Calculation of Heat Transfer Area of Injection Molding Cooling Circuit
During the injection molding process, the design of the cooling system directly impacts the molding cycle and dimensional accuracy of the product. The heat transfer area of the cooling circuit is a key parameter in cooling system design. Properly calculating this heat transfer area ensures that the product cools to the demolding temperature within the specified time, avoiding defects such as deformation and sink marks caused by insufficient cooling, while also improving production efficiency. The following, based on the principles of heat transfer, details the calculation method for the heat transfer area of the injection molding cooling circuit, including basic calculation formulas, parameter determination, and practical application considerations.
The calculation of the heat transfer area of the cooling circuit is based on the basic law of heat conduction. The core formula is: Heat transfer area ( A ) = heat to be dissipated ( Q ) ÷ (heat transfer coefficient ( K ) × Logarithmic mean temperature difference (ΔT ) . The heat to be dissipated, Q , is the amount of heat released when the plastic melt cools from the molding temperature to the demolding temperature, expressed in watts ( W ). The heat transfer coefficient, K , reflects the heat transfer efficiency between the cooling medium and the mold, expressed in W/ ( m²・°C). The logarithmic mean temperature difference, ΔT , is the average temperature difference between the cooling medium inlet and outlet temperatures and the mold cavity surface temperature , expressed in °C. This formula comprehensively considers various factors in the heat transfer process and serves as the theoretical basis for calculating the heat transfer area.
the amount of heat to be dissipated, Q , is the first step in calculating the heat transfer area. Its value can be determined based on parameters such as the plastic’s specific heat capacity, mass, and temperature difference. The formula is: Q = G × c × ( T1 – T2 ) ÷ t , where G is the mass of plastic injected per injection ( kg ), c is the plastic’s specific heat capacity ( kJ/ ( kg・°C)), T1 is the molding temperature of the plastic melt (°C), T2 is the demolding temperature of the finished product (°C), and t is the cooling time ( s ). For example, consider a product made of polyethylene ( PE ), with a mass of 0.5 kg per injection molding . The specific heat capacity of PE is c = 2.3 kJ/ ( kg · °C), the molding temperature T1 = 200 °C, the demolding temperature T2 = 40 °C, and the cooling time t = 30 seconds . Then, Q = 0.5 × 2.3 × (200 – 40) ÷ 30 = 0.5 × 2.3 × 160 ÷ 30 ≈ 6.13 kJ/s = 6130 W. It’s important to note that in actual production, mold heat absorption and ambient heat dissipation must also be considered. Therefore, the calculated Q value needs to be multiplied by a correction factor of 1.1 to 1.3.
the heat transfer coefficient (K) requires consideration of factors such as the cooling medium type, flow rate, mold material, and cooling circuit structure. Water is a commonly used cooling medium, and its K value is significantly affected by flow rate. When the water flow rate is 1-2 m/s , the K value is generally between 1000 and 3000 W/ ( m²・°C). If oil is used as the cooling medium, the K value is lower, typically between 300 and 800 W/ ( m²・°C). The thermal conductivity of the mold material also affects the K value. For example, steel has a higher thermal conductivity than cast iron, so steel molds have a relatively higher K value. Furthermore, factors such as the roughness and presence of rust in the cooling circuit can affect heat transfer efficiency. In practice, the appropriate K value must be selected based on the specific application. A range of 1500-2000 W/ ( m²・°C) is generally recommended .
To calculate the logarithmic mean temperature difference ΔT, you need to know the cooling medium’s inlet temperature (Tin), outlet temperature (Tout), and mold cavity surface temperature (Tm). The formula is: ΔT = ((Tm-Tin)-(Tm-Tout)) ÷ ln ((Tm-Tin)/(Tm-Tout)). For example, if the mold cavity surface temperature Tm = 60°C, the cooling water inlet temperature Tin = 20°C, and the outlet temperature Tout = 30°C, then ΔT = ((60-20)-(60-30)) ÷ ln ((60-20)/(60-30)) = (40-30) ÷ ln (40/30) ≈ 10 ÷ 0.2877 ≈ 34.76°C. In actual calculations, if the temperature difference between the inlet and outlet of the cooling medium is small (usually the temperature difference is required to be no more than 5°C), it can be simplified to the arithmetic mean temperature difference, that is, ΔT≈((Tm-Tin)+(Tm-Tout))÷2, to reduce the amount of calculation.
determining the values of Q , K , and ΔT , the heat transfer area A can be calculated using a basic formula . For the example above, with Q = 6130W , K = 1800W/ ( m²・°C), and ΔT = 34.76 °C, then A = 6130 ÷ ( 1800 × 34.76 ) ≈ 6130 ÷ 62568 ≈ 0.098m² = 980cm² . In actual design, to ensure effective cooling, a margin of 10% to 20% should be added to the calculated heat transfer area . Furthermore, the layout of the cooling circuits, such as the number, diameter, and spacing of the circuits, must be considered to ensure uniform distribution of the heat transfer area within the mold and avoid insufficient cooling in certain areas. Scientifically calculating the heat transfer area of the cooling circuits provides a reliable basis for optimizing the cooling system design, thereby improving injection molding efficiency and part quality.
