3. Steam Traps - Testing
• Recommended Steam Trap Testing Intervals
• High-Pressure (150 psig and above): Weekly to Monthly
• Medium-Pressure (30 to 150 psig): Monthly to Quarterly
• Low-Pressure (below 30 psig): Annually
• Four basic ways to test steam traps
• temperature
• sound
• visual
• electronic
www.energy.gov
4. Steam Trap Losses
Leaking Steam Trap Discharge Rate*
Trap Orifice Steam Loss, lb/hr
Dia (in) Steam Pressure, psig
15 100 150 300
1/32 0.85 3.3 4.8 –
1/16 3.4 13.2 18.9 36.2
1/8 13.7 52.8 75.8 145
3/16 30.7 119 170 326
1/4 54.7 211 303 579
3/8 123 475 682 1,303
www.energy.gov
Minimal or no maintenance for 3 to 5 years -
between 15% to 30% of the installed steam
traps may have failed
Regularly scheduled maintenance program -
leaking traps should account for less than 5%.
5. Steam Trap Losses - Example
Steam Costs - $10 per thousand pounds ($10.00/1,000 lb)
Steam Trap Leak – 150 psi (g) steam line is stuck open
Trap Orifice - 1/8th inch
Estimated steam loss as 75.8 pounds per hour (lb/hr)
Annual Savings = 75.8 lb/hr x 8,760 hr/yr x $10.00/1,000 lb = $6,640
www.energy.gov
6. The U.S. DOE Energy Efficiency & Renewable Energy Program provides a steam system modeler tool to determine the
amount of fuel energy required to produce steam, visit the website at:
https://www4.eere.energy.gov/manufacturing/tech_deployment/amo_steam_tool/equipBoiler?random=superSteam
Boiler Horsepower = F x M
0.83 x 34.5
F = Pounds of feed per hour
M = Percent of moisture to be added by steam
0.83 = An approximate correction factor for make-up water at 50 degrees F.
34.5 = The amount of water evaporated in one hour at 212°F, which equals one boiler horsepower.
Boiler Calculations
7.
8. Steam Pipe Insulation Losses
A survey of a plant steam system found 120 feet of bare half of an inch steam line and 70 feet of bare
1 inch steam line operating at 150 psi and 230 feet of bare 2 inch steam line operating at 30 psi. The
value of the steam was $2.40 per MBTU (from natural gas as boiler fuel). The amount of heat loss per
year was determined from the following figure. The heat loss as determined from chart is:
½ inch line = 120 x 300/100 = 360 MBTU/ yr.
1 inch line = 70 x 430/100 = 301 MBTU/ yr.
2 inch line = 230 x 370/100 = 851 MBTU/ yr.
Total Heat Loss = 1512 MBTU/ yr.
The amount of heat saved by insulating bare steam lines depends on the type of insulation and other
variables. If its assumed that a 95 percent efficiency—a value reasonably achieved in insulation
installations—is obtained, then:
The cost saving = 1512 MBTU/ yr. x .95
x $2.40/MBTU= $3,447 /yr.
9. Compressed Air Leaks
Recognized and Underappreciated
• Waste as much as 20%-30% of the
compressor’s output.
• Fluctuating system pressure, which can
cause air tools and other air-operated
equipment to function less efficiently,
possibly affecting production
• Excess compressor capacity, resulting in
higher than necessary costs
• Decreased service life and increased
maintenance of supply equipment
(including the compressor package) due to
unnecessary cycling and increased run time.
www.energy.gov
11. Compressed Air Leaks
www.energy.gov
Leakage ratesa(cfm) for different supply pressures and approximately equivalent orifice sizesb
Pressure Orifice Diameter (inches)
(psig) 1/64 1/32 1/16 1/8 1/4 3/8
70 0.29 1.16 4.66 18.62 74.4 167.8
80 0.32 1.26 5.24 20.76 83.1 187.2
90 0.36 1.46 5.72 23.10 92 206.6
100 0.40 1.55 6.31 25.22 100.9 227
125 0.48 1.94 7.66 30.65 122.2 275.5
a For well-rounded orifices, values should be multiplied by 0.97 and by 0.61 for sharp ones.
Leakage rates identified in cubic feet per minute (cfm) are proportional to the square of the orifice diameter.
13. Compressed Air Leaks - Example
100 leaks 1/32” at 90 psig
50 leaks of 1/16” at 90 psig
10 leaks of 1/4” at 100 psig.
7,000 annual operating hours
Electric rate of $0.05 kilowatt-hour (kWh)
Compressed air generation requirement ~ 18 kilowatts (kW)/100 cfm.
Cost savings = # of leaks x leakage rate (cfm) x kW/cfm x # of hours x $/kWh
Cost savings from 1/32” leaks = 100 x 1.46 x 0.61 x 0.18 x 7,000 x 0.05 = $5,611
Cost savings from 1/16” leaks = 50 x 5.72 x 0.61 x 0.18 x 7,000 x 0.05 = $10,991
Cost savings from 1/4” leaks = 10 x 100.9 x 0.61 x 0.18 x 7,000 x 0.05 = $38,776
Total cost savings from eliminating these leaks = $57,069
Note that the savings from the elimination of just 10 leaks of 1/4” account for almost 70% of the overall savings. As
leaks are identified, it is important to prioritize them and fix the largest ones first.
www.energy.gov
14. Compressed Air – Self Analysis
Identification and tagging, tracking, repair, verifying, and employee involvement. Set a reasonable target for cost-
effective leak reduction:
**5%-10% of total system flow is typical for industrial facilities.
Compressed air is one of the most expensive sources of energy in a plant. The over-all efficiency of a typical
compressed air system can be as low as 10%-15%. For example, to operate a 1 hp air motor at 100 psig, approximately
7-8 hp of electrical power is supplied to the air compressor.
Cost ($) = (bhp) x (0.746) x (# of operating hours) x ($/kWh) x (% time) x (% full-load bhp)
Motor Efficiency
bhp—Motor full-load horsepower (frequently higher than the motor nameplate horsepower—check equipment specification)
0.746—conversion between hp and kW
% time—percentage of time running at this operating level
% full-load bhp—bhp as percentage of full-load bhp at this operating level
Motor efficiency—motor efficiency at this operating level
www.energy.gov
15. Compressed Air – Self Analysis Example
200-hp compressor (which requires 215 bhp)
6800 operating hours annually.
Fully loaded 85% time (motor efficiency = 0.95)
Unloaded 15% time (25% full-load bhp and motor efficiency = 0.90)
Electric rate is $0.05/kWh.
Cost when fully loaded =
(215 bhp) x (0.746) x (6800 hrs) x ($0.05/kWh) x (0.85) x (1.0) =
0.95
=$48,792
Cost when unloaded =
(215 bhp) x (0.746) x (6800 hrs) x ($0.05/kWh) x (0.15) x (0.25) =
0.90
=$2,272
Annual energy cost = $48,792 + $2,272 = $51,064
www.energy.gov
16. A good example of excess costs from inadequate maintenance can be seen with pipeline filter elements. Dirty filters increase pressure
drop, which decreases the efficiency of a compressor. For example, a compressed air system that is served by a 100-horsepower (hp)
compressor operating continuously at a cost of $0.08/kilowatt-hour (kWh) has annual energy costs of $63,232. With a dirty coalescing filter
(not changed at regular intervals), the pressure drop across the filter could increase to as much as 6 pounds per square inch (psi), vs. 2 psi
when clean, resulting in a need for increased system pressure. The pressure drop of 4 psi above the normal drop of 2 psi accounts for 2% of
the system’s annual compressed air energy costs, or $1,265 per year. A pressure differential gauge is recommended to monitor the
condition of compressor inlet filters. A rule of thumb is that a pressure drop of 2 psi will reduce the capacity by 1%.
Although it varies by plant, unregulated usage is commonly as high as 30 to 50 percent of air demand. For systems in the 100 psig range
with 30 to 50 percent unregulated usage, a 2 psi increase in header pressure will increase energy consumption by about another 0.6 to
1.0 percent because of the additional unregulated air being consumed. The combined effect results in a total increase in energy
consumption of about 1.6 to 2 percent for every 2 psi increase in discharge pressure for a system in the 100 psig range with 30
to 50 percent unregulated usage.
17. Motors
Motors do not use energy when turned off. Reducing motor operating time by just 10% usually saves more energy than replacing a standard
efficiency motor with a premium efficiency motor. In fact, given that more than 97%1 of the life cycle cost of purchasing and operating a motor
in a typical installation is energy related, turning a motor off 10% of the time could reduce energy costs enough to purchase several new
motors.
Just one minute of additional running time consumes far more energy than a motor starting event.
Peak monthly demand charges are generally based on a facility’s average energy use over a fixed or rolling average window of 15 to 60 minutes
in duration. Only when starting large or large groups of motors at once should the user be concerned with peak demand charges. Check with
your utility to determine how it assesses peak demand charges.
Starting stresses a motor by:
Applying higher than rated full-load torque to the shaft during acceleration
Applying high magnetic forces to the rotor cage and winding end turns
Heating the stator winding and the rotor cage.
Frequent torque shocks from starting could shorten shaft life through metal fatigue. However, most shaft failures are attributed to bearing
failures, excessive belt tension, misapplication, or creep during storage (large motors). Overheating the stator winding and the rotor cage
occurs if the hourly number of starts exceeds the NEMA recommendations or the duration of rest time between starts is less than the NEMA
allowable value. Heat from exceeding these limits can degrade winding insulation and cause thermal stressing of the rotor cage, leading to
cracks and failed end-ring connections.
18. Annual Savings for Premium versus Old Standard Efficiency Motors
Motor Efficiency Annual Savings from Using a
at 75% Load Premium Efficiency Motor
Old Standard Premium
Horsepower Efficiency Efficiency Annual Energy Dollar Savings
Motor Motor Savings, kWh $/year
10 86.7 92.2 3,105 250
25 89.9 93.8 5,160 410
50 91.6 95.0 8,630 690
100 92.2 95.3 15,680 1,255
200 93.3 96.2 29,350 2,350
Note: Based on purchase of a 1,800 RPM TEFC motor in operation 8,000 hours per year (hrs/year) at 75% load at an electrical rate of $0.08/kilowatt-hour (kWh).
www.energy.gov
23. 1. Monthly cost per kW/hr. from power bill monthly kW/hr. usage = cost per kW.
2. Number of power meters ____________ .
3. If more than 1 meter, cost savings to eliminate additional meters savings.
4. Complete power factor tested under normal load conditions__________________.
5. Calculated power factor savings if less than 90 percent _________ savings
6. Peak demand less average demand in kW.
Peak__________ - Average _______ = Demand Opportunity ________________
7. Demand cost/Kw from supplier.
Demand Opportunity ______ x Cost/Kw ________ = Potential Savings__________
8. Air compressor leak test. Record pressure at compressor with no air users operating.
15 min_____ 30 min_____ 45 min_____ 60 min_____
9. If more than a 10 lb drop in 1 hour, identify all air leaks. Number __________
10. Does air compressor draw air from outside? Yes ___________ No__________
11. Record watts of lights used, but not necessary.
Watts per hr./day_______ / 1000 = kW/hr.________
12. List equipment systems and number of HP hours running day, but not being used.
HP. Running System Not Used Hr./Day
24. Steam Generation
1. Percentage of steam lines insulated ____________ %.
2. Percentage of condensate lines insulated _________
%.
3. Does Condensate return to the feedwater tank?
Yes __________ No ______
4. Complete an inspection of steam traps and note any
that are leaking.
Traps leaking_____ / total traps = ______percent leaking
5. List liquid tanks and heating source.
Is Tank Insulated? Product Stored?
Heating Method?
Temperature Control Range
Product Temperature
6. Natural gas/propane meter reading with all gas uses off.
Reading Time
________ 30 min
________ 60 min
7. Are the boiler insulation and shell in good condition?
Yes_______No_______
8. Identify any steam openings into sewer or into the
atmosphere.
Number of openings ________
9. Number of space heaters in service and not required.
_______________
10. Test boiler efficiency. ______(Should be above 82 percent).
11. Measure average thickness of scale on boiler tubes
(1/32 in = 7% fuel waste. 1/16 in. = 13% fuel waste).
12. Natural gas cost/1000 cu ft on propane cost/gal
NG__________ Propane_______
25. Compressed Air System Evaluation
Date: ___________________ Completed by:_______________________
1. Evaluate all equipment using air and determine the highest air pressure required in plant. Set compressor air to
that pressure.
2. Shut down entire plant so that every area is quiet.
3. Run air compressor so the lines are at maximum pressure.
4. Shut off compressor.
5. Walk through plant and mark all leaks with caution tape or anything to make it visible.
6. Write general location of leaks below.
7. Record time that it takes until pressure has decreased to 40psi.
Example:
If maximum required pressure is 100psi and it takes 20 minutes until pressure is 40psi.
100psi - 40psi = 60psi lost in 20 minutes.
60psi/20min = 5psi/min
Recording Time and Pressure Drop
End time_____________- Start time_______________=Total time
Bagging pressure________________psi - 40psi. = __________psi pressure drop
Pressure drop_____________psi/__________min (total time) =__________min
Repeat after leaks are fixed and record improvement.
1. ______________________________________________________
2. ______________________________________________________
3. ______________________________________________________
4. ______________________________________________________
5. ______________________________________________________
6. ______________________________________________________
