Super practical calculation formula of common injection molding process!
1. Formula of clamping force F (TON):
F=Am*Pv/1000
F: Clamping force: TON Am: projected area of mold cavity: CM2
Pv: filling pressure: KG/CM2
(Generally, the filling pressure of plastic materials is 150-350KG/CM2)
(The lower value shall be taken for good liquidity, and the higher value shall be taken for bad liquidity)
Ejection pressure=filling pressure/0.4-0.6
Example: projected area of mold cavity 270CM2, filling pressure 220KG/CM2
Clamping force=270 * 220/1000=59.4TON
2. Ejection pressure Pi (KG/CM2) formula:
Pi=P*A/Ao
Namely: injection pressure=pump pressure * effective area of injection cylinder ÷ sectional area of screw
Pi: injection pressure P: pump pressure A: effective area of injection cylinder
Ao: screw sectional area
A=π * D2/4 D: diameter π: pi 3.14159
Example 1: How to calculate the injection pressure when the pump pressure is known?
Pump pressure=75KG/CM2 Effective area of injection cylinder=150CM2
Screw sectional area=15.9CM2 (｜ 45mm) Formula: 2 ° R2, that is, 3.1415 * (45mm ÷ 2) 2=1589.5mm2
Pi=75*150/15.9=707 KG/CM2
Example 2: How to calculate the pump pressure when the injection pressure is known?
Required injection pressure=900KG/CM2 Effective area of injection cylinder=150CM2
Screw sectional area=15.9CM2 (｜ 45)
Pump pressure P=Pi * Ao/A=900 * 15.9/150=95.4 KG/CM2
3. Ejection volume V (CM3) formula:
V= π*(1/2Do)2*ST
Namely: ejection volume=3.1415 * radius 2 * ejection stroke
5: Ejection volume CM3 π: Pi 3.1415 Do: screw diameter CM
ST: ejection stroke CM
Example: screw diameter 42mm, injection stroke 165mm
V= π*(4.2÷2)2*16.5=228.6CM3
4. Ejection weight Vw (g) formula:
Vw=V* η*δ
Namely: ejection weight=ejection volume * specific gravity * mechanical efficiency
Vw: Ejection weight g V: Ejection volume η： Specific gravity δ： Mechanical efficiency
Example: Ejection volume=228.6CM3 Mechanical efficiency=0.85 Specific gravity=0.92
Ejection weight Vw=228.6 * 0.85 * 0.92=178.7G
5. Ejection velocity S (CM/SEC) formula:
S=Q/A
Namely: injection speed=pump output ÷ effective area of injection cylinder
S: Ejection velocity CM/SEC
A: Effective area of injection cylinder CM2
Q: Pump discharge CC/REV formula: Q=Qr * RPM/60 (per minute/L), that is, pump discharge=pump discharge per revolution * motor rotation/minute
Qr: pump output per revolution (per revolution/CC)
RPM: motor rotation/minute
Example: the motor speed is 1000 RPM/minute, the pump output per revolution is 85 CC/RPM, and the effective area of the injection cylinder is 140 CM2
S=85*1000/60/140=10.1 CM/SEC
6. Ejection rate Sv (G/SEC) formula:
Sv=S*Ao
Namely: injection rate=injection speed * screw sectional area
Sv: Ejection rate G/SEC S: Ejection speed CM/SEC Ao: Screw sectional area
Example: injection speed=10CM/SEC screw diameter ｜ 42
Area=3.14,159 * 4.2 * 4.2/4=13.85CM2
Sv=13.85*10=138.5G/SEC
2
1. Theoretical volume: (π/4=0.785)
(1) Screw diameter ²* 0.785 * ejection stroke=theoretical ejection volume (cm ³);
(2) Theoretical injection volume/0.785/screw diameter=injection stroke (cm)
2. Ejection weight
Theoretical injection volume * plastic specific gravity * injection constant (0.95) Ideal=injection weight (gr);
3. Ejection pressure
(1) Injection cylinder area ²/ Screw area ²* Maximum system pressure (140kg/cm ²)²= Ejection pressure (kg/cm ²);
(2) Diameter of injection cylinder ²/ Screw diameter ²* Maximum system pressure (140kg/cm ²)= Ejection pressure (kg/cm ²);
(3) Maximum injection pressure of material pipe combination * actual use pressure (kg/cm ²)/ Maximum system pressure (140kg/cm ²)= Ejection pressure (kg/cm ²).
4. Ejection rate
(1) Screw area (cm ²)* Ejection velocity (cm/sec)=Ejection velocity (cm ³/ sec);
(2) Screw diameter (cm ²)* 0.785 * ejection velocity (cm/sec)=ejection velocity (cm ³/ sec).
5. Ejection speed
(1) Ejection rate (cm ³/ Sec)/screw area (cm ²)= Ejection velocity (cm/sec);
(2) Pump capacity per revolution (cc/rev) * motor speed (rev/sec)/60 (s)/output area (cm ²)= Ejection velocity (cm/sec)
(Motor speed RPM: 60HZ ——- 1150, 50HZ —— 958)
6. Injection cylinder area
(1) Ejection pressure (kg/cm ²)/ Maximum system pressure (140kg/cm ²)* Material pipe area (cm ²)= Injection cylinder area (cm ²);
(2) Single cylinder — (cylinder diameter ²- Plunger diameter ²)* 0.785=injection cylinder area (cm ²);
Double cylinder — (cylinder diameter ²- Plunger diameter ²)* 0.785 * 2=area of injection cylinder (cm ²).
7. Pump single rotation volume
Injection cylinder area (cm ²)* Ejection speed (cm/sec) * 60 seconds/motor speed=pump single revolution volume (cc/sec), (motor speed RPM: 60HZ ——- 1150, 50HZ ——- 958)
8. Screw speed and single rotation volume of hydraulic motor
(1) Pump single rotation volume (cc/rec) * motor speed (RPM)/hydraulic motor single rotation volume=screw speed;
(2) Pump single rotation volume (cc/rec) * motor speed (RPM)/screw speed=hydraulic motor single rotation volume
9. Total injection pressure
(1) Maximum system pressure (kg/cm ²)* Injection cylinder area (cm ²)= Total injection pressure (kg);
(2) Ejection pressure (kg/cm ²)* Screw area (cm ²)= Total injection pressure (kg)
10. Conversion of ounces and related units
(1) 1 ounce (oz)=28.375 grams (gr);
(2) 1 pound (ib)=16 ounces (oz);
(3) 1 kg=2.2 lb (ib); That is, 1 jin=1.1 pounds;
(4) 1 pound (ib)=454 grams (gr)=0.454 kilograms (kg)
11. Mold closing force
(1) Hand curling:
Area of mold closing cylinder (cm ²)* Maximum system pressure (140kg/cm ²)/ 1000 * hand bending magnification (20-50)=mold closing force (Ton)
(2) Single cylinder direct compression type: focus on the world of injection molding
Area of mold closing cylinder (cm ²)* Maximum system pressure (140kg/cm ²)/ 1000=mold closing force (Ton)
12. Formula of track column diameter and mold closing force
Track column diameter ² (cm ²)* 0.785 * Young’s coefficient (about 1000kg/cm for scm4 ²)* 4=approximate closing force (Ton)
13. Relation formula between finished product arrangement projection area and mold closing force
Finished product arrangement projection area (inches ²)* Standard thickness (1.5mm)/average thickness of finished product (mm) * raw material constant used/PS raw material constant (1)=mold closing force (Ton);
(1) The diameter of the projection surface shadow of the finished product arrangement is calculated by taking the injection gate as the center and the long side as the radius; Arrangement diameter ² (Inches ²)* 0.785=finished product arrangement projection area (inch ²).
(2) The raw material constant is used, calculated by approximate empirical value, and the good/bad liquidity is less than 1; Those worse than PS are listed as more than 1. For example:
ABS 1.05; AS 1.2; PMMA 1.3; PC 1.6; PBT 0.9;
PP 0.7; PE 0.7-0.8; Plastic steel 0.8; NILON 0.7-0.9
All kinds of raw materials are also divided into different grades. It is advisable to know more about them and only pay for them
(3) And the requirements for the finished product part of the injection direction, such as the height of the cup is about 30% of the projected area
14. Supporting force
Area of mould supporting cylinder (cm ²)* Maximum system pressure (140kg/cm ²)/ 1000=supporting force (Ton)
15. Power unit
1 HP=0.754 kW;
1 kW=1.326 HP=1000 W;
1 kW=1 kWh (1KW/Hr)
16. Relation between pump size and horsepower
P=maximum service pressure (e.g. 125kg/cm ². 140kg/cm ²);
Q=one minute output of oil pump (L/min);
Q=pump capacity per revolution (cc/rec) * motor speed (RPM)/1000=pump output per minute (L/min)
Applicable horsepower:
P*Q/540=HP;
P*Q/612=KW.
Maximum pressure (LP) of the matched motor without speed reduction:
HP*450/Q=LP;
KW*612/Q=LP.
17. Calculation of electricity use
(motor capacity+electric heating capacity+dryer capacity) * power consumption constant (about 40%)=actual hourly power consumption (kWh/Hr)
3
1: What is the injection ability of injection molding machine?
Ejection capacity=Ejection pressure (kg/cm2) × Ejection volume (cm3)/1000
2: What is the power of injection molding machine?
Ejection horsepower PW (KW)=Ejection pressure (kg/cm2) × Ejection rate (cm3/sec) × nine point eight × 100%
3: What is the injection rate of injection molding machine?
Ejection rate V (cc/sec)=p/4 × d2 × g
D2:: material pipe diameter g: material density
4: What is the injection force of the injection molding machine?
Glue injection thrust F (kgf)=p/4 (D12-D22) × P × two
D1: inner diameter of oil cylinder D2: outer diameter of piston rod P: system pressure
5: What is the injection pressure of the injection molding machine?
Injection pressure P (kg/cm2)=[p/4 × (D12-D22) × P × 2]/(p/4 × d2)
6: What is the plasticizing ability of an injection molding machine?
Plasticizing capacity W (g/sec)=2.5 × (d/2.54)2 × (h/2.54) × N × S × 1000/3600/2
H=tooth depth at the front end of the screw (cm) S=raw material density
7: What is system pressure? What is the difference with injection pressure?
System pressure (kg/cm2)=the highest working pressure set in the oil pressure circuit
8: Shooting speed?
H: Speed=Distance/Time
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