Reaction torque in stationary parts of rotating machines 1

[electricpete]

In "Machine elements : life and design" by Boris M. Klebanov, David M. Barlam, Frederic E. Nystrom on pages 321-323 they seem to state that transmitted shaft torque in rotating machines always creates equal/opposite reaction torques on stationary parts.

The relevant portion of the text is attached to this message and can also be seen here:

http://books.google.com/books?id=B0...idered as made of one piece of metal.&f=false

That is, the rotor of pump 1 is connected with its housing by forces of hydrostatic pressure that create the same torque on the rotor and on the housing. The same is true of the electromagnetic field in the motor. We can separate the rotors from the stators and apply to each of them the additional torques +T or ?T. But there’s no need for such detail. From Newton’s law of action and reaction, it follows that, for calculation of external forces, any unit (gear, pump, motor, etc.), with its shafts or other moving parts can be considered as made of one piece of metal.

You can see the same claim in figure 11.2 and associated example calculation 11.1.

My opinion: It is true for the motor that the electromagnetic torque on the rotor is also exerted in equal/opposite fashion on the stator (it is action at a distance through fields). But it is not true for the pump – there are no fields to transmit the torque between pump rotor and stator. There may be some torque created by pressure forces and friction forces on pump casing, but I don't see any reason to suspect they would be the same as the torque transmitted from motor to pump by the shaft. Look at the scenario of axial flow pump or fan – very little torque on the casing.

Am I off base, or is the book off-base?


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[ivymike]

if the moments don't add up, then something "inside the box" is being rotationally accelerated...

in an internal combustion engine, crankshaft torque must be reacted at the piston-bore interface, via piston thrust force on the bore multiplied by the perpendicular distance to the crank. Seems crazy if you draw it, but newton's law doesn't work out otherwise.

 

[ivymike]

if you have a cross-section of a particular pump, we can probably point out where the moment is being reacted. If there's a rotor pushing some fluid resulting in a torque on the rotor, the fluid pressure is probably acting unevenly on the stator and/or housing too.

 

[ivymike]

take this picture of a wankel engine, for example.

http://en.wikipedia.org/wiki/File:Wankel-1.jpg

it's easy enough to picture pressure in the big open area acting on the rotor to create a moment about the crank. If you add up all the little force vectors, you'll find that there's an opposite moment exerted by the gas upon the housing about the same axis.

 

[electricpete]

Thanks.

I drew myself a very simple radial flow pump and convinced myself that what the book said is true for that case.

I can't picture it for an axial flow pump or fan.

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[electricpete]

I'm sure you are right and book is right. But it's not obvious to me that straightforward application of Newton's law leads to the book's conclusion, since there is interaction with the fluid.

Attached slide 1 is (overly?) simplified radial flow pump with single rotating vane and single stationary vane. It is obvious to me that both stationary and rotating vane see the same torque for this simplified radial flow pump.

Attached slide 2 is axial flow fan or pump. The stationary part is just a hollow cyclinder - has no surfaces for torque to act on (neglecting friction). Clearly it sees no counter torque. The fluid flow will be mostly axial, but will have some circumferential component. If we add ideal stationary flow straighteners downstream of the device, will the torque on these necessarily be equal/opposite the shaft torque (under idealized conditions - no friction and no circumferential component of flow at straightener exit)?

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[electricpete]

I guess the answer to my last questioner is the straighteners must see the same torque.

Now I see it does follow from application of Newton's law (it just took awhile). Either fluid must be accelerated circumferentially (as in axial flow fan without straightener) or there must be equal/opposite countertorque on stationary component.

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[electricpete]

I just had one more thought. Discharge pipe of centrifugal pump is often smaller than suction pipe. In that case, the fluid is accelerated in the circumferential direciton, isn't it? (In which case we don't expect that equal/opposite counter torque, because some of the torque went to accelerate the fluid.)

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[electricpete]

I think I can answer my last question. Yes, the fluid is accelerating as it passes through the machine. But if machine is in steady state, there is no change of momentum of fluid within the control boundary, so no time rate of change and no accelerating torque... must still be equal and opposite counter torque on stationary component for radial flow machine. (right?)

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[electricpete]

Still waffling on my last question... seems a lot trickier to know how to treat the system with mass flowing accross the boundary.... could use some help.

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[ivymike]

if memory serves, Force=mdot*deltaV when you have a change in momentum within a boundary. So, for example, if water comes in going straight down and leaves going straight to the right, you must have an upward force and a rightward force each equal to mdot (mass flow rate) * speed of flow. (since deltaV is |speed|up+|speed|right).

 

[zekeman]

You don't need a deep understanding of Newtons Law of reaction to understand that the housing mounts must provide the same but opposite torque that is delivered by the pump shaft.
The action and reaction forces of Newton are INTERNAL with respect to the pump system which in isolation has external torque of the shaft, housing restraint and if you cut the fluid part , the linear force along the line of the piping, which is negligible.
So a free body analysis of the system would eliminate the internal forces and only shaft torque and housing restraint ( ignoring the momentum forces of the fluid) Housing torque must equal to pump shaft torque.

 

[unclesyd]

Could you use a Tesla Turbine vs a DiscFlo Pump for the analysis?

 

[desertfox]

Hi electricpete

I think of it like this:- imagine you have a centrifugal pump running and if you were able suddenly to prevent the rotor from turning and take off the bolts holding the pump case down, wouldn't the pump case then rotate?

desertfox

 

[drawoh]

electricpete,

The pump can be considered as an item separate from the motor. Torque is applied to the pump shaft, therefore there is a moment resisting the torque at the pump's feet.

In the diagram, the motor and pump are connected through a coupling. Try to imagine that the motor has a pulley which operates something to the left of the mount, and that the pump is driven by a pulley from something to the right of the mount.

JHG

 

[electricpete]

The outlet fluid also exits near the outer radius of the casing while the inlet fluid enters near the center.

Let’s say the inlet radius is 0.

Then using Mike’s suggestion F= mdot * v….
the torque associated with bringing the initial fluid (no rotating inertia with respect to shaft centerline) to the final radius would be

F = mdot * v * R
Where
Mdot is mass flow rate
V = velocity of outlet fluid
R = radius of outlet fluid.
Correct?

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[electricpete]

Correction in bold

Torque = mdot * v * R
Where
Mdot is mass flow rate
V = velocity of outlet fluid
R = radius of outlet fluid compared to shaft centerline
Correct?

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[cbrn]

Hi Electricpete,
I may very well be very late on that topic, but just let me add my 2-cents...
It's for the axial pump: though the "visual" representation of the phenomenon is easier with the radial pump, it's about the same in an axial pump. The fluid is moved by increasing its energy by increasing its angular momentum. The difference with the radial pump is that the axial one lacks che centrifugal effect, so it bases its functionning only on the "out-of-meridian-plane" deviation of the flow.
in any case, the torque must be reacted by something: electric torque on motor's rotor core is reacted (in fact, created...) by motor's stator, electric torque from the motor's rotor core is trasfered to the pump impeller, torque of the impeller is transfered to the fluid and from there to the pump's stator. If pump's stator and motor's stator are connected together, then no "external" torque ; otherwise, two torques on the foundations.
Regards