Motor Open Circuit Time Constant

[philly]

I was looking at a motor datasheet and noticed that it has a specification called an "Open Circuit Time Constant" listed in seconds. What is this time constant? Does this have something to do with how long it takes the motor voltage or magnetic field to decay after the voltage is removed from the motor.

It also had an X/R ratio which I believe is the ration between the reactance and resistance of the motor?

The motor also listed that it was "No-Load" tested, and it appeared that as the motor voltage was lowered the current was lowered. This is opposite what you usually see with a motor under load. Does this have something to do with the slip staying constant and therfore the rotor resistance staying constant with no load? Would't the motor frictional losses still produce a required torque on the motor?

 

[Jraef]

No, the "open circuit time constant" is not the decay time of the magnetic field. It is a test value used in impedance calculations, essentially something they used to determine the X/R ratio as well. Both of those values are used in programming Vector Drives, although in reality, most Vector Drives now ave "Auto-tune" where the VFD teaches itself what it needs to know about the motor and its response to step changes in voltage and frequency. In some really super critical precision applications, some people choose to forgo auto-tune and enter motor parameters manually, but it's often difficult to get those values from motor mfrs. So in a way, it's nice that yours has it on the daat sheet. Probably an "Inverter Duty" motor?

Without a load, if you decrease voltage you decrease the iron and copper losses in the motor. You also decrease available torque by the square of the voltage reduction, but if it is unloaded, you would never know because it needs very little torque to keep its own mass rotating against a little friction and windage.

 

[weressl]

I was looking at a motor datasheet and noticed that it has a specification called an "Open Circuit Time Constant" listed in seconds. What is this time constant? Does this have something to do with how long it takes the motor voltage or magnetic field to decay after the voltage is removed from the motor.

It also had an X/R ratio which I believe is the ration between the reactance and resistance of the motor?

The motor also listed that it was "No-Load" tested, and it appeared that as the motor voltage was lowered the current was lowered. This is opposite what you usually see with a motor under load. Does this have something to do with the slip staying constant and therfore the rotor resistance staying constant with no load? Would't the motor frictional losses still produce a required torque on the motor?

IEEE141, the Red Book:

[FONT=Times-Roman+2][FONT=Times-Roman+2]

f)​

[/FONT][/FONT][FONT=Times-Italic+2][FONT=Times-Italic+2]

Reclosure and transfer switch operations​

[/FONT][/FONT][FONT=Times-Roman+2][FONT=Times-Roman+2]

[B26]. Under normal operating conditions,
the self-generated voltage of an ac motor lags the bus voltage by a few electrical
degrees in induction motors and by 25 to 35 electrical degrees in synchronous
motors. The operation of a recloser on the utility power supply or the transfer to an
alternate source will cause the power to be interrupted for a fraction of a second or
longer. When power is removed from a motor, the terminal voltage does not collapse
suddenly, but decays in accordance with the open-circuit machine time constant (time
for self-generated voltage to decay to 37% of rated bus voltage). The load with its
inherent inertia acts as a prime mover that attempts to keep the rotor turning. The frequency
or phase relationship of the motor self-generated voltage no longer follows
the bus voltage by a ?xed torque angle, but starts to separate farther from it (out-ofphase​

in electrical degrees) as the motor decelerates.
[/FONT][/FONT]

 

[skeshesh]

IEEE141, the Red Book:

[FONT=Times-Roman+2][FONT=Times-Roman+2]

(time
for self-generated voltage to decay to 37% of rated bus voltage).​

[/FONT][/FONT]

Weressl hit the nail right in the brains. The time contast is the time it takes for a system's response to the step function to reach 1 - (1/e) = 63% of its final value. The "open circuit" time constant for a motor is the time it takes for the value of the system (voltage) to reach 37% of the base value (rated bus voltage) when the motor is open circuited.

 

[philly]

IEEE141, the Red Book:

[FONT=Times-Roman+2][FONT=Times-Roman+2]

f)​

[/FONT][/FONT][FONT=Times-Italic+2][FONT=Times-Italic+2]

Reclosure and transfer switch operations​

[/FONT][/FONT][FONT=Times-Roman+2][FONT=Times-Roman+2]

[B26]. Under normal operating conditions,
the self-generated voltage of an ac motor lags the bus voltage by a few electrical
degrees in induction motors and by 25 to 35 electrical degrees in synchronous
motors. The operation of a recloser on the utility power supply or the transfer to an
alternate source will cause the power to be interrupted for a fraction of a second or
longer. When power is removed from a motor, the terminal voltage does not collapse
suddenly, but decays in accordance with the open-circuit machine time constant (time
for self-generated voltage to decay to 37% of rated bus voltage). The load with its
inherent inertia acts as a prime mover that attempts to keep the rotor turning. The frequency
or phase relationship of the motor self-generated voltage no longer follows
the bus voltage by a ?xed torque angle, but starts to separate farther from it (out-ofphase​

in electrical degrees) as the motor decelerates.
[/FONT][/FONT]

Thanks for the reference. Can you tell me what section of the Red book this is in, I cant seem to find it.

Why is 37% the magic number? At this point, is the voltage too low to cause any out of phase issues when reclosing?

So for a transfer switch that is not synchronized, should there be at least a delay equal to or greater than this time constant before switching supply or restarting motor?

Without a load, if you decrease voltage you decrease the iron and copper losses in the motor. You also decrease available torque by the square of the voltage reduction, but if it is unloaded, you would never know because it needs very little torque to keep its own mass rotating against a little friction and windage.

Even with the small friction and winding, wont this still corrospond to s very small torque demand on the motor. Since torque is the product of V/Hz and current, then wont lowering the V/Hz cause an increased current even to supply the very low torque required from the friction and windage at not load?

If I had a 6 lead motor where I was unsure of how it was to be connected but I have the no load current then if I wired it in wye and it when it was intended to be delta would I see a smaller no-load current due to the reduced voltage? (This is assuming the reduced torque is able to start the motor)

If it was supposed to be a wye and it was wired in delta, I would expect to see a higher inrush, as well as see a higher no-load current?

 

[weressl]

Thanks for the reference. Can you tell me what section of the Red book this is in, I cant seem to find it.

[FONT=Helvetica-Bold+2][FONT=Helvetica-Bold+2]5.6.3.1 Large alternating-current rotating apparatus [/FONT][/FONT]page 248

Why is 37% the magic number? At this point, is the voltage too low to cause any out of phase issues when reclosing?

Yes

So for a transfer switch that is not synchronized, should there be at least a delay equal to or greater than this time constant before switching supply or restarting motor?

Yes

 

[philly]

So if I had a motor intended for Delta but had it in wye, would I see a lower no load current then the manufacturer published?

 

[weressl]

So if I had a motor intended for Delta but had it in wye, would I see a lower no load current then the manufacturer published?

At the same voltage?

Please note that the wye/delta start motors are not intended or designed to run in the wye connected mode.

The motor still requires the same amount of flux to turn, regardless of widning connections. The purpose of the D/Y start is to reduce the inrush current.

In other words, there is no free lunch.

If the motor is designed to operate in either connection mode, then the corresponding NLA/FLA would be published. The required supply voltage in each connection mode would be different.

 

[philly]

At the same voltage?

Please note that the wye/delta start motors are not intended or designed to run in the wye connected mode.

The motor still requires the same amount of flux to turn, regardless of widning connections. The purpose of the D/Y start is to reduce the inrush current.

In other words, there is no free lunch.

If the motor is designed to operate in either connection mode, then the corresponding NLA/FLA would be published. The required supply voltage in each connection mode would be different.

Yes I understand this wye-delta start concept.

I have a large custom 700hp 4.16kV 6-lead motor. There is no designation on the motor datasheet or motor itself if it should be wye or delta connected. Usually I would assume a Delta connection however I have seen many of these large custom motor be connected in wye.

I have the motor datasheet which shows the no-load current. If I connect the motor in wye and it is intended to be run in wye, then the no load current after the motor gets up to speed should match that on the datasheet. If the motor was intended to be a delta connection then after the motor gets up to speed (assuming it does) then I would expect the no-load current to be much less than published, b/c as we discussed above the no load current changes in proportion with voltage. I.E. voltage goes down current goes down.

So in a wye I would have 2400V across windings instead of 4.16kv and therefore no-load current would be less at no-load full speed, and would indicate a wrong connection.

 

[winnie]

1) The reason that the _no-load_ current will decrease with decreasing supply voltage is because you are decreasing the _magnetizing_ current flow. The current drawn by the motor may be considered a composite of two elements: the 'torque producing' current and the quadrature 'magnetizing' current. When you reduce the supply voltage, the torque producing current will go up and the magnetizing current will go down. For a fully loaded motor this tends to mean that reduced supply voltage means increased load current. But for an unloaded motor the change in magnetizing current will dominate and the current draw will go down.

2) Assuming correct connection, if you connect your motor 'delta' then it will draw more no-load current than if you connect it 'wye'. I would expect that a motor intended for delta connection would be able to safely run continuously at reduced load if connected in wye. It would have reduced magnetic flux and would draw excessive current at full load, leading to overheating and failure; but as stated above at no-load the current draw would be reduced.

3) The magic 63% number comes from 'e' the base used for 'natural' logarithms. Basically there are lots of things that can be described in terms of exponential decay, for example the discharge of a capacitor through a resistor. An exponential decay is something that can be described in terms of F(t) = 1/a^t where a is any number. Say a capacitor decays to 1/2 its initial voltage in 1 second. In 2 seconds it will have reached 1/4 initial voltage, 3 seconds gets you to 1/8, and in general at any time t this capacitor will have decayed to 1/2^t of its initial voltage. You can describe such exponential decays using _any_ base, eg 2 (half life) or 10. It turns out that the base 'e' (2.71828....) has some very nice mathematical properties (which we can go into later), so it gets used as the base for describing lots of decay and periodic functions. The time constant is simply the time to get to 1/e of the original value. After 1 time constant has passed, there still may be enough energy in the system to cause problems.

-Jon