Question

The monthly readings of a consumer's meter are as follows;

Maximum demand = 50 kW

Energy consumed = 36,000 kW ⋅ h

Reactive energy = 24,000 kVAR ⋅ h

What are the average load and average reactive power?

1. 1000 kW and 1,500 kVAR
2. 33.3 kW and 50 kVAR
3. 1500 kW and 1,000 kVAR
4. 50 kW and 33.3 kVAR

Answer 1

Correct Answer - Option 4 : 50 kW and 33.3 kVAR

Concept:

Energy consumed, \(E = {{P} \times T}\)

Where P = Average load or Average power in kW

T = Time in hours

Reactive energy, \(E_R = {{Q} \times T}\)

Where Q = Reactive power in kVAR

T = Time in hours

Calculation:

Given that monthly E = 36,000 kWh, monthly E_R = 24,000 kVARh

The average load is given by

\(P = {{E} \over T} = {{36000} \over 30× 24} \)  (∵ 1 month = 30 × 24 hours)

∴ P = 50 kW

Now average reactive power is given by

\(Q = {{E_R} \over T} = {{24000} \over 30× 24} \)   (∵ 1 month = 30 × 24 hours)

∴ Q = 33.3 kVAR