CEM Electrical Power Systems & Motors 4 — Questions and Answers
Question 1: What is the purpose of IEEE 519 in industrial power systems?
- To define motor efficiency test procedures
- To set harmonic current and voltage distortion limits at the point of common coupling (Correct answer)
- To establish transformer kVA sizing guidelines
- To specify motor insulation class ratings
Correct answer: To set harmonic current and voltage distortion limits at the point of common coupling
IEEE 519 establishes recommended limits for harmonic currents injected into the utility and harmonic voltage distortion at the point of common coupling (PCC).
Question 2: What happens to motor efficiency when it is operated significantly below its rated load (e.g., at 25% load)?
- Efficiency increases proportionally with reduced load
- Efficiency drops sharply because fixed losses become a larger fraction of input (Correct answer)
- Efficiency remains constant down to zero load
- Only copper losses increase; core losses decrease
Correct answer: Efficiency drops sharply because fixed losses become a larger fraction of input
Fixed losses (core, friction, windage) remain constant, so at very low loads they dominate input power, significantly degrading efficiency.
Question 3: In a wye (star) connected 3-phase system with 480 V line-to-line voltage, what is the phase voltage?
- 480 V
- 240 V
- 277 V (Correct answer)
- 340 V
Correct answer: 277 V
Phase voltage in a wye system = Line voltage / √3 = 480 / 1.732 ≈ 277 V.
Question 4: What is the primary advantage of a synchronous motor over an induction motor for large industrial drives?
- Lower initial purchase cost
- Ability to operate at leading power factor, providing reactive power compensation (Correct answer)
- Higher slip and smoother speed variation
- Simpler starting circuit requiring no external controls
Correct answer: Ability to operate at leading power factor, providing reactive power compensation
Synchronous motors can be over-excited to operate at leading power factor, effectively acting as a capacitor bank and improving facility power factor.
Question 5: A 100 kW motor has a power factor of 0.80. What is the apparent power (kVA) drawn from the utility?
- 80 kVA
- 125 kVA (Correct answer)
- 100 kVA
- 110 kVA
Correct answer: 125 kVA
Apparent power (kVA) = Real power (kW) / Power factor = 100 / 0.80 = 125 kVA.
Question 6: Which loss component in a transformer increases with the square of the load current?
- Hysteresis loss
- Eddy current loss
- Copper (I²R) loss (Correct answer)
- Stray no-load loss
Correct answer: Copper (I²R) loss
Copper losses (I²R) vary with the square of load current, making them the dominant variable loss at high loading conditions.
Question 7: What is the recommended approach when performing a simple payback calculation for a motor replacement project?
- Use nameplate efficiency for both old and new motors
- Measure actual load and operating hours; use measured efficiency at that load point (Correct answer)
- Assume motors always operate at 100% of rated load
- Use the motor's synchronous speed instead of actual operating speed
Correct answer: Measure actual load and operating hours; use measured efficiency at that load point
Accurate payback analysis requires field-measured load levels and actual operating hours because motors rarely run at nameplate conditions.
CEM Electrical Power Systems & Motors 4
What is the purpose of IEEE 519 in industrial power systems?