Question

A conference room of size 25 m \(\times\) 12 m is illuminated by CFL having efficacy of 60 lumen/W. The lighting power density (LPD) of the room is 12 \(\text{W/m}^2\). If the utilization and maintenance factors are 0.8 and 0.5, respectively, then the estimated illumination level (in lux) of the conference room is (in integer).

Hint:

Illumination (lux) = LPD \(\times\) lamp efficacy \(\times\) utilization factor \(\times\) maintenance factor; the room area cancels out since LPD is already stated per unit area.

Answer 1

Correct Answer: 288

Another way to reach the same result is to work with the room's total quantities (total watts, total lumens) instead of per square metre values, then divide by the area only at the end.

1. Find the room area: The conference room measures 25 m by 12 m, so its floor area is $25 \times 12 = 300$ $m^2$.
2. Find the total installed lighting power: The LPD is $12$ $\text{W/m}^2$, so the total connected load is $12 \times 300 = 3600$ W.
3. Convert power to raw lumens using efficacy: Each watt gives 60 lumens, so the lamps produce $3600 \times 60 = 216000$ lumens in total.
4. Apply the utilization factor: Only a fraction of the emitted light lands on the working plane; with $UF = 0.8$, the usable lumens become $216000 \times 0.8 = 172800$ lumens.
5. Apply the maintenance factor: Accounting for dirt and lamp depreciation with $MF = 0.5$ gives maintained lumens as $172800 \times 0.5 = 86400$ lumens.
6. Divide by the floor area to get lux: Illumination level is lumens per square metre, so $E = \dfrac{86400}{300} = 288$ lux.

Working with total lumens and dividing by the actual area at the end gives exactly the same number as working per square metre throughout, confirming the result. So the illumination level inside the conference room works out to 288 lux.