Let's solve the injection molding problem step by step. --- ### **Given:** - **Number of plates:** 2 - **Plate dimensions:** \( 60 \times 120 \times 1 \) mm - **Material:** Polycarbonate (PC) - **Melt (fill) temperature:** \( T_{melt} = 300^\circ C \) - **Mold (ejection) temperature:** Typically for PC, ejection at ~90°C (assume unless otherwise given) - **Fill time:** \( t_{fill} = 2 \) s - **Gate:** Edge gate, diameter \( D_{gate} = 0.5 \) mm, length \( L_{gate} = 2 \) mm - **Sprue:** Length \( = 12 \) cm, \( D_1 = 4 \) mm (base), \( D_2 = 7 \) mm (top) - **Runners:** 2 circular, \( L = 1.5 \) cm, \( D = 5 \) mm #### **Assumed/Typical Properties for PC:** - Density (\( \rho \)): \( 1.2 \) g/cm³ - Viscosity: Use typical for PC at 300°C, estimate as needed - Heat capacity (\( C_p \)): \( 2.0 \) J/g·°C - Thermal conductivity (\( k \)): \( 0.2 \) W/m·K --- ## **a. Minimum Pressure to Fill the Cavity** ### **Step 1: Calculate Total Volume to Fill** Each plate: \( V_{plate} = 60\,\text{mm} \times 120\,\text{mm} \times 1\,\text{mm} = 7200\,\text{mm}^3 \) For 2 plates: \( V_{total,plates} = 2 \times 7200 = 14,400\,\text{mm}^3 = 14.4\,\text{cm}^3 \) #### **Add runner system volume:** - **Sprue** (truncated cone): \( V_{sprue} = \frac{\pi h}{3}(R_1^2 + R_1 R_2 + R_2^2) \) \( h = 120\,\text{mm},\ R_1 = 2\,\text{mm},\ R_2 = 3.5\,\text{mm} \) \( V_{sprue} = \frac{\pi \times 120}{3}(2^2 + 2 \times 3.5 + 3.5^2) = \frac{\pi \times 120}{3}(4 + 7 + 12.25) = \frac{\pi \times 120}{3}(23.25) \) \( = 40\pi \times 23.25 \approx 40 \times 3.14 \times 23.25 \approx 125.6 \times 23.25 \approx 2918\,\text{mm}^3 \) - **Runners** (2): Each: \( V = \pi r^2 L = \pi (2.5)^2 \times 15 = 3.14 \times 6.25 \times 15 \approx 3.14 \times 93.75 \approx 294.5\,\text{mm}^3 \) Two runners: \( 2 \times 294.5 = 589\,\text{mm}^3 \) - **Gate (edge, small volume):** \( V_{gate} = \pi r^2 L = \pi (0.25)^2 \times 2 = 3.14 \times 0.0625 \times 2 \approx 3.14 \times 0.125 = 0.39\,\text{mm}^3 \) Negligible compared to the rest. **Total volume to fill:** \( V_{total} = 14,400 + 2918 + 589 = 17,907\,\text{mm}^3 \approx 17.9\,\text{cm}^3 \) --- ### **Step 2: Calculate Required Flow Rate** \( Q = \frac{V_{total}}{t_{fill}} = \frac{17,900\,\text{mm}^3}{2\,\text{s}} = 8,950\,\text{mm}^3/\text{s} = 8.95\,\text{cm}^3/\text{s} = 8.95 \times 10^{-6}\,\text{m}^3/\text{s} \) --- ### **Step 3: Calculate Pressure Drop (Simplified)** Pressure drop is mainly due to flow through **gate** (smallest cross-section). - **Gate area:** \( A_{gate} = \pi r^2 = \pi (0.25\,\text{mm})^2 = 0.196\,\text{mm}^2 = 1.96 \times 10^{-7}\,\text{m}^2 \) - **Gate velocity:** \( v_{gate} = \frac{Q}{A_{gate}} = \frac{8.95 \times 10^{-6}}{1.96 \times 10^{-7}} \approx 45.7\,\text{m/s} \) Assume melt viscosity for PC at 300°C: \( \eta = 0.4\,\text{Pa}\cdot\text{s} \) (typical value). - **Pressure drop through short capillary (edge gate):** \( \Delta P = \frac{8 \eta L v}{r^2} \) - \( L = 2\,\text{mm} = 0.002\,\text{m} \) - \( r = 0.25\,\text{mm} = 0.00025\,\text{m} \) - \( v = 45.7\,\text{m/s} \) \( \Delta P = \frac{8 \times 0.4 \times 0.002 \times 45.7}{(0.00025)^2} \) \( = \frac{0.8 \times 45.7 \times 0.002}{6.25 \times 10^{-8}} \) \( = \frac{0.8 \times 45.7 \times 0.002}{6.25 \times 10^{-8}} \) \( = \frac{0.07312}{6.25 \times 10^{-8}} \approx 1,170,000\,\text{Pa} = 1.17\,\text{MPa} \) **Add runner, sprue losses (smaller compared to gate), multiply by a factor (say, 1.5) for total pressure:** **Total minimum pressure:** \( P_{min} \approx 1.5 \times 1.17 = 1.76\,\text{MPa} \) #### **Typical injection molding pressures for PC are 80–120 MPa.** Our calculation is an ideal minimum, actual machines use much higher pressure to account for viscosity rise, temperature gradients, etc. But as per the question, this is the calculated minimum. --- ## **b. Clamping Force** - **Projected area (A):** Largest area the cavity sees. - **For plates:** \( A = 2 \times (60\,\text{mm} \times 120\,\text{mm}) = 2 \times 7200 = 14,400\,\text{mm}^2 = 0.0144\,\text{m}^2 \) - **Clamping force:** \( F_{clamp} = P_{injection} \times A_{projected} \) - Use a safety factor (1.1–1.2×, use 1.2). \( F_{clamp} = 1.76 \times 10^6 \times 0.0144 \times 1.2 = 1.76 \times 0.0144 = 25,344\,\text{N} \) \( F_{clamp} = 25,344 \times 1.2 = 30,412\,\text{N} \approx 30.4\,\text{kN} \) (Again, this is for the *minimum* pressure.) **If using a typical PC injection pressure (say, 80 MPa):** \( F_{clamp} = 80 \times 10^6 \times 0.0144 \times 1.2 = 1,382,400\,\text{N} = 1,382\,\text{kN} \) **Final answer (minimum):** \( F_{clamp,min} \approx 30.4\,\text{kN} \) --- ## **c. Cooling Time (Fourier Heat Transfer)** For thin plates, cooling time can be estimated: \[ t_{cool} = \frac{\left( \text{thickness} \right)^2}{\pi^2 \alpha} \cdot \ln \left( \frac{8 (T_m - T_w)}{\pi^2 (T_e - T_w)} \right) \] Where: - **thickness (\( s \)):** 1 mm \( = 0.001\,\text{m} \) - **\( \alpha \):** Thermal diffusivity \( = \frac{k}{\rho C_p} \) Given: - \( k = 0.2\,\text{W/m·K} \) - \( \rho = 1200\,\text{kg/m}^3 \) - \( C_p = 2000\,\text{J/kg·K} \) \[ \alpha = \frac{0.2}{1200 \times 2000} = \frac{0.2}{2.4 \times 10^6} = 8.33 \times 10^{-8}\,\text{m}^2/\text{s} \] Let: - \( T_m = 300^\circ C \) (melt) - \( T_w \) (mold wall temp) = typically 80°C for PC - \( T_e \) (eject temp) = 90°C for PC \[ t_{cool} = \frac{(0.001)^2}{\pi^2 \cdot 8.33 \times 10^{-8}} \cdot \ln \left( \frac{8(300 - 80)}{\pi^2 (90-80)} \right) \] Calculate: - \( (0.001)^2 = 1 \times 10^{-6} \) - \( \pi^2 = 9.87 \) - \( \pi^2 \cdot 8.33 \times 10^{-8} = 8.22 \times 10^{-7} \) - \( \frac{1 \times 10^{-6}}{8.22 \times 10^{-7}} = 1.216 \) - \( 8 \times (300-80) = 8 \times 220 = 1760 \) - \( \pi^2 \times (90-80) = 9.87 \times 10 = 98.7 \) - \( \frac{1760}{98.7} = 17.84 \) - \( \ln(17.84) = 2.883 \) So, \[ t_{cool} = 1.216 \times 2.883 = 3.5\,\text{s} \] --- ## **d. Water Flow Rate to Cool the Plates** **Heat to be removed per cycle:** \[ Q = m C_p \Delta T \] - \( m = \text{mass of 2 plates} \) - \( V_{plates} = 14.4\,\text{cm}^3 = 14.4 \times 10^{-6}\,\text{m}^3 \) - \( m = 14.4 \times 10^{-6} \times 1200 = 0.0173\,\text{kg} \) - \( C_p = 2000\,\text{J/kg}\cdot K \) - \( \Delta T = 300 - 90 = 210\,^\circ C \) \[ Q = 0.0173 \times 2000 \times 210 = 0.0173 \times 420,000 = 7,266\,\text{J} \] **Cooling water:** \[ Q = \dot{m}_{water} C_{p,water} \Delta T_{water} \] Assume \( \Delta T_{water} = 5^\circ C \) (from 20 to 25°C), \( C_{p,water} = 4186\,\text{J/kg}\cdot K \) \[ \dot{m}_{water} = \frac{Q}{C_{p,water} \Delta T_{water}} = \frac{7,266}{4186 \times 5} = \frac{7,266}{20,930} \approx 0.347\,\text{kg} \] **Per cycle (say, 10 s):** \[ \dot{m}_{water,rate} = \frac{0.347}{10} = 0.0347\,\text{kg/s} \] **Convert to L/min:** - \( 1\,\text{kg} \approx

**Final Answers:** **a. Minimum pressure to fill the cavity:** Approximately **1.76 MPa** (theoretical minimum at gate; actual injection pressures are typically much higher, around 80–120 MPa for PC). **b. Clamping force:** Minimum **~30.4 kN** based on calculated pressure; practical clamping force for production is usually in the range of **1,382 kN** at typical injection pressures (~80 MPa). **c. Cooling time:** Approximately **3.5 seconds** for the plates to cool from 300°C to their ejection temperature (~90°C), considering Fourier heat transfer. **d. Water flow rate to cool the plates:** Approximately **0.035 kg/sec** (or about **2.1 L/min**, assuming water density ~1 kg/L), to remove the heat load in about 10 seconds. --- *Note:* These calculations involve approximations and typical property values; actual process parameters may vary based on equipment and specific material conditions.

**a. Minimum pressure to fill the cavity:** The pressure needed to inject the melt through the gate is about 1.76 MPa based on flow calculations. However, actual machines usually require much higher pressure (80–120 MPa) to overcome viscosity and ensure complete filling. **b. Clamping force:** The clamping force needed depends on the projected area of the mold. For the plates, this area is about 0.0144 m². Using the minimum pressure, the force is roughly 30 kN, but in practice, pressures are higher, so the force is around 1380 kN for typical operation. --- **c. Cooling time:** It takes approximately 3.5 seconds for the plates to cool from 300°C to about 90°C, based on heat conduction principles. This ensures the parts are cooled enough to eject without warping. **d. Water flow rate:** To cool the plates effectively, about 0.035 kg of water per second (around 2 liters per minute) is needed, which removes the heat in roughly 10 seconds. This maintains proper cycle times and part quality.