Electrical Power Systems Technology

Electrical Power Conversion Systems 295

Sample Problem:

Given: a factory has a peak demand of 12 MW and an average power

demand of 9.86 MX.

Find: the load (demand) factor for the factory.

Solution:

Avg. Demand

DF = ———————

Peak Demand

9.86 MW

= ————

12MW

DF = 0.82

The average demand of an industry or commercial building is the

average electrical power used over a specific time period. The peak de-

mand is the maximum amount of power (kW) used during that time pe-

riod. The load profile shown in Figure 11-1 shows a typical industrial de-

mand-versus-time curve for a working day. Demand peaks that far exceed

the average demand cause a decrease in the load factor ratio. Low load

factors result in an additional billing charge by the utility company.

Utility companies must design power distribution systems that take

peak demand time into account, and ensure that their generating capac-

ity will be able to meet this peak power demand. Therefore, it is inefficient

electrical design for an industry to operate at a low-load factor, since this

represents a significant difference between peak-power demand and aver-

age-power demand. Every industry should attempt to raise its load factor

to the maximum level it can. By minimizing the peak demands of indus-

Figure 11-1. Load profile for an industrial plant