"Motors, Power, and Data Loggers Greg Jourdan-Wenatchee Valley College Tuesday, May 8, Sessions Session 1-8:30-9:25 a.m. Motors 101 Session

"Motors, Power, and Data Loggers Greg Jourdan-Wenatchee Valley College Tuesday, May 8, 2018 3 Sessions Session 1-8:30-9:25 a.m. Motors 101 Session 2-9:30-10:25 a.m. Power and Motors Session 3-10:30-10:25 a.m. Plug Load and Data Loggers

Session 2 Topics and Activities This hands-on discussion will cover calculating techniques to determine how much your power your electric equipment, motors, and miscellaneous electric loads cost per day, month, and year. Calculating electric cost of motors and equipment (30 minutes) Introduce motor operating cost calculations How to determine HP load Nameplate vs actual Amp method and Slip Method How to determine operating hours using BAS Motor logger Calculating operating cost

Session 2 Topics and Activities This hands-on discussion will cover calculating techniques to determine how much your electric equipment, motors, and miscellaneous electric loads cost per day, month, and year. 15 minutes hands on activity: Calculating the operating cost of a standard efficiency and high efficiency motor Use BOC and reference materials to calculate the energy use and operating cost of an electric motor. 10 minutes question & Answer

Electric Motor Management A BOC continuing education class

Objectives Review Electrical Terms and Formulas Calculate motor operating costs Become familiar with a motor database for repair/replace decision making Review a model repair specification Review motor maintenance practices

Discussion Questions Do you know how many motors are in your facility? Do you have (and actively use) a motor management program? Do you have a repair/replace procedure for motors that fail?

Definitions for Common Electrical Terms Terms Description Symbol Unit Voltage or Electromotive Force Force that causes electrons to flow V or E Volt Current Flows of electrons A or I Ampere Resistance Opposition of flow of electrons R or Ω Ohms Watts Power used in a circuit W or kw Watts or kwatts

Ohms Law Power Wheels Resistive vs Inductive Circuits

Activity 1 Use Ohm s Law to solve problems 1. P = E = I = 10 R = 12 2. P = E = 480 I = 24 R = 3. P = E = 24 I = 24 R = 4. P = 4,500 E = 240 I = R = 5. P = E = 208 I = 44 R = 6. P = 15,000 E = I = 30 R = 7. P = E = 12,480 I = 60 R = 8. P = 95 E = I = R = 2.5

Activity 1 Use Ohm s Law to solve problems-answers 1. P = 1200 E =120 I = 10 R = 12 2. P = 11520 E = 480 I = 24 R = 20 3. P = 576 E = 24 I = 24 R = 1 4. P = 4,500 E = 240 I = 18.75 R = 12.8 5. P = 9152 E = 208 I = 44 R = 4.72 6. P = 15,000 E = 500 I = 30 R = 16.6 7. P = 748,8000 E = 12,480 I = 60 R = 208 8. P = 95 E = 15.41 I = 6.16 R = 2.5

Test Your Knowledge - Question 1 Using Ohm s Law, calculate the amps used by 1500 watt electric heater if the voltage is 120 Vac at the power outlet or at the power strip. 1. 0.50 amps 2. 2.00 amps 3. 2.40 amps 4. 12.5 amps

Test Your Knowledge - Question 2 What can happen if too many appliances are plugged into the same outlet on the same circuit? 1. Increased load in the circuit will increase voltage and cause the breaker to trip. 2. Increased load in the circuit will increase amperage in the circuit, and cause the breaker to trip. 3. A decrease in voltage will cause an increase in current and cause the breaker to trip. 4. Decreased load will increase the resistance and cause the breaker to trip.

Class Discussion 3 Portable space heater & power strip One (1) electric 1.5 kw portable heater connected to a power strip P T = E T = 120 Vac I T = R T = Totals L1 1. P T = 1,500 P 1 = 1,500 2. E T = 120 E 1 = 30 3. I T = 3 x I 1 = 3 4. R T = 0 R 1 = 10

Class Discussion 4 Portable space heater & power strip Complete the Parallel Circuits Assume all loads are equal Exercise with two (2) electric 1.5 kw portable heaters connected to same power strip P T = 3.9 kw E T = 120 Vac I T = 12.5 amps R T = 4.8 ohms Totals L1 L2 1. P T = 3,000 P 1 = 1,500 P 2 = 1,500 2. E T = 120 E 1 = 120 E 2 = 120 3. I T = 25 I 1 = 12.5 I 2 = 12.5 4. R T = 4.8 R 1 = 9.6 R 2 = 9.6

Class Discussion 5 Solve in Class Using Ohms Law to Determine: kw, kwh, and Cost of Operation Compressor Operating Electrical VAC = 480, 300 amps,.88 PF Formula for kw (3 phase) = Volts x Amps x 1.732 x PF / 1000 kw = Volts x Amps x 1.732 x PF = 1000 kwh/day = x 24 hours = Cost of electricity/day @.06/kW = x = $ Cost of electricity/month = $ x 30 days = $

Practice Exercise Solve in Class Using Ohms Law to Determine: kw, kwh, and Cost of Operation Compressor Operating Electrical VAC = 480, 300 amps,.88 PF Formula for kw (3 phase) = Volts x Amps x 1.732 x PF kw = 480 Volts x 300 Amps x 1.732 x.88 PF = 193.14 kw 1000 kwh/day = 193.14 x 24 hours = 4,635 Cost of electricity/day @.06/kW = 4,635 x.06 = $278.12 Cost of electricity/month = $278.12 x 30 days = $8,343/month

Motors: Choose Well $780 Car and Motor 15 HP Electric Motor $811 $9,000 Purchase Price Annual Fuel Cost $4,258 Purchase Price Annual Fuel Cost

Motor Efficiency Car Mileage Upgrade 25 mpg to 30mpg 20% increase in efficiency Life savings = $1,300 15hp Motor Upgrade 86% to 90% efficient 4% increase in efficiency Life savings = $3,867

Nameplate Nomenclature

Nameplate Definitions Manufacturer Catalog/Model Number Specification Number Serial Number

Nameplate Definitions Horsepower HP = T x S 5252 16' 6 " 33' T = Torque in FT. LBS S = Speed in RPM 2000 lb 1000 lb One horsepower equals 33,000 foot pounds per minute or lifting 2,000 lbs 16 1/2 ft. or 1,000 lbs 33 ft. in one minute

Motor Starting Current Code Letter Range Code Letter Range A 0.00-3.14 B 3.15-3.54 C 3.55-3.99 D 4.00 4.49 E 4.50 4.99 F 5.00 5.59 G 5.60 6.29 H 6.30 7.09 J 7.10-7.99 K 8.00 8.99 L 9.00 9.99 M 10.0 11.19 N 11.2 12.49 P 12.5 13.9 R 14.0 15.9 Single Phase Inrush Amperes = Code Letter x HP x 1,000 / rated voltage Three Phase Inrush Amperes = Code Letter x HP x 577 / rated voltage

Efficiency Typical 5HP, 4 Pole, 3 Phase 85% Efficient Motor 100% Power In 85% output Power Energy Losses Stator Res. Rotor Res. Iron Core Friction/ Windage Stray Load Total Percent of 25% 20% 5% 10% Losses 40% 100% Percent of Input 6.2% 3.6% 3.1% 0.7% 1.4% 15%

% Efficiency Efficiency Curve Typical 20 HP, 1725 RPM, 3 phase 100 80 60 Nameplate HP or 100% Rated Load 40 20 0 4 8 12 16 20 24 Output Horsepower Note, many motors peak or best efficiency is around 75% load

Efficiencies

Power Factor and Efficiency AMPS Typical Motor Characteristics 15HP, 1725 RPM, 200 Volt 100% 80% Efficiency 50 40 60% 30 40% 20% 20 10 0 2 4 6 8 10 12 14 16 18 Horsepower Nameplate HP or 100% Rated Load

Which uses more electricity? 75 HP, 1770 RPM 100 HP, 1770 RPM If these motors have similar performance characteristics, each coupled to loads requiring 75 HP and run the same amount of time, which uses more electricity less efficiently?

Apples for Apples NEMA standardizes performance data (full load values) and allows easy comparison and replacement of similar motors.

Power Factor Power Factor Efficiency

Power Factor Efficiency Current Power Power Factor

Power Factor Efficiency Current Power Power Factor Some motor current is used to create magnetic fields

Power Factor Efficiency Current Power Power Factor Some motor current is used to create magnetic fields It rushes in, and then rushes out

Power Factor Efficiency Current Power Power Factor Some motor current is used to create magnetic fields It rushes in, and then rushes out It does no useful shaft work

Power Factor Efficiency Current Power Power Factor Some motor current is used to create magnetic fields It rushes in, and then rushes out It does no useful shaft work HOWEVER, it burdens the utility distribution system

Power Factor Power Factor Efficiency Current Power Some motor current is used to create magnetic fields It rushes in, and then rushes out It does no useful shaft work HOWEVER, it burdens the utility distribution system The distribution system must move more current for a given customer load The utility wants this burden to be minimized

Power Factor Explained (For Electrical and Power Engineers)

Power Factor Explained (For Building Maintenance Staff)

Power Factor Explained (for Beer Drinkers) Apparent Power is not Real Power

Power Factor Can Be Corrected with VAR Managers or Better Motor Drives

Calculating Motor Operating Costs Formulas Tools Exercise using motor nameplate data

Electric Motor Operation Cost Formula (HP x Hours Of Operation) x.746 Motor Efficiency (Decimal) = Kilowatt Hours (kwh) kwh x Aggregate Cost per kwh = Approximate Operation Cost What values should be used?

Calculating Power on 3 Phase (from Motor Management Handout)

Solve this Real World Problem What is the kw?

Ingredients to Use For comparing motors in today s exercises motor nameplate (full load) Calculated Given Motor in question HP x Hours of Operation x.746 Motor Efficiency (Decimal) = Kilowatt Hours or kwh May or may not be on motor nameplate (full load) (kwh x Aggregate Cost per kwh) = Approximate Operating Cost Motor in question, from above Price from power bill The objective (at nameplate or full load)

Let us Practice Hours of Operation Determine annual hours of motor operation (HP x Hours of Operation) x.746 Motor Efficiency (Decimal) = Kilowatt Hours or kwh

Total Hours per Year For example, annual hours = 2,400 (HP x Hours of Operation) x.746 Motor Efficiency (Decimal) = Kilowatt Hours or kwh

Your Power Bill Simplified Electric Bill On-Peak kwh: 27,600 Cost: $888.72 On-Peak Demand kw: 407 Cost: $429.80 Off-Peak kwh: 14,000 Cost: $372.40 Off-Peak Demand kw: 311 Cost: $176.40 Power Factor Avg.: 0.84 Cost: $217.00 Please Remit (TOTAL): $2,084.32 kwh x Aggregate Cost per kwh = Approximate Operating Cost

Aggregate Cost of Power Simple Electric Bill On-Peak kwh: On-Peak Demand kw: Off-Peak kwh: Off-Peak Demand kw: Power Factor Avg.: 27,600 407 14,000 311 0.84 Cost: $888.72 Cost: $429.80 Cost: $372.40 Cost: $176.40 Cost: $217.00 Please Remit (TOTAL): $2,084.32 Total $ of Power Bill On Peak kwh + Off Peak kwh =? Price per kwh

Price per Kilowatt Hour Simple Electric Bill On-Peak kwh: On-Peak Demand kw: Off-Peak kwh: Off-Peak Demand kw: Power Factor Avg.: 27,600 407 14,000 311 0.84 Cost: $888.72 Cost: $429.80 Cost: $372.40 Cost: $176.40 Cost: $217.00 Please Remit (TOTAL): $2,084.32 $ 2,084.32 27,600 + 14,000 = $ 2,084.32 41,600 = $ 0.05 per kwh kwh x Aggregate Cost per kwh = Approximate Operating Cost

Percentage of Efficiency 88.0 87.5 87.5 88.5 89.0 90.0 91.0 92.0 93.0 94.0 Trends in Motor Efficiency 96 94 20 HP, 1800 RPM, TEFC EPAct Effective 1997 92 90 88 86 84 1945 1955 1965 1975 1985 1995 2001 Source Baldor Electric Solid Bar: Standard or Epact Efficiency Stripe Bar: Premium Efficiency

1800 RPM, TEFC 20 HP, 2400 Hrs, $0.05 per kwh Year 1945 1955 1965 1975 Standard Efficiency 88% 90% 88.5% 87.5% Year 1985 1995 2001 Standard Efficiency 87.5% 89% 91% Epact Premium Efficiency 92% 93% 94% ( HP x 2400 hrs) x.746 Efficiency (Decimal) = kwh kwh x $ 0.05 = $ Est. Operating Cost

20 HP, 2400 Hrs, $0.05 per kwh Year 1945 1955 1965 1975 1985 1995 2001 Efficiency 88% 90% 88.5% 87.5% 87.5% 89% 91% Epact Operation Cost $ 2,034.55 $ 1989.33 $ 2,023.05 $ 2046.17 $ 2,046.17 $ 2,011.69 $ 1,967.47 Efficiency 92% 93% 94% Operation Cost $ 1,946.09 $ 1925.16 $ 1,904.68 New Motor 2001 Epact Price New Motor 2001 Premium Price Complete Rewind (approx) Price $721.00 $828.00 $500.00 10 year life power cost of 2001 Epact $19,674.70 10 year life power cost of 2001 Premium $19,046.80

Motor Management Inventory all connected & stand-by motors Make pre-failure repair/replace decisions Use repair specifications with Repair Center Vendors

Motor Data Base Does anyone have motor information on file? Motor data entered in software?

Motor Data Base Benefits Saves Money and Time Performance- Motor and Systems Maintenance-Electrical and Mechanical Energy-Process and Plant-

Motor Database Software https://www.energy.gov/eere/amo/downloads /motormaster-tool

Data Collection Exercise How Many Motors in Class Today? Vs.

Data Collection Exercise Which is Easiest for Inventory Control? Vs.

Motor Maintenance

Bearing Lubrication Safety first priority What is enough? Follow specifications -- Remove relief plug -- Let motor run for 20-30 minutes before replacing plug

Should You Over Grease?

Over Greased?

Good Motor Maintenance Ventilation restriction problems corrected Corrected misalignment-use Laser Technology Lubricate per manufacturer s specs Test for/avoid imbalanced/improper voltage Avoid excessively high ambient temps Correct overloading beyond 100% Correct excessive vibration electronically Use a repair purchase specification

What to Look For Insulation Degradation Motor Temperatures Humidity Dust Dirt Insects and Animals

Failure Causes Misalignment

51.7% Bearings Motor Failures 50% 40% 35% 30% 25% 20% 15% 10% 5% 15.76% Stator Winding 15.58% External Device 15.20% Not Specific 4.70% Rotor Bars/End Rings 2.44% Shaft or Coupling Source EASA

Down Size the Task of Data Gathering Motor Information If you choose to collect motor information using the tools presented today, here s some advice: 1. Start with large motors and those that run long hours. 2. Add critical, replacement motors or production lines as time permits.

Hands On Exercise 15 minutes hands on activity: Calculating the operating cost of a standard efficiency and high efficiency motor in front of you or at the bench. Use reference materials to calculate the energy use and operating cost of an electric motor. Figure.10/kW/cost Figure your motor runs ½ of the year

List Things To Do to Improve your 1. Motors Operating Costs 2. 3. Put together and maintain a Motor Database Include Motor Operating Costs in decision making Commit to keeping the database up to date, easy to use and accessible Use a Repair Purchasing Specification

Summary Power and Motors Small changes to motor load, efficiency and run time make big differences in operating cost. Knowing your motor operating costs enables you to make effective repair/replace decisions when motors fail. A motor database is the key to effectively managing your motors and tracking operating costs. Free motor database tools are available.