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Electrical Systems and Motors Flashcards

7 cards from real CEM practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 7 Electrical Systems and Motors flashcards as text

  1. When performing a motor system energy audit, which field measurement tool is used to measure three-phase power, power factor, and harmonic content simultaneously?

    Answer: Power quality analyzer

    A power quality analyzer simultaneously measures voltage, current, real power, reactive power, power factor, and harmonic spectrum on all three phases.

  2. A motor rewind that reduces efficiency by 1-2% is most commonly caused by which practice?

    Answer: Burnout oven temperatures exceeding 700°F degrading the lamination insulation

    Excessive burnout oven temperatures damage the interlaminar insulation in the stator core, increasing iron losses and reducing efficiency by 1-2% or more per rewind.

  3. According to NEMA MG-1, what is the allowable voltage unbalance limit for motor operation without derating?

    Answer: 3%

    NEMA MG-1 recommends that motors not be operated with a voltage unbalance exceeding 1% without derating; above 1%, derating is required and above 5%, motor operation is not recommended.

  4. What does a 'motor management' or 'motor rewind vs. replace' analysis typically use as the financial decision threshold?

    Answer: The 40% rule: replace when rewind cost exceeds 40% of new motor cost

    The commonly cited '40% rule' suggests replacement when rewind cost exceeds 40% of the cost of a new premium efficiency motor, accounting for efficiency improvement value.

  5. In a lighting system energy audit, what is the efficacy of a light source measured in?

    Answer: Lumens per watt (lm/W)

    Luminous efficacy is expressed in lumens per watt (lm/W), representing how efficiently a light source converts electrical power into visible light.

  6. A transformer nameplate shows 1,000 kVA, 13,800V/480V, 60 Hz, with 5.75% impedance. What is the maximum available fault current at the secondary terminals?

    Answer: 25,070 A

    Full load secondary amps = 1,000,000 ÷ (√3 × 480) = 1,202.8 A; fault current = 1,202.8 ÷ 0.0575 ≈ 20,920 A — approximately 25,070 A when source impedance is also considered.

  7. Which type of electrical loss is reduced when conductors are upsized during an energy audit recommendation?

    Answer: I²R (resistive) losses

    Larger conductors have lower resistance, directly reducing I²R losses (copper losses) which are proportional to the square of current multiplied by conductor resistance.