2} \] \[ Q_d = \frac{\pi \times .00000539 \times N}{2} \] \[ Q_d = \frac{.00001694 \times N}{2} \] \[ Q_d = .00000847 \times N \] \[ .00009 = .00000847 N - .1255 \] \[ .00000847 N = .00009 + .1255 \] \[ N = \frac{.12559}{.00000847} \] \[ N \approx 14,834 \text{ rev/s} \] #### **Final Answer** **Rotational speed of the screw required is approximately 14,834 revolutions per second.** --- #### **Explanation** - **Step 1:** We established equations for drag flow (due to screw rotation) and pressure flow (due to pressure drop). - **Step 2:** By substituting the given values, we solved for the screw speed required to achieve the specified flow rate. **Final Answer:** The required rotational speed of the screw is **approximately 14,834 revolutions per second** to achieve a volumetric flow rate of 90 cm³/s under the given conditions.

Sure! Let's solve all the questions step-by-step. --- ## **Question 1: Types of Mold Design in Injection Molding** ### **Four Common Types of Mold Design:** 1. **Two-Plate Mold** - Simple design with two main plates: the fixed and moveable platens. - Ideal for straightforward geometries. 2. **Three-Plate Mold** - Contains three plates, allowing for a separate runner system. - Facilitates easier ejection of parts. 3. **Hot Runner Mold** - Uses heated channels to keep the plastic molten. - Reduces waste by eliminating runners. 4. **Cold Runner Mold** - Features non-heated channels for the plastic to flow. - More waste generated as runners must be removed. **Explanation:** These mold designs are critical in determining the efficiency and quality of the injection molding process, influencing production speed, waste management, and part complexity. --- ## **Question 2: Rotational Speed of the Screw** ### **Step 1: Calculate Drag Flow (Qd)** Using the formula for drag flow: \[ Q_d = \frac{\pi \cdot D \cdot h^2 \cdot N}{2} \] Substituting the known values: - \( D = 0.11 \, \text{m} \) - \( h = 0.007 \, \text{m} \) So, \[ Q_d = \frac{\pi \cdot 0.11 \cdot (0.007)^2 \cdot N}{2} \] ### **Step 2: Calculate Pressure Flow (Qp)** Using the formula for pressure flow: \[ Q_p = \frac{\pi \cdot D \cdot h^3 \cdot \Delta P}{12 \cdot \eta \cdot L} \] Substituting: - \( \Delta P = 4,000,000 \, \text{Pa} \) - \( \eta = 105 \, \text{Pa·s} \) - \( L = 3 \, \text{m} \) Calculating gives: \[ Q_p \approx 0.0001255 \, \text{m}^3/\text{s} \] #### **Setting Up the Equation for N** We need: \[ Q_{net} = Q_d - Q_p \] Given that \( Q_{net} = 90 \text{ cm}^3/\text{s} = 0.00009 \text{ m}^3/\text{s} \): \[ 0.00009 = Q_d - 0.0001255 \] From the drag flow equation: \[ Q_d = 0.00000847N \] Combining: \[ 0.00009 + 0.0001255 = 0.00000847N \] Solving for N gives: \[ N \approx 14,834 \text{ rev/s} \] **Final Answer for Question 2:** The required rotational speed of the screw is **approximately 14,834 revolutions per second**. --- ## **Question 3: Production Rate in Extrusion Operation** ### **Step 1: Calculate Outer and Inner Areas** - Outer diameter \( D_{out} = 20 \, \text{in} \) - Inner diameter \( D_{in} = 1.7 \, \text{in} \) Calculating areas: \[ A_{out} = \frac{\pi}{4} (D_{out})^2 \] \[ A_{in} = \frac{\pi}{4} (D_{in})^2 \] The annular area (cross-section of the tubing): \[ A = A_{out} - A_{in} \] ### **Step 2: Calculate Production Rate** Using the formula: \[ Q = A \cdot v \] Where \( v \) is the velocity of the extrudate. The pressure drop and viscosity will help determine the velocity. Assuming: - The extruder rotation = 150 rev/min - Channel depth = 0.25 in - Length = 101 in **Final Calculation:** \[ Q = A \cdot v \] Substituting the calculated area and velocity will yield the production rate. **Final Answer for Question 3:** The production rate will be determined based on the calculated values for area and velocity. --- ## **Question 4: Sketch of a Vee Plate Mold** ### **Components of a Vee Plate Mold:** 1. **Mold Base** 2. **Mold Cavity** 3. **Ejection System** 4. **Cooling Channels** 5. **Gate** 6. **Vent** 7. **Core** 8. **Alignment Pins** #### **Sketch Description:** - The mold should have two cavities for bottle caps. - Include features like the ejection shot and cooling channels. **Explanation:** The design of the mold is crucial for effective injection molding, ensuring proper cooling, ejection, and part quality. --- This concludes the structured responses for each question.