Hydraulic Motor Angular acceleration limitation 2

[Deaks]

I have a hydraulic motor that has an angular acceleration limit of 87,000 rad/s². The motor is driving a drill rod which is of cylindrical section and when it operating at full torque can be wound up by as much as 4 turns (length is 1000m). When the rod breaks into a void the energy is released and accelerates the motor. Does any one know a calculation method to ascertain if the motor limitation is exceeded? If you need more info let me know.
Cheers Ian

 

[GregLocock]

I'd have thought a simple model of a distributed mass/stiffness system to represent the shaft, and an inertia to represent the motor, would do. It almost sounds like a standard case.

Your big problem will be working out a damping value, for the motor, and for the shaft.

Is the moment of inertia of the motor significant compared with that of the shaft?

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[desertfox]

Hi Deaks

If the rod breaks into a void and the rod unwinds won't it
be opposing normal motor rotation in which case it would try to slow the motor down as opposed to accelerating it
or am I misunderstanding your problem?

Regards

desertfox

 

[zekeman]

Sounds like the torque on the rod is delivered by the motor and when the rod breaks, the torque in the shaft initially drives the motor backwards with the same torque at first and then as as the rod unwinds the energy in the shaft is partly sent to the motor and partly used to wind up the remaining part of the shaft attached to the motor in the opposite direction repeatedly until it gets damped out.
What you need is a partial differntial equation with interesting boundary conditions to describe the motion, since the energy is distributed along the shaft.
But the solution is only as good as the assumption of where the shaft breaks.

 

[desertfox]

Hi Deaks

Thinking further about your problem what your sayimg is the motor is under load when the drill breaks through the void the torque load is released and therefore the motor wants to accelerate.
If the motor is designed with an acceleration limit and is run under the conditions required by the manufacturer I can't see why it would exceed that acceleration limit unless your over driving the rotor to start with.

regards

desertfox

 

[sreid]

For electric servo motors, the maximum acceleration is a limit only in the sense that it is the maximum acceleration that can be acheived if maximum current is applied at zero speed to an unloaded motor (motor inertia only). This is usually a theoretical limit. I'll bet the same applies to hydraulic motors.

 

[JStephen]

It sounds to me like when the rod breaks through, it spins around 4 times but then it accelerates the motor as it decelerates.

 

[BobM3]

If the load is driving the motor then, to protect the motor from high delta-P, you could put a cross-port relief valve across the motor.

At what point of this dynamic are you worried about motor acceleration? Is it when the void is hit (and the load on the motor suddenly drops off) or is it the oscillating load on the motor as the unloaded bit is allowed to torsionally vibrate?

 

[MintJulep]

I think desertfox has noted an important point.

Where will the stored energy go? Towards accelerating the now unconstrainted end of the drill, or into the motor.

Do you know the drag losses of the drill in the bore?

It seems more likely that the near-instataneous drop in torque at the bit is faster than the motor can respond, so the motor accelerates as it sees the reduction in load.

 

[GregLocock]

This is a standard case. section 8.4 in Rao "Mechanical Vibrations"

The system is a flywheel on a continuous shaft. I suspect the inertia of the flywheel is irrelevant, so the mode shape becomes the first mode of a uniform shaft. Therefore the motor will oscillate with an initial semipeak amplitude of 8pi

Simple harmonic motion tells you the maximum acceleration is w^2*theta, where w is the first natural frequency of the system, w=1/L*sqrt(C*G/mu/Ip) from Blevins, table 8-19

C=pi*R^4/2
Ip=polar area moment of inertia in torsion
mu=mass density of shaft material



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[EdDanzer]

The plumbing and control valve allow the motor to over speed when the rod breaks. The hose and piping act like an accumulator storing volume as the pressure goes up. The control valve does not pressure compensate the flow out of the motor so when there is a rapid change in torque the motor will speed up. To solve the problem install pressure compensated flow control valves with reverse free flow checks (Sun CNGC-XXX or FDEA-XXX as examples) set just above the maximum flow for the maximum rpm the motor needs to operate at to meter the flow out of the motor.

 

[Deaks]

Thanks everyone. A few more details might clarify the situation. The motor is delivering about 735 Nm at 800-1000 rpm and in my mind the rods resemble a wound up elastic band, storing energy. When one end of the system is released the end of the rod will accelerate and the torque from the motor drops and it then goes into braking mode. The acceleration/deceleration to the new condition is where the rad/s² limit would be reached. The limitation is placed on the motor by a weak link, the syncronising shaft between the output shaft and the cylinder block (it's a bent axis motor). I've seen these twist under extreme loads.
Greg: motor MOI would low compared to the shaft I believe (and I did give the star for the second post)
Desert fox: how did you get a 2 x star when I didn't give them to you?
Ed: I don't believe hydraulic restraints will work in this as the problem is mechanical caused downstream of the oily bit
Cheers Ian

 

[EdDanzer]

Deaks,

A hydraulic motor will act as a brake if the flow and pressure out of the motor are limited. I have seen motor shafts broken by limiting flow out with over running loads.
If you would like to bet the flow limitation won’t slow the motor down contact me through

http://www.danzcoinc.com

 

[zekeman]

Your problem involves solving the partial diff eq
ud^2f/dt^2=GIpd^2f/dx^2
where:
u=moment of inertia per unit length of shaft
GIp=torsonal stiffness of shaft

This is the wave equation (I use "d" here as a partial derivitive)
with boundaries:
GIpdf/dx(0,x)=initial torsional condition in shaft
df/dt(0,x)=angular velocity of shaft at start
df/dx(t,L)=0 (stress at unconstrained end of shaft is zero
At motor end I would use
Jd^2f/dt^2=-udf/dx at (o,t)
The solution is probably in the form of the sum of
e^-kt(Asinkx+Bcoskx) for discrete values of k that satisfy the boundaries.

In the end you are looking for GIpdf/dx at x=0 which is the torque on the motor mass.

 

[zekeman]

Correction:
The form of the solution is approximately
(Asin(k1t)+Bcos(k1t))*(Csin(kx)+Dcos(kx))

 

[IceStationZebra]

My thinking is along the same lines as "Ed Danzer". For the motor to act as a brake you either have to be limiting the exhaust or inlet. Since I really doubt that you are throttling the exhaust the installation of a check valve that allows exhaust flow into the pump inlet will prevent this. (This is frequently done on motors driving a high inertia load.) The motor would still speed up but it won't be applying a braking load.

I can only think of two reasons for the angular acceleration limit. The first is torque related. The second is to prevent a vacuum in the inlet due to flow restrictions, which would cause havoc with the slippers. An overrunning check valve would cover both of these in your application.

A mechanical solution would be an overrunning clutch on the pump output.

ISZ

 

[IceStationZebra]

Also, you should post questions of this sort in the "fluid power professionals" forum. There are many people eager to help there. ISZ

 

[PNachtwey]

"Also, you should post questions of this sort in the "fluid power professionals" forum. There are many people eager to help there. ISZ"
I doubt there is anyone there that could help. That hasn't seen this thread. Go ahead. Give it a try.

Greg Locock has the answer. I would work the formula backwards and find the highest natural frequency permissible

sqrt(87,000/(8*PI))=58.8 rad/s or about 9.36 Hz. It doesn't seem likely that a pipe 1000m long could twist 8 turns in about .1 seconds, but I will leave that to the mechanical guys with the numbers to work out. What I do know is that .1 or even 0.05 seconds is an eternity to electronics. One can easily monitor the differential pressure across the motor and look for a sudden drop in differential pressure.
This would be the key for the controller to resist velocity changes by using a position differentiator and a derivative gain.

 

[zekeman]

I have worked out this problem and find that contrary to some opinions, if you assume that after the breakthru, conservatively, in the absence of damping and allowing the system to be FREE at the motor end, it would oscillate superimposed on the initial forward rotation, with torque on the motor starting out opposed and then forward, driving the motor in the original direction as you note, but its torque can NEVER exceed the initial motor torque, ti.
The max acceleration forward would be
ti/Im
I don't think your motor moment of inertia, Im, is small enough to get the 87000 rad/sec^2
If, however, you attempt to restrain the motor with braking, the torque could get exceedingly high as the total system energy will come into play.