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Parallel hybrid EV with direct torque coupling - torque ratings of components 1

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123MB

Electrical
Apr 25, 2008
265
Hi All

We are designing a parallel hybrid vehicle which has an ICE engine driving a transmission and an electric motor coupled to the transmission output shaft.

A direct torque coupling is used between the motor and transmission output shaft - the motor is basically driven via a belt drive where the driven pulley is directly connected to the transmission output.

The philosophy is that the motor speed will match the tranny output speed and be applying more or less torque than the output shaft ie to regen, assist, etc. If something goes wrong it will match speed and torque so in theory should present no load.

My confusion relates to the fact that the transmission output is capable of applying a lot more torque than the motor and belt can handle and above the motor rated torque.

For example, the motor might be able to apply a max of 100 Nm and the tranny output Being 1000 Nm.

Will this break something? What will
The torque on the motor shaft and belt be in this case?


Cheers
 
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Yes, if the rear wheels slip the entire output of the tranmsission could drive the motor. I suspect your belt will slip or break.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
It seems that the transmission cannot apply more torque to the motor than the motor is capable of reacting or else the motor will just speed up.
 
This is my main question whether the motor will be driven by the transmission torque but nothing will break… I am finding it hard to be convinced though..
 
The motor will only have a positive or negative torque that it can either generate electricity or supply torque by using electricity.

Your drive needs to be capable of the highest of those two. It matters not what the transmission can supply, the motor will only supply or take 100 Nm. Speed has nothing to do with it.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
Thanks all

Littleinch, so what do you think will happen when more than rated torque is applied the motor shaft while it is in the generating mode?

I suppose the torque at the motor shaft is generated by the regen action of the motor, up to rated torque, and this determines the torque transmitted through the belt and transmission output.
 
The motor/generator controller limit the amount of torque the motor can produce, so it will likely turn faster and the controller will lower the current to limit the voltage. At some point, if the motor starts to overheat, the controller will detect that condition and either lower the current further or cut it off altogether.

If no such protections are in place, then "bad things will happen" comes to mind.
 
Consider the freebody of the driveshaft, elec motor & wheels -
1000nm is applied to the system at the transmission output end of the driveshaft.
The wheels have just started slipping and so are free to spin. The applied torque results in an instantaneous rotational acceleration of the whole system (torque=angular accel*system rotational inertia). Assume for simplicity the belt is rigid as this will conservatively overestimate the belt drive load. The load share between the wheels and electric motor will be proportionate to the rotational inertia of each (since angular accel is constant for the whole system).
Consider adding a factor on the belt drive loads to account for dynamic loading.
Can the belt drive / motor handle the resulting load?
 
Thanks Ng2020

Fundamentally the belt drive/motor isn’t rated to the max load of the transmission output. I will need to check regarding dynamic loading as you describe.
 
And the speed is also limited by the rotational speed of the engine for which the motor will need to match.

Depending on the inertial mass of the motor under very high acceleration such as wheel slip you might get a bit more torque for a very short period, so you will need a bit more torque capacity than the motor max.

But in normal operation the motor will only exert the torque it can supply or import. You need a resistance to rotation to generate torque. If that resistance isn't there then neither is the torque.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
"the driven pulley is directly connected to the transmission output."

I think the output shafts of RWD automotive transmissions I'm somewhat familiar with are not going to enjoy the radial/side loading of a drive belt at all.
 
The transmission may be "capable of" producing much more output torque than the motor and drive in question is, but if I understand the mechanical layout correctly, that full output torque is never "passing through" the motor and drive in question.

What it means is that if the motor seizes a bearing or otherwise locks up then the main powertrain will be capable of breaking it, but if the motor seizes a bearing then it's already broken anyhow ... so what.

Very abrupt inertial loads can cause the momentary output torque of a motor (or whatever) to be exceeded, and sometimes by a lot. Maybe the driver floors it on an ice patch and then the car slowly rolls forward until the drive wheels contact dry pavement. Maybe the driver drives over a jump with the pedal to the metal leading to the wheels spinning much faster than the road speed, and then the car lands. Maybe the driver slams on the brakes as hard as possible and locks the brakes almost instantly.

If you're worried about the auxiliary motor and drive torque rating being exceeded and breaking something, perhaps you can design-in a failure point ... a shear pin, or some other such thing, that will intentionally break should something abnormal happen.
 
All this talk of slipping / spinning wgeels is old school.

I haven't been able to spin wheels for years now due to anti lock systems.

Only if you turn all that protection off can you do it, but then you take the consequences.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
We haven't the faintest clue of the type of vehicle that the original poster is designing (possibly something off-road and thus not subject to requirements for all the nannies that a developed-world road vehicle has), nor where in the world it is destined to be used (possibly something outside of the developed world and not subject to all of the regulations that the developed world has). Or maybe someone presses the "ESP OFF" button for long enough to actually turn all that stuff off ...

If it ends up as a line item on an FMEA with an exclusion "These circumstances will never happen because of X and Y and Z" then so be it.
 
Thanks all

Can anyone please describe the inertia concern in a bit more basic terms as to how it could break something (mech is not my strong suit)

Cheers guys appreciate it!
 
GregLocock said:
Yes, if the rear wheels slip the entire output of the tranmsission could drive the motor. I suspect your belt will slip or break.

I'm confused on why you and others seem to be suggesting that potentially 1000Nm of output from the transmission could be fed to the belt/electric motor. Just because a transmission/drive is capable of applying 1000Nm doesn't mean it CAN if there is minimal resistance aka from a freewheeling electric motor.
 
Thanks everyone for your responses I appreciate it

The vehicle is an on highway freight prime mover with traction control.
 
123MB said:
Can anyone please describe the inertia concern in a bit more basic terms as to how it could break something (mech is not my strong suit)

The rotor of an electric motor has a certain amount of rotational inertia, called the "moment of inertia" I. It's like a flywheel.

Accelerating (deceleration = negative acceleration) a flywheel to a certain rotational speed in a certain amount of time requires a certain amount of torque in the shaft. The standard symbols used are Greek which I have no idea how to type on my keyboard.

But, basically, it's analogous to F = m A (force = mass x acceleration) except for rotation rather than translation (linear motion, motion in a straight line).

Acceleration = dV / dT (acceleration is the amount of change in velocity divided by the amount of change in time)

So F = m dV / dT (Force = mass x amount of change in velocity divided by the amount of time it took for that change) - Analogous, except using rotations and torques instead of linear motions.

If the amount of time for a change in velocity to occur becomes small then A (acceleration) becomes big and F (force) becomes big. If this consists of an object smashing into an immovable object such that dT approaches zero then F approaches infinity.

Now ... if this is a rotating motion rather than a linear-translation motion then the forces become torques, and if it's being transmitted through a shaft and the amount of torque exceeds what the shaft can take, stuff breaks.

* * *

Spin up a flywheel. A big, heavy one. Spin it really, really fast. Then - Slow it down by pushing something lightly against it so that the flywheel slows down smoothly over a long time. Easy, right? Maybe whatever you're using as a brake gets warm (this is how the energy is being dissipated). Still ... Easy. No problem to slow the flywheel down gradually. Now, instead, after spinning it up, jam something into the spokes abruptly. What happens? Stuff goes flying and/or stuff breaks.

OK, back to your drivetrain. The motor has a heavy rotor. The driver has disabled the traction control and/or maybe a sensor has failed or some such thing with the same effect. Driver is stuck on an ice patch. Driver mashes the accelerator to the floor. The drive wheels, and your electric motor rotor, spin up really really fast. Then the truck gets to the edge of the ice patch and onto the concrete ... What's the rate of deceleration of your rotor? Big.
 
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