Common recommendations

Transcription

1 Motor Overview Why are motors important? What is a motor? Optimizing the motor component power factor efficiency and losses motor/load relationships standards Common recommendations Review of CD-ROM

2 Industrial Electricity Use Lighting/Other 8% Motor-Driven Systems 69% Process Heating 11% Electrolytics 12% Source: DOE presentation, March 1999: Introduction to Motor Systems Management

3 Industrial motor energy use by application Total 2.3 Quadrillion Btu/yr material handling 12% material processing 23% pumps 24% other 4% refrigeration 7% compressed air 16% fans 14% Source: United States Industrial Electric Motor Systems Market Opportunities Assessment, Office of Industrial Technologies, Office of Energy Efficiency and Renewable energy, US DOE, December 1998, Table 1-16, page 43

4 Motor size profile compared to energy use Total population 12.4 million units Total Energy Use: 2.3 Quadrillion Btu/yr < 5 hp 59% % hp 26% hp 9% % % % % < 5 5% >250hp 1% hp 2% hp 3% Source: OIT pg. 40

5 What is a motor? Source: Energy-Efficient Motor Systems: A Handbook on Technology, Programs, and Policy Opportunities, 1991, American Council for and Energy-Efficient Economy.

6 Types of Electric Motors AC induction/asynchronous Single-phase Three-phase AC synchronous DC motors

7 Induction Motors and Power Factor Active component responsible for torque and useful work Reactive component creates the rotating magnetic field Active part is small at no load and at its max at full load, while reactive is almost constant Ratio of active to total current is called power factor (PF)

8 Induction Motors and Power Factor At no load, active component is equal to motor inefficiencies At full load PF is typically between 70-95% Utilities penalize for PF lower than 85-95% AC synchronous motors can generate reactive power

9 Induction Motors and Power Factor Source: Energy-Efficient Motor Systems: A Handbook on Technology, Programs, and Policy Opportunities, 1991, American Council for and Energy-Efficient Economy.

10 Electrical Power In Losses Motor efficiency and losses 80-95% Shaft Power Out 5-20%

11 Motor losses and load relationship Source: Energy-Efficient Motor Systems: A Handbook on Technology, Programs, and Policy Opportunities, 1991, American Council for and Energy-Efficient Economy.

12 Motor efficiency standards NEMA (National Electrical Manufacturer s Association) defines energy efficient motors EPACT 92 (Energy Policy Act of 1992)

13 Motor h and costs Source: Energy-Efficient Motor Systems: A Handbook on Technology, Programs, and Policy Opportunities, 1991, American Council for and Energy-Efficient Economy.

14 Motor Recommendations 1. Replace motors with energy-efficient motors NEW HI-EFF MOTOR MOTOR REWIND 3. Use energy-efficient belts 2. Install variable speed drive control 4. Optimize motor size with load

15 Motor Recommendations Measure Avg. Savings ($/yr) Simple Payback (yr) % Implemented E2 Motors 1, VSD Controls 12, E2 Belts 2, Optimize Size 3,

16 Energy Efficient (E2) Motors NEW HIGH EFFIENCY MOTOR MOTOR REWIND

17 E2 Motors - Background About 20-40% less losses than standard motors Increase in efficiency of 1-10% Larger motors yield smaller efficiency increases Motormaster Program available at

18 E2 Motors - Background Horsepower Rating (hp) Estimated Existing Efficiency Proposed Efficiency Cost Premium ($ per motor)

19 E2 Motors - Background E2 motors may not always be beneficial Less slip can help or hurt energy conservation Not usually suited for variable torque loads because they require hp µ rpm 3 Increasing speed by 1% corresponds to a 3% increase in hp

20 To Rewind or Not to Rewind? What is motor rewinding? Commonly done with large motors Why is it done? Are there any drawbacks? When should it be considered?

21 Rewinding Source: Energy-Efficient Motor Systems: A Handbook on Technology, Programs, and Policy Opportunities, 1991, American Council for and Energy-Efficient Economy.

22 E2 Motors Anticipated Savings Ê 1 1 ˆ EC = HP N C 1 LF UF H Á - ç C ç Ë P Ê 1 1 ˆ DR = HP N C LF CF Á - 1 Ë ç ç C P

23 E2 Motors Anticipated Savings EC annual energy conservation, kwh/yr HP horsepower of motor considered, hp N number of motors of a given size, no units C1 conversion constant, kw/hp LF load factor, fraction of rated load at which motor normally operates UF usage factor, fraction of operating time when motor actually runs, no units H annual operating time of equipment driven by motor, h/yr _ C estimated efficiency of existing motor _ P estimated efficiency of proposed motor DR demand reduction, kw/mo CF coincidence factor, probability that the equipment contributes to the facility peak demand, per month

24 E2 Motors Load Factor Load Factor - Fraction of rated load at which motor NORMALLY operates. Estimating Load Factor Kilowatt Ratio Method Amperage Method Slip Method

25 E2 Motors-Kilowatt Ratio Method Load Factor = P i P ir = È V I PF Í Î 1000 È hp Í Î h P i = Measured 3f power in kw P ir = Input power at full-rated load in kw V = RMS voltage, line-to-line of 3 f I = RMS current, mean of 3 f PF = Power factor as a decimal hp = Nameplate rated horsepower h fl = Efficiency at full-rated load fl 3

26 E2 Motors-Amperage Method I Load Factor = I V r V r I = RMS current, mean of 3 f I r = Nameplate rated current V = RMS voltage, mean line-to-line of 3 f Vr = Nameplate rated voltage

27 E2 Motors-Amperage Method

28 E2 Motors Slip Method The speed of the rotating magnetic field is known as the synchronous speed. frequency of the applied voltage (Hz) 60 S s = number of pole pairs Thus, a three-phase induction motor with 4 poles (2 pole pairs) at 60 Hz has a synchronous speed of 1800 rpm.

29 E2 Motors Slip Method The difference between the synchronous and actual speeds is called the motor slip.

30 E2 Motors Slip Method Load Factor = S s S s - S - S m r S s = Synchronous speed in rpm S m = Measured speed in rpm S r = Nameplate full-load speed

31 E2 Motors Incremental Savings = Annual where EML f motor replacement fraction 1 EML = Estimated Motor Life ( ª 12 yrs) = th IS n = Incremental Savings in n year = n x f x S

32 E2 Motors Sample Problem Motor Driven Equipment Number of Motors Horsepower (hp) Load Factor Coincidence Factor Usage Factor Operating Time (h/yr) Hydrostatic test pump Oven fan Should either of these motors be replaced with high efficiency motors or should they simply be rewound? If replacement is recommended, what are the associated energy, demand, and cost savings? What is the payback period? Use the table of data extracted from the Motormaster program to help you decide. Given: Unit Electric Cost = $0.05/kWh Demand Cost = $10.00/kW

33 Adjustable Speed Drive (ASD) Control P2/E2 Site Assessment Grant Program Training

34 ASD - Background Provides the ability to precisely match motor output to process requirements. Power varies as the cube of speed Most effective for applications with varying loads. Common applications: fans, compressors, pumps P2/E2 Site Assessment Grant Program Training

35 ASD - Potential Benefits Improved product quality Improved process throughput Improved process control Energy savings Reduced maintenance

36 Common Problems with VFDs VFDs create dirty power (odd-power harmonics) which can upset sensitive equipment Early bearing failure Doesn t save energy when motor is at full load Low speeds can cause cogging

37 Flow Control Methods Use of outlet dampers and throttle valves Change head pressure to control flow rate Control motor speed

38 Flow Control without ASD Motor runs at constant speed Flow rate controlled by valve

39 Power Source Flow Control without ASD Signal Control Valve Controlled Flow Pump Motor

40 Flow Control with ASD Motor speed varies Flow rate controlled by motor speed Energy is saved

41 Power Source ASD Flow Control with ASD Signal Controlled Flow Pump Motor

42 Types of ASDs Mechanical Hydraulic clutches Adjustable belts and pulleys Electrical (Variable Frequency Drives) Pulse Width Modulated (PWM) Voltage Source Inverter (VSI) Current Source Inverter (CSI) Brushless DC Drive (BDCD) Switched Reluctance Drive (SR) Flux Vector Drive (FV)

43 Pulse Width Modulated (PWM) Uses complex computer software Motor frequency produced with sinusoidally varying pulses of voltage Advantages High power factor Can control several motors at once Smooth speed control, even at low speeds, which eliminates cogging Responds very quickly to speed changes Suitable for almost all applications

44 PWM Disadvantages Over voltage may lead to failure for motors under 50 hp Decrease cable length between motor and ASD Noisy operation Early bearing failure Use shaft grinding brush instead of graphite grease

45 Voltage Source Inverter (VSI) Use simple electronic circuit Advantages Able to control several motors at once Very high speed drives available Disadvantages Speed control close to zero unavailable Low frequency results in cogging Power factor proportional to speed No regenerative braking

46 Current Source Inverter (CSI) Mostly replaced with PWM Advantages Capable of regenerative braking Disadvantages Difficulty controlling more than one motor Poor performance below 50% of rated load

47 Brushless DC Drive (BDCD) Uses permanent magnets on rotor as poles Advantages Motor shaft speed and position controlled very accurately Able to quickly reaccelerate to high speeds Disadvantages Limited to motors with permanent magnets

48 Switched Reluctance Drive (SR) Configured specifically for switched reluctance motors Advantages Wide speed range Dynamic response Disadvantages Expensive Cogging occurs close to zero speed

49 Flux Vector Drive (FV) Best available control Uses same basic control as PWM Advantages Allows maximum torque performance Calculates exact direction and magnitude of motor flux Disadvantages Very expensive should only be used if superior control is needed

50 ASD Efficiency Efficiency with which ASDs convert constant-frequency power into adjustablefrequency power decreases with speed Uses 4-6% more of the full-load nameplate horsepower while running at full-load Cogging may occur

51 ASD Efficiency Variable Speed Drive Rating (hp) Percent of Full Operating Speed 25% 50% 75% 100% 1 9.4% 44.2% 70.5% 82.5% % 74.7% 88.3% 92.4% % 79.0% 90.3% 93.5% % 79.4% 90.6% 93.8% % 83.5% 92.1% 94.4% % 89.1% 95.0% 96.6% % 91.3% 96.1% 97.3%

52 ASD Efficiency 100% 80% 60% 40% 20% 0% ASD Efficiency % of Full Operating Speed 1 hp 10 hp 100 hp

53 Fraction of Flow Rate Rated Load % % Flow Rate % Fraction of Rated Load % ASD Power Draw 100 hp motor

54 Screening Methodology High annual operating hours Variable load characteristics Moderate to high horsepower rating

55 Percent Operating Hours Load Duty Cycle Example of an Excellent VFD Candidate Percent Rated Flow

56 Percent Operating Hours Load Duty Cycle Example of a Poor VFD Candidate Percent Rated Flow

57 Anticipated Savings EC = CEU - PEU ECS = EC x unit electric cost EC annual energy conservation, kwh/yr CEU current energy usage, kwh/yr PEU proposed energy usage, kwh/yr ECS energy cost savings, $/yr

58 Anticipated Savings CEU HP C 1 UF LF H = ç M HP horsepower of motor considered, hp C1 conversion constant, kw/hp UF usage factor, no units LF load factor, no units H annual operating time of equipment driven by motor, h/yr _ M efficiency of the driving motor, no units

59 Anticipated Savings PEU HP C 1 UF i FRL ç Â ( H ) = i M Hi annual operating time at corresponding flow rate, h/yr FRL i fraction of the rated load at which motor will operate for corresponding operating time, no units

60 Anticipated Savings FRL i = Ê Q Á Ë Q ç pi ci ASD ˆ 3 Qpi proposed flow rate percentage, no units Qci one-speed flow rate percentage, 100% _ ASD adjustable speed drive efficiency, no units

61 Sample Problem A plastics plant uses a fan with a 100 hp motor that currently runs at constant speed. The air flow rate could be manipulated with an adjustable speed drive. The required air flow and parameters are shown below: Fraction of Flow Required (%) Operating Hours at Flow (h/yr) , , LF UF Motor efficiency % Operating hours - 6,000 h/yr Unit electric cost - $0.05/kWh What is the energy conservation and cost savings realized by installing an ASD?

62 ASDMaster Developed by the Electric Power Research Institution (EPRI) Adjustable Speed Drive Demonstration Office (ASDO) and Bonneville Power Administration Automated Windows based software package Total System Approach

63 ASDMaster Instruction on basic ASD operation Determines economic benefits Helps select the appropriate ASD for the application and environment Locates ASD manufacturers

64 Step 1 Screening Initial Data: Size of motor (hp) Type of equipment Operating Hours Variability of load on motor Feedback: High probability of energy savings Low probability of energy savings Inconclusive

65 Step 2 Data Collection Load Duty Cycle Determined from historical operating data Equipment Curve Obtained from equipment manufacturer Describes capability of machinery as pressure rise at various flow rates System Curve Quadratic function describes how system pressure changes with flow rate Input 2 points: Design point (static pressure) which is usually shown on equipment curve and any other point

66 Step 3 Calculations (Performed by ASDMaster) 1. Affinity Point 2. Shaft Speed 3. Fluid Power 4. Shaft Power 5. Motor Power 6. ASD Power 7. Annual Energy Consumption

67 ASDMaster Information Best Practices Website: /software_tools.shtml ASD link includes: Order form Overview and demonstrations Software available for $150

68 Energy Efficient (E2) Belts

69 E2 Belts - Background Standard V-Belts Efficiency of 90-96% Losses from flexing and slippage Cogged V-Belts 1-3% more efficient Lower flexing losses Synchronous Belts Efficiency of 98-99% No slippage Most expensive Not suitable for all applications

70 E2 Belts Anticipated Savings EC = Á Ë Ê N x HP x C ç 1 x LF ˆ x UF x H x FS Ê N HP C 1 LF ˆ DR = Á CF FS Ë ç ECS = EC x (effective energy rate) DS = DR x C 5 x (effective demand rate)

71 E2 Belts Anticipated Savings EC annual energy conservation, kwh/yr HP horsepower of motor considered, hp N number of motors of a given size, no units C 1 conversion constant, kw/hp LF load factor UF usage factor H annual operating time of equipment driven by motor, h/yr FS fractional energy savings _ efficiency of the driving motor

72 E2 Belts Anticipated Savings DR demand reduction, kw/mo CF coincidence factor ECS energy cost savings, $/yr DS demand savings, $/yr C 5 number of months per year equipment contributes to the peak demand, generally this constant is 12 months/yr

73 E2 Belts Sample Problem A manufacturer of canned fruit is interested in reducing energy usage on its 50-1 hp exhaust fans by retrofitting with cogged V-belts. The plant manager stated that these fans run continuously throughout the year at full load. What will be the electric conservation, demand reduction, and total cost savings realized by using cogged V-belts on all of the motors? Given: Motor Efficiency = 80% Unit Electric Cost = $0.05/kWh Demand Cost = $10.00/kW

74 Optimize Motor Size

75 Motor Size - Background Motors commonly oversized to allow for growth and safety Motor efficiency and PF drops sharply after about 40% At low loads larger motors still maintain better h and PF Focus on oversized motors with high operating hours Know the typical and maximum load characteristics

76 Motor Size - Background Source: Energy-Efficient Motor Systems: A Handbook on Technology, Programs, and Policy Opportunities, 1991, American Council for and Energy-Efficient Economy.

77 Motor Size - Background Source: Energy-Efficient Motor Systems: A Handbook on Technology, Programs, and Policy Opportunities, 1991, American Council for and Energy-Efficient Economy.

78 Motor Size Anticipated Savings EC = CEU - PEU CEU PEU = = HP HP c p N LF c C 1 H UF ç c N LF p C 1 H UF ç p Ê HP c LF c HP p LF p ˆ DR = Á - C 1 Ë ç c ç p N CF

79 Motor Size Anticipated Savings EC annual energy conservation, kwh/yr CEU current energy usage, kwh/yr PEU proposed energy usage, kwh/yr HP C horsepower of current motor, hp HP P horsepower of proposed motor, hp LF C fraction of rated load at which current motor normally operates LF P fraction of rated load at which proposed motor will operate N number of motors of a given size, no units

80 Motor Size Anticipated Savings C 1 conversion constant, kw/hp UF usage factor H annual operating time of equipment driven by motor, h/yr _ C estimated efficiency of existing motor _ P estimated efficiency of proposed motor DR demand reduction, kw/mo CF coincidence factor

81 Motor Size Sample Problem During an assessment, two 100 hp vacuum pump motors were metered throughout the day. It was determined that the North side vacuum pump was continuously loaded at 80 hp. The metering of the South side vacuum pump showed that it never needed more than 40 hp and that the average load was 35 hp. Both pumps run continuously throughout the year. What do you suggest they do and what are the associated savings? Given: Efficiency of 100 hp 35% rated load = 85% Unit electric usage cost = $0.05/kWh Unit demand cost = $10.00/kW Horsepower Rating (hp) Efficiency at 75% rated load

82 Motor Resources OIT BestPractices Program: Motors - Baldor Motor and Drives - Energy-Efficient Motor Systems: A Handbook on Technology, Programs, and Policy Opportunities, 1991, American Council for an Energy-Efficient Economy