Unity Power Factor: Impact on Motor Performance

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Unity power factor improves motor efficiency by reducing current and minimizing losses in the circuit. Adding capacitors to achieve a unity power factor does not affect the motor's performance directly, as the motor cannot detect the presence of capacitors. However, significant losses can lead to voltage sag, which may adversely affect motor operation under certain conditions. In practical applications, such as with a two-speed motor, achieving a unity power factor allows for lower current draw, resulting in cost savings on electricity. Overall, while unity power factor is beneficial, its impact can vary based on specific circumstances.
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hi,

iam wondering about the unity power factor ..

for example if we have a motor that will cause the power factor to be lagging by 0.3

what will happen if we add capacitor to make the power factor 1 (unity)

now we will not have any reactive power and we reduced the current to the minimum so that real power will equal the apparent power.

will that affect the motor ?
 
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It will not affect the motor at all. The motor cannot even determine whether the capacitor is present.
 
Unity pf will affect the current and thereby the losses in the circuit feeding your motor. If this losses are significant they will sag the voltage feeding your motor. In rare conditions this lower voltage could adversly affect your motor.
For example, I have a two speed motor with 11kW rated output at pf near to 1. The same
motor can be used at half speed 7KW output but with a reduced pf of only 0.68.
By compensating the lousy pf I am able to run this motor on a 380V 3phase 25A supply. instead of the otherwise required 35A. Since in my country there is a huge difference
in electricity base price delivered to 25A mainfuses compared to 35A (hundreds of $ per year) it makes sense to compensate. Under other conditions things may be different.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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