Reactive Power compensation - Essay Example

This is due to the fact that it has no reactive power at all. As a matter of fact, its reactive power is equivalent to zero. In this case, the power triangle mimics and horizontal line. This should logically be so noting that the opposite side which represents reactive power has a length of 0 cm. inappropriate power factor can be rectified, paradoxically, through addition of an extra load to the circuit. In essence, the added load is equivalent reactive power acting in an opposite direction. The addition cancels the effects resulting from a load's inductive reactance.

Notably, only capacitive reactance can cancel the inductive reactance and hence a parallel capacitor is added to the provided circuit to act as the extra load. As a result of the impact resulting from the two reactance acting in opposite directions, and parallel to each other, the circuit's total impedance becomes equivalent to the entire resistance. This assists in making the impedance phase angle equivalent, or in the least tends towards zero. Having the knowledge that that the un-rectified reactive power is 561.

724 VAR (inductive), there is a need to derive the right size of a capacitor to generate an equivalent amount of reactive power. Given that the identified capacitor will act in a direction parallel to the source, the following formula is applied in calculation and it begins with identification of voltage and reactance: But And hence, The simulation is done using a rounded of capacitor value of 29, yielding the following results, True power = 447.002 Apparent power = 447.008 For case 2, where capacity improves power factor to 0.

95 lagging, Circuit sketch The circuit has both inductance and resistance and hence the two are combined to form, Given that, P = True power, Q= Reactive power, and S = Apparent power P is given as, S is given as, Q is given as, Redrawing the circuit, we have Resistive/reactive load: For power factor = 0.95 Consequently, This indicates the capacitive reactance XC must be Original XL - Improved PF XL = 80.2986 – 16.434 = 63.8646 ohms Simulating this, a 20 is used, as shown True power = 447.