PRIME MOVERS OF ENERGY: PUMPS

6 PRIME MOVERS OF ENERGY

6.1. PUMPS
Pumps are widely used for transfer of liquids from one place to another. Pumps are usually
driven by electric motors, thus some of the considerations about pumps and electric motors might
overlap. For some specific applications, pumps can be driven by compressed air or hydraulically.
There are a lot of different types of pumps used in the industry. Pump selection depends upon
the application. For example, centrifugal pumps are used predominantly for transfer of large volumes,
metering pumps are used for precise delivery of liquids, ensuring constant discharge regardless of back
pressure in the lines, and progressive cavity pumps or peristaltic pumps are used for delivery of very
viscous materials.
It is a norm that a pump manufacturer also provides pump curves at the time of the sale. They
are essential for establishing the operation range and if any changes for pumping systems are considered
the curves have to be looked at.
6.1.1. Operation
Opportunities for savings in pump operation are often overlooked because pump inefficiency is
not readily apparent.
The following measures can improve pump efficiency:
1.
2.
3.
4.
5.
6.
7.
8.

Shut down unnecessary pumps.

Restore internal clearances if performance has changed significantly.
Trim or change impellers if head is larger than necessary.
Control by throttle instead of running wide open or bypassing the flow.
Replace oversized pumps.
Use multiple pumps instead of one large pump.
Use a small booster pump.
Change the speed of the pump for the most efficient match of horsepower requirements with output.
Pumps can run inefficiently because:

1. Present operating conditions differ from the design conditions. This change often occurs after a
plant has undertaken a water conservation program.
2. Oversized pumps were specified and installed to allow for future increases in capacity.
3. Conservative design factors were used to ensure the pump would meet the required conditions.
4. Other design factors were chosen at the expense of pump efficiency when energy costs were lower.

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Pump Survey
A survey of pumps should concentrate on the following conditions associated with inefficient
pump operation. These are discussed in order of decreasing potential for energy savings in existing
installations. For the survey to produce worthwhile savings, only pumps above a certain size, such as
25 horsepower, need be checked.
1.
a)
b)
c)
d)
2.

Excessive pump maintenance. This problem is often associated with:

Oversized pumps that are heavily throttled.
Pumps in cavitation.
Badly worn pumps.
Pumps that are misapplied for the present operation.
Any pump system with large flow or pressure variations. When normal flows or pressures are less
than 75 percent of their maximum, energy is probably being wasted from excessive throttling, large
bypassed flows, or operation of unneeded pumps.
3. Bypassed flow. Bypassed flow, either from a control system or deadhead protection orifices, is
wasted energy.
4. Throttled control valves. The pressure drop across a control valve represents waster energy, which
is proportional to the pressure drop and flow.
5. Fixed throttle operation. Pumps throttle at a constant head and flow indicate excess capacity.
6. Noisy pumps or valves. A noisy pump generally indicates cavitation from heavy throttling or excess
flow. Noisy control valves or bypass valves usually mean a high pressure drop with a
corresponding high energy loss.
7. A multiple pump system. Energy is commonly lost from bypassing excess capacity, running
unneeded pumps, maintaining excess pressure, or having a large flow increment between pumps.
8. Changes from design conditions. Changes in plant operating conditions (expansions, shutdowns,
etc.) can cause pumps that were previously well applied to operate at reduced efficiency.
9. A low-flow, high-pressure user. Such users may require operation of the entire system at higher
pressure.
10. Pumps with known overcapacity. Overcapacity wastes energy because more flow is pumped at a
higher pressure than required.
Once the inefficient pumps have been identified, the potential savings and the cost of
implementing the changes should be analyzed. Comparison of the actual operating point with the pump
performance curve will facilitate the analysis. Actual performance may differ from the original design
because of process changes, faulty basic data, conservative safety margins, or planned expansions never
realized.

Energy Conservation Measures

Energy may be saved in pump operation in a number of ways, including the following techniques
arranged in approximate increasing order of investment cost.

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1. Shut Down Unnecessary Pumps

This obvious but frequently overlooked energy-saving measure can often be carried out after a
significant reduction in the plants water usage. If excess capacity is used because flow requirements
vary, the number of pumps in service can be automatically controlled by installing pressure switches on
one or more pumps.
2. Restore Internal Clearances
This measure should be taken if performance changes significantly. Pump capacity and efficiency are
reduced as internal leakage increases from excessive backplate and impeller clearances and worn throat
bushings, impeller wear rings, sleeve bearings, and impellers.
3. Trim or Change Impellers
If head is excessive, this approach can be used when throttling is not sufficient to permit the complete
shutdown of a pump. Trimming centrifugal pump impellers is the lowest cost method to correct
oversized pumps. Head can be reduced 10 to 50 percent by trimming or changing the pump impeller
diameter within the vendors recommended size limits for the pump casing.
4. Control by Throttling
Controlling a centrifugal pump by throttling the pump discharge wastes energy. Throttle control is,
however, generally less energy wasteful than two other widely used alternatives: no control and bypass
control. Throttles can, therefore, represent a means to save pump energy.
5. Replace Oversized Pumps
Oversized pumps represent the largest single source of wasted pump energy. Their replacement must
be evaluated in relation to other possible methods to reduce capacity, such as trimming or changing
impellers and using variable speed control.
6. Use Multiple Pumps
Multiple pumps offer an alternative to variable speed, bypass, or throttle control. The savings result
because one or more pumps can be shut down at low system flow while the other pumps operate at
high efficiency. Multiple small pumps should be considered when the pumping load is less than half the
maximum single capacity.

7. Use a Small Booster Pump

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The energy requirements of the overall system can be reduced by the use of a booster pump to provide
the high-pressure flow to a selected user and allow the remainder of the system to operate at a lower
pressure and reduced power.
8. Change Pump Speed
Variable-speed drives yield the maximum savings in matching pump output to varying system
requirements. However, variable speed drives generally have a higher investment cost than other
methods of capacity control. Several types of variable-speed drives can be considered:

Variable-speed motors, either variable frequency or DC.

Variable-speed drives such as traction drives, for constant-speed motors.
Two-speed motors when low speed can satisfy the requirements for significant portion of the time.

As an example of the savings from the use of a smaller pump, assume 300 tons of refrigeration
is required during the summer months but only 75 tons for the remaining nine months. One of two 700gpm chilled-water pumps, equipped with 40-horsepower motors, is operated during the winter, with
two thirds of the flow bypassed. A new 250-gpm pump designed for the same discharge head as the
original two units consumes only 10 horsepower. The electric savings from operating the small pump
during the winter is:
Annual Savings

= (40 hp - 10 hp) x 6,000 hrs/yr x 9 mos/12 x $0.041/hp-hr

= $5,540

The installation cost of a new pump is about $5,000.

The following example illustrates the possible savings from trimming an impeller. A double
suction centrifugal pump with a 13.75-inch diameter impeller pumps process water. The demand is
constant, 2,750 gpm, and the pump is controlled by a manual throttle valve. The pump operates at
164-feet head, 2,750 gpm and 135.6-brake horsepower (point A in Figure 6.1). A 16-psig (37-foot)
pressure drop occurs across the partially closed throttle valve, with only a 6-foot drop across the
completely open valve.
If the pump were exactly matched to the system requirements, only 127 feet of head would be
required without the valve. Because even the fully open valve has a 6-foot pressure drop, the minimum
head required becomes 133 feet. To this, a 5 percent allowance should be added as a tolerance for the
accuracy of the field measurements and impeller trimming. The minimum total head required, therefore,
is 140 feet. Based on the pump affinity laws, the trimmed impeller diameter should be 13 inches, as
shown in step 1 below.
With a trimmed 13-inch impeller, the pump will operate slightly throttled at 140-feet head,
2,750 gpm and 115.7-brake horsepower, as shown by point B in Figure 6.1. The trimmed impeller
reduces power consumption by 19.9-brake horsepower and saves $5,440 per year (see steps 2-4).

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Trimming and balancing an impeller usually cost less than $1,000, and payback, therefore, is less than
three months.

FEET

1. Determine the impeller diameter to reduce head from 164 feet to 140 feet while maintaining 2,750gpm flow. Apply the affinity laws and note that both the head and flow are reduced as the impeller
is trimmed.

NPSH

6'
8'

220

50 55

14 1/2

60 65

10'

12'

70
75

200

160
140

18'

22'

83 84
84 83
82
80

13

12 1/4

78

11 1/2

150
HP

120
100
100
HP

80

P
0H
20

TOTAL HEAD

180

13 3/4

14'
78 80 82

125
HP

75
HP

60
40
0

500

1000

Figure 6.1:

1500

2000

2500

3000

3500

4000

GPM

Typical Centrifugal Pump Characteristics

a) H1 / H2 = D12 / D22 and Q1 / Q1 = D1 / D1

H1Q1/ H1Q1 = D13 / D13

b) Holding Q constant = D1 / D2 = (H1 / H2)1/3 = (164 / 140)1/3 = 1.054

c) D1 = (D2 / 1.054) = 13.0 inches
H = head in feet; H1, before the reduction, and H2, after the reduction
D = diameter of the impeller in inches
Q = flow in gpm
E = pump efficiency
2. Point A: Oversized pump (13.75-inch impeller) throttle back to 2,750 gpm.
bhp = (H x Q x (SG)) / (3,960 x E) = (164 x 2,750) / (3,960 x 0.84) = 135.6

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3. Point B: Trimmed impeller (13 inches) throttled back to 2,750 gpm.

bhp = (140 x 2,750) / (3,960 x 0.84) = 115.7
4. Annual Savings = 135.6 - 115.7 = 19.9 bhp
$/yr = 19.9 x (1/0.90) motor eff. x 6,000 hrs/yr x $0.041/hp-hr
= $5,440
As with other equipment, energy conservation for pumps should begin when the pump is
designed. Nevertheless, the savings from modification of an existing system often justify the cost.
The following example illustrates the application of affinity laws for variable frequency drive
pump savings. With fans the affinity laws can be applied directly because the system resistance is purely
flow-related. With pumps or fans having a static head offset, the system resistance curve also changes
with pump speed.
A typical centrifugal pump curve in Figure 6.2 shows that by throttling the 1,750 rpm motor
the pump delivers 2,500 gpm at 236 ft. head. Given a system analysis showing that 150 ft. of head is
required to deliver 2,500 gpm with no throttling, the savings for operating the pump at reduced speed
without throttling can be determined by the following trial-and-error method.
The affinity laws are: S1 / S2 = Q1 / Q2 = (H1 / H2)1/2 = (BHP 1 / BHP2)1/3
where
S1 = original pump speed, rpm
S2 = new pump speed, rpm
Q1 = flow on original pump curve, gpm
Q2 = system flow required, gpm
H1 = head on original pump curve, ft.
H2 = head required by system for Q2, ft.
BHP1 = pump horsepower at Q1 and H1
BHP2 = pump horsepower required for operation at Q2, H2
1. Assume a new pump speed, try 1,500 rpm.
2. Calculate the speed ratio, S1 / S2 = 1,500/1,700 = 0.8571
3. Calculate Q1 from the affinity laws.
Q1 = Q2 / (S1 / S2) = 2,500/0.8571= 2,917 gpm
4. Determine H1 from the original curve at Q1 = 233 ft.
5. Calculate H2 from the affinity laws:
H2 = (S1 / S2)3 x H1 = 0.8571 x 233 = 199.7 ft.
6. Compare H2 from step 5 with the desired H2. Since H2 at 199.7 ft. is greater than the desired H1 at
150 ft., the calculation must be repeated using a lower rpm. Several iterations of this procedure
give:
S1 = 1,405 rpm, Q1 = 3,114 gpm, and H1 = 232.5 ft.

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From Q1 and H1 above a new operating point 1 is determined. The important concept here is that point
1 is not the original system operating point (2,500 gpm, 236 ft.). Rather it is the one and only point on
the original pump curve that satisfies the affinity law equations at the new operating point 2 (2,500 gpm,
150 ft.). It must be determined before BHP 2 can be calculated from the affinity laws.
10" x 18" Radial Flow Pump
15 1/2" Diameter Impeller
1750 RPM
250

A
.

400
BHP
300

200

BHPA

175

200
TDH, ft.

Pump Brake Horsepower

Total Dynamic Head, ft.

225

100

150
Speed2 = ?

1000

2000

3000

4000

5000

6000

Flow, gpm

Figure 6.2:

Centrifugal Pump Curve

7. From the pump curve determine BHP1 for Q1 at 3,114 gpm.

BHP1 = 258
8. Calculate BHP2 using affinity law
BHP2 = BHP1 x (S2 / S1)3
= 258 x (1,405/1,750)3 = 258 x 0.5175
= 133.5 BHP
9. From the pump curve determine the actual BHP (BHPA) for the original operating point at 2,500
gpm. BHPA = 230 BHP
10. Determine reduction in horsepower: BHP savings = 230 - 133.5 = 96.5 BHP
Note the savings are not found from BHP1 - BHP2, but BHP A - BHP2
Manual calculation of savings for variable speed drives will be tedious if they must be
determined for a number of conditions. Computer programs can simplify the task.

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PRIME MOVERS OF ENERGY: PUMPS

Total Head (ft)

Pump Curve, Speed 1

System
Curve
Operating
Points

Pump Curve, Speed 2

Flow (gpm)
Figure 6.3:

Typical Pump and System Curves, Driven by Adjustable Speed Drive

Total Head (ft)

Pump Curve

Operating Points

System Curve
Throttled

System Curve
Open

Flow (gpm)

Figure 6.4:

178

Typical Pump and System Curves for Pump with Throttling Valve

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PRIME MOVERS OF ENERGY: PUMPS

25
Throttling
Method

Input Power (hp)

20

Power
Savings

15

Adjustable
Speed
Method

10

40%

60%

80%

100%

Flow (%)
Figure 6.5:

Pump Power Requirements for Throttling and Adjustable Speed Motors

6.1.2. Considerations for Installation Design

The position of the pump with respect to the reservoir from which the liquid is to be taken is of
utmost importance. If the pump is higher than the tank from which the fluid is being pumped the boiling
of the fluid at local temperature can occur. The formation of the bubbles is called cavitation. The
bubble collapse can happen at the higher pressure region (tips of the impeller), thus causing cavitation
erosion. The efficiency of the pump is very low and the damage of the impeller will follow soon. The
fatigue failure is the problem. In order to avoid cavitation in the pump, the installation has to satisfy a
condition of net positive suction head (NPSH). The manufacturer of the pump supplies the net positive
suction head required and that is the minimum pressure head at the inlet for the type and model of the

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PRIME MOVERS OF ENERGY: FANS

pump which has to be maintained in order to avoid cavitation inside the pump. The net positive suction
head required accounts for pressure drop inside the pump. The pressure head at the inlet has to be
calculated for each installation. The conventional tools for pressure losses in pipes are commonly used
and adequate. Since the occurrence of bubble forming inside the housing of the pump is absolutely
forbidden the backpressure of the system is of the same importance as NPSH. The adequate
backpressure will prevent the formation and can be achieved, if not available by the nature of the
system, by installation of backpressure valve.

OPERATING
SITUATION

HOURS OF
OPERATION

AVERAGE
kW USAGE

Constant Operation at
Full Capacity
Single Speed Fan
Cycling
Two Speed Fan
Cycling
Variable Control at
Constant Speed
Variable Speed
Control

1202.2

P = 16.2
B = 32.4
P = 16.2
B = 32.4
P = 4.3
B = 8.55
P = 2.72
B = 5.44
P = 1.99
B = 3.98

P = 765.3 (*)
B = 852.7
P = 1132 (*)
B = 1146
1202.2
1202.2

PROPELLER
FAN ENERGY
[kWh]
19475.6

BLOWER FAN
ENERGY
[kWh]
38951.2

12397.9

27627.5

4867.6

9798.3

3270

6540

2392.4

4704.8

(*) The propeller fan will operate slightly fewer hours in these modes because of the cross towers
cooling effect with the fan off.
Table 6.1.

Comparative Energy Usage with Various Methods of Control

6.2. FANS

Fans provide the necessary energy input to pump air from one location to another while they
overcome the various resistances created by the equipment and the duct distribution system. Fans have
been classified in a general way as either centrifugal fans or axial-flow fans, according to the direction of
airflow through the impeller. There are a number of subdivisions of each general type. Generally, the
subdivisions consist of different styles of impellers and the strength and arrangement of construction.
Because of the type of impeller dictates fan characteristics, it influences the amount of energy
(horsepower) the fan needs to transport the required volume of air. The centrifugal fan has four basic
types of impellers--airfoil, backward curved, radial, and forward-curved. Table 6.2 gives the nominal
efficiency of the various types of fans at normal operating conditions.

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Type of Fan
Axial Fan
Centrifugal Fans
Airfoil Impeller
Backward-Curved Impeller
Radial Impeller
Forward-Curved Impeller
Table 6.2:

Efficiency
%
85-90
75-80
70-75
60-65
55-60

Nominal Efficiency of Fans at Normal Operating Conditions

Reductions in exhaust airflows are usually obtained by adjustment of dampers in the duct.
Damper control is a simple and low-cost means of controlling airflow, but it adds resistance, which
causes an increase in fan horsepower. Accordingly, if fan output is heavily throttled or dampered, the
savings opportunity of alternate methods of volume control should be investigated.
More efficient methods of volume control are to:
1. Install inlet vane control.
2. Reduce the speed of the fan.
3. Provide variable-speed control.
Figure 6.6 shows the reduction in horsepower realized by reducing fan speed.
Before alternate methods of volume control are considered, the condition of the existing fan and
duct system should be checked. Some factors that can reduce fan efficiency are:
1. Excessive static-pressure losses through poor duct configuration or plugging.
2. Duct leakage from poor joints or flange connections, access doors left open, damage or corrosion,
etc.
3. An improperly installed inlet cone, which inadequately seals the fan inlet area and allows excessive
air recirculation.
4. Excessive fan horsepower caused by poor fan maintenance, such as bad bearings, shaft
misalignment, worn impeller blades, or corroded fan housing.
5. Dirt and dust accumulations on fan blades or housing.
6. Buildup of negative pressure.
Once the existing system operates as efficiently as possible, alternate methods to control flow can be
evaluated.

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PRIME MOVERS OF ENERGY: FANS

100

s
per
am
D
tlet
Ou

60
t
Inle

s
per
Dam
e
Van

Sp
ee
d

40
Va
ria
ble

Percent Horsepower

80

20

0
0

20

40

60

80

100

Percent Volume

Figure 6.6:

Effect of Volume Control on Fan Horsepower

6.2.1. Inlet Vane Control

Inlet vane control is the most commonly used device for automatic control of centrifugal or inline fan output after damper control. Pre-spinning as well as throttling the air prior to its entry into the
wheel reduces output and save power. Fans must be of sufficient size to permit retrofitting; the wheel
diameter should be larger than 20 inches.
6.2.2. Reduced Speed
When fan output can be reduced permanently, an economical method is to change belt sheaves.
A slower-speed motor can also be used if the first approach is not suitable. A two-speed motor is

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another alternative if the fan operates at low volume for a significant portion of the time but full capacity
is still required part-time.
As an example of the savings to be realized from a reduction in fan speed, assume the exhaust
airflow requirements have been reduced 50 percent on a 20-horsepower centrifugal fan. Reducing fan
rpm 50 percent by changing belt sheaves will halve fan output. Figure 6.6 shows a horsepower
comparison of various methods of centrifugal fan control for typical fans. A 50 percent reduction with
an outlet damper requires 80 percent of rated power; with a slower-speed motor, only 25 percent of
rated power is required. (Refer to the variable speed control curve on the figure.) Therefore:
Annual Savings = (20 hp x 80% - 20 hp x 25%) x 6,000 hrs/yr x $0.041/hp-hr
= $2,700
The reduction in fan output will result in operation of the electric motor at less than rated
capacity. If the horsepower required at the reduced flow is less than about one third of rated
horsepower, the potential savings for substitution of a smaller motor should also be investigated.

6.2.3. Variable Speed

If fan output must be varied but operates at reduced capacity much of the time, a variable drive
should be evaluated. (See separate discussion on variable-speed drives.) Automatic variation of fan
speed through fluid or magnetic couplings or variable-speed motors has limited application because of
the high initial cost.

6.3. AIR COMPRESSORS

Air compressors in manufacturing facilities are often large consumers of electricity. There are
two types of air compressors: reciprocating and screw compressors. Reciprocating compressors
operate in manner similar to that of an automobile engine. That is, a piston moves back and forth in a
cylinder to compress the air. Screw compressors work by entraining the air between two rotating
augers. The space between the augers becomes smaller as the air moves toward the outlet, thereby
compressing the air.
6.3.1. General
Screw compressors have fewer moving parts than reciprocating compressors have and are less
prone to maintenance problems. However, especially for older types of screw compressors, screw
compressors tend to use more energy than reciprocating compressors do, particularly if they are
oversized for the load. This is because many screw compressors continue to rotate, whereas
reciprocating compressors require no power during unloaded state.

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This section includes demand-side management measures for increasing outside air usage,
reducing air leakage around valves and fittings in compressor air lines, recovering air compressor
cooling water, recovering air compressor waste heat, pressure reduction, adding screw compressor,
controls, compressor replacement, and adding low-pressure blowers.
6.3.2. Typical Performance Improvements
1. Use Outside Air For Compressor Intakes
If compressor power is measured:
ES = 3 V LFA H WR

LFA = LA LPF LFT + UA UPF UFT

If compressor power is not measured:
ES =

HP FR LF C 2 H WR
EFF
k 1

FR =

Pdo kN

P 1
i
k 1

Pdm kN

P 1
i

LF = FL FTL + FU FTU
WR =

WI WO
WI

WR =

TI TO
TI + 460

OR

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2.

Compressor Controls
ES =

3.

HP C H FR ( LFC LFP)
EFF

Install Small Compressor

P ( FU L TU L + FLL TLL ) P (FU S TU S + FLS TL S )

ES =

H C FR
EffL
EffS

HPS is based on cfm loading during off peak periods

4.

Reduce Compressor Air Pressure

ES =

5.

(1 FR ) HP LF UF C H
EFF

Reduce Compressed Air Leaks

Leak diameter estimated on site

ES = L H C5

k 1

k
P0 N

Pi C2 V f
N C4
1
P

k 1

L=
Ea Em

P
D
NL (Ti + 460) l C1 C2 C d
V=

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C3 T + 460

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6. Install a Lower Pressure Blower

For replacing plant compressed air used in plating tanks for agitation
ES = (PC PB ) H C1

P
k
Pi C 2 V f
N C3 0

k 1
P
i
PC =
Eac Emc

k 1
k N

k 1

k N

0
Pi C2 Q
N C3
1
Pi

k 1

PB =
Eab E mb

7.

Recover Compressor Waste Heat

Flow and temperature of compressor exhaust air measured on site

ES =

Q C p C1 (Te Tr ) HH
EFF

Waste Heat Recovery

Description
For both screw and reciprocating compressors, approximately 60% to 90% of the energy of
compression is available as heat, and only the remaining 10% to 40% is contained in the compressed
air. This waste heat may be used to offset space heating requirements in the facility or to supply heat to
a process. The heat energy recovered from the compressor can be used for space heating during the
heating season. The amount of heat energy that can be recovered is dependent on the size of the
compressor and the use factor. For this measure to be economically viable, the warm air should not
have to be sent very far; that is the compressor should be located near the heat that is to be used.
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Definitions
Use Factor - The fraction of the yearly hours that the compressor is used.
Applicability
Facility Type - Any facility that uses an air compressor and has a use for the waste heat.
Climate - Wherever space heating is required for a significant portion of the year.
Demand-Side Management Strategy - Strategic conservation.
For More Information
Varigas Research, Inc., Compressed Air Systems, A Guidebook on Energy and Cost Savings,
Timonium, MD, 1984.

Compressor Waste Heat Recovery: Costs and Benefits 1

Options
Waste Heat
Recovery

Installed Costs
($)2
2,098

Energy Savings
(MMBtu/yr)
676

Cost Savings
($/yr)3
2,786

Simple Payback
(yr)
0.8

1. Tabulated data were taken from the Industrial Assessment Center (IAC) data base. All values are averages
based on the database data. The implementation rate for this measure was 34%.
2. One example from the IAC data base to further clarify the costs is as follows: The waste heat from a 75 hp
screw compressor was used to heat the plant. The energy savings were 417 MMBtu/yr, the cost savings were
$2,594/yr, and the implementation cost was $1,530 - giving a simple payback of seven months.
The energy cost savings are based on proposed dollar savings as reported to IAC from the center, usually almost
identical to actual savings reported from the facility.

Operating Pressure Reduction

Description
Demand and energy savings can be realized by reducing the air pressure control setting on an air
compressor. In many cases, the air is compressed to a higher pressure than the air-driven process
equipment actually requires. By determining the minimum required pressure, one may find that the
pressure control setting on the compressor can be lowered. This is done by a simple adjustment of the
pressure setting and applies to both screw and reciprocating compressors. The resulting demand and
energy savings depend on the power rating of the compressor, the load factor, the use factor, the
horsepower reduction factor, the current and proposed discharge pressures, the inlet pressure, and the
type of compressor. This measure should only be considered when the operating pressure is greater
than or equal to 10 psi higher than what is required for the equipment (with exception to situations with
extremely long delivery lines or high pressure drops).

Definitions

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Power Rating - The power indicated by the air compressor manufacturer-usually shown on the
nameplate.
Power Reduction Factor - The ratio of the proposed power consumption to the current power
consumption, based on operating pressure.
Inlet Pressure - The air pressure at the air intake to the compressor, usually local atmospheric pressure.
Applicability
Facility Type - Any facility that has an air compressor.
Climate - All.
Demand-Side Management Strategy - Strategic conservation.
For More Information
National Technical Information Service, Compressed Air Systems, A Guidebook on Energy and Cost
Savings, #DOE/CS/40520-T2, March 1984.

Pressure Reduction: Costs and Benefits 1

Options
Pressure
Reductions

Installed Costs
($)2
864

Energy Savings
(MMBtu/yr)
187

Cost savings
($/yr)3
2,730

Simple Payback
(yr)
1.0

1. Tabulated data were taken from the Industrial Assessment Center (IAC) database. All values are averages based
on the data base data. The implementation rate for this measure was 48%.
2. One example from the IAC data base to further clarify the costs is as follows: Reducing the air pressure control
setting on a 75 hp air compressor from 115 psig to 100 psig resulted in energy savings of 22,500 kWh and cost
savings of $1,180/yr. The implementation cost was $270, resulting in a simple payback of three months.
The energy cost savings are based on proposed dollar savings as reported to IAC from the center, usually almost
identical to actual savings reported from the facility.

Elimination of Air Leaks

Description
Air leaks around valves and fittings in compressor air lines may represent a significant energy
cost in manufacturing facilities. Sometimes up to 20% of the work done by the compressor is to make
up for air leaks. The energy loss as a function of hole diameter at an operating pressure of 100 psi is
shown in the following table:

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Hole Diameter

Free Air Wasted

[ft3/yr]
by a Leak of Air
at 100 psi
90 400 000
40 300 000
10 020 000
2 580 000
625 000

[in]
3/8

1/8
1/16
1/32

Energy Wasted
Per Leak
[kWh/h]
29.9
14.2
3.4
0.9
0.2

Source: National Bureau of Standards

Rule of Thumb --=5%-10% of total energy consumption

Table 6.3:

Fuel and Air Losses Due to Compressed Air Leaks

Definitions
Gage Pressure - The system pressure supplied by the compressor.
Absolute Pressure - The sum of the gage pressure and the atmospheric pressure. The gage and the
absolute pressures are used in calculating the amount of air lost due to air leaks.
Applicability
Facility Type - Any facility that has an air compressor.
Climate - All.
Demand-Side Management Strategy - Strategic conservation and peak clipping.
For More Information
American Consulting Engineers Council, Industrial Market and Energy Management Guide, SIC 32
Stone, Clay and Glass Products Industry, Washington, DC, 1987, P. III-30.
Turner, et. al., Energy Management Handbook, John Wiley and Sons, New York, NY, 1982, pp. 424425.

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Leakage Reduction: Costs and Benefits 1

Options
Leakage
Reduction

Installed Costs
($)2
934

Energy Savings
(MMBtu/yr)
230

Cost Savings
($/yr)3
3,540

Simple Payback
(yr)
0.3

1. Tabulated data were taken from the Industrial Assessment Center (IAC) data base. All values are averages
based on the data base data. The implementation rate for this measure was 79%.
2. One example from the IAC data base to further clarify the costs is as follows: Repairing air leaks in a
compressed air system having air compressors of 150 hp, 60 hp and 25 hp-all operating at 110 psig-resulted in energy
savings of 35,750kWh and cost savings of $2,760/yr. The implementation cost was $500.
3. The energy cost savings are based on proposed dollar savings as re ported to IAC from the center, usually
almost identical to actual savings reported from the facility.

Case Study
Estimated Energy Savings = 408.0 MMBtu/yr
Estimated Cost Savings = $5,730/yr
Estimated Implementation Cost = $460
Simple Payback = 1 month

Recommended Action
Leaks in compressed air lines should be repaired on a regular basis.

Background
The cost of compressed air leaks is the energy cost to compress the volume of lost air from
atmospheric pressure to the compressor operating pressure. The amount of lost air depends on the line
pressure, the compressed air temperature at the point of the leak, the air temperature at the compressor
inlet, and the estimated area of the leak. The leak area is based mainly upon sound and feeling the
airflow from the leak. The detailed equations are included at the end of the AR. An alternative method
to determine total losses due to air leaks is to measure the time between compressor cycles when all air
operated equipment is shut off.
The plant utilizes one 75hp compressor that operates 8,520 hrs/yr. Measurements taken during
the site visit showed the compressor to continuously draw 77.7 hp. Approximately 24% of this load is
lost to air leaks in the plant. The majority of the air leaks are due to open, unused lines. There are
several plant locations where pneumatic machinery could be connected to the primary air line, but at the
time of the site visit, no machines were connected. These open lines were typically found on or near Ibeams. The terms I-beam #1, #2, and #3 are used in the tables of this AR to label the leaks. In

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order to allow for correct location of these open lines, a list of the terms and their approximate locations
follow:

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Terms

Description

I-Beam #1

Leak located on I-beam near rotary automatic #2.

I-Beam #2

Leak located on I-beam near catalogue machine.

I-Beam #3

Leak located on hose attached to I-beam near Machine 6700.

Anticipated Savings
Values for all factors affecting the cost of compressed air leaks were determined during the site
visit, and are listed in the following tables. Because of long piping runs to the equipment, the
compressed air temperature is estimated to be the same as room temperature.

Condition of Pneumatic System at Time of Site Visit

Variable
Air temperature at compressor inlet, F
Atmospheric pressure, psia
Compressor operating pressure, psig
Air temperature at the leak, F
Line pressure at the leak, psig
Compressor motor size, hp
Compressor motor efficiency
Compressor type
Number of stages
Compressor operating hours, per year
Electric cost, per MMBtu

92
14.7
115
72
115
75
91.5%
Screw
1
8,520
$14.05

Using these values, the volumetric flow rate, power lost due to leaks, energy lost and cost for
leaks of various sizes were calculated specifically for the conditions at this plant. The results are shown
in the following table.
As the table shows, the cost of compressed air leaks increases exponentially as the size of the
leak increases. As part of a continuing program to find and repair compressed air leaks, the table or
graph can be referenced to estimate the cost of any leaks that might be found.

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Cost of Compressed Air Leaks At This Plant

Hole
Diameter
_______

Flow
Rate
Cfm
_______

Power
Loss
hp
______

Energy
Lost
MMBtu/yr
________

Energy
Cost
per year
________

1/64
1/32
1/16
1/8
3/16
1/4
3/8

0.5
1.8
7.2
29.0
65.2
115.8
260.6

0.1
0.4
1.7
6.9
15.4
27.4
61.7

0.2
8.7
36.9
149.7
334.1
594.4
1,334.8

$31
$122
$518
$2,103
$4,694
$8,351
$18,805

The estimated energy savings and corresponding cost savings for the air leaks found during the
site visit are listed in the table below:
Summary of Savings

Machine

Cardboard Boxes Area

Cardboard Boxes Area
Hand Dye
Straight Knife
Web
I-beam #1
I-beam #2
I-beam #3

TOTALS

Leak
Diameter
In

Power
Loss
hp

Energy
Savings
MMBtu/yr

Cost
Savings
per year

1/16
1/16
1/16
1/8
1/16
1/16
1/16
1/16

1.7
1.7
1.7
6.9
1.7
1.7
1.7
1.7

36.9
36.9
36.9
149.7
36.9
36.9
36.9
36.9

$518
$518
$518
$2,103
$518
$518
$518
$518

18.8

408.0

$5,729

From the table above, the total estimated energy savings from repairing the air leaks are 408.0
MMBtu./yr and the total cost savings are $5,730/yr.

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Implementation Costs
In general, implementation of this AR involves any or all of the following:
1) replacement of couplings and/or hoses;
2) replacement of seals around filters;
3) shutting off air flow during lunch or break periods; and
4) repairing breaks in lines, etc.
Specific repairs and implementation costs for the leaks found during the site visit are given in the table
below.
Implementation Costs
Machine
____________________
Cardboard Boxes Area
Cardboard Boxes Area
Hand Dye
Straight Knife
Web
I-beam #1
I-beam #2
I-beam #3
____________________

Repair Needed
_________________
_
install shut-off valve
install shut-off valve
install shut-off valve
replace coupling
change 0.5 tube
install shut-off valve
install shut-off valve
replace coupling
_________________
_

TOTALS

Parts
_____
_
$50
$50
$50
$2
$9
$50
$50
$2
_____
$263

Labor
_____
_
$25
$25
$25
$25
$25
$25
$25
$25
_____
_
$200

Total Cost
_________
_
$75
$75
$75
$27
$34
$75
$75
$27
_________
_
$463

Assuming that this work can be done by facility maintenance personnel, these leaks can be eliminated
for approximately $460. Thus, the cost savings of $5,730 would pay for the implementation cost of
$460 in about 1 month.

Equations for Air Flow, Power Loss, and Energy Savings

The volumetric flow rate of free air exiting the hole is dependent upon whether the flow is
choked. When the ratio of atmospheric pressure to line pressure is less than 0.5283, the flow is said to
be choked (i.e., traveling at the speed of sound). The ratio of 14.7 psia atmospheric pressure to 129.7
psia line pressure is 0.11. Thus, the flow is choked. The volumetric flow rate of free air, Vf, exiting the
leak under choked flow conditions is calculated as follows:

NL (Ti + 460)
Vf =

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Pl
D 2
C 4 C5 Cd
Pi
4

C 6 Tl + 460

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where
Vf
NL
Ti
Pl
Pi
C4
C5
Cd
D
Cb
Tl

h a g o r e a n

c o n s t a n t ,

3 . 1 4 1 6

= volumetric flow rate of free air, cubic feet per minute

= number of air leaks, no units
= temperature of the air at the compressor inlet, F
= line pressure at leak in question, psia
= inlet (atmospheric) pressure, 14.7 psia
= isentropic sonic volumetric flow constant, 28.37 ft/sec-R0.5
= conversion constant, 60 sec/min
= coefficient of discharge for square edged orifice1, 0.8 (no units)
= leak diameter, inches (estimated from observations)
= conversion constant, 144 in2/ft2
= average line temperature, F

The power loss from leaks is estimated as the power required to compress the volume of air lost from
atmospheric pressure, Pi, to the compressor discharge pressure, P, as follows2:

P
k
Pi C6 V f
N C 7 0
Pi
k 1

L=
Ea Em

k 1

kN

Where
L
k
N
C7
Po
Ea
Ea
Ea
Ea
Ea
Ea
Ea
Ea

= power loss due to air leak, hp

= specific heat ratio of air, 1.4, no units
= number of stages, no units
= conversion constant, 3.03 x 10-5 hp-min/ft-lb
= compressor operating pressure, psia
= air compressor isentropic (adiabatic) efficiency, no units
= 0.88 for single stage reciprocating compressors
= 0.75 for multi-stage reciprocating compressors
= 0.82 for rotary screw compressors
= 0.72 for sliding vane compressors
= 0.80 for single stage centrifugal compressors
= 0.70 for multi-stage centrifugal compressors
= 0.70 for turbo blowers

1. A.H. Shapiro, The Dynamics and Thermodynamics of Compressible Fluid Flow, Vol 1, Ronald Press, NY,
1953, p. 100.
2. Compressed Air and Gas Institute, Compressed Air and Gas Handbook, Fifth Edition, New Jersey, 1989, Chapters
10 and 11.
3. Pneumatic Handbook, 7th ed., Anthony Barber, Trade and Technical Press, 1989, p. 49.

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Ea
Em

= 0.62 for Roots blowers3

= compressor motor efficiency, no units

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The annual energy savings, ES, are estimated as follows:

ES = L H C8
where
H
C8

= annual time during which leaks occurs, h/yr

= conservation factor, 0.002545 MMBtu/hp-h

The annual cost savings, CS, can be calculated as follows:

CS = ES x unit cost of electricity
Quantifying air leaks is relatively simple if the system can be shut down for 10 to 15 minutes and
if there is an operating pressure gage in the system.
It is a good idea to ask plant personnel to shut down their compressors briefly (and close a
valve near the compressor if the compressor begins to relieve the system pressure through and
automatic bleed). It is important to assure that there are no plant processes taking air from the system
at the time of this test--the only thing relieving the pressure should be leaks. If there is not an operating
plant pressure gage in the system, a cheap one and a collection of bayonet fittings should be at hand so
the gage can be attached to the end of one of the plants supply hoses.
One should monitor the pressure decay as a function of time for about a 10 psi drop and then
measure the sizes of the major receivers/accumulators and major air headers. The pressure drop test
never takes more than 15 minutes, and usually less. Measuring the size of major receivers and air lines
is a short job for an experienced student. Small lines (1.5 inch or less) can be ignored and leaving them
out makes the result conservative.
Application of the perfect gas law will yield the leak rate in scfm. Then one can turn to a
reference like the DOE/C/40520-TZ by Varigas Research to get compressor hp required per scfm. It
is possible to correct the 100 psig data there to other pressures.
This is a much better procedure than listening for leaks and quantifying them by ear as to such
things as roar, gush, whisper, etc. because the leak rate is reasonably quantified in a conservative
way. It has the disadvantage of leaving the assessment team clueless about the cost of repair, which
must then be estimated. It is a good practice to listen for the big leaks and to try to see what is causing
them to aid in eliminating costs.

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This procedure, along with a couple of other common projects is covered in two publications:
Darin W. Nutter, Angela J. Britton, and Warren M. Heffington, Five Common Energy Conservation
Projects in Small- and Medium-Sized Industrial Plants, 15th National Industrial Energy Technology
Conference, Houston, TX, March 1993, pp. 112-120.
The same article was rewritten for Chemical Engineering. The reference is:
Conserve Energy to Cut Operating Costs, Chemical Engineering, September 1993, pp. 126-137

Cooling Water Heat Recovery

Description
Air compressors, 100 hp and larger, are often cooled by water from a cooling tower. The
temperature of the water after leaving the cooling coils of the compressor may be sufficiently high that
heat can be extracted from the water and used in a process. For example, boiler feedwater could be
preheated by the water used to cool the compressor. Preheating make-up water displaces boiler fuel
that would ordinarily be used to heat the make-up water.
Definitions
Cooling Coil - Finned tubes on a water-cooled compressor through which water flows and across
which air flows.
Applicability
Facility Type - Any manufacturing facility that has a large, water-cooled air compressor.
Climate - All.
Demand-Side Management Strategy - Strategic conservation.

Waste Water Heat Recovery: Costs and Benefits 1

Options
Waste Water
Heat Recovery

Installed Costs
($)2
16,171

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Energy savings
(MMBtu/yr)
3,306

Cost Savings
($/yr)3
14,676

Simple Payback
(yr)
1.1

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1. Tabulated data were taken from the Industrial Assessment Center (IAC) data base and represent HOT waste
water IN GENERAL, not just cooling water. All values are averages based on the data base data. The
implementation rate for this measure was 41%.
2. One example from the IAC data base to further clarify the costs is as follows: Installing a heat exchanger to
recover heat from waste water to heat incoming city water resulted in energy savings of 145 MMBtu/yr, cost savings
of $777/yr, and an imp lementation cost of $2,600, giving a simple payback of 3.4 years.
3. The energy cost savings are based on proposed dollar savings as reported to IAC from the center, usually almost
identical to actual savings reported from the facility.

Compressor Controls
Description
Screw compressors may consume up to 80% of their rated power output when they are running
at less than full capacity. This is because many screw compressors are controlled by closing a valve; the
inlet throttling valve on a typical throttled-inlet, screw-type compressor is partially closed in response to
a reduced air system demand. The pressure rise across the compression portion of the unit does not
decrease to zero, and thus power is still required by the unit. Accordingly, an older unit will continue to
operate at 80% to 90% and a new unit at 40% to 60% of its full load capacity horsepower. When
several screw-type air compressor are being used, it is more efficient to shut off the units based on
decreasing load than to allow the units to idle, being careful not to exceed the maximum recommended
starts/hour for the compressor. Modular systems that conserve energy by operating several small
compressors that are brought on line as needed instead of operating one large compressor continuously
are often found in retrofit and new installations.

Definitions
None.
Applicability
Facility Type - Any facility that has screw-type air compressors.
Climate - All.
Demand-Side Management Strategy - Strategic conservation.

Screw Compressor Controls: Costs and Benefits 1

Options
Screw
Compressor
Controls

200

Installed Costs
($)2

Energy Savings
(MMBtu/yr)

Cost Savings
($/yr)3

Simple Payback
(yr)

3,463

342

5,074

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1. Tabulated data were taken from the Industrial Assessment Center (IAC) data base. All values are averages
based on the data base data. The implementation rate for this measure was 48%.
2. One example from the IAC data base to further clarify the costs is as follows: Installing controls on a 100 hp
compressor resulted in energy savings of 128,600 kWh and a cost savings of $6,750/yr, at an implementation cost of
$1,500.
3. The energy cost savings are based on proposed dollar savings as reported to IAC from the center, usually
almost identical to actual savings reported from the facility.

Outside Air Usage

Description
The amount of work done by an air compressor is proportional to the temperature of the intake
air. Less energy is needed to compress cool air than to compress warm air. On average, outside air is
cooler than in inside a compressor room. This is often the case even on very hot days. Piping can often
be installed so that cooler outside air can be supplied to the intake on the compressor. This is
particularly simple and cost-effective if the compressor is located adjacent to an exterior wall.
The energy and cost savings are dependent on the size of the compressor, the load factor, and
the number of hours during which the compressor is used. The payback period is nearly always less
than two years. The load factor is fairly constant for compressors that operate only when they are
actually compressing air. Most reciprocating compressors are operated in this manner. When that are
on, they operate with fairly constant power consumption, usually nearly equal to their rated power
consumption; when they are cycled off, the power consumption is zero. Screw compressors are often
operated in a different manner. When loaded (i.e., actually compressing air), they operate near their
rated power, but when compressed air requirements are met, they are not cycled off but continue to
rotate and are unloaded. Older screw compressors may consume as much as 85% of their rated
power during this unloaded state. Therefore, if a screw compressor is to be operated continuously, it
should be matched closely to the compressed air load that it supplies. Often, plant personnel purchase
compressors having several times the required power rating. This may be done for a variety of reasons,
but often in anticipation of expansion of the facility and a commensurate increase in the compressed air
requirements.

Definitions
Rated Load - The power usage indicated by the air compressor manufacturer; usually shown on the
nameplate.
Load Factor - The average fraction of the rated load at which the compressor operates.
Applicability
Facility Type - Any facility that uses compressed air in its operations. The savings increase as the size
of the compressor and the hours of use increase for both types of compressors.
Climate - Any climate in which the average outdoor air temperature is less than the air temperature in
the compressor room.
Demand-Side Management Strategy - Strategic conservation and peak clipping.
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For More Information

Witte, L.C., P.S. Schmidt, D.R. Braun, Industrial Energy Management and Utilization, Hemisphere
Publishing Corp., Washington, DC, 1988, pp.433, 437.
Baumeister, T., L.S. Marks, eds., Standard Handbook for Mechanical Engineers, 7th Edition,
McGraw-Hill Book Co., New York, NY, 1967, pp. 14.42-14.61.

Outside Air Usage: Costs and Benefits 1

Options
Outside Air
Usage

Installed Costs
($)2
593

Energy Savings
(MMBtu/yr)
82

Cost Savings
($/yr)3
1,246

Simple Payback
(yr)
0.5

1. Tabulated data were taken from the Industrial Assessment Center (IAC) data base. All values are averages
based on the data base data. The implementation rate for this measure was 52%.
2. One example from the IAC data base to further clarify the costs is as follows: Supplying outside air to the
intakes of three air compressors (100 hp, 75 hp, and 50 hp) resulted in energy and cost savings of 10,050 kWh and
$490/yr. The implementation cost was $780.
3. The energy cost savings are based on proposed dollar savings as reported to IAC from the center, usually
almost identical to actual savings reported from the facility.

Compressor Replacement
Description
It is often advantageous to install a smaller compressor to more closely match the compressed
air requirements normally met by oversized or large compressors, for processes that have periods of
low compressed air usage. A smaller compressor will reduce energy usage and associated costs
because the smaller compressor will operate at a better efficiency than the larger compressor when air
requirements are low. Generally pre-1975 stationary screw-type compressors, if oversized for the
load, will run unloaded much of the time when the load is low. They are unloaded by closing the inlet
valve and hence are referred to as modulating inlet type compressors. Based on manufacturers data,
these compressors can consume as much as 85% of the full load horsepower when running unloaded.
Some pre- and post-1975 compressor manufacturers have developed systems that close the inlet valve
but also release the oil reservoir pressure and reduce oil flow to the compressor. Other strategies have
also been developed but are not usually found on older (pre-1975) screw-type compressors. The
unloaded horsepower for screw compressors operating with these types of systems typically ranges
from 80% to 90% of the full load horsepower for older compressors and from 40% to 60% for newer

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compressors, depending on the particular design and conditions. In any event, if the compresses air
requirements are reduced during particular periods (such as a third shift), but are not eliminated entirely,
then installing a smaller compressor to provide the air requirements during these periods can be costeffective.

Definitions
None.
Applicability
Facility Type - Any facility that has a screw compressors and in which there are time periods during
which the compressed air requirements are low.
Climate - All.
Demand-Side Management Strategy - Strategic conservation.
Optimum Sized Equipment: Costs and Benefits 1

Options
Compressor Replacement

Installed Costs
($)2
11,826

Energy Savings
(MMBtu/yr)
975

Cost Savings
($/yr)3
9,828

Simple Payback
(yr)
1.2

1. Tabulated data were taken from the Industrial Assessment Center (IAC) data base. All values are averages
based on the data base data. The implementation rate for this measure was 39%.
2. One example from the IAC data base to further clarify the costs is as follows: A manufacturer of computer
peripheral equipment replaced a 200 hp air compressor with a 75 hp air compressor. The energy savings were $61,850
kWh and the cost
savings were $2,725. The implementation costs were $4,000.
3. The energy cost savings are based on proposed dollar savings as reported to IAC from the center, usually
almost identical to actual savings reported from the facility.

Low-Pressure Blowers
Description
Compressed air is sometimes used to provide agitation of liquids, to control vibration units for
material handling (as air lances), and for other low-pressure pneumatic mechanisms. For such
purposes, it is more efficient to use a blower to provide the required low-pressure air stream. Use of
low-pressure air from the blower would reduce energy consumption by eliminating the practice of
compressing air and then expanding it back to low pressure for use.

Definitions
Plating Tanks - Tanks containing chemicals used in plating operations, such as chrome plating.
Applicability
Facility Type - Any facility having plating tanks.
Climate - All.

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Demand-Side Management Strategy - Strategic conservation and peak clipping.

Reduce Compressed Air Usage: Costs and Benefits 1

Options
Low-Pressure Blowers

Installed Costs
($)2
3,023

Energy Savings
(MMBtu/yr)
404

Cost Savings
($/yr)3
5,677

Simple Payback
(yr)
0.5

1. Tabulated data were taken from the Industrial Assessment Center (IAC) data base. All values are averages
based on the data base data. The implementation rate for this measure was 54%.
2. One example from the IAC data base to further clarify the costs is as follows: A plating facility added a low
pressure blower. The energy savings were $41,000 kWh/yr and the cost savings were $3,200/yr. The implementation
cost was
$5,000.
3. The energy cost savings are based on proposed dollar savings as reported to IAC from the center, usually
almost identical to actual savings reported from the facility.

Case Study
Estimated Energy Savings = 428.7 MMBtu/yr
Estimated Cost Savings = $5,720/yr
Estimated Implementation Cost = $8,500
Simple Payback = 18 months
Recommended Action
A low pressure blower should be installed to provide agitation air for 3 plating tanks. Use of
low pressure air from a blower, as compared to use of compressed air, would reduce electrical
consumption by eliminating the current practice of compressing air and the expanding it back to the
lower pressure.

Background
A 100 hp compressor is currently in use at this facility, and a significant amount of the power
consumed by the compressor (31%) is used to provide air to agitate 3 plating tanks. This compressor
produces compressed air at 117 psig, but less pressure is actually needed to provide effective agitation.
The pressure and flow rate requirements for effective agitation are calculated from the following
equations:

Q = AF F
and
Pa = (0.43 SD SG ) + 0.75

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where
Q
AF
A
Pa
SD
SG

= flow rate required for agitation, cfm

= agitation factor
= surface area of agitation tanks, 63.5 sq. ft.
= pressure required for agitation, psig
= depth of solution, 3 ft.
= specific gravity of water, 1.0

For agitation tanks containing water, the agitation factor is 1.0 cfm/sq. ft.1 The effective surface area of
the tanks is 63.5 sq. ft. Thus, the flow rate required for agitation is calculated as follows:
Q = 1..0 63.5 = 63.5cfm

The pressure required for effective tank agitation is calculated as follows:

1.

Serfilco '91-'92 Catalog U p. 118

P = 0.43 3.0 1.0 + 0.75 = 2 psig

Because the difference between the pressure delivered by the compressor and the pressure required for
effective tank agitation, the compressor is doing a large amount of unnecessary work. By implementing
a blower that has a pressure output more closely matched to the agitation requirement, significant energy
savings can be realized.
Anticipated Savings
Energy savings due to use of air at reduced pressure, ES, are estimated as follows1:

ES = (PC PB) H C1
where
PC
PB
H
C1

= power consumed by compressor to agitate tank, hp

= power consumed by blower to agitate tank, hp
= operating hours, 5,746 h/yr
= conversion factor, 0.756 kW/hp

The volume of free air used for agitation Vf at this plant as obtained from the plant personnel is 130 cfm.
The power PC that is required to compress the volume of free air Vf needed for agitation from
atmospheric pressure Pi to the compressor discharge pressure Po can be calculated as follows2:

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P
k
Pi C 2 V f
N C3 0
k 1
P
i

PC =
E ac E mc

k 1

kN

where
P
C2
Vf
k

= inlet (atmospheric pressure), 14.7 psia

= conversion constant, 144 in2/ft2
= volumetric flow rate of free air, 130 cfm
= specific heat ration of air, 1.4 (no units)

_________________________________________________
1.

Compressed Air and Gas Handbook, 1961.

2. Compressed Air and Gas Handbook, Fifth Edition, Compressed Air and Gas Institute, New Jersey, 1989, Chapters
10 and 11.

N
C3
Po
Eac
Eac
Eac
Eac
Eac
Eac
Eac
Emc

= number of stages, 1 stage

= conversion constant, 3.03 x 10-5 hp-min/ft-lb
= pressure at the compressor outlet, 131.7 psia (117 psig)
= air compressor isentropic (adiabatic) efficiency, 82%
= 0.88 for single stage reciprocating compressors
= 0.75 for multi-stage reciprocating compressors
= 0.82 for rotary screw compressors
= 0.72 for sliding vane compressors
= 0.80 for single stage centrifugal compressors
= 0.70 for multi-stage centrifugal compressors
= compressor motor efficiency, 92% for a 100 hp motor

Thus, the power that is currently consumed by the compressor to provide air for tank agitation is
calculated as follows:
0 .4

1. 41
1 .4
131
.
7

14.7 144 130

1 3.03 10

14.7

0 .4

PC =
0.82 0.92

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Similarly the power required by the blower to provide the same amount of air for agitation, PB, can be
calculated as follows:

P
k
Pi C 2 Q
N C3 b
k 1
P
i

PB =
E ab E mb

k 1
k

where
Pb
Eab
Eab
Eab
Emb

= pressure at the blower outlet, 17.7 psia (3 psig). This value accounts for Pa plus
losses in the air lines.
= blower isentropic (adiabatic) efficiency, 60%
= 0.70 for turbo blowers
= 0.62 for Roots blowers1
= compressor motor efficiency, 92% for a 100 hp motor

Thus, the power that would be consumed by the blower to provide air for tank agitation is estimated as
follows:
_______________________________________
1. Anthony Barber, Pneumatic Handbook, 7th ed., Trade and Technical Press, 1989, p.49

0 .4

1 .4
1 .4
17
.
7

14.7 144 63.5

1 3.03 10
1
14.7

0 .4

PB =
0.60 0.80

For this facility, the energy savings, ES, that can be realized by installing a blower to provide agitation air
for the three tanks are estimated as follows:
ES = (33.7 1.6) 5746 0.746 = 137,597

kWh
MMBtu
= 469.7
yr
yr

The annual cost savings, CS, can be estimated as follows:

CS = ES X unit cost of electricity

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PRIME MOVERS OF ENERGY: AIR COMPRESSORS

MMBtu $13.34

CS = 469.7
= $6,265 / yr
yr MMBtu

Implementation Cost
Implementation of this AR involves purchase and installation of a low pressure blower and
corresponding controls. The purchase price for a blower that will provide a 3 psig air at a flow of 63.5
cfm, including controls, is estimated as $7,500. The installation cost is estimated as $1,000, including
modifications to tanks described below, giving a total implementation cost of $8,500. Thus, the cost
savings of $5,720/yr would have a simple payback of about 18 months.
In order for a 3 psig blower to deliver 63.5 cfm of air, the size of the air outlets in the tanks may
have to be modified. Assuming that there are 12 total outlets (4 outlets per tank), the required outlet
diameter is calculated from the equation for unchoked flow (less than the speed of sound) as follows:

D=

4 Q Tl + 460
P
NL C5 C 6 C 7 C db (Ti + 460 ) l
Pi

2 (k l )
k

P
l
Pi

(k 1)
k

where
T
NL
C5
C6
C7
Cbd
Ti
Pl

= average line temperature, F

= number of outlets used for agitation, 12
= conversion constant, 60 sec/min
= conversion constant, 1/144 in2/ft2
= isentropic subsonic volumetric flow constant, 109.61 ft/sec-R0.5
= coefficient of discharge for subsonic flow through a square edged orifice, 0.6
= temperature of the air at the compressor inlet, 101F
= line pressure at the agitation tanks, 17.7 psia

Thus, the required diameter of the air outlets is calculated as follows:

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4 63.5 75 + 460

D=

1
17.7
12 60
109.61 0.6 (101 + 460 )

144
14.7

2 0 .4
1. 4

0 .4

17.7 1. 4

14.7

D = 0.20 inches
Therefore, if the current diameter of the air outlets is not equal to 0.20 inches, the outlets should be
enlarged.

6.3.3. General Notes on Air Compressors

1. Screw units use 40-100% of rated power unloaded.
2. Reciprocating units are more efficient, more expensive.
3. About 90% of energy consumption becomes heat (10%).
4. RULE OF THUMB: Roughly 20 hp per 100 cfm @ 100 psi.
5. Synchronous belts generally are not appropriate (cooling fins, pulley size).
6. Use low pressure blowers vs. compressed air whenever possible (agitation, heat guns, pneumatic
transfer, etc.).
7. Cost of sir leaks surprisingly high.
8. Second, third, weekend shifts may have low air needs that could be served by smaller compressor.
9. Outside air is cooler, denser, easier to compress than warm inside air.
10. Friction can be reduced by using synthetic lubricants.
11. Older compressors are driven by older, less efficient motors.
12. Compressors may be cooled with chilled water or have reduced condenser capacity.

REFERENCES
1.

Compressed Gas Association, Handbook of Compressed Gases, Reinhold, 1966

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2.
White, F.G., Industrial Air Compressors, Foulis, 1967
3.
Janna, W.S., Introduction to Fluid Mechanics, PWS Publishing Company, 1993
4.
Wolanski, W., Negoshian, J., and Henke, R., Fundamentals of Fluid Power, Houghton
Miffin, 1977
5.
Anderson B., The Analysis and Design of Pneumatic Systems, John Wiley and Sons, 1967
6.
Fluid Power Handbook and Directory, Hydraulics And Pneumatics, 1994
7.
Marks Standard Handbook for Mechanical Engineers, McGraw-Hill, 1987
8.
Vacuum and Pressure Systems Handbook, Gast Manufacturing Corporation, 1986

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