Super practical common injection molding process calculation formulas!
Time：2022-08-11 08:42:36 / Popularity： / Source：
1
1. Clamping force F(TON) formula:
F=Am*Pv/1000
F: Clamping force: TON Am: Cavity projected area: CM2
Pv: Filling pressure: KG/CM2
(General plastic material filling pressure is 150-350KG/CM2)
(The lower value for good liquidity, the higher value for poor liquidity)
Injection pressure=filling pressure/0.4-0.6
Example: Cavity projected area 270CM2 Filling pressure 220KG/CM2
Clamping force=270*220/1000=59.4TON
F: Clamping force: TON Am: Cavity projected area: CM2
Pv: Filling pressure: KG/CM2
(General plastic material filling pressure is 150-350KG/CM2)
(The lower value for good liquidity, the higher value for poor liquidity)
Injection pressure=filling pressure/0.4-0.6
Example: Cavity projected area 270CM2 Filling pressure 220KG/CM2
Clamping force=270*220/1000=59.4TON
2. Injection pressure Pi (KG/CM2) formula:
Pi=P*A/Ao
That is: injection pressure = pump pressure*effective area of injection cylinder/screw cross-sectional area
Pi: Injection pressure P: Pump pressure A: Effective area of injection cylinder
Ao: screw cross-sectional area
A=π*D2/4 D: diameter π: pi 3.14159
Example 1: Knowing pump pressure, find injection pressure?
Pump pressure=75KG/CM2 Effective area of injection cylinder=150CM2
Cross-sectional area of screw=15.9CM2(∮45mm) Formula: 2〒R2 is: 3.1415* (45mm/2)2=1589.5mm2
Pi=75*150/15.9=707KG/CM2
Example 2: Knowing injection pressure, find pumping pressure?
Required injection pressure=900KG/CM2 Effective area of injection cylinder=150CM2
Cross-sectional area of screw=15.9CM2(∮45)
Pump pressure P= Pi*Ao/A=900*15.9/150=95.4 KG/CM2
That is: injection pressure = pump pressure*effective area of injection cylinder/screw cross-sectional area
Pi: Injection pressure P: Pump pressure A: Effective area of injection cylinder
Ao: screw cross-sectional area
A=π*D2/4 D: diameter π: pi 3.14159
Example 1: Knowing pump pressure, find injection pressure?
Pump pressure=75KG/CM2 Effective area of injection cylinder=150CM2
Cross-sectional area of screw=15.9CM2(∮45mm) Formula: 2〒R2 is: 3.1415* (45mm/2)2=1589.5mm2
Pi=75*150/15.9=707KG/CM2
Example 2: Knowing injection pressure, find pumping pressure?
Required injection pressure=900KG/CM2 Effective area of injection cylinder=150CM2
Cross-sectional area of screw=15.9CM2(∮45)
Pump pressure P= Pi*Ao/A=900*15.9/150=95.4 KG/CM2
3. Formula of injection volume V(CM3):
V= π*(1/2Do)2*ST
That is: injection volume = 3.1415 * radius 2 * injection stroke
V: Injection volume CM3 π: Pi 3.1415 Do: Screw diameter CM
ST: Injection stroke CM
Example: screw diameter 42mm, injection stroke 165mm
V= π*(4.2÷2)2*16.5=228.6CM3
That is: injection volume = 3.1415 * radius 2 * injection stroke
V: Injection volume CM3 π: Pi 3.1415 Do: Screw diameter CM
ST: Injection stroke CM
Example: screw diameter 42mm, injection stroke 165mm
V= π*(4.2÷2)2*16.5=228.6CM3
4. Injection weight Vw(g) formula:
Vw=V*η*δ
That is: injection weight = injection volume * specific gravity * mechanical efficiency
Vw: Injection weight g V: Injection volume η: Specific gravity δ: Mechanical efficiency
Example: Injection volume=228.6CM3 Mechanical efficiency=0.85 Specific gravity=0.92
Injection weight Vw=228.6*0.85*0.92=178.7G
That is: injection weight = injection volume * specific gravity * mechanical efficiency
Vw: Injection weight g V: Injection volume η: Specific gravity δ: Mechanical efficiency
Example: Injection volume=228.6CM3 Mechanical efficiency=0.85 Specific gravity=0.92
Injection weight Vw=228.6*0.85*0.92=178.7G
5. Formula of injection speed S(CM/SEC):
S=Q/A
That is: injection speed = pump discharge volume ÷ effective area of injection cylinder
S: Injection speed CM/SEC
A: Effective area of injection cylinder is CM2
Q: pump discharge CC/REV formula: Q=Qr*RPM/60 (per minute/L) namely: pump discharge = pump discharge per revolution * motor revolutions per minute
Qr: pump output per revolution (per revolution/CC)
RPM: Motor revolutions per minute
Example: Motor speed is 1000RPM/min, pump output per revolution is 85 CC/RPM, and effective area of injection cylinder is 140 CM2
S=85*1000/60/140=10.1 CM/SEC
That is: injection speed = pump discharge volume ÷ effective area of injection cylinder
S: Injection speed CM/SEC
A: Effective area of injection cylinder is CM2
Q: pump discharge CC/REV formula: Q=Qr*RPM/60 (per minute/L) namely: pump discharge = pump discharge per revolution * motor revolutions per minute
Qr: pump output per revolution (per revolution/CC)
RPM: Motor revolutions per minute
Example: Motor speed is 1000RPM/min, pump output per revolution is 85 CC/RPM, and effective area of injection cylinder is 140 CM2
S=85*1000/60/140=10.1 CM/SEC
6. Ejection rate Sv(G/SEC) formula:
Sv=S*Ao
That is: injection rate = injection speed * screw cross-sectional area
Sv: Injection rate G/SEC S: Injection speed CM/SEC Ao: Screw cross-sectional area
Example: Injection speed=10CM/SEC Screw diameter∮42
Area=3.14159*4.2*4.2/4=13.85CM2
Sv=13.85*10=138.5G/SEC
That is: injection rate = injection speed * screw cross-sectional area
Sv: Injection rate G/SEC S: Injection speed CM/SEC Ao: Screw cross-sectional area
Example: Injection speed=10CM/SEC Screw diameter∮42
Area=3.14159*4.2*4.2/4=13.85CM2
Sv=13.85*10=138.5G/SEC
2
1. Theoretical output volume: (π/4=0.785)
(1) Screw diameter²*0.785*Injection stroke = Theoretical injection volume (cm³);
(2) Theoretical injection volume/0.785/screw diameter = injection stroke (cm).
(2) Theoretical injection volume/0.785/screw diameter = injection stroke (cm).
2. Shot weight
Theoretical injection volume * plastic specific gravity * injection constant (0.95) ideal = injection weight (gr);
3. Injection pressure
(1)Injection cylinder area²/screw area²*system maximum pressure (140kg/cm²)²=injection pressure (kg/cm²);
(2)Injection cylinder diameter²/screw diameter²*system maximum pressure (140kg/cm²)=injection pressure (kg/cm²);
(3) Maximum injection pressure of material tube combination * actual pressure (kg/cm²) / system maximum pressure (140kg/cm²) = injection pressure (kg/cm²).
(2)Injection cylinder diameter²/screw diameter²*system maximum pressure (140kg/cm²)=injection pressure (kg/cm²);
(3) Maximum injection pressure of material tube combination * actual pressure (kg/cm²) / system maximum pressure (140kg/cm²) = injection pressure (kg/cm²).
4. Injection rate
(1) Screw area (cm²)*Injection speed (cm/sec)=Injection rate (cm³/sec);
(2) Screw diameter (cm²)*0.785*Injection speed (cm/sec)=Injection speed (cm³/sec).
(2) Screw diameter (cm²)*0.785*Injection speed (cm/sec)=Injection speed (cm³/sec).
5. Injection speed
(1) Injection rate (cm³/sec) / screw area (cm²) = injection speed (cm/sec);
(2) Pump single-turn volume (cc/rev)*motor speed (rev/sec)/60(sec)/injection area (cm²)=injection speed (cm/sec).
(motor speed RPM:60HZ------1150,50HZ-----958)
(2) Pump single-turn volume (cc/rev)*motor speed (rev/sec)/60(sec)/injection area (cm²)=injection speed (cm/sec).
(motor speed RPM:60HZ------1150,50HZ-----958)
6. Injection cylinder area
(1) Injection pressure (kg/cm²)/maximum system pressure (140kg/cm²)*material tube area (cm²)=injection cylinder area (cm²);
(2) Single cylinder---(Injection cylinder diameter²-plunger diameter²)*0.785=Injection cylinder area (cm²);
Double Cylinder---(Injection Cylinder Diameter²-Plunger Diameter²)*0.785*2=Injection Cylinder Area(cm²).
(2) Single cylinder---(Injection cylinder diameter²-plunger diameter²)*0.785=Injection cylinder area (cm²);
Double Cylinder---(Injection Cylinder Diameter²-Plunger Diameter²)*0.785*2=Injection Cylinder Area(cm²).
7. Pump single revolution volume
Injection cylinder area (cm²)*Injection speed (cm/sec)*60 seconds/motor speed = pump single revolution volume (cc/sec), (motor speed RPM: 60HZ------1150,50HZ--- --958)
8. Screw speed and hydraulic motor single revolution volume
(1) Pump single-turn volume (cc/rec) * motor speed (RPM) / hydraulic motor single-turn volume = screw speed;
(2) Pump single-turn volume (cc/rec)*motor speed (RPM)/screw speed = hydraulic motor single-turn volume.
(2) Pump single-turn volume (cc/rec)*motor speed (RPM)/screw speed = hydraulic motor single-turn volume.
9. Total injection pressure
(1) Maximum pressure of system (kg/cm²) * area of injection cylinder (cm²) = the total injection pressure (kg);
(2) Injection pressure (kg/cm²) * screw area (cm²) = total injection pressure (kg).
(2) Injection pressure (kg/cm²) * screw area (cm²) = 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 kilogram (kg) = 2.2 pounds (ib); namely: 1 catties = 1.1 pounds;
(4) 1 pound (ib) = 454 grams (gr) = 0.454 kilograms (kg).
(2) 1 pound (ib) = 16 ounces (oz);
(3) 1 kilogram (kg) = 2.2 pounds (ib); namely: 1 catties = 1.1 pounds;
(4) 1 pound (ib) = 454 grams (gr) = 0.454 kilograms (kg).
11. Mold closing force
(1) Curved hand:
Mould-closing cylinder area (cm²)*system maximum pressure (140kg/cm²)/1000*crank hand magnification (20-50) = mould-closing force (Ton)
(2) Single-cylinder direct pressure type: pay attention to world of injection molding
Mould-closing cylinder area (cm²) * system maximum pressure (140kg/cm²)/1000 = mould-closing force (Ton)
Mould-closing cylinder area (cm²)*system maximum pressure (140kg/cm²)/1000*crank hand magnification (20-50) = mould-closing force (Ton)
(2) Single-cylinder direct pressure type: pay attention to world of injection molding
Mould-closing cylinder area (cm²) * system maximum pressure (140kg/cm²)/1000 = mould-closing force (Ton)
12. Formulas for pillar diameter and closing force
Column diameter²(cm²)*0.785*Young's coefficient (scm4 is about 1000kg/cm²)*4=probable value of closing force (Ton).
13. Formula of relationship between projected area of finished product arrangement and closing force
Projected area of finished product arrangement (inch²) * standard thickness (1.5mm) / average thickness of finished product (mm) * raw material constant / PS raw material constant (1) = closing force (Ton);
(1) Projected surface shadow of finished product arrangement takes injection gate as center and long side as radius to calculate diameter; arrangement diameter²(inch²)*0.785=finished product arrangement projected area(inch²).
(2) Using raw material constant, calculated based on rough empirical value, flowability is better than PS and listed as less than 1; poorer than PS is 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¨¨¨etc.
All kinds of raw materials are also divided into different grades, it is advisable to learn more and only pay for participation.
(3) Requirements for finished part of injection direction, such as height of cup, can be calculated as about 30% of projected area.
(1) Projected surface shadow of finished product arrangement takes injection gate as center and long side as radius to calculate diameter; arrangement diameter²(inch²)*0.785=finished product arrangement projected area(inch²).
(2) Using raw material constant, calculated based on rough empirical value, flowability is better than PS and listed as less than 1; poorer than PS is 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¨¨¨etc.
All kinds of raw materials are also divided into different grades, it is advisable to learn more and only pay for participation.
(3) Requirements for finished part of injection direction, such as height of cup, can be calculated as about 30% of projected area.
14. Molding force
Moulding cylinder area (cm²) * system maximum pressure (140kg/cm²)/1000 = moulding force (Ton)
15. Electricity Units
1 horsepower (HP) = 0.754 kilowatts (KW);
1 kilowatt (KW) = 1.326 horsepower (HP) = 1000 watts (W);
1 kilowatt (KW) = 1 kilowatt-hour unit of measure (1KW/Hr).
1 kilowatt (KW) = 1.326 horsepower (HP) = 1000 watts (W);
1 kilowatt (KW) = 1 kilowatt-hour unit of measure (1KW/Hr).
16. Relationship between pump size and horsepower
P=Maximum working pressure (eg: 125kg/cm².140kg/cm²);
Q=One minute spit out volume of oil pump (L/min);
Q=Volume of oil pump per revolution (cc/rec)*motor speed (RPM)/1000=One minute discharge volume of oil pump (L/min).
Applicable horsepower:
P*Q/540=HP;
P*Q/612=KW.
Maximum pressure (LP) that matched motor can reach without deceleration:
HP*450/Q=LP;
KW*612/Q=LP.
Q=One minute spit out volume of oil pump (L/min);
Q=Volume of oil pump per revolution (cc/rec)*motor speed (RPM)/1000=One minute discharge volume of oil pump (L/min).
Applicable horsepower:
P*Q/540=HP;
P*Q/612=KW.
Maximum pressure (LP) that matched motor can reach without deceleration:
HP*450/Q=LP;
KW*612/Q=LP.
17. Calculate electricity usage
(motor capacity + electric heating capacity + dryer capacity) * electricity consumption constant (about 40%) = actual hourly electricity consumption (degree, KW/Hr).
3
1: What is injection capacity of injection molding machine?
Injection capacity = injection pressure (kg/cm2) * injection volume (cm3)/1000
2: What is injection horsepower of injection molding machine?
Injection horsepower PW (KW) = injection pressure (kg/cm2) * injection rate (cm3/sec) * 9.8 * 100%
3: What is injection rate of injection molding machine?
Ejection rate V(cc/sec)=p/4*d2*g
d2: diameter of material tube g: density of material
d2: diameter of material tube g: density of material
4: What is injection thrust of injection molding machine?
Injection thrust F(kgf)=p/4(D12-D22)*P*2
D1: Cylinder inner diameter D2: Piston rod outer diameter P: System pressure
D1: Cylinder inner diameter D2: Piston rod outer diameter P: System pressure
5: What is injection pressure of injection molding machine?
Injection pressure P(kg/cm2)=[p/4*(D12-D22)*P*2]/(p/4*d2)
6: What is plasticizing capacity of 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 screw (cm) S = raw material density
h = tooth depth at the front end of screw (cm) S = raw material density
7: What is system pressure? What is difference with injection pressure?
System pressure (kg/cm2) = the highest working pressure set in hydraulic circuit
8: Injection speed?
H: speed = distance/time
Recommended
Related
- A brief introduction to current status and development trend of die-casting aluminum alloy industry12-28
- Technical points for injection molding of transparent products.12-28
- Influence of external factors on quality of die castings in die casting production and countermeasur12-27
- Injection mold 3D design sequence and design key points summary12-27
- Effect of heat treatment on structure and mechanical properties of die-cast AlSi10MnMg shock tower12-26