Voltage variation effect on motors and lighting

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mull982

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I understand that a slighty lower voltage variation than that given by a motor nameplate voltage will result in greater full load current, and conversley a slightly higher voltage variation will result in a lower full load current. Why is the starting current just the opposite of this, with a slightly lower voltage resulting in a decrease in starting current, and a slightly higher voltage resulting in an increase in starting current. Does this have something to do with the starting torque?

I had a similar question in regards to lighting. I read that a slightly lower than rated voltage (voltage rating of light) variation of a light will reslult in the light illuminating less light, and conversley a slightly higher voltage variation in a light will result in it illuminating brighter. Why is this? Does this have something to do with the current draw through the light?
 
When a motor is running in its normal operating range, it is a very high efficiency converter of electrical power to mechanical power. Thus the input power is equal to the output power plus a little bit of loss. For _small_ variations in input voltage, the output mechanical power will remain pretty much constant, and the change in losses will be slight. So you essentially have a machine that is consuming _constant_ input power. For a constant power load, if you reduce the input voltage, the load will draw more current.

When a motor is just starting, however, it is extremely inefficient. In fact, at zero speed, the efficiency is by definition _zero_. Right at startup, _all_ of the input power is going into heating the motor up. If you think of the motor as a transformer with a shorted secondary, you will see that a stalled (zero speed) motor is pretty much a resistive load, with a low resistance. Over a rather wide range, starting current will scale with input voltage.

Unless they have regulating ballasts, lights will in general draw less current as the supply voltage is lowered. Less power is delivered to the light, and less light is produced.

-Jon
 
To add to Jon's reply, you can reduce the starting current by reducing the voltage, as a Soft Starter does, because motor torque follows the square of the voltage, and current follows the torque output. So if, for instance, you reduced the motor terminal RMS voltage to 50% at startup, the torque, and therefore the current, would be reduced to .5 x .5 = .25 (25%) of what it would have been at full voltage. So if you take Jon's "inefficient motor" which would normally pull 600% FLA at startup, it will pull only 150% FLA at 50% voltage. Because the torque is reduced, the time to accelerate will be greater (assuming 25% starting torque is enough to accelerate it at all). However if you plotted time and power (a.k.a. energy) to accelerate to full speed, once at full voltage and once at 1/2 voltage, the areas of the two curves would be identical. The 1/2V curve would be low and wide, the FV curve would be high and narrow.
 
It becomes very important to determine if the reduced torque capability of the motor will be sufficient to overcome the starting torque combined with the load torque. Motor torque curves must be used to verify this.

Jon, you stated during motor starting the motor is pretty much resistive load with low resistance? Motor starting power factors are very low, which would indicate the load is predominantly reactive during starting.
 
Sorry about that. Kingpb is correct. I was thinking about the motor equivalent circuit when I wrote the above, and did my thinking too quickly.

When the motor is stalled, the rotor acts as a very low resistance, much lower than when the rotor at normal operating speed. Thus there is substantially more _real_ power flow, and this real power flow is delivered to the motor as heat.

However the motor still has all of its inductive aspects, which remain pretty well constant. Lower resistance, same inductance; total impedance goes down but gets proportionately more inductive.

-Jon
 
Actually, both statements are technically correct if you just get the semantics correct. "INRUSH" current refers to that resistive load model Jon put forth, but is only applicable to the first instant power is supplied until the magnetic fields are created. After that, it is "STARTING CURRENT" and power factor comes in to play due to the inductance and movement of the stator / rotor fields relative to each other; the high reactance model that kingpb put forth.
 
Jraef said:
Actually, both statements are technically correct if you just get the semantics correct. "INRUSH" current refers to that resistive load model Jon put forth, but is only applicable to the first instant power is supplied until the magnetic fields are created. After that, it is "STARTING CURRENT" and power factor comes in to play due to the inductance and movement of the stator / rotor fields relative to each other; the high reactance model that kingpb put forth.
Click to expand...

So that I understand this correctly I am going to reference a motor TCC for the regions that you described above.

Is what you are calling "Inrush", the region between .01 and .1 seconds on a motor TCC, where the curve slopes lightly upwards until it reaches the first vertical or 6xFLA? For this inrush portion, you are saying that the load is a resistive load as put forth by Jon? Higher voltage leads to higher current in this region?

Is the second region which you are calling the "Starting Current" the region between .1s and 5-10s (where the curve drops to FLA) on a motor TCC. In this region you are saying that the load is mostly reactive due to inductive properties and that the power factor is extremely low (typically .17 or so)?

So I am assuming that the final region of "FLA" region is the region in which the load has both resistance and inductive properties (or resistance and reactive) and thus has a better power factor typically around .85 or so? Higher voltage leads to lower current in this region?

Am I looking at all the motor starting and operating characteristics correctly?

I appreciate the help.
 
I don't think that it is fair to try to characterize the startup transient in terms of power factor. Normally for electrical power systems, we talk about resistance or inductance in the context of a single fixed supply frequency. But during the startup transient, you have a wild mix of supply frequencies, simply because the 'switching the voltage on' has its own frequency characteristics. The inductance of the winding is changing over fractions of the AC cycle.

The startup transient may push the core into saturation, and for the fraction of a cycle where it is saturated, the inductance of the coil is reduced. For part of the AC cycle, you see essentially normal inductance. For other parts of the cycle you see greatly reduced inductance, so that the coil looks more like a resistor. This is not because of higher resistance, but because of lower inductance, and lower total impedance. But within a very few AC cycles, the magnetization will have settled into proper synch with the supply voltage, and the normal inductance will be apparent.

During the startup transient, instantaneous current can substantially exceed the locked rotor current.

Locked rotor current describes how the motor responds if the rotor is _fixed in place_, and the core magnetization has stabilized in the steady state. Often this will be measured with the rotor blocked and at reduced voltage

As the rotor speeds up, power factor will increase until a maximum (at a speed somewhere around 150% to 300% of normal full load slip), and then start decreasing again, as the current going to mechanical output decreases and the stator magnetizing current starts to dominate. When the motor is at synchronous speed (zero load), current is pretty much at a minimum, and power factor is zero.

-Jon

(edited to clarify: the peak power factor happens at a particular speed, not that the peak power factor is 1.5 to 3x the normal power factor)
 
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