Sổ tay tính toán cơ khí P9 ppsx

9.1
SECTION 9
AIR AND GAS COMPRESSORS
AND V ACUUM SYSTEMS
Estimating the Cost of Air Leaks in
Compressed-Air Systems 9.1
Selecting an Air Motor for a Known
Application 9.4
Air-Compressor Cooling-System Choice
for Maximum Coolant Economy 9.10
Economics of Air-Compressor Inlet
Location 9.14
Power Input Required by Centrifugal
Compressor 9.16
Compressor Selection for Compressed-
Air Systems 9.18
Sizing Compressed-Air System
Components 9.24
Compressed-Air Receiver Size and
Pump-Up Time 9.26
Vacuum-System Pump-Down Time
9.27
Vacuum-Pump Selection for High-
Vacuum Systems 9.30
Vacuum-System Pumping Speed and
Pipe Size 9.33
Determining Air Leakage in Vacuum
Systems by Calculation 9.34
Checking the Vacuum Rating of a
Storage Vessel 9.36
Sizing Rupture Disks for Gases and

Liquids 9.39
Determining Airflow in Pipes, Valves,
and Fittings 9.40
System Economics and Design Strategies
ESTIMATING THE COST OF AIR LEAKS IN
COMPRESSED-AIR SYSTEMS
Find the cost of compressed air leaking through a 0.125-in (0.3175-cm) diameter
hole in a pipe main of a typical industrial air piping system, Fig. 1, to the atmo-
sphere at sea level when the air pressure in the pipe is 10 lb /in
2
(gage) (68.9 kPa),
the plant, Fig. 2, operates 7500 h/yr, air temperature is 70
ЊF (21.1ЊC), and the cost
of compressed air is $1.25 per 1000 ft
3
(28.3 m
3
). What is the cost of the leaking
air when the pipe pressure is 50 lb/in
2
(gage) (344.5 kPa) and the other variables
are the same as given above?
Calculation Procedure:
1. Find the volume of air discharged to the atmosphere
Air flowing through an orifice or nozzle attains a critical pressure of 0.53 times the
inlet or initial pressure. This reduced pressure occurs at the throat or vena contracta,
which is the point of minimum stream diameter on the outlet side of the air flow.
If the outlet or back pressure exceeds the critical pressure then the vena contracta
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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
9.2 PLANT AND FACILITIES ENGINEERING
FIGURE 1 Typical compressed-air system main and branch pipes (Factory Manage-
ment and Maintenance).
or throat pressure rises to equal the backpressure. Air flow through a hole in a pipe
or tank replicates the flow through an orifice or nozzle.
When an inlet air pressure of 10 lb/in
2
(gage) ϩ 14.7 ϭ 24.7 lb/in
2
(abs) (170.2
kPa), the critical pressure is 0.53
ϫ 24.7 ϭ 13.09 lb/in
2
(abs) (90.2 kPa). Since
13.09 lb /in
2
(abs) is less than the atmospheric backpressure of 14.7 lb/in
2
(abs)
(101.3 kPa), the throat pressure equals the backpressure, or 14.7 lb/in
2
(abs) (101.3
kPa). Knowing this, we can compute weight of the escaping air from W
ϭ 1.06
A(P
1
[P Ϫ P
1

]/T)
0.5
, where W ϭ leakage rate, lb /s (kg/s); A ϭ area of leakage
hole, in
2
(cm
2
); P ϭ pipeline or initial air pressure, lb /in
2
(abs) (kPa); P
1
ϭ outlet
or backpressure, lb/in
2
(abs); T ϭ absolute temperature of the air before leakage
ϭ ЊF ϩ 460.
Substituting, using the values given above, W
ϭ 1.06 ϫ 0.012272(14.7[24.7 Ϫ
14.7]/530)
0.5
ϭ 0.006851 lb /s (0.0031 kg/s). Converting this air leakage rate to
lb/h (kg/h), multiply by 3600 s/h, or 0.006851
ϫ 3600 ϭ 24.66 lb /h (11.19 kg/
h). Since the cost of compressed air is expressed in $/ft
3
, the flow rate of the
leaking air must be converted. Since air at 14.7 lb/in
2
(abs) (101.3 kPa) weighs
0.075 lb per ft

3
(1.2 kg/m
3
), the rate of leakage is 24.66/0.075 ϭ 328.8 ft
3
/h (9.31
m
3
/h).
2. Determine the annual cost of the air leakage
This compressed-air plant operates 7500 h/yr. Since the leakage rate is 328.8
ft
3
/h, the annual leakage through this opening is 7500 ϫ 328.8 ϭ 2,466,000 ft
3
(58,691 m
3
). At a cost of $1.25 per 1000 ft
3
, the annual total cost of this leak is
$1.25
ϫ 2,466,000/1000 ϭ $3,082.50. This is a sizeable charge, especially if there
are several leaks of this size, or larger, in the system.
3. Find the rate of leakage at the higher line pressure
When the backpressure is less than the critical pressure, a different flow equation
must be used. In the second instance, the critical pressure is 0.53 (50 lb/in
2
(gage)
ϩ 14.7) ϭ 34.29 lb/in
2

(abs) (236.26 kPa). Since 34.29 lb/in
2
(abs) (236.26 kPa)
is greater than the atmospheric backpressure of 14.7 lb/in
2
(abs) (101.3 kPa), the
critical pressure is greater than the backpressure. Air leakage through the hole is
now given by W
ϭ 0.5303(ACP)/(T)
0.5
, where C ϭ flow coefficient ϭ 1.0; other
symbols as before.
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS
9.3
FIGURE 2 Typical compressed-air plant showing compressor and its associated piping and accessories (Power).
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS
9.4 PLANT AND FACILITIES ENGINEERING
Substituting, W ϭ 0.5303(0.012272)(1.0)(64.7)/(530)
0.5
ϭ 0.01829 lb/s (0.0083
kg/s). Converting to an hourly flow rate as earlier, 3600
ϫ 0.01828 ϭ 65.84 lb/h
(29.89 kg/h).
4. Compute the annual cost of air leakage at the higher pressure

Following the same steps as earlier, annual leakage cost
ϭ 65.84 lb/h (7500
h/yr)(1.25 /1000 ft
3
)/0.075 lb/ft
3
ϭ $8,230.00 per year. Again, this is a significant
loss of revenue. Further, the loss at the higher pressure is $8230 /3082.50
ϭ 2.67
times as great. This points out the fact that higher pressures in a compressed-air
system can cause more expensive leaks.
Related Calculations. Compressing air requires a power input to raise the air
pressure from atmospheric to the level desired for the end use of the air. When
compressed air leaks from a pipe or storage tank, the power expended in compres-
sion is wasted because the air does no useful work when it leaks into the atmo-
sphere.
In today’s environment-conscious world, compressed-air leaks are considered to
be especially wasteful because they increase pollution without producing any ben-
eficial results. The reason for this is that the fuel burned to generate the power to
compress the wasted air pollutes the atmosphere unnecessarily because the air pro-
duces only a hissing sound as it escapes through the hole in the pipe or vessel.
SELECTING AN AIR MOTOR FOR A KNOWN
APPLICATION
Show how to select a suitable air motor for a reversible application requiring 2 hp
(1.5 kW) at 1000 rpm for an industrial crane. Determine the probable weight of
the motor, its torque output, and air consumption for this intermittent duty appli-
cation. An adequate supply of air at a wide pressure range is available at the
installation.
Calculation Procedure:
1. Assemble data on possible choices for the air motor

There are four basic types of air motors in use today: (1) radial-piston type; (2)
axial-piston type; (3) multi-vaned type; (4) turbine type. Each type of air motor has
advantages and disadvantages for various applications. Characteristics of these air
motors are as follows:
(1) Radial-piston air motors, Fig. 3, have four or five cylinders mounted around
a central crankshaft similar to a radial gasoline engine. Five cylinders are preferred
to supply more horsepower with evenly distributed power pulses. In such a unit
there are always two cylinders having a power stroke at the same time. The radial-
piston motor is usually a slow-speed unit, ranging from 85 to 1500 rpm. It is suited
for heavy-duty service up to 20 hp (15 kW) where good lugging characteristics are
needed. Normally they are not reversible, though reversible models are available at
extra cost.
(2) Axial-piston air motors, Fig. 4, are more compact in design and require less
space than a four- or five-cylinder radial-piston motor. Air drives the pistons in
translation; a diaphragm-type converter changes the translation into rotation. This
arrangement supplies high horsepower per unit weight. Axial-piston motors are
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS
AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS 9.5
FIGURE 3 (a) Five-cylinder piston-type radial air motor used in sizes from about 2 hp (1.5
kW) to 22 hp (16.4 kW) and at speeds from 85 to 1500 rpm. (b) How five-cylinder air motor
distributes power. Two cylinders are always on power stroke at any instant (Gardner-Denver
Company).
available in sizes from 0.5 to 2.75 hp (0.37 to 2.1 kW). They run equally well in
either direction. To make the motor reversible, a four-way air valve is inserted in
the line.
(3) Multi-vaned motors, Fig. 5, are suitable for loads from fractional hp (kW)
to 10 hp (7.5 kW). They are relatively high-speed units which must be geared down

for usable speeds. The major advantages of multi-vaned motors is light weight and
small size. However, if used at slow speed, the gearing may add significantly to
the weight of the motor.
(4) Air-turbine motors deliver fractional horsepowers at exceptionally high
speeds, from 10,000 to 150,000 rpm, and are an economical source of power. They
are tiny impulse-reaction turbines in which air at 100 psi (689 kPa) impinges on
buckets for the driving force. Force-feed automatic lubrication sprays a fine film of
oil on to bearings continuously, minimizing maintenance.
Based on the load requirements, 2 hp (1.5 kW) at 1000 rpm, a reversible radial-
piston air motor, Table 1, would be a suitable choice because it delivers up to 2.8
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9.6 PLANT AND FACILITIES ENGINEERING
Diaphragm converts
piston translation
to rotation
Pistons
(four or five)
Output
shaft
FIGURE 4 Axial-piston air motor available in various output sizes (Keller Tool Com-
pany).
FIGURE 5 Typical multi-vane type air motor, available in fractional hp sizes and up to
some 10 hp (7.5 kW) (Gast Manufacturing Company).
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS

AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS 9.7
TABLE 1 Specifications of Typical Air Motors
Rated hp*
(kW)
Speed at rated
hp—rpm
Free speed
rpm
Weight
lb (kg)
Stall torque
ft—lbs
Air consumption
at rated hp ft
3
free air/min
Radial piston motors (non-reversible)*
2.9 (2.2) 1,500 3,200 130 (59)
3.3 (2.5) 1,300 3.000 130 (59)
3.8 (2.8) 1,200 2,700 130 (59)
Radial piston motors (reversible)*
2.5 (1.7) 1,200 2,200 135 (61.3)
2.8 (2.1) 1,000 1,950 135 (61.3)
3.2 (2.4) 900 1,600 135 (61.3)
5.2 (3.9) 750 1,600 200 (90.8)
*at 90 lb /in
2
(620 kPa).
Ingersoll-Rand.
hp (2.1 kW) at 1000 rpm with air delivered to the motor at 90 lb/in

2
(620 kPa).
The weight of this motor, Table 1, is 135 lb (61.3 kg).
2. Compare the advantages of air motors to other types of motive power
Air motors have a number of advantages over their usual competitors—electric
motors. These advantages are: (1) In explosive or gaseous environments, air motors
are lower in cost than larger, heavier, explosion-proof electric motors. Air motors
operate relatively trouble-free in moist, humid environments where the electric mo-
tor may suffer from a buildup of fungus and corrosion. And since the air motor
requires little maintenance, it can be mounted in inaccessible locations. (2) With
an air motor, the output speed can be varied from zero to free-speed no-load rotation
by merely changing the volume of air supplied to the motor. Controls are simple
in design and use. (3) Air motors can weigh as little as one-quarter that of electri-
cally-powered units; their physical dimensions are about 50 percent those of elec-
trical devices. Further, air motors do not spark; they cannot burn out from over-
loading; the air motor is not injured by stalling. Air motors start and stop positively;
they have a consistent output torque which can be changed by varying the inlet air
pressure.
Air motors do, however, have limitations. Thus: (1) Compared to electric motors,
air motors are inefficient. An air motor requires about 5 hp (3.7 kW) input to the
air compressor to produce one horsepower (0.7 5kW) at the motor outlet. (2) Air
motors are rarely practical in sizes greater than 20 hp (15 kW). Their most efficient
range is 1 /20 to 20 hp (0.04 to 15 kW). (3) The initial cost of an air motor is high;
in larger sizes, above 1 hp (0.75 kW), air motors cost up to five times that of
equivalent electric motors.
3. Check the motor duty cycle and load against the unit’s characteristics
When selecting an air motor, the first factor to be considered is the type of duty
cycle, intermittent or continuous. A crane, for which this motor will be used, does
have an intermittent duty cycle because it is not normally used continuously. There
is a rest period while the crane load is being put on the crane and again while

being off-loaded from the crane.
The great majority of air-motor applications have a low-load cycle; the air mo-
tors are used for only a few seconds continuously and have long off-duty periods.
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9.8 PLANT AND FACILITIES ENGINEERING
Torque
Horsepower
Governor
controlled
curves
Rated performance
Horsepower
Torque
Speed RPM
1
Free speed
0
Stall speed
FIGURE 6 Performance curve of a typical air motor. Note how a built-in governor can change
the shape of the curve by limiting the maximum speed of the air motor (Product Engineering).
The duty cycle will usually determine the type of motor and the size of compressor
that must be used.
4. Check the horsepower and speed required
Performance curves, Fig. 6, show an air motor’s torque and horsepower (kW) output
at various rpm. Such curves can be varied somewhat by using governors or by
modifying the air intake or exhaust ports. However, the basic shape of the perform-
ance curve depends on the fundamental design of the air motor. It is common

practice to rate an air motor at its maximum output, i.e., at the top of the dome-
shaped performance curve. The reversible radial-piston motor chosen here has ad-
equate horsepower and speed for the anticipated load.
5. Determine the effect of air pressure and quantity on the air motor output
Table 2 shows how the air pressure available at the motor inlet affects both the
power output and rpm of typical air motors. For the motor being considered here,
the output would be sufficient at the lowest air pressure listed. Thus, the motor
choice is acceptable.
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS
9.9
TABLE 2 Effect of Air Pressure on Motor Performance
Motor style
no.
Rated hp (kW)
1234 5
rpm at rated hp
1 2 345
Free speed—rpm
12345
At 60 psi lb
/in
2
1.9 2.5 3.2 6.4 0.4 1,200 1,100 850 800 900 2,650 2,250 2,030 1,900 1,600
(413 kPa) (1.4) (1.9) (2.4) (4.8) (0.29)
Gardner-Denver Company.
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS
9.10 PLANT AND FACILITIES ENGINEERING
Related Calculations. When using tables of air-motor performance, it is im-
portant to keep in mind that the stall torque and air consumption vary for each
motor. Hence, these values are not listed in the usual performance tables. There are
so many variables in air-motor choice that stall torque and air consumption are
unique for each application and are supplied by the motor manufacturer when the
motor choice is made.
As a general rule of thumb, stall torque ranges between 2 and 2.5 times the
torque developed when operating at maximum horsepower output.
In small motors, up to 2.5 hp (1.9 kW), air consumption varies from 35 to 40
ft
3
/min (0.99 to 1.1 m
3
/min) of free air per hp (0.746 kW). Larger air motors
consume 20 to 25 ft
3
/min (0.57 to 0.70 m
3
/min) of free air per hp. These con-
sumption rates apply to non-reversible motors. Reversible air motors consume 30
to 35 percent more air.
The data, tables and illustration in this procedure are from Product Engineering
magazine.
AIR-COMPRESSOR COOLING-SYSTEM CHOICE
FOR MAXIMUM COOLANT ECONOMY
Select a suitable cooling system for a two-stage 5000-hp (3730-kW) engine-driven
air compressor, Fig. 7, installed in a known arid hard-water area when the rated

output of the compressor is 25,000 ft
3
/min (708 m
3
/min) at 100 lb/in
2
(abs) (689
kPa). Water conservation is an important requirement for this compressor because
of the arid nature of the area in which the unit is installed. Use standard cooling-
water requirements in estimating the capacity of the cooling system.
Calculation Procedure:
1. Assess the types of cooling systems that might be used
Several types of cooling systems can be used for air compressors such as this.
Because the air compressor is used in an arid area subject to water shortages, a
recirculating system of some type is immediately indicated. Since both the engine
and air-compressor cooling water require temperature reduction in such an instal-
lation, the two requirements are usually combined in one cooling system.
The first arrangement that might be chosen, Fig. 8, combines a heat exchanger
for engine power-cylinder cooling and a cooling tower for raw-water cooling for
the compressor. Either a natural-draft cooling tower, such as that shown, or a me-
chanical-draft cooling tower might be used. The cooling-tower choice depends on
a number of factors. In an arid area, however, natural-draft towers are known to
perform well in dry climates. Further, they require much less piping and electric
wiring than mechanical-draft towers.
Another possible cooling-system arrangement uses a closed coil in the cooling
tower for both the power and air cylinders, Fig. 9. This totally closed system does
not allow contact between the compressor and engine cooling water with the at-
mosphere. This means that the compressor and engine cooling water can be treated
to reduce scale formation. Raw water recirculated through the cooling tower does
not contact the compressor coolant.

Where installation costs are critical, raw water can be used to cool the air-
compressor cylinders, Fig. 10. The engine power cylinders, which usually operate
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS
9.11
FIGURE 7 Gas-engine driven compressor has oil-cooled power pistons. Compressor, left, uses metallic piston-rod packing (Cooper-
Bessemer Corp.).
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS
9.12 PLANT AND FACILITIES ENGINEERING
FIGURE 8 Heat exchanger, center, for power-cylinder cooling and raw-water cooling
for the compressor (Ingersoll-Rand Co.).
FIGURE 9 Closed cooling system for power and air cylinders utilizing pipe coil in
the cooling tower (Ingersoll-Rand Co.).
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS 9.13
FIGURE 10 Raw water cools the air cylinders; power cylinders use
closed system protected by thermostatic valve (Ingersoll-Rand Co.).
110°F (43.3°C) 130°F (54.4°C)
FIGURE 11 Cooling-tower recirculating system is not recommended because of the
possibility of scale and impurities buildup (Ingersoll-Rand Co.).
at a higher temperature, are cooled by a closed system protected by a thermostatic
valve.

An open cooling-tower system, Fig. 11, is not recommended for installations
such as this because of the possible heavy scale buildup. However, such an open
cooling system might be used where the economics of the installation permit it and
scale buildup is unlikely to occur.
2. Determine the air-compressor cooling load
Use flow rates given in Table 6, page 9.24 to estimate the cooling water flow rate
for this air compressor. Thus, with the intercooler and jacket in series, using the
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS
9.14 PLANT AND FACILITIES ENGINEERING
higher flow rate of 2.8 gal/min (364.3 L/ s) per 100 ft
3
/min (100 m
3
/s), the cooling-
water flow required for this 25,000 ft
3
/min (708 m
3
/s) air compressor is (25,000/
100)(2.8)
ϭ 700 gal/min (44.2 L/s).
3. Compute the engine jacket-water cooling load
The engine jacket water cooling load is computed separately using a cooling-water
temperature rise of 10 to 20
Њ F (5.6 to 11.1ЊC) during passage through the engine,
as given in the Internal-Combustion Engine section of this handbook. Further, the
usual jacket-water flow rate is 0.25 to 0.60 gal /(min bhp) (0.02 to 0.05 kg/kW).

For a 5000-hp (3730-kW) engine, using the maximum flow rate, (5000)(0.6)
ϭ
3000 gal/min (189.3 L/s).
Additional cooling water may be used for the turbocharger, if fitted, and for
aftercooling. Steps for calculating these cooling-water flows are given in the section
cited above. Such cooling-water flows are usually additive to the jacket-water flow,
depending on the cooling arrangement used.
Related Calculations. Cooling systems for air and gas compressors are im-
portant for reliable and safe operation of these units. Hence, great care must be
exercised in choosing the most reliable and economic cooling system.
Today, both mechanical-draft and natural-draft cooling towers are popular
choices. An economic study is needed to determine the best choice when the cool-
ing effectiveness of both types of towers are about equal. Data given on cooling
towers elsewhere in this handbook can be helpful to the designer in choosing the
best type of tower to use for a given installation of air or gas compressors.
Straight flow-through cooling of small compressors and their drive engines is
often used where adequate water supplies are available. Thus, in large cities the
cooling water may be taken from the water main and discharged to the sewer after
passage through the compressor and engine. Cost of the water may be small com-
pared to the investment in a cooling tower. But with increased environmental con-
cerns, this scheme of cooling may soon be extinct.
ECONOMICS OF AIR-COMPRESSOR INLET
LOCATION
A plant designer has the option of locating an air-compressor inlet pipe either inside
the compressor building or outside the structure. The prevailing average indoor
temperature is 90
ЊF (32.2ЊC) while the average outdoor temperature is 50ЊF (10ЊC).
Air requirements for this plant from the compressor are: 1000 ft
3
/min (28.3 m

3
/
min) of free air at 70
ЊF (21.1ЊC) at 100 lb/in
2
(gage) (689 kPa) for 7500 h/yr; the
200-hp (149.1 kW) compressor drive motor operates at full load throughout the
7500 hr load year. Determine which is the best location for the compressor intake
based on power savings with an electric power cost of $0.04/kWh.
Calculation Procedure:
1. Determine the power savings possible with cooler intake air
Compute the intake volume, I, required to deliver 1000 ft
3
/min of free air at each
of the possible intake temperatures from I
ϭ 1000(density of air at 70 ЊF, lb/ft
3
/
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS 9.15
density of air at inlet temperature, lb/ft
3
), where I ϭ intake volume, ft
3
(m
3
), re-

quired at the air inlet temperature.
For inlet air at 50
ЊF (21.1ЊC), the outside intake air temperature, using a table
of air properties, I
ϭ 1000(0.07493/0.7785) ϭ 962.49 ft
3
(27.2 m
3
). With an inlet
temperature of 90
ЊF (32.2ЊC), using the inside-of-the-building air intake, I ϭ
1000(0.07493/0.07219) ϭ 1037.95 ft
3
(29.37 m
3
), say 1038 ft
3
(29.37) cu m).
The power saving from using lower-temperature intake air is then hp saving
ϭ
100(intake volume required at the higher intake temperature Ϫ intake volume re-
quired at the lower intake temperature)/intake volume required at the higher intake
temperature. Substituting, hp saving
ϭ 100(1038 Ϫ 962)/1038 ϭ 7.32 percent.
2. Find the annual power saving with the lower intake temperature
The annual power saving, P kWh, can be found from: P ϭ (hp saving/ 100)(motor
hp)(0.746 kW /hp)(annual operating hours). Since the compressor operates at full
load 7500 h/yr using 200 hp, the annual power saving is P
ϭ (7.32/100)(200
hp)(0.746)(7500)

ϭ 81,910.8 kWh.
The annual cost saving, A
ϭ (kWh/ h per yr saved)(power cost, $/ kWh). With
a power cost of $0.04/kWh, the annual cost saving, A
ϭ (81,910.8)(0.04) ϭ
$3276.43. If an outside inlet were more expensive than an indoor inlet, this saving
could be used to offset the increased cost.
Related Calculations. As a general rule, an outside air intake, Fig. 12, is more
economical than an inside air intake when the air in the building is at a higher
temperature than the outside air. The only time an outside air intake might be less
desirable than an indoor air intake is when the outside air is polluted with corrosive
vapors, excessive dust, abrasive sand, etc., which would be injurious to people or
machines. Under these circumstances the designer might elect an indoor air intake.
However, before choosing an indoor intake, review the efficacy of outdoor air filters
of various types, Fig. 13.
Air in industrial districts may contain from 1 to 4 grains of dirt per 1000 ft
3
(28.3 m
3
). If the intake air for a compressor contains only 1 gram per 1000 ft
3
(28.3 m
3
), 7200 grains of dirt will pass into the air compressor in 1 week’s oper-
ation. With the higher level of 4 grains per 1000 ft
3
(28.3 m
3
), 28,800 grains, or
over 4 lb (1.8 kg) will be carried into the compressor during 1 week’s operation.

Frequently, much of the dirt carried in the air is abrasive. If this dirt is allowed to
get into the compressor cylinders it will mix with the lubricating oil and cause rapid
wear of piston rings, cylinder walls, valves, and other parts.
Intake-air filters, Fig. 13, can reduce much of the danger of abrasive particles
in the supply air. Each type has its favorable features. Viscous coated wire filters,
Fig. 13a, are often used for small- and medium-size compressors. Centrifugal air-
flow units, Fig. 13b, and traveling-curtain oil-bath filters, Fig. 13c, are popular for
larger air compressors. The final choice of an intake filter is a function of com-
pressor capacity, intake-air quality, annual operating hours, and expected life of the
compressor installation. Filter manufacturers can be most helpful to the plant de-
signer in evaluating these factors.
The general procedure given here is valid for air compressors of all types: cen-
trifugal, reciprocating, vane, rotary, etc., and the procedure can be used for air
compressors in plants of all types—chemical, petroleum, manufacturing, marine,
industrial, etc. This procedure has universal application because it is based on the
properties of air, the compressor power input, the annual operating hours, and the
cost of power. These values can be found in any application of air compressors in
industry today.
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9.16 PLANT AND FACILITIES ENGINEERING
(a)
(b)
10-ft (3-m)
FIGURE 12 (a) Outside intake-air filter for air compressor should have intake pipe as
short as possible and be fitted with long-radius elbows (Ingersoll-Rand Co.). (b) Glazed-
tile tunnel for outdoor-air intake.
Compressed-Air and -Gas System

Components and Layouts
POWER INPUT REQUIRED BY CENTRIFUGAL
COMPRESSOR
A centrifugal compressor handling air draws in 12,000 ft
3
/min (339.6 m
3
/min) of
air at a pressure of 14 lb/in
2
(abs) (96.46 kPa) and a temperature of 60ЊF (15.6ЊC).
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS 9.17
(a)
(b) (c)
FIGURE 13 (a) Viscous coated-wire intake-air filter (Air-Maze Corp.). (b) Centrifugal
air-flo w oil-bath intake-air filter also acts as a silencer. (c) Traveling-curtain oil-bath intake-
air filter cleans itself in the oil (American Air Filter Co., Inc.).
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS
9.18 PLANT AND FACILITIES ENGINEERING
The air is delivered from the compressor at a pressure of 70 lb/in
2
(abs) (482.4
kPa) and a temperature of 164

ЊF (73.3ЊC). Suction-pipe flow area is 2.1 ft
2
(0.195
m
2
); area of discharge pipe is 0.4 ft
2
(0.037 m
2
) and the discharge pipe is located
20 ft (6.1 m) above the suction pipe. The weight of the jacket water, which enters
at 60
ЊF (15.6ЊC) and leaves at 110ЊF (43.3ЊC), is 677 lb/min (307.4 kg/min). What
is the horsepower required to drive this compressor, assuming no loss from radia-
tion?
Calculation Procedure:
1. Determine the variables for the compressor horsepower equation
The equation for centrifugal compressor horsepower input is,
22
w (t Ϫ t ) ϩ R
wV
Ϫ VZϪ Z
j 0 ic
2121
hp ϭ c (t Ϫ t ) ϩϩϩ
ͫͬͫͬ
p 21
0.707 50,000 778 0.707
In this equation we have the following variables: w
ϭ weight, lb (kg) of unit flow

rate, ft
3
/s (m
3
/s) through the compressor, lb (kg), where w ϭ (P
1
)(V
1
)/R(T
1
), where
P
1
ϭ inlet pressure, lb/in
2
(abs) (kPa); V
1
ϭ inlet volume flow rate, ft
3
/s (m
3
/s);
R
ϭ gas constant for air ϭ 53.3; T
1
ϭ inlet air temperature, degree Rankine.
The inlet flow rate of 12,000 ft
3
/min ϭ 12,000/60 ϭ 200 ft
3

/s (5.66 m
3
/s); P
1
ϭ 14.0 lb/in
2
(abs) (93.46 kPa); T
1
ϭ 60 ϩ 460 ϭ 520 ЊR. Substituting, w ϭ
14.0(144)(200)/53.3(520) ϭ 14.55 lb (6.6 kg).
The other variables in the equation are: c
p
ϭ specific heat of air at inlet tem-
perature
ϭ 0.24 Btu /lb ЊF (1004.2 J/kg ЊK); t
2
ϭ outlet temperature, R ϭ 624 ЊR;
t
1
ϭ inlet temperature, R ϭ 520ЊR; V
1
ϭ air velocity at compressor entrance, ft /
min (m/min); V
2
ϭ velocity at discharge, fpm (kg /min); Z
1
ϭ elevation of inlet
pipe, ft (m); V
2
ϭ elevation of outlet pipe, ft (m); w

j
ϭ weight of jacket water
flowing through the compressor, lb/min (kg /min); t
i
ϭ jacket-water inlet temper-
ature,
ЊF(ЊC); t
0
ϭ jacket outlet water temperature, ЊF(ЊC).
The air velocity at the compressor entrance
ϭ (flow rate, ft
3
/s)/ (inlet area, ft
2
)
ϭ 200/2.1 ϭ 95.3 ft/s (29 m/s); outlet velocity at the discharge opening ϭ 200/
0.4
ϭ 500 ft/s (152.4 m/s).
2. Compute the input horsepower for the centrifugal compressor
Substituting in the above equation, with radiation losses, R
c
ϭ 0,
22
14.55 500 Ϫ 95.3 20
hp ϭ 0.24(624 Ϫ 520) ϩϩ
ͫͬ
0.707 50,000 778
ϩ [677 /60 ϫ (110 Ϫ 60)]/0.707
ϭ 20.6(24.95 ϩ 4.8 ϩ 0.0256) ϩ 797 ϭ 1,409 hp (1051 kW)
Related Calculations. This equation can be used for any centrifugal compres-

sor. Since the variables are numerous, it is a wise procedure to assemble them
before attempting to solve the equation, as was done here.
COMPRESSOR SELECTION FOR COMPRESSED-
AIR SYSTEMS
Determine the required capacity, discharge pressure, and type of compressor for an
industrial-plant compressed-air system fitted with the tools listed in Table 3. The
plant is located at sea level and operates 16 h/day.
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS 9.19
TABLE 3 Typical Computation of Compressed-Air Requirements
Calculation Procedure:
1. Compute the required airflow rate
List all the tools and devices in the compressed-air system that will consume air,
Table 3. Then obtain from Table 4 the probable air consumption, ft
3
/min, of each
tool. Enter this value in column 1, Table 3. Next list the number of each type of
tool that will be used in the system in column 2. Find the maximum probable air
consumption of each tool by taking the product, line by line, of columns 1 and 2.
Enter the result in column 3, Table 3, for each tool.
The air consumption values shown in column 3 represent the airflow rate re-
quired for continuous operation of each type and number of tools listed. However,
few air tools operate continually. To provide for this situation, a load factor is
generally used when an air compressor is selected.
2. Select the equipment load factor
The equipment load factor
ϭ (actual air consumption of the tool or device, ft

3
/
min)/(full-load continuous air consumption of the tool or device, ft
3
/min). Load
factors for compressed-air operated devices are usually less than 1.0.
Two variables are involved in the equipment load factor. The first is the time
factor, or the percentage of the total time the tool or device actually uses com-
pressed air. The second is the work factor , or percentage of maximum possible
work output done by the tool. The load factor is the product of these two variables.
Determine the load factor for a given tool or device by consulting the manufac-
turer’s engineering data, or by estimating the factor value by using previous ex-
perience as a guide. Enter the load factor in column 4, Table 3. The values shown
represent typical load factors encountered in industrial plants.
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS
9.20 PLANT AND FACILITIES ENGINEERING
TABLE 4 Approximate Air Needs of Pneumatic Tools
3. Compute the actual air consumption
Take the product, line by line, of columns 3 and 4, Table 3. Enter the result, i.e.,
the probable air demand, in column 5, Table 3. Find the sum of the values in column
5, or 601 ft
3
/min. This is the probable air demand of the system.
4. Apply allowances for leakage and future needs
Most compressed-air system designs allow for 10 percent of the required air to be
lost through leaks in the piping, tools, hoses, etc. Whereas some designers claim
that allowing for leakage is a poor design procedure, observation of many instal-

lations indicates that air leakage is a fact of life and must be considered when an
actual system is designed.
With a 10 percent leakage factor, the required air capacity
ϭ 1.1(601) ϭ 661
ft
3
/min (18.7 m
3
/min).
Future requirements are best estimated by predicting what types of tools and
devices will probably be used. Once this is known, prepare a tabulation similar to
Table 3, listing the predicted future tools and devices and their air needs. Assume
that the future air needs, column 5, are 240 ft
3
/min (6.8 m
3
/min). Then the total
required air capacity
ϭ 661 ϩ 240 ϭ 901 ft
3
/min (25.5 m
3
/min), say 900 ft
3
/min
(25.47 m
3
/min) ϭ present requirements ϩ leakage allowance ϩ predicated future
needs, all expressed in ft
3

/min.
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS
AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS 9.21
TABLE 5 Air Compressor Brake Horsepower (kW) Input*
5. Choose the compressor discharge pressure and capacity
In selecting the type of compressor to use, two factors are of key importance:
discharge pressure required and capacity required.
Most air tools and devices are designed to operate at a pressure of 90 lb /in
2
(620 kPa) at the tool inlet. Hence, usual industrial compressors are rated for a
discharge pressure of 100 lb/in
2
(689 kPa), the extra lb /in
2
providing for pressure
loss in the piping between the compressor and the tools. Since none of the tools
used in this plant are specialty items requiring higher than the normal pressure, a
100-lb/in
2
(689-kPa) discharge pressure will be chosen.
Where the future air demands are expected to occur fairly soon—within 2 to 3
years—the general practice is to choose a compressor having the capacity to satisfy
present and future needs. Hence, in this case, a 900-ft
3
/min (25.5-m
3
/min) com-

pressor would be chosen.
6. Compute the power required to compress the air
Table 5 shows the power required to compress air to various discharge pressures at
different altitudes above sea level. Study of this table shows that at sea level a
single-stage compressor requires 22.1 bhp/(100 ft
3
/min) (5.8 kW/m
3
) when the
discharge pressure is 100 lb /in
2
(689 kPa). A two-stage compressor requires 19.1
bhp (14.2 kW) under the same conditions. This is a saving of 3.0 bhp /(100 ft
3
/
min) (0.79 kW /m
3
). Hence, a two-stage compressor would probably be a better
investment because this hp will be saved for the life of the compressor. The usual
life of an air compressor is 20 years. Hence, by using a two-stage compressor, the
approximate required bhp
ϭ (900 /100)(19.1) ϭ 171.9 bhp (128 kW), say 175 bhp
(13.1 kW).
7. Choose the type of compressor to use
Reciprocating compressors find the widest use for stationary plant air supply. They
may be single- or two-stage, air- or water-cooled. Here is a general guide to the
types of reciprocating compressors that are satisfactory for various loads and ser-
vice:
Single-stage air-cooled compressor up to 3 hp (2.2 kW), pressures to 150 lb/in
2

(1034 kPa), for light and intermittent running up to 1 h/day.
Two-stage air-cooled compressor up to 3 hp (2.2 kW), pressures to 150 lb/in
2
(1034 kPa), for 4 to 8 h/day running time.
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS
9.22 PLANT AND FACILITIES ENGINEERING
Single-stage air-cooled compressor up to 15 hp (11.2 kW) for pressures to 80
lb/in
2
(552 kPa); above 80 lb /in
2
(552 kPa), use two-stage air-cooled compres-
sor.
Single-stage horizontal double-acting water-cooled compressor for pressures to
100 lb /in
2
(689 kPa) hps of 10 to 100 (7.5 to 75 kW), for 24 h /day or less
operating time.
Two-stage, single-acting air-cooled compressor for 10 to 100 hp (7.5 to 75 kW),
5 to 10 h/day operation.
Two-stage double-acting water-cooled compressor for 100 hp (75 kW), or more,
24 h/day, or less operating time.
Using this general guide, choose a two-stage double-acting water-cooled recipro-
cating compressor, because more than 100-hp (75-kW) input is required and the
compressor will operate 16 h/day.
Rotary compressors are not as widely used for industrial compressed-air systems
as reciprocating compressors. The reason is that usual rotary compressors discharge

at pressures under 100 lb/in
2
(68.9 kPa), unless they are multistage units.
Centrifugal compressors are generally used for large airflows—several thousand
ft
3
/min or more. Hence, they usually find use for services requiring large air quan-
tities, such as steel-mill blowing, copper conversion, etc. As a general rule, ma-
chines discharging at pressures of 35 lb/in
2
(241 kPa) or less are termed blowers;
machines discharging at pressures greater than 35 lb/in
2
(241 kPa) are termed
compressors.
Using these facts as a guide enables the designer to choose, as before, a two-
stage double-acting water-cooled compressor for this application. Refer to the man-
ufacturer’s engineering data for the compressor dimensions and weight.
8. Select the compressor drive
Air compressors can be driven by electric motors, gasoline engines, diesel engines,
gas turbines, or steam turbines. The most popular drive for reciprocating air com-
pressors is the electric motor—either direct-connected or belt-connected. Where
either dc or ac power supply is available, the usual choice is an electric-motor drive.
However, special circumstances, such as the availability of low-cost fuel, may dic-
tate another choice of drive for economic reasons. Assuming that there are no
special economic reasons for choosing another type of drive, an electric motor
would be chosen for this installation.
With an ac power supply, the squirrel-cage induction motor is generally chosen
for belt-driven compressors. Synchronous motors are also used, particularly when
power-factor correction is desired. Motor-driven air compressors generally operate

at constant speed and are fitted with cylinder unloaders to vary the quantity of air
delivered to the air receiver. A typical power input to a large reciprocating com-
pressor is 22 hp (16.4 kW) per 100 ft
3
/min (2.8 m
3
/min) of free air compressed.
Air compressors are almost always rated in terms of free air capacity, i.e., air
at the compressor intake location. Since the altitude, barometric pressure, and air
temperature may vary at any locality, the term free air does not mean air under
standard or uniform conditions. The displacement of an air compressor is the vol-
ume of air displaced per unit of time, usually stated in ft
3
/min. In a multistage
compressor, the displacement is that of the low-pressure cylinder only.
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS 9.23
FIGURE 14 Central system for compressed-
air supply.
FIGURE 15 Unit system for compressed-air
supply.
9. Choose the type of air distribution system
Two types of air distribution systems are in use in industrial plants: central and
unit. In a central system, Fig. 14, one or more large compressors centrally located
in the plant supply compressed air to the areas needing it. The supply piping often
runs in the form of a loop around the areas needing air.
A unit system, Fig. 15, has smaller compressors located in the areas where air

is used. In the usual plant, each compressor serves only the area in which it is
located. Emergency connections between the various areas may or may not be
installed.
Central systems have been used for many years in large industrial plants and
give excellent service. Unit systems are used in both small and large plants but
probably find more use in smaller plants today. With the large quantity of air
required by this plant, a central system would probably be chosen, unless the air
was needed at widely scattered locations in the plant, leading to excessive pressure
losses in the distribution piping of a central system. In such a situation, a unit
system with the capacity divided between compressors as necessary would be
chosen.
Related Calculations. Where possible, choose a larger compressor than the
calculations indicate is needed, because air use in industrial plants tends to increase.
Avoid choosing a compressor having a free-air capacity less than one-third the
required free-air capacity.
When choosing a water-cooled compressor instead of an air-cooled unit, remem-
ber that water cooling is more expensive than air cooling. However, the power input
to water-cooled compressors is usually less than to air-cooled compressors of the
same capacity. For either type of cooling, a two-stage compressor, with intercooling,
is more economical when the compressor must operate4hormore in a 24-h period.
Table 6 shows the typical cooling-water requirements of various types of water-
cooled compressors.
When the inlet air temperature is above or below 60
ЊF (15.6ЊC), the compressor
delivery will vary. Table 7 shows the relative delivery of compressors handling air
at various inlet temperatures.
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS

9.24 PLANT AND FACILITIES ENGINEERING
TABLE 6 Cooling Water Recommended for Intercoolers, Cylinder Jackets, Aftercoolers
TABLE 7 Effect of Initial Temperature on
Delivery of Air Compressors [Based on a
nominal intake temperature of 60
ЊF
(15.6
ЊC)]
TABLE 8 Air-Consumption Altitude
Factors (100-lb / in
2
or 689.5-kPa air supply)
SIZING COMPRESSED-AIR-SYSTEM
COMPONENTS
What is the minimum capacity air receiver that should be used in a compressed-
air system having a compressor displacing 800 ft
3
/min (0.38 m
3
/s) when the intake
pressure is 14.7 lb/in
2
(abs) (101.4 kPa) and the discharge pressure is 120 lb/ in
2
(abs) (827.4 kPa)? How long will it take for this compressor to pump up a 300-ft
3
(8.5-m
3
) receiver from 80 to 120-lb/in
2

(551.6 to 827.4 kPa) if the average volu-
metric efficiency of the compressor is 68 percent? For how long can an 80-lb/in
2
(abs) (551.6-kPa) tool be operated from a 120-lb/ in
2
(abs) (827.4-kPa), 300-ft
3
(8.5-
m
3
) receiver if the tool uses 10 ft
3
/min (0.005 m
3
/s) of free air and the receiver
pressure is allowed to fall to 85 lb/in
2
(abs) (586.1 kPa) when the atmospheric
pressure is 14.7 lb/in
2
(abs) (101.4 kPa)? What diameter air piston is required to
produce a 1000-lb (4448.2-N) force if the pressure of the air is 150 lb /in
2
(abs)
(1034.3 kPa)?
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AIR AND GAS COMPRESSORS AND VACUUM SYSTEMS 9.25

Calculation Procedure:
1. Compute the required volume of the air receiver
Use the relation V
m
ϭ dp
1
/p
2
, where V
m
ϭ minimum receiver volume needed, ft
3
;
d
ϭ compressor displacement, ft
3
/min (use only the first-stage displacement for
two-stage compressors); p
1
ϭ compressor intake pressure, lb /in
2
(abs); p
2
ϭ com-
pressor discharge pressure, lb/ in
2
(abs). Thus, for this compressor, V
m
ϭ 800(14.7
/120)

ϭ 97 ft
3
(2.7 m
3
). To provide a reserve capacity, a receiver having a volume
of 150 or 200 ft
3
(4.2 or 5.7 m
3
) would probably be chosen.
2. Compute the receiver pump-up time
Use the relation t
ϭ V(p
f
Ϫ p
i
)/(14.7de), where t ϭ receiver pump-up time, min;
p
f
ϭ final pressure, lb/in
2
(abs); p
i
ϭ initial receiver pressure, lb/in
2
(abs); d ϭ
compressor piston displacement, ft
3
/min; e ϭ compressor volumetric efficiency,
percent. Thus, t

ϭ 300(120 Ϫ 80) /[14.7(800)(0.68)] ϭ 1.5 min. When the com-
pressor discharge capacity is given in ft
3
/min of free air instead of in terms of
piston displacement, drop the volumetric efficiency term from the above relation
before computing the pump-up time.
3. Compute the air supply time
Use the relation, t
s
ϭ V(p
max
Ϫ p
min
)/(cp
a
m), where t
s
ϭ time in minutes during
which the receiver of volume V ft
3
will supply air from the receiver maximum
pressure p
max
lb/in
2
(abs) to the minimum pressure p
min
lb/in
2
(abs); c ϭ ft

3
/min
of free air required to operate the tool; p
a
ϭ atmospheric pressure, lb/in
2
(abs). Or,
t
s
ϭ 300(120 Ϫ 85)/[(10)(1.47)] ϭ 7.15 min.
Note that in this relation p
min
is the minimum air pressure to operate the air tool.
A higher minimum tank pressure was chosen here because this provides a safer
estimate of the time duration for the supply of air. Had the tool operating pressure
been chosen instead, the time available, by the same relation, would be t
s
ϭ 81.5
min.
This calculation shows that it is often wise to install an auxiliary receiver at a
distance from the compressor but near the tools drawing large amounts of air. Use
of such an auxiliary receiver, particularly near the end of a long distribution line,
can often eliminate the need for purchasing another air compressor.
4. Compute the required piston diameter
Use the relation A
p
ϭ F /p
m
, where A
p

ϭ required piston area to produce the desired
force, in
2
; F ϭ force produced, lb; p
m
ϭ maximum air pressure available for the
piston, lb/in
2
(abs). Or, A
p
ϭ 1000 /150 ϭ 6.66 in
2
(43.0 cm
2
). The piston diameter
d is d
ϭ 2(A
p
/
␲
)
0.5
ϭ 2.91 in (7.4 cm).
Related Calculations. The air consumption of power tools is normally ex-
pressed in ft
3
/min of free air at sea level; the actual capacity of any type of air
compressor is expressed in the same units. At locations above sea level, the quantity
of free air required to operate an air tool increases because the atmospheric pressure
is lower. To find the air consumption of an air tool at an altitude above sea level

in terms of ft
3
/min of free air at the elevation location, multiply the sea-level
consumption by the appropriate factor from Table 6. Thus, a tool that consumes 10
ft
3
/min (0.005 m
3
/s) of free air at sea level will use 10 (1.310) ϭ 13.1 ft
3
/min
(0.006 m
3
/s) of 100 lb/in
2
(689.5-kPa) free air at an 8000-ft (2438.4-m) altitude.
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