Electrical 80549

The factory hall is lit by 76 light bulbs, each of which has an input of 200W. Determine how much electrical work will be saved annually if the light is on an average of 100 minutes less per day. Consider 260 business days.

Correct answer:

E =  6.6 MWh

Step-by-step explanation:

P=200 W kW=200:1000  kW=0.2 kW  n=76 t=100 min h=100:60  h=1.66667 h d=260  E1=t P=1.6667 0.2=310.3333 kWh  E2=n d E1=76 260 0.3333=3197606586.6667 kWh  E=E2/1000=6586.6667/1000=6.6 MWh



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