You may be right, I'll not challenge the numbers at this time. I think that my point is still correct, if not completely accurate.

If I understand you correctly, that 600-800% inrush does not all go to torque. We only get maybe 300% torque at starting and the rest of the current is lost.

Bill Sturm

 

Correct, the 600-800% current does not all go to generating torque. Unlike VSD operation, line starting creates instantaneous high slip. To explain my previous reply another way (line operation perspective) - The frequency the rotor "sees" is the slip freq. At high slip frequencies, rotor circuit impedance is much higher creating an additional phase shift that causes the natural (optimum for torque generation) 90 degrees angle between magnetizing
and torque producing current vectors to be worse. If the angle is worse the only way for the motor to generate a certain torque is to draw much more
current. But this much higher current also creates more losses in the stator and rotor! A viscous cycle. This is why, with line operation, so much current is wasted i.e. not producing torque.

 

> If I understand you correctly, that 600-800% inrush
> does not all go to torque. We only get maybe 300%
> torque at starting and the rest of the current is lost.

My background is mechanical, not electrical, so take this with a grain of salt. It has been my understanding that for 2-phase squirrel-cage motors, the high inrush current is due to the instantaneous short circuit created by the start and/or run capacitor(s). If true, it would expect
this current to be lost to heat, probably in the capacitor.

D. Leese

 

Ken,

Regarding boosting magnetizing current I agree if the motor is not saturated at the rated Im you can boost torque even more with higher Im. I didn't know Nema B had this margin so I learned
something here.

I'm not sure I understand your other trick - variation of above to suppress BEMF on PM brushless motors to increase speed. I know you could phase advance PM brushless motors if all you
want is more speed in ONE direction (other direction suffers). We did this on blower motors for computer peripheral applications. You can't do the so-called field weakening because the field is fixed i.e. permanent magnets. Or do you mean something else?

 

At 12:14 PM 7/7/2000 -0400, you wrote:
>Correct, the 600-800% current does not all go to generating torque. Unlike
>VSD operation, line starting creates instantaneous high slip.

I guess the original idea was that if the motor can handle 600-800% current for line starts, then why not "crank" it with several hundred percent from a drive for short durations to get better dynamic response. You might want to
purposely oversize a drive for these situations.

This was explained to me several years ago by a person with lots of drive experience, but I have never tried it. It was suggested that a 5 HP motor might be paired with a 20 HP drive for a roll feed application, for example.

Does this idea have any merit?

Bill Sturm

 

Responding to David Leese's comment:

If you were referring to operation of a 3-phase motor on a single-phase source, then, the capacitor is placed across the winding which is not directly connected to the source?

If you truly meant a 2-phase motor (servo?), then, the capacitor is usually in series with the fixed-phase winding.

Anyway, the VSD can't alter a squirrel-cage motor's electrical parameters. Its stator and rotor parameters are fixed by design.

Phil Corso, PE
Trip-A-Larm Corp
Deerfield Beach, FL

ps: I accept your GOS reference to your discipline. I started in my earlier life wanting to be a "big" bridge designer.

 

Of course the drive salesman wants you to get a bigger drive . . . he has his kids college education to pay for!

Realistically, once you saturate the iron in a motor . . . adding current to the rotor only results in creating heat. The breakdown torque of a motor is pretty much the point at where the magnetic fields are at maximum magnitude applied at optimum force angle (90 degrees) in the motor. This limitation is typically in the 300 - 350 percent range.

Now . . . if you wanted to take a 5HP motor and tap it on the 230Volt leads and connect a 20HP Drive to it running at 460Volts . . . set the
drive up for a 120Hz motor with a Full load speed of about 3500 rpm . . . then, you will be able to get up to 20HP out of the motor for brief periods of time and 10HP continuous at 3600 rpm without motor damage.

Ken

 

You can't get unlimited torque out of an induction motor (or any motor for that matter) just because you have a drive with unlimited current. Basic IM torque generating components are magnetizing current or V/Hz (sets-up strength of rotating stator magnetic field) and slip (rotor speed drop vs. stator, phase angle, losses). A) An IM can only utilize so much magnetizing current. Beyond that it goes into magnetic saturation and is wasted. B) As torque increases so does slip. At a certain point (breakdown) negative effects of high slip (impedance/phase angle, losses) start to decrease
torque even though current is increasing, as in line operation.

 

Ken,

You hit the nail on the head. You and I told him the same thing. This confirms my belief there are a lot of people (not only sales types) spreading mis-information about applying VSDs. Others
should know besides this list I know you as a linear motor customer. I see you also know the double motor nameplate HP trick.

 

You probably have a higher starting current too, but that might not be what you measure as your inrush current. Depending on how the motor is made the starting current might not be as high as on a "standard" induction motor (higher rotor resistance = lower starting current and higher
starting torque, but is less efficient at normal speed) An induction motor with a really high inertia load might take a minute or so starting, and doing so with a 10 times too high current does of course heat the motor a lot, but that is what causes the problem and if you motor don't get
overheated then that part isn't any problem (for the motor). Note: a normal motor have a fan attached to cool the motor, during start the fan isn't rotating at full speed yet - and isn't cooling the motor enough.

/Johan Bengtsson

----------------------------------------
P&L, the Academy of Automation
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: [email protected]
Internet: http://www.pol.se/
----------------------------------------

 

Responding to the recent discussion on the effect of inrush current. Much of what has been said is misleading.

Starting Time.
Typically, a standard squirrel-cage motor has a no-load runup time of a fraction of a second to about 6 seconds, depending on type, but,
regardless of size (except for sub-fractional motors). Example, runup time for a 22,000 Hp, 2-pole machine used to drive an ethylene compressor was just 5-seconds.

Stator Temperature-Rise.
The principle source of heat is not stator winding losses, but the heat transferred from the rotor. During startup, at high slip, and until the
breakdown-torque speed is reached, virtually all of the electrical input is converted to heat stored in the rotor. Very little of the input power is converted to mechanical power. That heat is eventually transferred to the stator.

Starting Duty.
The number of starts and cooldown duration is specified to cover the stator temperature-rise problem, for moderate to very large machines.

Cooling Fan.
The cooling fan has no real impact on cooling during startup. The contact time between the coolant media and the hot internals for an
"open-type" motor is nil. The contact time between the internal coolant media, and the "carcass" of "closed-type" motor is also nil.

Capacitor Analogy
Wrong way to go.

Transformer Analogy.
Comparing the heating of a motor to the effects of a transformer's shorted secondary is akin to comparing Apple(s) to IBM-clones.

If any of the List is interested in a "BITE" story related to motor starting time and large scale failure, let me know.

Regards,
Phil Corso, PE
Trip-A-Larm Corp
(Deerfield Beach, FL)

 

To the List,

I wish to correct a statement I made in my 28-Jul response. Runup time IS dependent on size.

The general relationship turns out to be logarithmic:

t ~ M x Ns / Tm, where,

t = time, in seconds.
M = Moment of Inertia (that inertia stuff again).
Ns = rated full load speed.
Tm = mean accelerating torque.
K = a dimensional constant.

I apologize for my Technical In-correctness, or is it Non-correctness!

Regards,
Phil Corso, PE
Trip-A-Larm Corp
(Deerfield Beach, FL)

 

<clip>
>Starting Time.
>Typically, a standard squirrel-cage motor has a no-load runup time of a
>fraction of a second to about 6 seconds, depending on type, but,
>regardless of size (except for sub-fractional motors). Example, runup
>time for a 22,000 Hp, 2-pole machine used to drive an ethylene
>compressor was just 5-seconds.

What do you plan to do when the load can not be
disconnected?


<clip>
>Transformer Analogy.
>Comparing the heating of a motor to the effects of a transformer's
>shorted secondary is akin to comparing Apple(s) to IBM-clones.

During the first part of the startup it is a quite
good analogy since a motor with a locked rotor and
a transformer is behaving quite identical if you
look at them electrically and magetically. There is an air gap in the magnetic field but that is about all the difference there is to the analogy (a transformer will work even if there is a air gap in the magnetic field).
The things heating both a transformer and a motor is:
* Power losses due to the current flowing thru the
windings.
* Iron losses.

The analogy doesn't fit when the slip get closer to normal running values but at the first part of the startup (when there is a high slip and a high current) it is close enough.


/Johan Bengtsson

----------------------------------------
P&L, the Academy of Automation
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: [email protected]
Internet: http://www.pol.se/
----------------------------------------

 

Responding to Johan's comments, I understand your "over-simplification" approach, but it is misleading to those who are trying to understand
basics:

Starting Time
Please note that I said "No-Load" runup time. I was not referring to the situation where the load and motor are uncoupled or disconnected.

No-Load means that the coupled load imparts no appreciable load torque to the motor's developing torque during runup. However, the load's moment of inertia is taken into account.

For most applications like compressors and pumps "No-Load" means that their suction (and/or discharge valves) can be closed during runup. For
fans then their inlet louvres are closed.

Transformer Analogy
There is a world of difference between starting a motor and "energizing" a transformer.

1) A transformer is seldom intentionally energized with its secondary connected. First the primary source is connected, then the secondary
load is applied. The most notable exception is recovery from a source voltage sag or momentary interruption. This is commonly called
"Reacceleration" or "restoration" following a power source transient.

2) The transient inrush current to a transformer is more influenced by its magnetizing current. Closing or energizing with a shorted secondary
can be considered as having some equivalence. But its time variation is not as complex as is the case of a motor.

3) The electrical input to a motor is influenced by the energization current and both the stator and rotor resistance.... from its "breakaway" torque point at time zero, to the point where its
"breakdown" torque is reached. The inrush current has both an active and a magnetizing component, and does show appreciable variation with time.

The "BITE" stories that I referred to were examples where "runup time" was a major factor in some catastrophic events.

Regards,
Phil Corso, PE
Trip-A-Larm Corp
(Deerfield Beach, FL)

 

Johan,
I'm not sure if you were asking the question or not - but it's a good one.

With a recip or other positive displacement compressor, (such as a screw compressor) the flow is essentially proportional to speed and the power is therefore dependent on the developed pressure. To minimise starting load on the compressor, it should be started with a bypass or recirculation valve allowing flow from discharge back to suction fully opened.

On the other hand, with a centrifugal pump, fan or compressor, the pressure is dependent on speed and the flow variable - so these should be started in the no-flow condition to keep load on the motor, and starting time, to a minimum.

Of course, with centrifugal compressors, this directly conflicts with the need to worry about surge. Ain't Murphy wonderful?.

Cheers,

Bruce

On 01 August, 2000 13:19, Johan Bengtsson P&L Automatik AB
[SMTP:[email protected]] wrote:
> <clip>

> What do you plan to do when the load can not be
> disconnected?

 

Sorry if I didn't express myself clear enough.
When I referred to a transformer I wasn't thinking of the inrush current to a transformer during energization. I was thinking of the steady state with a short circuited secondary coil and the currents and heat losses in that
case.

Ok, I realize what you mean by No-load, but the load inertia could be *very* high in some cases, mening the torque needed during normal load would need a long time to accelerate it to normal operating speed. This would of course be made worse if the selected motor type is unable to even produce this torque during startup (it would be a stupid choice to have that motor type there). What I mean is that it is depending on
the application too, not just the motor. Example: something transporting things horisontally don't need much torque once it is up to normal speed (mostly some friction), but it would need all it's torque to get there.

I think we basically agree to most of this anyway, it is probably mostly a communication problem...


/Johan Bengtsson

----------------------------------------
P&L, the Academy of Automation
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: [email protected]
Internet: http://www.pol.se/
----------------------------------------

 

Actually I was more concerned about the loads inertia, if the load can not be mecanically disconnected from the motor you will always have the loads ineria at startup, regardless of any bypass valves and other. But of course it helps the motor to reduce as much load as possible during startup. Sorry about not expressing myself clear enough....

(See the other posts too...)


/Johan Bengtsson

----------------------------------------
P&L, the Academy of Automation
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: [email protected]
Internet: http://www.pol.se/
----------------------------------------

 

Johan, variation is why God made us Unjineers.

Anymore requests for my startup/runup BITE stories?

Regards,
Phil Corso, PE
Trip-A-Larm Corp
(Deerfield Beach, FL)

 

Here is a great article for you: http://www.belden.com/products/capwmtp.htm

 

> Beware of the brake, if there is one. It must be externally <
--- snip ---

This statement is correct.

> Don't run it too slowly so that the fan isn't cooling the motor. <
--- snip ---

This statement is correct.

> Lastly, startup torque on a basic induction motor is essentially 6X the full load rating. When you put a VFD rated at 2HP you will get around 1.5X the rating to start it up, this can cause problems. The newer style vector drives do better (I don't remember the actual number) but they don't make 6X. <

Not true. A Variable Frequency Drive varies the frequency. Startup torque on a basic induction motor draws a lot of current (ie 6X) but it does not make 6X rated torque. Typically around 80% to 160% of rated torque. A VFD can typically produce around 80% of pullout torque so that would be around 180% to 250% or rated torque at a current of around 2x to 3x rated motor torque. All you need to do is size the drive appropriately to the starting torque requirements and commission it properly.

A standard motor sees a slip frequency of 50Hz at startup. The same applies if you use a soft start (hence a soft start has limited starting torque). But a VFD will apply just the slip frequency at startup and hence produce a lot more controlled torque at much lower current.