Energy Efficiency in Lighting Questions and Answers

Answer: A

Correct answer: raises the utilisation factor

Light striking a pale surface is reflected back into the room instead of being absorbed, so a greater proportion of the flux emitted reaches the working plane. The lamps are unchanged; what improves is how much of their output is put to use, and in a room with high walls the effect is substantial.

Answer: B

Correct answer: intermittently used spaces such as stores and washrooms

A sensor saves the hours during which the space is lit and empty, so the saving is proportional to how much of the lit period is unoccupied. A store lit ten hours and used two offers eighty per cent; a production hall occupied throughout the shift offers nothing at all.

Answer: C

A photocell measures the working plane and trims the luminaires so the total stays at the design level, so the perimeter of a daylit building may need very little electric light for much of the day. Abrupt switching is disliked by occupants, which is why dimming is preferred.

Answer: D

Correct answer: switching areas separately so unused areas can be dark

Zoning means the switching follows the way the space is used, so a single occupant working late does not light the whole floor. It costs only wiring and switches at the outset and is nearly impossible to retrofit cheaply, which is why it belongs in the original design.

Answer: A

Lighting a whole workshop to the level a fine inspection bench needs is enormously wasteful, and a local luminaire at the bench with a lower ambient level gives better seeing for less energy. Some general lighting is always retained for circulation and comfort.

Answer: B

Correct answer: replacing magnetic ballasts with electronic ones

A better ballast reduces the watts drawn while the lamp is on, which is an efficiency measure. Controls reduce the hours the lamp is on - zoning, occupancy sensing and daylight linking all work that way. Both save energy, but by different means, and a good scheme uses both.

Answer: C

Correct answer: 1500 lm

\[ 100 \times 15 = 1500\ \mathrm{lm}. \] Efficacy multiplied by power gives flux, and the same 1,500 lumens would come from about 12.5 watts of a modern diode luminaire - which is the whole case for the replacement in one comparison.

Answer: D

Correct answer: 12.5 W

\[ \frac{1500}{120} = 12.5\ \mathrm{W}. \] Flux divided by efficacy gives power, which is the previous calculation run backwards. Multiplying instead of dividing gives 180,000, and inverting the efficacy carelessly gives 125 - the latter offered here.

Answer: A

Correct answer: 26 100 kWh a year

\[ (100 - 13) \times 100 \times 3000 = 26{,}100{,}000\ \mathrm{Wh} = 26{,}100\ \mathrm{kWh}. \] The saving is the difference in wattage, not the new wattage or the old one, so 3,900 and 30,000 are what those two mistakes produce - and both are offered.

Answer: B

Correct answer: 22 per cent

\[ \frac{46 - 36}{46} = \frac{10}{46} = 21.7\ \mathrm{per\ cent}. \] Taking the loss against the lamp wattage instead gives 28 per cent, which is offered here. This is why an installation must be counted at circuit watts rather than at the figures marked on the tubes.