Motor Management at BASF By: T. R. Tom Theising, C.E.M. Energy System Manager BASF Corporation (979) 709-6389 thomas.theising@basf.com
BASF worldwide Ohio Facilities: Beachwood (2) Cincinnati Elyria Greenville Independence Streetsboro Whitehouse 2012 Sales: $98.1 Billion Employees: 110,000
BASF North America 2012 sales Employees Production sites R&D sites $18.5 billion 16,665 99 27 Key customer industries Agriculture Automotive Chemicals Coatings Construction Health & Beauty Packaging 3
Why develop a motor management policy? * Reduce costs * Simplify the repair/replace decision * Reduce down-time * Make decisions before motors fail * Select the right motor when replacement is the decision * Improve the quality of motor repair * Develop procedure for auditing repair shops 4
Electric Consumption by Motors 90% 80% 70% 60% 50% 40% 30% 20% 10% 0% 81% 81% 74% 73% 63% TEXTILE CHEMICAL PAPER FURNITURE TOBACCO 1994-1996 AEC
Advantage of Motor Management example $100k to $175k Annual Savings* 10 MW of actual load X 75% industry average = 7.5 MW 7.5 MW X 8400 Hrs of annual operation = 63,000 MWH 63,000 MWh X $55/MWh = $3.5M ann. cost to operate $3.5 M X 3% eff. improvement = $100k potential savings $3.5 M X 5% eff. improvement = $175k potential savings * These savings would not be immediate 6
Motors Team
Goal Develop a corporate Motor Management Guideline to encourage and facilitate the practical, energy efficient, and cost effective management of BASF s NEMA frame motor population resulting in reduced operating costs. 8
What should a motor management policy include? * Efficiency specifications * Criteria for motor purchase * Inventory management program * Historical tracking system * Repair vs. replace decision methodology * Motor repair guidelines * Repair shop audit procedures * Implementation procedures 9
Motor Management Guideline To fully implement change all sites must participate, and therefore must agree with the concept of motor management. Motor management must be implemented at the site level to be effective. 10
Why Are Site Specific Motor Management Policies Necessary? Existing Policies are: Replace all failed motors below 20 HP and rewind everything above 20 HP Replace everything with Premium Efficiency motors High Efficiency motors are not worth the extra cost Replace everything with IEEE-841 motors 11
Contents of Technical Reference Manual * Motor Repair * Repair vs. Replacement * New Motor Purchases * Documenting Your Motor Inventory * Maintenance of Operating Motors * Maintenance of Spare Motors * Site Procedure Development 12
Final Product * Technical Reference Manual * Tri-Fold Brochure 13
How Can the Tri-fold Brochure help? Simplify the repair/replace decision Make the decision before the motor fails Improve new motor purchasing Improve the quality of motor repair
How Do I Make the Repair Vs. Replace Decision? Calculate your average cost/kwh from utility bills Use your operating hours and cost/kwh to find your horsepower breakpoint on Chart 1 Repair motors larger than breakpoint horsepower 1994-1996 AEC
Rewinding is Economically Viable for: Large Motors Low operating hour motors Specialty motors Metric frames Energy-efficient motors 1994-1996 AEC
Questions? We create chemistry to help power, move, house and feed a growing world population: Wind and solar power Biotechnology QUESTIONS? Energy-efficiency Sustainable construction Water treatment Fuel cell technology About $550 million of BASF s annual research expenditures go into energy efficiency, climate protection, and resource conservation
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Chart 1 Exercise Facility B Runs 3 Shifts, 7 Days Electricity rate is $0.04/kWh If a 100 HP standard efficiency motor fails, should it be: -repaired OR -replaced with a new energy-efficient motor? -Why? 1994-1996 AEC
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Inventory Tracking Trends Decreasing motor life with each repair cycle Repair cost increasing with each repair cycle Increased no-load watts with each repair cycle Repetitive failure mode, ie, bearings or single phasing Repeated failure on one machine 1994-1996 AEC
Personal Logic Rule of Thumb ($0.04/kWh x 24 hr/dy x 0.746 kw/hp)/.90 = $0.80/Hp-Day Therefore: ($0.01/kWh x 24 hr/dy x 0.746 kw/hp)/.90 = $0.20/Hp-Day i.e. @ $0.06/kWh $1.20/Hp-Day; @ $0.08/kWh $1.60/Hp- Day