INTERESTING 2

[vanskat]

Investigations into a recent 400 hp motor failure uncovered that the pump dedicated to the motor was serviced and was to be put back in service. Incorrect procedures were followed in terms of opening suction and discharge valves. As a result of the system architecture the impeller began to rotate in the opposite direction subsequent to which the motor was started.
Curiously on examination the stator windings looked fairly ok based on what happened. No excess dirt or moisture, insulation OK, no major burns/soot etc. I expected more severe damage though.
(Magnetic overload was defective)

What I am curious about though is the mathematical analysis of this situation. Considerations, equations, conditions and theories.....
Where to start? I know I would have to consider the locked rotor current etc....

 

[electricpete]

Here's the way I look at the situation. (Open to suggestions).

Using the equivalent circuit to develop the mathematic expression for the torque speed curve, we would predict that the torque speed curve extends fairly smoothly past speed=0 to negative speed. In fact, the faster the motor rotates in reverse direction, the lower the starting torque. So, contrary to what might be intuitive, there should be no excessive torques on the shaft from this situation.

The same should hold true for starting current. I'm guessing that current should not exceed LRA by more than a few percent assuming that motor is rotating in reverse direction at only a few percent of forward syncronous speed.

The basis for the above statements is that the electrical behavior depends on the relative speed between the rotor and stator. There is nothing special about the standstill condition (relative speed=syncronous speed) wherein an abrupt change in the equivalent circuit would occur. Therefore the torque vs speed and current vs speed curves should be smooth curves on both sides of the zero speed.

The biggest factor will be that the load will have to accelerate longer before it comes up to speed, since it is starting at a negative speed. For large inertia loads (applies more to fans than pumps), acceleration of the load is the major contributor to heating in the rotor and the stator.

It can be mathemtatically shown that the rotor I^2*R during startup of unloaded motor (no load torque, only inertia) which is exactly equal to the change in kinetic energy of the rotor. (It's not based on simple conservation of energy... it's a long derivation which I can supply if requested). If you add a mechanical load, then the rotor heating will be the same quantity multiplied by a quantity which resembles the average value of Te/(Te-Tm) over all values of speed during the startup. Stator heationg will be the same amount as rotor heating multipled by a correction factor Rs/Rr.

So bottom line is your starting current transient will start at a current only slightly higher than LRA, but will last longer than a normal start due to extra acceleration of the load inertia required.

I would expect that this MAY result in a trip of the motor on overload. Although I'm not sure what you meant by magnetic overload... I usually associated "magnetic" with instantaneous trip features, not overload features.

So you have more heating than normal, in an amount you may be able to estimate.

Possibly more important is the fact that the rotor fan cooling is NOT as effective in removing that heat when the speed is in the vicinity of zero (cooling is directly related to speed of the rotor fan). Since you spent much more time than normal near the zero-speed region, you generated all of the additional heat at a period of time when the motor was least able to effectively remove it (during the time before speed even got to zero).

Thinking about it a little more, the current and heating production rate would be exactly the same as for a normal start from the time when speed=0 on. But the difference would be that you're starting from a higher temperature before you even get to speed=0.

That's my first cut. Let me know if you want the equations that form the basis for my statements.

 

[electricpete]

By the way, what was the speed of the motor. If it's a low-speed it might be fair to look at it as high inertia load even though it's driving a low-inertia pump.

I forgot to mention that for large motors, the most severe stress during starting can be on the rotor, rather than the stator. The reason is based on the fact that the rotor currents tend to be concentrated on the outside of the rotor bars due to skin effect which is highest during starting (because relative speed betw rotor and stator is highest => rotor current frequency is highest). This would be even more pronounced for reverse rotation. (higher relative speed, more rotor skin effect). The skin effect was NOT taken into account in the above discussion of rotor heating where rotor resistance was assumed constant.

 

[gordonl]

An experience was shared with me were a fan windmilled in reverse, and when started the shaft was snapped. Would the motor apply positive torque while the load enertia applies a negative torque, to add up to higher than normal starting torque? Would this be higher than a straight locked rotor torque?

I'm obviously seeking answers more than I'm providing.

 

[electricpete]

Good comment Gord. It contradicts with my impression based on a simple physical model which I assume to be pretty close to reality. But I can't argue with real-world experience.

The motor applies a torque T. The load has an moment of inertia J and rotating momentum J*w (where w=2*pi*speed). The result of applying known motor torque T should be an acceleration: dw/dt = T/J (similar to a=F/m). The inertia or momentum does not generate any torque of it's own. It only changes in a manner such that rate of change of momentum will balance the applied motor torque. We might therefore consider that a reaction torque (or force) is associated with a change in momentum. But that reaction torque cannot exceed the torque that caused the change in momentum in the first place.

With all that said, can anyone explain why a shaft would break due to energizing motor while rotating in the reverse direction?

 

[electricpete]

Perhaps it is mildly interesting to provide an analogy in linear motion.

Consider a train riding at high-speed on a low-friction track. A man is riding on top of the train with his shoes glued to the train. He turns on a fire extinguisher spraying forward to provide a jet-propulsion force in the reverse direction which will start to slow down the train. The force transmitted through the man's shoes will depend only on the fire extinguisher force, not the speed of the train (or even the direction of motion of the train).

Here is the correspondence between the linear analogy and the rotary quantities:

LINEAR ROTARY
------ ------
Force T orque
Distance Angle
Speed Rotational Speed (RPM).
Fire Ext. Motor
Train Inertia Pump/Motor rotating inertia
Train speed Pump/Motor initial reverse RPM speed
Shoes Shaft

 

[GordS]

I think a motor shaft shaft may break when started and already rotating in the reverse direction because of a slight variation to electricpete's equation - dw/dt = T/J. Wouldn't it actually be dw/dt = dT/J? Since the outside system is supplying a torque negative to what is produced by the motor, wouldn't the resulting dw/dt be greater than that of a normal start where there is no reverse torque to overcome? Since F=ma and a dw/dt then there is more force being transmitted through the shaft which may exceed its capabilities.

 

[gordonl]

They same thought later came to me after pete's post which pointed out where I was wrong, its not the fan inertia that supplies more torque to the shaft but the actual windmilling. In the case previously quoted I beleive the windmilling was due to a parallel fan running without the appropriate dampers being totally closed forcing air back through the "down" fan. This would supply a torque to oppose motor torque.

 

[gordonl]

I reread the original post and would make the following comments to vanskat to supplement electricpete's.

As Pete stated the damage during starting is typically rotor damage. Typical damage to rotors is usually cracking and/or opening of rotor bars. There are online monitoring systems which monitor motor vibration, supply harmonics, or even magnetic field changes over time to detect these faults.

 

[electricpete]

Wow, I didn't realize that Gord and Gordon were two different people. Sorry about that guys!

Regardless of rotating inertia/momentum OR loading conditions imposed by the pump windmilling, I still believe the shaft torque during starting cannot exceed the torque developed by the motor, which will likely not exceed locked rotor torque. (with the possible exception of torsional oscillations which might create higher torque but are not common). I'll have to think a little more about the arguments that you two made.

 

[gordonl]

Before I disappear for the long weekend I'd like to through this out for commparison:

If two motors were connected shaft to shaft in opposing directions wouldn't the shaft torque be double, or the addition of the two opposing torques. Now replace one of these motors with a fan which is providing torque through windmilling. I suppose the same would apply to the pump as long as the head of water is applying a torque to the pump.

On a tangent, Multilin has a new motor protection relay (369)which has an option for detecting backspin and preventing starting until it stops spinning.

 

[electricpete]

Gordon.
I'd like to address your example.

If you have two motors on the same shaft, assuming equal torque produced by each in opposite direction, then shaft will not spin (assuming it was initially at rest).

I contend that the torque applied to the shaft will be the same as torque produced by either of the two motors (but not the sum of both).

Here's an easy thought experiment that proves it. Replace one of the two motors with a rigid immovable clamp (clamped to the floor,let's say). The shaft still does not move. From the standpoint of the remaining motor, it doesn't know the difference between the clamp on the other end of the shaft and the motor on the other end of the shaft. It is still putting out it's original torque and the clamp is providing equal and opposite counter-torque.

In summary, it seems easy to see that the 1-motor system with clamp does not develop 2x motor torque in the shaft, and that this is equivalent to the two-opposing motor system in terms of shaft torque.

 

[electricpete]

Gordonl - I have to backtrack a little bit on my statement that windmilling cannot cause torque to exceed locked rotor torque under any circumstances. In a very extreme case of windmilling I can see where windmilling might cause higher torque on shaft than the motor alone could deliver. I spent quite awhile working thinking about it. If I get some more time I'll try to form it into a coherent post.

 

[jbartos]

Suggestions to vanskat (Electrical) Oct 5, 2001 marked by ///\\\:
Investigations into a recent 400 hp motor failure uncovered that the pump dedicated to the motor was serviced and was to be put back in service. Incorrect procedures were followed in terms of opening suction and discharge valves. As a result of the system architecture the impeller began to rotate in the opposite direction subsequent to which the motor was started.
Curiously on examination the stator windings looked fairly ok based on what happened. No excess dirt or moisture, insulation OK, no major burns/soot etc. I expected more severe damage though.
(Magnetic overload was defective)
///Could you clarify the "Magnetic overload"?\\What I am curious about though is the mathematical analysis of this situation. Considerations, equations, conditions and theories.....
///References:
1. Gordon R. Slemon, "Magnetoelectric Devices Transducers, Transformers, and Machines," John Wiley and Sons, Inc., New York, 1966
2. M. G. Say, "Alternating Current Machines," John Wiley & Sons, New York, 1978
3. Any manufacturer who manufactures plugging duty motors since these motors must withstand a certain shaft dynamic load reversal.\\Where to start? I know I would have to consider the locked rotor current etc....
///Perhaps, a good start would be to contact the motor manufacturer and inquire about the motor design. It appears that the motor for a pump application does not normally require to be of plugging duty type; however, it may. Additional what may be considered is shaft metal fatigue or a defect, and the motor designed for its lightest shaft duty, i.e. the regular motor.\\

 

[vanskat]

Really good feedback by all.
1. By reference to magnetic overload I refer to a current-sensitive device with an iron core or plunger, like a solenoid, placed inside a coil. The tripping current level is adjusted by altering the initial position of the plunger with respect to the coil. When the load current becomes excessive, the plunger is drawn towards the stronger magnetic field created by excessive current, interrupting the circuit. (Would react before thermal..)

2. MOTOR DATA
SYNCH SPEED: 1490 RPM HP: 400
VOLTAGE: 460V LUBE TYPE: G
FLA: 452A MANUFACTURER: RELIANCE
FUSE SIZE: 500A STARTER TYPE: WESTINGHOUSE (6)
SWITCH SIZE: 800A OVERLOAD PROTECTION: N26

3. Question Pete? In your post you mentioned that for large motors the most severe stress during starting can be for the rotor rather than the stator. However, the damage was seen on the stator windings. Furthermore, the situation was that 4 pumps fed a common header and one was removed to be repacked. On returning it to service with the suction and discharge valves open wouldn't a greater force (negative torque) be applied to the shaft? Very much dissimilar to gordonl's statement about the windmilling rather than the fan inertia.

 

[electricpete]

To compare and contrast rotor heating vs stator heating:

Under load, the amp-turns of the stator will always equal the amp-turns of the rotor (neglecting exciting current supplied by the stator). Another way to say this is that the actual stator and rotor current magnitudes will always follow a fixed ratio dependent upon the physical turns ratio (regardless of slip). [Asides - #1 the voltage ratio between stator and rotor does vary with slip - #2 the rotor frequency varies with slip]

Based on the above you would conclude that relative heating between rotor and stator is equal at all speeds... with approx 5 times current (25 times I^2*R) during starting as compared to running.

The additional effect which compounds the situation is the skin effect which increases the rotor resistance during starting. This increases the rotor I^2*R still further beyond 25 times, and worse yet that increased I^2*R is released in a smaller area of the rotor near the surface. Skin effect depends on the depth of the rotor bar... not a big concern for small bars in small motors but grows with motor size.

For your specific situation, even though rotor heating increase by more than stator heating (starting compared to running), stator heating during starting is still approx 25x that during running. Also the larger current creates larger magnetic force on the conductors during starting which put mechanical stress on both end turns and slot area.

Now to the question of torque, inertia/momentum etc.

Imagine that you have a shaft with torque applied on two ends. If the same torque is applied on both ends, no acceleration/decel occurs and the torque felt on the shaft is the same as that applied on either end. Not too interesting. (If this case corresponded to our motor we would see the shaft could never undergo more torque than put out by the motor).

Now apply two different torques Ta and Tb on each end with Ta>Tb. What is the torque seen by the shaft.

If all of the mass is located at the B end of the shaft, then the shaft will see the greater Torque Ta Throughout it's lenght.

If all of the mass is located at the A end of the shaft, then the shaft will see the lower Torque Tb Throughout it's lenght.

If there is Ma located at the A end and Mb located at the B end, then the shaft will see an intermediate torque which might be viewed as a weighted average of Ta and Tb.

Tshaft = Ta*Mb/(Mb+Ma) + Tb*Ma/(Mb+Ma).

If the motor torque plays the role of Ta (the greater torque), then the shaft can never see a torque exceeding the motor torque. We can't break the shaft unless we had a motor capable of breaking the shaft if you tried to start the motor with rotor bound up (due to seized bearing for example). I believe this is not common.

The only way to exceed motor torque is if the windmilling effect is capable of generating a TORQUE greater than motor torque. It is very important to distinguish at this point that this windmilling effect would be a torque imposed on the rotor by the fluid system and NOT simply the rotating momentum of the mass (which has already been accounted for by Ma and Mb).

So once again we need Tpump > Tmotor. (Even so the shaft torque will be limited by weight distribution as follows:
To the extent that weight is shifted from motor towards pump, Tshaft will be less than Tpump as given by the above equation.)

Still, let's assume we have this huge reverse Torque due to windmilling. If you draw a freebody diagram you can label three Torques on the shaft. Twindmill in reverse direction, Tmotor in forward direction, and Tnet=Twindmill-Tmotor in the reverse direction which will be proportional to the acceleration. But wait a second... this is telling us that the shaft still ACCELERATES (not just rotates) in the REVERSE direction when I turn the motor on. This is a very bizarre situation which represents EXTREME windmilling torque much higher than I would expect from simply.

Maybe the fluid guys can tell me there is something I have missed by modeling the windmilling effect as a rotating inertia and a torque. One complication is that the impeller torque will presumably be a function of speed which I have not allowed for above. But still my gut feel is it just isn't a likely scenario.
I'm welcome to other suggestions.

 

[electricpete]

For completeness...

I mentioned additional stator stresses during starting of
#1 - Heating, #2 - Magnetic forces.

I failed to mention that voltage stress during starting may be higher due to possible steep wavefront of the applied voltage (creates high turn-to-turn stress of first few coils) and possible partial reflection of wave at interface from cable to motor lead and motor lead to stator which provides possible increase in voltage at those locations.

 

[electricpete]

In my discussioin of Torques above, you should substitute the "rotational inertia" for mass.

I analysed the system as a linear system using masses and forces. Before posting on the board I converted Torques to Forces but forgot to convert masses to rotatiional inertia's.