Question

An industrial consumer has a load pattern of 2000 kW at 0.8 lagging for 12 hours and 1000 kW at unity power factor for 12 hours. Its load factor is
1. 0.6
2. 0.8
3. 0.5
4. 0.75

Answer 1

Correct Answer - Option 4 : 0.75

Concept:

Load factor: The ratio of average load (AL) to the maximum demand (MD) during a given period is known as the load factor.

\(Load factor =\dfrac{AL}{MD}\)

If the plant is in the operation of T hours

\(Load factor = \dfrac{{AL\times T}}{{MD \times T}}\)

Note: To find Load factor, Demand Factor, Diversity Factor, etc, we used unit of Power in kW or W, and unit of Energy in kWh or Wh, not in kVA or kVAh

Application:

Given,

P₁ = 2000 kW for 12 hr,

P₂ = 1000 kW for 12 hr,

Since, maximum power is P₁ hence,

MD = 2000 kW

Now, Average load (AL) can be calculated by the ratio of total energy consumed in kWh to the total time

\(AL=\dfrac{(2000\ kW\ \times\ 12\ hr)+(1000\ kW \times\ 12\ hr)}{24\ hr}=1500\ kW\)

From above concept,

\(Load \ factor=\dfrac{AL}{MD}=\dfrac{1500}{2000}\)

Load Factor = 0.75