How synchronous motors are used for power factor correction

Synchronous Phase Modifier:

To control the voltage of a transmission line, a synchronous motor is used at the receiving end. In this application, the synchronous motor is operated to run without load and draw a main current. Synchronous motor when used for the above purpose is called synchronous phase converter. As the power factor of the system is improved by the action of the synchronous capacitor, the voltage drop between the transmitting and receiving ends of the line is reduced and line regulation is improved.

Control % = {(EV)/V} x 100

Power factor correction with synchronous capacitor:

The power factor of a load can be improved from cosθ to 1 or to the desired value. ₁ to cosθ ₂ by installing and operating a synchronous capacitor in the system. Losses in the motor can be neglected and the motor will be around 90 ^Ó The figure shows the phasor diagram for the operation of a synchronous motor to improve the power factor.

Synchronous motor for power factor correction:

In the figure above, V is the reference pointer I _C is the load current with lagging power factor cosθ ₁ . OA is the active component of the charging current and AC is the reactive component. If the synchronous motor is operated as a synchronous capacitor and losses are neglected, OD represents the current consumed, which is advanced by 90 ^Ó . When this value is equal to the AC reactive load current component, the resultant of the currents consumed by the load and the synchronous motor is just OA, giving the load the same output power in KW, but improving the load power factor for the load. unit. since OA is in phase with V. So L _C = L _M sinθ ₁ is the prerequisite for improving the load power factor to unity under certain operating conditions.

The KVA rating of the synchronous capacitor required for this purpose is I _C x V volt-amperes per phase or √3VI _C /1000 KVA would be the three-phase power.

The rated power of the synchronous capacitor is √3VI _C /1000KVA. If the power factor is improved to less than 1, a smaller synchronous capacitor capacity will be required.

In the figure, cosθ is the initial power factor of the load:

OA – Eu _M cosθ ₁ is the active component of the current AC – I _M sinθ ₁ is the reactive component of the load current (delayed) OD – reactive (main) component of the synchronous capacitor

If this value is equal to BC, then the resulting load current vector is OB and the new load power factor is cosθ ₂ . Then EU _C = I _M sinθ ₁ – I _M sinθ ₂ The active load component is reduced from AC to AB. The rated power of the synchronous capacitor required for this purpose is = √3VI _C /1000KVA The power component of the OA load remains the same and the same load is powered, but now with an improved power factor.