Electric Supercharger

[Frank Lee]

Nobody has throttled an engine with a tc/sc have they?

I wonder if there is enough energy in the intake airstream to be worth harvesting.

[Christ]

Originally Posted by Frank Lee

Nobody has throttled an engine with a tc/sc have they?

I wonder if there is enough energy in the intake airstream to be worth harvesting.

Apparently, one of my posts didn't post...

The way to throttle an engine with a fan would be to run it at a high speed for low throttle, and run it at zero speed for max throttle... the problem with this, is that it wouldn't be anything close to instant throttle response, and the amount of energy scavenged would directly affect the VE of the engine - according to the laws of conservation of energy, when you extract energy from a mass, the mass would then have that much less energy to give.

Since you're extracting energy from air flow, any energy extracted would cause that flow to slow down, negatively impacting the engine's intake charge.


Per this site:

We can substitute the intake fan for a wind turbine here, and substitute the influx of air for wind.

We can determine the power that can be extracted from the intake air by using the following formula:
P = 0.5 x rho x A x Cp x V3 x Ng x Nb

We'll replace the variables with nominal figures, assuming we're at sea level, and the fan has a 4" diameter, using the Betz limit, which is considered universally unreachable, as our coefficient of performance Cp, assuming our generator is 80% efficient (Ng), and assuming there is no gear or drive losses (Nb):

P = 0.5 x 1.225 kg/m3 x .008 m2 x .59 x V3 x .8 x 1

The only plug left to make is the actual volume of air in M/S.

50 CFM works out to 0.0235973722 m/s3, which replaces V3 in the equation:

P = 0.5 x 1.225 kg/m3 x .008 m2 x .59 x 0.024 m/s3 x .8 x 1

So, given the best know circumstances, the power generation capability would be: 0.0000555072watts at idle speed using 50 CFM of airflow.

Yep - Five one-hundred-thousandths of one watt, under nearly perfect circumstances.

And that's extracting an enormous amount of the energy from the airflow, so much so that the engine couldn't idle any more, with perhaps the most efficient equipment currently known to exist.

Anyone care to check my math? I think I messed up somewhere.

[Frank Lee]

I'd say you messed up. It would be based on the engine's displacement for one thing so there is no one answer.

[Christ]

Originally Posted by Frank Lee

I'd say you messed up. It would be based on the engine's displacement for one thing so there is no one answer.

I was using the figure given by TomO for intake flow of the VX engine (1.5 liters) at idle, 50CFM.

I think there's something wrong with my conversion math, though. I had to convert CFM to M/S^3, and convert 12.56 IN^2 to M^2.

[pgfpro]

I still think by just using a properly sized turbo you will get the same gains or better.(living proof is my engine)

Turbo technology has vastly improved in the last five years.

[TomO]

Originally Posted by Christ

I was using the figure given by TomO for intake flow of the VX engine (1.5 liters) at idle, 50CFM.

I think there's something wrong with my conversion math, though. I had to convert CFM to M/S^3, and convert 12.56 IN^2 to M^2.

Just a correction: the 50CFM seen in the VX motor is at 2200rpm, not idle. At idle (550 RPM) it's 12.32CFM.

[Christ]

Originally Posted by TomO

Just a correction: the 50CFM seen in the VX motor is at 2200rpm, not idle. At idle (550 RPM) it's 12.32CFM.

Woops, thanks!

Anyone care to check my math?

I did a similar calculation for someone claiming that they made a difference with a squirrel cage fan in a fog light hole in another thread, it came to less than 100 Watts produced at highway speed...

Tonight, I'll redo the formula with 100% VE at 6k RPM. We'll see the max theoretical figure that the 1.5 L engine could produce.

Or, I could do the same calculation for a 12 liter engine at 100% VE, if that would suit anyone's thoughts better.

Either way, the number is ridiculously low, and the engine still wouldn't be able to run properly with that much energy being extracted from the intake stream.

Remember, the Betz limit is there for a reason... the theoretical max extraction of energy from wind is 59%, because at 60%, equilibrium is reached (according to Betz, anyway.) If this weren't true, a brick wall would be 100% efficient at harnessing the wind's energy, since it would not allow any of the wind to pass.

With a fan, it's rotational speed is inversely proportional to the amount of air allowed to pass. The fan would provide more resistance to air passage at higher speeds, and less resistance at lower speeds, to the extent that (unlike I said earlier) full throttle would not be a stopped fan blade, but instead, a blade that spun only enough to allow an unrestricted passage for fluid (air).

Without actually doing the calculations, I can theorize that the fan in the intake would reach equilibrium, in a power generation sense, at almost all times. There wouldn't be enough energy to extract at high fan speeds, due to low intake volume, and the fan wouldn't extract enough of the available energy at low fan speeds, due to the low generator armature speed.

[user removed]

Half the displacement (4 cycle inducts only half the displacement per revolution), minus the manifold vacuum percentage of atmospheric pressure, times the RPM.

.5DXvac/ATPXRPM=volume of air flow.

750 CC (per revolution inducted) at 550 RPM at 22 inches of manifold vacuum which leaves 8 inches available (assumption). That is assuming atmospheric pressure is 28 inches, slightly low but easy to calculate.

Manifold vacuum at 22 inches would leave 8 inches of available atmospheric pressure for induction or about 25%, of the volume if it was unrestricted.

46 cubic inches---X---550 RPM---X25% of atmospheric pressure volume of air.

That works out to 6325 cubic inches of air at idle or 3.66 cubic feet of air.

Christ, I am not engineer by any stretch of the imagination, but if the power is so low then conversely it would take very little power to boost the engine to 1 atmosphere pressure. I think the quote was over 200 Amps for the power necessary to provide boost. It would seem to me that you should be able to get much more than .0005 out of the same system.

Basically the basis of my thought was the fact that most engines operate at a certain average of manifold vacuum.

I was not talking about an propeller of any type. I was talking about a positive displacement pump that would have to rotate if any air passed through it.

It could not be reciprocating and would by design only restrict the air available to the engine in the same way a throttle plate does.

That is a loss that already exists.

Exhaust heat already exists.

Both do represent energy losses, but in the same way that aero drag on your vehicle represents an existing energy loss.

Please do not consider this as any rebuttal to any of you who are interested in this thread and have taken there precious time to post a response.

My line of thinking (which could certainly be miles off, merely conceptual) is if you could extract the energy lost due to throttle plate restrictions which are always there, then you could use that recaptured energy to provide boost when necessary. As long as the boost was very limited it would be an energy neutral condition.

Now it is certain that you could never extract the same energy you would need to apply to reverse the situation and provide boost to the engine, but that expenditure would only be a very small amount of the total running time of the same engine.

A decent analogy would be the way we tested tire-wheel combinations on the car when we had a vibration problem that did not respond to balancing.

We took one of the small angle grinders, backed the brake pads off the front rotors and used the angle grinder to spin the wheel up to a high speed. The angle grinder produced very little power. You could hold the disc and pull the trigger and it would not spin over.
However that same low power source would spin a wheel-tire assembly up to close to 100 MPH. That's enough energy to rip your arm off if you grabbed that rotating tire.

My calculation looks a lot lower that Tom's. I would assume he figured the air going into the engine would be at atmospheric pressure, when the volume represented only 25% of what it would be if the throttle was wide open.

Not sure of anything, just thinking out loud, no need to slam me as a moron.

We do know how much power it takes to boost the sir into the engine. I would think you could extract about a third of that energy with a proper setup.

I'll show a picture of my small demo prototype on a variable displacement, positive displacement pump, that is what I would use in this theoretical application if you like.

Its the basis of my pending patent.

What I really like about the electric supercharger is that it does not add to the engines loads unless it is being used. Even then it is using battery energy, which could be recovered when you were in DFCO or at other times when it would not be a direct cost of operation all the time.

regards
Mech

[micondie]

In all this discussion one thing has not been addressed. The electricity to run the supercharger has to be generated by the alternator.
With the less than 100% efficiency of both the alternator and the supercharger motor you end up with a net loss of energy.

[TomO]

Originally Posted by micondie

In all this discussion one thing has not been addressed. The electricity to run the supercharger has to be generated by the alternator.
With the less than 100% efficiency of both the alternator and the supercharger motor you end up with a net loss of energy.

In my scenario, I would have the fan be on it's own circuit running off a deep cycle battery stowed in the trunk.