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MODULE TITLE : APPLICATIONS OF PNEUMATICS AND

HYDRAULICS

TOPIC TITLE : SPECIFICATION, SELECTION AND

MAINTENANCE OF EQUIPMENT

LESSON 2 : EFFICIENT AIR DISTRIBUTION

APH - 3 - 2

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 ________________________________________________________________________________________

We have previously looked at the production of a compressed air supply

suitable for use in pneumatic power and control systems. The next stage of the

process is to deliver the air supply to its point of usage in as efficient a manner

as possible. In this lesson we deal with some of the design requirements of an

efficient air distribution system.

________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

On completion of this lesson you should be able to:

• appreciate the factors that affect the size of the pipe used to distribute

the air supply

• calculate the pipe diameter of a distribution main to fulfil a system

requirement

• appreciate the advantages and disadvantages associated with the use

of differing pipe materials and methods of assembly.

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________________________________________________________________________________________

THE COMPRESSED AIR DISTRIBUTION SYSTEM ________________________________________________________________________________________

The compressed air is conveyed from its point of storage, the air receiver, to its

point of usage, the air tool or pneumatic circuit, by a series of interconnecting

pipes and fittings. Placed at strategic points in the pipework system will be

other components, such as filters, drain traps and pressure-reducing valves,

whose function is to ensure that the air supply reaches its point of usage in a

satisfactory condition with regard to pressure, flowrate and quality.

One of the most important factors when designing a new air main is to ensure

that the pipe carrying the air is of the correct size. If too small a pipe is used,

this will impose an increased resistance to flow in the system, and result in an

unacceptably high pressure drop, which results in increased costs of operation.

If a pipe is used that is far too large, then this becomes uneconomical due to

the increased cost of the installation. The correct sizing of an air main is not as

easy as it first may seem: a number of factors must be determined before the

main distribution pipe can be correctly sized. For example, we need to

determine:

• the maximum flowrate of air to be transmitted, Q, in free air (m3 min–1)

• the maximum pressure of the air to be transmitted (bar)

• the maximum permissible pressure drop ∆p (mbar)

• the pipe length (m).

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THE MAXIMUM FLOW RATE (Q) OF AIR TO BE DISTRIBUTED

To calculate the maximum flowrate Q, you would think that you simply need

to add together the total air consumption (in free air terms) of every piece of

pneumatic equipment to be used on the plant.

This, however, could result in a figure that is unrealistically high, by virtue of

the fact that it is unlikely that every piece of equipment is being used all of the

time.

To obtain a more probable figure, it is necessary for the engineer to investigate

the duty cycle of each pneumatic device to be used, to identify exactly how

much air is being used continually, and what the peak demands for air are.

This assessment will usually result in a figure which is appreciably lower than

the initial calculation. However, some allowance should always be made for

possible additions to the plant capacity: it is usual to add 25% to the calculated

flow rate Q to cater for such.

To try and estimate total air usage is not by any means an easy task, but

information, concerning the air requirements of specific air tools, is available

from manufacturer's literature and a chart showing the general capacity

requirements of a selection of common pneumatic tools is shown in TABLE 1.

To use the chart in TABLE 1 simply identify the particular tool for which you

wish to find the consumption and read off the given value in either cfm (cubic

feet per minute) or m3 min–1.

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TABLE 1 Free air consumption of common pneumatic tools

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Tool cfm m3 min–1

Adhesive guns (average) 10 0.28 Air hoists

up to 1000 kg 1 cu.ft./ft.lift 0.09 m3/m lift 1000 kg to 5000 kg 5 cu.ft./ft.lift 0.45 m3/m lift 5000 kg and larger 15 cu.ft./ft.lift 1.38 m3/m lift

Air winches (1000 kg) 123 3.5 Blow guns (average) 18 0.5 Chipping hammers

1 kg to 2 kg weight 12 0.34 2 kg to 7 kg weight 28 0.8

Concrete vibrators 30 0.85 Drills

6 mm chuck capacity (1/4") 6 0.17 10 mm chuck capacity (3/8") 20 0.57 12.5 mm chuck capacity (1/2") 35 1.0 38 mm chuck capacity (1.1/2") 95 2.7

Air feed drill and tappers 6 mm capacity (1/4") 21 0.6 12.5 mm capacity (1/2") 33 0.93

Grinders die 6 mm collet (1/4") (average) 10 0.28 disc 100 mm wheel (4") 20 0.57

175 mm wheel (7") 40 1.13 Jack hammers, pavement breakers etc. 180 5.1 Paint sprays (average) 7 0.2

varies from 2 to 20 0.06 to 0.57 Polishers

175 mm diameter buff (7") 20 0.57 Sand-blasting units

small 67 1.9 medium 113 3.2 large 300 8.5

Sanders disc 125 mm disc (5") 14 0.4

175 mm disc (7") 20 0.57 orbital average 10 0.28

Screw drivers (clutch type) bit size 4 mm (3/16") 5 0.14 bit size 6 mm (1/4") 10 0.28 bit size 10 mm (3/8") 21 0.6

Sheet metal shears and nibblers 10 0.28 Sump pumps (at 30 m head)

2 1/s Diaphragm actuation 60 1.7 7 1/s Air motor actuation 109 3.1 13 1/s Air motor actuation 205 5.8

Wrenches ratchet 1/4" drive 7 0.2

3/8" drive 10 0.28 1/2" drive 12 0.34

impact 3/8" drive 9 0.25 1/2" drive 17 0.48 3/4" drive 25 0.71

1" drive 28 0.79

Using TABLE 1 calculate the total amount of air required in m3 min–1 to supply the

following equipment:

2 ×× 2000 kg air hoists + 3 ×× pneumatic drills of 10 mm capacity + 1 small sand blasting unit.

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

________________________________________________________________________________________

Two air hoists require 2 × 0.45 m3 min–1 = 0.9 m3 min–1

Three drills require 3 × 0.57 m3 min–1 = 1.71 m3 min–1

Small sand blaster 1 × 1.9 m3 min–1 = 1.9 m3 min–1

Total air demand per min = 4.5 m3 min–1

The air consumed by the actuation of pneumatic cylinders can also be

determined quite easily with the aid of the chart shown in TABLE 2 overleaf.

This chart lists the air consumed per centimetre of piston stroke for the

customary pressures and cylinder diameters used in pneumatic control

engineering. The air consumption on the chart is stated in litres volume at

suction conditions (free air).

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T A

B L

E 2

ir c

on su

m pt

io n

of p

ne um

at ic

c yl

in de

(e xp

re ss

ed i

n li

tr es

p er

c m

s tr

ok e

un de

r su

ct io

n (f

re e

ai r)

c on

di ti

on s

as a

f un

ct io

n of

p is

to n

di am

et er

a nd

w or

ki ng

p re

ss ur

To calculate the amount of air consumed per minute, it is necessary to use one

of two formulae, dependent upon the cylinder being of a single or a double-

acting design.

A single-acting cylinder is one which is powered pneumatically in one

direction only and returned by some other means, either by external force or by

a spring installed within the cylinder.

A double-acting cylinder is one which is powered pneumatically in both

directions.

For a single-acting cylinder use:

Air consumption

For a double-acting cylinder use:

Air consumption

where Q = air consumption in litres per minute.

q = air consumption per centimetre of piston stroke.

s = piston stroke in centimetres

n = the number of operating cycles per minute.

Note that no allowance is made in the calculation for the volume of the piston

rod. In the case of a double-acting cylinder one cycle of the actuator consists

of two strokes, one to move it from its original position and one to return it.

Q s n q= × ×( ) ( )2 l min litres per minute–1

Q s n q= × × ( ) l min litres per minute–1

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Example

A machine has a pneumatic clamp operated by a double-acting cylinder with a

piston diameter of 40 mm and a stroke length of 100 mm. The cylinder cycles

twice a minute and operates at a pressure of 6 bar. Calculate the air consumed

per minute in its operation.

Solution

For a double acting cylinder Q = 2 (q × s × n) litres min–1.

where q = 0.085 litres cm–1 (from TABLE 2)

s = 10 cm

n = 2 cycles min–1

Q = 2 × (0.085 × 10 × 2) Q = 3.4 litres min–1.

To convert litres min–1 into m3 min–1 divide by 1000

then 3.4 l min–1 = 0.0034 m3 min–1

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A machine uses two double-acting cylinders in its operating cycle. Cylinder 1 is 40 mm

diameter with a stroke length of 65 mm and is cycled 4 times per minute. Cylinder 2

has a diameter of 100 mm, a stroke length of 200 mm and is cycled twice per minute.

The operating pressure of the machine is 7 bar. Calculate the air required in

m3 min–1 to supply the machine.

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From TABLE 2, q1 = 0.097 l cm –1 and q2 = 0.611 l cm

–1.

Air consumed by cylinder 1

Air consumed by cylinder 2

Total air consumed by machine

The tables and formulae given will assist the engineer in estimating the air

consumed by common items of pneumatic equipment, and will allow him/her

to determine the overall compressed-air requirements for relatively small

installations or additions to existing plant. However, the estimation of the

compressed-air requirements of large installations is a very complex task and

the assistance of specialists should be sought.

Q = + =0 005044 0 04888 0 05392. . . m min3 –1

Q s n q2 2 2 22

2 20 2 0 611

48 88

= × × ×( )

. l min–11

3 –1 m min= 0 04888.

Q s n q1 1 1 12

2 6 5 4 0 097

5 044

= × × ×( )

. .

. l min––1

3 –1 m min= 0 005044.

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MAXIMUM PRESSURE TRANSMITTED

The maximum pressure to be transmitted by the air main also needs to be

determined. Air leaving the system air-receiver at a pressure of 7 bar will, on

being transmitted through the pipework system, experience a reduction in

pressure; the size of this reduction is dependent upon a number of variables

and is unavoidable. This means that, if we want the air to arrive at the point of

usage at a specific pressure, we must put it into the system at a higher pressure.

Hence the compressed air in the distribution system could be at an appreciably

higher pressure than that required to operate the plant equipment. Estimating

this maximum pressure in the distribution main means taking into account the

following factors that will all affect the pressure in the system:

• the minimum satisfactory operating pressure required to operate all of the

equipment connected to the system: this figure is normally between 6 and

7 bar for most common pneumatic equipment

• the expected system pressure drops which will occur during distribution

due to turbulence, pipe friction, and number and type of fittings used and

so on: this pressure drop should not exceed 5% of the working pressure

• the cut-in/cut-out pressure-differential setting of the compressor capacity

control system, which is normally between 0.5 and 1.0 bar

• a contingency allowance, normally 10% of the minimum operating

pressure, to cater for unforeseen circumstances.

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To illustrate these points, it can be calculated that for a system with a minimum

operating pressure of 6 bar and a cut-in/cut-out differential of 1 bar, the

maximum pressure in the distribution system could be:

min. operating pressure 6.0 bar

system pressure drop 0.3 bar (5% of 6 bar)

cut-in/cut-out pressure differential 1.0 bar

contingency allowance 0.6 bar (10% of 6 bar)

Total system pressure 7.9 bar

From this calculation it can be seen that, to distribute air at a minimum

pressure of 6 bar throughout the system, the pressure required for the pipe

sizing calculation would be 8 bar.

Why do you think that the number and type of pipe fittings will affect the pressure

drop in the system?

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

________________________________________________________________________________________

The fittings in the system will increase the frictional resistance of the pipework, and will

also disrupt the flow of air passing through the system, causing turbulence. Both of these

factors will increase the pressure energy expended by the air in passing through the pipe

system, resulting in less pressure being available to do work in an actuator at the point of

usage.

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THE MAXIMUM SYSTEM PRESSURE DROP

As previously stated, some pressure drop in a compressed-air system is

unavoidable. The prime causes are pipe friction, increased resistance to flow

due to valves, bends and fittings, and the speed at which the air travels through

the pipework system.

To minimise pressure drop, the inside surfaces of the pipe should be clean and

free from obstructions, no matter how small. The number of pipe fittings

should be kept to a minimum and, where possible, the pipe should be bent to

change direction gradually, rather than use a sharp-radius fitting.

The velocity of the air travelling through the pipework system is dependent

upon the amount of air being distributed and the internal diameter of the pipe.

The speed at which the air travels should be strictly controlled, as velocities in

excess of 10 m s–1 will result in more turbulent conditions and a subsequent

loss in pressure. Where possible, velocities should be kept between 6 and

10 m s–1 to minimise pressure drop due to turbulence.

It is usual to try to restrict pressure drop in a distribution main to a maximum

of 5% of the system working pressure.

Compressed air is travelling through an air main of 65 mm diameter at a speed of

8 m s–1. If the pipe diameter is reduced to 50 mm diameter, what effect do you think

this will have on the air velocity?

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

________________________________________________________________________________________

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For the flowrate of air to be maintained through the main, it is necessary for the velocity of

the air to increase.

If the velocity of the air increases beyond 10 m3 s–1 then it could result in excessive

pressure drop.

PIPE LENGTH

The length of the pipe run will also affect the choice of pipe diameter, as the

further the air has to travel then the greater will be the pressure drop

experienced. The total length of the pipe in the distribution system is made up

of the actual length of pipe-work, plus a theoretical equivalent length of pipe

which would give the same resistance to flow as that imposed by the valves

and fittings used in the system.

The actual length of the pipeline is measured from the air receiver to the closed

end of the distribution system in the case of a single line radial system as in

FIGURE 1.

FIG. 1 Single Line Radial Air Distribution System

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In the case of a ring main, such as that shown in FIGURE 2, the length of the

ring is measured and divided by two to give the actual length of pipe imposing

the resistance to flow in the system.

FIG. 2 Ring Air Distribution System

It is advantageous to use ring distribution as this imposes far less pressure drop

at the remote points from the receiver than are experienced using a single pipe

radial system.

To obtain the equivalent pipe length of bends, fittings and valves, it is usual to

use some form of nomogram or table as in TABLE 3.

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TABLE 3 Equivalent Pipe Lengths for Valves and Fittings

To use this chart we need initially to identify the size of the pipe fitting in

millimetres nominal bore and the type of fitting being used, then simply read

off the chart the equivalent length of pipe, of that diameter which would

impose the same pressure drop in the system.

Type of fitting

3-6

1.2

0.3

1.5

0.3

0.15

0.5

5-10

2.0

0.5

2.5

0.5

0.25

0.7

7-15

3.0

0.7

3.5

0.6

0.3

1.0

10-25

4.5

1.0

0.5

2.0

100

15-30

1.5

0.8

2.5

125

20-50

2.0

1.0

3.5

150

25-60

2.5

1.5

4.0

Seat valve

Diaphragm valve

Gate valve

Elbow

Bend R=d

Bend R=2d

Hose conne- ction T-piece

Reducer

Equivalent pipe length in m.

Inner pipe diameter in mm.

d2 d

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What length of 40 mm diameter pipe will be required to impose a pressure drop

equivalent to 4 tee pieces, 2 diaphragm valves and 4 elbows fitted in a distribution

main?

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________________________________________________________________________________________

Since

4 × 40 mm tee ≡ 4 × 3 m pipe = 12 m 2 × 40 mm D/V ≡ 2 × 2 m pipe = 4 m 4 × 40 mm elb ≡ 4 × 2.5 m pipe = 10 m

Total pipe length equivalent = 26 m

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PIPE SIZING

Once we have determined the amount of air to be distributed, the maximum

pressure in the system, and the allowable pressure drop, and then considered

the factors that influence the effective total length of pipework, we can get to

the true purpose of the whole exercise, which is to size the pipe system.

Once again it is usual to use some form of pipe-sizing nomogram as in

FIGURE 3 to simplify the calculation. The nomogram shown in FIGURE 3 is

for steel pipe between 15 mm and 100 mm internal diameter.

It should be noted that for copper or ABS (Acrylonitrile Butadiene Styrene

plastic) pipes, the pressure drop is about 20% less than that for a steel pipe of

the same internal diameter.

To use the nomogram to identify a pipe of suitable size, we plot the maximum

air pressure in bar and the pressure drop per metre pipe run in mbar; then we

extend the connecting line until it crosses the reference line. (To calculate the

pressure drop per metre pipe run, divide the allowable pressure drop in mbar

by pipe length in metres.)

Connect the point on the reference line to the maximum air-flowrate (Q) value

expressed in l s–1, extending this line to indicate the internal pipe diameter

required.

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FIG. 3 Pipe Sizing Nomogram for Steel Pipe

100 100 90

5000

2000

1000

500

200

100

5.0

4.0

3.0

2.0

1.0

7 8 9

Reference line

Internal dia of pipe in mm

Flow of free air in l s–1

Pressure drop through pipe in mbar m–1

Air pressure in bar

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To convert m3 min–1 into l s–1 multiply by 16.7.

Example

Determine the diameter of the distribution main to carry air at a flowrate of

18 m3 min–1 FAD. The maximum system pressure is 8 bar and the total actual

length of the pipework is 300 metres. However, fitted into the system are

8 bends whose radius is twice the pipe diameter, 4 pipe tees and two gate

valves. The maximum permitted pressure drop in the system is 0.4 bar.

Solution

To start with, the information given above, ignoring equivalent pipe lengths for

valves, fittings and so on, has to be plotted on the nomogram as shown in

FIGURE 3.

Actual pipe length is 300 metres:

0.4 bar divided by 300 = 0.0013 bar m–1 pressure drop = 1.3 mbar m–1

Flowrate = 18 m3 min–1 = 18 × 16.7 = 300.6 l s–1

Maximum system pressure = 8 bar.

(

1 1 60

1000 60

16 7

m min m s

l s

3 –1 3 –1

–1

= ss–1 )

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Plotting this information on the nomogram, illustrated by the broken lines

joining at point X as shown in FIGURE 3, will indicate a pipe diameter of

70 mm. However, this does not take into account the effect of the pipe fittings,

and can only be classed as an 'interim' diameter.

The commercially-available pipe, closest to (but larger than) the 70 mm

interim diameter, is 80 mm. This is the size we will use in referring to

TABLE 3 to determine the equivalent pipe lengths for all the fittings.

8 bends (r = 2d) = 8 × 0.5 = 4 m 4 pipe tees = 4 × 7.0 = 28 m 2 gate valves = 2 × 1.0 = 2 m

total equivalent pipe length = 34 m.

This equivalent pipe length can now be added to the actual pipe length to give

a total length of 300 + 34 = 334 metres, which we can now use to recalculate

the allowable pressure drop per metre of pipe run. This indicates a new value

of 1.2 mbar m–1. When re-plotted on the nomogram, illustrated by the solid

lines with the joint point Y as shown in FIGURE 3, this requires a pipe of

about 72 mm diameter. This would then suggest that we use a pipe of 80 mm

diameter, which could cope with the extra resistance to flow imposed by the

bends and fittings without increasing the pressure drop, and would therefore be

satisfactory for this installation.

Now that the internal diameter of the pipe has been determined, a remaining

consideration regarding choice of pipe material can be discussed.

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Choice of Pipe Material

For a factory air main, galvanised or black steel pipe connected together by

flanges and screwed or welded fittings tends to be the norm.

The main reason for the popularity of screwed fittings is the ease with which

systems can be assembled. The ends of the pipe are screwed usually with a

B.S.P (British Standard Pipe) taper or parallel thread. There is a wide selection

of screwed fittings available which allow bends, reducers, tees and unions to

be installed in the system very easily and quickly. In the case of galvanised

pipe, where welding would destroy the zinc coating, screwed fittings become

the only jointing method.

Unfortunately, even galvanised pipework has a tendency to corrode at the

screwed connections, due to the condensed moisture in the pipe system. This

rusting will introduce contamination into the circuit, which will cause

problems as it is carried further into the system.

The ease with which the system can be assembled can also lead to over-

enthusiasm in the use of screwed fittings. It must always be borne in mind that

for every fitting used in the system a price has to be paid, in terms of pressure

drop leading to power loss.

Great care should be exercised when assembling pipework, using screwed

fittings, to ensure that all joints made are air-tight. This will involve the use of

some form of thread-sealing compound or PTFE (polytetrafluoroethylene)

sealing tape, applied to the male threads of the connection prior to assembly.

When the screwed joint is assembled, the sealing compound will seal all

clearances between the threads and prevent air leakage. The finished system

can be checked for leaks by charging the system with air, applying a soapy

water solution to each joint in turn, and looking for any bubbles that would

indicate an escape of air.

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Steel pipe systems with welded joints are preferable to screwed connections, as

there is usually less leakage from an all-welded system and the cost of the

installation is cheaper. However, one drawback associated with the use of

welded fittings is the formation of scale on the inner surface of the weld, which

in time will tend to rust.

Copper tubing has been used for many years wherever rigid connections are

required. It is favoured for its corrosion resistance and the ease with which it

can be softened and bent. However, when bending copper tube, it is essential

that proper bending tools are used to avoid kinking, as this would result in a

high resistance to flow and subsequent pressure drop. Copper pipe can be

joined by soldering or, more usually, by the use of compression fittings or bite

sleeve unions. Copper is becoming increasingly popular for distribution

mains. Contrary to expectation copper piping can, in certain situations, be

much more economical than a galvanised steel system, because of its excellent

pressure-drop characteristics, due to the need for fewer pipe fittings and its

reduced resistance to flow. It is also virtually corrosion free, which reduces the

risk of system contamination due to the entry of rust into pneumatic

equipment.

Plastic piping is also being used for compressed air mains of various kinds.

Plastic pipes can be joined by plastic welding or by the use of specially

designed compression unions. Although the cost of the installation is greater

than for steel, it has the advantages of being virtually maintenance free, light in

weight, and low resistance to flow.

Where do you think it would not be possible to use plastic piping as an air main?

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

________________________________________________________________________________________

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In situations where the main comes into close contact with hot surfaces it could damage or

weaken the surface of the pipe.

Whichever pipe material is selected great care should be taken to remove

foreign material from the pipework prior to commissioning. This will include

swarf left over from the screw-cutting operation, burs left on the end of the

pipe after it has been cut, any internal dirt or scale in the pipe, and any excess

jointing materials. If this potential source of system contamination is not

removed, it can result in premature component failure and increased machine

downtime due to pneumatic control failure.

Now try to answer the following Self-Assessment Questions.

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________________________________________________________________________________________

NOTES ________________________________________________________________________________________

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

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. List four factors that have to be determined before it is possible to select

a pipe of suitable diameter for a compressed-air distribution main.

2. State two advantages and two disadvantages associated with the use of

threaded pipe fittings.

3. What are the effects of incorrectly sizing an air main?

4. Determine the size of an air main made from steel pipe, using the

necessary nomograms and tables, for the distribution of 9 m3 min–1 of air

at a system working pressure of 7 bar. The length of the pipe run is 200

metres and a maximum pressure drop of 0.5 bar is allowed. The

distribution main must include the following fittings and valves:

6 bends each with a radius twice the pipe diameter

2 elbow fittings

4 tee connectors

2 gate valves.

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________________________________________________________________________________________

NOTES ________________________________________________________________________________________

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

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. The factors that must be determined before a pipe can be correctly sized

are:

(i) flowrate of air to be distributed

(ii) maximum working pressure

(iii) length of the pipe run

(iv) allowable pressure drop in the system.

2. Two advantages associated with the use of screwed pipe fittings are:

(i) a wide selection of fittings available

(ii) easy assembly.

Two disadvantages are:

(i) increase in system pressure drop

(ii) susceptibility to corrosion

3. The effects associated with undersizing a distribution main are increased

pressure drop, resulting in power loss due to increased resistance to flow,

and high air velocity producing more turbulence.

The effects of oversizing a distribution main are increased cost of raw

materials: but other than that no negative effects will be experienced. It is

always better to oversize than undersize.

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4. Information to be plotted on nomogram (FIGURE 3)

Flowrate Q = 9 m3 min–1 = 150.3 l s–1

Max. pressure = 7 bar

Allowable pressure drop = 0.5 bar = 500 mbar

Length of pipe main = 200 m

∴ Allowable ∆p m–1 pipe run = = 2.5 mbar m–1.

This would result in an interim pipe size of 49 mm which is

approximately equal to a 50 mm nominal bore pipe. We will therefore

use 50 mm bore to obtain the equivalent pipe lengths for all fixtures and

fittings included.

From TABLE 3 this gives:

6 bends (r = 2d) = 6 × 0.3 = 1.8 m 2 elbows = 2 × 3.5 = 7.0 m 4 tees = 4 × 4.0 = 16.0 m 2 gate valves = 2 × 0.7 = 1.4 m

∴ Total equivalent pipe length = 26.2 m

Add this to the original length:

200 + 26.2 = 226.2 m

This will give an allowable ∆p m–1 pipe run of 2.2 mbar m–1, which when re-plotted on nomogram FIGURE 3 will give a revised pipe diameter of

51 mm. This will then require the next commercially available pipe to be

used which is 60 mm.

500 200

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

Prior to being able to size an air main, you have to determine the amount of air

to be distributed, at what pressure, over what distance, and what is the

allowable pressure drop.

Fittings assembled into the system will impose a pressure drop, leading to a

power loss and therefore their use should be restricted to absolute necessity.

We have also presented a method of correctly sizing a distribution main, and

given you some indication of the advantages and disadvantages associated

with the use of different pipe materials.

Teesside University Open Learning (Engineering)

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