LE0393finalreport

LE0393 – Page ii
1INDUSTRIAL ASSESSMENT REPORT
Prepared by
Industrial Assessment Center
Department of Mechanical Engineering and Mechanics
Lehigh University
Bethlehem, Pennsylvania 18015
A PROGRAM SPONSORED BY THE
U. S. DEPARTMENT OF ENERGY
ADVANCED MANUFACTURING OFFICE
(AMO)
Report Number: LE0393
Assessment Date: February 6, 2015
Plant Location: Bristol, PA 19007
Plant Principal Product: Garage doors and track
SIC Code: 3442
NAICS Code: 332321
Report Date: April 4, 2016
ASSESSMENT PARTICIPANTS
Dr. Alparslan Oztekin - Lead Assessor
Nipun Goel - Lead Student
Albert Woody– Safety Student
Hang Chen
Ruolei Han
Ben Latson
Man Gao
Kaiyuan Peng
CONTRIBUTORS TO LEHIGH UNIVERSITY

LE0393 – Page iii
INDUSTRIAL ASSESSMENT CENTER REPORT LE0393
Dr. Alparslan Oztekin
Nipun Goel
Laura Solomon
Hang Chen
Ruolei Han
Ben Latson
Albert Woody
Benjamin Latson
Brett Barber
Jeffrey Kasle
Marcus Khoury
"Mohammed Alrehili
Sung Rok Choo
Man Gao
Kaiyuan Peng
LEHIGH UNIVERSITY INDUSTRIAL
ASSESSMENT CENTER PERSONNEL
Dr. Alparslan Oztekin
Dr. Sudhakar Neti
Nipun Goel
Laura Solomon
Hang Chen
Ruolei Han
Ben Latson
Albert Woody
Benjamin Latson
Brett Barber
Jeffrey Kasle
Marcus Khoury
"Mohammed Alrehili
Sung Rok Choo
Man Gao
Kaiyuan Peng

LE0393 – Page iv
Preface
The work described in this report was performed under the direction of the Industrial
Assessment Center (IAC) at Lehigh University. The IAC program is managed by Rutgers, The
State University of New Jersey, under agreement with the U. S. Department of Energy through
the Advanced Manufacturing Office in the office of Energy Efficiency and Renewable Energy,
which financially supports the program.
The objective of the IAC is to identify, evaluate, and recommend – through analyses of
industrial plant operations – opportunities to conserve energy, minimize waste, and reduce the
overall cost of operations. Our recommendations are based upon observations and
measurements made at your plant. As our time was limited, we do not claim to have complete
detail on every aspect of the plant's operations. At all times, we try to offer specific and
quantitative recommendations of cost savings, energy conservation, and waste minimization of
the plants we serve. However, we do not attempt to prepare engineering designs or otherwise
perform services that you would expect from an engineering firm, a vendor, or a manufacturer's
representative. When the need for that kind of assistance arrives, we urge you to consult them
directly. If, however, you would like to discuss the contents of this report, or if you have other
questions about energy use and/or waste minimization, please feel welcome to contact us at the
IAC.
Disclaimer
The contents of this report are offered as guidance. The U. S. Department of Energy,
Rutgers, The State University of New Jersey, Lehigh University, and all technical sources
referenced in this report do not: (a) make any warranty or representation, expressed or implied,
with respect to the accuracy, completeness, or usefulness of the information contained in this
report, or that the use of any information, apparatus, method, or process disclosed in this report
may not infringe on privately owned rights; (b) assume any liabilities with respect to the use of,
or for damages resulting from the use of, any information, apparatus, method or process
disclosed in this report. This report does not reflect official views or policy of the above-
mentioned institutions. Mention of trade names or commercial products does not constitute
endorsement or recommendation of use.

LE0393 – Page v
Executive Summary
An assessment was performed at your plant on February 6, 2015. Energy consumption
for a twelve-month period (December 2013 to November 2014) of electricity and gas bills
consisted of the following:
Energy Quantity Total Cost
Electricity 986,096kWh or 10,199MMBtu $ 116,110
Natural Gas 7,152 MMBtu $ 23,079
Total 17,351 MMBtu $139,189
A summary of assessment recommendations (ARs) described in this report is contained
in Table 1. If all of the recommendations shown in Table 1 were implemented, the total annual
cost savings would be $30,301. The total implementation cost for these recommendations is
$63,587, with an average payback period of about 2.1 years. This would save approximately
3,767 MMBtu/yr.
Total Annual Savings Total Implementation Cost Average Payback Period
$30,301 $63,587 2.0 Years
All of the assessment recommendations are described in detail in this report. The annual
cost savings and implementation costs represent our best estimates. You may want to consult
other sources to verify these estimates before a final decision for implementation of these
recommendations is made. We welcome inquires (610-758-5742)and further discussion on any
information or data contained in this report. As previously discussed, we will contact you in the
near future regarding implementation of the assessment recommendations.
Assessment Recommendations (ARs)
The assessment recommendations, based on the plant visit with associated possible
annual cost savings or profit generation of $30,301, are given in Table 1.
For additional information, please contact the Lehigh University Industrial Assessment
Center [phone: 610-758-5742; email: inluiac@lehigh.edu].

LE0393 – Page vi
ARC No. Description
Type
of AR
Annual
Savings
Total
Annual
Cost
Savings
($)
Implementation
Cost
($)
Pay
Back
Period
(yrs)
AR 1
Reduce Compressed
Line Pressure from 120
psi to 95 psi
Electric
Demand
13,793 kWh/yr
(142 MMBtu/yr)
80 kW/yr
$1,443 $25 0.1
AR 2
2.4236.2
Install Setback
Thermostats
Natural
Gas
1,313 MMBtu/yr $4,240 $1,125 0.3
AR 3
2.4236.2
Repair Leaks in
Compressed Air System
Electricity
Demand
26,987 kWh/yr
( 279 MMBtu/yr)
156 kW/yr
$2,823 $1,000 0.5
AR 4
2.7121.3
Clean Skylights Electricity
1,712 kWh/yr
(18 MMBtu/yr)
$129 $120 0.9
AR 5
2.4221.2
Draw Compressor
Intake Air From
Outside
Electricity
Demand
1,597 kWh/yr
( 16 MMBtu/yr)
12 kW/yr
$183 $365 2.0
AR 6
2.4221.2
Install a demand
monitor
Demand 242 kW/year $1,198 $3,052 2.5
AR 7
2.7135.3
Replace MH Fixtures
with T-5 Fixtures with
Dimmers
Electricity
Demand
147,230kWh/yr
(1,522
MMBtu/yr)
774 kW/yr
$15,376 $40,880 2.7
AR 8
2.7261.3
Replace T-12 Lamps
with T-5 Lamps
Electricity
Demand
11,748kWh/yr
(121 MMBtu/yr)
68 kW/yr
$1,275 $4,018 3.2
AR 9
2.4144.1
Install Variable
Frequency Drive on
2x30 HP Compressor
Electric
Demand
31,155 kWh/yr
( 332 MMBtu/yr)
180 kW/yr
$3,259 $11,400 3.5
AR 10
2.4236.2
Install Air Induced
Nozzles on Air Guns
Electricity
Demand
372 kWh/yr
(4 MMBtu/yr)
57 kW/yr
$310 $1,313 4.2
AR 11
2.4221.2
Use Comp exhaust for
heating Warehouse in
winter
Natural
Gas
20 MMBtu/yr $65 $289 4.4
Total 3767 MMBtu/yr $30,301 $63,587 2.1
Table 1: Summary of Assessment Recommendations (ARs).

LE0393 – Page vii
ARC No. Description
Type
of AR
Annual
Savings
Total
Annual
Cost
Savings ($)
Implementation
Cost
($)
Pay
Back
Period
(yrs)
AAR 1
2.7447.3
Install Weather Seals
on Dock Doors
Natural
Gas
649 MMBtu/yr $2,096 $10,800 5.2
AAR 2
2.7442.3
Install automatic
sideways door in the
docking area
Natural
Gas
93 MMBtu/yr $302 $2,960 9.8
AAR 3
2.9112.2
Install Array of Solar
Panels on Roof
Electricity
2,101,209 kWh/yr
(21731
MMBtu/yr)
265,814 $2,962,960 11.5
AAR 4
2.4236.2
Redesign Compressed
Air Line in Plant Area
Electricity
Demand
4,655 kwh/yr
(48 MMBtu/yr)
27 kW/yr
$487 $15,000 30.8
Total
22,521 MMBtu
/yr
$268,699 $2,991,720 11.2
Table 2: Summary of Additional Assessment Recommendations (AARs).

LE0393 – Page viii
Table of Contents
Page
Preface ................................................................................................................................iv
Disclaimer...........................................................................................................................iv
Executive Summary............................................................................................................ v
Table of Contents .............................................................................................................viii
List of Tables...................................................................................................................... x
List of Figures.....................................................................................................................xi
Introduction ........................................................................................................................1
Assessment Recommendations (ARs)................................................................................2
Additional Assessment Recommendations (AARs)...................................................................7
AR 1 - Reduce Compressed Line Pressure from 120 psi to 95 psi ..................................10
AR 2 - Install Setback Thermostats..................................................................................13
AR 3 - Repair Leaks in Compressed Air System.............................................................16
AR 4 - Clean Skylights.....................................................................................................21
AR 5 - Draw Compressor Intake Air From Outside.........................................................23
AR 6 - Install a demand monitor ......................................................................................26
AR 7 - Replace MH Fixtures with T-5 Fixtures with Dimmers.......................................32
AR 8 - Replace T-12 Lamps with T-5 Lamps..................................................................36
AR 9 - Install Variable Frequency Drive on 2x30 HP Compressor.................................40
AAR 1 - Install Weather Seals on Dock Doors................................................................53
AAR 2 - Install automatic sideways door in the docking area .........................................57
Summary of Estimated Savings and Implementation Costs.....................................................57
Current Practice and Observations.....................................................................................57
Anticipated Savings...........................................................................................................57
Current Practice and Observations .......................................................................................64
Summary of Plant Statistics..............................................................................................68
Plant Background .............................................................................................................69
General Information .........................................................................................................69

LE0393 – Page ix
Advanced Manufacturing Office Tools and Guidelines...........................................................71
Energy Resources and Management ........................................................................................76

LE0393 – Page x
List of Tables
Page
Table 1: Summary of Assessment Recommendations (ARs).....................................................vi
Table 2: Summary of Additional Assessment Recommendations (AARs)...............................vii
Table 3: Compressed Air System Parameters. .........................................................................17
Table 4: Cost of Compressed Air Leaks...................................................................................17
Table 5: Summary of Savings for Compressed Air Leaks. ......................................................18
Table 6: 24 hour Average Monthly Temperature for Bucks County, PA ................................24
Table 7: Potential Demand Savings. ........................................................................................30
Table 8: Power Consumption of Motor with Flow Rate. .........................................................41
Table 9: Compressed Air System Parameters. .........................................................................45
Table 10: Cost of Regular Nozzles...........................................................................................45
Table 11: Summary of Savings. ..............................................................................................46
Table 12 : Location and Dimension of the Door Gaps in the Building....................................54
Table 13: Total Energy Consumption. .....................................................................................68
Table 14: Electricity Usage by Month (kWh): December 2013 to November 2014................80
Table 15: Natural Gas Usage by Month (MMBtu): December 2013 to November 2014. .......81

LE0393 – Page xi
List of Figures
Page
1INDUSTRIAL ASSESSMENT REPORT ................................................................................ii
Figure 2: EXAIR Air Induced Nozzle. .....................................................................................44
Figure 3 : Annual Heat Lost from Doors*................................................................................55
Figure 4 : a satellite image of the facility top view ..................................................................61
Figure 5: Compressed Air Loop Sample. .................................................................................66
Figure 6: Process Flow Diagram. .............................................................................................69
Figure 7: Plant Layout..............................................................................................................70
Figure 8: Electricity Usage by Month (kWh): December 2013 to November 2014. ...............82
Figure 9: Electricity Cost by Month ($):December 2013 to November 2014..........................83
Figure 10: Peak Demand by Month (kW): December 2013 to November 2014......................84
Figure 11: Demand Cost by Month ($):December 2013 to November 2014. ..........................85
Figure 12: Gas Usage by Month ($):December 2013 to November 2014. ..............................86
Figure 13: Gas Cost by Month ($):December 2013 to November 2014. .................................87
Figure 14: Annual Energy Usage Pie Chart. ............................................................................88
Figure 15: Annual Energy Cost Pie Chart. ...............................................................................89
Figure 16: Total Energy Cost vs. Billing Month. .....................................................................90

LE0393 – Page 1
Introduction
An assessment was performed at your plant on February 6, 2015. Energy consumption
for a twelve-month period of electricity bills (December 2013 to November 2014) and natural
gas bills (December 2013 to November 2014) consisted of the following:
 Electricity: 986,096 kWh or 10,199 MMBtu
 Natural Gas: 7,152 MMBtu
The total equivalent energy consumption is 17,351 MMBtu, with a total cost of
$139,189. The energy costs for the plant, and those used for the energy savings, are as follows.
 Electricity
o Average Electricity Rate $0.076/kWh
o Average Demand Rate $4.96/kW
 Natural Gas $3.23/MMBtu
Energy bills for the plant for the twelve-month periods above are listed in tabular form
at the end of this report, followed by graphical representations of the energy usage and cost.
A summary of assessment recommendations (ARs) described in this report is contained
in Table 1, and a summary of suggested recommendations can be found in Table 2.
Additionally, a brief description of each recommendation is given below. If all of the
recommendations shown in Table 1 were implemented, the total annual cost savings would
be $30,301. The total implementation cost for these recommendations is $63,587, with an
average payback period of about 2.1 years.This would save approximately 3,767 MMBtu/yr.
All of the assessment recommendations and suggested recommendations are described
in detail in this report. The annual cost savings and implementation costs represent our best
estimates. You may want to consult other sources to verify these estimates before a final
decision for implementation of these ARs is made. We welcome inquires (610-758-5742) and
further discussion on any information or data contained in this report. As previously discussed,
we will contact you in the near future regarding the implementation of our recommendations.

LE0393 – Page 2
Assessment Recommendations (ARs)
The assessment recommendations, based on the plant visit with associated possible
annual cost savings or profit generation of $30,301 are given below.
AR 1: Reduce Compressed Line Pressure from 120 psi to 95 psi
A common source of energy waste is compressing air to a higher pressure than required by air-
driven equipment, as is the case with this company. The minimum required pressure for the
building is determined to be 95 psi, not 120 psi. The compressors’ settings should be reduced
to 95 psi for both 30 HP compressors. The annual electricity savings and demand savings for
this AR is 13793 kWh, and 80 kW, respectively. The annual cost savings is $1443 and, with
an implementation cost of about $25, the payback period will be 7 or 8 days or .24 months.
AR 2: Install Setback Thermostats
Comfortable temperatures are maintained within the office 24 hours per day all year. The office
and administrative areas only operate 8 hours per day, 5 days per week, and 52 weeks per year.
Much energy is wasted maintaining the set temperatures during unoccupied hours. Annual
natural gas savings for this AR is 1,313 MMBtu/yr. The estimated annual cost savings for this
AR is $4,240 and, with an implementation cost of about $1,125, the payback period will be 3
months.
AR 3: Repair Leaks in Compressed Air System
A perfectly sealed compressed air system can never be fully achieved; therefore, it is
appropriate to assume that there are leaks in the compressed air lines in any plant. There is a
total of 135 HP (1 × 75 HP, 2 × 30 HP) worth of compressor power at this plant, with difference
in piping schemes, therefore, a conservative estimate for the amount and size of leaks has been
given in the calculations below. 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 estimation is based on the velocity of the air from the leak through a restricted air

LE0393 – Page 3
anemometer with negligible compressibility. The annual electricity savings for this AR is
26,987 kWh, and the annual demand savings is 156 kW. The annual cost savings is $2,823 and,
with an implementation cost of $1,000, the payback period will be approximately 4 months.
AR 4: Clean Skylights
It was observed that the 6 skylights in the Spring and Spot Welding production area are
either dirty or beginning to yellow. As a result less light was being allowed through these
building features and additional lighting was required inside of the building to make up for the
loss during daylight hours. With the current Metal Halide lamps being used in the area the
realized savings will not be as great as with T-5 lamps equipped with dimmers. The annual
energy savings for this AR is 1,712 kWh. The estimated annual cost savings is likely to be $129
and, with $120 in implementation costs, the payback period will be about 0.93 years.
AR 5: Draw Compressor Intake Air From Outside
Currently, this plant has 2 air compressors, one each of 30 HP. The compressors are
located in a common isolated room and have their intake from within this room. The annual
electricity savings for this AR is 1597 kWh, and the annual demand savings is 12.28 kW. The
cost savings is likely to be $183 /yr and, with an implementation cost of $365, the payback
period is about 2 years.
AR 6: Install a demand monitor
Electric demand is the average rate at which electric energy (measured in kWh) is used
during a specific metered period. This period is called the demand interval and is fifteen minutes
in length at this plant. Demand is defined as the average energy used during the interval, as
follows:
hperiod,demandofLength
intervaldemandduringusedkWh
=kWDemand,
h1/4
kWh
 .

LE0393 – Page 4
The peak demand is the highest average load (measured in kW) reached over all of the
demand intervals within a given billing period. A high demand charge, therefore, results from a
large usage of power during any demand interval of the billing period.
High demand charges can result from a high rate of energy usage for short periods during
production hours. This problem may have one or more solutions. Plant production schedules and
the economics of each situation should be considered when assessing the options. One possible
solution may be to distribute the facility's electrical usage over alternate shifts. Thus, if high peak
demand occurs during one shift, while several equipment stations with heavy electrical usage are
used, it may be possible to move usage of one or more stations to a shift with less peak usage,
resulting in a lower overall peak usage.
An alternate solution is to interlock specific pieces of equipment, thereby preventing them
from consuming power at their peak rates at the same time. This is not always feasible when the
natural operating interval of the equipment is much shorter than the demand interval, or when the
machines must be in continuous operation to maximize production. Controlling electrical
resistance heaters and other heating and ventilating equipment during periods when process
requirements are peaking is another demand control strategy. This is sometimes referred to as duty
cycling or load shedding. This concept is feasible if the thermal storage capacity of the facility is
large enough or if slight temperature changes can be tolerated. Often load shedding is
accomplished by installing a demand controller.
Demand controllers are devices that can be connected to an electric meter in order to
monitor electric demand through the meter. Different warning levels can be selected so that when
the peak reaches specified levels, an alarm sounds and displays the level of demand that would
need to be shed to remain below a desired level. Demand controllers can also automatically turn
off unnecessary equipment to shed some of the electrical load if necessary. A demand controller
is much more reliable than a manual control and is often used to ensure the overall reliability of
equipment scheduling.
Another possibility is to schedule the operation of high consumption electrical
equipment to specific times and to stagger the scheduled employee breaks. Coordinating these
times could reduce the amount of equipment operating at one time, thus decreasing the demand.
As was mentioned earlier, some equipment must operate continuously to achieve maximum
production, thus no rescheduling of that equipment could be done.

LE0393 – Page 5
The annual demand savings for this AR is 242 kW. The annual cost savings is $1,198
and, with $3,025 in implementation costs, the payback period will be approximately 2.5 years.
AR 7: Replace MH Fixtures with T-5 Fixtures with Dimmers
Replacing the metal halide fixtures with T-5 fixtures in the main manufacturing areas is
necessary for energy savings. There are 224 MH fixtures that can be replaced. Higher efficiency
lighting has been a focus for many lighting manufacturers in recent years. New technology has
led to innovative lamps that have a longer rated life and require less wattage, with minimal
reduction in overall lumen output. The annual electricity savings for this AR will be 147,230
kWh, and the annual demand savings will be 774 kW. The estimated annual cost savings for
this AR is $15,376 and, with $40,880 in implementation costs, the payback period will be about
2.66 years.
AR 8: Replace T-12 Lamps with T-5 Lamps
Replacing the T-12fixtures with T-5fixtures in these areas is necessary for energy savings.
There are a total of 65 T-12 fixtures in the facility that can be replaced. Higher efficiency lighting
has been a focus for many lighting manufacturers in recent years. New technology has led to
innovative lamps that have a longer rated life and require less wattage, with minimal reduction
in overall lumen output. The annual electricity savings for this AR will be 11,748 kWh, and
the annual demand savings will be 68 kW/yr. The estimated annual cost savings for this AR is
$1275/yr with $4,018 in implementation costs. The payback period will be about 3.2 years.
AR 9: Install Variable Frequency Drive on 2 x 30 HP Compressor
In many commercial and industrial environments, the application of variable speed control is cost
effective. Energy savings result from reduced power consumption by the motors. As the system
power requirements are reduced, the power consumed by the equipment can be reduced by an
amount significantly greater than can be achieved with the existing controls. The plant has two 30
HP compressors that is not running at full load all the time, and it is a good candidate for VFD and
keep other compressor as a backup. The total annual electricity savings for this AR is 31155

LE0393 – Page 6
kWh. The cost savings is likely to be $3259 and, with an implementation cost of $11400, the
payback period is about 3.5 years.
AR 10: Install Air Induced Nozzles on Air Guns
Twenty-five air guns are used for the cooling of motors during testing. However, only 1
gun is used at a time for 1.5 hours per work day. To use high pressure compressed air directly for
cleaning purposes is not safe, but also not efficient energy usage. It is recommended that air
induced nozzles be used in this application. Air induced nozzles use the coanda effect, or small
directed nozzles, to amplify compressed airflow 25 times or more. As illustrated in Figure 1,
compressed air is ejected through a series of nozzles on the outer perimeter. As the air travels along
the outer wall of the nozzle, surrounding air is entrained into the stream. The airstream that results
is a high volume, high velocity blast of air at minimal consumption. The air is always ejected so it
can vent safely. The annual electricity savings for this AR is 372 kWh and the annual demand savings
will be 56.83kW. The annual cost savings is likely to be $310 and, with an implementation cost of
$1,313, the payback period will be about 4.2 years.
AR 11: Use Comp exhaust for heating Warehouse in winter
The exhaust air from two 30 HP compressors, Kaeser ASD 30T Screw (233CFM), are released to
the atmosphere. This warm air could be redirected during winter months to help heat up the facility.
Currently, gas space heaters are used for heating. The total annual natural gas savings for this AR
is 20.3 MMBtu. The implementation cost of this recommendation is $288.6, and with annual cost
savings of $65.54, the payback period is 4.4 years.

LE0393 – Page 7
Additional Assessment Recommendations (AARs)
The additional assessment recommendations, based on the plant visit with associated
possible annual cost savings or profit generation of $268,699 are given below.
AAR 1: Install Weather Seals on Dock Doors
There are a total of 22 dock doors with 2 that are unused and 2 that open to an
unheated warehouse which leaves 18 doors that could benefit from the installation of weather
seals. The total gap area from all the 18 doors, assuming a 2 inch gap on both sides with no gap
on the bottom, was calculated at 59.32 ft2. In addition, it was assumed that the indoor
temperature was 70°F, while the outside temperature is averaged at 45°F during the winter
months. The annual gas savings for AR 11 is 639.04 MMBtu. The annual cost savings is
$2096.38 and, with $10,800 in implementation costs, the payback period will be about 62
months.
AAR 2: Install automatic sideways door in the docking area
Currently the loading dock has one 20 ft x 8 ft door. This door fits two trucks
side by side; often only one truck is present in the loading dock. This door is open for
approximately 2 hours a day in the winter. During this time the facility loses heat due to
infiltration. By replacing the current door with a door that opens sideways the open area can be
reduced to 10 ft x 8ft when only one truck is present. This will decrease the amount of heat loss
since it is a function of area. The shipping dock is heated using two steam space heaters with
an efficiency of approximately 60%. Annual energy savings for this AR is 93.5 MMBtu. Annual
cost savings is $302 and, with Implementation Cost of $2960, the payback period will be 9.8
years.
AAR 3: Install Array of Solar Panels on Roof
As a rule of thumb for most solar installations, there is approximately 1 kW of solar capacity
per 100 ft2. Taking this to be one of our assumptions, your facility stands to gain 1628 kW (from
162800 ft2 total) of installed solar generation.

LE0393 – Page 8
Most solar installations in Pennsylvania produce on average 1,200 kWh/yr for every 1
kW of capacity, giving an initial estimate of approximately 1953600 kWh/yr in energy savings.
A more accurate analysis was done for your specific location using NREL’s PVWatts calculator
that estimated 2,101,209 kWh/yr of energy savings and $159,692/yr. Also, we estimated that at
systems above 10 kW of capacity, the price per watt stays around $2.6 which gives a system
cost of approximately $4,232,800. On top of all of this are state and federal incentives that
drastically lower the overall price and were also used in the calculations. The annual electricity
savings for this AR will be 2,101,209 kWh. The estimated annual cost savings with incentives
for this AR is $265,814 after the first year and, with $2962960 in implementation costs, the
payback period will be about 11.5years.
AAR 4: Redesign Compressed Air Line in Plant Area
The plant’s main compressed air line is built of copper tubing with a two inch
diameter. The system also has to cope with the many air leaks. In order to avoid extra pressure
drop without needing stainless steel piping, a three inch diameter pipe would be the optimum
pipe size to supply the amount of compressed air for the process. The annual electricity savings
for this AR is 4,655 kWh, and the annual demand savings is 27kW. The annual cost savings is
likely to be $487 and, with an implementation cost of $15,000, the payback period will be less
than 31 years.

LE0393 – Page 9
Assessment Recommendations
&
Calculation of Cost Savings

LE0393 – Page 10
AR 1 - Reduce Compressed Line Pressure from 120 psi to 95 psi
Recommended Action
Reduce the line pressure from 120 psi to 95 psi in the building, as too much energy is
wasted compressing the air to a higher pressure than required.
Summary of Estimated Savings and Implementation Costs
Annual Cost Savings $1,443
Implementation Cost $25
Payback Period .24 months
Electricity Savings 13,793 kWh/yr
Demand Savings 80 kW/yr
ARC Number N/A
Current Practice and Observations
A common source of energy waste is compressing air to a higher pressure than required
by air-driven equipment, as is the case with this company. The minimum required pressure for
the building is determined to be 95 psi, not 120 psi. The compressors’ settings should be reduced
to 95 psi for both 30 HP compressors.
Anticipated Savings– 30 HP Compressors
An estimate of the savings can be realized by reducing the pressure settings on the
compressors. Although the motor size and air temperature affect the energy usage in the plant,
the reduction in air pressure will be calculated as a fraction of energy saved as a result of
reducing the pressure setting of the compressor. The reduction in the horsepower output of the
compressor by reducing the pressure set point from 120 psig to 95 psig is calculated as follows:
POW =
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APCCP
1
AP
APRCP
1
Nk
1k
Nk
1k
,
where
POW = Compressor power reduction; %
RCP = Recommended compressor operating pressure; 95 psig

LE0393 – Page 11
CCP = Current compressor operating pressure; 120 psig
k = Specific heat ratio; 1.4
N = Number of stages; 1
AP = Atmospheric pressure; 14.7 psig.
The percent reduction in horsepower output is:
POW =
1-
95+14.7
14.7
æ
è
ç
ö
ø
÷
1.4-1
1.4´1
æ
è
ç
ö
ø
÷
-1
120+14.7
14.7
æ
è
ç
ö
ø
÷
1.4-1
1.4´1
æ
è
ç
ö
ø
÷
-1
é
ë
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
=
= 12.15%.
The power usage is reduced by 12.15 percent when the pressure set point is lowered
from 120 to 95 psig. The energy savings resulting from lowering the compressor set point is
calculated as follows:
ES = HP  OH  LF  RF  C1  POW / Eff,
where
HP = Horsepower of compressors; 60 HP
OH = Annual operating hours; 2,080 hr/yr
LF = Load factor; assumed 100 percent
RF = Run time as fraction of operating time; assumed 100 percent
C1 = Constant; 0.746 kW/HP
POW = Power reduction; 12.15 percent
Eff = Efficiency of compressor; 82 percent.
The energy savings, ES, is:
ES = 60 HP  2,080 hr/yr  1.0  1.0  0.746 kW/HP  .1215 / 0.82
= 13794.81 kWh/yr
The demand savings, DS, is calculated as:
DS = HP  C1 LF  C2  CF  POW / Eff,
where
CF = Coincidence factor  probability that the equipment contributes to the facility

LE0393 – Page 12
peak demand; assumed 1.0 per month
C2 = Months during which peak demand can be reduced; 12 months/yr
DS = 60 HP  0.746 kW/HP  1.0  12 mo/yr  1.0  0.1215 / 0.82,
= 79.6 kW/yr.
The annual cost savings, ACS, is calculated as:
ACS = ES  Average unit cost of electricity + DS  Average unit cost of demand
= 13794.81 kWh/yr  $0.076/kWh + 79.6 kW/yr  $4.96/kW
= $1443/yr.
Implementation Cost
It is estimated that the only implementation costs will be associated with labor charges
to turn the compressors to a lower setting. The hourly wage is considered to be $25 per hour.
The annual electricity savings and demand savings for this AR is 13793 kWh, and
80 kW, respectively. The annual cost savings is $1443 and, with an implementation cost
of about $25, the payback period will be 7 or 8 days or .24 months.

LE0393 – Page 13
AR 2 - Install Setback Thermostats
Recommended Action
Install programmable setback thermostats to set office and plant temperatures back by
8 oF at night and on weekends.
Comfortable temperatures are maintained within the office 24 hours per day all year.
The office and administrative areas only operate 8 hours per day, 5 days per week, and 52 weeks
per year. Much energy is wasted maintaining the set temperatures during unoccupied hours.
Anticipated Savings
The majority of the natural gas used that tge facility is used for heating both the plant
and office area in winter. It is proposed that the thermostats in the office and administrative
areas be programmed to setback temperature by 8oF at night and on weekends to save energy.
Timer based thermostats could turn down the AC units to a very low load when no one is in
these areas and turn up the AC units one hour before the start of the workday. Installation costs
are generally low for these units since only the thermostat on the wall need be updated.
Savings from Setback Temperature during the Heating Season
An estimate of the savings which could be realized through installation of the setback
timers can be made by using the following approach. The percent of time during the week when
the office is not occupied is:
Po = (HPW-OOH)  100%/HPW hrs/wk
= (168  40) hrs not operating/wk  100%/168 hrs/wk
= 76.19%,
where
Implementation Cost $1,125
Payback Period 3 months
Gas Savings 1,313 MMBtu/yr
ARC Number 2.7261.3

LE0393 – Page 14
Po = Percent of time during the week when the plant is not operating
HPW = Total hours per week; 24hr/day  7 days/week, 168 hrs/week
OOH = Total hours per week the plant is occupied; 8 hr/day  5days/week, 40
hrs/week
The average temperature difference between the plant and the outdoors during the
winter months can be determined by:
T = Tp – (65 – DDY/HD),
where
T = Average temperature difference
Tp = Temperature maintained in the areas during winter; assumed to be 75oF
DDY = Heating degree days for the year; assumed at 4,725
HD = Number of days per year when the average temperature drops
significantly below 65oF; 210 days
Therefore, the average temperature difference during the winter months is:
T = 75 – (65 –4,725/210)
= 32.5 oF.
The energy loss from the building is proportional to the temperature difference between
the inside and outside. If the temperature in the building is lowered 8oF (typical levels of 64oF
in unoccupied buildings are normal) during non-working hours, the resulting energy savings
can be calculated using the following formula:
NGS = (PRT/T)  Po  YU,
where
NGS = Natural gas energy savings in units consumed
Po = Percent of time during the week when the areas are not occupied; 76.19%
PRT = Current reduction in temperature during off hours; 8oF
T = Average temperature difference between inside and outside during
winter months; 32.5oF
YU = Yearly usage for heating in the office area; 7,000 MMBtu/yr.
Therefore,
NGS = (8oF/32.5oF)  0.7619  7,000
= 1,313 MMBtu/yr

LE0393 – Page 15
The natural gas cost savings, NGCS, is:
NGCS = NGS  Unit cost natural gas
= 1,313 MMBtu/yr  $3.23/MMBtu
= $4,240/yr.
Implementation Cost
The cost for purchasing a weekly programmable set back thermostat is about $200.
There are a total of 5 thermostat units installed. Each thermostat is expected to take one hour
to install, with an in-house labor rate of $25 per hour. Materials are therefore expected to cost
$1,000 and labor is expected to cost $125. The total implementation cost is estimated to be
$1,125.
Annual natural gas savings for this AR is 1,313 MMBtu/yr. The estimated annual
cost savings for this AR is $4,240 and, with an implementation cost of about $1,125, the
payback period will be 3 month

LE0393 – Page 16
AR 3 - Repair Leaks in Compressed Air System
Recommended Action
It is recommended that an ultrasonic leak detection device be purchased to identify leaks
in the compressed air lines.
Payback Period 4.8 months
ARC Number 2.4236.2
A perfectly sealed compressed air system can never be fully achieved; therefore, it is
appropriate to assume that there are leaks in the compressed air lines in any plant. There is a
total of 135 HP (1 × 75 HP, 2 × 30 HP) worth of compressor power at this plant, with difference
in piping schemes, therefore, a conservative estimate for the amount and size of leaks has been
given in the calculations below. 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 estimation is based on the velocity of the air from the leak through a restricted air
anemometer with negligible compressibility.
Anticipated Savings
The parameters listed below affect the cost of compressed air leaks.
Parameter Value
Air temperature at compressor inlet, °F 92
Air temperature at point of leak, °F 72
Compressor operating pressure, psig 120
Line pressure at point of leak, psig 120
Compressor motor size, HP (total) 135
Compressor motor efficiency 0.82

LE0393 – Page 17
Compressor type Screw
Compressor annual operating hours 2,080
Table 3: Compressed Air System Parameters.
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 these conditions. The results are
shown in Table 2.
Hole
Diameter
(inches)
Flow
Rate
(cfm)
Power
Loss
(hp)
Demand
Loss
(kW/yr)
Energy
Loss
(kWh/yr)
Leak
Cost
($/yr)
1/64 0.35 0.09 0.81 141 $ 15
1/32 1.42 0.36 3.24 562 $ 59
1/16 5.67 1.45 12.97 2,249 $ 235
1/8 22.67 5.80 51.90 8,996 $ 941
3/16 51.01 13.04 116.77 20,240 $ 2,117
1/4 90.69 23.19 207.59 35,982 $ 3,764
Total 171.81 43.93 393.29 68169 $ 7,132
Table 4: Cost of Compressed Air Leaks.
As the table above shows, the cost of compressed air leaks increases exponentially as
the size of the leak increases. This can be seen even more clearly in Figure 1. 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.

LE0393 – Page 18
Figure 1: Cost of Energy Lost to Compressed Air Leaks.
The estimated demand reduction, energy conservation, and corresponding cost savings
for the six air leak are listed in Table 3. These air leaks are given as a sample to show the amount
of savings that can be achieved by a having a rigorous air audit throughout the year.
Leak Location
Number
of
Leaks
Leak
Diameter
(inches)
Demand
Reduction
(kW/yr)
Energy
Conservation
(kWh/yr)
Cost
Savings
($/yr)
Location 1 4 1/32 12.97 2,249 $235
Location 2 3 1/16 38.92 6,747 $706
Location 3 2 1/8 103.79 17,991 $1,882
TOTAL 9 155.69 26,987 $2,823
Table 5: Summary of Savings for Compressed Air Leaks.
Implementation Cost
The cost savings for this recommendation assumes that there are only 9 leaks in the
plant. Additional leaks are probable and can most easily be detected and repaired by plant
maintenance personnel during non-production periods, when the hiss of the air leaks will be
easier to detect. Plant personnel should also be equipped with ultrasonic leak detection
equipment which will make finding leaks very simple. We recommend that periodic detection
0
500
1,000
1,500
2,000
2,500
3,000
3,500
4,000
4,500
0 1/16 1/8 3/16 1/4 5/16
EnergyCost($/yr)
Leak Diameter (inches)
Cost of Energy Lost to Air Leaks

LE0393 – Page 19
and repair of air leaks be performed during non-production periods. A suggested maintenance
schedule would be a one-time thorough inspection, followed by less intense monthly or weekly
inspections of all of the air lines in the plant.
An ultrasonic leak detection device cost approximately $1,000 and can be purchased
from any industrial supplier.
The annual electricity savings for this AR is 26,987 kWh, and the annual demand
savings is 156 kW. The annual cost savings is $2,823 and, with an implementation cost of
$1,000, the payback period will be approximately 4 months.
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 125 psia line pressure is 0.1176. Thus, the flow is choked. The volumetric flow rate
of free air, Vf, exiting the leak under choked flow conditions is calculated as follows:
Vf =
460TC
4
D
CCC
P
P
)460T(NL
13
2
d21
i
1
i



,
where
Vf = Volumetric flow rate of free air; 0cubic feet per minute
NL = Number of air leaks; 0
Ti = Temperature of the air at the compressor inlet; 0F
Pl = Line pressure at leak in question; 0
Pi = Inlet (atmospheric) pressure; 0
C1 = Isentropic sonic volumetric flow constant; 28.37 ft/sec-R0.5
C2 = Conversion constant; 60 sec/min
Cd = Coefficient of discharge for square edged orifice*; 0.8
 = Pythagorean constant; 3.1416
D = Leak diameter; 0
C3 = Conversion constant; 144 in2/ft2
* A.H. Shapiro, The Dynamics and Thermodynamics of Compressible Fluid Flow, Vol. 1, Ronald Press, N.Y. 1953, p.
100.

LE0393 – Page 20
Tl = Average line temperature; F.
The power loss from leaks, estimated as the power required to compress the volume of
air lost from atmospheric pressure, Pi, to the compressor discharge pressure, Po, is as follows**.
L = ,
EE
1-
P
P
CN
1-k
k
VCP
ma
Nk
1k
i
o
4f3i




















where
L = Power loss due to air leak; 0HP
k = Specific heat ratio of air; 1.4,
C4 = Conversion constant; 3.03  10-5 HP-min/ft-lb
Po = Compressor operating pressure; psia
Ea = Air compressor isentropic (adiabatic) efficiency,
Ea = 0.82 for single stage screw compressors
Ea = 0.75 for multi-stage reciprocating compressors
Ea = 0.82 for rotary screw compressors
Ea = 0.72 for sliding vane compressors
Ea = 0.80 for single-stage centrifugal compressors
Ea = 0.70 for multi-stage centrifugal compressors
Em = Compressor motor efficiency; 0.8.
The demand reduction, DR, and annual energy conservation, EC, are estimated as
follows:
DR = L  C5  CF
EC = L  H  C5,
where
C5 = Conversion factor; 0.746 kW/ HP
CF = Coincidence factor – probability that the equipment contributes to
The facility peak demand; 1
H = Annual time during which leak occurs; 0hr/yr.
** Chapters 10 and 11, Compressed Air and Gas Handbook, Fifth Edition, Compressed Air and Gas Institute, New Jersey,
1989.

LE0393 – Page 21
AR 4 - Clean Skylights
Recommended Action
Clean the existing skylights in the Spot Welding production area.
It was observed that the 6 skylights in the Spring and Spot Welding production area are
either dirty or beginning to yellow. As a result less light was being allowed through these
building features and additional lighting was required inside of the building to make up for the
loss during daylight hours. With the current Metal Halide lamps being used in the area the
realized savings will not be as great as with T-5 lamps equipped with dimmers. For the purpose
of providing useful calculations it will be assumed that such an upgrade has been made to
the facility.
Anticipated Savings
Cleaning the skylights in a production area will result in an increase in the level of
light in the area. Assuming the current light levels are appropriate for the tasks being
performed this will result in the potential for decreasing the amount of electrical lighting
necessary to maintain the same level of light necessary for the tasks being performed. The
energy conservation, EC, and energy cost savings, ECS, for such a scenario can be calculated
with the following.
EC =
1
)()01.0()01.0(
C
WNLNFPLPHOH 
EC = Energy conservation
OH = Operating hours, 2,080 hrs/yr
PH = percent of hours in which lighting reduction will occur, 50%
Annual Cost Savings $129
Payback Period 0.93 years
Electricity 1,712 kWh/yr
ARC Number 2.7121.3

LE0393 – Page 22
PR = percent reduction in lighting during reduction hours, 30%
NF = number of lighting fixtures, 49 fixtures
NL = number of lamps per fixture, 4
W = wattage per lamp, 28 watts
C1 = 1000 watts/kilowatt
The Annual cost savings is calculated by
ACS = EC  Effective energy rate,
ACS = Annual Cost Savings
Light levels measured in the Shaping production area and the Spot Welding production
area were 400 lux and 315 lux respectively when taken from areas beneath the skylights.
Rounding to a difference of ~100 lux generated by the presence of cleaned skylights, it can be
seen that a 30% reduction in electrical light output can be achieved while maintaining the same
level of ~300 lux in the production area. Assuming that useful daylight hours that will allow
effective use of the skylights represents 50% of operating hours the energy conservation and
energy cost savings are calculated as follows.
EC =

2,080  (50  0.01)  (30  0.01)  49  (4  28)
1000
= 1,712 kWh/yr.
ACS =

1,7120.076
= $129/yr.
Implementation Cost
The implementation cost associated with cleaning the skylights in the Spot Welding
production area would be minimal. Assuming that a crew is capable of cleaning 2 skylights per
hour and that the labor cost associated is $40/hr the total implementation cost comes out to
$120.
The annual energy savings for this AR is 1,712 kWh. The estimated annual cost
savings is likely to be $129 and, with $120 in implementation costs, the payback period
will be about 0.93 years.

LE0393 – Page 23
AR 5 - Draw Compressor Intake Air From Outside
Recommended Action
Use ducting to bring outside air to the intake of air compressors.
Payback Period 2 years
ARC Number 2.4221.2
Currently, this plant has 2 air compressors, one each of 30 HP. The compressors are
located in a common isolated room and have their intake from within this room.
Anticipated Savings
A compressor does less work to compress colder air. The fractional reduction in
compressor work (CWR) resulting from using outside air is
CWR = (CWI – CWO) / (CWI)
where
CWI = compressor work with inside air, HP
CWO = compressor work with outside air, HP
The compressor work reduction can be written in terms of temperature as follows
CWR = (TI  TO) / (TI + 460)
where
TI = Average indoor temperature, 75oF
TO = Average outdoor temperature, 64.6oF
Since the compressors are in operation 8 hours a day, therefore, 24 hour average
monthly outdoor temperature was used. The data is obtained from www.worldclimate.com for
Bristol, PA and is provided below in Table 6: 24 hour Average Monthly Temperature for Bucks
County, PA

LE0393 – Page 24
. Since the compressors are kept in a closed room, the heat generated by these
compressors is not vented out which increases the temperature of the room. The temperature
within the room was assumed to be around 75°F
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Year
°C -1.38 0.55 5 10.8 16.3 21.9 24.4 23.6 19.1 12.2 6.6 1.3 11.7
°F 29.5 33.0 41 51.5 61.5 71.5 76.0 74.5 66.5 54.0 44.0 34.5 53.1
Table 6: 24 hour Average Monthly Temperature for Bucks County, PA
The annual energy savings, ES, associated with reduction in compressor work is given
by the following equation:
ES = (HP) x (C1) x (OH) x (RF) x (LF) x (CWR) / (EFC)
where
HP = compressors rating; 60 HP
C1 = 0.746 = conversion constant; kWh/HP-hr
OH = As per the information obtained from the plant personnel, the compressors work
8x5x52 weeks; 2,080 hrs/yr
RF = estimated run time as a fraction of operating time; no units (assumed 0.75)
LF = load factor; 100%
EFC = estimated efficiency of compressor; % (assumed at 0.85)
Therefore, for the 2 compressors,
ES = (60)  (0.746)  (2080)  (0.75)  (1)  [(75 64.6) / (75+ 460)] / (0.85)
= 1596.89 kWh/yr
The reduced energy usage will also result in a lower demand charge; the reduction in
demand is as follows:
RD = (HP)  (C1)  (LF)  (C2)  (CF)  (CWR) / (EFC)
Where
CF = coincidence factor - probability that the equipment contributes to the facility,
peak demand, per month (assumed 1)
C2 = conversion constant, 12 months/yr
RD = (60)  (0.746)  (1)  (12)  [(7564.6) / (75 + 460)] /(0.85)
RD = 12.28 kW/yr

LE0393 – Page 25
The annual cost savings, ACS, is
ACS = (ES) x (Unit cost of electricity) + (RD) x (Unit cost of demand)
= (1596.89 kWh/yr) x ($0.076/kWh) + (12.28 kW/yr) x ($4.96 /kW)
= $183/yr
Therefore, the total savings in energy is 1596.89 kWh/yr, with a demand savings of
12.28 kW/yr and total cost savings of $183/yr.
Implementation Cost
The compressors in this plant are located fairly close to the outside. Implementation cost
is estimated to be about $36.50/ft for material and installation costs. Therefore the total
implementation cost is $365 for 10ft of ductwork. Ductwork must be provided from the intake
of the compressor to the outside of the plant. The damper should remain open when the outside
temperature is lower than the temperature near the compressors and should be closed during
summer when outside air is relatively warmer than the air in the plant.
It is recommended that during any contact with vendors for the implementation of this
AR, any effects of moisture content of the outside air on the compressor be taken into account.
During the colder months, outside air is dryer, but during the summer months that may not be
true.
The annual electricity savings for this AR is 1597 kWh, and the annual demand
savings is 12.28 kW. The cost savings is likely to be $183 /yr and, with an implementation
cost of $365, the payback period is about 2 years.

LE0393 – Page 26
AR 6 - Install a demand monitor
Recommended Action
An overall energy management system for this facility should be determined and
implemented. This would reduce the electric demand of the facility, thus reducing the overall
cost of electric bills. One suggested strategy is to stagger the start of the large electric consuming
devices, such as the air conditioning units and lighting arrays, to prevent an excessive amount
of equipment from operating concurrently.
Payback Period 2.5years
ARC Number 2.3131.1
Electric demand is the average rate at which electric energy (measured in kWh) is used
during a specific metered period. This period is called the demand interval and is fifteen minutes
in length at this plant. Demand is defined as the average energy used during the interval, as
follows:
hperiod,demandofLength
intervaldemandduringusedkWh
=kWDemand,
h1/4
kWh
 .
The peak demand is the highest average load (measured in kW) reached over all of the
demand intervals within a given billing period. A high demand charge, therefore, results from a
large usage of power during any demand interval of the billing period.
High demand charges can result from a high rate of energy usage for short periods during
production hours. This problem may have one or more solutions. Plant production schedules and
the economics of each situation should be considered when assessing the options. One possible
solution may be to distribute the facility's electrical usage over alternate shifts. Thus, if high peak
demand occurs during one shift, while several equipment stations with heavy electrical usage are

LE0393 – Page 27
used, it may be possible to move usage of one or more stations to a shift with less peak usage,
resulting in a lower overall peak usage.
An alternate solution is to interlock specific pieces of equipment, thereby preventing them
from consuming power at their peak rates at the same time. This is not always feasible when the
natural operating interval of the equipment is much shorter than the demand interval, or when the
machines must be in continuous operation to maximize production. Controlling electrical
resistance heaters and other heating and ventilating equipment during periods when process
requirements are peaking is another demand control strategy. This is sometimes referred to as duty
cycling or load shedding. This concept is feasible if the thermal storage capacity of the facility is
large enough or if slight temperature changes can be tolerated. Often load shedding is
accomplished by installing a demand controller.
Demand controllers are devices that can be connected to an electric meter in order to
monitor electric demand through the meter. Different warning levels can be selected so that when
the peak reaches specified levels, an alarm sounds and displays the level of demand that would
need to be shed to remain below a desired level. Demand controllers can also automatically turn
off unnecessary equipment to shed some of the electrical load if necessary. A demand controller
is much more reliable than a manual control and is often used to ensure the overall reliability of
equipment scheduling.
Another possibility is to schedule the operation of high consumption electrical
equipment to specific times and to stagger the scheduled employee breaks. Coordinating these
times could reduce the amount of equipment operating at one time, thus decreasing the demand.
As was mentioned earlier, some equipment must operate continuously to achieve maximum
production, thus no rescheduling of that equipment could be done.
Possible Demand Control Strategies
The determination of an overall energy management systemthat would be best for
this facility requires engineering analysis that is beyond the scope of this report.
Additional advice should be obtained from a reputable controls supplier or engineering firm if
a more detailed study is required. However, following are some suggestions garnered from
experience with other IAC clients:
 Contact the customer service representative at the local utility to discuss the facility's
electricity usage in order to gain some insight as to the times when the peak demand occurs

LE0393 – Page 28
each month. By matching this information with production records, patterns may emerge
that would indicate strategies for control. For example, if the peak demand is consistently
set at a given time that corresponds to running a particular piece of equipment, some
rescheduling may be in order.
 Consider the impact that the time of day has on the peak demand. In many plants, the peak
demand is set during one of the following periods: (1) in the morning when some of the
equipment is first turned on; (2) just before or just after breaks or lunch; (3) during summer
afternoons due to cooling; and (4) just before or after shift changeovers. If the peak occurs
during these periods, observe the equipment and operations to identify opportunities to
reduce unnecessary equipment loading.
 Consider staggering employee breaks. Employees returning from a break period will use
process-related equipment that has been idle or turned off. Initial use of idle equipment
often increases demand usage for a short period of time.
 Consider moving some operations that occur during the typical peak period to the off-peak
period. For example, if an operation can be shifted past office hours, when the office lights
are typically turned off, then the load during the peak demand period will be reduced. Some
process equipment may also be moved to a swing shift in order to reduce peak demand. At
this site, every kilowatt reduction in the peak demand translates into $4.96/month,
correlating to a savings of $59.52/yr.
 Consider staggering the start-up of equipment. Motors typically experience an in-rush
current, which is much higher than their rated current, when they are first energized. This
in-rush current varies depending on the starting torque of the motor, but is at least five times
as great as the rated current. This higher in-rush current does not last long (perhaps twenty
seconds), but averaged over fifteen minutes with a lot of equipment, this additional load
could be significant. For example, a 100 HP motor (74.6 kW) with a 500% inrush current
over a twenty second period would result in a demand of 81.2 kW (when averaged over
fifteen minutes). This power requirement is about 9% higher than the level load of 74.6
kW, which is expected for this motor.
 Develop a strategy to turn off or reduce the load on selected equipment when the demand
starts to rise. Identify which equipment is critical to the production process first, and then
consider ways to reduce the load from other equipment.

LE0393 – Page 29
 Install equipment to monitor demand onsite in real time. Demand managers and controllers
are available to monitor and/or automatically turn off equipment (or "shed" equipment) as
the demand rises. This will be more instructive than the utility's demand plots, which are
good for understanding when the peak occurred, but not if it is occurring. This equipment
will also make it easier to match which equipment is on when the peak demand is set.
 Consider interlocking equipment. Interlocking similar equipment prevents several units
from turning on and consuming power at the same time. For example, consider a building
cooled by two roof top units (RTUs) with air conditioning compressors that are rated 2.5
tons and one 5 tons whose compressors are rated to draw 3 and 8 kW, respectively. If the
two units ran at the same time, the combined load would be 11 kW, which is 3 kW higher
than either unit individually. Although it may not be likely that both units would run
simultaneously during the same interval of time coinciding with the demand period for each
of the cooling months, it may occur during the summer months, or about four months per
year. Therefore, interlocking the two RTUs would result in an estimated 12 kW per year
demand savings.
 Involve production personnel. While equipment causes the peak electric demand, facility
personnel typically turn on the equipment. Enlist the support and assistance of production
personnel to help identify necessary and unnecessary equipment when the peak occurs.
These same people will also have a role in implementing the demand control strategy, so it
is important that they understand what peak demand is and how much it costs.
 Be vigilant! Demand control requires daily monitoring. It only takes one motor or one set
of equipment to be on for thirty minutes to set an excessive peak for the month.
It is noted that some of these suggestions may not be feasible demand control strategies at
this plant. All of these suggestions, however, do demonstrate the philosophy behind demand
control principles.
Anticipated Savings
For this facility, one reasonable solution would be staggering the start-up of equipment
and monitoring demand with a demand controller connected to the electric meter. Motors
typically experience an in-rush current when they are first energized, which is much higher than
their rated current. This in-rush current varies depending on the starting torque of the motor,
but is at least five times as great as the rated current. This higher in-rush current does not last

LE0393 – Page 30
long (perhaps twenty seconds), but averaged over fifteen minutes with a lot of equipment, this
additional load could be significant. For example, a 100 HP motor (74.6 kW) with a 500%
inrush current over a twenty second period would result in a demand of 81.2 kW (when
averaged over fifteen minutes). This power requirement is about 9% higher than the level load
of 74.6 kW, which is expected for this motor.
By monitoring demand with a demand controller, plant personnel would be aware of
the decrease by rescheduling equipment at that instant, and the effectiveness of that practice
should be improved. The cost savings associated with this analysis is dependent upon the amount
of equipment that can be rescheduled. These savings are summarized below.
Potential Annual Savings
due to Reduced Peak Usage
Power Shifted
From Peak
KW for 12 months
Annual
Savings
10 kW $50
25 kW $124
50 kW $248
75 kW $372
100 kW $496
Table 7: Potential Demand Savings.
The average monthly on-peak demand value for this facility is 402.7 kW. The major
equipment in this plant consists of lights, molders, shapers, air compressors. If the start-up of some
of this equipment can be rescheduled (half of the equipment started at least 15 minutes later than
the other half), peak demand should be decreased. It is estimated that at least 5 percent of peak
demand can be saved with this practice. In this report, a 5 percent value (the minimum value) was
considered for calculations. The yearly demand and demand cost savings, DS and DCS, that can
be expected from managing this equipment can be estimated as follows:
DS = MDS  S  C1
DCS = DS  Average cost of electric demand,
where

LE0393 – Page 31
MDS = Monthly average peak demand for this facility; 402.7 kW/month
S = Percentage of demand that can be saved; 5 percent
C1 = Conversion constant; 12 months/yr.
The demand, DS, and associated cost savings, DCS, are thus estimated as follows:
DS =402.7 kW/mo  0.05  12 months/yr
= 242kW/yr
DCS = 242 kW/yr  $4.96/kW
= $1,198/yr.
Implementation Cost
The first step in implementation of this assessment recommendation is an understanding
of the rate structure applied to electrical energy usage at the facility. This information is publicly
available online, but not provided on monthly bills.
If this energy management strategy were implemented on a manual basis, there would
be negligible implementation costs. However, with manual control, the possibility of working
the equipment considered together during a peak period could still raise the monthly demand to
an undesirable level. This would reduce the cost savings that could be realized. Installation of
equipment to monitor and/or automatically turn off equipment could be purchased to solve this
problem.
As mentioned previously, the determination of an overall energy management system
that would be best for this facility requires engineering analysis that is beyond the scope of this
report. Additional advice should be obtained from a reputable controls supplier or engineering
firm if a more detailed study is required. Due to the size of this plant, one demand monitor may
be necessary; a demand monitor is about $3,000. The installation cost is estimated to be $25
based on a labor rate of $25/hr. The total cost is about $3,025.
The annual demand savings for this AR is 242 kW. The annual cost savings is
$1,198 and, with $3,025 in implementation costs, the payback period will be
approximately 2.5 years.

LE0393 – Page 32
AR 7 - Replace MH Fixtures with T-5 Fixtures with Dimmers
Recommended Action
T-5 fixtures with dimmers should replace existing Metal Halide fixtures in various
locations throughout the facility should be replaced with 4 lamp T-5 fixtures. The replacement
should include a 1:1 ratio of T-5 to MH fixtures to ensure proper lighting conditions.
ARC Number 2.7135.3
Replacing the metal halide fixtures with T-5 fixtures in the main manufacturing areas is
necessary for energy savings. There are 224 MH fixtures that can be replaced. Higher efficiency
lighting has been a focus for many lighting manufacturers in recent years. New technology has
led to innovative lamps that have a longer rated life and require less wattage, with minimal
reduction in overall lumen output.
Dimmers
Dimmers are important for reducing the energy consumption of lighting fixtures when
the full output of the fixture is not necessary to illuminate the area where work is being done.
Dimmers in conjunction with other building features such as skylights make sense as the output
of lights can be reduced during daylight hours when more light is reaching the work area
through the skylights. Alternatively, dimmers can also be used to lower the light level in an area
when the full fixture output is not necessary such as when foot traffic is present but more
technical or complicated tasks is not being done.
Fluorescent Lighting
Electronic ballasts are currently available which, when used with the proper T-5
fluorescent lamps (the T rating refers to lamp tube diameter in 1/5ths of an inch), provide a very
high quality light while using significantly less energy than the existing metal halide lights and T-

LE0393 – Page 33
12 or T-8 fluorescent lamps. The T-5 lamps provide a high quality light that renders color
significantly better than the existing MH lamps, thus providing excellent lighting for office and
production areas. An added benefit to electronic ballasts is the high frequency at which they
operate, eliminating the flicker often associated with standard fluorescent lighting.
Anticipated Savings
The savings results from replacing the MH fixtures in this area with T-5 lamp fixtures
with dimmers, as outlined in this section. The MH fixtures power ratings and replacement bulbs
power ratings are outlined in the table below. The 400 W MH fixtures should be replaced with
fixtures for 4 × 28 Watt lamps. After checking the T-5 light descriptions, this recommendation
should supply the same amount of light as the present fixtures. Thus, electricity use could be
reduced further.
Dimmers, the percent operating energy reduction, PER, is dependent upon the amount
of alternative light coming into the area. This value can be measured using light sensors or
estimated by using electric utility studies that estimate the fraction of lighting energy wasted.
The areas of concern have access to illumination from skylights approximately 50% of the time
during the total plant operating hours. The estimated energy conservation, EC, and energy cost
savings, ECS, resulting from installing dimmers and light sensors are calculated using the
following relations:
EC =
1
21
C
)PHRHPFWN()HCFWN( 
ECS = EC  Average cost of unit electricity,
where
N1 = Number of Metal Halide fixtures in plant area; 224 at 400 W
N2 = Number of proposed T-5 fixtures in plant area; 224 at 4 × 28 W
CFW = Power rating of current fixtures in area; 400 W
PFW = Power rating of proposed fixtures in area; 4 × 28 W = 112 W
H = Operating hours of lamp; 8 hrs × 5 days × 52 weeks = 2,080 hrs/yr
PER = Reduced percent operating energy reduction due to dimmers and light sensors
in this area; estimated to be 75% (50% florescent lighting during 50% of operating hours).
C1 = Conversion constant; 1,000 W/kW.

LE0393 – Page 34
The following relation gives the demand reduction, DR, if the lights in a specific area were
lowered and replaced with T-5 lamps:
DR = ,
C
1 2CF))PFWN()CFWN((
1
21 
where
peak demand, per month, assumed to be 1.0.
The lamp replacement cost savings, LRS, is calculated using the following relation:
LRS = OH
AL
LCL
AL
LCL
2
22
1
11





 


,
where
LC1 = Existing lamp cost; $26
LC2 = Proposed lamp cost; $5 (when purchased in bulk)
AL1 = Existing average lamp life; 20,000 hrs
AL2 = Proposed average lamp life; 36,000 hrs
L1 = Number of existing lamps, i.e., bulbs; 224
L2 = Number of proposed lamps, i.e., bulbs; 224 × 4
Therefore, the annual cost savings, ACS, can be estimated as follows:
ACS = (EC  Energy rate) + (DR  Demand rate) + Lamp replacement cost savings.
The estimated energy conservation, EC, for replacing all of these Metal Halide lamps
with T-5 lamps with motion sensors is calculated as follows:
EC =

((224  400W ) - (224  4  28W )  0.75)  2,080hrs/yr
1,000 W/kW
,
= 147,230 kWh/yr.
The lights will likely be operating at their rated power when the peak demand is set each
month, so CF = 1.0/month. Thus, the demand reduction is calculated as follows:
DR =

((224  400W)- (224  4  28W )) 1.0 12months/yr
1,000 W/kW
,
= 774 kW/yr.
The lamp replacement cost savings, LRS, are as follows:
LRS =

224  26.00
20, 000

224  4  5.00
36, 000





 2,080

LE0393 – Page 35
= $347/yr.
The annual cost savings, ACS, in the plant production area is:
ACS = 147,230 kWh/yr  $0.076/kWh + 774 kW/yr  $4.96/kW + $347/yr
= $15,376
Therefore, this recommendation results in an energy savings of 147,230 kWh/yr and a
demand savings of 774 kW/yr, with a total cost savings of $15,376.
Implementation Costs
The implementation cost for this recommendation includes the equipment and labor costs
required for the new lamps, ballasts and occupancy sensors. The implementation cost for each T-
5 fixture with a dimmer and light sensor is about $150, and each T-5 lamp costs about $5.
Replacing 2 fixtures per hour at a labor rate of $25/hr, the total labor cost is $2,800. Therefore,
the total implementation cost for this AR is approximately $40,880.
The annual electricity savings for this AR will be 147,230 kWh, and the annual
demand savings will be 774 kW. The estimated annual cost savings for this AR is $15,376
and, with $40,880 in implementation costs, the payback period will be about 2.66 years.

LE0393 – Page 36
AR 8 - Replace T-12 Lamps with T-5 Lamps
Recommended Action
T-5 fixtures should replace existing T-12 fixtures in the welding and springs section, and
woodshop section. This recommendation takes into account the elimination of work bench T-12
lamps in the rolling and hardware area (11) and welding and springs area (11).
Annual Cost Savings $1275/yr
ARC Number 2.7142.3
Replacing the T-12fixtures with T-5fixtures in these areas is necessary for energy savings.
There are a total of 65 T-12 fixtures in the facility that can be replaced. Higher efficiency lighting
has been a focus for many lighting manufacturers in recent years. New technology has led to
innovative lamps that have a longer rated life and require less wattage, with minimal reduction
in overall lumen output.
Fluorescent Lighting
Electronic ballasts are currently available which, when used with the proper T-5
fluorescent lamps (the T rating refers to lamp tube diameter in 1/5ths of an inch), provide a very
high quality light while using significantly less energy than the existing metal halide lights and T-
12 or T-8 fluorescent lamps. Also, the T-5 lamps render color significantly better than the existing
T-12 lamps, thus providing excellent lighting for office and production areas. An added benefit to
electronic ballasts is the high frequency at which they operate, eliminating the flicker often
associated with standard fluorescent lighting.
Anticipated Savings
The savings results from replacing 43 T-12 fixtures (41 rated at 120W and 2 rated at
360W) with 43 T-5 fixtures (41 with 2 × 28 Watt lamps and 2 with 6 × 28 Watt lamps) and

LE0393 – Page 37
removing 22 T-12 fixtures (22 rated at 120W). After checking the T-5 light descriptions, this
recommendation should supply the same amount of light as the present fixtures.
The estimated energy conservation, EC, is given by the following relation:
EC =
1
21
C
H)PFW(NH)CFW(N 
ECS = EC  Average cost of unit electricity,
where
N1 = Number of T-12 fixtures; 65
CFW = Power rating of current fixtures; 63 at 120W, 2 at 360W
N2 = Number of T-5 fixtures; 43
PFW = Power rating of proposed fixtures; 41 at 56W, 2 at 168W
OH = Operating hours of lamp; 8 hrs × 5days ×52 weeks = 2,080 hrs/yr
C1 = Conversion constant; 1,000 W/kW.
The following relation gives the demand reduction, DR, if the lights in a specific area
were replaced with T-5 lamps:
DR = ,
C
PFW)(NCFW)(N
1
221 C
where
peak demand, per month, assumed to be 1.0.
C2 = 12mo/yr
The lamp replacement cost savings, LRS, is calculated using the following relation:
LRS = OH
AL
LCL
AL
LCL
2
22
1
11





 


,
where
LC1 = Existing lamp cost; $5
LC2 = Proposed lamp cost; $5 (when purchased in bulk)
AL1 = Existing average lamp life; 20,000 hrs
AL2 = Proposed average lamp life; 36,000 hrs
L1 = Number of existing lamps; 138
L2 = Number of proposed lamps; 94

LE0393 – Page 38
Therefore, the annual cost savings, ACS, can be estimated as follows:
ACS = (EC  Energy rate) + (DR  Demand rate) + Lamp replacement cost savings
The estimated energy conservation, EC, for replacing all of these T-12 lamps with T-5
lamps with motion sensors is calculated as follows:
EC =
1,000W/kW
/2080168W))(2-360W)(2)56(41)120((63 yrhrWW 
= 11,748 kWh/yr.
The lights will likely be operating at their rated power when the peak demand is set each
month, so CF = 1.0/month. Thus, the demand reduction is calculated as follows:
DR =
1,000W/kW
/120.1168W))(2-360)(2)56(41120W)((63 yrmoW 
= 68 kW/yr.
The lamp replacement cost savings, LRS, are as follows:
LRS = 080,2
36,000
00.594
20,000
00.5138





 


LRS = $45/yr.
The annual cost savings, ACS, in the plant production area is:
ACS = 11,748 kWh/yr  $0.076/kWh + 68 kW/yr  $4.96/kW + $45/yr
= $1275/yr
Therefore, this recommendation results in an energy savings of 11,748 kWh/yr and a
demand savings of 68 kW/yr, with a total cost savings of $1275/yr.
Implementation Costs
The implementation cost for this recommendation includes the equipment and labor
costs required for the new lamps and ballasts. Each fixture requires 2 ballasts, which cost about
$35 each. Each fixture also requires either 2 or 6 lamps, which cost about $5 each. Replacing 2
fixtures an hour at a labor rate of $25/hr the total labor cost are $538. Therefore, the total
implementation cost for this AR is about $4,018.

LE0393 – Page 39
The annual electricity savings for this AR will be 11,748 kWh, and the annual
demand savings will be 68 kW/yr. The estimated annual cost savings for this AR is
$1275/yr with $4,018 in implementation costs.The payback period will be about 3.2 years.

LE0393 – Page 40
AR 9 - Install Variable Frequency Drive on 2x30 HP Compressor
Recommended Action
Install a variable frequency drive on the 30 HP air compressor.
Summary of Estimated Savings and Implementation Cost
Electricity Savings 31155 kWh/yr
ARC Number 2.4144.1
In many commercial and industrial environments, the application of variable speed control
is cost effective. Energy savings result from reduced power consumption by the motors. As the
system power requirements are reduced, the power consumed by the equipment can be reduced by
an amount significantly greater than can be achieved with the existing controls. The plant has two
30 HP compressors that is not running at full load all the time, and it is a good candidate for VFD
and keep other compressor as a backup.
Anticipated Savings
The affinity laws estimate that the change in the power of the motors varies as the cube of
the speed of the motor, or flow, changes, as per the following:
3
p
c
3
p
c
proposed
current
Flow
Flow
Speed
Speed
Power
Power

















 .
This relationship is used to estimate the energy use of a given motor with a variable
frequency drive.
The table below shows the relative power consumption of a motor using VFD control,
compared to a motor with standard controls*. Notice that the affinity laws are not exactly
followed for VFD power consumption. This is a result of losses incurred by the variable
frequency drive, which reduces the motor's efficiency. Therefore, with VFD control, as the flow
* Electric Power Research Institute, Adjustable Speed Drives Directory, Table 3.1, p. 18, 1991.

LE0393 – Page 41
rate decreases, the VFD/motor system efficiency decreases. Consequently, the actual power
consumption is higher than the theoretical power consumption estimated by the affinity laws,
with more deviation at lower flow rates. More accurate power consumption estimates can be
obtained for varying flows if pump or fan curves from the manufacturers are available.
Load
%
Power Consumption of Motor
No Control
%
VFD
%
100 100 105
95 100 86
90 100 73
85 100 64
80 100 57
75 100 50
70 100 44
65 100 38
60 100 32
55 100 26
50 100 21
45 100 17
40 100 14
35 100 11
30 100 8
25 100 6
20 100 5
Table 8: Power Consumption of Motor with Flow Rate.
The annual energy savings, ES, and corresponding cost savings, ECS, can be calculated as
follows:
ES = CEU  PEU
ECS = ES  Avoidable cost of electricity,
Where,
CEU = Current time weighted energy consumption for a given motor, kWh
PEU = Projected time weighted energy consumption for a given motor, kWh.
The current energy consumption, CEU, and proposed energy consumption, PEU, can
be estimated as follows:
CEU =
Existing,m
2 OHLFCHP



LE0393 – Page 42
And,
PEU =
oposedPr,m
2CHP


 (FR1  H1  OH),
Where,
HP = Horsepower of motor, 30 HP*2
OH = Annual operating hours of motor; 8*5*52=2080 total hr/yr
C2 = Conversion constant, 0.746 kW/HP
FR1 = Power consumption of variable frequency drive motor, originally at 70% load,
44%
H1 = Fraction of time the motor will operate at 70% load: 100%
LF = Load factor, average fraction of rated power at which motor operates; 0.7
Existing,m = Efficiency of motor, 0.85
oposedPr,m = Efficiency of motor, 0.90 (average efficiency of full load and part loads).
For this calculation, it is assumed that the motor will run at 100 percent for the operating
hours.
CEU =
0.85
20807.00.746302 
= 76671 kWh/yr
PEU =
0.90
0.746302 
 (0.44 1  2080)
= 45516 kWh/yr
ES = 76671 kWh/yr  45516 kWh/yr
= 31155 kWh/yr.
The annual demand savings, DS, for a given piece of equipment can be estimated as
follows:
DS = 



OH
ES
 C3  CF  DUF
DCS = DS  Avoidable cost of electric demand,
Where,
C3 = Conversion constant, 12 months/yr
CF = Coincidence factor – probability that the equipment contributes to the

LE0393 – Page 43
facility peak demand per month, 1.00
DUF = Fraction of year equipment contributes to peak demand, 1.00
DS =
hrs/yr2080
kWh/yr31155
 12 mo/yr  1  1
= 180 kW/yr.
The total cost savings, TCS, is:
TCS = ECS + DCS
= 31155 kWh/yr  $0.076/kWh + 180 kW/yr  $4.96/kW
= $3259 /yr.
Implementation Cost
The cost for a variable frequency controller for one 30 HP compressor is estimated to be
$5200, with an additional $500 in installation costs.
And the total implementation fee should be
2*($5200+$500) = $11400
The plant could contact a contractor for a more accurate price for the VFD and installation.
The total annual electricity savings for this AR is 31155 kWh. The cost savings is
likely to be $3259 and, with an implementation cost of $11400, the payback period is about
3.5 years.

LE0393 – Page 44
AR 10 - Install Air Induced Nozzles on Air Guns
Recommended Action
Use induction air nozzles equipped with regulators for air lines used in the plant to reduce the
usage of the compressed air.
Electricity Savings 372 kWh/yr
Demand Savings 57 kW
ARC Number 2.4236.2
Twenty-five air guns are used for the cooling of motors during testing. However,only 1 gun is
used at a time for 1.5 hours per work day. To use high pressure compressed air directly for cleaning
purposes is not safe, but also not efficient energy usage. It is recommended that air induced nozzles be
used in this application. Air induced nozzles use the coanda effect,or small directed nozzles, to amplify
compressed airflow 25 times or more. As illustrated in Figure 1, compressed air is ejected through a
seriesof nozzles on the outer perimeter. Asthe air travels along the outer wall of the nozzle, surrounding
air is entrained into the stream. The airstream that results is a high volume, high velocity blast of air at
minimal consumption. The air is always ejected so it can vent safely.
Figure 2: EXAIR Air Induced Nozzle.
Anticipated Savings
 from www.exair.com

LE0393 – Page 45
Table 1 shows the values of the parameters pertinent to this calculation. Line pressure
at the point of the air guns is set to 40 psi.
Parameter Value
Air temperature at compressor inlet, °F 75
Air temperature at point of nozzle °F 70
Compressor operating pressure, psig 110
Line pressure at point of nozzle psig 40
Compressor motor size, HP (total) 75
Compressor motor efficiency 0.92
Compressor type Screw
Annual operating hours of nozzles 390
Table 9: Compressed Air System Parameters.
Using regular air guns, the volumetric flow rate, power lost due to leaks, energy lost,
and cost for leaks of various sizes were calculated specifically for these conditions. The
details of calculations are presented at the end of this recommendation. The results are shown
in Table 3.
Nozzle
Diameter
(inches)
Flow
Rate
(cfm)
Power
Usage
(hp)
Demand
Usage
(kW/yr)
Energy
Usage
(kWh/yr)
Nozzle
Cost
($/yr)
1/64 0.11 0.03 1.18 8 $6
1/32 0.46 0.11 4.74 31 $26
1/16 1.83 0.43 18.94 124 $103
1/8 7.34 1.71 75.77 496 $414
3/16 16.51 3.84 170.48 1,117 $930
1/4 29.35 6.83 303.08 1,986 $1,654
Table 10: Cost of Regular Nozzles.
From the table above, it is recommended that 1/16 inch nozzle diameters be used for
the guns.
NOTE: Only 1 nozzles is used at a specific time, but the total number of nozzles is 25. Thus,
we use 1 nozzles to calculate the energy savings.

LE0393 – Page 46
Nozzle
Location
Number
of
Nozzles
Existing
Nozzle
Diameter
[in]
Proposed
Nozzle
Diameter
[in]
Demand
Reduction
[kW/yr]
Energy
Conservation
[kWh/yr]
Cost
Savings
[$/yr]
Production
Area
1 1/8 1/16 56.83 372 $310
Table 11: Summary of Savings.
The electricity savings, ES, is calculated as:
ES = N1  EU1  N2  EU2,
where
N1 = Number of air guns currently in use; 1 (plant owns 25 in total)
EU1 = Electricity consumption of one air gun every year; 496 kWh/yr
N2 = Number of induced air nozzles proposed; 1(plant should replace all 25)
EU2 = Electricity consumption of one induced air nozzle every year; 124 kWh/yr
Therefore,
ES = 1  496  1  124
= 372 kWh/yr.
The demand savings, DS, is:
DS = (N1  DU1  N2  DU2)  CF,
where
DU1 = Demand consumption of one air gun; 75.77 kW/yr
DU2 = Demand consumption of one induced air nozzle; 18.94kW/yr
CF = Coincident factor for air guns and air nozzles contributed to demand usage;
1.0.
Therefore,
DS = (1  75.77  1 18.94)  1.0
= 56.83kW/yr.
The total cost savings, TCS, is calculated as follows:
TCS = 372 kWh/yr  $0.076/kWh + 56.83 kW/yr  $4.96/kW
= $310/yr.
Implementation Cost

LE0393 – Page 47
Each induction air nozzle costs about $50, and the estimated labor rate for in-house
maintenance is $25 per hour. It is estimated that replacing the 25 regulators will take 2.5 hours,
so the total cost of this recommendation is about $1,313.
The annual electricity savings for this AR is 372 kWh and the annual demand savings will
be 56.83kW. The annual cost savings is likely to be $310 and, with an implementation cost of
$1,313, the payback period will be about 4.2 years.
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 100 psia line pressure is 0.147. Thus, the flow is choked. The volumetric flow rate
of free air, Vf, exiting the leak under choked flow conditions is calculated as follows:
Vf =
460TC
4
D
CCC
P
P
)460T(NL
13
2
d21
i
1
i



,
where
Vf = Volumetric flow rate of free air; 7.34 cubic feet per minute
NL = Number of air leaks; 1
Ti = Temperature of the air at the compressor inlet; 75 °F
Pl = Line pressure at leak in question; 40psia
Pi = Inlet (atmospheric) pressure; 14.7 psia
C1 = Isentropic sonic volumetric flow constant; 28.37 ft/sec-R0.5
C2 = Conversion constant; 60 sec/min
Cd = Coefficient of discharge for square edged orifice*; 0.8
 = Pythagorean constant; 3.1416
D = Leak diameter; 1/8 in
C3 = Conversion constant; 144 in2/ft2
Tl = Average line temperature; 70°F
* A.H. Shapiro, the Dynamics and Thermodynamics of Compressible Fluid Flow, Vol. 1, Ronald Press, N.Y. 1953, p. 100.

LE0393 – Page 48
Vf =
46070144
4
)8/1(*1416.3
8.06037.28
7.14
40
)46075(1
2


Vf = 7.34cubic feet per minute
The power loss from leaks, estimated as the power required to compress the volume of
air lost from atmospheric pressure, Pi, to the compressor discharge pressure, Po, is as follows**.
L = ,
EE
1-
P
P
CN
1-k
k
VCP
ma
Nk
1k
i
o
4f3i




















where
L = Power loss due to air leak; 1.7 hp
k = Specific heat ratio of air; 1.4
C4 = Conversion constant; 3.03  10-5 HP-min/ft-lb
Po = Compressor operating pressure; 110 psi
Ea = Air compressor isentropic (adiabatic) efficiency;
Ea = 0.82 for single stage screw compressors
Ea = 0.75 for multi-stage reciprocating compressors
Ea = 0.82 for rotary screw compressors
Ea = 0.72 for sliding vane compressors
Ea = 0.80 for single-stage centrifugal compressors
Ea = 0.70 for multi-stage centrifugal compressors
Em = Compressor motor efficiency; 0.92
** Chapters 10 and 11, Compressed Air and Gas Handbook, Fifth Edition, Compressed Air and Gas Institute, New Jersey,
1989.

LE0393 – Page 49
L = ,
0.920.82
1-
14.7
110
103.031
1-1.4
1.4
7.3414414.7
11.4
11.4
5



















L = 1.7
The demand reduction, DR, and annual energy conservation, EC, are estimated as
follows:
DR = L  C5  CF
EC = L  H  C5,
where
C5 = Conversion factor; 0.746 kW/ HP
CF = Coincidence factor – probability that the equipment contributes to
the facility peak demand; 1.0
H = Annual time during which leak occurs; 390 hr/yr.
DR = 1.7  0.746 1
DR = 1.27
EC =1.7  390  0.746
EC= 494.6

LE0393 – Page 50
AR 11 - Use Comp exhaust for heating Warehouse in winter
Recommended Action
Divert warm air from the compressor exhaust to save in heating costs during the
winter
Annual Cost Savings $65.54
Energy Savings 20.3 MMBtu/year
ARC Number 2.4221.2
Current Practice and Observation
The exhaust air from two 30 HP compressors, Kaeser ASD 30T Screw (233CFM), are
released to the atmosphere. This warm air could be redirected during winter months to help
heat up the facility. Currently, gas space heaters are used for heating.
Anticipated Savings
To calculate the amount of energy in the exhaust air, we use the change in temperature
of the air in the compressor and the flow of air, both obtained from the manufacturer’s manual.
The enthalpy gain of the exhaust air, EEA, is obtained by following equation:
EEA = (FA) (60 min /hr.) (DN) (CP) (DT)
Where
FA = air flow rate; 13.2 m3/min, (doubled because there are two compressors, 6.6
m3/min each)
DN = density of air; 1.293 kg/m3
CP = specific heat of air; 1.005 kJ/kg*K
DT = change in temperature; 7oC
EEA = (13.2 m3/min) (60 min /hr.) (1.293kg/m3) (1.005 kJ/kg*K) (10oC)
= 10291.76 kJ/hr.
This energy can be diverted into the plant to save space-heating dollars during the winter
months. Annual energy savings, ES, can be obtained with the following equation:
ES = (EEA)(HR) (Conv)

LE0393 – Page 51
Where
HR =hours of operation per year
Conv =Conversion factor from kJ to MMBtu
ES = (10291.76) (8 hours/day) (5 days/week) (52 weeks/year) (9.48x10-7)
ES =20.3 MMBtu/year
The annual cost savings, CS, is dependent on the unit cost of heating energy. Since gas
radiant heaters are in use, the unit cost is $3.23/MMBtu
CS = (ES) (Unit cost of heating $ /MMBtu)
= (20.3) (3.23)
= $65.54/year
Implementation Cost
Ductwork to divert the warm air from the heat exchanger to the plant needs to be installed.
The cost of such ductwork depends on geometry, available access and other factors. A
conservative estimate of the material costs is $ 163.6 ($4.09/ft with 40ft of ducting), it was
estimated that 5 hours of labor are required for the installation and so the installation cost
amounts to $125.
The total annual natural gas savings for this AR is 20.3 MMBtu. The implementation
cost of this recommendation is $288.6, and with annual cost savings of $65.54, the
payback period is 4.4 years.

LE0393 – Page 52
Additional Assessment Recommendations
&
Calculation of Cost Savings

LE0393 – Page 53
AAR 1 - Install Weather Seals on Dock Doors
Recommended Action
Install insulation on dock doors to reduce heat loss during winter months.
Annual Cost Savings $2096.38
Payback Period 61.8 months
Natural Gas Savings 649.04 MMBtu
ARC Number 2.7447.3
Current Practice and Observation
There are a total of 22 dock doors with 2 that are unused and 2 that open to an unheated
warehouse which leaves 18 doors that could benefit from the installation of weather seals. The
total gap area from all the 18 doors, assuming a 2 inch gap on both sides with no gap on the
bottom, was calculated at 59.32 ft2. In addition, it was assumed that the indoor temperature was
70°F, while the outside temperature is averaged at 45°F during the winter months.
Anticipated Savings
It was generally noticed that there is a large amount of heat loss through doors that are
not properly protected. By installing seals on and around doors, the heat loss can be
significantly reduced. The heat gain or loss through an open door, HLF, can be estimated from
the accompanying graph (Figure 3) just for the 8 hours of operation. However, the heat loss
should be modified for the number of days the plant works, DW, and the average outside
temperature. The annual heat loss, AHL, is obtained from the following equation:
AHL = (HLF)  (DW/5)  (TI – TO) / 13,
where
HLF = Heat lost through door, 300 MMBtu/yr
DW = Number of operating days per week, 5 days/week
TI = Ambient or room temperature of heated room, 70oF

LE0393 – Page 54
TO = Average temperature of outside air during winter, 45oF.
This plant operates five days per week. There are 3 large doors in the bay that have
gaps and their locations and dimensions are given in the table below.
Location
Number
of
Doors
Number
of Gaps
Height
(ft)
Width of
Gap (ft)
Area,
each gap
(ft2)
Total
Area
(ft2)
Loading Dock Doors
4 8 9
0.167 (2
inches)
1.5 12
2 4 9 1.5 6
1 2 9 1.5 3
Track Department Doors 1 2 10 1.67 3.33
Storage/Track Area
Doors
1 2 10 1.67 3.33
3 6 10 1.67 10
2 4 9 1.5 6
1 2 9 1.5 3
Warehouse Doors
2 4 14 2.33 9.33
1 2 10 1.67 3.33
Total 18 36 59.32
Table 12 : Location and Dimension of the Door Gaps in the Building.
The open areas around the doors were calculated as 59.32 ft2. It is estimated that these
gaps are open for 8 hour each day, and that the inner room is heated for 24 hours each day.
From the attached graph, the heat lost during the winter, HLF, is assumed to be 300 MMBtu/yr
for the 59.32 ft2 gaps.
The temperature near the door was measured to be 70oF. The average outside
temperature (for the heating season only) is about 45oF. For the plant conditions of interest, the
annual heating loss, AHL, is:
AHL = (300)  (5/5)  (70 – 45) / 13
= 576.92 MMBtu/yr.

LE0393 – Page 55
The door seals being considered are generally approximately 90 percent efficient.
Thus, energy savings, ES, is calculated to be:
ES = (AHL)  (EDS) / (ECS),
where
AHL = Annual heat loss, 576.92 MMBtu/yr
EDS = Efficiency of door seals, 90%
ECS = Heating system efficiency, assumed to be 80%.
The energy savings is:
ES = (576.92 MMBtu/yr)  (0.90) / (0.80)
= 649.04 MMBtu/yr.
The annual cost savings, ACS for the doors is:
ACS = (649.04 MMBtu/yr)  ($3.23/MMBtu)
= $2096.38/yr.
Figure 3 : Annual Heat Lost from Doors*
Implementation Cost

LE0393 – Page 56
The cost of installing door seals was given by plant personnel as approximately
$600/door. The total implementation cost for this AR is $10,800.
The annual gas savings for AR 11 is 639.04 MMBtu. The annual cost savings is
$2096.38 and, with $10,800 in implementation costs, the payback period will be about 62
months.

LE0393 – Page 57
AAR 2 - Install automatic sideways door in the docking area
Recommended Action
In the shipping area, there is a shipping door 20 ft x 8 ft. It is opened 2 hours a day.
This door should be replaced with a sideways door.
Annual Cost Saving $302
Gas Savings 93.5 MMBtu/yr
ARC Number 2.7442.3
Currently the loading dock has one 20 ft x 8 ft door. This door fits two trucks side by
side; often only one truck is present in the loading dock. This door is open for approximately
2 hours a day in the winter. During this time the facility loses heat due to infiltration. By
replacing the current door with a door that opens sideways the open area can be reduced to 10
ft x 8ft when only one truck is present. This will decrease the amount of heat loss since it is a
function of area. The shipping dock is heated using two steam space heaters with an
efficiency of approximately 60%.
Anticipated Savings
The annual heating energy savings, HES, and corresponding cost savings, HECS, which
could be achieved by reducing the heating load due to infiltration, can be estimated from the
following relationship:
𝐻𝐸𝑆 =
𝐾𝐻∗𝐴∗𝐸∗𝐹𝐴∗𝐹𝐻
𝐸𝐹𝐹∗𝐶
HECS = HES * available rate of natural gas ($3.23)
Where
KH = infiltration heat loss constant, Btu/yrin2
A = area of infiltration opening, 23,040 in2
E = effectiveness of opening (0.5 for perpendicular, north or south, winds;

LE0393 – Page 58
0.25 for diagonal, east or west winds)1, no units, taken 0.25 for this case
FA = fraction of area reduced, no units, taken 50% for that case
FH = fraction of hours/year that infiltration occurs when doors are open, no units;
.025
(2 hr/day * 109 days/yr (90% of winter) / (8760 hr/year))
EFF = efficiency of heating system, no units, taken 0.6 for this case
C = conversion constant, 1,000,000 Btu/MMBtu
The infiltration heat loss constant is determined from the following relationship:
C
CDHHVρC
KH
2
1p 
 ,
Where
Cp = specific heat of air, 0.24 Btu/lbF
ρ = density of air, 0.07 lb/ft3
V = velocity of average wind, 1050 ft/min
DHH = degree heating days2 (68 F base), 4,416F d/yr
C1 = conversion constant, 1440 min/day
C2 = conversion constant, 144 in2/ft2.
The heating set point temperature for the areas considered is 68F. Thus, the heat loss
coefficient is calculated as follows:
𝐾𝐻 =
(0.24
𝐵𝑡𝑢
𝑙𝑏℉
)(0.07
𝑙𝑏
𝑓𝑡3)(1050
𝑓𝑡
min
)(4,417
℉ 𝑑
𝑦𝑟
)(1440
𝑚𝑖𝑛
𝑑
)
144
𝑖𝑛2
𝑓𝑡2
KH = 779,159 Btu/yr/in2
Energy Savings
Heating energy savings can be calculated as follows:
𝐻𝐸𝑆 =
(779,159
𝐵𝑡𝑢
𝑦𝑟 𝑖𝑛2)(23,040 𝑖𝑛2)(0.25)(0.5)(0.025)
(0.60)(1,000,000
𝐵𝑡𝑢
𝑀𝑀𝐵𝑡𝑢
)
1 ASHRAE Handbookof Fundamentals 1989, p.23.8, Equation 18.
2 Calculated using TypicalMeteorologicalYear (TMY) data.

LE0393 – Page 59
HES = 93.5 MMBtu/yr
HECS = (93.5 MMBtu/yr) * ($3.23/MMBtu)
HECS = $302/yr
Implementation Cost
Typical cost for a 10 x 12 loading dock door is approximately $1000. The facility will
need two of these doors to cover loading dock opening. Therefore material cost will be $2000.
Assuming that installation will take two men a full day to replace the each opening at the in-
house labor of $32/hr. The total labor cost will be $960 giving a total implementation cost of
$2960. The payback period associated with this cost is 2.9 years.
Annual energy savings for this AR is 93.5 MMBtu. Annual cost savings is $302 and,
with Implementation Cost of $2960, the payback period will be 9.8 years.

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