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.
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The total load taken from an a.c. supply consists ofA- a heating load ...
Solution:
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 improve the power factor to unity, a capacitor is connected in parallel with the load.
Let the kVAR rating of the capacitor be X.
Now, the total load on the supply after connecting the capacitor in parallel will be:
15 kW + 40 kVA at a power factor of 1 (as kVA = kW/Power factor) + 20 kW at a power factor of 1
= 75 kVA
The total current drawn by the load before connecting the capacitor is:
I = Power/Voltage
For the heating load:
I1 = 15 kW/230V = 65.22 A
For the motor load:
I2 = 40 kVA/(0.6 x 230V) = 120.97 A
For the load:
I3 = 20 kW/(0.8 x 230V) = 108.7 A
Total current before connecting capacitor = I1 + I2 + I3 = 294.89 A
The total current after connecting the capacitor in parallel will be:
I = Power/Voltage
For the heating load:
I1 = 15 kW/230V = 65.22 A
For the motor load:
I2 = 40 kVA/(1 x 230V) = 173.91 A
For the load:
I3 = 20 kW/(1 x 230V) = 86.96 A
Total current after connecting capacitor = I1 + I2 + I3 = 326.09 A
Now, the reactive power supplied by the capacitor is given by:
Q = P*tan(arccos(PF))
Where, PF is the power factor after connecting the capacitor.
For a power factor of 1, the reactive power supplied by the capacitor is zero.
So, the total reactive power required to improve the power factor to unity is:
Q = S*tan(arccos(PF))
Where, S is the apparent power of the load.
S = 75 kVA
PF = cos(arctan(0.6)) = 0.8 lagging
Q = 75*tan(arccos(0.8)) = 47.03 kVAR
Therefore, the kVAR rating of the capacitor should be 47 kVAR.