Air Compressors Good

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Air Compressors Two Types • Positive Displacement – Reciprocating – Rotary Screw • Dynamic – Centrifugal – Axial flow Rotary Screw Compressors Rotary Screw Compressors • “Direct Injection”, “Oil-injected”, “Directcooled”, “Flooded” – Various names are suggestive of the direct contact of lubricating/cooling oil onto the rotating screw parts – Requires a separator just after compression to remove oil from the compressed air • “Oil-free”, “Oil-less”, “Dry” compressors also available (food industry, hospitals, etc.) Animations • http://www.youtube.com/watch?v=g8VpnTR WESQ • http://www.youtube.com/watch?v=WFZ1bhF Eh2U • http://www.youtube.com/watch?v=zQdBTxNQHU Due to overlapping and continuous compression cycles, the rotary screw design generates virtually no vibration. Disturbing noise is minimized, providing a wider choice for the unit's location on the vehicle. The ends of the rotors uncover the inlet: air enters the compression chamber. The air is entrapped in the 'compartment' formed by a male lobe and a female flute. As the rotors turn, Compressed air the compartment leaves through the becomes outlet port progressively smaller, thereby compressing the entrapped air. Centrifugal Compressors Centrifugal Compressors • https://www.youtube.com/watch?v=rtIF1LdZT ak http://www1.eere.energy.gov/industry/bestpractices/pdfs/compressed_air_sourcebook.pdf Compressor Controls • • • • • On/Off (Start/Stop) Modulated Load/Unload Variable Displacement Variable Speed Drive (VSD) 100 200 300 400 500 600 Tu Amps Interval data (4 seconds) for System (Not Assigned) and Periods (Not Assigned) 12/15/2008 11:54:06 PM to 12/17/2008 12:32:05 AM e 0 Tu 1:0 0 e 02 Tu :0 0 e 0 Tu 3:0 0 e 0 Tu 4:0 0 e 05 : Tu 00 e 06 Tu :0 0 e 07 Tu :0 0 e 0 Tu 8:0 0 e 09 Tu :0 0 e 1 Tu 0:0 0 e 11 Tu :0 0 e 12 : Tu 00 e 13 Tu :0 0 e 14 Tu :0 0 e 1 Tu 5:0 0 e 16 Tu :0 0 e 1 Tu 7:0 0 e 18 Tu :0 0 e 1 Tu 9:0 0 e 2 Tu 0:0 0 e 21 Tu :0 0 e 2 Tu 2:0 0 e 2 W 3:0 ed 0 00 :0 0 Curr_250_hp_rent (Amps) Curr_250_hp (Amps) Curr_100_hp (Amps) 10 : 100 200 300 400 Amps 0 Interval data (4 seconds) for System (Not Assigned) and Periods (Not Assigned) 12/16/2008 10:24:15 AM to 12/16/2008 10:33:08 AM Curr_250_hp_rent (Amps) Curr_250_hp (Amps) Curr_100_hp (Amps) 24 10 :20 :2 4 10 :40 :2 5 10 :00 :2 5 10 :20 :2 5 10 :40 :2 6 10 :00 :2 6 10 :20 :2 6 10 :40 :2 7 10 :00 :2 7 10 :20 :2 7 10 :40 :2 8 10 :00 :2 8 10 :20 :2 8 10 :40 :2 9 10 :00 :2 9 10 :20 :2 9 10 :40 :3 0 10 :00 :3 0 10 :20 :3 0 10 :40 :3 1 10 :00 :3 1 10 :20 :3 1 10 :40 :3 2 10 :00 :3 2 10 :20 :3 2 10 :40 :3 3: 00 Typical Compressor Controls/Characteristics 100 90 80 70 60 50 40 30 20 10 0 0 Power, percent of full load Start/Stop) Mod (Load/Unload) Mod/Unload 50 Capacity (Flow), percent 100 Compressors - Saving Energy • • • • • • Reduce run time – turn off when not needed Lower system pressure to lowest possible level Repair leaks Recover waste heat Additional system volume (load/unload only) Reduce use of pneumatic tools Lower Pressure • Rule of thumb: For systems in the 100 psig range, every 2 psi decrease in discharge pressure results in approximately 1 percent power decrease at full output flow Lower Pressure – Unregulated Usage • Rule of thumb: For systems with 30 to 50 percent unregulated usage, a 2 psi decrease in header pressure will decrease energy consumption by about 0.6 to 1.0 percent because of unregulated air • Total is 1.6% to 2% power decrease for every 2 psi drop Lower Pressure • Calculations – Estimate annual energy usage (assume compressor runs fully loaded unless you know otherwise) KWh = HP/ηm x 0.746 x LF x H – Compute reductions: every 2psi reduction is a 1.6% to 2.0% reduction in power, so each 2psi reduction gives 0.984 to 0.98 of the previous value kWhnew=kWhold x (1 – PCT )∆P/2 ≈kWholdx(1 -PCTx∆P/2) Example – Reduce Pressure • A 150hp compressor runs 90% loaded 12 hours per day 5 days per week at a delivery pressure of 110psi. Estimate the energy and cost savings if the pressure is reduced to 90psi. Assume $0.05/kWh for electricity and the motor is 95% efficient • Solution: Assume a 1.8% reduction for every 2psi to account for unregulated usage. First estimate the annual energy usage: Example/Solution – Reduce Pressure kWh= HP /ηmx 0.746 x LF x H = 150hp/0.95 x 0.746kW/hp x 0.9 x 3120 hrs = 330,753 kWh/yr Compute reductions kWhnew = 330,753 kWh x ( 1 – 0.018 ) 20/2 = 275,816 kWh Energy Savings are 54,937 kWh/yr Net reduction is 16.6% At $0.05/kWh, this amounts to $2,747 / yr Reduce Air Leaks • A typical plant that has not been well maintained will likely have a leak rate equal to 20 percent of total compressed air production capacity. • Proactive leak detection and repair can reduce leaks to less than 10 percent of compressor output Reduce Air Leaks • Calculations – Savings realized depend on the type of compressor controls – Input power decreases linearly with decrease in airflow Control Slope (∆kW%/∆cfm%) Start/Stop 100/100 Mod 35/100 Unload 80/100 – So for a 10% reduction in flow by repairing leaks Control ∆kW% Start/Stop 10% Mod 3.5% Unload 8% Example - Reduce Leaks • The compressor from the previous example uses modulating controls. The system reduces flow by 10% through a leaks program. Estimate energy and monetary savings at $0.05 per kWh. • Solution: For a 10% reduction in flow, a 3.0% reduction in power results: Example - Reduce Leaks Savings = 275,816 kWh/yr x 0.035 = 9,654 kWh/yr At $0.05/kWh, this comes to Cost Savings = 9,654 kWh/yr x $0.05/kWh = $ 83 Recover Waste Heat • As much as 80 to 93 percent of the electrical energy used by an industrial air compressor is converted into heat. • In many cases, a properly designed heat recovery unit can recover anywhere from 50 to 90 percent of this available thermal energy and put it to useful work heating air or water. • Net potential is 40% to 84% recovery Example - Recover Waste Heat • The compressor from the previous example is air cooled. Estimate the amount of natural gas heating that could be displaced during twelve weeks of winter operation, and the cost savings at 85% combustion efficiency and $4.00 /MMBtu fuel cost. • Solution: Assume 50% of the input power can be recovered: Example - Recover Waste Heat Savings = HP x LF x 2545Btu/hp-hr x 50% x H = 150hp x 0.90 x 2545Btu/hp-hr x 0.50 x (12wks x 5days/wk x 12hrs/day) x 1 MMBtu / 1E6Btu = 123.7 MMBtu Cost Savings = Savings / EFF x FuelCost = 123.7 / 0.85 x $4.00 = $ 582 Examples - Summary Measure Baseline operation Reduce Pressure Repair Leaks Recover Waste Heat* TOTAL SAVINGS - Energy TOTAL SAVINGS - Cost Energy 330,753 kWh/yr 54,937 kWh/yr 9,654 kWh/yr 36,254 kWh/yr (123.7 MMBtu/yr) 100,845 kWh/yr $ 3,812 Cost/Savings $16,538 /yr $ 2,747 /yr $ $ 483 /yr 582 /yr 16.6% 2.9% 11.0% 30.5% 23.0% *Assumes plant can use all available heat over a 12-week period Additional Volume Additional Volume