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ID number:298130
 
Author:
Evaluation:
Published: 22.04.2009.
Language: Latvian
Level: Secondary school
Literature: n/a
References: Not used
Extract

Telpas funkcija
Priekšnams/ halle
Telpas platība m2
5
Jauda W
75
Normatīvs –nepieciešamā apgaismojuma jauda W/ m2
75 : 5 = 15 W/m2

Kopējā apgaismojuma jaudas aprēķins konkrētai telpai
Ja griestu augstums nepārsniedz 300 cm, tad rēķina sekojoši –

Piemēram, ja telpas platība ir 12 m2, tad –
15 W/m2 x 12 m2 = 180 W, (nepieciešams 3 gab. x 60 W kvēlspuldzes)

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