Kers rev matching?

All that has to do with the power train, gearbox, clutch, fuels and lubricants, etc. Generally the mechanical side of Formula One.
Post Reply
Mahfuz
0
Joined: 30 Apr 2015, 16:40

Kers rev matching?

Post

Hi all,

Firstly, i am completetly new to the forum (my first forum membership period) so cut me some slack plese.
If this topic has been covered please post a link to the appropriate forum, although i have looked myself and havent had any success.

Anyway the question-again forgive me if it sounds stupid...or maybe i am just over thinking.

I want to know how Kers systems (both flywheel and elec) couple/link/synchronise with the drive shaft of an ICE or its gearbox. In my mind, the shaft from the flywheel/motor must be spinning at either the same rpm or faster, than the engine driveshaft , to actually 'add' torque.
(Also can someone confirm if the KERS flywheel/elec motor actually has to be spinning at the same rate or faster)

Thanks in advanced
Maf

gruntguru
563
Joined: 21 Feb 2009, 07:43

Re: Kers rev matching?

Post

"The MGU-K must be solely and permanently mechanically linked to the powertrain before the main clutch."
je suis charlie

bergie88
8
Joined: 25 Aug 2014, 12:20

Re: Kers rev matching?

Post

The rules say that there must be a permanent connection, which does not mean that they have to run at the same speed. I think that the power unit manufacturers use a certain ratio with a fixed gear between the engine and the electric motor. This might not be necessary because the electric motor can easily run at the same speed of the engine, but it have some benefits with efficiency of the electric motor and the size of it, reducing weight and required space.

Mahfuz
0
Joined: 30 Apr 2015, 16:40

Re: Kers rev matching?

Post

Thank you both for the reply.
I did think it would be fairly simple to get a motor to spin at the same rate as the engine, so I guess with elec Kers its a matter of programming that aspect of the motor.

However, I'm sure the Kers system still has to spin at the same rate as the main shaft (taking into account any gearing) to be able to 'add' torque.

Let me give an example. If a cars engine has 100Nm of torque and is spinning at 3000rpm, a Kers system with an extra 50Nm needs to be spinning at 3000rpm (at the output- so gearing included)to actually add the 50Nm
If the Ker's system is actually only spinning at 2000rpm will it not just create resistance for the main shaft to over come and essentially 'take away' some of the engines 100Nm of torque.

Another example-more of an analogy...
If I'm running at 20m/s whilst pushing a heavy trolly, and my 'kers' system in this instance is a bodybuilder running at 10m/s. We are both connected by rope. In this case the bodybuilder can provide a huge amount of 'torque' but its useless if he can't even catch up to me to help me push the trolley, whilst I'm still pulling a round his weight due to the rope.

marcush.
159
Joined: 09 Mar 2004, 16:55

Re: Kers rev matching?

Post

your thinking allows only for a special case of driving the electric motor but is this really what´s going on ? Will the electric motor try to stick to a preprogrammed rpm (sweep)? If you are driving the Motor with PWM it is more likely you will feed a certain pulse width and the revs will adjust simply to the load the motor can carry with the power supplied .Sure you can programme this the other way round and try to maintain a certain rpm but for power boosting this may be a bad idea for strategy

bergie88
8
Joined: 25 Aug 2014, 12:20

Re: Kers rev matching?

Post

Lets assume there is a 1:1 connection between the engine's driveshaft and the electric motor. In that case the electric motor will always run at the same speed as the engine, as a result of the fixed mechanical connection between them. When there is no electrical power fed to or withdrawn from the electric motor, the electric motor is not inducing any friction (please correct me if this is wrong). When a certain amount of electrical power is supplied to the electric motor it will produce a certain amount of mechanical power, based on its efficiency on that operating point as a function of speed and torque. Off course this operating point must be within its maximum torque line.

Lets give an example. When the engine rotates at 2000 rpm the electric motor will also rotate at the same speed. When there is for instance 50 Nm of torque requested from the electric motor and the efficiency of it is 90% (at 2000 rpm and 100 Nm), you will need 2000x50/0.9=111kW of electrical power, resulting in 100kW of electrical power.

Post Reply