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Old 11th February 2014, 02:20   #211
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Default Re: Torque generation and distribution

Samurai san.

(P= K*N*T) this is valid only at a steady state which is constant rpm.
if the Rpm changes equation changes to
P= K*N* T + I( moment of inertia) * angular accelaration* angular diaplacement/t

The first part is power to sustain a torque. The second part is to accelerate .
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Old 11th February 2014, 08:03   #212
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Default Re: Torque generation and distribution

But I am saying it will be in steady state, say at 1000rpm, with zero acceleration. Therefore it is P=(K*N*T) in my case.
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Old 11th February 2014, 14:07   #213
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When one refers to a constant torque drive, it is assumed that the load is the same and the speed is the same. Hence the torque requirement would be constant. The drive would take a speed sense as a feedback to regulate the torque to maintain at rated speed.

Example: A Mixer. For a given load, it has set 3 to 4 speed settings. When one sets one of the settings, and turns it on, the drive would regulate the torque to maintain it at same speed. And due to this feedback the steady state is : rated speed with constant torque. For such a system we should not be looking at transients.

PS : I did some textbook reading over the weekend to refresh myself. Its primarily a DC motor with Armature Voltage control for speed control and sensing is the back EMF based on Rotor speed.
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Old 11th February 2014, 23:50   #214
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Quote:
Originally Posted by Samurai View Post
You know, you reminded me of something. When I was a teenager, I read these books on Physics (The Book Thread). These books were written in 1913, and had numerous example of inventors trying to create perpetual motion machines....

Not really, not unless you are assuming very theoretical situation like no friction, no gravity, etc.
I really dont understand why this talk of perpetual motion and newtons second and first law coming into picture. When I have already said that 10 Nm is the torque required to overcome system resistance, this should answer your above point.

Ok lets modify the example, that the shaft at one end has a fan blade attached to it and is dipped in a viscous fluid (resistance). So once the clamp is released there is still resistance of 10 Nm after the clamps are released.

So now does it makes sense...

Just for you to get a feel I will give another analogy (I have learnt something from you!)

You have a bicycle and you are pedalling hard read standing on the pedals (torque) to climb a hill (resistance) and suddenly the chain comes off the sprocket.

What happens?

Your leg were applying torque and suddenly the load is removed which makes your legs go down with violent speed (read increase in rpms).

After that depending on your body you will keep on pedaling at increased rpms which suit your body (system design/inertia)

Well ultimately the person riding the bicycle will fall down and have a nasty bump between the legs. Telling from my own experience

Quote:
It will induce acceleration, but not steady, in fact the acceleration will be decreasing until the equilibrium is reached, where acceleration is zero.
No not as per your definition of constant torque drives.
Again I repeat for maintaining rpms power is required not torque. Torque is required only to overcome the resistance.

If torque was required to maintain rpms then we should have rpms in definition of torque which is not there anywhere.

Its already mentioned in my earlier posts that
T=F*r formula to be used for static equilibrium (that also applies to steady state ie a shaft rotating at constant rpm with constant load)

T=I*alpha (For transient state)

and P=K*N*T applies for steady state again.

Let me explain the T=I*alpha with another analogy..

Lets take a shaft without any load, and suddenly you apply torque to it. What happens it starts rotating agreed, but if you observe at a microscopic level the portion of the shaft (fibres)near the torque source twists (due to inertia even without load) and the other end of the same shaft lags behind by a very minute amount. This twist leads to the increase in torque because torque is measured by strain or twisting of the shaft.

Another example in an accident your seat belts hold your body but your neck gets a whiplash beating right. The same thing happens with a shaft microscopically(as the shaft is not perfectly rigid practically in real world).

The more the impact speed more the de-acceleration (g forces) the more the damage to the neck, same for a shaft.

Quote:
Now you have rightfully said power is required to maintain that 1000rpm. But if you reduce T, power will go down. To keep that power, you need the torque that got you there. There you go, I am using your formula.
How easily you forgot the rpms and kept it constant.
If you reduce torque the power will go down, agreed but then it doesnot because there is a increase in rpms to balance the power.Thats what we see in the bicycle example right?

Quote:
Let's take your scenario. You say the rpm will go on increasing if we sustain the 50Nm of torque, that means P will also go on increasing... possible? Not possible, the drive can't possibly provide infinite power, while we know constant torque drives are real.
To keep torque constant the Power will go on increasing. Yes, the drive should provide infinite power (as per the definition of constant torque drives which would maintain constant torque on shaft irrespective of load). This tendency to reach infinite power should destroy the system.

If you have seen Constant Torque drives in real then it should have a speed limiting mechanism also.
I think ampere has also talked about the same.

Last edited by amit_purohit20 : 11th February 2014 at 23:58. Reason: Some corrections done
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Old 12th February 2014, 00:18   #215
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Default Re: Torque generation and distribution

Quote:
Originally Posted by amit_purohit20 View Post

If torque was required to maintain rpms then we should have rpms in definition of torque which is not there anywhere.

Its already mentioned in my earlier posts that
T=F*r formula to be used for static equilibrium (that also applies to steady state ie a shaft rotating at constant rpm with constant load)


.
Hmm IMO No.
Torque is required to maintain rpm against friction.

There will be frictional forces on surface in contact ( which is by mass*g*friction coefficient) which gives an opposing frictional torque to the supplied torque. This torque = frictional force * shaft radius

This torque needs to be overcome to maintain rpm.

Another way to this about this is energy conservation. If power (energy/time) is used for spinning, that should go somewhere.

If you look at the system as a box with only power going in ( and don't know about what is happening inside the box), you would assume that there is destruction of power ( or energy/time). we don't want to do that , do we?? In reality that power is used in overcoming friction and comes out of the box as heat ( because of friction). That is the only way in which energy can be conserved.

In an ideal case when there is no friction ( can be achieved in places where there is no gravity)- if you supply a torque continuously (even 1 Nm) your Rpm will keep increasing till the inertial forces break the part apart. But the moment you stop applying the torque, the rpm remains where it was ( after acceleration up to that point)


Actually once the motion starts , the friction coefficient drops - remember static friction v/s Kinetic friction. So the breakaway torque is higher than what is normally needed to keep it spinning.

I hope this helps.
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Old 12th February 2014, 00:27   #216
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Default Re: Torque generation and distribution

Quote:
Originally Posted by Jomz View Post
Hmm IMO No.
Torque is required to maintain rpm against friction.

I hope this helps.
If you read my entire post we both talk the same, so no need to disagree!

If you quote/read only a portion of my post, it would seem that we have different view points but I agree to what you said here.

Last edited by amit_purohit20 : 12th February 2014 at 00:50. Reason: Correction
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Old 12th February 2014, 00:43   #217
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Quote:
Originally Posted by amit_purohit20 View Post
If you read my entire post we both talk the same, so need to disagree!

If you quote/read only a portion of my post, it would seem that we have different view points but I agree to what you said here.
My apologies.... I thought otherwise
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Old 12th February 2014, 01:12   #218
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Default Re: Torque generation and distribution

Quote:
Originally Posted by ampere View Post
When one refers to a constant torque drive, it is assumed that the load is the same and the speed is the same. Hence the torque requirement would be constant.
No, that is not my understanding. If load and rpm are constant, then torque will be constant in any drive, not just constant torque drive. The constant torque drive is something where load can vary, and rpm will adjust to maintain the same torque. If load halves, the rpm will double, provided the drive has the power.

Quote:
Originally Posted by amit_purohit20 View Post
I really dont understand why this talk of perpetual motion and newtons second and first law coming into picture. When I have already said that 10 Nm is the torque required to overcome system resistance, this should answer your above point.
Say you are cycling up an incline. Say you are applying constant amount of torque through the pedals, and have reached 20kmph of speed. Now, if you drop the torque, will you sustain the same speed? Won't you drop the speed?

Quote:
Originally Posted by amit_purohit20 View Post
Ok lets modify the example, that the shaft at one end has a fan blade attached to it and is dipped in a viscous fluid (resistance). So once the clamp is released there is still resistance of 10 Nm after the clamps are released.
Oh, I never assumed that shaft has no load. Sutripta did mention here (Torque generation and distribution) that the shaft has nonzero MOI. That is why I am unable think about a runaway shaft rotating too fast.

Quote:
Originally Posted by amit_purohit20 View Post
Just for you to get a feel I will give another analogy (I have learnt something from you!)
.
.
Your leg were applying torque and suddenly the load is removed which makes your legs go down with violent speed (read increase in rpms).

After that depending on your body you will keep on pedaling at increased rpms which suit your body (system design/inertia)

Well ultimately the person riding the bicycle will fall down and have a nasty bump between the legs. Telling from my own experience
Nice example for visualizing, except people are not constant torque source. None of us have to power to increase the rpms to maintain the torque.

Quote:
Originally Posted by amit_purohit20 View Post
If torque was required to maintain rpms then we should have rpms in definition of torque which is not there anywhere.
Huh? You mean if rpm is a function of torque, then rpm should be in torque definition? Why?

No, that is not how I look at it. Take a regular car, to drive the car on road you may need 50Nm of torque pushing the engine to 2000rpm. If you put the gearbox in neutral, and push the A-pedal all the way down to generate same 50Nm of torque, the engine may hit redline to provide the same resistance. At any point, if you let go the pedal, the rpm will drop.

Anyway, I am at the edge of my understanding with this stuff, not really my field. I am really getting out of my depth here.
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Old 12th February 2014, 03:20   #219
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Default Re: Torque generation and distribution

Quote:
Originally Posted by amit_purohit20 View Post
I really dont understand why this talk of perpetual motion and newtons second and first law coming into picture.
This is my doing. I assumed that this, from you, ....


Quote:
Originally Posted by amit_purohit20 View Post
T=I*Angular acceleration

Torque is dependent on acceleration and not angular velocity (rpm) so that means to get the torque of 100 Nm the angular acceleration will have to be constant and not the angular velocity (rpm).
...was your attempt to have your understanding conform with Newton's Laws of Motion...... and his F = MA where Force = Mass times Acceration....as opposed to an older Greek formulation wherein they assumed that F = MV where V is velocity.

What? You mean you didn't know you are a Newtonian?.......this may mean that you are way late on your membership dues.

Last edited by DirtyDan : 12th February 2014 at 03:33.
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Old 12th February 2014, 07:40   #220
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Default Re: Torque generation and distribution

Quote:
Originally Posted by Samurai View Post
No, that is not my understanding. If load and rpm are constant, then torque will be constant in any drive, not just constant torque drive. The constant torque drive is something where load can vary, and rpm will adjust to maintain the same torque. If load halves, the rpm will double, provided the drive has the power.
I said earlier I meant for steady state. As per my understanding a constant torque drive does not provide speed control as a feature to the user. If you take the mixer the speed settings are a given. So the rating is constant torque for a max load for a max speed setting. And we see that as max power rating for the device. If the load changes, the torque will change to a different value to run it at the rated speed. And this is; provided the load is lower than the max limit, such that the rated speed can be attained.

As in the case of the mixer, if we stuff beyond capacity, it does reduce in speed. Because there is a max rating for torque that the motor can handle.
Similarly if we try to run at no load, it will attain a max constant speed for which the the torque needed will be attained. Same is the case (I think) with the motor driving conveyor belts also.
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Old 12th February 2014, 10:32   #221
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Default Re: Torque generation and distribution

I am still assuming a theoretical constant torque source, as mentioned by Sutripta. So, when the clamp is released, it can only get back the load by raising the rpm. That is all I am trying to establish.
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Old 12th February 2014, 11:25   #222
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You have to understand if you have a source which can deliver constant torque, or constant force, and you use it either the load has to be there, or if the load is not there, it will cause accleration.
Imagine a solar sail in space.
Photons are constant, and in absence of any resistance in space, the spaceship with such a sale will keep accelerating. The rate of acceleration will reduce as speed approaches "c" as the mass increases.
Therefore the speed will approach "c", never reaching c, as the mass will tend to infinity as speed tends to "c".

Now same thing if you attach a wheel to an ideal source.
The wheel will keep accelerating(rpm will keep increasing) until the frictional losses (shaft/air/etc.,) match the source.
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Old 12th February 2014, 21:26   #223
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Default Re: Torque generation and distribution

Guys,
Lets get back to the very basics, and start with a very basic system.
Of interest to us here is not the nature of the driving (power) source, but rather the behaviour of the system. Thus the specifying of a (theoretically perfect) constant torque source. It removes a variable. It can be reintroduced once our understanding is clear. (For the time being think of it as a weight supported by a pulley/ sheave. Assume weightless superstrong string supporting the mass, weightless sheave etc.)

Let this sheave be supported on an axle, the other end of which is our wheel. (Once again, for initial simplicity, let the axle/ shaft be weightless etc.)

The wheel needs to have a finite MoI.

We are interested in the
a) torque as seen by/ transmitted by the axle/ shaft.
b) The behaviour (speed/ acceleration) of the wheel.

The whole system can be devoid of any other forces.
or
the wheel can have a fixed (ie. not a function of velocity) load on it.
or
the wheel can have a load which is a (monotonically increasing) function of the speed.

Case 1: the wheel is clamped.
Torque transmitted by shaft = torque generated by source.
Velocity =0. Acc = 0.

Let the clamp be released.
If the system has no other force acting on it,
Torque transmitted by shaft = torque generated by source.
The wheel will accelerate using Newtons F= M * A (well, the rotating,rather than linear interpretation).
And will continue to accelerate. To infinite velocity if the force is maintained long enough.

If the wheel can have a load which is a (monotonically increasing) function of the speed.
The same reasoning and equations hold as before, except that at any point of time, part of the force will be used to overcome the drag, the rest to accelerate the wheel.
Since our force (torque) is constant, and the drag increases as a function of velocity, at some velocity the drag will equal the force supplied. This will be the terminal velocity.

(My comments)
At all times the shaft carries the torque supplied by the source.
If the source is a perfect constant torque source, the the shaft will always see that constant torque.
Even if the torque source is not perfect, over a short interval (of time and rpm) we can take it to be so.
Thus my reasoning that if clamps are released at t=0, torque as seen by the shaft at - delta, 0, and + delta will remain the same.
There will be no torque spike (seen by the shaft) at + delta (when the clamps are released.)
Nor will the torque (seen by the shaft) fall to 0 at + delta (when the clamps are released.)

To all those who are wondering about the 'resistance' of the free running wheel in a perfect system: the resistance comes from the acceleration of the wheel.

Regards
Sutripta

PS. And can we please keep loops out of this discussion.
For, Do While, Repeat Until loops
Ground loops
Open and closed loops,
Loops in mountain railways.
Loops in knitting.
.....
.....
Far too many of those. Far more than I can handle.

Last edited by Sutripta : 12th February 2014 at 21:34.
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Old 12th February 2014, 22:47   #224
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Now same thing if you attach a wheel to an ideal source.
The wheel will keep accelerating(rpm will keep increasing) until the frictional losses (shaft/air/etc.,) match the source.
If the frictional losses are always less than the applied torque the shaft will go on accelerating infinitely (because the source is a theoretical source here). This will cause the whole system to blow apart due to induced centrifugal forces.

@Samurai - I hope this point helps to understand the difference between velocity and acceleration due to the applied torque.

So what happens in case of an engine which has a shaft attached to it, other end of the shaft carries a fanblade which is dipped in viscous fluid?
The accelerator position is fixed at a certain level.

Now as per the system inertia the engine settles down after some time at a certain rpm.

What are the conditions that make the engine settle down at a certain rpm. And how do we limit the engine rpms? One point is the fuelling, any other ways?
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Old 12th February 2014, 23:34   #225
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Quote:
Originally Posted by amit_purohit20 View Post
If the frictional losses are always less than the applied torque the shaft will go on accelerating infinitely (because the source is a theoretical source here). This will cause the whole system to blow apart due to induced centrifugal forces.

@Samurai - I hope this point helps to understand the difference between velocity and acceleration due to the applied torque.

So what happens in case of an engine which has a shaft attached to it, other end of the shaft carries a fanblade which is dipped in viscous fluid?
The accelerator position is fixed at a certain level.

Now as per the system inertia the engine settles down after some time at a certain rpm.

What are the conditions that make the engine settle down at a certain rpm. And how do we limit the engine rpms? One point is the fuelling, any other ways?
The engine will settle down at the rpm dependent on frictional losses.
You also need to understand there is an ECU.
ECU reads the throttle position sensor, and sends fuel.
At idle, it just maintains enough fuel to maintain 1000rpm or so.

At idle, even a small blip on throttle can make rpms rise.
At load even flooring the pedal can mean just 2000 rpm(eg climbing an incline)
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