Ohio Energy. Workshop G. Energy Efficiency Through Better Control: Maximizing the Energy Efficiency by Minimizing Waste

Transcription

1 Ohio Energy Workshop G Energy Efficiency Through Better Control: Maximizing the Energy Efficiency by Minimizing Waste Tuesday, February 16, :45 a.m. to Noon

2 Biographical Information J. Kelly Kissock, Ph.D., P.E. Professor and Chair Dept. of Mechanical and Aerospace Engineering / Renewable and Clean Energy University of Dayton Kettering Laboratories 361-B, 300 College Park, Dayton, OH kkissock@udayton.edu Phone: Fax: Dr. Kissock is a Professor and Chair of the Mechanical and Aerospace Engineering and Director of the Renewable and Clean Energy program at the University of Dayton. He is also Director of the University of Dayton Industrial Assessment Center. He is a Registered Professional Engineer in the State of Ohio. Dr. Kissock works in the fields of building, industrial and renewable energy systems. He has published over 100 technical papers on energy efficiency and renewable energy and conducted seminars on energy efficiency across the world. Dr. Kissock served as Associate Editor of the ASME Journal of Solar Energy Engineering and has chaired several technical committees and conferences. His work has been recognized with two Distinguished Educator Awards by Who s Who Among America s Teachers, the 2003 U.S. Department of Energy Center of Excellence Award, the 2006 Ohio Governor s Award for Excellence in Energy, the 2009 University of Dayton Alumni Award for Scholarship and the 2011 Champion of Energy Efficiency award from the American Council for an Energy Efficient Economy. UD School of Engineering: Celebrating 100 Years of Engineering Education and Research

3 Energy Efficiency as Flow and Force Kelly Kissock Director: UD Industrial Assessment Center Chair: Department of Mechanical and Aerospace Engineering / Renewable and Clean Energy

4 Industrial Assessment Center (IAC) Program Sponsored by U.S. Department of Energy Program began during 1970s energy crisis 24 centers at universities throughout the U.S. 20 no cost assessments per year for mid sized industries Goals Help industry be more energyefficient and competitive Train the next generation of energy engineers Develop new, innovative solutions for energy efficiency

5 University of Dayton Industrial Assessment Center Energy Assessment Analyze data before visit Visit site for one day Work with client to identify and quantify saving opportunities Deliver semi confidential report Results ( ) 13% average energy savings identified 7% average implemented energy savings 64% average return on investment ~970 energy audits since

6 Energy: Flow and Force Energy is composed of both flow and force quantity and quality When considering energy efficiency opportunities, think about both

7 Electrical Power Pin (kw) = Current (A) x Voltage (kv) Flow Force

8 Fluid Power W f = V P Flow Force

9 Pump/Fan and System Curves P Pump/Fan Curve System Curve P 1 Operating Point W f = V P V 1 V

10 Inefficient Flow Control By-pass Valve Fan w/ Inlet Vanes By-pass loop (No savings) By-pass damper (No savings) Outlet valve/damper or Inlet Vanes (Small savings) Intermittent Flow (Small savings)

11 Efficient Flow Control Trim impellor for constant-volume pumps Slow fan for constant-volume fans VFD for variable-volume pumps or fans

12 Minimize Elevation Gain: Increase Initial Reservoir Level W elev = V P elevation difference between inlet and outlet Reducing elevation difference reduces work to overcome elevation by: (20 15) / 20 = 20%

13 Minimize Friction: Use Large Diameter Pipes/Ducts W friction = V P friction P friction =k / D 5 W friction = V k / D 5 Work to overcome friction varies inversely with 5 th power of pipe diameter Doubling pipe diameter reduces work to overcome friction by: 1- (1/2) 5 = 97% 2 inch 18 ftwg 4 inch 0.6 ftwg

14 Minimize Friction: Use Smooth Pipes/Ducts W friction = V P friction P friction ~ friction factor f f steel = f plastic = Plastic pipes reduce work to overcome friction by: ( ) / = 14% Material Roughness (m) Glass, plastic Smooth Copper, brass, lead 1.5 x 10 6 (tubing) Cast iron uncoated 2.4 x 10 4 Cast iron asphalt 1.2 x 10 4 coated Commercial steel or 4.6 x 10 5 welded steel Wrought iron 4.6 x 10 5 Riveted steel 1.8 x 10 3 Concrete 1.2 x 10 3

15 Minimize Friction: Use Gradual Elbows W friction = V P friction P friction ~ loss coef k k sqr = 1.24 k long rad = 0.10 Long-radius elbows reduce work to overcome friction by: ( ) / 1.24 = 92%

16 Minimize Friction: Minimize Turns, Fittings and Distance Layout pipes then equipment: shorter runs, fewer turns, fewer valves Use 3D design software to reduce interferences from 35% to 0% Example: Original design: 14 pumps and 95 hp Re-design: 7 hp P

17 Reduce Force with VAV Fan Control Strategies Fan Outlet Control Critical Zone Reset Control Supply Duct Control 16

18 Fan Speed Control Strategies Fan Outlet Control Supply Duct Control Critical Zone Reset

19 Reduce Force with Variable Flow Pumping Control Strategies Most systems use pressure as control variable Savings depend on location and setpoint of pressure sensor Savings maximized when pressure is dynamically reset based on valve position

20 Variable Flow Pumping Control: Pressure Sensor at Discharge

21 Variable Flow Pumping Control dp at Intermediate End Use

22 Variable Flow Pumping Control dp at Most Remote End Use

23 Variable Flow Pumping Control Maximum Open Valve Pressure Reset

24 Variable Flow Pumping Control Summary P A B C D A: Pset,discharge (Worst) B: Pset,midway C: Pset,remote D: Pset,max valve (Best) V 2 = V 1 / 2 V 1 V

25 Energy Efficient Compressed Air Systems

26 Reduce Compressed Air Flow

27 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 Cost Savings = 150 scfm / (4.2 scfm/hp x 0.90) x 0.75 kw/hp x (1 0.50) x 6,000 hr/yr x $0.10 /kwh = $8,933 /yr Cost of 3/8 inch solenoid valve = $100 Plant manager taking charge!

28 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 Cost Savings = 10.4 scfm / (4.2 scfm/hp x 0.90) x 0.75 kw/hp x (1 0.50) x 6,000 hr/yr x $0.10 /kwh = $620 /yr Nozzles cost about $10 each

29 Reduce Flow by Regulating Pressure After installing nozzles, reduce air pressure to end uses with regulator to minimize flow. Example Regulate pressure to 1/8 in tube from 100 to 50 psig Flow Savings = 11.6 (scfm/lbf) x [1/8 (in)] 2 x (115 65) psia = 9.1 scfm Cost Savings = 9.1 scfm / (4.2 scfm/hp x 0.90) x 0.75 kw/hp x (1 0.50) x 6,000 hr/yr x $0.10 /kwh = $539 /yr

30 Control Air Based on Demand Instead of Timer Timers usually set for peak conditions Demand based control follows 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 Cost Savings = 20.4 cfm / (4.2 scfm/hp x 0.90) x 0.75 kw/hp x (1 0.50) x 6,000 hr/yr x $0.10 /kwh = $1,214 /yr

31 Identify and Fix Leaks

32 Employ Efficient Volume Control FP = FP 0 + (1 FP 0 ) FC FP = P / P rated FC = C / C rated

33 Reduce Distribution System Pressure Drop

34 Use Looped Piping to Reduce P If P < 10 psi at last end use, use looped rather than linear piping Design Guidelines Main line: size from average cfm to get P = 3 psi Branch line: size from cfm peak to get P = 3 psi Feed lines: size from peak cfm to get P = 1 psi Select hose with P <1 psi

35 Eliminate Collision Fittings to Reduce P P = 5 psi P = 0 psi

36 Maintain Filters to Reduce P Place filter upstream of dryer to protect dryer P filter < 1 psi

37 Size Dryer so P < 5 psi Low Flow: P = 1 psi High Flow: P = 6 psi

38 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 Cost Savings = 140 scfm x (1/4.2 1/7.2) hp/scfm / 0.90 x.75 kw/hp x (1 0.50) x 6,000 hr/yr x $0.10 /kwh = $3,472 /yr

39 Energy Efficient Steam Systems

40 Steam: Flow and Force Q = m (h2 h1) Flow Force h = h (P, T) When using steam for: Power: Force = (P2 P1) Heat: Force = (T2 T1)

41 Insulate hot surfaces at end use Reduce Steam Demand Cover uninsulated tanks

42 Fix Steam Traps Steam traps are automatic valves that discharge condensate from a steam line without discharging steam. If trap fails open, steam by-passes heat exchanger and releases heat in condensate return system If trap fails closed, condensate fills the heat exchanger and chokes-off heat to process. Fixing failed steam traps is highly costeffective.

43 Insulate Insulate Pipes and Tanks Steam pipes Condensate return pipes Condensate return tanks Deaerator tank Valves

44 Minimize Steam Pressure Reducing boiler steam pressure to match the highest required end-use temperature increases boiler efficiency decreases heat loss from boiler and pipes decreases flash loss Force

45 Energy Efficient Process Heating

46 Heat Transfer From Hot Fluid 1 2 Q Q = m cp [T1-T2] Flow Force

47 Energy Balance on Furnace Flow Flow Flow Flow

48 Heat Transfer Between Hot and Cold Materials h c Q Q = UA [Th-Tc]m Flow Force

49 Heat Transfer From/To Fluid Parallel-flow Counter-flow h c Q Q = UA [Th-Tc]m,pf h Q c Q = UA [Th-Tc]m,cf [Th-Tc]m,pf < [Th-Tc]m,cf Qpf < Qcf

50 Counter-Flow Reduces Exhaust Temperature and Maximizes Heat Transfer Q T Parallel Flow T x Q Counter Flow x

51 Replace Reverb with Counter-flow Stack Furnace Reverb cross-flow furnace Stack counter-flow furnace

52 Relocate Exhaust Ports To Increase Counter-flow

53 Set Oven Exhaust Dampers for Counter-flow Product In Product Out 100% open 75% open 50% open 25% open 12% open

54 Pre-heat Load Using Counter-flow

55 Preheat Combustion Air T = 95 F Exhaust Gases T = 950 F Preheated Combustion Air T = 615 F Recuperator T = 1,465 F Burner Furnace

56 Cascade Heat to Lower-Temperature Process Exhaust Stack Damper High Temperature Oven Low Temperature Oven

57 Energy-Efficient Process Cooling

58 Ammonia Refrigeration Energy Saving Opportunities 1. Reduce refrigeration load 2. Reclaim heat 3. Increase suction pressure 4. Employ dual suction pressures 5. Reduce minimum head pressure set-point 6. Improve evaporative condenser effectiveness 57

59 Reduce Refrigeration Load 58

60 Reduce Refrigeration Load 59

61 Reclaim Heat Reclaimed Heat Reclaimed Heat 60

62 Increase Suction Pressure 61

63 Increase Suction Pressure 62

64 Employ Dual Suction Pressures 63

65 Employ Dual Suction Pressures 64

66 Reduce Minimum Head Pressure 65

67 Reduce Minimum Head Pressure 66

68 Bigger Condensers Reduce Head T,P SEER 10 to 20

69 Use Cooling Tower When Temps Allow Tc (F) Twb (F) Fyr (%) % % % % Fraction of year cooling tower can deliver water at Tc (Assume Tr = 10 F in Dayton OH)

70 Use Water Cooled Chillers for Year Round Loads E/Q (Air-cooled) = 1.0 kw/ton E/Q (Water-cooled) = 0.8 kw/ton

71 Data Centers Improve Air Flow Over Circuits To Reduce Tc and Energy

72 Improve Cooling Heat Transfer Cross flow cooling of extruded plastic with 50 F chilled water from chiller

73 Improve Cooling Heat Transfer Cross Flow: = 0.69 Tw1 = 50 F Tp = 300 F Mcp min = 83.2 Btu/min-F Q = mcp min (Tp Tw1) Q = (300 50) Q = 14,352 Btu/min Counter Flow: = 0.78 Q = 14,352 Btu/min Tp = 300 F Mcp min = 83.2 Btu/min-F Q = mcp min (Tp Tw1) 14,352 Btu/min = (300 Tw1) Tw1 = 79 F Cooling tower delivers 79 F water much of the year using 1/10 as much energy as chillers!

74 Combined Heat and Power

75 Heat Engines Carnot discovered that engines that use heat to make power must reject some heat to the environment. First statement of 2 nd Law of Thermodynamics

76 Electrical Power Generation 45% (cool towers)+ 20% (exhaust gas) = 65% of all energy!

77 Combined Heat and Power (CHP) Systems CHP systems find productive uses for low temperature heat Increases energy utilization / efficiency Application

78 Energy and Availability Energy can t be created or destroyed. Energy crisis? And yet, when we burn a tank of gas it feels like something useful is gone That useful energy is called availability

79 Energy Analysis Ewaste Ein Euseful Energy approach is to minimize Ewaste to increase Euseful

80 Availability Analysis Qin Th Qout Ein Eout Tc Ain Th Aout Ain Ades Aout Ades Tc Energy conserved but availability destroyed through heat transfer ( T) and friction ( P) Force

81 Energy Efficiency is Both Energy and Availability Analysis Flow Flow savings were result of minimizing energy waste Force savings were result of minimizing availability destruction

82 UD IAC Energy Efficiency Guidebook (EEG))z 12 Energy Systems 7Principles of Energy Efficiency

83 University of Dayton Industrial Assessment Center

84 Thank you!