Throttle wins system

[mort]

Originally Posted by Old Mechanic

Lets assume that throttle restriction is energy costly (it is), when an engine uses a throttle plate to control air flow into the engine. I guess the question is can you extract enough energy from the restriction to make it worthwhile?

I never really thought it was that much of an energy loss since the most pressure differential you can have is 14.7 PSI (give or take depending on atmospheric pressure).
Would any system be sufficient to do the job of the alternator?

regards
Mech

Intake throttling losses are less than 5% of output power under most conditions. And there isn't any energy to be gotten at full throttle, and almost nuttin at idle. So you're kinda stuck trying to get a small fraction of your part throttle at cruise. Maybe your car cruises at 65 mph using 20 hp. So you have about 1 hp to spin the alternator. That would be fine for fair weather.

I remember seeing a patent application for a turbine throttle with variable vanes and a clutch which revved up a flywheel. Moving the vanes acted as the throttle. At cruise you store energy in the flywheel. When you need 100% power the vanes "reverse" and the flywheel powers the turbine. The math is right on the edge of working.

Also looking at jakobnev's graph. Well I think he's missed something. There's hardly any air flow at idle - "full vacuum" so there can't be much power. Not 1kw.

In fact you can try back of the envelope figurin' The power lost to pumping the air into an engine is W=dP * Q, for me, using conventional dimensions, that's horsepower = pressure drop in psi multiplied by flow in cubic feet per second. Well, there's a bunch of conversions in there. Assuming the engine is in the 20% efficiency range and putting out 1 horsepower at idle, that works out to about .66 lb of gasoline per hour. The engine will use 129 cubic feet of air to burn that gasoline. Or 2.2 cu ft per minute, which is about 3700 cu in per minute. The throttle is nearly shut, so the pressure drop might be .95 bar - 14 psi. 14 psi * 3700 cu in per minute is 52000 in - lb per minute or 50 ft-lb per sec. or almost 0.1 hp. about 75 watts.

-mort

[jakobnev]

W=dP * Q is woefully inaccurate for expansion ratios much greater than one tho.

Any chance you can redo that in SI-units?

[user removed]

Originally Posted by drmiller100

throttle pumping loss is apparently measurable, and noticable. If you have a manual transmission, you can feel it yourself - coasting down the road, turn off the ignition and note your rate of deceleration. Then floor the throttle - you should feel deceleration lessoning.
This is the reason diesel engines have such crappy compression braking with out "jake brakes" and the like.

Actually if you are coasting with the engine engaged to the power train, and you floor it (engine off) you will increase the resistance.

Why?

Because when you have the throttle closed the in cylinder compression is very low since the air flow is restricted by the closed throttle (high manifold vacuum). With the throttle wide open you have 100% of the available compression, which in most engines is about 10 times atmospheric pressure.

You can see it with a vacuum gauge. 0 measured vacuum means highest possible compression while highest vacuum reading means lowest possible compression. Obviously it takes more energy to compress 10 atmospheres into one than it does to suck 1 atmosphere into an almost perfect vacuum (emphasis on almost). This is why most engines shut the fuel off during deceleration. The compression is so low the fuel will not burn anyway and you have huge amounts of unburned hydrocarbons coming out of the engine.

regards
Mech

[dcb]

I could see using a vvt on the exhaust stream, to add a flexible amount of torque to the drivetrain or something.

But since throttle losses are not necessary (direct injection) it would be best to eliminate them alltogether.

The heat energy on the intake would have to be via heat exchanger, which is going to lose significant energy in the process, the exhaust gas specific heat is not going to be room temperature, and the turbine intake specific heat is not going to be near what left the exhaust valve.

[mort]

Originally Posted by jakobnev

W=dP * Q is woefully inaccurate for expansion ratios much greater than one tho.

Any chance you can redo that in SI-units?

Hi jakobnev,
You must be thinking of some other situation. If your p1 is atmospheric the power to pump something is the sucking force times the flow. The equation is exact.

I guess I'll leave the SI conversion for somebody else. I never got the feel of 100000 Pascals. The only numbers you need to make up are A/F (14.7:1) energy density of the fuel (about 12 kwh/kg maybe for petrol) and engine efficiency.

-mort

[drmiller100]

Originally Posted by Old Mechanic

Actually if you are coasting with the engine engaged to the power train, and you floor it (engine off) you will increase the resistance.


Mech

Did you try it, or are you just talking your perception of theory?

If you tried it, you will find you are wrong.

Then to rationalize, when you have WOT (or a diesel) you compress 10:1 or 20"1 compression on the compression stroke, but you get all that energy back on the power stroke - kind of like a big pneumatic spring.

[drmiller100]

Originally Posted by mort

I Assuming the engine is in the 20% efficiency range and putting out 1 horsepower at idle, that works out to about .66 lb of gasoline per hour. The engine will use 129 cubic feet of air to burn that gasoline.
-mort

how many pounds per hour does a 2 liter engine consume at idle?

where does all that energy go?

ever wonder why diesels get such better mileage then gasoline engines?

[dcb]

fyi, direct injection gasoline thread:
http://ecomodder.com/forum/showthrea...gas-17460.html

so better to use the hot exhaust gas directly on a turbine wheel, no? a-la power recovery turbine http://ecomodder.com/forum/showthrea...tml#post204535

as far as I know adding a heat exchanger will at most equalize the heat between the flows, vs using the exhaust directly. If hot is good, hotter is better.

re: throttle open/closed coast. I never noticed a significant deceleration difference (seat of pants) except open is a little louder.

[jakobnev]

Originally Posted by mort

The equation is exact.

It seems you and i disagree on whether air is compressible or not.





(Clue: Your formula is accurate for hydro power)

[user removed]

Originally Posted by drmiller100

Did you try it, or are you just talking your perception of theory?

If you tried it, you will find you are wrong.

Then to rationalize, when you have WOT (or a diesel) you compress 10:1 or 20"1 compression on the compression stroke, but you get all that energy back on the power stroke - kind of like a big pneumatic spring.

You never get all of the energy back in any IC engine, unless you wish to ignore carnot's law, and yes I have tried it. I recommended it on this site years ago as a way to separate compression-suction losses from mechanical losses.
Coast in neutral
Coast with engine engaged and downshift at exactly the same points where you would normally up shift.
Do the same with the throttle wide open.
Push the car to the same speed and downshift at the same points as the previous test.

The "you get it all back statement" basically assumes there are not other losses involved when there are always losses involved. Moving air through an engine requires more work than sucking against a restricted intake. The rationale of you get it all back would apply equally to the vacuum scenario as it does to the compression scenario.

I have tried it, have you?

Tell you what, have you ever tried a compression test with the throttle closed, then realized you needed to open the throttle to get a true reading.
When you do that (open throttle) the cranking speed slows down because compression increases the total resistance.

Tried it on my Echo years ago and it slowed down faster with the throttle wot.

regards
Mech