Starting I in induction motor at synchronous speed 2

[ccdubs]

If I have an induction motor (500 kW) being driven at exactly synchronous speed, what sort of starting currents would I be likely to get if I start it DOL?

What happens if the speed slightly varies from synchronous?

I have been told by an engineer that at synchronous speed the starting transient won't be higher than the rated current of the machine and would be less than 1 sec.

I don't understand why the starting current is lowered just because the machine is synchronous. At time 0+ the stator windings would still basically appear as a short and as there is little remanence in the core the steel still needs to be magnetised.

 

[dpc]

The difference is the slip. Starting a motor from a standstill, the slip is essentially 1, causing the effective rotor resistance to be very low - hence high induced currents. Starting near synchronous speed, the slip is much closer to zero and this increases the effective rotor resistance dramatically.

I don't have quantitative answers for you, but induction generators are typically started in this fashion and the resulting inrush is quite brief - similar to energizing a transformer, I suppose. Energizing an induction motor from a standstill is a little like energizing a transformer with the secondary shorted.

Hope that helps.

 

[electricpete]

During normal starting the impedance seen is approx Znormalstart = j * (X1+X2) + R2

Current in a given phase will look like
I(t) = sqrt(2)* V/|Znormalstart|* [ A*exp(R/(L1+L2)*t) + cos(2Pi*f*t) ]
It rmeains like this until speed begins to increase close to sync, at which time sinusoidal current drops further.

It has a sinusoidal component and an exponential component.
|A| <=1, depending on phase of voltage at time of closing.
V/|Znormalstart| = LRC

At sync speed the R2/s term will be an open circuit. Now the magnetizing branch becomes more important. The impedance seen is Zsyncstart = j * (X1+Xm)

Where Xm>>X2 or R2
Now the current will look like:
I(t) = sqrt(2)* V/|Zsyncstart|* [ A*exp(R/(L1+L2)*t) + cos(2Pi*f*t) ]


The current has the same general initial shape with exponentially decaying dc plus sinusoidal. The magnitudes of both of these components are lower by ratio of (Zsyncstart/Znormalstart) >>1.
Since V/|Zsyncstart| ~ no-load current, the sinusoidal starting current has magnitude approximately same as no-load current (20-60% of FLA) which can increase by factor of up to 2 during the initial period before the dc decays.

 

[electricpete]

I should finish the thought:
For motors 6-pole speed or faster,it is safe to can assume no-load current <50% of FLA. Assume you have one of these faster motors.

Then the ac component of your current will be <50%FLA after start. When you add in the exponentially decaying component the peak can increase by factor of 2 to a peak current equivalent to that seen under full load.

 

[jbartos]

Suggestion to the previous posting: A clarification is needed regarding those exponentially decaying components. They appear to be exponentially growing.
Also, a clarification is needed regarding R2 and X2 whether they are on the rotor side or the stator side.

 

[jbartos]

Suggestions to ccdubs (Electrical) Nov 10, 2003 marked ///\\If I have an induction motor (500 kW) being driven at exactly synchronous speed, what sort of starting currents would I be likely to get if I start it DOL?
///Clarification is needed. Normally, the induction motors run at subsynchronous speeds since they experience a small slip (for all practical purposes, the slip is different from zero).\\What happens if the speed slightly varies from synchronous?
///It depends whether you mean synchronous machine or induction machine.\\I have been told by an engineer that at synchronous speed the starting transient won't be higher than the rated current of the machine and would be less than 1 sec.
///Some synchronous machines have a rotor cage that enables them to be started as a squirrel-cage induction motor.\\I don't understand why the starting current is lowered just because the machine is synchronous.
///It depends on the design of the synchronous motor starting arrangement, i.e. rotor cage.\\ At time 0+ the stator windings would still basically appear as a short and as there is little remanence in the core the steel still needs to be magnetised.
///Not quite, it depends on the rotor cage resistance.\\\

 

[bigbillnky]

The best way to understand this is to first understand how an induction motor starts and runs. We know from basic theory that a conductor cutting the flux of a magnetic field induces a current flow in the conductor and the amount of flow is dependent on the angle of the conductor cutting the field and the speed at which the field is cut. That conductor is the rotor bars in the standard squirrel cage induction motor. The magnetic field is the stator windings that surround the rotor. At startup, with the rotor stationary, the inrush is high due to the fact that the stator field is revolving around the rotor at synchronous speed. This induces a high current in the rotor, and that current creates a magnetic field that is attracted to the stator field and causes the rotor to start in motion. Since there is maximum current flow in the rotor, the field created is very strong. As the rotor speeds up, it is cutting less flux from the stator field. That means there is less current flowing in the rotor bars. As the motor nears synchronous speed, even less flux is cut, meaning even lower rotor current. At a point the rotor reaches slip speed, that is the point that the rotor can not overcome losses to pull into synchronism, usually 3 to 5% of synchronous speed. Now it is easy to see if the motor is already turning at near synchronous speed the initial inrush is extremely brief, probably within 6 cycles. Also, you can understand how a DC excited synchronous motor is pulled to synchronous speed once it has reached slip speed by applying DC current to the rotor to create a strong magnetic attraction to the stator field. Most synchronous motors start as induction motors. I hope this helps..........bigbill

 

[jbartos]

Suggestion: Several clarifications to the original posting are overdue.

 

[ccdubs]

Clarifications.

I mentioned in the original post that the induction motor is being DRIVEN at synchronous speed. This is being achieved with a VSD and a smaller motor coupled to the large motors through shaft.

This is probably enough clarification to answer your other points.

What may help is a description of what is being done. This design is for a gearbox test rig. The induction motor drives a step down GB which then drives the step up GB being tested. The tested GB is connected to a synchronous generator.

The induction motor has a through shaft, one end coupled to the GB the other end to a smaller induction motor. The small motor has a VSD which enables us to rotate the system to synchronous speed and put the generator online. Once online we then intend to power the large drive motor to put 500 kW through the system. Hence the question on starting currents at synchronous speed.

Torque will be applied to the generator via the first step down gearbox which has a variable and controllable mechanical slip charecteristic.

 

[jbartos]

Suggestion: The major item needed a clarification in the original posting was:
&quot;&quot;I don't understand why the starting current is lowered just because the machine is synchronous.&quot;&quot;
There are big differences between the synchronous machine (e.g. motor) and induction machine (e.g. motor).

 

[RAMConsult]

cc, the reason the starting current is so low at exactly synchronous speed is that the line is only supplying the initial inrush and steady-state magnetizing current plus copper losses while the small drive motor is supplying all the power that is required to overcome the inertia plus friction and windage losses in the larger machine.

jb, other posters realize the context implies that the sentence reads as &quot;...the induction machine is at synchronous speed.&quot;

 

[jbartos]

Suggestion to the previous posting: It is not that clear from the original posting since the machine is at synchronous speed when it rotates at RPM corresponding to electrical synchronous speed. The starting machine whether induction motor or synchronous motor is not synchronous since the synchronous means the mechanical speed is in synchronism with the electrical speed. This is physically impossible at starting machines.
Compare with the original posting and my remarks ///\\\:
&quot;I don't understand why the starting current is lowered just because the machine is synchronous. At time 0+ the stator windings would still basically appear as a short
///There is not exactly short since many starting machines have Zm~0.17p.u. to about 0.28p.u.\\ and as there is little remanence in the core the steel still needs to be magnetised.&quot;

 

[RAMConsult]

Sorry jb, I commented on the cc's opening question where he posits &quot;If I have an induction motor (500 kW) being driven at exactly synchronous speed,...&quot;; i.e., if this condition were achieved what would be the result. Furthermore, at your prodding, cc clarified that his setup was capable of synchronous operation through the use of a VFD. I simply reinforced dpc's and billinky's explanations with a different perspective; that's my value-add.

Perhaps I should have written my sentence to you as &quot;the context of the opening sentence implies that the sentence in question reads as...&quot;, but I'm more interested in engineering rather than semantical precision.

 

[jbartos]

Suggestion to ccdubs (Electrical) Nov 10, 2003 marked ///\\Clarifications.
I mentioned in the original post that the induction motor is being DRIVEN at synchronous speed. This is being achieved with a VSD and a smaller motor coupled to the large motors through shaft.
This is probably enough clarification to answer your other points.g
///Thank you for clarification.\\What may help is a description of what is being done. This design is for a earbox test rig. The induction motor drives a step down GB which then drives the step up GB being tested. The tested GB is connected to a synchronous generator.
///This was not so apparent in the original posting.\\The induction motor has a through shaft, one end coupled to the GB the other end to a smaller induction motor. The small motor has a VSD which enables us to rotate the system to synchronous speed and put the generator online.
///Please, clarify the relationship of the generator to the induction motor with the through shaft.\\ Once online
///What is online? The generator started by the induction motor and rotating the generator at synchronous speed?\\ we then intend to power the large drive motor
///What kind of motor is it synchronous or induction one?\\ to put 500 kW through the system. Hence the question on starting currents at synchronous speed.
///Once, the synchronous machine is driven to be rotating at synchronous speed, one may connect power supply to its terminals. If the machine is synchronous, the field winding also has to produce sufficient field flux to have the rotating machine (motor) rotating on its own without the propulsion through the shaft.\\Torque will be applied to the generator via the first step down gearbox which has a variable and controllable mechanical slip characteristic.
///Essentially, there is a Motor-Generator Set, sometimes called M-G set that will be powering the 500kW to the system. Or am I missing anything?\\

 

[ccdubs]

jbartos,

///Please, clarify the relationship of the generator to the induction motor with the through shaft.\\Assume that the ratio is 1:1, i.e., the induction motor is coupled to the synch genny.

///What is online? The generator started by the induction motor and rotating the generator at synchronous speed?\\In this case the generator would be put online (using auto sync). The 500kW induction motor would still be offline and driven by the small motor with VSD. So if I take the VSD offline, the sync genny will drive the induction genny at sync speed.

///What kind of motor is it synchronous or induction one?\\Both motors are induction and the generator is synchronous.

///Essentially, there is a Motor-Generator Set, sometimes called M-G set that will be powering the 500kW to the system. Or am I missing anything?\\No this is probably very similar to what you are talking about except that we will apply load through the gearbox. We are sure the test rig will work but I needed to supply starting current values to the network owner to determine if we were going to effect the voltage for other LV customers.

 

[jbartos]

Revisiting the original ccdubs (Electrical) Nov 10, 2003 posting by ///\\
If I have an induction motor (500 kW) being driven at exactly synchronous speed, what sort of starting currents would I be likely to get if I start it DOL?
///The only current supplied to the motor will be the motor magnetizing branch mostly inductive current that is relatively small fraction of the motor rated current, e.g. 10% or so depending on the motor design. This magnetizing branch current is often neglected in the full load motor analysis.\\What happens if the speed slightly varies from synchronous?
///The supplied current to the motor terminals will be slightly higher than the current supplied to the motor driven at synchronous speed.\\I have been told by an engineer that at synchronous speed the starting transient won't be higher than the rated current of the machine and would be less than 1 sec.
///Yes, the starting current transient exponentially increases on the mostly inductive load, which the motor magnetizing branch is, and the motor magnetizing branch current will level off at the steady state magnetizing branch current level.\\I don't understand why the starting current is lowered just because the machine is synchronous.
///There is no load on the synchronously rotated motor to be powered over the motor terminals, only the motor magnetizing branch current.\\ At time 0+ the stator windings would still basically appear as a short and as there is little remanence in the core the steel still needs to be magnetised.
///To the contrary. The inductor is starting with the essentially zero current. The capacitor is starting with the short circuit current. See the basic equations:

http://www.eng.hull.ac.uk/teaching/57022/13207_Handouts01.pdf

etc.\\\