Add capacitors from line to ground.
No-one ever aims to reduce the power factor, the ideal power factor is equal to 1, and that is the maximum possible value. A load with a power factor of 0.7 draws 40% more current along the supply wires compared to a equal-power load with a power factor of 1. That means that the power loss in the resistance of the supply wires is doubled in the case of the poor power factor. Since the supply company receives no extra revenue for the lost power, it does not like this situation and sometimes penalises users with poor power factors with extra tariffs. The power factor can often be improved by placing a passive reactor in parallel with the load to draw off the reactive volt-amps (VAR or kVAR) so that the supply wiring sees a load with a good power factor. Normally a bad load like a motor draws inductive VARs and in this case it can be corrected with a parallel capacitor that draws an equal number of capacitive VARs. Looked at another way, the added capacitor 'tunes' the load to resonate at the supply frequency.
The power in watts equals the VA times the power factor. For a resistive load like a convector heater or an iron the power factor is 1 For other things like motors the power factor might be 0.7. A poor power factor is not a good thing because more current is needed from the supply to produce a given amount of power, so that requires thicker wires (more expensive). For a power factor of 1, 70 kVA = 70 kW For a power factor of 0.7, 70 kVA = 50 kW.
Watts = Volts x Amps x Power Factor. To answer your question requires that the Power Factor be know. The Power Factor ranges from zero to one and is one for a pure resistive load. If your device is resistive the answer is 1500 watts.
Technically, it depends on the power factor of the 50 KVA circuit, which is not stated.If the load is purely resistive, then the power factor is ' 1 ', and50 KW = 67.05 Horsepower .If the power factor is not ' 1 ', then 50 KVA = (67.05 HP) x (power factor).
It depends on the power factor, which depends on the reactance of the load.For a typical power factor of 0.92, 150 KVAR translates to 383 KVA, which translates to 352 KW.Power factor is the cosine of the phase angle (theta) between voltage and current. KVA times cosine (theta) is KW, while KVA times sine (theta) is KVAR.
The most common method of improving the power factor of a load is to connected a capacitor or capacitor bank, of appropriate reactive power (expressed in reactive volt amperes), in parallel with the load.
Improving power factor does nothing to improve a circuit. It merely affects the amount of current drawn by the load.
Power factor doesn't necessarily 'improve with the load', but it is determined by the load.
yes power capacitar ust inprove the power factor and sae the kvah reding
Simple technique by adding PP capacitor with automatic power factor controller.
To improve the power factor
In a circut we use capacitor in series for improving power factor
To improve the power factor
To increase capacitive load and decrease inductive loadAnswerThe most common method is to add a capacitor, or a capacitor bank, in parallel with the load. In practise, the reactive power of the capacitor (they are not rated in farads, but in reactive volt amperes) must be a little short of being equal to the reactive power of the load, so that the power factor approaches, but does not equal, unity.
The 0.8 Power Factor provided by generator manufacturers is not the load power factor, but it is the nominal power factor used to calculate the kW output of an engine to supply the power for a particular alternator kVA output. Alternators are therefore designed to supply their rated kVA at 0.8 lagging power factor.
The capacitors must be sized according to the reactive power being used, not the real power being used, to improve power factor.
The simplest method of power-factor improvement is by using appropriate capacitors, connected in parallel with the load. Power-factor improvement capacitors are rated in reactive volt amperes, not farads.