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Energy Efficiency in Lighting Questions and Answers

Answer: C

Its own loss is a few watts against ten or more, and the high frequency operation raises the lamp's own efficacy and removes the flicker. Its life is generally shorter and its cost higher, and it needs its power factor and harmonics attended to.

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Answer: D

A 36 W tube on a magnetic ballast draws about 46 W at the circuit, so an audit that counts lamp wattages alone understates the load by a fifth. Circuit watts, measured or taken from the gear's data, is the figure to use.

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Answer: A

Diode output depends steeply on current, so a constant-current supply is what keeps the light output and the life predictable. Its efficiency and its own life are frequently what limit the luminaire rather than the diodes themselves.

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Answer: B

A design that ignored it would fall below the required illuminance within a year or two, so the installation is designed to exceed the standard when new. Cleaning and timely replacement are what keep the assumed factor honest.

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Answer: C

A fluorescent tube may lose a fifth of its output by the end of its rated life and a diode luminaire is rated to the point where it has lost thirty per cent. Measuring the illuminance in an old installation therefore often shows it below standard.

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Answer: D

By the end of life the tubes are giving well below their rated output while consuming full power, and replacing a whole bay at once costs far less per lamp than answering individual failures. Cleaning at the same visit compounds the benefit.

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Answer: A

Correct answer: absorbs light that has already been paid for

The lamp draws the same power whether the luminaire is clean or dirty, so every lumen absorbed by dirt is energy bought and thrown away. Cleaning restores it for the price of the labour alone, which is why it is the first item in any lighting audit and why the maintenance factor exists in the design.

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Answer: B

It depends on the luminaire's distribution, the room's proportions and the reflectances of its surfaces, and it is read from a table against the room index. Light absorbed by dark walls never reaches the task and is counted here.

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Answer: C

Correct answer: length, width and mounting height above the working plane

\[ K = \frac{L \times W}{H_m(L+W)}. \] The index expresses how tall and narrow the room is relative to its plan, since a tall narrow room loses more light to the walls before it reaches the working plane. It is used with the surface reflectances to look up the utilisation factor.

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Answer: D

Total flux required is the illuminance times the area divided by the product of the two factors, and dividing by the flux per luminaire gives the count. It is a design method rather than an audit measurement, but an auditor uses it to judge whether an installation is right.

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