Understanding Power Quality Impact of Motor Starting

AI Thread Summary
Starting large motors causes voltage sag due to their high inrush current, which exceeds normal operational levels. This high current draw leads to increased resistance in the power source and wiring, resulting in a voltage drop at the motor terminals. When the motor stalls, all input power converts to heat instead of mechanical energy, risking overheating. The reactive components of the motor's inductive load complicate the relationship between voltage and current, as the effective resistance varies during startup. Understanding these dynamics is crucial for managing power quality in motor applications.
nafas
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Hello,

In Power Quality, how does starting of large motor will cause voltage sag? How does voltage sag cause the motor to stall and overheat?

A friend mentioned that it has something to do with sustaining the source power, but I simply don't get it.

Thank you in advance.
 
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Motors draw more current during start up (or if you stall it) than when rotating.
In normal operation, most of the input power gets converted to mechanical energy.
In the stopped/stalled condition all the input power gets converted to heat.

All real world power sources and connecting wiring have impedance (or resistance).
If you had a power connection with a zero impedance then there would be no voltage sag.
 
Hi,

If starting a motor consumed high current, then why doesn't the voltage increase too; since V=IR?

Thanks.
 
Motor is an inductive load, so it draws high initial current.
voltage or the potential is not absorbed, rather electrons are ( current).
you just need to supply more electrons to take your motor coil to operating level with the same potential.
the R in ohms law corresponds to resistive real value where as an inductive element possesses reactive component as well.
Hope this helps.
 
nafas said:
Hi,

If starting a motor consumed high current, then why doesn't the voltage increase too; since V=IR?

Thanks.

Due to the Conservation of Energy.

CS
 
nafas said:
Hi,

If starting a motor consumed high current, then why doesn't the voltage increase too; since V=IR?

Thanks.
It depends on you're viewpoint or which components of V=IR you are looking at.
The effective "R" of the motor is variable and very low at startup.
V is a constant defined by the construction of the source.

However, the wires connecting the voltage source (and the voltage source) have a value of R associated with them, separate from the motor.
The voltage drop due to R_source and R_wire does increase with current and must be subtracted from V_source.
This is seen as a "voltage sag" at the terminals of the motor.
 
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