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Volume 8 – Mechanical and Electrical Services - Malaysia Geoportal

Volume 8 – Mechanical and Electrical Services - Malaysia Geoportal

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Chapter 7 ELECTRICAL SERVICES<br />

Room Index , K is given as:<br />

L x W (7.3)<br />

Hm( L+W)<br />

Where:<br />

L = length of room<br />

W = width of room<br />

Hm = mounting height of fitting above working plane.<br />

Maintenance factor (MF) is estimated for the effective delivery of average illumination reaching the<br />

working surface. Dirt on the fitting has the effect of reducing its light output from it. The<br />

conventional assumption is that on average, a lighting installation delivers 80% of the light it would<br />

do if it were perfectly clean. Thus the average maintenance factor, MF= 0.8. A higher maintenance<br />

factor, say 0.9, can be assumed if the fittings are cleaned regularly, or it could be as low as 0.5 in a<br />

foundry. Taking dirt into account, the modified illumination formula is<br />

E = CU x MF x installed flux per unit area. (7.4)<br />

Practical Design Example by considering a classroom of floor dimension 7.5m by 9m <strong>and</strong> ceiling<br />

height of 3m. The lumunaire chosen is 1.2m, 36W fluorescent lamps. IES st<strong>and</strong>ards recommended<br />

an illumination of 300 lux for reading room.<br />

To determine how many lamps are needed to achieve this illumination level<br />

Assume the table top is 0.76 m high, The mounting height is 2.1m, then L= 9m, W= 7.5m,<br />

Hm=2.1m.<br />

Room Index, K = L x W<br />

Hm( L+W)<br />

= 9 x7.5<br />

2.1x (9+7.5)<br />

= 1.95<br />

= 2 (approximately)<br />

As the ceiling <strong>and</strong> walls are painted white, reflection factor of ceiling of 70% <strong>and</strong> walls of 50% can<br />

be reasonably assumed. From technical data obtainable from manufacturer, or IES guide book,<br />

Co-efficient of utilization, CU = 0.6<br />

Take maintenance factor, MF = 0.8<br />

Therefore, installed flux required = illumination flux x area<br />

CU x MF<br />

= 300(lux) x (9 x7.5)<br />

0.6 x 0.8<br />

= 42,187.5 lm<br />

The lighting design lumens (LDL) = 3650 lm<br />

of 1 x36W fluorescent lamp<br />

Therefore, number of fluorescent = 42,187.5<br />

lamp required. 3650<br />

= 11.6<br />

= 12 nos.(approximately)<br />

7-4 March 2009

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