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D I D A C T I C L I N E

PNEUMATICS MANUAL

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METAL WORK DIDACTIC LINE was born as support to the informative meetings about our products and their relevant applications.Its use is to be intended only in this context.

Absolute pressure and relative pressure 1 How a pressure gauge works 3 Pressure and rate of flow - part 1 4 Pressure and rate of flow - part 2 6 Boyle’s law 8 The effects of temperature on gases 10 Gay-Lussac’s law 12 Relationship between pressure, volume and temperature 15

Actuators 23 Effect of back pressure 24 Single-acting cylinders 25 Double-acting cylinders 26 Cushioning 27 Through-rod cylinders 28 Tandem cylinders 28 Multi-position cylinders 29 Anti-rotation cylinders 30 Rodless cylinders 30 Magnetic and non-magnetic cylinders 32 Rotary actuators 33Single-rack actuators 33 Double-rack actuators 34 Blade actuators 34 Chain actuators 34 Points to bear in mind 35

Valves 45

3/2 Valves 46 5/2 Valves 46 5/3 Valves 47Control systems 48 Manual operation 48 Mechanical operation 49 Pneumatic operation 49 Electrical operation 50Valve flow rate 50Standard valves 54Valve modules 55

ABSOLUTE PRESSURE AND RELATIVE PRESSURE

ACTUATORS

Index

VALVES

I

Air treatment 63Filter 63 Pressure regulators 66Lubricators 67Venturi principle 68Shut-off valves 70Soft start valves 71

Basic logic functions 79OR function 80And function 81Yes function 82Not function 83Memory function 84Timer function 84Bimanual safety function 85

Problems: causes and remedies 91 1. Cylinders 92 2. Sensors 93 3. Valves 94 4. Regulators 95 5. Filters/depurators 95 6. Lubrificator 96 7. V3V shut off valve 96 8. APR soft start valve 97 9. Skilltronic 98

Conversion tables 99Degree of protection 101Comparison of factors for various forms of power 102

UNITS

LOGIC ELEMENTS

TECHNICAL DATA

PROBLEMS: CAUSES AND REMEDIES

II III

PNEUMATICS MANUAL

II III

IV 1

To understand this subject, it is important to remember that all aeriform substances (gases) tend to expand.

The aim of the experiment we are about to conduct is to verify this property and to observe the effect of free air, i.e. atmospheric pressure, on a certain quantity of air enclosed in a vessel.

Here, too, we will use a diagram: (Fig.1)

Let us suppose that the vessel is a non-inflated balloon sealed by a valve. The balloon is slack although there is some air inside. It is clear that the quantity of air inside the balloon with the valve open

or closed is the same, i.e. the pressure inside is the same as the pressure outside.

Let us now place the balloon under the transparent bell of a “pneumatic machine”, a machine used to extract air, commonly called a vacuum machine. The balloon is surrounded by the air at atmospheric pressure inside the bell.

When the vacuum machine is switched on, the air in the bell is drawn out. This emptying is shown by the balloon filling slowly until it has reached the

maximum volume allowed by its elasticity.

Why does this happen?

because air tends to expand; because there is no longer any resistance to expansion as there is no air

outside.

ABSOLUTE PRESSUREAND RELATIVE PRESSURE

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From a laboratory experiment, let us now move on to a simpler one using air at atmospheric pressure and a balloon:

Imagine we are at sea level, i.e. zero altitude, holding a balloon. Everyone knows that atmospheric pressure at zero altitude is 1 bar.

We move up a mountain to increasingly higher altitudes. Something we notice as we climb is that the balloon gradually inflates

although it is sealed.

Why is this?

Because atmospheric pressure gradually decreases as we move higher up. The balloon inflates because of the difference in pressure inside and outside. If we open the valve, a certain quantity of air flows out until the pressure inside

is the same as the pressure outside. The balloon is now slack again. If we close the valve again and descend to sea level, we will notice that the

balloon is slacker than when we started to descend because the pressure outside (at zero altitude) is higher than that inside (lower pressure because the air inside entered at a greater height).

The following conclusions can now be drown:

1. The air inside a vessel can only be reduced to zero pressure using a vacuum machine.

2. The air inside a vessel communicating with the atmosphere is always at the same pressure as that outside.

3. A vessel containing air at a certain pressure can release only a part of that pressure.

Now let us now look at the two examples given in Figs. 2 and 3.

A) The “outside” is atmosphere at a pressure of 1 bar (Fig. 2). Although the volume of air in the vessel is much less than the volume that

surrounds it, the air is at a greater pressure. When a link with the outside is established, the compressed air in the vessel is freed because it does not encounter any resistance and releases some molecules until an equilibrium is established, determined by the greater volume of the atmosphere.

B) The “outside” is another vessel at a pressure of 1 bar, i.e. atmospheric pressure (Fig. 3). When the valve between the two vessels is opened, the air flows from the one

containing the higher pressure to the one with the lower pressure until they are the same at an average value of 5+1 divided by 2 = 3 bar.

In this case, since no air has entered the atmosphere, the quantity of molecules released from the first vessel is smaller as they are acquired by the second one.

If we connect a gauge to the 4-bar vessel in any position, it will give the following values depending on the altitudes to which the vessel can hypothetically be taken.

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In conclusion we can say that:

1.The pressure of the air inside any sealed vessel is referred to as absolute pressure.

2.The pressure of the air available for use outside the vessel is referred to as relative pressure.

3.Relative pressure is the difference between the absolute pressure and the atmospheric pressure outside the vessel.

It uses the elastic deformation of an elliptically-shaped metal pipe.Fig. 5 shows the pressure gauge in the home position, i.e. when the difference in

pressure between the inner and the outer pipe is zero. End B is open; end A is closed and connected to a system of levers that converts its deformation into rotation of a pin with an index mounted on it.

A pressure gauge uses the elastic deformation of a elliptical metal tube.

Figure 4 shows the gauge at rest, i.e. when the difference in pressure inside and outside the tube is zero. End B is open and the other end A is connected to a system of levers that converts the deformation of end A into the rotary movement of a pivot which has a needle attached to it.

In Figure 5 we can see that with air under pressure, end A of the tube tends to straighten under the force of the air inside obstructed by the lesser force of the atmospheric pressure outside. The difference between the two pressures is indicated on a scale by the needle of the calibrated gauge.

HOW A PRESSURE GAUGE WORKS

Pressure on the gauge (es 6bar)

atmospheric pressure (1 bar)

Absolute vacuum (0 bar)

Relative pressure(6 bar)

Absolute pressure (7 bar)

atmospheric pressure pressure inside the vessel pressure on the gauge

-at sea level p=1 bar 5 bar 5-1 = 4 bar-at 1,000 metres p=0,9 bar 5 bar 5-0,9 = 4,1 bar-at 5,000 metres p=0,5 bar 5 bar 5-0,5 = 4,5 bar

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The two basic quantities in pneumatics are pressure and rate of flow.To help us understand the difference between these two quantities and how they

depend on each other, we will take water as a reference.

Let us look at Fig. 1.This is a vessel – now empty – fitted with two valves, A (outlet valve) and B (inlet valve).

We can represent the situation using a system of Cartesian axes, i.e. a diagram. On the horizontal axis (x-axis), we indicate the time required to empty the tank

when it is partially or completely filled with water. On the vertical axis (y-axis) we indicate the water level L. It is clear for the moment that level L is at the point of intersection of the two

axes (time zero corresponds to level zero).

Let us move on to Fig. 2.With valve A (outlet) closed and valve B (inlet) open, the vessel fills with water up to

level L. In the diagram, level L is at a precise point on the y-axis, but the time on the

x-axis is still zero because the water cannot flow out since valve A is closed.

Now we come to Fig. 3.Valve B is closed but valve A is open. This causes the water level to drop (and the

volume to decrease) until it reaches level zero, which is exactly when the vessel is completely empty. Clearly the emptying time will depend on whether valve A is fully or partially open. On the graph, we can follow the stages of emptying in terms of both level

and time. The plotted curve represents the change in water level as the vessel empties.

Figure 4 shows the same vessel.The two valves have the same cross section and – much more importantly – are open

far enough to allow the water leaving the vessel to be replaced simultaneously by other water entering it. It is clear that the level of water in the vessel remains constant. The graph is an accurate representation of this situation. The water level is

represented by a horizontal section showing that it does not change when the cross section of the two valves is identical, and of course when the two valves are open by the same amount.

Considerations:

The level of water in the vessel represents the pressure. It remains constant when the following two conditions occur: 1. valves A and B are closed; 2. valves A and B, of equal cross section, are open by the same amount.

Therefore, when the degree of opening of one of the valves is changed, this causes a change in the water level: the more water that is discharged compared to the amount entering the vessel, the lower the water level.

Conversely, if more water enters than goes out, it is necessary to close the inlet valve

PRESSURE AND RATE OF FLOW - Part 1

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B temporarily, while valve A can remain open. This is because the vessel has a fixed volume and cannot contain more water than its capacity.

Conclusions

The pressure of the air in a sealed vessel: depends on the quantity of air to be enclosed in the vessel; can be used at any time with minimal variations compared to the initial value

if the quantity of outgoing air is replaced by the same quantity of incoming air.

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In Part 1 we used figures and diagrams to show the changes in water level L in the vessel in relation to time and the closed/open positions of valves A and B.

Here we will deal with the rate of flow, i.e. the quantity of water discharged not from one but from two pipes connected to the vessel, at the same pressure and hence same water level L.

Let us look at the figure, which now has another element added to it – a basin to measure the amount of water that has come out of the vessel.fig. 1):

The vessel with water up to level L has two pipes and valves connected to it, cross section s.

Only valve A2 is completely open to allow the passage of the fluid. For level L in the vessel to remain constant, it is necessary to open and

regulate the flow through valve B, which is larger than valves A1 and A2. At a set time X, the water in the basin reaches a height h, which can be

quantified in volume by multiplying by the internal surface area of the bottom of the basin.

fig. 2):Valves A1 and A2 are opened completely and simultaneously, which means the same

amount of water pours into the two basins. Now we ask ourselves a question:In the same time X, do the two heights h reach level h in Fig. 1? The obvious answer is no, unless level L in the vessel is adjusted to compensate for the

quantity of water discharged by the two pipes. In more technical terms: When there is a change in the outflow rate there must be an equivalent

change in the inflow rate to keep level L and hence the pressure the same. If this condition is met, it is possible to branch from a single vessel more than

one outlet pipe of the same section (including their respective valves) and convey the same quantity of water, i.e. the same rate of flow, to each utility.

Clearly this is true as long as the sum of all the outflow rates does not exceed the inflow rate, in which case level L (the pressure) is no longer constant.

We should also consider a situation in which the outlet pipes do not have the same cross section, while level L in the vessel to which they are connected is the same.

Under this condition: the pressure of the flow is constant, irrespective of the cross section; the flow rate is less in pipes with a smaller cross section.fig 3):

Here the vessel has a single pipe with cross section 2s (plus valve), i.e. equal to the sum of the two pipes shown in Fig. 2. If we make a clear distinction between the cross section and the diameter of a

pipe, we can state that through cross section 2s and time X passes a quantity of water the height of which is equal to the sum of the heights of water in the basins shown in Fig. 2.

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PRESSURE AND RATE OF FLOW - Part 2

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Therefore, the rate of flow in the pipe with cross section 2s is double that of the pipe with cross section s.

The CONCLUSIONS that can be drawn are fairly intuitable.

1. At the same pressure, the flow rate depends on the difficulty the fluid encounters as it flows through a pipe or a valve: the greater the cross section, the more quickly the fluid reaches its destination.

2. With the same pipe cross section, a drop in the initial pressure causes an immediate reduction in the rate of flow, i.e. the amount of fluid moving in a unit of time.

3.The possible consequences of these considerations on a compressed air system will be dealt with more fully further on. For now, it is enough to remember that without pressure there can be no flow rate, i.e. movement of the fluid.

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E.g. in a conduit of a given section, there is a mass flow of 1 litre of air (1 dm3) at 7 bar absolute pressure. This value expressed as volume of air corresponds to 7 litres of air (7 dm3) at the ambient atmospheric pressure (1 bar).

1 dm3

7 bar absolute1 dm3

1 bar absolute1 dm3

1 bar absolute1 dm3

1 bar absolute1 dm3

1 bar absolute

1 dm3

1 bar absolute1 dm3

1 bar absolute1 dm3

1 bar absolute

Mass flow

Volumetric flow rate(referring to absolute pressure)

=

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A particular feature of gases is their ability to expand. When there is a variation in the pressure acting on them (like the inflating balloon), they always tend to occupy the maximum volume available.

The physicist Robert Boyle formulated an important law on the behaviour of gases, according to which the volume of gas varies with the pressure acting on it, provided that the temperature is constant.

The law states that:

The pressure of a gas varies inversely with its volume at constant temperature.

This means that if the pressure doubles, the volume of gas is reduced by half; and if the pressure is reduced to a third, the volume of gas triples.

This law can be verified by means of a simple experiment as shown in the following series of pictures.

fig. 1): We have a pipe that is transparent, rigid, constant in cross section and bent in

the shape of a U. It is positioned vertically with the longer end open and the shorter one connected to a valve.

The valve is left open and enough mercury to fill part of the U section is poured into the pipe.

The level of mercury in the two branches is now the same because the two surfaces are at atmospheric pressure.

If we close the valve, this traps air at atmospheric pressure. The volume is indicated by value X.

fig. 2): More mercury is now poured into the pipe until there is a 76 cm difference

between the levels of mercury in the two branches. The pressure on the trapped air doubles. The volume of air inside the sealed part of the pipe halves.

fig. 3): Lastly we add more mercury to get a difference in level of 76+76=152 cm,

i.e. we triple the pressure. The volume of trapped air reduces even further until it becomes a third of the original volume.

Let us look at the first two figures again and assign the two quantities, pressure and volume, the following values:

initial pressure (fig. 1) = p1 final pressure (fig. 2) = p2 initial volume (fig. 1) = V1 final volume (fig. 2) = V2

What we have described above shows that the two quantities are inversely proportional, meaning that when one doubles the other halves.

p1 : p2 = V2 : V1

BOYLE’S LAW

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As we know, in a proportion the product of extremes is equal to the product of the means. P1 x V1 = p2 x V2

It follows that Boyle’s law can also be expressed as follows:

The product of a volume of gas times its pressure is a constant at fixed temperature.

Boyle’s law can be used to solve problems such as the two that follow:

1. Consider a 2m³ vessel containing air (or another gas) at a pressure of 5 bar.

How much air does the vessel contain at atmospheric pressure?

initial pressure p1=5bar initial volume V1=2m³ final pressure p2=1bar final volume V2=?

Write down the formula and replace the symbols with the numeric values provided.

The unknown value, i. e. the final volume, is represented by the letter X

p1xV1 = p2 x V2 5 bar x 2m³ = 1 bar x X X = (5 x 2)/1 = 10 m³

2. A mass of air at a pressure of 2.5 bar occupies a volume of 0.5 m³. Due to an increase in pressure, the volume is reduced to 0.1 m³. What is the new pressure? initial pressure p1 = 2,5 bar initial volume V1 = 0,5 m³ final pressure p2 = ? final volume V2 =0,1 m³

p1xV1 = p2xV2 2,5 bar x 0,5 m³ = X x 0,1 m³ X= (2,5 x 0,5)/0,1 = 12,5 bar

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All substances, whatever their state of aggregation (solid, liquid or gaseous) undergo changes in volume when subjected to a change in temperature.

This phenomenon occurs with different characteristics in the case of a gas because, as we know, it does not have its own volume or form, but rather it assumes that of the (sealed) vessel containing it.

Vessels of different dimensions can be “full” with equal volumes of gas, i.e. the same quantity of molecules. This is because gases tend to occupy the entire space available.

Furthermore, we know from Boyle’s law that at fixed temperature there is the same dependence between the volume of the vessel and the quantity of gas molecules inside, i.e. between the volume and the pressure of the gas.

We can therefore deduce that a variation in temperature will have an effect not only on the volume as well as the pressure.

By means of two separate experiments we can “isolate” this dependence and verify the behaviour of the same quantity of gas that is first heated and then cooled.

Fig. 1):Heating at constant pressure (i.e. the gas is free to expand).

A vessel full of air and sealed by a plug is connected to a transparent tube leading to a basin of water.

Under normal temperature conditions, the pressure of the air in the vessel is the same as atmospheric pressure on the surface of the water

The air cannot escape, nor can the water enter the tube. As soon as the heat produced by the flame of a candle, or merely the

heat generated by a hand, warms the air in the vessel, the air expands significantly and, being free to expand, bubbles out through the pipe into the atmosphere.

This phenomenon can be seen with the naked eye from air bubbling through the surface of the water in the basin.

When the source of heat is removed, the opposite occurs – a drop in temperature causes the volume of air in the vessel to contract.

This can be seen by the water entering the tube under the thrust due to the external pressure.

The water occupies the space left free by the air molecules that dispersed in the atmosphere during heating.

Fig. 2):Heating at constant volume (the gas is forced to stay at the same volume).

The same vessel full of air at a pressure of, say, 2 bar is connected to a pressure gauge which, under normal temperature conditions, reads 1 bar, i.e. the difference between the absolute pressure of 2 bar inside the vessel and atmospheric pressure which is 1 bar.

As the air in the vessel is heated, it dilates as explained in the previous experiment, but here it is forced to maintain the same volume (that of the vessel).

The gauge instantly shows an increase in pressure.

THE EFFECTS OF TEMPERATURE ON GASES

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As the air is allowed to cool, there is a drop in pressure which returns to the primitive value because there has been no drop in the amount of air molecules inside the vessel.

To recapitulate, in both experiments the increase in temperature causes an increase in the volume of gas.

This phenomenon has two effects: 1. If the gas is allowed to expand, i.e. occupy a greater volume, the pressure

does not change. This is because an increase in the volume of gas corresponds to the same increase in the volume of the vessel containing it (in the example in figure 1, the increase in the volume of gas is absorbed by the atmosphere, which is separate from the water in the basin).

2. Conversely, if the gas is prevented from expanding, each molecule’s tendency to dilate (i.e. increase its volume due to heating) is hindered, resulting in a significant increase in pressure due to the difficulty for the same number of molecules of greater volume to exist in a vessel of a constant volume.

At the end of these two experiments, it is natural to pose this question:Is it possible to determine the increase in the volume and pressure of the gas in

advance – i.e. prior to heating (or cooling)?The answer is undoubtedly yes, and the calculation can be done using a coefficient

that is common to all gases. The answer, however, will be provided at the end of the following section where there is a simple formula to perform the calculation.

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In the previous section we conducted some simple experiments to demonstrate the effects that a change in temperature has on gases in general.

We also referred to a coefficient that could be used to calculate the variations in volume and pressure.

Now we will use this coefficient to introduce a new quantity, temperature, needed to apply Gay-Lussac’s law.

What is a coefficient? It is a ratio used to measure a certain effect.In our example, the value is 1:273 and the effect is on pressure.What does it mean? A series of experiments performed by Gay-Lussac on the variation

in pressure inside a sealed vessel proved the following:

if the initial pressure is, say, 10 bar and the temperature is reduced by 1º C, the pressure will drop to the following:

10 – (1/273 x 10) x 1 = 10 – 0.0366 x 1 =9.96 bar

with a decrease of 100 °C, the pressure drops to:

10 – (1/273 x 10) x 100 = 10 – 0.0366 x 100 = 6.34 bar

with a decrease of 273° C, the pressure drops to:

10 – (1/273 x 10) x 273 = 10 – 0.0366 x 273 = 0 bar

i.e. at a temperature of –273°C the pressure of a gas is reduced to zero.

The temperature of –273°C is the true zero on the thermometric scale, and the temperature starting from this value is referred to as absolute temperature, indicated by a capital T.

The relationship between T in degrees Kelvin (°K) and t in degrees centigrade (°C) is as follows:

T = t + 273 examples: t = 20 °C T = 20 + 273 = 293 °K t = 0 °C T = 0 + 273 = 273 °K t = -20 °C T = -20 + 273 = 253 °K

Gay-Lussac’s law can be demonstrated using the figures above, but we prefer to do it by comparison with Boyle’s law using the following figures.

fig. 1): Consider a vessel having a volume V, comprised of two cylinders of different diameters. A piston of a negligible weight is in equilibrium between the absolute pressure (inside the vessel) and atmospheric pressure.

The value zero shown on the gauge is the difference between two equal pressures.

Let us assume that the ambient and the air temperature is T = T = 293°K, which correspond to 20°C

The temperature T is kept constant (Boyle’s law) and we assume that:

GAY-LUSSAC’S LAW

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1. a force draws the piston outwards, causing an increase in the volume of

the vessel; the pressure on the gauge drops below zero since the pressure inside V2 is

less than atmospheric pressure.

2. a force pushes the piston inwards, causing a decrease in the volume of the

vessel; the pressure on the gauge increases since the pressure inside V2 is greater

than atmospheric pressure.

In both cases, the movement of the piston alters the volume of the vessel, which causes a variation in pressure according to the following inverse proportion:

V1 : V2 = p2 : p1 Boyle‘s law

fig. 2): Now, instead of keeping the temperature constant and applying an external force to the piston, we increase the room temperature to T = 313°K, i.e. we cause the temperature of the gas to rise by 20°C compared to the previous situation. As the molecules expand, following dilation, the piston moves outwards until the pressure has dropped back to zero.

fig. 3): The temperature is now reduced. The effect it has inside the vessel is to contract the molecules, creating a sort of vacuum. The piston is drawn inwards until the gauge reads zero and a balance has been restored.

In Figures 2 and 3, the position of the piston changes as in the example in Fig. 1, but the pressure on the gauge remains the same.

Now let us consider the other variables in the experiment, absolute temperate T and the volume of the vessel.

- Under Gay-Lussac’s law, these quantities are directly proportional, i.e.

V1 : V2 = T1 : T2 with equal pressure

- The same law can be used for pressure if the volume of the vessel is kept constant, i.e.

p1 : p 2 = V1 : V2 with equal volume

Gay-Lussac’s law can be applied to solve the following problems:

1) A gas occupies a volume of 0.5 m³ at a temperature of 283°K. What will the volume be at 323°K if the pressure remains constant?

V1 = 0.5 m³ T1 = 283°K V2 = ? T2 = 323°K V1 : V2 = T1 : T2

If we enter the known variables, we get: 0.5 : V2 = 283 : 323 V2 = (0.5 x 323)/ 283 = 0.57 m³

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2) We have a cylinder of gas at a temperature of 283°K, pressure 2 bar. The cylinder is left in the sun and heats up to 50°C. What is the pressure inside the cylinder now?

p1 = 2 bar T1 = 283°K p2 = ? T2 = T1 + 50°c = 283 + 50 = 333°k p1 : p2 = T1 : T2

Substituting the known variables, we get:

2 : p2 = 283 : 333 p2 = (2 x 333)/ 283 = 2.35 bar

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Another look at the known formulae

- Boyle’s law V1 : V2 = p2: p1 constant temperature

- Gay-Lussac’s law V1 : V2 = T1 : T2 constant pressure p1 : p2 = T1 : T2 constant volume

reveals that pressure, volume and temperature never occur all together: each quantity is always considered constant, the others become variables. But it often happens that the quantities are variables and only one can be unknown. This is how to tackle a similar situation in practical terms. The problem can be explained by examining the figure.

At a room temperature of 20°C a gas compressed in a cylinder with an inside diameter of 50 mm occupies a volume of 0.98 dm³ when a load of 980 N is placed on the piston. The load must be doubled at a room temperature of 50°C. What is the displacement of the piston?

It is clear that the displacement of the piston is caused by the variation in the volumes of gas with the two loads: the value of one volume is known (0.98 dm³), the other is unknown.

To find a relationship between all variables, we have to imagine that the variation in the volume of gas occurs separately in two stages:

(Case A)

Stage 1: The gas is heated from room temperature T = 20°C, corresponding to an absolute temperature T1 = 20 + 273 = 293°K, to temperature T2 = 50 + 273 = 323°K.

If the pressure on the piston is kept constant at 980N, the volume of gas increases. This is due to the phenomenon of dilation and resulting expansion of the gas molecules.

According to Gay-Lussac’s law V1 : Vx = T1 : T2 i.e. V1 x T2 = Vx x T1 Substituting the known variables, we get

0.98 x 323 = Vx x 293 Vx = (0.98 x 323)/293= 1.08 dm³ (provisional figure)

Stage 2: When the gas reaches Vx = 1.08 dm³ following the increase in temperature, the value of T2 remains the same but the pressure increases to:

p2 = 980 x 2 = 1960 N

RELATIONSHIP BETWEEN PRESSURE,VOLUME AND TEMPERATURE

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The result is a drop in the volume of gas due to the increased pressure on the piston.

According to Boyle’s law:

Vx : V2 = p2 : p1 i.e. Vx x p1 = V2 x p2

Substituting the known variables, we get

1.08 x 980 = V2 x 1960 V2 = (1.08 x 980)/1960 = 0.54 dm³ (final figure)

We substituted pressure p1 and p2 for the loads on the piston. Was it a mistake? No! To prove this, we need to

calculate the surface area of the piston in cm² diameter 50 mm= 5 cm S = (5² x 3.14)/4 = 19.6 cm² calculate the pressure on each cm² of surface, which corresponds to the

pressure of the gas in bar

initial pressure p1=initial load / surface area = 980N/19.6cm² = 50N/cm² = 5 kg/cm²

final pressure p2=final load / surface area=(980 x 2)N /19.6cm²=100N/cm²=10 kg/cm²

With an equal surface area of the piston, doubling of the load corresponds to doubling of the pressure. So, if we substitute these values in

Vx x p1 = V2 x p2

this gives

V2=(1.08 dm³ x 50 N/cm²)/100 N/cm²=(1.08 dm³ x 5 kg/cm²)/10kg/cm²= 0.54 dm³

The result is therefore the same as above.

This result can be obtained directly from the following formula, which is a combination of the two initial ones:

(p1 x V1)/ T1 = (p2 x V2)/T2

In our case we need to know volume V2 to display the displacement of the piston.

V2 = (p1 x V1 x T2) / (T1 x p2) = (5 x 0.98 x 323) / (293 x 10) = 0.54 dm³

Now that we also know the other volume, we can calculate the change in the position of the piston with the aid of geometry:

volume = base surface area x height height in cm = volume in cm³/base surface area in cm² initial height = (980 cm³)/19.6 cm² = 50 cm final height = (540 cm³)/19.6 cm² = 27.5 cm piston displacement = 50 – 27.5 = 22.5 cm

In our problem we assumed that the gas was heated by a change in room temperature.

The bicycle pump experiment reminds us that when air is compressed and cannot

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expand, it generates heat - i.e. an increase in temperature – itself and transmits it externally.

It is clear that when the gas expands as the piston of the pump is pulled outwards, the opposite phenomenon occurs, i.e. a drop in temperature, which is as not as obvious as the heating. These variations in air temperature cause:

-a generation of heat during compression

-absorption of heat during expansion

We will now calculate these variations by representing the same problems in the following terms:

A quantity of gas at a temperature of 293°K occupies a volume V1 = 0.98 dm³ at a pressure of 5 bar. When the pressure is increased to 10 bar, the volume drops to V2 = 0.54 dm³.

What is the temperature of the gas?

Remember that Boyle’s law only applies at a constant temperature. So it follows that at 293°K, an increase in pressure from p1 to p2 causes the following drop in the volume of gas from V1 to Vx:

V1 : Vx = p2 : p1 i.e. V1 x p1 = Vx x p2

Substituting the known variables, we get Vx = (0.98 x 5)/ 10 = 0.49 dm³

According to Gay-Lussac’s law and taking the constant pressure p2, to which volume Vx refers, we can state that:

Vx : V2 = T1 : T2 i.e. Vx x T2 = V2 x T1

Substituting the known variables, we get

T2 = (0.54 x 293)/0.49= 323°K

The result is the same as temperature T2 in the initial problem.

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The conversion of pneumatic power into useful mechanical work takes place by means of organs, called actuators, that move in a linear direction or rotate.

The most common linear actuators are cylinders. In the field of pneumatic it is convenient to have on/off movements, with the piston rod fully extended or retracted. Intermediate positions are imprecise and unstable due to the elasticity of the air.

If several intermediate positions are required, it is possible to opt for an air cylinder or a mixed air/oil system or a system with a closed-loop proportional electronic control with feedback.

Basically a cylinder is made up of the following elements:

Body with cylinder heads and lining Piston Piston rod Static and dynamic gaskets

The operating principle is simple: air under pressure pushes against a surface of the piston and moves it inside the lining. It is necessary to provide relief in the opposite chamber for the back pressure generated by the movement of the piston, which reduces the volume inside the chamber. Dynamic gaskets mounted on the piston prevent air from passing from one chamber to the other; static gaskets prevent it from escaping outside.

The bore of the cylinder corresponds to the inside diameter of the lining. The stroke is the maximum movement the piston rod is capable of performing.

ACTUATORS

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If there is no back pressure relief, the surface of the piston in the opposing chamber acts as a compressor.

Initially there is a progressive deceleration of the stroke, then cylinder movement stops as the back pressure is greater than the pressure of the air fed into the cylinder.

This phenomenon, if correctly regulated, can help the cylinder make intermediate stops or can regulate its speed.

To limit the cylinder speed, it is therefore necessary to prevent the back pressure from being discharged quickly from the cylinder. Flow regulators are used for this purpose.

EFFECT OF BACK PRESSURE

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Their operation is simple: the air from the cylinder is conveyed through a bore the size of which is regulated by a pin. This slows down relief of the back pressure, which takes place in a single direction (relief) by means of a one-way gasket that allows the air to flow freely. There are also versions with two-way gaskets in which the flow of air is slowed down in both directions.

In a single-acting (S-A) cylinder, the pressurised air only acts on one face of the piston. Movement in either directions takes place by means of external elements, such as the force of gravity or a spring.

SINGLE-ACTING CYLINDERS

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S-A spring cylinders can only be used if the return stroke required a limited force. This is because the springs must be soft so as not to waste energy. For this reason they also have limited strokes and are normally used to drive turntables for such operations as locking or punching.

Single-acting cylinders also play an important role in systems which, for safety reasons, required instant return to the home position when there is an interruption in the compressed air supply.

In double-acting (D-A) cylinders, compressed air is used to move the piston forwards and backwards. This type of cylinder is used when a force in both directions is required. They are also called differential cylinders when the face of the piston to which the rod is not connected generates a force superior than the other one.

DOUBLE-ACTING CYLINDERS

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Impact between the cylinder heads and piston at the end of each movement reduces the life of the cylinder and generates excessive noise. To avoid this, the cylinders can be equipped with mechanical and pneumatic end-of-stroke cushions.

Pneumatic cushions are generally placed in the cylinder heads and they operate according to the back pressure relief principle. At the end of the stroke, the conical element connected to the piston enters the cushioning chamber and obstructs the main back pressure relief port. The back pressure is relieved through a small orifice, the bore of which can be adjusted by means of a pin to slow down the piston stroke.

If the piston heads are not large enough to hold the cushioning chamber, they are equipped with rubber rings called shock-absorbers.

The pneumatic cushions are influenced by the load applied to the piston rod and the velocity of the piston. If the load or velocity is too high, cushioning is not effective. There exists tables like the ones below showing what mass can be cushioned in relation to the rate of translation.

CUSHIONING

1.1/03

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S-A or D-E cylinders can be fitted with a through rod, one which projects from the cylinder into both cylinder heads. The piston generates an equal force in both directions since the surface area on which the compressed air acts is the same.

Furthermore, within certain limits, through rods allow the application of higher-than-standard radial loads.

In a standard cylinder, the rod acts on a single guide bushing and tends to bring it out of alignment. This causes the piston to move at an angle, which reduces the life of the gaskets.

With a through rod, the radial load is supported by two bushings, which share the load.

Tandem cylinders must be used when high forces are required but there is limited space available.

The cylinder lining is divided into two-four chambers. The same number of pistons as there are chambers are mounted on the piston rod. When compressed air enters, it acts simultaneously on the faces of all the pistons, generating a force which is in theory two or three times greater than that of a single piston. The only condition is that the stroke of each stage must be the same as that of all the others.

Retraction takes place when compressed air acts on the face of the last piston connected to the piston rod nearest the front cylinder head. During this stage, the force generated is that of a single cylinder.

THROUGH-ROD CYLINDERS

TANDEM CYLINDERS

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To operate, cylinders use air, which is a compressible fluid. It is therefore impossible to achieve precise, repeatable intermediate stops. The only way to do this is by using multi-position cylinders, which combine two or three stages. The stroke of the first stage must be equal to or greater than that of the next stage. The piston rods are separate.

This version allows intermediate stops that are precise and repeatable. They cannot be modified as they depend on the stroke of each stage.

MULTI-POSITION CYLINDERS

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As a normal piston rod is cylindrical in shape, it tends to rotate around itself as it moves. This rotation sometimes causes problems.

For example, if an organ is fixed on the piston rod to pick up objects in the same position, it must always have the same angle of approach and the piston rod must absolutely not rotate around itself.

The simplest method is to fix onto the piston rod a square or rectangular plate guided by two lateral rods. This means the piston rod is prevented from rotating. Another method is to fix two piston rods onto the piston. In this case, it is the actual cylinder head that prevents rotation. The latter type is called a twin-rod cylinder.

So far we looked at cylinders in which piston-generated movement is used to extend and retract the piston rod. This type of cylinder requires a certain amount of space, because when the rod is extended, the cylinder takes up nearly twice as much room as when the piston rod is retracted.

Rodless cylinders are used when there are space limitations. The piston is connected straight to a carriage that moves along the cylinder lining.

ANTI-ROTATION CYLINDERS

RODLESS CYLINDERS

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This type of cylinder generates the same force in both directions of the piston and very high speeds (up to 2 m/sec.) with very long strokes (up to 5700 mm).

Due to the design features, the loads applied must be calculated correctly. The correct type of carriage must be selected according to the radial force exercised by the mass applied – standard version - or connected to a ball-recirculation guide.

Tables can be used to make the best choice according to the mass applied to the carriage.

N.B.: when the cylinder is subjected simultaneously to torque and force, it is advisable to keep to the following equationsMa = F x ha Mr = L x hv + G x hr Mv = F x hv

Mv L Ma Mr Mv L ≤ 1; ≤ 1; + + 0.22 x + 0.4 ≤ 1;

Mv max L max Ma max Mr max Mv max L max

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Ma Mr Mv L G + + + + ≤ 1;

Ma max Mr max Mv max L max G max

DIMENSIONING - FORCES AND TORQUES Bore Centre Distance Actual force Cushioning stroke Max load Ma max Mr max Mv max Ø Y F at 6 bar [N] [mm] L [N] [Nm] [Nm] [Nm] 16 9 110 15 120 4 0.3 0.5 25 14 250 21 300 15 1 3 32 18 420 26 450 30 2 4 40 22 640 32 750 60 4 8 63 1550 40 1650 200 8 24

DIMENSIONING - FORCES AND MOMENTS Ø Actual force Cushioning K X Y Z Max load Max load Max Ma Max Mr Max Mv F a 6 bar [N] stroke [mm] [mm] [mm] [mm] [mm] L [N] G [N] [Nm] [Nm] [Nm] 16 110 15 35 16 29 33 500 500 16 15 16 25 250 21 50.5 21 44 51.5 1500 1500 100 50 100 32 420 26 59 22.5 53.5 70 3000 3000 200 100 200 40 640 32 68 24.7 58 73 4000 4000 200 140 200

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The piston operating system and the pneumatic cushioning are basically the same as with rodless cylinders, the only difference being the type of pneumatic seal between the inside of the cylinder and the outside. Unlike other cylinders, rodless cylinders have a lining with a slot allowing mechanical connection between the carriage and the piston. Pneumatic seal is provided by two straps fixed onto the cylinder heads. The lower strap is made of plastic or steel and is used to create an air seal; the upper one is usually made of steel and prevents dirt from getting into the cylinder. The carriage is equipped with dust scrapers which, besides keeping the outer strap clean, operate a strap guiding system.

Internally, after the carriage has passed, the lower strap is brought back into position by the gaskets mounted on the pistons and magnetic elements inside the lining.

All the cylinders considered so far can be fitted with a system for determining the position of the piston. If a magnetic element is placed inside the piston, the position can be determined by a magnetic sensor.

MAGNETIC AND NON-MAGNETIC CYLINDERS

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So far we have considered actuators that convert power into linear movement. Sometimes, the movement needs to be rotary, for example with turntables or positioners. This is why rotary actuators have been designed. Using various systems, they convert the piston’s linear movements into circular motion. The energy used is transformed into torque and not a linear force. The unit of measurement used is Nw/m and is proportional to the dimensions of the moving organs.

Let us know look at certain types of rotary actuators:

In this version the piston is mounted at the end of a rack and is moved longitudinally by compressed air. The rack transmits movement to a cog wheel which has a pinion splined onto it. The angle of rotation depends on the length of the rack and ranges from 90° to 360°. The actuator rotates alternately clockwise and anticlockwise.

As there is no system to pick up the mechanical play between the rack and the rotary element, this version is not suitable for applications requiring considerable accuracy in positioning. There are, however, versions with external mechanical stops that can be adjusted to stop rotation and take up any play.

ROTARY ACTUATORS

SINGLE-RACK ACTUATORS

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In this version, the pinion splined onto the rotary elements, is rotated by two opposed racks, each driven by a pneumatic piston. The system used to take up mechanical play is automatic and therefore suitable for applications in which positioning accuracy is important. These actuators are particularly compact in size as the racks are short even for 180° rotation.

One or two blades housed in a cylindrical chamber are connected to the rotary element. The element is rotated by air pressing on one of the sides of the blade. These very compact cylindrical actuators are suitable for applications that do not require particular accuracy or torque.

The pinion is rotated by a chain stretched at either end by a pneumatic piston. High precision is not possible with this type of actuator, either.

DOUBLE-RACK ACTUATORS

BLADE ACTUATORS

CHAIN ACTUATORS

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It is always good practice to use micro-regulators at the ports. When running the system for the first time, these micro-regulators should be closed completely to prevent them from getting damaged by rapid or unintentional movement.

If the connected force generated by the rotating mass is too high, it is advisable to use end-of-stoke hydraulic decelerators. Some models come ready fitted with decelerators inside the actuators. For those without, the user can mount decelerators externally.

If the axis of rotation is horizontal and distribution of the masses is asymmetrical, it may be difficult to regulate the flow rate by means of micro-regulators, in which case it is advisable to use external decelerators.

POINTS TO BEAR IN MIND

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π x Ø2

cyl xP line x 0,9 = F thrust 4

THRUST FORCE CALCULATING

π x (Ø2

cyl - Ø2

rod) xP line x 0,9 = F traction4

TRACTION FORCE CALCULATING

Ø = cmP = barF = Kg0,9 = efficiency

CYCLE AIR CONSUMPTION

π x (2Ø2

cyl - Ø2

rod) xPabsolutexC = Nl / cycle4

Ø = cylinder and piston rod (dm)P = absolut pressure (bar)C = stroke (dm)

MAXIMUM ABSORBABLE LOAD

Ø1,65

v=0,0075x M0,5

Ø = cylinder (mm)M = absorbable load (Kg)v = speed (m/s)

CYLINDER TIME PER STROKE CALCULATING

Max speed till Ø 50:2 m/s, over 1,5 m/s.Max absorbable load:

Mmax = 0,140 xØ1,8

Ø = cylinder (mm)M = absorbable load (Kg)

Mv2 = 56,25x10-6 xØ3,3

60xVolume Nl/cycletCycle sec

= Qelement Nl/min

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Cylinder Piston rod Useful Air consumption during thrust and traction in Nl/cm of stroke, depending on the working pressure P in bar at 20°C bore D diameter Motion area mm d mm cm2 1 bar 2 bar 3 bar 4 bar 5 bar 6 bar 7 bar 8 bar 9 bar 10 bar

12 4 thrust 1,13 0,0023 0,0034 0,0045 0,0057 0,0068 0,0079 0,0090 0,0102 0,0113 0,0124 traction 1,00 0,0020 0,0030 0,0040 0,0050 0,0060 0,0070 0,0080 0,0090 0,0100 0,0110

16 6 thrust 2,01 0,0040 0,0060 0,0080 0,0100 0,0121 0,0141 0,0161 0,0181 0,0202 0,0221 traction 1,73 0,0035 0,0052 0,0069 0,0086 0,0104 0,0121 0,0138 0,0156 0,0173 0,0190

20 8 thrust 3,14 0,0063 0,0094 0,0126 0,0157 0,0188 0,0220 0,0251 0,0283 0,0314 0,0346 traction 2,64 0,0053 0,0079 0,0106 0,0132 0,0158 0,0185 0,0211 0,0238 0,0264 0,0290

25 12 thrust 4,91 0,0098 0,0147 0,0196 0,0245 0,0295 0,0344 0,0393 0,0442 0,0491 0,0540 traction 3,78 0,0076 0,0113 0,0151 0,0189 0,0227 0,0264 0,0302 0,0340 0,0378 0,0415

32 12 thrust 8,04 0,016 0,024 0,032 0,040 0,048 0,056 0,064 0,072 0,080 0,088 traction 6,91 0,014 0,021 0,028 0,035 0,042 0,049 0,058 0,063 0,070 0,076

40 16 thrust 12,56 0,025 0,038 0,050 0,063 0,076 0,088 0,100 0,113 0,126 0,138 traction 10,55 0,021 0,032 0,042 0,053 0,063 0,074 0,088 0,095 0,106 0,116

50 20 thrust 19,63 0,039 0,059 0,079 0,098 0,118 0,137 0,157 0,177 0,196 0,216 traction 16,49 0,033 0,050 0,066 0,082 0,099 0,115 0,132 0,149 0,165 0,181

63 20 thrust 31,16 0,062 0,093 0,125 0,156 0,187 0,218 0,249 0,280 0,312 0,343 traction 28,02 0,056 0,084 0,112 0,140 0,168 0,196 0,224 0,252 0,280 0,308

80 25 thrust 50,24 0,100 0,150 0,200 0,250 0,301 0,351 0,402 0,452 0,502 0,552 traction 45,36 0,091 0,138 0,181 0,227 0,272 0,318 0,363 0,408 0,454 0,500

100 32 thrust 78,54 0,157 0,238 0,314 0,382 0,471 0,549 0,628 0,706 0,785 0,862 traction 70,50 0,141 0,211 0,282 0,352 0,423 0,493 0,564 0,635 0,705 0,775

125 32 thrust 122,66 0,245 0,368 0,490 0,613 0,736 0,859 0,981 1,104 1,226 1,349 traction 114,67 0,229 0,344 0,459 0,573 0,688 0,803 0,917 1,032 1,147 1,262

160 40 thrust 201,06 0,402 0,603 0,804 1,005 1,206 1,407 1,608 1,809 2,010 2,211 traction 188,49 0,377 0,565 0,754 0,942 1,130 1,319 1,508 1,696 1,884 2,673

200 40 thrust 314,15 0,628 0,942 1,257 1,571 1,885 2,199 2,513 2,827 3,145 3,456 traction 301,59 0,603 0,905 1,206 1,508 1,810 2,111 2,413 2,714 3,016 3,318

(P2 – P1) P = P1 + ––––––––––– • Cx CmaxP1 = Force with spring extendedP2 = Force with spring compressedCx = Required strokeCmax = Max stroke

ISO 6431 Single-acting cylinders Bore Force with spring Max stroke Force with spring mm compressed N mm extended N 32 63 250 35 40 88 250 51 50 102 250 64 63 102 250 64

ISO 6432 Single-acting cylinders Bore Force with spring Max stroke Force with spring mm compressed N mm extended N 8 3 50 1 10 5 50 1 12 7 50 3 16 20 50 5 20 22 50 12 25 28 50 17

SSC single-acting cylinders Bore Force with spring Max stroke Force with spring mm compressed N mm extended N 12 6 25 1,5 16 7 25 3 20 12 25 4 25 14 25 5 32 33 50 6 40 45 50 15 50 70 50 20 63 81 50 25

CONSUMPTION OF AIR IN THE CYLINDERS

FORCES OF SPRINGS IN SINGLE-ACTING CYLINDERS (THEORETICAL)

Round single-acting cylinders Bore Force with spring Max stroke Force with spring mm compressed N mm extended N 32 86 250 34 40 95 250 50 50 108 250 62

Single-acting cartridge cylinders Bore Force with spring Max stroke Force with spring mm compressed N mm extended N 6 5.03 15 – 10 7.05 15 – 16 7.05 15 –

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Cylinder Piston rod Useful Thrust and traction force in N depending on the operating pressure in bar bore D diameter Motion area mm d mm cm2 1 bar 2 bar 3 bar 4 bar 5 bar 6 bar 7 bar 8 bar 9 bar 10 bar

8 4 thrust 0.50 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 traction 0.38 0.4 0.8 1.1 1.5 1.9 2.3 2.6 3.0 3.4 3.8

10 4 thrust 0.79 0.8 1.6 2.4 3.1 3.9 4.7 5.5 6.3 7.1 7.9 traction 0.66 0.7 1.3 2.0 2.6 3.3 4.0 4.6 5.3 5.9 6.6

12 6 thrust 1.13 1.1 2.3 3.4 4.5 5.7 6.8 7.9 9.0 10.2 11.3 traction 0.85 0.8 1.7 2.5 3.4 4.2 5.1 5.9 6.8 7.6 8.5

16 6 thrust 2.01 2.0 4.0 6.0 8.0 10.1 12.1 14.1 16.1 18.1 20.1 traction 1.73 1.7 3.5 5.2 6.9 8.6 10.4 12.1 13.8 15.6 17.3

16 8 thrust 2.01 2.0 4.0 6.0 8.0 10.1 12.1 14.1 16.1 18.1 20.1 traction 1.51 1.5 3.0 4.5 6.0 7.5 9.0 10.6 12.1 13.6 15.1

20 8 thrust 3.14 3.1 6.3 9.4 12.6 15.7 18.8 22.0 25.1 28.3 31.4 traction 2.64 2.6 5.3 7.9 10.6 13.2 15.8 18.5 21.1 23.8 26.4

20 10 thrust 3.14 3.1 6.3 9.4 12.6 15.7 18.8 22.0 25.1 28.3 31.4 traction 2.36 2.4 4.7 7.1 9.4 11.8 14.1 16.5 18.8 21.2 23.6

25 8 thrust 4.91 4.9 9.8 14.7 19.6 24.5 29.5 34.4 39.3 44.2 49.1 traction 4.41 4.4 8.8 13.2 17.6 22.0 26.4 30.8 35.2 39.7 44.1

25 10 thrust 4.91 4.9 9.8 14.7 19.6 24.5 29.5 34.4 39.3 44.2 49.1 traction 4.12 4.1 8.2 12.4 16.5 20.6 24.7 28.9 33.0 37.1 41.2

32 12 thrust 8.04 8.0 16.1 24.1 32.2 40.2 48.3 56.3 64.3 72.4 80.4 traction 6.91 6.9 13.8 20.7 27.6 34.6 41.5 48.4 55.3 62.2 69.1

40 12 thrust 12.57 12.6 25.1 37.7 50.3 62.8 75.4 88.0 100.5 113.1 125.7 traction 11.44 11.4 22.9 34.3 45.7 57.2 68.6 80.0 91.5 102.9 114.4

40 16 thrust 12.57 12.6 25.1 37.7 50.3 62.8 75.4 88.0 100.5 113.1 125.7 traction 10.56 10.6 21.1 31.7 42.2 52.8 63.3 73.9 84.4 95.0 105.6

50 16 thrust 19.63 19.6 39.3 58.9 78.5 98.2 117.8 137.4 157.1 176.7 196.3 traction 17.62 17.6 35.2 52.9 70.5 88.1 105.7 123.4 141.0 158.6 176.2

50 20 thrust 19.63 19.6 39.3 58.9 78.5 98.2 117.8 137.4 157.1 176.7 196.3 traction 16.49 16.5 33.0 49.5 66.0 82.5 99.0 115.5 131.9 148.4 164.9

63 16 thrust 31.17 31.2 62.3 93.5 124.7 155.9 187.0 218.2 249.4 280.6 311.7 traction 29.16 29.2 58.3 87.5 116.6 145.8 175.0 204.1 233.3 262.5 291.6

63 20 thrust 31.17 31.2 62.3 93.5 124.7 155.9 187.0 218.2 249.4 280.6 311.7 traction 28.03 28.0 56.1 84.1 112.1 140.2 168.2 196.2 224.2 252.3 280.3

80 20 thrust 50.27 50.3 100.5 150.8 201.1 251.3 301.6 351.9 402.1 452.4 502.7 traction 47.12 47.1 94.2 141.4 188.5 235.6 282.7 329.9 377.0 424.1 471.2

80 25 thrust 50.27 50.3 100.5 150.8 201.1 251.3 301.6 351.9 402.1 452.4 502.7 traction 45.36 45.4 90.7 136.1 181.4 226.8 272.1 317.5 362.9 408.2 453.6

100 25 thrust 78.54 78.5 157.1 235.6 314.2 392.7 471.2 549.8 628.3 706.9 785.4 traction 73.63 73.6 147.3 220.9 294.5 368.2 441.8 515.4 589.0 662.7 736.3

125 32 thrust 122.72 122.7 245.4 368.2 490.9 613.6 736.3 859.0 981.7 1104.5 1227.2 traction 114.68 114.7 229.4 344.0 458.7 573.4 688.1 802.7 917.4 1032.1 1146.8

160 40 thrust 201.06 201.1 402.1 603.2 804.2 1005.3 1206.4 1407.4 1608.5 1809.6 2010.6 traction 188.50 188.5 377.0 565.5 754.0 942.5 1131.0 1319.5 1508.0 1696.5 1885.0

200 40 thrust 314.16 314.2 628.3 942.5 1256.6 1570.8 1885.0 2199.1 2513.3 2827.4 3141.6 traction 301.59 301.6 603.2 904.8 1206.4 1508.0 1809.6 2111.1 2412.7 2714.3 3015.9

FORCES GENERATED DURING THRUST AND TRACTION (THEORETICAL)

38

39

AC

TUA

TOR

S

SE magnetic cylinder

Hydraulic brake withadjustment in onedirection only

Hydraulic brakewith adjustmentin both directions

Cushion

Pressure multiplierfor fluids withidentical characteristics

Pressure multiplierfor fluids withdifferent characteristics

Pneumatic/hydraulictransducer

Constant volumecompressor

Constant volumepneumatic motor,unidirectional flow

Constant volumepneumatic motor,bidirectional flow

Variable volumepneumatic motor,unidirectional flow

Variable volumepneumatic motor,bi-directional flow

Rotary pneumatic motorCylinder with adjustablesingle cushioning

Y X

Y X

DE magnetic cylinderwith adjustablebilateral cushioning

DE magnetic twin-rodcylinder with adjustablebilateral cushioning

DE magnetic twin-rodcylinder with adjustablebilateral cushioning

DE magnetic twin-rodcylinder with adjustablebilateral cushioning-singlethrough rod

DE magnetic cylinder withadjustable bilateral cushioning+ DZB mechanical lock

DE magnetic cylinder withadjustable bilateral cushioning+ DZBA mechanical lock

DE cylinder withadjustable bilateralcushioning, through-rod

DE through-rod cylinder

DE magnetic cylinderwith adjustable bilateralcushioning, through-rod

DE magnetic cylinder,through-rod

DE cylinder

DE cylinder with cushioning

DE magnetic cylinder

SE cylinder

TRANSFORMATION

38

39

AC

TUA

TOR

S

40

41

AC

TUA

TOR

S

40

41

AC

TUA

TOR

S

42

43

AC

TUA

TOR

S

42

43

AC

TUA

TOR

S

44

45

An elementary pneumatic circuit is generally comprised of three basic parts:

a source of energy (compressed air) a utility (a unit converting energy into mechanical work) connecting pipes

For the circuit to be functional, there must also be a directional valve, namely and organ that enables the user to start, stop or reverse of the utilities.

Valves are defined by the following:

Number of ways Number of positions Control system Operating system Dimensions Size

Number of ways: the number of external ports in the valve which, depending on the organ of distribution (spool), are open, closed or in communication. The number of ways is established by ISO standard 11727:

1. supply 2. utility 3. port 2 outlets 4. utility 5. port 4 outlets

Number of positions: the number of different positions the spool can take up inside a valve. There are two-way two-position (2/2) valves, three-way two-position (3/2) valves, five-way two-position (5/2) valves and five-way three-position (5/3) valves.

Control system: the means controlling the spool and activating its positions. The control system can be manual, mechanical, pneumatic, electric or electro-pneumatic, depending on the type of force.

Reset system: the means to reposition the spool when the control signal ceases. Resetting can be automatic by means of a system using a mechanical or pneumatic spring. When a valve has automatic reset it is defined as monostable. If reset is effected by a system similar to the control system, the valve is defined as bistable because both positions can remain active even in the absence of a control signal.

The need for bistable or monostable valves depends on the type of application and the degree of safety required.

VALVES

44

45

VA

LVE

S

Dimensions: the dimensions of the threaded external ports on the valve body expressed in inches.

Size: only applies to valves referring to ISO 5599. This standard determines the dimensions of the valve body, the centre distances for the fixing holes and everything concerning the interface with the sub-base.

3/2 VALVES

3/2 valves are divided into normally-open (NO) or normally-closed (NC) valves. In the home position, NO valves allow air to flow from the inlet port 1 to utility port 2. NC valves prevent air flow from the inlet port 1 and connect utility port 2 to its outlet port 3.

The graphic symbol for an NC valve is:

The graphic symbol for an NO valve is:

5/2 VALVES

Unlike 3/2 valves, 5/2 valves have two utility ports (2 and 4) that are connected alternately to the inlet port 1. This means that these valves cannot be NO or NC. They can be monostable or bistable only, depending on the type of repositioning applied.

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46

47

VA

LVE

S

5/3 VALVES

5/3 valves are usually the monostable type. The stable part of the spool can assume 3 different configurations:

CLOSED CENTRES (CC). The spool is positioned so that it closes all the ports.

OPEN CENTRES (OC). The spool is positioned so that it connects utilities 2 and 4 to their respective outlets 3 and 5, whereas the inlet port 1 is closed.

PRESSURE CENTRES (PC). The spool is positioned so that it connects utilities 2 and 4 with the inlet port 1, whereas outlets 3 and 5 are closed.

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47

VA

LVE

S

In small valves, the spool can be operated directly by the pilot. With larger valves, direct control will require the use of large electric coils (electric control) or levers (mechanical or manual control) or compressed air (pneumatic control). This is why most valves use a pneumatic servo-piloted control. Compressed air from the inlet port 1 is conveyed to the control system which, when operated, sends air to one face of the spool, suitably proportioned and referred to as piston, and causes it to move. It can be deduced that operation takes place if the pressure of the air is enough to move the spool. If the pressure is too low (1-1,5 bar or a vacuum), this type of servo pilot cannot be used. In this case it is necessary to use direct piloting or a servo pilot. In the latter case, the air required to move the spool does not come from the inlet, its is conveyed from the outside straight to the pilot ports.

Let us now take a look at the principal types of operation.

MANUAL OPERATION

This can be performed directly with small valves, or by emission or interception of a pilot signal. The operating force required can be provided by the operator.

The following actuators can be used: Monostable or bistable buttons (rounded or flat) Rocking levers Lever-operated selectors Key-operated selectors Monostable or bistable pedals

Manual control valves are normally supplied in separate parts – the body is always the same but the actuator depends on the specific requirements.

CONTROL SYSTEMS

48

49

Manual valves Foot valves

VA

LVE

S

MECHANICAL OPERATION

On the top of this type of valve is a small unit designed to be operated by moving parts such as cams, pins or brackets. It is usually called a pneumatic limit switch.

The following versions are available: with rods or cylinders or push rod one- or two-way roller levers sensitive aerials

PNEUMATIC OPERATION

In these valves the spool is operated by compressed air, which is conveyed from the outside to a piston of a suitable size anchored to the spool. During the return stage, the air is conveyed to the pilot outlet.

48

49

Mechanical valves

Pneumatic valve

VA

LVE

S

ELECTRICAL OPERATION

These valves can be operated directly or with a servo pilot. In the former case, it is the electromagnetic force generated by the electrical coil that attracts the spool and causes it to move. The coil must be large and powerful enough to generate the required force. In servo-piloted (or electro-pneumatic) valves, the coil is only used to open the passage of air (from the inlet port 1) required to operate the spool.

In this case the electrical coils are small in size and have reduced power. If the supply pressure is not high enough (0-2 bar) to create the force required to operate the spool, pilot-assisted electro-pneumatic valves have to be used. In this case, the compressed air required to operate the spool does not come from the inlet port 1 but rather from the outside with connection to threaded holes in the pilot itself.

We have seen that valves are used to convey or cut off the supply of compressed air to an actuator. This means that the quantity of air required to operate the actuator passes through the valves. This is called the rate of flow, i.e. the quantity of air passing through a given cross section in a unit of time, depending on certain factors including the following:

the size of valve the shape of the internal passages the difference In pressure between the inlet port and outlet the temperature of the air.

If we call the flow rate Q, the velocity v, and the cross section A, this will give:

Q = v x A

The velocity v depends on the difference in pressure between the inlet port and the utility port. If the difference is zero the velocity is nil, which means there is no flow

VALVE FLOW RATE

50

51

Servo pilot valve

VA

LVE

S

rate. An increase in the pressure difference (∆P) will cause the velocity and flow rate to increase accordingly. The maximum velocity is 340 m/sec (sonic velocity). When this limit is reached, a further increase in ∆P will not cause an increase in velocity or rate of flow.

The rate of flow of a valve (expressed in Nl/min) is usually indicated at an inlet pressure of 6 bar with a ∆P of 1 bar.

The flow rate of a valve can be calculated using the following (and other) parameters:

Flow factor Kv Conductance C Flow coefficient Cv.

The table below shows the relations between these and other parameters, and the rate of flow Qn expressed in Nl/min.

CALCULATING THE FLOW RATE OF A VALVE USING FLOW COEFFICIENT KV

Coefficient kV gives approximate values when used for compressed air.The flow rate QN at a normal volume through a valve is:

P1 P1 Subsonic flow: P2 > Supersonic flow: P2 < 2 2

293 293QN = 28,6 · kv · P2 · ∆P QN* = 14,3 · kv · P1 · 273 + t 273 + t

whereQN = flow rate at a normal volume [Nl/min]

QN* = critical flow rate at a normal volume [Nl/min]

kV = hydraulic coefficient in

P1 = absolute upstream pressure [bar]

P2 = absolute downstream pressure [bar]

∆P = difference in pressure P1 – P2 [bar]

t = input air temperature [°C]

50

51

VA

LVE

S

CALCULATING THE FLOW RATE OF A VALVE USING FLOW COEFFICIENTS C AND b

The flow rate QN at a normal volume through a valve is:

Subsonic flow: P2 > b · P1 Supersonic flow: P2 < b · P1

r-b 293 293 QN = C · P1 · 1 - · QN* = C · P1 · 1-b 273 + t 273 + t where

QN = flow rate at a normal volume [Nl/min]

QN* = critical flow rate at a normal volume [Nl/min]

C = conductance in [Nl/min · bar]

P1 = absolute upstream pressure [bar]

P2 = absolute downstream pressure [bar]

r = upstream pressure : downstream pressure ratio P2 /P1

b = critical pressure ratio b = P2* /P1

t = input air temperature [°C]

CALCULATING THE FLOW RATE OF A VALVE USING FLOW COEFFICIENTS CV

The flow rate QN at a normal volume through a valve is:

Subsonic flow: P2 > 0,528 · P1 Supersonic flow: P2 < 0,528 · P1

273 273 QN = 400 · CV · P2 ∆P · QN* = 200 · CV · P1 · 273 + t 273 + t where

QN = flow rate at a normal volume [Nl/min]

QN* = critical flow rate at a normal volume [Nl/min]

CV = coefficient of flow [US · GMP / p.s.i.]

P1 = absolute upstream pressure [bar]

P2 = absolute downstream pressure [bar]

t = input air temperature [°C]

( )2

52

53

VA

LVE

S

( )1/2

( )2

CALCULATING THE NOMINAL FLOW RATE

The nominal flow rate QNm of a valve, i.e. the flow at normal volume passing through a valve with p1=6 [bar] (P1=7 [bar] absolute) and ∆P= 1 [bar], can be obtained from the previous formula as follows:

QNn = 66 · kv

QNn = 943,8 · CV

0,857 - bQNn = 7 · C · 1 - 1 -b

Equalising the first two formulae gives: kv = 14,3 · CV

REACTIONS BETWEEN QNn - CV - kv - KV - S - de2

QNn = flow rate in [Nl/min] with p1=6 [bar] (P1=7 [bar] absolute) and ∆P=1 [bar]

l kgkv hydraulic coefficient in min dm3 · bar

m3 kgKV hydraulic coefficient in h dm3 · bar

CV coefficient of flow [US · GPM / p.s.i.]

Se equivalent cross section [mm2]

4de

2 = S · through diameter2 in [mm2] obtained from the equivalent cross section π

( )1/2

Cv Se de2

QNn kv Kv

0.055

18

0.785

1.273

943.8

0.001

1.259

0.794

66

0.0151

16.66

0.060

0.019

52.4514.3

0.070

52

53

VA

LVE

S

More and more often the manufacturers of machinery and equipment are demanding that the components they use be fully interchangeable. In this regard, the German certifying body ISO has issued standard 5599/1 which defines the interface between valves and the sub-bases to which they are attached.

Other valves refer to VDMA 24563-02 which, like ISO, defines the interface but with

entirely different dimensions.

There are differences in the construction technology. ISO valves are slide valves in which the dynamic gaskets providing the spool’s pneumatic seal are mounted inside and separated by spacers. In VDMA valves, the gaskets are mounted on the spool.

Valves can be used separately or combined to form an island. In the latter solution, the quantity of pneumatic wiring is reduced and a more rational use is made of the space available. There are fewer pipes connecting to the compressed air supply since each item delivers air to the associated valves. Each valve can be supplied at a different pressure compared to the other ones by means of diaphragms positioned between the bases.

STANDARD VALVES

54

55

Valve ISO 5599/1

Valve VDMA 24 563-02

VA

LVE

S

Other valves are made to Namur standards. They are designed to be fixed straight onto the blocks of pneumatic motors. Namur standards describe the fixings and centre distances of the utility ports.

The increasing need to rationalise spaces and connections has led to the development

of more compact valves. These new generation of valves combines the advantages of assembly in series, extremely reduced dimensions and wiring common to several pilots. Other advantages are a single pneumatic compressed air supply pipe and a single multi-pole power cable.

These islands are usually fixed onto Omega rails of the type used in control cubicles by means of threaded screws or adapters.

This allows enormous flexibility in the use of these systems, particularly with serial communications such as Field Buses.

With the Field Bus systems it is possible to connect various islands to the PLC using a single cable, applying serial technology (e.g. RS 232 or RS 485) to data transmission and reception.

This reduces the cost of the wiring even further.

MODULES VALVE

54

55

Valve NAMUR

VA

LVE

S

CONTROLS

Manual control

Manual pushbuttoncontrol

Manual levercontrol

Manual controlwith 2-position lever

Manual controlwith 3-position lever

Manualpedal-operated control

Mechanical controlwith ferrule

Mechanical controlwith sensitive ferrule

Mechanical controlwith spring

Mechanical controlwith roller lever

Mechanical control withsensitive roller lever

Mechanical control withunidirectional roller lever

Mechanical controlwith drawer

Electrical control

Solenoid control

Solenoid,pilot-assisted control

Piezoelectric control

Pneumatic control

Mechanical stop

Release device

56

57

VA

LVE

S

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DISTRIBUTION

2-way/2 positions valve (2/2)normally closed

2-way/2 positions valve (2/2)normally open

3-way/2 positions valve (3/2)normally closed

3-way/2 positions valve (3/2)normally open

3-way/2 positions valve (3/2)NC-NO

5-way /2 positionsvalve (5/2)

5-way/3 positions valve (5/3)pressurized centres

5-way/3 positions valve (5/3)open centres

5-way/3 positions valve (5/3)closed centres

56

57

VA

LVE

S

58

59

VA

LVE

S

58

59

VA

LVE

S

60

61

VA

LVE

S

60

61

VA

LVE

S

62

63

When compressed air finally reaches the system, having been piped from the compressor room, it is dirty and damp and of very bad quality. The quality of the air greatly affects the various components of the system. First of all, the air must be filtered to remove the dust and water present. Also, the pressure may be too high for the products used. For these reasons, the system is equipped with filters and regulators.

A simple air treatment unit has the following component parts:

a filter to clear the air of particles of liquids and solids a regulator a pressure gauge or display

The size of the unit depends on the dimensions of the system and the amount of air used. To calculate the correct size for an air treatment unit, it is necessary to add together the quantity of air used by each actuator. Taking it for granted that not all actuators operate simultaneously with the same frequency, the number of Nl/min found is multiplied by a coefficient of contemporaneity, usually equal to 0.75.

Let us now consider the separate parts making up a compressed air treatment unit.

The filter has the job of removing from the air particles of solids and liquids, such as dust, rust, moisture and used oil.

A compressor that delivers 50 litres per second puts into the circuit in one year about 4000 litres of water and 8 litres of oil. The damage caused by this waste, if it gets into the system, can be very serious. Abrasive dust damages the dynamic seals of the actuators in the valves; water and oil as they pass through the circuit remove from all the organs the lubricating grease applied during assembly.

To separate the solid particles, it is standard practice to use a (usually sintered bronze) filter with a filtering capacity of 5 or 20 or 50 µm.

The method for separating water and oil is rather more complex.

FILTER

AIR TREATMENT

Filter

Condensate separatorwith manual discharge

Condensate separatorwith automatic discharge

Filter with condensateseparator withmanual discharge

Filter with condensateseparator withautomatic discharge

Graphic symbols about filters

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UN

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The incoming air is rotated by a centrifuge (3) so that the heavier particles are projected against the sides of the container (4) and forced to adhere to it. As they accumulate, they create drops to fall to the bottom by gravity. The remaining solid particles are held back by the filter (6).

The area in which the condensate accumulates is kept still by means of a screen (7) to prevent the impurities from getting back into the circuit. The condensate is drained out through a valve (10). Condensate can be discharged by opening the valve by hand, or semi-automatically - when the air pressure in the system is low, the valve is opened by a spring and the liquid drains out. There are also fully automatic systems in which the valve is opened and closed by means of floats.

1

7

2

3

4

5

6

8

9

10

1) BODY2) END PLATE3) CENTRIFUGE4) BOWL5) BAFFLE6) FILTER CARTRIDGE7) SCREEN8) GLASS9) GASKETS10) CONDENSATE DRAIN (RMSA)

CENTRIFUGE

SCREEN

64

65

UN

ITS

If the system requires a degree of filtration of less than 5µm, it is necessary to use a filtering element called a depurator. Depurators have a degree of filtration of up to 0.01µm, and are fitted with coalescent cartridges.

Coalescent cartridges can also use active carbon, which leaves a residue of only 0.003p.p.m.

These filters are used in the food, pharmaceutical and cosmetics industries.

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When the required pressure setting has been made, a pressure regulator is used to keep it constant even with changes in the rate of flow required by the system and the pressure upstream. The choice of regulator therefore depends first and foremost on the rate of flow, and also on the degree of accuracy required.

A knob (3), connected to an adjusting screw (5) and scroll (6), is operated to compress an adjusting spring (7). As the spring presses on the diaphragm (9) it pushes a rod (11) and a valve (12), which creates an opening for the compressed air. As the air flows into the secondary chamber, it also passes under the diaphragm, which generates a force opposing the force of the spring. When the former is greater than the latter force, the diaphragm rises and closes the valve.

If there is a pressure drop in the secondary chamber, such as when an actuator starts up, the force exercised by the diaphragm is less than that of the spring. The valve is pushed downwards, which frees the opening between the primary and secondary chamber, until an equilibrium is re-established.

If the pressure increases in the secondary chamber (hammering of an actuator) the force generated by the over-pressure on the diaphragm increases considerably. This causes the diaphragm to compress the spring even more, which frees a small hole that was previously covered by the diaphragm–rod coupling. Compressed air flows out through the hole and is discharged into the atmosphere through a relief valve, a hole in the body of the regulator. When an equilibrium is restored, the spring pushes against the diaphragm and closes the outlet.

PRESSURE REGULATORS

1) BODY2) END PLATE3) KNOB4) BELL5) ADJUSTING SCREW6) SCROLL7) ADJUSTING SPRING8) RING NUT9) ROLLING DIAPHRAGM10) REELIEVING GASKET11) STEM12) VALVE13) VALVE SPRING14) PLUG15) GASKET

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UN

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With latest-generation regulators, the control knob can be replaced by an electronic system that regulates the force of a pneumatic spring according to the voltage.

The filter and the regulator can be combined in a single element called a filter-regulator, in which the air is first filtered and then regulated.

Nowadays all pneumatic products are self-lubricated, meaning that the grease applied during assembly retains its lubricating features through time. There are, however, certain products that require constant lubrication, in which case a lubricator must be included in the air treatment unit downstream of the filter.

To understand how it operates, we will refer to a principle of physics, the Venturi tube. This principle describes the behaviour of a fluid as it flows through a constricted passage. According to Bernoulli’s principle, the rate of flow Q does not change, the velocity increases and the pressure decreases.

Pressure in the throat is therefore less than that in the larger section. Lubricators operate on this principle.

LUBRICATORS

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66

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As air enters the lubricator it is forced through a constricted passage that increases the velocity and reduces the pressure. The air is also conveyed into a bowl (2) which has the same input pressure. The bowl contains oil and is connected to the throat by means of a suction tube (6) and a metering pin (9). Remember that the oil is at the same pressure as the incoming air but the pressure is less at the throat. This means that the air in the constricted section draws in oil, which nebulizes when it comes into contact with the high-velocity flow. The quantity of oil drawn in by the air is determined by the metering pin.

It is also important to remember that the oil in the air takes the place of the lubricant inside the pneumatic elements. This means that there must never be a lack of oil in the bowl.

In ordinary speech, the word “phenomenon” refers to a remarkable occurrence; in the field of physics it means a change in the position or state of a body. Therefore the result of the experiment proposed below can also be considered a phenomenon.

Let us suppose we are holding between the fingers a thin metal disk of cross section S and move it through the air at a constant velocity perpendicular to the surface of the disk. The movement causes a perceivable reaction due to the resistance of the air. This reaction lasts as long as there is movement.

If we repeat the test with a smaller disk (e.g. S/2) at the same speed, the reaction is less. Figure 1 represents this simple experiment in graphic form and gives a fairly good idea of the air flows. It also shows that at the same (atmospheric) pressure, the resistance of the air is proportionate to the cross section of the moving body.

If we now consider a tube with an internal cross section corresponding to that of the first disk, we have the reverse situation, i.e. the air now flows through it (Fig. 2).

Let us suppose that the pressure of the air is uniform. In this case, the quantity of air molecules passing through section AA is equal to that passing through section BB, and remains constant irrespective of the distance between the two sections (Fig. 3). We can therefore state that the fluid moves at a uniform velocity since there is no obstacle

VENTURI PRINCIPLE

5

11

9

810

1

12

2

6

7

3

4

1) BODY2) BOWL3) PLUG4) GLASS5) VENTURI DIAPHRAGM 6) OIL SUCTION PIPE7) FILTER8) INSPECTION DOME 9) OIL FLOW REGULATING NEEDLE10) OIL FILLING PLUG11) END PLATE12) GASKETS

68

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preventing it from moving (Fig. 4).If the tube through which the tube flows is connected to one of a smaller cross section

(Fig. 5) and the pressure of the liquid is the same as in the previous experiment, the resistance to the passage offered by the system will be greater in section DD. Figure 6 shows the situation described above with a smaller number of air molecules.

It is important to highlight the analogy between the air’s resistance to the movement of the two disks of different cross section and the resistance encountered by the air when the cross section of the pipe through which it flows is not constant.

A very important law of physic (Bernoulli’s principle) states that with the same time interval and pressure, the same quantity of air passes through pipes of different cross section. If we look at the two cross sections CC and DD in Figure 6, we can deduce that by the same law the velocity of the air in section DD is higher (Fig. 7) than in section CC in order to maintain the same rate of flow, i.e. the quantity of air passing through in a unit of time. An increase in velocity corresponds to a drop in pressure (Fig. 8), starting from the section of the pipe with a smaller cross section.

This phenomenon was discovered by the Italian physicist, G.B. Venturi, and has numerous applications in pneumatics, particularly as regards lubricators.

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Fig7 Fig8

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Though not belonging to traditional units of FRLs, some components are being used more and more in air treatment units. One of this is the system’s relief valve, or “sectioning” valve. It is basically a 3/2 bistable normally-closed valve.

This type of valve can be operated manually, electrically or pneumatically, but the operating principle is always the same.

When the valve is not in use or is not operated by the manual, pneumatic or electropneumatic control (2), the intake is closed and the discharge is connected to the outlet.

When the valve is operated, air is sent to the piston (8) which closes the outlet and allows air to flow freely to the utility. When the valve is deactivated, the spring returns the piston to the home position, freeing the outlet.

At the same time, helped by the spring (6), the valve (7) closes the intake. This causes all the air downstream of the valve to be discharged.

SHUT-OFF VALVES

2

5

4

311

8

1 7 9 6

10

1) BODY2) ELECTRICAL PILOT3) PLATE4) GASKETS5) PISTON ROD GASKET6) VALVE SPRING7) VALVE8) PISTON ROD9) BOTTOM PLUG10) TOP PLUG11) PISTON ROD SPRING

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This type of valve is often required in air treatment units. Unlike a circuit relief valve, it allows the system to be pressurized gradually following an emergency shutdown or stoppage. Without a valve of this type, the increase in pressure would be very rapid leading to sudden movements of the actuators, which could cause damage or injury.

The operating principle of a progressive start-up valve is similar to that of a relief valve.

When the air supply is switched on, air operates the piston (8) by means of the metering pin (13).

The reduced dimensions of the piston and the reduced quantity of air merely cause the outlet to close. When the valve is activated by means of a pneumatic or electropneumatic control (2), the air also flows over the piston gasket (5). This force, together with that of the piston, presses on the valve (7) and causes it to open. The retaining spring (6) prevents it from opening completely. Air from the metering pin is compressed gradually as the pressure in the downstream chamber increases. […]

When the downstream pressure is about 60% of the upstream pressure, the force is enough to compress the spring completely and open the passage. During release, the principle is identical to that of a circuit selecting valve.

SOFT START VALVES

2

5

10

11

8

3

4

1 6 9 7

12

13

1) BODY2) ELECTRICAL PILOT3) PLATE4) GASKETS5) PISTON ROD GASKET6) VALVE SPRING7) VALVE8) PISTON ROD9) BOTTOM PLUG10) TOP PLUG11) PISTON ROD SPRING12) BASE FOR SOLENOID CNOMO 13) REGULATION NEEDLE

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Pneumaticpressure source

Operating line

Pilot line

Discharge line

Flexible lineconnection

Electric cable

Line connection(welding, screwing)

Line connection(welding, screwing)

Crossing ofunconnected lines

Discharge point

Discharge holewithout connection

Discharge holewith connection

Power pick-up pointwith closing cap

Power pick-uppoint with port

Quick-fit coupling withoutunidirectional valve

Quick-fit coupling withunidirectional valve

Quick-fit coupling(de-coupling with openterminal section)

Quick-fit(de-coupling with closedterminal section)

1-way swivelcoupling

3-way swivelcoupling

Silencer

Tank

Filter

Condensate separatorwith manual discharge

Condensate separatorwith automatic discharge

Filter with condensateseparator withmanual discharge

Filter with condensateseparator withautomatic discharge

Lubricator

Pressure gauge

Pressure switch

Optical tester

FRL+pressure gaugemaintenance unit

FRL+pressure gaugesimplifiedmaintenance unit

FR+pressure gaugemaintenance unit

TRANSMISSION AND PREPARATION

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Pressure reducerwithout blowoff valve

Pressure reducerwith blowoffrelief valve

Pressure pilotedreducer withblowoff relief valve

Shutoff valve

Progressivepneumaticstarter (APR)

Progressivesolenoidstarter (APR)

Progressivepneumatic starter(APR) (SK 100 only)

Progressivesolenoid starter(APR) (SK 100 only)

3-way shutoffvalve (V3V)with lockable control

3-way shutoffvalve (V3V)with pneumatic control

3-way shutoffvalve (V3V)with solenoid control

2/2 progressivepneumatic valve(VAP) (SK 100 only)

2/2 progressivesolenoid valve(VAP) (SK 100 only)

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At present, it is preferable for complex automation systems to be controlled electronically, but this is not always possible. Electric actuation is prohibited where there is a high risk of fire, and control based entirely on pneumatic logic is used instead. The main limitation of this type of logic is that it is impossible to develop more than simple on/off systems.

Despite being very complex, these circuits can be made with pneumatic logic elements that are both reliable and affordable.

These elements that go to make up logic functions are based on binary logic, i.e. they alternately assume two opposing states.

The presence of a pneumatic output signal is indicated by the number 1, its absence by 0. Small letters a and b refer to the input signals, S1 and S2 the outputs, and P the air supply. Logic elements can be represented as follows.

LOGIC ELEMENTS

BASIC LOGIC FUNCTIONS

OR

a b

S

AND

a b

S

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a

S

P

NOT

a

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S1

S2

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Logic basic functions symbols

Two outputs memory

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When pneumatic input signal a or b is activated, this activates output S. When signals a and b are present simultaneously, the higher pressure signal prevails. It is important to note that the pressure of air in S is determined by input signal a or b.

This mode of operation can be represented schematically as follows:

a b S 0 0 0 0 1 1 1 0 1 1 1 1

This element can be used when a pneumatic valve needs to be activated from two separate remote positions.

OR FUNCTION

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In this case output S is only active when input signals a and b are present simultaneously. If either is lacking, the output is deactivated. These signals can arrive together or separately, but both of them must be present.

This mode of operation can be represented schematically as follows:

a b S 0 0 0 0 1 0 1 0 0 1 1 1

This element can be used when a pneumatic valve needs to be activated from two separate positions that are close together and have the same function.

AND FUNCTION

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This function corresponds to that of a 3/2 normally-closed valve but with a simplified structure. In the presence of input signal a, output S is active and P supplies compressed air, otherwise output S is connected to its outlet port.

This mode of operation can be represented schematically as follows:

a S 0 0 1 1

This element amplifies the signal. Even if input signal a is at low pressure (min. 1.5 bar), the pressure at output S depends on the air supply P, which may be higher.

YES FUNCTION

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This function corresponds to that of a 3/2 normally-open valve but with a simplified structure. In the presence of input signal a, output S is not active and connected to the outlet port. If this is not the case, output S is active and P supplies compressed air.

This mode of operation can be represented schematically as follows:

a S 0 1 1 0

Below is a typical application.

With valve V in the rest position, the piston rod is in the extended position. This is because without a signal, the NOT logic element allows communication between output S and the air supply and activates the 5/2 valve. When valve V is activated, a signal is sent to the NOT element, which activates output S, deactivates the 5/2 valve and allows the piston rod to retract.

NOT FUNCTION

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The S signals of the logic functions described above are the monostable type. To keep them active even when signal a or b is not present, they need to be relayed to the Memory element.

Input signal x activates output S1 and deactivates S2. Conversely, signal y activates S2 and deactivates S1.

If the output signal needs to be deactivated/activated for a short time only, a timer should be used. This function activates (if NC) or deactivates (if NO) the output signal. The state can be maintained for the length of time set on a graduated knob. A pneumatic timer can be thought of as a small tank with a flow regulator controlling the rate of discharge. The wider the regulator is open, the shorter the activation time and vice versa.

MEMORY FUNCTION

TIMER FUNCTION

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We analysed the AND logic function above. This function activates the output only when the two inputs are active. The two inputs may reach the element separately and still activate the output. In certain applications this could cause a hazard.

Let us imagine that two hand-operated valves control the descent of a press. The operator needs to press both pneumatic buttons to operate the press. This requires the use of both hands together, thus preventing them from getting crushed.

The diagram includes an AND logic element controlled by two buttons. Since the input signals may not reach the AND element simultaneously, the operator

may use an object to operate one button and one hand to operate the other. In this case the operator is not using both hands, and the safety requirements are then not complied with.

This requires the use of a bimanual safety element that only activates the output if both pneumatic inputs are activated at the same time. If more than half a second passes between one input and the other, the output is not activated.

BIMANUAL SAFETY FUNCTION

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PROBLEMS: CAUSES AND REMEDIES

Below is a description of some of the problems that you may encounter when using Metalwork products. This document does not take into account any manufacturing defects or faults resulting from worn components. The tables below list the possible causes and remedies.

The indications given under REMEDIES are not to be taken as interventions that may affect the safety, but they may help engineers in trouble-shooting.

For correct connection and use of the products, always refer to the Metal Work catalogue.

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1. CYLINDERSPROBLEMS CAUSES REMEDIES1.1 The cylinder does not move. Excessive and/or out-of-alignment load Disconnect the load and check operation. External causes prevent Remove the obstacle. the cylinder from moving Pressione insufficiente per Adeguare la pressione al carico muovere il carico applicato facendo riferimento alla tabella delle forze di pag. 38

Insufficient pressure to move Adapt the pressure to the load, the load applied making reference to table on page 1.1/05 of the MW catalogue.

No pressure Check the flow regulators are completely closed. Check the control valve (see chapter on Valves). Check that the pressure of the air treatment unit is set correctly (see Chapter 4 – REGULATORS).

Excessive air leak from one chamber Service the cylinder. to the other due to the presence of dirt from the circuit or environment depositing on the piston gaskets.

1.2 The cylinder does not cover Cushioning needles in the closed position Make a fine adjustment.the entire stroke.

1.3 The cylinder moves slowly. Flow regulators regulated incorrectly Adjust regulation as required.

Wrong type of flow regulators Mount the correct type. (e.g. type B instead of C) Low pressure Increase the pressure.

Insufficient air flow Adjust the flow rate to the system’s requirements (air treatment unit, valve, pipes).

Excessive radial load Use the guide unit.

1.4 The cylinder jerks. Check the cause in previous paragraphs Speed too low If the application requires a low speed, use NO STICK SLIP cylinders.

1.5 The cylinder is not cushioned. Cylinder not cushioned Use a cushioned cylinder.

Cushioning needles too wide Adjust them as required. Load exceeds maximum Refer to the Max Cushionable cushionable load Load-Speed diagram on page 1.1/03 of the MW catalogue.

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2. SENSORSPROBLEMS CAUSES REMEDIES2.1 The sensor does not work. Cylinder not the magnetic type Replace the cylinder.

Sensor not powered on Check the power supply, voltage and connection of the sensor. (Reed: brown + / blue live) (Hall: brown + / blue - / black OUT).

Wrong type of PNP / NPN circuit used Replace with the correct type.

Presence of magnetic metal masses Use non magnetic material. nearby affecting the magnetic field User stainless steel for tie rods. of the cylinder Several sensors connected in series Each in-line sensor causes a slight drop in voltage. Several in series cause a considerable drop in voltage. Restrict or avoid this type of connection.

2.2 The sensor works but the Reed sensor connected wrongly Connect the brown wire to theindicator light does not come on. + and the blue wire to live

2.3 The sensor detects more Sensor positioned wrongly The sensor can normally be positionedthan one signal. to receive a single signal but sometimes it may be necessary to use special sensors with reduced sensitivity.

2.4 The sensor’s working life is Excessive load exceeding the maximum Interface the load with a relay.very short switching current

Inductive load not protected from Apply a protection circuit (diode, VDR, etc.). overvoltages Voltage too high or subject to fluctuation Insert a voltage stabiliser in the circuit.

Ambient temperature too high or too low Refer to the MW catalogue to determine the temperature range for the particular type of sensor.

Excessive vibration Replace the Reed sensors with Hall sensors, which are less sensitive to vibration as they have no mechanical contacts.

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3. VALVESPROBLEMS CAUSES REMEDIES3.1 The valve does not switch over. No electric or pneumatic signal Check operation in manual mode. Check the electric or pneumatic connection.

Supply voltage does not correspond Replace the coil or adapt the voltage. to the coil’s rated voltage Supply voltage out of tolerance Refer to the MW catalogue to determine the correct voltage range.

Both coils of a bistable valve Check the electric or pneumatic activated simultaneously connection.

Bistable manual control active Check, and deactivate if necessary.

Supply pressure too low Refer to the MW catalogue to determine the valve’s minimum operating pressure. If a lower operating pressure is required, use pilot-assisted valves.

No pilot pressure in pilot-assisted valves Increase the pressure to the required value. Excessive air consumption Use pilot-assisted valves. (e.g. free-flowing blow valves) Air supply connected wrongly Check that the air supply is via port 1.

Plugged outlets (ports 3 and 5 or Remove the plugs or adjust the flow electrical sleeve) regulators. If silencers are mounted, make sure they are not obstructed.

3.2 The valve leaks. Air supply connected wrongly Check that the air supply is via port 1.

Ports 2 and/or 4 not connected to utility Check the connections.

3.3 The circuit does not operate Wrong type of valve Refer to the valve diagram.properly.

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4. REGULATORSPROBLEMS CAUSES REMEDIES4.1 The regulator releases Regulator mounted back-to-front Check the arrows showing theair through the relief valve. direction of flow and reconnect the regulator.

Downstream pressure higher When the two pressures equalise, than pressure setting discharge from relief valve ceases.

4.2 The regulator does not reach Upstream pressure lower than Regulate the upstream pressure.the desired pressure. pressure setting

Rated pressure too low Check the regulator rating.

Excessive drawing-in of air Refer to the MW catalogue for the regulator pressure and flow rate graphs.

4.3 Regulation not very sensitive. Rated pressure too high

4.4 After air has been drawn in, Regulator down-regulated – The pressure must always be setthe pressure is less than the setting. from higher to lower setting from a lower to a higher pressure.

4.5 The regulated pressure increases Air drawn in through gauge port With Skillair® regulators, it is not to the upstream pressure (Skillair® series). possible to draw in air through gauge ports because this affects operation.

4.6 The knob does not turn. Knob in locked position Pull the knob upwards to release it and make the setting.

5. FILTERS/DEPURATORSPROBLEMS CAUSES REMEDIES5.1 The filter does not discharge RMSA valve closed Turn anticlockwise to release. condensate.

Filter always under pressure Press the valve to discharge condensate by hand.

5.2 The flow rate is reduced. Filter clogged Replace the filtering element.

5.3 The depurator clogs frequently. Solid particles in the circuit Install a 5 µm filter before the depurator.

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6. LUBRICATORPROBLEMS CAUSES REMEDIES6.1 The lubricator does not work. Lubricator mounted wrongly Check the direction arrows and reconnect the lubricator.

Oil flow regulating pin in the closed Regulate as required. position

No air consumption downstream The lubricator only works if air passes of the lubricator through it to draw in the oil.

Oil too viscous Refer to the MW catalogue for the recommended grade of oil.

Ambient temperature too low If the temperature is too low, this will cause the lubricant to thicken, which would lead to a malfunction.

6.2 Too much oil enters the circuit. Flow regulating pin in the open position Regulate as required.

7. V3V SHUT OFF VALVEPROBLEMS CAUSES REMEDIES7.1 The V3V discharges permanently. Valve mounted wrongly Check the direction arrows and reconnect the valve.

7.2 The air does not passes downstream. Button not operated in the manual version Press the activation button. No pneumatic signal in the Check that the pneumatic control is pneumatic version present. No electrical signal in the Check that the electrical coil is electro-pneumatic version energised. Input pressure too low Refer to the MW catalogue to determine the minimum input pressure and use the electro-pneumatic pilot-assisted version.

No pilot pressure in the Supply air to the pilot. electro-pneumatic pilot-assisted version.

7.3 The V3V does not deactivate. Bistable manual control of electric Check, and deactivate if necessary. pilot active Electrical coil energised Remove the electric signal.

Pneumatic signal active Remove the pneumatic signal.

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8. APR SOFT START VALVEPROBLEMS CAUSES REMEDIES8.1 The air does not flow downstream. Flow regulating pin in the closed position Regulate as required.

No pneumatic signal in the Check the presence of a pneumatic pneumatic version signal.

No electrical signal in the Check that the electrical coil is electro-pneumatic version energised. Input pressure too low Refer to the MW catalogue to determine the minimum input pressure and use the electro-pneumatic pilot-assisted version.

No pilot pressure in the electro-pneumatic Supply air to the pilot. pilot-assisted version. 8.2 The actuator does not supply Flow pins in the open position Regulate as required.air progressively (air passage fullyopen instantly). 8.3 The valve does not provide full rate. Considerable amount of air drawn in Eliminate any air bleed or free-release before the APR has completed its function ports until the APR has completed its function. 8.4 The APR discharges permanently. Valve mounted wrongly Check the direction arrows and reconnect the valve.

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9. SKILLTRONICPROBLEMS CAUSES REMEDIES9.1 The regulator does not switch on Wrong wiring Consult the wiring diagram.(red light on right off). Wrong supply voltage Check voltage 24 Vdc +/- 10%.

9.2 The regulator does not regulate Wrong wiring Consult the wiring diagram.the pressure. Wrong setting of the parameter Choose the type of available input to identifying the input type set parameter d.

Set pressure higher than input pressure Regulate the input pressure.

Pressure regulation threshold too high Parameter E defines the pressure range (parameter E) within which pressure is not regulated. Select and enter a parameter.

No input pressure Restore the supply pressure.

9.3 Feed and relief solenoid valves Air drawn from downstream circuit or In this case operation is regular becauseare continuously active. downstream pressure greater than the solenoid valves must be active to set pressure maintain the set pressure constant. No or low input pressure The feed solenoid valve is activated until the protection activates (if provided). Restore correct air supply pressure.

9.4 The two red lights flash alternately. No or low input pressure The two flashing red lights indicate activation of the feed valve protection, set via parameter L. Restore the correct pressure. 9

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Technical system and International system British system CGS system Multiply by Multiply by

Length m 1 m 0,0254 in (inch) m 0,3048 ft (foot)Time s 1 s 1 sArea m2 1 m2 0,000645 in2

m2 0,0929 ft2

Volume m3 1 m3 16,39·10–4 in2

m3 0,02832 ft2

Speed m·s–1 1 m·s–1 0,3048 ft·s–1

Acceleration m·s–2 1 m·s–2 0,3048 ft·s–2

Mass kg·s2·m–1 9,81 kg 0,4536 lb (pound) kg 14,594 slug = lb ƒ ·s2·ft–1

Force kg o kp 9,81 N 4,4483 lb ƒ (pound) kg 0,981 da N = 10 NTorque kg·m 9,81 N·m 1,356 lb ƒ ·ftDensity kg·s2·m–1 9,81 kg·m–3 16,02 lb·ft–3

Specific weight kg·m–1 9,81 N·m–3 157,16 lb ƒ ·ft–3

Work, energy kg·m 9,81 J 1,356 lb ƒ ·ft KWh=3,6·106JHeat Cal 4186 J 1055,1 BTUPower kg·m·s–1 9,81 W 1,3558 lb ƒ ·ft·s–1

CV 735 W 745,7 HPPressure kg·m–2 9,81 Pa 6,8948·10 p.s.i.=lb ƒ ·in–2

kg·cm–2 9,81·10 Pa kg·cm–2 0,981 bar = 105PaMass flow kg·s·m–1 9,81 kg·s–1 0,4536 lb·s–2

Volume flow m3·s–1 1 m3·s–1 0,02832 ft·s–1

Dynamic viscosity Nl/min–1 0,0000167 Nm3 · S–1 0,000472 SCFM kg·s·m–2 9,81 Pa·s 6,896 lb ƒ ·s·in–2

Kinematic viscosity Po (poise-system CGS) 0,1 Pa·s m2·s–2 1 m2·s–2 0,0929 ft2·s–1

St (stokes-system CGS) 10–4 m2·s–2

Technical system and International system British system CGS system Divide by Divide by

TABLE 1 - CONVERSION BETWEEN SYSTEMS OF MEASUREMENT

°F = [1,8 · °C] + 32°C = [°F - 32] · 0,55°K = °C + 273

°C = Celsius degrees

°K = Kelvin degrees

°F = Fahrenheit degrees

TABLE 2 - TEMPERATURE CONVERSION TABLE 3 - MULTIPLES AND SUB-MULTIPLESName Symbol Valuetera T 1012

giga G 109

mega M 106

kilo k 103

etto h 102

deca da 10deci d 10–1

centi c 10–2

milli m 10–3

micro µ 10–6

nano n 10–9

pico p 10–12

98

99

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ICA

L D

ATA

To obtain the pressure for the following units, multiply the number given for the source units by the coefficient shown

Source Pa kPa MPa bar mbar kp/cm2 cm H2O mm H2O mm Hg p.s.i.units

Pa 1 10–3 10–5 10–5 10–2 10,1972·10–6 10,1972·10–3 101,972·10–3 7,50062·10–3 0,145038·10–3

kPa 103 1 10–3 10–2 10 10,1972·10–3 10,1972 101,972 7,50062 0,145038

MPa 106 103 1 10 104 10,1972 10,1972·103 101,972·103 7,50062·103 0,145038·103

bar 105 102 10–1 1 103 1,01972 1,01972·103 10,1972·103 750,062 14,5038

mbar 100 0,1 10–4 10–3 1 1,01972·10–3 1,01972 10,1972 0,750062 14,5038·10–3

kp/cm2 98.066,5 98,0665 98,0665·10–3 0,989665 980,665 1 1000 10.000 735,559 14,2233

cm H2O 98,0665 98,0665·10–3 98,0665·10–6 0,98665·10–3 0,98665 10–3 1 10 0,735559 14,2233·10–3

mm H2O 9,80665 9,80665·10–3 9,80665·10–6 98,0665·10–6 98,0665·10–3 10–4 0,1 1 73,5559·10–3 14,2233·10–3

mm Hg 133,322 133,322·10–3 133,322·10–3 1,33322·10–3 1,33322 1,35951·10–3 1,35951 13,5951 1 19,3368·10–3

p.s.i. 6.894,76 6,89476 6,89476·10–3 68,9476·10–3 68,9476 70,307·10–3 70,307 703,07 51,7149 1

TABLE 4 - PRESSURE UNIT CONVERSION FACTORS

Maximum recommended flow rate in Nl/min for pneumatic circuit piping. Flow rate values are calculated as follows:• pipes Ø 2 to Ø 12 with a pressure drop equal to 0.3% of operating pressure per metre of pipe.• pipes Ø 15 to Ø 40 with a pressure drop equal to 0.15% of the operating pressure per metre of pipe.

Inside diameter in mm - Nominal diameter in gas inches Pressure 1/8’’ 1/4’’ 3/8’’ 1/2’’ 3/4’’ 1’’ 1 1/4’’ 1 1/2’’bar Ø 2 Ø 4 Ø 6 Ø 8 Ø 10 Ø 12 Ø 15 Ø 20 Ø 25 Ø 32 Ø 40

2 3,5 19 53 110 190 300 370 750 1350 2500 43004 6,2 35 97 200 350 550 700 1400 2400 4500 78006 9 50 140 290 500 800 1000 2000 3500 6500 115008 11,8 66 185 380 660 1050 1300 2600 4500 8500 1500010 14,5 82 230 470 820 1300 1600 3250 5700 10500 18500

TABLE 8 - RECOMMENDED FLOW RATE10

0

10

1

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Consumption ConsumptionType of equipment at full load Nl/min. Type of equipment at full load Nl/min. 6 mm Ø drill 300 Bench tamper 35012 mm Ø drill 500 8 kg tamper 70020 mm Ø drill 1150 10 mm Ø riveting machine 45045 mm Ø drill 1650 20 mm Ø riveting machine 1000M6 screwdriver or bolt screwer 300 4 kg chisel 380M10 screwdriver or bolt screwer 400 6 kg chisel 500M16 impulse screwer 1150 Small paint-spray gun 160M25 impulse screwer 1650 Industrial paint-spray gun 5001” Ø wheel grinder 350 1 mm Ø cleaning bellows 656” Ø disk grinder 1500 2 mm Ø cleaning bellows 2509” Ø disk grinder 2100 5 mm Ø nozzle sandblasting machine 1600Polishing machine 1200 8 mm Ø nozzle sandblasting machine 42001000 kg hoist 2150 Plaster sprayer 500Spot welder 300 Heavy-duty concrete vibrator 2500 35 kg concrete breaker 1650 18 kg breaker 1850 30 kg breaker 2850

TABLE 9 - INDICATIVE AIR CONSUMPTION FOR DIFFERENT TYPES OF EQUIPMENT

1st No DESCRIPTION 1st No DESCRIPTION0 Not protected 0 Not protected1 Protected against solid bodies 1 Protected against water falling greater than Ø50 mm vertically (condensate)2 Protected against solid bodies 2 Protected against drops of water greater than Ø12 mm falling up to 15° off the vertical 3 Protected against solid bodies 3 Protected against rain water up greater than Ø2.5 mm to 60° off the vertical4 Protected against solid bodies 4 Protected against sprays of water greater than Ø1 mm from any direction.5 Protected against dust 5 Protected against jets of water fired from any direction6 Totally protected against dust 6 Protected against sea waves or the like 7 Protected against the effects of immersion

DEGREE OF PROTECTION

DEGREE OFPROTECTIONAGAINST THEPENETRATIONOF LIQUIDS

DEGREE OF PROTECTIONAGAINST THE PENETRATIONOF FOREIGN BODIESCOMING INTO CONTACTWITH LIVE PARTS.

IP 6 5

Norma EN 60529 and CEI 529

10

0

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Power production

Power storage

Power conveyance

Loss

Power costs

Environmental factors

Linear motion

Oscillating motion

Rotary motion

Linear force

PNEUMATICIn fixed or moving compression systems driven by an electric motor or internal combustion.Choice of compression system according to volume and pressure requirements. Air is available for compression anywhere in an unlimited quantity.

Enormous quantities of power can be stored at low cost. Stored compressed air is transportable in gas cylinders.

Easily transportable in pipes up to 1000 metres in length (pressure loss).

No negative consequences apart from power loss. Compressed air disperses in the atmosphere.

High for hydraulic and electrical systems: 1 m3 of compressed air at 6 bar costs 1-2 �cents, depending on the system and the rate of use.

Compressed air is unaffected by changes in temperature. No risk of fire or explosion, even without further safety measures. Risk of freezing with high air humidity, high flow rates and low environmental temperature.

Easy with cylinders up to stroke 200 mm, with high acceleration and deceleration. Speed range 10-1500 mm/sec.

Easy up to 360° with a cylinder, rack and pinion (linear motion figures) or with directional cylinder.

Compressed air motors of various types, wide rotation speed range (over 500,000 rpm), easy reverse.

Low power density due to low pressure, can be supercharged up to stopping, no power consumption for maintenance forces (stop), forces economically possible according to the air pressure and cylinder size (1-50,000 N).

HYDRAULICIn fixed or moving compression systems driven by an electric motor or exceptionally by an internal combustion engine with a generator and electric motor.Very small systems can be hand-operated. Mobile systems are an exception. Choice of pumping systems according to volume and pressure requirements.

Only limited storage, with the aid of air, economically justifiable with very small quantities.

Transportable in pipes up to 100 metres in length (pressure loss).

Energy loss and severe contamination of the environment with hydraulic fluids (risk of injury).

Sensitivity to temperature variations. Risk of fire in the event of leaks.

Easy with cylinders, good adjustment at low speeds.

Easy up to 360° and over with an oscillating cylinder.

Hydraulic motors of various types, rotation speed range less than that of compressed air motors, easier adjustment at low speeds.

High power density due to high pressure, can be supercharged up to the safety limit (overpressure valve), permanent consumption of energy for maintenance forces.

ELECTRICALUsually only superior to a region level. Conditioned by location (water, coal, nuclear power).

Storage very difficult and expensive, mostly in small quantities (accumulators, batteries).

Easy to transport over any distance.

If not in contact with other objects (conductivity) no energy loss. Risk of fatal accident with high voltages.

Minimal costs.

Not sensitive to temperature variations (normal values, insulating material). Hazardous areas require other safety devices for protection against fire and explosion.

With translating electromagnets or linear motors for short strokes only, otherwise with rotary motion and mechanical transmission.

Obtained from rotary motion via mechanical transmission.

Best performance in the field of rotary actuators.

Low efficiency due to mechanical elements downstream, cannot be supercharged, high power consumption when running empty.

COMPARISON OF FACTORS FOR VARIOUS FORMS OF POWER

10

2

10

3

TEC

HN

ICA

L D

ATA

Rotary force

Adjustability

Handling

Noise

HYDRAULICTotal torque during stopping without power consumption, can be supercharged up to stopping without negative consequences, low-power density, maximum power consumption when running empty.

Force: by pressure (pressure-regulating valve) in the 1-10 interval, simple, depending on the load.Speed: simple, using flow regulator or quick-relief valve; difficult at low speeds.

Good results even with little training; simple layout and installation of open systems; good teaching aid.

Loud discharge, can be reduced considerably using silencers.

HYDRAULICTotal torque even during stopping, but with maximum power consumption, can be supercharged up to the safety limit (overpressure valve), high power density.

Force: simple in a wide range via pressure, does not depend on the load.Speed: excellent and precise at low speed.

More difficult than pneumatic, safety at high pressure, leak and return piping, problems of hermetic seal.

Pump noise at high pressures spread by rigid piping.

ELECTRICALMinimum torque during stopping, cannot be supercharged, low power density.

Only limited possibilities, expensive.

Requires technical training, risk of accident and short-circuiting, risk of damaging the equipment and controls if wrongly connected.

High noise level from switches and moving electromagnets during switching, otherwise usual workshop noise.

COMPARISON OF FACTORS FOR VARIOUS FORMS OF POWER

10

2

10

3TE

CH

NIC

AL

DA

TA

10

4

10

5

10

4

10

5

10

6

10

7

10

6

10

7

10

8

10

9

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