Energy Efficient Compressed Air Systems Compressed Air Fundamentals

Energy Efficient Compressed Air Systems

Compressed Air Fundamentals § d. Welec = ∫V d. P / [Effcompressor Effmotor Effcontrol] § V = volume flow rate § DP = pressure rise § Eff = efficiencies of compressor, motor, and control Internal cooling decreases electrical power

Compressed Air Savings Opportunities § d. Welec = ∫V d. P / [Effcompressor Effmotor Effcontrol] § Reduce volume flow rate § Reduce pressure rise § Increase cooling during compression § Increase compressor efficiency § Increase motor efficiency § Increase control efficiency

Compressed Air System

Screw Compressor Operation

Compressed Air System Savings Opportunities • End use – Eliminate inefficient uses of compressed air – – • Eliminate air pumps, agitation, cooling and suction Use blower for low-pressure applications Install solenoid valves to shut off air Install air saver nozzles Install differential pressure switches on bag houses • Distribution – Fix leaks – Decrease pressure drop in distribution system • Compressor System – – – Compress outside air Use refrigerated dryer Direct warm air into building during winter Use load/unload control with auto shutoff or VSD for lag compressor Stage compressors with pressure settings or controller Add compressed air storage to increase auto shutoff

Plant and Compressor Data For Calculating Example Savings • • • Plant operates 6, 000 hours per year Electricity cost including demand = $0. 10 /k. Wh Air compressor produces 4. 2 scfm/hp Air compressor motor is 90% efficient Air compressor runs in load/unload control with FP 0 = 0. 50

Eliminate Inefficient Uses of Compressed Air

Replace Air Pump with Electric Pump § Air motors use 7 x more electricity than electrical motors § Example § Replace 10 x 1 -hp air pumps with electric pumps § Cost Savings = 10 hp / 0. 90 x 6/7 x 0. 75 k. W/hp x 6, 000 hr/yr x $0. 10 /k. Wh = $4, 300 /yr

Replace Compressed Air with Mechanical Agitation § Compressed air agitation uses ~10 x more electricity than mechanical agitation § Example § Replace air agitation from 0. 5 -in pipe at 50 psig with mechanical agitator. § Flow Savings = 11. 6 (scfm/lbf) x [1 (in)]2 x 55 psia = 150 scfm § Power Savings = 150 scfm / (4. 2 scfm/hp x 0. 90) x 0. 75 k. W/hp x (1 -0. 50) = 15 k. W § Cost Savings = 15 k. W x 6, 000 hr/yr x $0. 10 /k. Wh = $9, 000 /yr

Replace Compressed Air Cooling with Chiller § ‘Vortex’ coolers use 4 x more electricity than electric chillers § Example § Replace 10, 200 Btu/hr cooler using 150 scfm with chilled water § Cost Savings = 150 scfm / (4. 2 scfm/hp x 90%) x 75% x 0. 75 k. W/hp x (1 -0. 50) x 6, 000 hr/yr x $0. 10 /k. Wh = $6, 700 /year

Replace Air With Heat Pipe Cabinet Coolers § Air coolers use 8 x more electricity than (heat pipe + small fan) coolers § Example § Replace 2, 400 Btu/hr compressed air cooler with heat pipe cooler § Cost comp air = 2, 400 Btu/hr / 80 (Btu/hr)/cfm / (4. 2 cfm/hp x 90%) x 0. 75 k. W/hp x (1 -0. 50) x 8, 760 hr/yr x $0. 10 /k. Wh = $2, 607/year § Cost heat pipe = 0. 6 A x 120 V / 1, 000 VA/k. W x 8, 760 hr/yr x $0. 10 /k. Wh = $ 63 /year § Implementation Cost = $1, 000 Source: www. airtxinternational. com and www. systemdirections. com

Replace Pneumatic Suction Cups with Magnets § Suction cups use 6. 0 scfm while holding while magnets use average of 0. 3 scfm § Example § Replace cups with magnets if holding 3, 000 hours per year. § Power Savings = 5. 7 scfm / (4. 2 scfm/hp x 0. 90) x 0. 75 k. W/hp x (1 -0. 50) = 0. 57 k. W § Cost Savings = 0. 57 k. W x 3, 000 hr/yr x $0. 10 /k. Wh = $167 /yr

Use Blowers for Low-Pressure Applications § Blowers generate 7. 2 scfm /hp at 20 psig while compressors generate 4. 2 scfm/hp at 100 psig § Example § Install low-pressure blower for application needing 140 scfm of 20 psig air § Power Savings = 140 scfm x (1/4. 2 – 1/7. 2) hp/scfm / 0. 90 x. 75 k. W/hp x (1 -0. 50) = 5. 8 k. W § Cost Savings = 5. 8 k. W x 6, 000 hr/yr x $0. 10 /k. Wh = $3, 472 /yr

Reduce End Use Compressed Air Demand

Reduce Blow-off with Solenoid Valves § Flow from open tube (scfm) = 11. 6 (scfm/lbf) x [Diameter (in)] 2 x Pressure (psia) § Example § Install solenoid to shut-off blowoff from 3/8 -in pipe at 100 psig 80% of time § Flow Savings = 11. 6 (scfm/lbf) x [3/8 (in)]2 x 115 psia x 80% = 150 scfm § Power Savings = 150 scfm / (4. 2 scfm/hp x 0. 90) x 0. 75 k. W/hp x (1 -0. 50) = 15 k. W § Cost Savings = 15 k. W x 6, 000 hr/yr x $0. 10 /k. Wh = $8, 933 /yr § Cost of 3/8 -inch solenoid valve = $100

Reduce Blow off with Air-Saver Nozzles § Nozzles maximize entrained air and generate same flow and force with ~50% less compressed air § Example § Add nozzle to 1/8 -in tube at 100 psig § Flow Savings = 11. 6 (scfm/lbf) x [1/8 (in)]2 x 115 psia x 50% = 10. 4 scfm § Power Savings = 10. 4 scfm / (4. 2 scfm/hp x 0. 90) x 0. 75 k. W/hp x (1 -0. 50) = 1. 0 k. W § Cost Savings = 1. 0 k. W x 6, 000 hr/yr x $0. 10 /k. Wh = $620 /yr § Nozzles cost about $10 each

Activate Bag House Air Pulses Using Pressure Differential Instead of Timer § Timers are designed for peak conditions, where demand-based control matches actual conditions § Example § Install differential pressure control to reduce timed pulse from 34 cfm by 60% § Flow Savings = 34 cfm x 60% = 20. 4 cfm § Power Savings = 20. 4 cfm / (4. 2 scfm/hp x 0. 90) x 0. 75 k. W/hp x (1 -0. 50) = 2. 0 k. W § Cost Savings = 2. 0 k. W x 6, 000 hr/yr x $0. 10 /k. Wh = $1, 214 /yr

Fix Leaks

Fix Leaks § Leaks continuously drain power § Leakage rate increases exponentially with leak diameter § Example § Fix one 1/64” leak § Power Savings =. 25 cfm / (4. 2 scfm/hp x 0. 90) x 0. 75 k. W/hp x (1 -0. 50) = 0. 025 k. W § Cost Savings = 0. 025 k. W x 6, 000 hr/yr x $0. 10 /k. Wh = $15 /yr

Fix Leaks § Many plants lose ~20% of compressed air to leaks § Example § Fix leaks in plant with fully loaded 100 -hp compressor and 20% leakage § Power Savings = 100 hp / 0. 90 x 0. 20 x 0. 75 k. W/hp x (1 -0. 50) = 8. 3 k. W § Cost Savings = 8. 3 k. W x 6, 000 hr/yr x $0. 10 /k. Wh = $5, 000 /yr

Identify Leaks Using Ultrasonic Sensor

Quantify Leakage By Logging Flow or Compressor Power

Fix Leaks Frequently § Leak Loss = Rate x Time § Repairing leaks frequently cuts leak load at same implementation cost § Example § Fix leaks every 2 weeks instead of every 4 weeks if one new 1 -cfm leak per week § Reduces leakage by 40%

Starve Leaks by Shutting off Branch Headers § Valve on branch header can starve all downstream leaks when area is not in use § Example § Install valve to shut off header with 200 cfm leak load for 4, 000 hr/yr § Power Savings = 200 scfm x 50% / (4. 2 scfm/hp x 0. 90) x 0. 75 k. W/hp x (1 -0. 50) = 9. 9 k. W § Cost Savings = 9. 9 k. W x 4, 000 hr/yr x $0. 10 /k. Wh = $3, 969 /yr

Use Rubber In Place of Braded Hose Braided hoses dry-rot and develop leaks that can’t be detected with ultrasonic sensors.

Reduce Distribution System Pressure Drop

Use Looped Piping System §If DP < 10 psi at farthest end use, use looped rather than linear design §Design Guidelines §Main line: size from average cfm to get DP = 3 psi §Branch line: size from cfm peak to get DP = 3 psi §Feed lines: size from peak cfm to get DP = 1 psi §Select hose with DP < 1 psi

Avoid ‘Collision’ Connections

Maintain Filters • Place filter upstream of dryer to protect dryer • DP filter < 1 psi

Size Dryer for DP< 5 psi Low Flow: DP = 1 psi High Flow: DP = 6 psi

Then, Reduce Compressed Air Pressure § Work = V DP, thus compressor requires less work to produce air at lower outlet pressure § Fraction savings from reducing pressure = 1% per 2 psi pressure reduction § Example § Reduce pressure setting of fully-loaded 100 -hp compressor from 110 to 100 psig § (P 2 high/P 1)0. 286 = [(110 psig +14. 7 psia) / 14. 7 psia]0. 286 = 1. 84 § (P 2 low/P 1)0. 286 = [(100 psig +14. 7 psia) / 14. 7 psia]0. 286 = 1. 80 § Fraction savings = (1. 84 – 1. 80) / (1. 84 – 1) = 5. 2 % § Cost Savings = 100 hp x 5. 2% / 90% x 0. 75 k. W/hp x 6, 000 hr/yr x $0. 10 /k. Wh = $2, 582 /yr

Reduce Pressure To Maximum End Use Plus Friction Loss

Dry Air Efficiently

Refrigerated and Desiccant Dryers § Refrigerated dryer: § Dries air by cooling § Cools to Tdew-point = 35 F § Uses 6 W/scfm § Desiccant dryer: § Dries air by passing through desiccant, then purging desiccant of water § Cools air to Tdew-point = -40 F to -100 F § Uses 16 to 30 W/scfm Desiccant dryers “should one be applied to portions of compressed air systems that require dew points below 35 F. Because desiccant dryers require a higher initial investment and higher operating costs, Kaeser strongly recommends using refrigerated dryers whenever practical. ” Kaeser Regenerative Desiccan Dryers

Desiccant Dryer Purging § Three types of purge § Compressed air purge § Uses 15% of compressed air for purging § Total is about 30 W/scfm § Heated compressed air purge § Uses 7% of compressed air for purging § Plus 7 W/scfm for heating § Total is about 22 W/scfm § Heater blower air purge § Uses 3 W/scfm for blower § Plus 13 W/scfm for heating § Total is about 16 W/scfm § Purge cycle can be timed or demand-controlled Source: www. aircompressors. com

Use Refrigerated Rather than Desiccant Dryer § Example: § Replace desiccant dryer using compressed air purge with refrigerated dryer for 200 hp (840 scfm) compressor § Desiccant Power = 840 scfm x 15% / (4. 2 scfm/hp x 90%) x 0. 75 k. W/hp x (1 -0. 50) = 12. 5 k. W § Refrigerated Power = (840 scfm x 85% x 0. 006 k. W/scfm x (1 -0. 50) = 2. 1 k. W § Cost Savings = (12. 5 k. W – 2. 1 k. W) x 6, 000 hr/yr x $0. 10 /k. Wh = $6, 215 /yr

Use Demand-Control Rather than Timed Purge § Summer air 4 x wetter than winter air. § Timed purge set for peak (summer) conditions § Example § Switch from timed to demandcontrol purge and reduce purge by 50% on dryer for 200 hp (840 scfm) compressor § Timed-Purge Power = 840 scfm x 15% / (4. 2 scfm/hp x 90%) x 0. 75 k. W/hp x (1 -0. 50) = 12. 5 k. W § Cost Saving = 12. 5 k. W x 50% x 6, 000 hr/yr x $0. 10 /k. Wh = $3, 250 /yr . 016 lbw/lba. 004 lbw/lba

Replace Timed-solenoid with No-loss Drains § Winter air holds 50% less water than summer air § Timers designed for peak conditions, where demand-based control matches actual conditions § Example § Replace 3/8 -inch timed-solenoid drain that opens 3 seconds every 30 seconds with noloss drain that eliminates <90% of air losses. § Flow Savings = 11. 6 (scfm/lbf) x [3/8 (in)]2 x 115 (psia) x 10% x 90% = 16. 9 scfm § Power Savings = 16. 9 scfm / (4. 2 scfm/hp x 0. 90) x 0. 75 k. W/hp x (1 -0. 50) = 1. 68 k. W § Cost Savings = 1. 68 k. W x 6, 000 hr/yr x $0. 10 /k. Wh = $1, 005 /yr § 3/8 -inch no-loss drains costs $600

Optimize Compressor Cooling

Compress Outdoor Air § Compressing cool dense air reduces compressor work: § Fraction Savings = (Thi - Tlow) / Thi § Fraction Savings = ~ 2% per 10 F § Example § Install PVC piping to duct outside air at 50 F to compressor rather than inside air at 80 F. § Fraction Savings = [(80 + 460) - (50 + 460)] / (80 + 460) = 5. 9% § Cost Savings = 20 k. W x 5. 9% x 6, 000 hr/yr x $0. 10 /k. Wh = $706 /yr

Direct Warm Air Into Plant During Winter § 75% of compressor input power lost as heat § Example § Add duct work to direct warm air into plant during winter for compressors drawing 105 k. W if heating system operates 2, 000 hours per year and is 80% efficient § Heat Load Savings = 105 k. W x 75% x 3, 413 Btu/k. Wh x 2, 000 hours/year = 537 mm. Btu/yr § Cost Savings = 537 mm. Btu/year / 80% x $10 /mm. Btu = $6, 719/year

Employ Efficient Compressor Control FP = FP 0 + (1 -FP 0) FC and P = Prated FP

Power Signatures from Modulation and Load/unload Control

Savings From Switching To Efficient Control § FP = FP 0 + (1 -FP 0) FC § Typical Intercepts § FP 0 bypass = 1. 00 § FP 0 modulation = 0. 70 § FP 0 load/unload = 0. 50 § FP 0 variable-speed = 0. 10 § FP 0 on/off = 0. 00 § Example § Switch 100 -hp compressor at 50% capacity from modulation to variable-speed control. § FP (modulation) = 0. 70 + (1 - 0. 70). 50 =. 85 § FP (variable-speed) = 0. 10 + (1 - 0. 10). 50 =. 55 § Savings = 100 hp x (. 85 -. 55) /. 90 x 0. 75 k. W/hp x 6, 000 hr/yr x $0. 10 /k. Wh = $15, 000 /yr

Savings Penalty for Inefficient Control § Consider savings from reducing fraction capacity (FC) § P 1 = Prated x FP 1 = Prated x [FP 0 + (1 -FP 0) FC 1] § P 2 = Prated x FP 2 = Prated x [FP 0 + (1 -FP 0) FC 2] § Psave = P 1 – P 2 § Psave = Prated x [FP 0 + (1 -FP 0) FC 1] - Prated x [FP 0 + (1 -FP 0) FC 2] § Psave = Prated x (FC 1 - FC 2) x (1 -FP 0) § Psave = Unadjusted savings x (1 -FP 0) § Actual Savings = Unadjusted Savings x (1 – FP 0) § Example § Calculate actual savings for reducing leaks by 100 scfm if compressor operates in modulation control. § Unadjusted savings = 100 scfm / (4. 2 scfm/hp x 0. 90) x 0. 75 k. W/hp x 6, 000 hr/yr x $0. 10 /k. Wh = $11, 905 § Actual savings = $11, 905 /yr x (1 – 0. 70) = $3, 571 /yr

Modulation to Load/Unload with Autoshutoff Reduced power 35% and saved $17, 000 /yr

Centrifugal Compressor Control

Energy-Efficient Centrifugal Compressor Control • Minimize/eliminate by-pass using effective multi-compressor control • Operate compressor with throttling/unload control if available • Adjust throttling/surge pressure points to widen throttling range

Throttling/Surge Pressure Set Points Narrow or Widen Throttling Band

Poor Throttling/Surge Pressure Set Points Inlet butterfly valve throttles to 95%, then bypass

Excellent Throttling/Surge Pressure Set Points Inlet guide vane throttles to 65%, then bypass

Multi-compressor Control • Cascading pressure set-point control • Single-pressure network control • Smart PLC control

Cascading Pressure Set-Point Control § “Stage” compressors into lead and lag compressor(s) by sequentially reducing load/unload pressures. § Designate compressor with best control efficiency as final lag compressor § Staging allows Lead to run fully loaded and the Lag compressor(s) to turn off or run with smaller partload penalty if they have better control efficiency. § Simple, effective, and inexpensive § However: § Increases pressure § Only applicable for compressors in same location

Single-Pressure Network Control • When compressors in different locations, staging compressors using cascading pressure set points doesn’t work, since the compressors see different pressures. • When more than three compressors are staged using cascading pressure set points, pressure increased and pressure range is large. • In these cases, use a sequencer with common pressure sensor to stage compressors.

Smart PLC Control • Monitors pressure, flow and power • Most flexible (controls different compressor types, locations, manufacturers) • Trends data • Accommodates changes Source: Taming Multiple Compressors, Niff Ambrosino and Paul Shaw, www. plantservices. com

Two Compressors Properly Staged

Savings From Staging Compressors Example: Stage load/unload 100 -hp and 50 -hp compressors producing 400 scfm § Cr, 100 = 100 hp x 4. 2 scfm/hp = 420 scfm § Cr, 50 = 50 hp x 4. 2 scfm/hp = 210 scfm § Unstaged § FC = 400 scfm / (420 scfm + 210 scfm) = 0. 63 § P 100 = Pr, 100 x FP 100 = 100 hp x [0. 50 + (1 - 0. 50). 63] = 81. 7 hp § P 50 = Pr, 50 x FP 50 = 50 hp x [0. 50 + (1 - 0. 50). 63] = 40. 9 hp § Ptotal = P 100 + P 50 = 81. 7 hp + 40. 9 hp = 122. 6 hp § Staged with auto shutoff § FC 100 = 400 scfm / 420 scfm = 0. 95 § P 100 = Pr, 100 x FP 100 = 100 hp x [0. 50 + (1 - 0. 50). 95] = 97. 6 hp § P 50 = 0 hp (with auto shutoff) § Ptotal = P 100 + P 50 = 97. 6 hp + 0 hp = 97. 6 hp § Savings = (122. 5 hp – 97. 6 hp) /. 90 x 0. 75 k. W/hp x 6, 000 hr/yr x $0. 10 /k. Wh = $12, 500 /yr §

Savings From Staging Compressors Before Staging After Staging § Example: Stage unstaged load/unload 100 -hp and 50 -hp compressors producing 400 scfm

Compressed Air Storage • Necessary for load/unload compressors • Lengthens load/unload cycle time: – Reduces compressor wear – Allows sump to completely blowdown, which reduces unloaded power – Increases auto-shutoff, which reduces unloaded power • Acts as additional compressed air capacity to meet demand spikes – Reduces pressure variation to process – Allows average pressure to be lowered, which reduces compressor energy use

Locating Compressed Air Storage • Locating storage after dryer protects dryer from compressed air demand spikes that can prevent dryer from effectively removing moisture.

Sizing Compressed Air Storage § Relationship between volume, flow and pressure from mass balance. § Mflowin – Mflowout = d. M/dt § Vflowin r – Vflowout r = (RT/PV) / dt § 2. 7 gal/rated scfm (0. 36 ft 3/rated scfm) or 11. 3 gal/rated hp (1. 5 ft 3/rated hp) of trim compressor guarantees load/unload cycle time > 1 minute with 10 psi pressure band § Example § Calculate storage for 100 hp compressor so load/unload cycle time > 1 minute with 10 psi pressure band. § 100 hp x 4. 2 scfm/hp x 2. 7 gal/scfm =1, 134 gal

Power During Load and Sump Blowdown § Initial rapid power increase when compressor loads § Subsequent slow power increase as pressure builds from load to unload pressure set points. § Initial rapid power decrease as compressor unloads § Subsequent slow power decrease as pressure in sump is bled down to near atmospheric pressure to reduce back pressure. § Blowdown time ~ 30 seconds

Add Storage to Lengthen Load/Unload Cycle and Enable Full Blowdown Example • Adding storage reduces average power by 2. 9% • If 100 -hp compressor at 50% load, savings are: 50 hp / 0. 90 x 2. 9% x 0. 75 k. W/hp x 6, 000 hr/yr x $0. 10 /k. Wh = $725 /yr

Power Savings From Adding Storage (to Achieve Full Blowdown) For flooded compressor in load/unload mode at FC = 50% and blowdown = 30 sec

Add Storage to Enable Auto Shutoff Example • Adding storage enables auto shutoff and reduces average power by 14. 7% • If 100 -hp compressor at 50% load, savings are: 50 hp / 0. 90 x 14. 7% x 0. 75 k. W/hp x 6, 000 hr/yr x $0. 10 /k. Wh = $3, 675 /yr

Add Primary Storage and Reduce Pressure §Plant requires 100 psig §Small storage doesn’t dampen L/UL swing §Pset = 100107 §Large storage dampens L/UL swing §Pset = 97107

Add Secondary Storage and Reduce Pressure Example • Adding secondary storage allows reducing pressure by 6 psi which reduces compressor energy by 3% • If 100 -hp at 50% load, compressor savings are: 50 hp / 90% x 3% x 0. 75 k. W/hp x 6, 000 hr/yr x $0. 10 /k. Wh = $750 /yr

U. D. Air. Sim Software • Air. Sim simulates compressed air systems to understand the dynamics between compressed air control, capacity, storage and demand. • The software is available free of charge at: http: //academic. udayton. edu/kissock/h ttp/RESEARCH/Energy. Software. htm

D. O. E. Air. Master+ Software • AIRMaster+ provides a systematic approach for assessing the supply-side performance of compressed air systems. • AIRMaster+ evaluates the energy savings potential of any or all of the following eight energy efficiency actions: • • Reduce air leaks Improve end-use efficiency Reduce system air pressure Use unloading controls Adjust cascading set points Use automatic sequencer Reduce run time Add primary receiver volume • http: //www 1. eere. energy. gov/industry/ bestpractices/

Summary of Key Equations and Relations • • • Input power (k. W) = Voltage (V) x Current (A) x 1. 73 x Power factor (k. W/k. VA) / 1, 000 VA/k. VA Annual energy use (k. Wh/yr) = Input power (k. W) x Operating hours (hr/yr) Annual electricity cost ($/yr) = Annual energy use (k. Wh/yr) x Unit electricity cost ($/k. Wh) Flow from open tube (scfm) = 11. 6 (scfm/lbf) x Pressure (psig) x [Diameter (in)] 2 Input power from flow (k. W) = Flow (scfm) x 0. 75 k. W/hp / (Specific output (scfm/hp) x Motor efficiency) x (1 -FP 0 ) Typical compressor/blower specific output: 4. 5 scfm/hp at 100 psig 7. 2 scfm at 20 psig Savings from reducing operating pressure ~ 0. 5% per psi Savings from reducing intake air temperature ~ 2% per 10 F Refrigerated dryer electricity use ~ 4 -6 W/scfm Unheated desiccant dryer air use ~ 15% of flow Recoverable heat from air compressors ~ 75% of electrical power (k. W) x 3, 412 (Btu/k. Wh) Fraction Power = [(Fraction Capacity x (1 – Fraction Power at No Load)] + Fraction Power at No Load – Typical Fraction Power at No Load (Modulation Control) = 0. 70 – Typical Fraction Power at No Load (Load/unload Control) = 0. 30 - 0. 60 – Typical Fraction Power at No Load (Variable Speed Drive) = 0. 10 – Typical Fraction Power at No Load (On/Off) = 0. 0

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Compressed Air Storage • Lengthens load/unload cycle time: – Reduces compressor wear – Allows sump to completely blowdown, which reduces unloaded power – Increases auto-shutoff, which reduces unloaded power • Reduces pressure variation to process – Allows average pressure to be lowered, which reduces compressor energy use • 2. 7 gal/rated scfm of trim compressor (1. 5 ft 3/hp) guarantees load/unload cycle time > 1 minute

Dryer Pressure Drop Low Flow DPDRYER = 1 psi DPCOMP = 8 psi High Flow DPDRYER = 6 psi DPCOMP = 6 psi

Dryer Pressure Drop Causes Short Cycling • 10 psi DP at compressor reduced to 5 psi “effective DP”

Locating Primary Storage • If DP across dryer is small, locate primary storage downstream of dryer to reduce flow variation through dryer and thoroughly dry air • If DP across dryer is large, locate primary storage upsteam of dryer to enable compressor to realize full load/unload pressure range

Inlet Guide Vane vs Inlet Butterfly Valve Throttling Source: Ingersol Rand Centac Compressor Manual

Stage Compressors for Efficient Control • “Stage” compressors into lead and lag compressor(s) by sequentially reducing load/unload pressures of the lag compressors. • Designate compressor with best control efficiency as Lag compressor • Staging allows Lead compressor to run fully loaded and the Lag compressor(s) to turn off or run with smaller part-load penalty if they have better control efficiency.

Use Efficient Compressed Air Pumps • Some pumps use ~30% less air than others

• DOE 16% of our industrial motor system energy use. Seventy percent of our manufacturing facilities use compressed air in their production process. • • Compressed Air Audit “Studies indicate that as much as 35% of the compressed air produced in the market today is wasted to leaks, and everyone has leaks. ” Wayne Perry, technical director, Kaeser Compressors “It has been our experience that plants which have no formal, monitored, disciplined, compressed air leak-management program will have a cumulative leak level equal to 30% to 50% of the total air demand, ” Henry van Ormer, Air Power USA • “The compressed air system had to run at 98 psi because the grinding area. The header pressure was lowered to 85 psi. Results after 18 months showed that tool repair went down for the grinders, production increased by 30% and total air demand fell from 1, 600 to 1, 400 cfm. “ Henry van Ormer, Air Power USA • “The easy answer to many system problems is to jack up the pressure. Unfortunately, the leaks will leak more, and unregulated users will waste more air and more energy. ” Norm Fischer, Centrifugal Equipment Service • “More important, the back pressure sends a false unload signal to the controls, causing premature unloading or extra compressors to be on line, ” van Ormer says. “Using a 30 degree to 45 degree directional angle entry instead of a tee will eliminate this pressure loss. The extra cost of the directional entry is usually negligible. ” • Remember, pressure costs money in two ways — power to produce increased pressure costs one half of one percent per psi, and excess pressure produces excess flow that must be compressed. Van Ormer Source: The top 10 targets of a compressed air audit, Rich Merritt, www. plantservices. com

• • • “More important, the back pressure sends a false unload signal to the controls, causing premature unloading or extra compressors to be on line, ” van Ormer says. “Using a 30 degree to 45 degree directional angle entry instead of a tee will eliminate this pressure loss. The extra cost of the directional entry is usually negligible. ” “Upgrading to copper or aluminum piping provides excellent value for money and ideal delivery characteristics, ” Perry says. “When upgrading, ensure that the physical piping diameter is sized to deliver the required air flow with minimum pressure drop. ” “Open blow, refrigeration and vortex cooling may all be replaceable with heat tube cabinet coolers with a potential savings of 3. 5 k. W to 4 k. W each on a 30 - by 24 - by 12 -inch average cabinet, ” van Ormer says. “The initial cost is usually in the $700 to $750 range with a potential resultant power savings of $1, 000 to $2, 000 per year each. ” • 14. 2 at inlet • “Open blow, refrigeration and vortex cooling may all be replaceable with heat tube cabinet coolers with a potential savings of 3. 5 k. W to 4 k. W each on a 30 - by 24 - by 12 -inch average cabinet, ” van Ormer says. “The initial cost is usually in the $700 to $750 range with a potential resultant power savings of $1, 000 to $2, 000 per year each. ”

• Use boost er comp ressor for highpress ure applic ations

Single Pressure Network Control • Narrower pressure band • However: – Compressors from same manufacture – Compressors in same location Source: Taming Multiple Compressors, Niff Ambrosino and Paul Shaw, www. plantservices. com

Cascading Pressure Set-Point Control • Simple, effective, and inexpensive • However: – Increases pressure – Only applicable for compressors in same location Source: Taming Multiple Compressors, Niff Ambrosino and Paul Shaw, www. plantservices. com

Cascading Pressure Set-Point Control • “Stage” compressors into lead and lag compressor(s) by sequentially reducing load/unload pressures of the lag compressors. • Designate compressor with best control efficiency as Lag compressor • Staging allows Lead compressor to run fully loaded and the Lag compressor(s) to turn off or run with smaller part-load penalty if they have better control efficiency.

Sizing Primary Storage §Required Storage Per Blow Down Time (gal) §Compressor Size §Blowdown Time (sec) §(hp) § 30 § 45 § 60 § 90 § 124 § 186 § 247 § 371 §Pressure Band § 50 § 619 § 928 § 1, 237 § 1, 856 §(psi) § 30 § 45 § 60 § 90 § 100 § 1, 237 § 1, 856 § 2, 474 § 3, 711 § 5. 5 § 8. 2 § 11. 0 § 16. 5 § 150 § 1, 856 § 2, 783 § 3, 711 § 5, 567 § 10 § 2. 7 § 4. 1 § 5. 5 § 8. 2 § 200 § 2, 474 § 3, 711 § 4, 948 § 7, 422 § 15 § 1. 8 § 2. 7 § 3. 7 § 5. 5 § 250 § 3, 093 § 4, 639 § 6, 185 § 9, 278 § 20 § 1. 4 § 2. 1 § 2. 7 § 4. 1 § 300 § 3, 711 § 5, 567 § 7, 422 § 11, 133 § 350 § 4, 330 § 6, 494 § 8, 659 § 12, 989 § 400 § 4, 948 § 7, 422 § 9, 896 § 14, 844 §Volume Storage Required per Rated Scfm (gal/rated-scfm) §Blowdown Time (sec)