Pneumatic and hydraulic maintenance questions.

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MODULE TITLE : APPLICATIONS OF PNEUMATICS AND

HYDRAULICS

TOPIC TITLE : SPECIFICATION, SELECTION AND

MAINTENANCE OF EQUIPMENT

LESSON 3 : PNEUMATIC SYSTEM FEATURES

APH - 3 - 3

Published by Teesside University Open Learning (Engineering)

School of Science & Engineering

Teesside University

Tees Valley, UK

TS1 3BA

+44 (0)1642 342740

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

In the previous lesson we dealt with the analysis of plant air consumption, and

the sizing of the pipe distribution system using a maker's chart. In this lesson

we will consider the distribution system, and show the other common method

of sizing pipework by calculation.

________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

On completion of this lesson you should be able to:

• correctly calculate a suitable pipe bore diameter for a given set of

conditions

• correctly calculate the pressure drop in a pipe system for a given set

of conditions

• appreciate the advantages of using a ring-type distribution system

• understand the reasons for using special pipe fittings.

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________________________________________________________________________________________

PIPE DIAMETER ________________________________________________________________________________________

We have seen that the pressure drop in pipework systems is an important

consideration. The amount of pressure drop in any system is dependent upon

various factors, the most important of which are listed below:

• the velocity of the air

• the length of the pipework

• the number and type of fittings used

• the roughness of the pipe walls.

The velocity of the air is a critical factor with regard to pressure drop. It has

been found that if the velocity in a main distribution pipe is kept below

10 m s–1, the pressure drop is normally within acceptable limits. The velocity

of air in branch lines is not quite as critical. As long as the line is no more than

a few metres long the velocity can be as high as 20 m s–1.

The only way we have of controlling the velocity of air in a distribution system

for any given flowrate is to size the pipework correctly, the larger the diameter,

the lower the velocity.

We investigated the correct selection of pipe diameter using charts in a

previous lesson. While the use of these charts is common practice and they

produce accurate results, they are not always available: so we will look at a

method of correctly sizing pipework by calculation.

Consider the following example.

A compressor delivers 200 � s–1 of free air into a pipe system at a pressure of

6 bar gauge. The velocity of the air is to be limited to 9 m s–1. Calculate a

suitable diameter of pipe to satisfy these requirements.

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The first thing to consider is the fact that the air entering the pipe will not be

free air; it will be compressed to 6 bar gauge and hence its volume will be

reduced. To calculate this value of compressed air, we use the gas laws.

Assuming that the temperature change is negligible:

Dividing by time t on the both sides of the equation above, we have

Since

then

Working in absolute values:

Q2 1 01 200

6 1 01

28 82

= ×

. .

. s–1�

Q V

p Q p Q

Q p Q

1 1 2 2

2 1 1

p V

t 1 1 2 2=

p V p V1 1 2 2=

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There is a standard formula which equates the relationship between:

flowrate Q in m3 s–1

velocity v in m s–1

area A in m2

The relationship is:

We can use this formula to calculate a value for A which represents the cross

sectional area of the inside of the pipe.

Re-arranging the formula for , it can be seen that the value for Q

above is given in � s–1. This must be converted to m3 s–1. This is done by

dividing the value by 1000. Also, the maximum velocity of the air in the pipe

is 9 m s–1.

Substituting the values:

This value is the cross-sectional area of the pipe bore required. However,

pipes are sized by bore diameter, therefore another calculation is required.

A = ×

28 82 1000 9

0 0032

. m2

A Q

v =

Q v A= ×

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To calculate the diameter when the area is known is simply a matter of

rearranging the formula for the area of a circle.

The selection of a suitable pipe would entail looking through manufacturers'

catalogues to find the nearest size above 63.8 mm.

A compressor delivers 350 �� s–1 of free air into a pipe system at a pressure of 7.5 bar

gauge. The velocity of the air is not to exceed 8 m s–1. Calculate a suitable bore

diameter of pipe to satisfy these conditions.

...................................................................................................................................................

________________________________________________________________________________________

A d

d A

= = ×

π π

4 4 0 0032

0 0638

63 8

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Convert free air to compressed air:

and

Working in absolute values:

Calculate the area required using Q = v × A

Note that we convert Q to m3 s–1

Calculate the diameter:

A d

d A

= ×

= =

4 0 00519

0 0813 81 3

. . m mm

A = ×

= 41 54

1000 8 0 00519

. . m2

A Q

v =

Q2 1 01 350 7 5 1 01

41 54

= ×

. . .

. s–1�

p V p V

Q p Q

1 1 2 2

2 1 1

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________________________________________________________________________________________

PRESSURE DROP ________________________________________________________________________________________

In most situations this method of calculating pipe diameter will give acceptable

results. However, in situations where very long runs are encountered it is

advisable to check what the pressure drop at the end of the system is going to

be. The pressure drop within a system can be calculated using the following

formula:

where

If we apply this calculation to the exercise question, where the pipe diameter

was calculated to be 81.3 mm, there are two other pieces of information that

are required:

• the equivalent length of pipe; assume 200 m

• the ratio of compression; this is found by dividing the absolute pressure at

the start of the pipe by the absolute pressure at the compressor inlet.

Thus R p

p = =

7 5 1 01 1 01

8 43

. . .

l Q

= =

the equivalent length of pipe in metres tthe flow through the pipe in s free ai–1� rr

the ratio of compression at the stR p

p = =2

aart of the pipe

the internal diameter of d = tthe pipe in mm.

pressure drop = 800 2

5 31

Rd .

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Substituting the values:

It can be seen that this is a low value for the pressure drop within the system.

An acceptable pressure drop for most systems is no more than 5% of the

system pressure.

If the pipe system in the previous example was extended to 300 m in length, calculate a

new value for the pressure drop.

...................................................................................................................................................

________________________________________________________________________________________

It can be seen from this example that the addition of extra pipework increases

the pressure drop within the system.

pressure drop = × ×

800 300 350 8 43 81 3

0 2

5 31. .

551 bar

pressure drop = × ×

800 200 350 8 43 81 3

0 1

5 31. .

667 bar

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RADIAL PIPE AND RING MAIN SYSTEMS

In small workshops a simple radial system, as shown in FIGURE 1, is often

adequate. As system size increases, so does the distance that the air has to

travel to its point of use, reducing its pressure. If extra equipment is added to

the system, particularly at the end of the line, the pressure drop can become

unacceptable.

FIG. 1 Radial Pipe System

To reduce this effect the system should be connected in the form of a ring, as

shown in FIGURE 2.

Points of use

Compressor

Highest pressure drop at the

point furthest away

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FIG. 2 Ring Main System

Why should the ring system have lower pressure drop than the radial system?

...................................................................................................................................................

________________________________________________________________________________________

The air can now travel to the point of use from two directions, effectively using two pipes of

the same diameter to distribute the air instead of one. This has the effect of reducing the air

velocity and hence reduces the pressure drop.

It should be noted that the use of two pipes to distribute the air as in the ring

system does not give the same cross-sectional area as doubling the pipe

diameter.

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________________________________________________________________________________________

AIR VELOCITY AND MOISTURE ________________________________________________________________________________________

Under ideal conditions, no moisture should enter the distribution system:

however, this is very rarely the case and moisture is nearly always present.

This moisture is going to condense out at some point in the system and cause

problems if its removal is not catered for.

We have seen the importance of controlling air velocity to prevent pressure

drops. It is also important that the velocity is kept below 10 m s–1 to allow

moisture to settle out and be drained away. If air velocity is excessive the

moisture will be carried to the point of use causing problems in the machine.

It is normal practice to provide a fall in the direction of flow of approximately

1% to aid drainage. Low points within the system are fitted with drop legs to

drain moisture away; these are normally connected to automatic drain traps.

FIGURE 3 shows correct moisture drainage when the air velocity is low.

FIG. 3 Drainage of Moisture

The moisture is travelling along the bottom of the pipe, falls into the drop leg,

and is then removed.

Water

To auto drain trap

Fall in direction of flowLow air velocity

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FIGURE 4 shows the same drainage arrangement, but when the air velocity is

excessive.

FIG. 4 Effect of High Air Velocity

The moisture is carried in suspension and passes over the drop leg further into

the system.

FIGURE 5 illustrates a typical distribution system of the ring main type,

showing a fall in the direction of flow and the drainage arrangements.

FIG. 5 System Drainage Arrangement

Air from receiver

Branch main

Fall

Fall Fall

DP = Drain point

DPDP

Water is whipped up and carried in suspension

The water passes over the drain leg

into the system

High air velocity 1%

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Why do you think the branch main, which is being used to supply a machine or some

other part of the plant, is taken off the top of the pipe?

...................................................................................................................................................

________________________________________________________________________________________

If the air velocity is low, any moisture present will be travelling along the bottom of the

pipe: therefore taking any supplies to feed machines etc. off the top of the pipe will reduce

the risk of supplying wet air at the point of use. Supplies which are taken off the top of the

distribution pipe should use a sweeping bend and not 90° elbow fittings: this will minimise

pressure drops.

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________________________________________________________________________________________

SPECIAL FITTINGS ________________________________________________________________________________________

There are several special pipe fittings used in air distribution systems, aimed at

either reducing pressure losses or assisting the removal of water: some of the

most common types will now be explained.

THE SWEPT TEE

This fitting is often used to reduce pressure losses in systems. It can be used to

reduce pressure losses when the flow of air is only entering from one direction.

FIGURE 6 shows a swept tee being used to supply air to a branch line on a

radial system: it can be seen that the air has an unrestricted passage into the

branch line.

FIG. 6 Swept Tee Reducing Pressure Losses

Air flow

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ECCENTRIC REDUCERS

It is often necessary to reduce the diameter of a main pipe; this is normally

done using a reducing fitting. There are two kinds of reducer available:

• the concentric reducer

• the eccentric reducer.

The eccentric reducer is the best choice to aid water removal. FIGURE 7

shows a concentric reducer being used: it can be seen that water will collect in

the bottom of the pipe, and does not flow freely on to the drain point.

FIG. 7 Incorrect Method of Pipe Reduction

FIGURE 8 shows an eccentric reducer in the same situation: water flows freely

through it to the point of removal.

Water trapped in the bottom

of the pipe

Air flow

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FIG. 8 Correct Method of Pipe Reduction

ISOLATION VALVES

It is good practice to install sufficient isolation valves within a distribution

system so that, in the event of failure of part of the system, it can be isolated

and the rest of the system run as normal. It is important that each piece of

equipment being served with air can be isolated, so that maintenance work can

be undertaken. To skimp on isolation valves is false economy and can lead to

major plant disruption.

However, if valves are fitted in the main system it is important that they have

low pressure-drop characteristics.

That concludes this lesson on air distribution. Now attempt the Self-

Assessment Questions.

Moisture carried to removal point

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________________________________________________________________________________________

NOTES ________________________________________________________________________________________

...................................................................................................................................................

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. State three factors which would affect the amount of pressure drop in a

distribution system.

2. A 75 mm compressed-air main has a 25 mm supply branch taken off to

feed a machine. The main then reduces to 50 mm diameter. Make a

sketch of this arrangement illustrating the essential design features.

3. Using the pressure drop formula:

Calculate the pressure drop at the end of a 65 mm pipe of equivalent

length 150 m carrying 250 � s–1 free air. The pressure at the start of the

pipe is 6.5 bar gauge and at the compressor inlet is 1.01 bar absolute.

4. Explain the effect high air velocities have on moisture removal in an air

main.

5. Explain why eccentric reducers are preferred to concentric reducers in air

mains.

pressure drop = 800 2

5 31

Rd .

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________________________________________________________________________________________

NOTES ________________________________________________________________________________________

...................................................................................................................................................

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. The features which would affect the amount of pressure drop are:

• the length of pipework

• the air velocity

• the diameter of the pipe.

2. The answer is shown in FIGURE 9.

FIG. 9

The compression ratio in absolute values

R p

p = =

+ =

= ×

6 5 1 01 1 01

7 44

800 1

. . .

pressure drop 550 250

7 44 65 0 238

5 31

× ×

= .

. .

bar

Pressure drop = 800 2

5 31

Rd .

Air flowφ 75 mm

φ 25 mm

φ 50 mm

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4. If the air velocity is excessive, moisture will be carried in suspension,

passing over drain points and up into supply lines to be delivered at points

of use causing machine problems.

5. The concentric reducer creates a step in the bottom of the pipe which

hinders the free flow of water to collecting points. The eccentric reducer

alleviates this problem.

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

It should now be apparent that correct pipe sizing and good distribution

practice are essential, if compressed air is to be delivered at its point of use at

the correct pressure and containing the minimum amount of moisture.

It can be seen that correct pipe sizing by calculation is a more complex and

lengthy procedure than the use of nomograms. The advantage of being able to

apply both methods is that results found by one method can be checked by the

other.

The main points to remember with regard to distribution systems are:

1. The air velocity in mains should be kept below 10 m s–1 to reduce

pressure drops and aid moisture removal.

2. The minimum number of pipe fittings should be used.

3. A fall in the direction of flow of approximately 1% should be allowed.

4. A ring main is preferred to a radial system to reduce pressure drops.

5. The pipe diameter is correctly calculated.

6. Sweeps and bends are preferred to 90° elbows.

7. Sufficient isolation valves are fitted.

8. If extra pipework is fitted to provide service to more machines, the

pressure drop is re-calculated.

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setdistillerparams << /HWResolution [2400 2400] /PageSize [612.000 792.000] >> setpagedevice