SSC JE Electrical Engineering 2014 Set 1 Paper

Question : 192 of 200

Marks: +1, -0

An industrial consumer has a daily load pattern of 2000 kW, 0.8 lag for 12 hours and 1000 kW UPF for 12 hours. The load factor is

Solution:

\Load \factor=

{\Sum \of \individual \Maximum \Demand}/{\Station \Maximum \Demand}

{2000 ×12+ 1000 ×1 ×12}/{2000×24}

= 0.6

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