A 33kVA 240V single-phase alternator supplies a small factory operating at a power factor of 0,75 lagging. If the alternator operates at an efficiency of 90%. Calculate
A) The true power delivered at the factory
B) The current drawn by the factory
C) the active and reactive components of this current
D) the mechanical power required to drive this factory
To solve these calculations, we will use the following formulas:
A) True Power (P) = Apparent Power (S) * Power Factor (PF)
B) Current (I) = Apparent Power (S) / Voltage (V)
C) Active Power (P) = Current (I) * Voltage (V) * Power Factor (PF)
Reactive Power (Q) = Current (I) * Voltage (V) * Square Root(1 - Power Factor (PF)^2)
D) Mechanical Power (Pm) = True Power (P) / Efficiency (η)
Given:
Apparent Power (S) = 33kVA
Voltage (V) = 240V
Power Factor (PF) = 0.75
Efficiency (η) = 90%
A) True Power (P) = 33kVA * 0.75 = 24.75kW
B) Current (I) = 33kVA / 240V = 137.5A
C) Active Power (P) = 137.5A * 240V * 0.75 = 24.75kW
Reactive Power (Q) = 137.5A * 240V * √(1 - 0.75^2) = 18.75kVAR
D) Mechanical Power (Pm) = 24.75kW / 0.90 = 27.50kW