How does recovery of braking energy improve I.C. engine performance?

[krish1988]

I have a report to write on "recovery of braking energy to enhance the performance of I.C. Engine".
i can calculate the energy lost during braking but i don't understand how that energy is restored in batteries and how much energy can be stored and etc.
Can anyone explain fundamental working of whole situation or what books or links should i refer.

 

[Shane-O]

Don't Formula One cars use this?... I don't know the physics behind it.

 

[xxChrisxx]

Can't think of any books, or technical articles.

There are loads and loads of different systems, but typically it's not actually the brakes that recover the energy. A motor generator is put in between the engine and the wheels.

When slowing down the motor then acts as a brake (in a similar manner to engine braking) which slows the driven wheels. So in effect the wheels are turning the motor, which makes it generate power.

I'd start by going a google search for how regenerative wraking and KERS. How stuff works is usually good at explaining the basics. They'll have pictures too!

 

[johnbbahm]

Breaking a car the energy of the moving mass is converted to heat (by way of a break pad)
The heat is dissipated over time. Any regenerative system would store that energy some way to help get the care moving again. The improvement of such a system would be offset by any added weight.

 

[BobKerns]

OK, here's how regenerative braking works.

First, a little background information about motors and generators. This is all directly derived from the basic electromagnetic physics of a conductor in a magnetic field.

Basically, a motor and a generator are the same thing, used differently. Every motor acts as a generator, and generates voltage as it turns, proportional to the rotational speed. Every generator acts as a motor, giving torque opposing the driving torque, proportional to the current.

The opposing voltage is sometimes referred to as "back EMF", but let's just simplify the picture, and consider them both to be the same thing. By thinking of just one device (as it actually is), we have a very simple and accurate model.

So we can describe a motor very simply (as a slightly idealized model ignoring resistance and similar effects), thus:

a) The voltage is proportional to the speed. Reverse direction, and you get reversed voltage. If you control the voltage, you control the speed. If you control the speed, you control the voltage.

b) The current is proportional to the torque. Reverse the current, and you reverse the torque. If you control the current, you control the torque. If you control the torque, you control the current.

You can't control both voltage and current at the same time. If you change the voltage, how much current is determined by how much torque is required to speed up or slow down the motor and whatever it is attached to.

Likewise, if you control the current, you apply whatever torque is indicated by that current, and the voltage will reflect what the load does, speeding up, slowing down, or staying the same.

So, I'm driving my Prius (I actually have two Gen 1 Prius's), and I step on my brake, but not hard enough to engage the friction brakes (it does have friction brakes).

Or, I brace for a stop on my Segway, and it begins to decelerate accordingly. (It does NOT have friction brakes, and relies entirely on regenerative braking. Friction brakes would be incompatible with its self-balancing function).

To do this, both devices draw power from the drive motor. That is, they take current in the same direction as the motor's voltage is pushing it. Power = voltage x current, so this is power out. Power integrated over time is energy -- and this is the energy that's stored in the battery.

Now, there's a problem. Batteries operate at an approximately fixed voltage, based on the electrochemical reactions that go on inside.

But voltage varies with speed. So a regenerative system, to work at a low speed, has to accept the current at a low voltage, and yet stuff it into the battery at a high voltage.

To do this, we can use a bit of switching-transistor magic, in a circuit called a voltage multiplier. Let's take, say, four capacitors. We accept charge quickly from the motor (high current) and use it to charge the capacitors in parallel, matching the motor's output voltage at this speed.

Then, we flip a few transistor switches, and stack the capacitors in series, giving 4x the voltage, but only 1/4 the current (so it's the same amount of power out). This then pushes the charge into the batteries.

To take the energy back out on starting up again, we just reverse the process. Charge up the capacitors in series, dump them into the motor at a low voltage but high current for a good high starting torque -- but carefully, so we don't spin the tires. Note that this gives us high torque right where gasoline engines are at their worst -- low speeds.

That's about it, really. The batteries to hold the charge aren't very big, if you're only holding the energy from braking. Both a Segway and a Prius have batteries that are much larger than that, and can store enough energy to drive for a considerable distances on battery alone. (24 miles, for a Segway). Electric motors are relatively light for a given output, compared to an internal combustion engine.

If you also use battery power to satisfy peak demand, you can save weight by using a lighter internal combustion engine AND add to its life by operating it in a lower-stress operating region. Extreme hybrids only need to size the engine to the average power demand.

Of course, in an electric vehicle, like the Segway, you have electric motors, motor controllers, and batteries anyway. You need ZERO additional parts, and ZERO additional weight to do regenerative braking. It's free energy. The same is true if you have a hybrid for other reasons (overall efficiency). There is never a reason to NOT do regenerative braking, if you have motors and batteries (or ultracapacitors) in place.

 

[BobKerns]

Oh, a point about the motor controller and voltage multiplier circuitry.

This is typically done at a rate ranging from around 100 hz to maybe 50 khz, depending on the application. Fast enough that, to the motor, it is seeing essentially a continuous load or source of current or voltage. This switching is very efficient, because switching transistors switch VERY quickly, spending almost no time in a resistive half-on/half-off state.

Modern batteries are also very efficient at storing energy, and even electric motors are pretty efficient. If you have 90% efficiency storing the energy, and 90% efficiency getting it back out, you end up recovering 81% of your original energy. A particular application might have better or worse than those numbers, but that's ballpark what we're talking about.

 

[krish1988]

Thank you BobKerns, it has helped me a lot to clear my concept and I really appreciate. I also have another topic regarding air/fuel ratio and BSFC.

Q1. why bsfc increases at low engine output?
Q2. why bsfc improves as engine output increases?
Q3. I want to know what are the influence of the air/fuel ratio on engine output and Brake specific fuel consumption (BSFC).

I am looking at Ricardo E6 C.I.Engine.

 

[BobKerns]

I'm afraid I can't be as much help when discussing internal combustion engines. But let's look at it mathematically for a moment.

At low engine RPM, the torque is low -- there is a whole torque curve, low at both the low and the high end.

Power is the product of the angular velocity and the torque. So you're going to have the the product of two small numbers for the output, but you're still going to consume a full cylinder of fuel/air mixture -- plus you have the fixed frictional losses.

Similarly, the high end of the RPM range, torque drops. I'm not sure what reasons predominate, but I expect it's harder to get fuel/air into the cylinders, and harder to get complete combustion in a very limited time. Ultimately, the speed of propagation of the flame front presents a hard limit on how fast an engine can run.

So it makes sense to me that at low output, and especially at low RPM, the engine will be much less efficient, and at high RPM as well. What what we observe is higher efficiency when an engine is operated near its maximum torque.

As for mixture -- too rich, and you're wasting unburned gasoline. Too lean, and the engine won't be running at its most efficient; in fact, if it's too lean, it won't run at all.

I hope that help you get started, but I bet you can find a better explanation online.

 

[krish1988]

Thank again for your help and it did got me bit started but i haven't understood the influence of air/fuel ratio on engine output and brake specific fuel combustion. Could you please help me out to make the concept more clearer.