### Pneumatic Cylinders

#### 1. Design of pneumatic cylinders (double acting)

A pneumatic cylinder is designed for converting pressure energy into useful work (motion). A double acting cylinder is operated by the reciprocal input of compressed air. When compressed air is applied to the rear port of the cylinder while the other side is open to the atmosphere, the cylinder starts to advance. To return the piston to its initial position the air supply has to be connected to the front port while the rear chamber of the cylinder has to be exhausted. The switching of air is done by means of a directional control valve.If large masses are moved by a cylinder, cushioning is used in the end positions. Before reaching the end position, a cushioning piston interrupts the direct flow of air to the outside. The air is forced to flow through flow control valve. Therefore the speed of the piston is slowed down for the last part of the stroke to reduce impact on the cylinder.

· Pneumatic end position damping is adjustable via two regular screws.

· The Piston of the cylinder can be fitted with a permanent magnet which can be used to operate a proximity switch for position sensing

· Cylinder barrel is usually made from seamless drawn steel tubing and can be coated to withstand harsh environments.

· Piston rod is preferably made form heat-treated, rust-proofed type of steel.

· The two caps are fastened to the cylinder barrel by tie rods

· The seals are made from either Perbunan (up to +80 °C), Viton (up to +190°C) or Teflon (up

to +200 °C)

#### 2. What is pressure?

The surface of the globe is entirely covered by a mantle of air.The following terms and units are required for definitions in pneumatics:

Unit | Symbol | Units and unit symbol |

Length | L | Meter (m) |

Mass | m | Kilogram (kg) |

Time | t | Second (s) |

Temperature | T | Kelvin (K) |

Unit | Symbol | Units and unit symbol |

Force | F | Newton (N) = 1 Kg*m/s^{2} |

Area | A | Square meter (m^{2}) |

Volume | V | Cubic meter (m^{3}) |

Flowrate | Q | (m^{3}/s) |

Pressure | p | Pascal (Pa) 1 Pa = 1 N/m^{2} 1 bar = 10 ^{5} Pa |

_{at}), this pressure cannot be felt. The prevailing atmospheric pressure is therefore regarded as the base and any deviation is termed p

_{e}.

A pressure of p

_{e}= 3 bar is a pressure which is 3 bar above the atmospheric pressure.

A pressure of p

_{e}= -0.5 bar is a pressure which is 0.5 below the atmospheric pressure.

Is the pressure compared to the zero point (a room with absolutely no air inside is without any pressure) it is termed as absolute pressure p

_{abs}.

_{at}is therefore defined as p

_{at}= 1,013 bar (standardised atmospheric pressure).

The international unit for pressure is Pascal (Pa).

**1 Pa = 1 N/m**

^{2}In pneumatics the more common unit for pressure is bar.

**1 bar = 10 N/cm**

^{2}Thus the conversion between bar and Pascal is as following:

1 Pa = 10

^{-5}bar

1 bar = 10

^{5}Pa

#### 3. What is force?

Law of Newton:Force = Mass x Acceleration |

**F**Force (Newton)

**m**Mass (Kg)

**a**Acceleration (m/s

^{2})

The acceleration on earth due to the gravity is

**g = 9.81 m/s**» 10 m/s

^{2}^{2}.

Example: On a box with a mass of 50 Kg acts a force of 490.5 N.

#### 4. Cylinder piston force

The**theoretically available force**is derived from the working pressure and the respective effective piston surface area. This available force is reduced by the proportion of system friction; generally, system friction amounts to between 5 and 10 % of the theoretically available force. The system friction is expressed through the efficiency factor.

The theoretical piston force is calculated as the following:

Force = Effective piston area x operating pressure x Efficiency |

**A**Effective piston area (m

^{2})

**p**Operation pressure (Pa)

**h**

**Efficiency factor**

**Example:**

Cylinder diameter d

_{1}= 100 mm

Piston rod diameter d

_{2}= 25 mm

Pressure p

_{e}= 6 bar

Efficiency factor h = 0.85

**1. Effective area for advance stroke:**

A

_{a}= d

_{1}

^{2}x p /4 = (100 mm)

^{2}x p /4 = 7853 mm

^{2}=

__78.5 cm__

^{2}**2. Force for advance stroke:**

F

_{a}= p x A

_{a}x h = 60 N/cm

^{2}x 78.5 cm

^{2}x 0.85 =

__4003.5 N__

**3. Effective area for return stroke:**

A

_{r}= (d

_{1}

^{2}x p/4) - (d

_{2}

^{2}x p/4) = ((100 mm)

^{2}x p/4) - ((25mm)

^{2}x p/4) = 7363.1 mm

^{2}=

__73.6 cm__

^{2}**4. Force for return stroke**

F

_{r}= p x A

_{r}x h = 60 N/cm

^{2}x 73.6 cm

^{2}x 0.85 =

__3753.6 N__

**Note:**A double acting cylinder has on its advance stroke a greater force than on its return stroke. This is due to different effective piston areas for. For the return stroke the area of the rod has to be subtracted form the area of the piston to get the effective area.

**Exercise:**

Cylinder diameter d

_{1}= 63 mm

Piston rod diameter d

_{2}= 20 mm

Pressure p

_{e}= 6 bar

Efficiency factor h = 0.9

What is the piston force for advance and return stroke?

**Theoretically available force in N at p = 6 bar**

Piston [mm] | 32 | 40 | 50 | 63 | 80 | 100 | 125 | 160 | 200 | 250 | 320 |

Piston rod [mm] | 12 | 16 | 20 | 20 | 25 | 25 | 32 | 40 | 40 | 50 | 63 |

Advancing | 483 | 754 | 1178 | 1870 | 3016 | 4712 | 7363 | 12064 | 18850 | 29452 | 48255 |

Retracting | 415 | 633686 | 9901056 | 16821750 | 27212847 | 4418 | 6881 | 1131011110 | 1809617895 | 28274 | 46385 |

Important: In order to ensure a high service life and sufficient force to accelerate the piston, the cylinder should not be loaded with more than 80% of its theoretical piston force! |

**Example:**

A cardbox with a mass of 30 Kg has to be lifted vertically by a pneumatic cylinder. The supply pressure is 6 bar. What cylinder size (diameter) has to be selected when the utilisation factor of the cylinder should not exceed 80% ? It is supposed that there is no friction in the cylinder.

**1. What force is necessary to lift the cardbox?**

F

_{g}= m x g = 30 kg x 10 m/s

^{2}=

__300 N__

**2. What effective piston area is necessary (at 6 bar) to lift the cardbox?**

F = p x A

A = F/p = 300 N / 60 N/cm

^{2}=

__5 cm__

^{2}**3. Select cylinder diameter**

A = d

^{2}x p/4

= = 2.52 cm =

__25.2 mm__

Since there is no cylinder with a diameter of 25.2 mm the next available size has to be selected.

Selected cylinder size:

__32 mm__

**4. Control of utilisation factor:**

Theoretical piston force for diameter 32 mm at 6 bar: F

_{th}= 483 N (see table)

Utilisation = 100 x F

_{g}/F

_{th}= 100 x 300 N/483N =

__62.1 %__

The utilisation factor is not exceeded!

**Exercise:**

A pneumatic cylinder is used to lift an oil barrel on a ramp (see sketch). The mass of a full barrel is 150 Kg. The supply pressure is 6 bar. What cylinder size (diameter) has to be selected when the utilisation factor of the cylinder should not exceed 80% ? It is supposed that there is no friction in the cylinder.

#### 5. Cylinder air consumption

For the preparation of the air and to obtain facts concerning the costs, it is important to know the air consumption of a system.The air consumption depends on operation pressure, piston diameter and stroke length. It is calculated by the following formula:

Q = |

**Q**Air consumption of double acting cylinder (l/min)

**A**Effective piston area advance stroke (cm

_{1}^{2})

**A**Effective piston area return stroke (cm

_{2}^{2})

**s**Piston stroke (cm)

**n**Cycles per minute

**p**atmospheric pressure (bar)

_{at}**p**operating pressure (bar)

_{e}**Example:**

Double acting cylinder with d = 50 mm (piston rod diameter = 20 mm)

Stroke length s = 100 mm

Operating pressure p

_{e}= 6 bar

Cycles: n = 20 1/min

Atmospheric pressure p

_{at}= 1 bar

**1. Effective piston area advance stroke**

A

_{1}= (50 mm)

^{2}x p/4 = 1963.4 mm

^{2}=

__19.63 cm__

^{2}**2. Effective piston area return stroke**

A

_{2}= A

_{1}- (20 mm)

^{2}x p/4 = 1963.4 mm

^{2}- 314.2 mm

^{2}= 1649.2 mm

^{2}=

__16.5 cm__

^{2}**3. Air consumption**

Q = =

__50.6 l/min__

**Exercise:**

Calculate the air consumption for a DNC-50-100-PPV-A. Piston rod diameter is 25 mm. The number of cycles is 15 per minute.

**Theoretical air consumption per 10mm stroke, operating pressure 6bar**

Piston [mm] | 32 | 40 | 50 | 63 | 80 | 100 | 125 | 160 | 200 | 250 | 320 |

Piston rod [mm] | 12 | 16 | 20 | 20 | 25 | 25 | 32 | 40) | 40 | 50 | 63 |

Advancing | 0.056 | 0.088 | 0.137 | 0.218 | 0.352 | 0.550 | 0.859 | 1.407 | 2.99 | 3.436 | 5.630 |

Retracting | 0.048 | 0.0740.080 | 0.1150.123 | 0.1960.204 | 0.3170.330 | 0.515 | 0.803 | 1.3191.296 | 2.1112.088 | 3.299 | 5.412 |

Air consumption in normal litres is obtained on the basis of a standard operating pressure pe=6bar. Leakage is therefore not taken into account.

#### 6. Traversal force on piston rod

Traversal forces on the piston rod will cause a surface pressure on the bearing bush of the cylinder. The piston rod acts like a lever. The longer the stroke the higher the surface pressure on the bush. High surface pressure will cause excessive wear of the bush. If the traversal force exceeds the maximum permissible load of the cylinder bush, a external guide or support has to be provided (e.g. Festo FEN guide unit).**Example:**

A mass of m = 10 Kg has to be moved horizontally for 600 mm by a pneumatic cylinder.

1. What cylinder size has to be selected if there is no external guidance provided?

2. What cylinder/guide unit combination would be capable to do the same job?

#### 7. Allocation of cylinder and valves

The speed of a pneumatic cylinder is dependent on the load, air pressure, length of tubing between cylinder and control valve, diameter of tubing and the flow rate of the control valve. The average speed of a standard cylinder is about 0.1 - 1.5 m/sec. As a guideline for size the following table can be used for sizing the appropriate valve for a certain cylinder diameter. The cylinder speeds which can be achieved in practise, are adequate for most cases.Piston diameter | Valve port size | Nominal width (mm) | Flow rate (l/min) |

to 12 | M3 | 1.5 | 80 |

> 12 - 25 | M5 | 2.5 | to 200 |

> 25 - 50 | G 1/8 | 3.5 | to 500 |

> 50 - 100 | G ¼ | 7.0 | to 1140 |

> 150 - 200 | G ½ | 12.0 | to 3000 |

> 200 - 320 | G ¾, G 1 | 18.7 | to 6000 |

#### 8. Recommendations

The following recommendations are to observed for pneumatic cylinders.**Compressed air quality:**

Humidity: Pressure dew point up to 10 °C

Temperature: Higher than the dew point of the compressed air, lower than the maximum permitted

ambient temperature.

Solids: Particle size up to 40 mm

Lubrication: Festo pneumatic cylinders have been already given a basic lubrication from the

factory, thus they can be operated with unlubricated as well as lubricated air.

Note: If a cylinder has once been operated with lubricated air, lubrication is always needed, as additional oil flushes out the basic lubrication. |

**Speed:**

Up to 1 m/s the factory set lubrication is adequate. From speeds upward 1 m/s lubricated air is recommended for operating cylinders.

**Frequency:**

If cylinders are operated at the maximum possible speed, pauses between stroke movements have to be arranged. For unlubricated operation, the maximum frequency is to be based on a mean speed of 1 m/s.

Furthermore the limiting values indicated in the catalogue, as regarding to pressures, speeds, masses, moments, lateral forces, buckling loads, actuating forces and temperatures in compliance with technical notes must be adhered to by the user under all circumstances.

If used as recommended, Festo cylinders and combinations of such cylinders are suitable for as good as maintenance-free operations.