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Shaft torque from air gap torque 10

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kevd

Electrical
Nov 1, 2002
18
Hopefully a very simple question!

How do you calculate shaft torque from air-gap torque under transient conditions?

If my load inertia is J1 and my motor inertia is J2, is it as simple as Shaft Torque = J1/(J1+J2)*air-gap torque?

Thanks
 
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First you use air-gap torque I believe to mean motor torque. (term air-gap torque sometimes carries a connotation that rotor losses are included but I don't think you are concerned with that. I believe transient torque on the shaft depends on both the motor torque and the load torque. (I am not opsitive). Here is my rough thought process.

Imagine that you have a shaft with opposite-direction torques 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.

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

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

If the inertia is split with inertia Ja located at the A end and Jb 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*Jb/(Jb+Ja) + Jb*Ja/(Jb+Ja).

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Kevd
Your equation for shaft torque is correct IF you can ignore the springiness of the shaft. If this is not true, the shaft torque is
Ts = TL + (Tag-TL)*(JL/(Jm + JL)*[1-cos (wn*t)]
where
TL = load torque
Tag = motor electrical torque
JL, Jm = load, motor moment of interia respectively
wn = mechanical first natural frequency
=sqrt(K*(Jm+JL)/(Jm*JL))
K = shaft spring constant

Reference "Bus Transfer of AC Induction Motors, A Perspective", paper PCIC-89-07 from the IEEE.
 
I consider that the air gap torque, load torque and inertias follow the equation:

Tg = Tl + (Jm) d^2Q/dt + (Jl) d^2Q/dt---1
Tg = Tl + (Jm +Jl) d^2Q/dt
and then;
d^2Q/dt=(Tg-Tl)/(Jm+Jl)---2

Were:
Tg = Motor Torque at the air gap or rotor bars.
Tl = Load Torque
Q = Shaft angular position.
d^2Q/dt = second derivate of angular position ( angular acceleration)
Jm = Motor rotor inertia.
Jl = Load Inertia

Ts = Tl + (Jl)d^2Q/dt---3
Ts = Torque at the motor shaft extension.

by inserting (2) in (3)


Ts= Tl + Jl/(Jm+Jl)*(Tg-Tl)

Note that at steady speed d^2Q/dt =0
 
Thanks everyone who took the trouble to answer so fully.
The reason I ask was this topic is not clearly covered in many electro or mechanical text books.
 
Comment on electricpete (Electrical) Feb 24, 2004 marked ///\\
If the inertia is split with inertia Ja located at the A end and Jb 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*Jb/(Jb+Ja) + Jb*Ja/(Jb+Ja).
///Please, would you clarify the equation in terms of Tb?\\\
 
There is a typo in my equation which should be obvious from the context (weighted average of Ta and Tb). It should be:

Tshaft = Ta*Jb/(Jb+Ja) + Tb*Ja/(Jb+Ja).

If you re-arrange aolalde's equation it is exactly the same result.

Thanks to aolalde for posting the proof.

It gives instantaneous shaft torque as function of instantaneous motor and load torque during the transient.

I would be interested to hear the assumptions and proof behind Gord's equation. It is the same as aolalde's execpt the multiplier [1-cos (wn*t)] after the 2nd term. I think maybe it applies for a step change in torque.

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Suggestion to the previous posting:
1. Reference:
Daugherty, Roger H., "Bus Transfer of AC Induction Motors: A Perspective," IEEE Transactions on Industry and Applications, Vol. 26, No. 5, pp 935-942, Sept/Oct 1990.

Comment: Reference 1 does not include the detail derivations; however, it refers to previously published literature.

2. Another related reference:
Gill G. Richards, M.A. Laughton, "Limiting Induction Motor Transient Torques Following Source Discontinuities," IEEE Transactions on Energy Conversion, Vol 13, No 3, pp 250-256, September 1998
 
jb - what makes you suggest I need those two references (you did suggest them specifically to me after all). Derivation of this result was provided by aolalde. Do you have some disagreement?

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jb - I apologize for misunderstanding your comment. I believe you are addressing my question about Gord's result.

If anyone can provide the assumptions and derivation here (not reference to a hardcopy textbook) I would be interested.

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Derivation based on the Daugherty paper:
Ts=JL/(JM+JL)xTe....(3)
d(dTs)/dt^2 + K x [(JM+JL)/(JM x JL)] x Ts = K x Te/JM - K x Tl/JL .... (4)
wn=2 x pi x Fn = Sqrt[(K x JM x JL / (JM x JL)] ....(5)
(wn is the first natural torsional frequency)
Assuming that the load torque Tl in (4) remains constant at A while the motor electrical torque Te abruptly changes from A to B, as would happen to the average value of the electromagnetic torque when switching operation takes place, the resulting shaft torque transient response is:
Ts=A+(B-A)x[JL/(JM+JL)]x[1-cos(wnxt)] ....(6)
Ts is a solution of the second order differential equation (4).
 
The derivation and assumptions behind my response seem to be the same as those identified by other responders in this thread. The only difference is that it includes a factor that varies from +1 to -1 to include the fact that shaft torque will not be constant but will osscilate at a frequency determined by the system natural frequency.
 
Question to the previous posting: Have you set the constant K equal to 1?
 
Jbartos...no, the K constant in not "1". It is in the expression for the system natural frequency - exactly as in the Daugherty reference.
 
Question to the previous posting: Please, are you claiming that:
Constant A in my equation is equal to TL in your equation?
Constant B in my equation is equal to Tag in your equation?
 
jbartos...yes. In fact the reference I used also used the designations "A" and "B...I changed the designations so they more closely look like the quantities they represent in reality.
 
Correction (I beg your pardon)
wn=2 x pi x Fn = Sqrt[K x( JM + JL) / (JM x JL)] ....(5)
is the correct form.
 
Suggestion: Solving (4) for its constants A and B specifically leads to:
Ts(t)=(K/wn^2)x[(Te/JM)-(Tl/JL)]x[1-cos(wnxt)]=
=[(JLxTe-JMxTl)/(JM+JL)]x[1-cos(wnxt)]....(7)
which is essentially what electricpete derived except the [1-cos(wnxt)] term and the sign that is - in (7) and + in electricpete equation.
The ordinary second order differential equation (4) may be solved quickly by Laplace or Fourier Transform.
 
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