Lighting Systems 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 Lighting Systems flashcards as text
A facility has 500 luminaires each with two 32W T8 lamps and a ballast consuming 6W. What is the total connected lighting load in kW?
Answer: 35 kW
Power per luminaire = 2×32 + 6 = 70W; total = 500 × 70W = 35,000W = 35 kW.
Which photometric measurement describes luminous intensity distribution in a plane containing the luminaire's axis, used to generate isolux diagrams?
Answer: Candlepower distribution curve
Candlepower distribution curves (polar intensity plots) show how a luminaire distributes light in specific planes, enabling isolux calculations.
Stroboscopic effect in lighting is a concern primarily because:
Answer: It can make rotating machinery appear stationary, creating safety hazards
When light flickers at frequencies matching the rotation speed of machinery, equipment can appear stationary, posing serious workplace safety risks.
A lighting energy audit reveals that a building operates lights 24/7 in spaces unoccupied at night. Installing occupancy sensors reduces operating hours from 8,760 to 5,000 hours/year. If lighting uses 50 kW, what is the annual energy savings at $0.10/kWh?
Answer: $18,800
Hours saved = 8,760 – 5,000 = 3,760 h; savings = 50 kW × 3,760 h × $0.10 = $18,800.
Which commissioning activity verifies that installed lighting controls perform as specified after construction is complete?
Answer: Functional performance testing
Functional performance testing (a commissioning process) verifies that sensors, dimmers, and scheduling controls operate correctly under real conditions.
When comparing LED to high-pressure sodium (HPS) for outdoor area lighting, LED provides an advantage in:
Answer: Instant restrike capability and better color rendering
LEDs strike instantly with no warm-up and offer CRI 70–90+, while HPS requires minutes to restrike after outage and has CRI ~20–25.
In the context of energy auditing, a Lighting Power Density (LPD) calculation for a classroom that is 900 sq ft with 14 fixtures at 54W each yields approximately:
Answer: 0.84 W/ft²
LPD = (14 × 54W) / 900 ft² = 756 / 900 = 0.84 W/ft².