How can regenerative braking be used to save energy in trains?

[trideon]

Hello, I'am new to this forum and this is actually my first post. And I have some questions about regaining braking energy.

We are thinking of using an electric motor to speed up and slow down a new and innovative train. The idea is to regain energy while the train is slowing down with the electric motor.

I've read some articles on the net about braking with electric motors and regaining energy, but I can't seem to find any good formula's for calculating the regained energy. The formula's I found are hard to understand and they differ a lot with other websites.

Maybe you could help me out. I would like to hear some ideas and formula's from you.

Thanks,

some specs of the train:
20 tons, 20.000kg
top speed 250 km/h , 70m/s
it rides on wheels, r=0,5m (max)

some formula's I found:
pdf url document from an electric motor manufacturer.
http://www.reliance.com/prodserv/standriv/appnotes/d7743.pdf
The formula's who are in this pdf document arent clear to me. Maybe because I'am familiar
with the metric system

http://img693.imageshack.us/img693/3609/formule.th.jpg

 

[Bob S]

Your kinetic energy is 49 MJ (megajoules) or about 13.6 kilowatt-hours. Where do you plan to store the recovered energy? (Note:the battery in the Toyota Prius holds about 1.2 kWh).
Bob S

 

[OmCheeto]

Hello, I'am new to this forum and this is actually my first post.

greetings trideon

And I have some questions about regaining braking energy.

We are thinking of using an electric motor to speed up and slow down a new and innovative train. The idea is to regain energy while the train is slowing down with the electric motor.

I've read some articles on the net about braking with electric motors and regaining energy, but I can't seem to find any good formula's for calculating the regained energy. The formula's I found are hard to understand and they differ a lot with other websites.

Energy recoverable from braking equals 0.5 times the mass of the vehicle times the initial velocity squared.

Until you understand this, all other formula's will only lead to confusion.

 

[trideon]

Your kinetic energy is 49 MJ (megajoules) or about 13.6 kilowatt-hours. Where do you plan to store the recovered energy? (Note:the battery in the Toyota Prius holds about 1.2 kWh).
Bob S

First thanks for responding.

I haven't really thought about the energy storing. Maybe a lot of big battery's or give it back to the electricity net. And I'am thinking of the usage of multiple electric motors.

Energy recoverable from braking equals 0.5 times the mass of the vehicle times the initial velocity squared.

Until you understand this, all other formula's will only lead to confusion.

I know that the net regenerated energy is E_k=1/2×m×v² reduced with friction losses and efficiency losses in the system.

My question is: Is there any easy or understandable way to calculate the losses. What values/data do I need.
Oh and the Deceleration is 2m/s^2 = a, in F=m×a. So I know the force F (40.000N) and a = 2m/s^2.

 

[OmCheeto]

First thanks for responding.

I haven't really thought about the energy storing. Maybe a lot of big battery's or give it back to the electricity net. And I'am thinking of the usage of multiple electric motors.

I know that the net regenerated energy is E_k=1/2×m×v² reduced with friction losses and efficiency losses in the system.

My question is: Is there any easy or understandable way to calculate the losses. What values/data do I need.

In the situation where one comes to a full stop:

Initial Kinetic Energy = Change in Stored Energy + Energy Lost

therefore

Losses = Initial Kinetic Energy - Change in Stored Energy

 

[Topher925]

Your losses and the behavior of the train doesn't depend on an equation, it depends on the regenerative braking system. There is no way to just simply calculate these losses accurately, or even ballpark the numbers really, unless you have detailed information about the braking system itself.

Regen braking systems used in cars are very complex and require a lot more than just switching a few resistors on a motor or generator.

 

[Bob S]

My question is: Is there any easy or understandable way to calculate the losses. What values/data do I need.
Oh and the Deceleration is 2m/s^2 = a, in F=m×a. So I know the force F (40.000N) and a = 2m/s^2.

The weight of the train is 20,000 Kg x 9.81 m/sec² = 196,000 N. The desired stopping force is 40,000 N. So the needed wheel friction coefficient on the rails is about 40,000/196,000 = 0.2. Even with regenerative braking on every steel wheel on steel rails, you will not be able to achieve half of this. So you will need to switch to rubber tires.
Also, the stopping regenerative power is

P = Force x velocity = mass x deceleration x velocity = 20,000 Kg*2 m/sec² x 70 m/sec = 2.8 MW = 3,800 HP.

So your regenerative motor/generators (brushless dc motor/generators) are significant.

Bob S

 

[trideon]

The weight of the train is 20,000 Kg x 9.81 m/sec2 = 196,000 N. The desired stopping force is 40,000 N. So the needed wheel friction coefficient on the rails is about 40,000/196,000 = 0.2. Even with regenerative braking on every steel wheel on steel rails, you will not be able to achieve half of this. So you will need to switch to rubber tires.]

I was forgotten to put this with it: The train rides frictionless on the rails, because it uses an aquaplaning effect by his advantage. The train is riding on a thin sheet of water. A bit hard to explain everything but the point is the friction caused by wheels touching the rails can be neglected.
Anyway thanks for the formula's, I will use and apply them.

Thanks every1 for responding I think I have enough to make some good conclusions about regenerative braking. I was hoping there was a simple way to calculate the regenerated energy but I came to the conclusion that its very hard to do en there are very variables which should not be neglected.

 

[Ivan Seeking]

...Note that dynamic brakes are used by electric multiple unit trains as well. In these designs, careful blending with air braking is required to maintain a smooth braking profile. Electronic control is used to determine that the brake effort demanded by the brake controller is matched by the brake effort achieved by the train. Preference is given to the dynamic brake to save wear on brake blocks (shoes) or pads and air braking is added if necessary to achieve the braking rate required.

Dynamic braking can be used on electric railways to convert the energy of the train back into usable power by diverting the braking current into the current rail or overhead line. This is known as regenerative braking. It is used in the same way as rheostatic braking but the energy can be used by other trains requiring power. The power developed by a braking train may not be accepted by the line if no other trains are drawing power so trains equipped with regenerative braking will usually have resistor grids as well to absorb the excess energy. The balance between regenerated current and rheostatic current is also controlled electronical. See also our Electric Traction Pages Page and under Dynamic Brakes in North American Freight Train Brakes...

http://www.railway-technical.com/brake1.shtml#DynamicBraking