The total load taken from an a.c. supply consists ofA- a heating load ...
The parallel loads draw a total reactive power of 47 kVAR lagging. In order to raise the p.f. to unity, the capacitor should supply leading reactive power of 47 kVAR.
View all questions of this testThe total load taken from an a.c. supply consists ofA- a heating load ...
Given:
- Heating load = 15 kW
- Motor load = 40 kVA at a power factor of 0.6 lagging
- Load = 20 kW at a power factor of 0.8 lagging
To find:
The kVAR rating of the capacitor to raise the power factor to unity.
Solution:
Step 1: Calculate the total power consumed by the load:
Power consumed by the heating load = 15 kW
Power consumed by the motor load = Apparent power * Power factor = 40 kVA * 0.6 = 24 kW
Power consumed by the load = 20 kW
Total power consumed = 15 kW + 24 kW + 20 kW = 59 kW
Step 2: Calculate the reactive power consumed by the load:
Reactive power consumed by the motor load = Apparent power * sin(θ) = 40 kVA * sin(arccos(0.6)) = 32 kVAR
Reactive power consumed by the load = 20 kW * tan(arccos(0.8)) = 9.798 kVAR
Total reactive power consumed = 32 kVAR + 9.798 kVAR = 41.798 kVAR
Step 3: Calculate the reactive power provided by the capacitor:
Reactive power provided by the capacitor = Total reactive power consumed - Reactive power consumed by the load = 41.798 kVAR - 9.798 kVAR = 32 kVAR
Therefore, the kVAR rating of the capacitor should be 32 kVAR.
Hence, the correct answer is option 'D' (47 kVAR).