SIZING PNEUMATIC
SYSTEMS
Introduction
• Pneumatic systems are sized to meet output power
requirements.
• The air distribution system is sized to carry the required
air flow with minimum friction losses through sections of
pipe and various fittings.
• the compressor and receiver are sized as one unit
• Pneumatic valves are used to regulate the pressure, air
velocity, and flow rate, and to manage the direction of
flow.
• Cylinders are sized to meet the demand of the load
resistance . The bore is determined from the force
requirement and the stroke from the length of travel.
Pressure
Pressure is defined as the force exerted by a fluid per
unit area.
Units in SI are Pa=N/m2. The pressure unit Pascal is too
small for pressure encountered in practice.
Therefore, kPa and MPa are commonly used.
Units in British are : psf = lbf/ft2, psi = lbf/in2
You have to convert from psi to psf ( 144 in2 = 1 ft2)
5
1
bar

10
Pa

0
.
1
MPa

100
kPa
1
atm

101
,
325
Pa

101
.
325
kPa

1
.
013
ba

14
.
6
p
3
Pressure (Continued)
Absolute pressure, is
measured relative to
absolute vacuum (i.e.,
absolute zero pressure.)
Gauge pressure, is
measured relative to
atmospheric pressure


P

P

P
for
press
abo
P
gage
abs
atm
at


P

P

P
for
press
bel
P
vac
atm
abs
atm
4
Fluid Flow in Pipes
• Laminar (Re below 2000)
– Flow is consistent and streamline
• Turbulent (Re above 4000)
– Flow is mixed up and disorganized
• However, there is a range where the fluid
is neither laminar nor turbulent
– (2000 < Re < 4000)
Friction Losses
PNEUMATIC DISTRIBUTION
SYSTEMS
PNEUMATIC DISTRIBUTION SYSTEMS
Air systems are plumbed to minimize losses between the
receiver located near the compressor and the point of use.
A loss of 10% pressure is allowable under normal operating
conditions with less than half the loss being attributed to the
main transmission line.
Loop systems provide air through more than one path to the
air drops, thereby reducing run length restrictions.
Where loops are particularly long, installing more than one
compressor at convenient locations reduces the run length
and resulting pressure drop.
Provision also is made to pitch air lines slightly from 0.1- .25
in./ft, to collect moisture accumulation in lines at drain points
where water can be removed periodically.
Loop air distribution system
Types of Losses
• Major losses are calculated directly using the
overall length
• Minor losses are converted into equivalent
piping (straight) lengths then calculated using
the Major losses formula.
– Major Losses: Pipes, hoses, and tubing
– Minor Losses: Valves, fittings, bends,
enlargements/contractions, and orifices
Harris Formula
• Pressure losses in transmission lines can
be calculated using the Harris formula, or
convenient tables.
• From the Harris formula, pressure loss
due to friction (Pf) is computed from
pressure drop
•
•
•
•
•
•
•
Where
Pf= pressure drop due to friction (psi)
L = length of pipe (ft)
Q = ft3/ sec of free air
CR = ratio of compression at the pipe entrance
d = actual internal diameter of the pipe (in.)
C = experimental coefficient
CALCULATED
VALUES FOR d 5.31 FOR SCHEDULE
40 PIPE SIZES
• The friction loss through fittings is determined for each size and type
of fitting experimentally, and expressed as an equivalent length of
straight pipe of the same size.
• The equivalent length s of all fittings in the system then are added
together and combined with the length term in the Harris formula to
determine the pressure drop for the system.
• The friction loss for a number of common fittings of various sizes is
given in Table 16-2. (equivalent length)
• It should be noticed that for any particular fitting the equivalent
• length increases dramatically with pipe size.
COMPRESSORS
Air compressors are sized to supply present equipment, with a 25%50% capacity built in for future expansion.
First the pressure range is selected , usually 80 to 140 lbf/in ", Then the
free air demands of all tools and equipment using the air are totaledboth the continuous demand and the average air demand.
The continuous demand is the amount of air required if all the air tools
and equipment were operated continuously; whereas the average
demand is the free air consumption rate multiplied by the
percentage of time the equipment will be in use.
The compressor cannot be sized below the average value of the air
consumption because, even with a large receiver supplying air
during peak demands, the compressor could not recover.
• Air compressors are selected by their capacity to supply free
air , even though air tools are commonly rated at air
consumption values for a given pressure range.
COMPRESSORS
Piston displacement does provide information
about the size of the unit, but because
volumetric efficiency decreases both with
pressure and speed, it does not give an
accurate estimate of the available delivery from
the compressor.
For normal use , the compressor can be sized
from the average demand, which is 30% to 60%
of the continuous demand, depending upon the
range of operating pressure.
COMPRESSORS
Theoretical
power
PinQ  Pout

HP 
65.4  Pin

PinQ  Pout

kW 
17.1  Pin




0.286

 1




0.286

 1

• Table 16-4 matches the compressor
horsepower with the average and
continuous demand of pneumatic
equipment in commonly used pressure
ranges .
COMPRESSORS
RECEIVERS
• The receiver dampens pressure pulses from the compressor at the
inlet and supplies air at substantially constant and steady pressure
at the outlet.
• Pulsation generated in the outlet line from valve shifting and
component operation is transmitted back to the receiver and
damped.
• During times of excessive demand, the receiver supplies an output
flow in excess of the compressor input delivery capability.
• Receivers with large volumes also reduce the frequency of
compressor operation, thus reducing operating costs and wear
associated with excessive starting and stopping of the unit.
• Moisture that accumulates in the receiver is condensed and drained
off through the petcock or automatic drain provided at the bottom of
the receiver.
RECEIVERS
• The time that a receiver can supply air between a
maximum operating pressure (P1) and minimum
acceptable pressure (P2) is computed from
vr 
14.7tQ r
p1  p 2
• If the compressor is running and delivering an input to
the receiver (Qc), the formula becomes
vr 
14.7t (Q r  Qc )
p1  p 2
Increasing the size of the reservoir by 25% for unexpected overload and
25% for future expansion of production capacity is common.