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Energy Efficient Compressed Air Systems

Energy Efficient Compressed Air Systems. Compressed Air Fundamentals. dW elec = ∫V dP / [ Eff compressor Eff motor Eff control ]. V = volume flow rate D P = pressure rise Eff = efficiencies of compressor, motor, and control. Internal cooling decreases electrical power.

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Energy Efficient Compressed Air Systems

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  1. Energy EfficientCompressed Air Systems

  2. Compressed Air Fundamentals dWelec= ∫V dP/ [EffcompressorEffmotorEffcontrol] • V = volume flow rate • DP = pressure rise • Eff = efficiencies of compressor, motor, and control Internal cooling decreases electrical power

  3. Compressed Air Savings Opportunities dWelec = ∫V dP / [EffcompressorEffmotorEffcontrol] • Reduce volume flow rate • Reduce pressure rise • Increase cooling during compression • Increase compressor efficiency • Increase motor efficiency • Increase control efficiency

  4. Compressed Air System

  5. Screw Compressor Operation

  6. 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

  7. Plant and Compressor Data For Calculating Example Savings • Plant operates 6,000 hours per year • Electricity cost including demand = $0.10 /kWh • Air compressor produces 4.2 scfm/hp • Air compressor motor is 90% efficient • Air compressor runs in load/unload control with FP0 = 0.50

  8. Eliminate Inefficient Uses of Compressed Air

  9. Replace Air Pump with Electric Pump • Air motors use 7x 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 kW/hp x 6,000 hr/yr x $0.10 /kWh = $4,300 /yr

  10. Replace Compressed Air with Mechanical Agitation • Compressed air agitation uses ~10x 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 kW/hp x (1-0.50) = 15 kW • Cost Savings = 15 kW x 6,000 hr/yr x $0.10 /kWh = $9,000 /yr

  11. Replace Compressed Air Cooling with Chiller • ‘Vortex’ coolers use 4x more electricity than electric chillers • Example • Replace 10,200 Btu/hr cooler using 150 scfmwith chilled water • Cost Savings = 150 scfm / (4.2 scfm/hp x 90%) x 75% x 0.75 kW/hp x (1-0.50) x 6,000 hr/yr x $0.10 /kWh = $6,700 /year

  12. Replace Air With Heat Pipe Cabinet Coolers • Air coolers use 8x 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 kW/hpx (1-0.50) x 8,760 hr/yr x $0.10 /kWh = $2,607/year • Cost heat pipe = 0.6 A x 120 V / 1,000 VA/kW x 8,760 hr/yr x $0.10 /kWh = $ 63 /year • Implementation Cost = $1,000 Source: www.airtxinternational.com and www.system-directions.com

  13. Replace Pneumatic Suction Cups with Magnets • Suction cups use 6.0 scfmwhile 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 kW/hpx (1-0.50) = 0.57 kW • Cost Savings = 0.57kW x 3,000 hr/yr x $0.10 /kWh = $167 /yr

  14. 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 scfmx (1/4.2 – 1/7.2) hp/scfm / 0.90 x .75 kW/hpx (1-0.50) = 5.8 kW • Cost Savings = 5.8kW x 6,000 hr/yr x $0.10 /kWh = $3,472/yr

  15. Reduce End Use Compressed Air Demand

  16. 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 psiax 80% = 150 scfm • Power Savings = 150 scfm/ (4.2 scfm/hp x 0.90) x 0.75 kW/hp x (1-0.50) = 15 kW • Cost Savings = 15 kW x 6,000 hr/yr x $0.10 /kWh = $8,933/yr • Cost of 3/8-inch solenoid valve = $100

  17. 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 psiax 50% = 10.4 scfm • Power Savings = 10.4 scfm / (4.2 scfm/hp x 0.90) x 0.75 kW/hp x (1-0.50) = 1.0kW • Cost Savings = 1.0 kW x 6,000 hr/yr x $0.10 /kWh = $620 /yr • Nozzles cost about $10 each

  18. 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 kW/hp x (1-0.50) = 2.0kW • Cost Savings = 2.0 kW x 6,000 hr/yr x $0.10 /kWh = $1,214/yr

  19. Fix Leaks

  20. 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 kW/hp x (1-0.50) = 0.025 kW • Cost Savings = 0.025 kW x 6,000 hr/yr x $0.10 /kWh = $15 /yr

  21. 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 kW/hp x (1-0.50) = 8.3kW • Cost Savings = 8.3 kW x 6,000 hr/yr x $0.10 /kWh = $5,000 /yr

  22. Identify Leaks Using Ultrasonic Sensor

  23. Quantify Leakage By Logging Flow or Compressor Power

  24. 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 ifone new 1-cfm leak per week • Reduces leakage by 40%

  25. 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 kW/hp x (1-0.50) = 9.9 kW • Cost Savings = 9.9 kW x 4,000 hr/yr x $0.10 /kWh = $3,969/yr

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

  27. Reduce Distribution System Pressure Drop

  28. 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

  29. Avoid ‘Collision’ Connections

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

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

  32. 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 = • 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 • (P2high/P1)0.286 = [(110 psig +14.7 psia) / 14.7 psia]0.286 = 1.84 • (P2low/P1)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 hpx 5.2% / 90% x 0.75 kW/hp x 6,000 hr/yr x $0.10 /kWh = $2,582 /yr

  33. Reduce Pressure To Maximum End Use Plus Friction Loss

  34. Dry Air Efficiently

  35. 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

  36. 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

  37. 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 kW/hp x (1-0.50) = 12.5kW • Refrigerated Power = (840 scfm x 85% x 0.006 kW/scfm x (1-0.50) = 2.1kW • Cost Savings = (12.5 kW – 2.1kW) x 6,000 hr/yr x $0.10 /kWh = $6,215/yr

  38. Use Demand-Control Rather than Timed Purge • Summer air 4x wetter than winter air. • Timed purge set for peak (summer) conditions • Example • Switch from timed to demand-control 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 kW/hp x (1-0.50) = 12.5 kW • Cost Saving = 12.5 kW x 50% x 6,000 hr/yr x $0.10 /kWh = $3,250 /yr .016 lbw/lba .004 lbw/lba

  39. 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 no-loss 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 kW/hp x (1-0.50) = 1.68kW • Cost Savings = 1.68kW x 6,000 hr/yr x $0.10 /kWh = $1,005 /yr • 3/8-inch no-loss drains costs $600

  40. Optimize Compressor Cooling

  41. 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 kW x 5.9% x 6,000 hr/yr x $0.10 /kWh = $706 /yr

  42. 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 kW if heating system operates 2,000 hours per year and is 80% efficient • Heat Load Savings = 105 kW x 75% x 3,413 Btu/kWh x 2,000 hours/year = 537 mmBtu/yr • Cost Savings = 537 mmBtu/year / 80% x $10 /mmBtu = $6,719/year

  43. Employ Efficient Compressor Control FP = FP0 + (1-FP0) FC and P = Prated FP

  44. Power Signatures fromModulation and Load/unload Control

  45. Savings From Switching To Efficient Control • FP = FP0 + (1-FP0) FC • Typical Intercepts • FP0bypass = 1.00 • FP0modulation = 0.70 • FP0load/unload = 0.50 • FP0 variable-speed = 0.10 • FP0on/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 kW/hp x 6,000 hr/yr x $0.10 /kWh = $15,000 /yr

  46. Savings Penalty for Inefficient Control • Consider savings from reducing fraction capacity (FC) • P1 = Prated x FP1 = Prated x [FP0+ (1-FP0) FC1] • P2= Prated x FP2 = Prated x [FP0 + (1-FP0) FC2] • Psave = P1 – P2 • Psave = Prated x [FP0 + (1-FP0) FC1] - Prated x [FP0 + (1-FP0) FC2] • Psave= Prated x (FC1 - FC2) x (1-FP0) • Psave= Unadjusted savings x (1-FP0) • Actual Savings = Unadjusted Savings x (1 – FP0) • 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 kW/hp x 6,000 hr/yr x $0.10 /kWh = $11,905 • Actual savings = $11,905 /yr x (1 – 0.70) = $3,571 /yr

  47. Modulation to Load/Unload with Auto-shutoff Reduced power 35% and saved $17,000 /yr

  48. Centrifugal Compressor Control

  49. 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

  50. Throttling/Surge Pressure Set Points Narrow or Widen Throttling Band

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