CEM Compressed Air and Power Systems 2 — Questions and Answers
Question 1: Power factor correction in industrial facilities is important because low power factor:
- Reduces the voltage available to equipment
- Increases current draw for the same real power load, increasing I²R losses and potentially incurring utility penalties (Correct answer)
- Reduces the efficiency of electric motors at full load
- Eliminates the ability to use variable frequency drives
Correct answer: Increases current draw for the same real power load, increasing I²R losses and potentially incurring utility penalties
Low power factor means more current flows for the same real (kW) load, increasing resistive losses in conductors and transformers and often triggering utility demand or power factor penalty charges.
Question 2: The nameplate efficiency of an electric motor represents its efficiency at:
- 25% of rated load (quarter load)
- 100% of rated full load under standard test conditions (Correct answer)
- 50% of rated load (half load)
- The minimum load the motor is designed to handle
Correct answer: 100% of rated full load under standard test conditions
Motor nameplate efficiency is measured at full rated load per NEMA standards; actual efficiency varies with load and is typically lower at very light loads.
Question 3: NEMA Premium efficiency motors save energy compared to standard efficiency motors primarily because they have:
- Smaller frame sizes requiring less copper
- Reduced electrical (I²R) and core (iron) losses through better materials and design (Correct answer)
- Higher operating speeds at the same frequency
- Lower starting torque reducing inrush current
Correct answer: Reduced electrical (I²R) and core (iron) losses through better materials and design
Premium efficiency motors use higher grades of silicon steel, more copper in windings, closer tolerances, and improved designs to reduce core and copper losses, achieving 2-8% better efficiency than standard motors.
Question 4: Demand response programs allow large energy users to save money by:
- Generating their own electricity to avoid grid purchases
- Curtailing or shifting electricity use during grid peak events in exchange for utility payments or rate credits (Correct answer)
- Installing on-site storage to level demand peaks
- Negotiating lower base rates with the utility for year-round use
Correct answer: Curtailing or shifting electricity use during grid peak events in exchange for utility payments or rate credits
In demand response programs, facilities agree to reduce load when called upon by utilities or grid operators during high-demand events, receiving payments or bill credits in return.
Question 5: Transformer losses consist of two components. Which component varies with load?
- Core (no-load) losses, which are constant regardless of load
- Copper (load) losses, which vary with the square of the load current (Correct answer)
- Both core and copper losses vary linearly with load
- Neither component varies; total transformer losses are fixed
Correct answer: Copper (load) losses, which vary with the square of the load current
Copper losses (I²R heating in windings) vary with the square of load current; core losses (hysteresis and eddy currents in iron) are essentially constant whenever the transformer is energized.
Question 6: An energy manager evaluating electricity tariffs should focus on the structure of:
- Only the energy charge (cents/kWh) on the bill
- Both the energy charge (kWh), demand charge (kW), and any power factor or time-of-use components (Correct answer)
- The monthly customer charge only
- Only the fuel adjustment clause
Correct answer: Both the energy charge (kWh), demand charge (kW), and any power factor or time-of-use components
Electricity bills often contain multiple components — energy, demand, time-of-use rates, power factor charges — and effective rate optimization requires understanding all components to target the right efficiency measures.
CEM Compressed Air and Power Systems 2
Power factor correction in industrial facilities is important because low power factor: