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Quote:

My previous estimate had been that it should require 10 horsepower or less. Have I missed something?




Yup, your calculations are extremely simplistic. You don't take into account the flow. What you think is taking into account the flow is just taking the volume of air every second seperately, and compressing it to 5psig and then moving on to the next volume of air, and compressing it.
You are not actually taking the account the continual flow. Imagine if you will you have your ordinary shop compressor that is controlled by a variable speed (and hence power) compressor. The tank connected to it is a given volume (the actual # doesn't really matter). Now, say you have the tank pressurized to 5 psig, and you open the outlet valve just a little bit, and the compressor turns on to regulate things inside to 5 psig. Say the pump turns using just enough power to continuously maintain that 5psig. It won't take much power to keep the pressure up while the leak is small. Now, double the size of the leak, how much more power is required? A lot more. Ok, then try opening up the leak to 50% of the end area of the storage tank, now you are talking supercharger like pressurized flow.
Your calcs don't even assume there is a small leak, just that it keeps filling up new tanks every second . . .
The actual calculations are far from trivial, and far more than I even want to be reminded of this late at night (I hated fluids . . .)


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Originally posted by warmonger:
hey, before you waste your money, Starter motors are not capable of sustained use. The duty cycle on them sucks! More like 10% If you crank for 15-20 seconds you have to wait 1-2 minutes to cool it down.



That's about as much as I would plan to use it, generally. And if a starter blows, replacements are cheap.

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Originally posted by Rara:
Quote:

My previous estimate had been that it should require 10 horsepower or less. Have I missed something?




Yup, your calculations are extremely simplistic. You don't take into account the flow. What you think is taking into account the flow is just taking the volume of air every second seperately, and compressing it to 5psig and then moving on to the next volume of air, and compressing it.
You are not actually taking the account the continual flow.




Yes I am. The math I used is for a continuous situation, the kind you can do calculus on. It's similar to the basic equation for air drag on a moving body, in that the faster the air flows through the intake pipe, the more power it takes to apply a given pressure to it.

Originally posted by Rara:
Imagine if you will you have your ordinary shop compressor ... Your calcs don't even assume there is a small leak, just that it keeps filling up new tanks every second . . .



Yes they do. You're mistaken. If it weren't for the rate of flow being accounted for, the power consumption over time would average out to zero.

Originally posted by Rara:
The actual calculations are far from trivial, and far more than I even want to be reminded of this late at night (I hated fluids . . .)




Calculating it exactly is complicated, but it is simple enough to calculate the minimum amount of power that must be consumed regardless of the details of a particular design -- namely, the amount corresponding to the true work done on the fluid itself in compressing it. It's the number below which the compressor would have to be a perpetual motion machine. And since compressor makers advertise an efficiency figure which is measured relative to that ideal minimum (at least, I hope that's what they're doing if they're honest)... then just dividing by that efficiency figure gives you a shortcut to the genuine power consumption. Centrifugal devices seem to claim numbers in the range of 70 to 80 percent.

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Paul,

You calculated that you need 0.15m^3 of air at 5psi every second. Lets assume you have an intake system that actually has two 0.15m^3 tanks with a valve that seals either one tank or the other from the intake, and that the system switches between them every second. So the first second, your compressor fills tank A to 5psi, then the compressor switches to tank B and fills that to 5psi. While the compressor was filling tank B, the engine took the air out of the tank A, then while the compressor is filling tank A, the engine takes the air out of tank B, which was just filled to 5psi. The cycle constantly repeats like this, as long as the engine is running. Sounds like a great system. Except that the engine will only be fed air at 5psi from tank A the instant that the valve on tank A opens. After the valve opens, the pressure gradually decreases as the air leaves the tank. So you may have 5psi of boost at the start of each second, at the 0.5 second mark you'll be at 2.5psi of boost, at the end of each second you are NA.

This is why, as Rara stated, your calculations for power are much lower than they should be. You were calculating how much power is required to pressurise 0.15m^3 of air to 5psi every second, not how much power is required to keep a flow of air of 0.15m^3/sec at 5psi.

Hopefully this makes sense,
Bob

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Originally posted by Cougar Bob:
Paul,

You calculated that you need 0.15m^3 of air at 5psi every second. Lets assume you have an intake system that actually has two 0.15m^3 tanks with a valve that seals either one tank or the other from the intake, and that the system switches between them every second.



Why should we assume that? It has nothing at all to do with anything previously discussed.


Originally posted by Cougar Bob:
So the first second, your compressor fills tank A to 5psi, then the compressor switches to tank B and fills that to 5psi. While the compressor was filling tank B, the engine took the air out of the tank A, then while the compressor is filling tank A, the engine takes the air out of tank B, which was just filled to 5psi. The cycle constantly repeats like this, as long as the engine is running. Sounds like a great system. Except that the engine will only be fed air at 5psi from tank A the instant that the valve on tank A opens. After the valve opens, the pressure gradually decreases as the air leaves the tank. So you may have 5psi of boost at the start of each second, at the 0.5 second mark you'll be at 2.5psi of boost, at the end of each second you are NA.



That's true, if that were the kind of system under discussion. But none of the math discussed so far has anything to do with this bizarre setup, which actually has much more complicated math than the continuous situation does.

Originally posted by Cougar Bob:
This is why, as Rara stated, your calculations for power are much lower than they should be. You were calculating how much power is required to pressurise 0.15m^3 of air to 5psi every second, not how much power is required to keep a flow of air of 0.15m^3/sec at 5psi.

Hopefully this makes sense,
Bob




That is not why at all. If you can point to any reason at all why this applies to anything I said, and explain it to me with a mathematical equation that means something, go ahead.

Now let us review the calculation I used, and what it's based on: It presumes a continuous flow which travels at a given constant rate -- though actually the math works fine for a variable rate, if you're willing to do a little calculus. That rate can be expressed as a volume per unit of time, which by the engine displacement and max revs I took as 0.15 m^3/s on the upstream side of the compressor. (It's less at lower RPMs.) Now to pass through an intake tube, that volume of fluid has to pass through a specified cross sectional area. Passing a volume through a given area at a given rate determines a linear velocity for how fast it is flowing through. We are attempting to produce a given pressure in that stream of air. Multiplying the desired pressure by the cross sectional area gives us a given force that has to be constantly applied to the fluid.

Now when you apply force to a stationary object, you don't have to consume any power to do so. But if you apply force to an object moving away from you, it requires continuous expenditure of energy. This is why an accelerating car produces less and less G's of acceleration, the faster it goes: because it takes more and more power to maintain the same force against the road, as the difference in speed between the car and the road increases.

Note that this has nothing to do specifically with compression -- it is a generality about any case of applying force to something in motion.

The way to calculate the energy expenditure is with "work = force times distance". If you push something with 90 N of force for 6 meters, you've expended 540 J of energy. If you keep pushing it constantly, the energy expended in a given span of time is the force times the distance traveled in that same span of time. This span can be a long average, or it can be an instantaneous differential of a constantly varying situation -- the same rule applies.

The distance per second equals the volume per second over the cross section. The force equals the desired pressure times the cross section. The cross section cancels out, leaving us with power consumption equal to pressure times volume per second. It really is that simple. This is not a formula for the work of filling a tank, but for how power consumption relates to the current rate of flow at each instant.

Of course in a real compressor, the total power consumption will be significantly higher due to inefficiencies, but that extra will be some fairly constant fraction of this basic amount. Whatever the actual consumption, it will rise and fall the same as this ideal figure does. The ratio of that difference is the efficiency of the compressor.

Now how does this apply specifically to compressing air? Well, there's one factor that isn't accounted for, and that is that the volume decreases after it's pressurized. I took my figure of 0.15 m^3/s from the upstream side, but the force figures apply more to the downstream side, which means that the flow rate is lower than what I used, meaning that I actually overestimated the minimal necessary power consumption.

I'm sorry to go on at such length to lay this out, but if the short form doesn't get the job done, sometimes you gotta do what you gotta do.

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I have done some work on automotive power electronics. Cant say much, but if you are serious, look into power conversion. You will achieve much higher efficiency at higher voltages. Most automotive power engineers agree high power systems cannot be run on less than a 42v core. FYI the next gen Toyota hybrid will be running a 42v battery to a sealed voltage converter to a 300-500v core ISA.

Now I would love to see it when its done, but every application of e-turbos I've seen is supplimental. Turbos alone are very very efficent, however they tend to have lag. E-turbos have no lag, but use lots of power. Once combined the system works very well, the e gets the engine to speed and starts the turbo, which then takes over. Some systems are even so advanced that the e-turbo can extract power and prevent over-boost without a bov.



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Originally posted by geoffct:
Cant say much, but if you are serious, look into power conversion. You will achieve much higher efficiency at higher voltages.




Yes, that option is definitely in my mind as something to look into. It's hard to argue with the inexpensiveness of an ordinary starter motor, though...


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Paul, you are still missing the point between continously compressing a given amount of air, and compressing a continuous flow. It sounds like its the same thing, but the difference is huge. If you don't believe me, try it.


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Originally posted by Rara:
Paul, you are still missing the point between continously compressing a given amount of air, and compressing a continuous flow. It sounds like its the same thing, but the difference is huge. If you don't believe me, try it.




If you think I've got it wrong, show me where with math.

I think you're more confused than me on this point... come on, if you read through what I laid out, how can that even be STRETCHED to apply to "continuously compressing a given amount of air" as opposed to working on a flow? There's no way that stuff could even be used that way, without running it through a fancy integration first. What I stated applies only to flow. Compression as such (change of volume) is not even in there! Whether you're filling a tank, squirting through a nozzle, or pushing through a thick filter, the work done on the fluid will equal the flow rate times the back-pressure.

So to me, the criticism that I didn't consider flow sounds so backwards that I can only assume you didn't follow the reasoning at all. If anything, a valid criticism would be completely opposite: that I considered only flow and disregarded actual compression.

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I have a fairly decent grasp on the laws of physics, and
all I can see from all these posts is that Rara has a
pretty good grasp too. You maybe need to study this stuff
better before you waste alot of money on your prototypes.

First of all, a starter motor doesn't come anywhere near
even close to 10 hp. The only reason it can turn an engine
over is because it is gear reducted (ie. big flywheel vs.
tiny starter gear). If you were to gear it the other way
to try and turn a compressor impeller the speed necessary
to feed an engine running at 5000rpms with 4psi say, I
don't even think the starter would barely turn over (the
vacuum from the engine's intake across the impeller would
probably help the starter turn a little). I don't even think
you could get .001 psi at 5000rpm with a typical starter motor
off a car. A guy I know has slow-acting hydralics on his
Impala (not even fast acting mind you), and he requires 6
Optima batteries in the trunk to run just 2 motors and they
don't last very long at all. You are going to find that
electricity is not very efficient for trying to run
compressors. Do you know how hard it is to even turn a
compressor by hand? I have a customer that we put a Paxton
supercharger kit on his truck a while back, and when you
take the belt off it and try to turn the pulley with your
hand, it is VERY hard to turn. You cannot even make it
spin by it self (you'd have to mount a huge flywheel to it to
make it do that). You are going to find that you are
trying to accomplish the unsolvable task doing this on a
backyard budget. Believe that Ford, Chevy, Chrysler,
Kenworth, Peterbuilt, John Deere, Case, Caterpillar, and
others have experimented with this idea for years, and
who knows,
maybe in 10-20 years they will be able to make this
workable, but don't hold your breathe. I'm not trying to
discourage you, I'm only trying to save you $$$. You don't
think people have tried this crap for years? People have
tried every way to convert the engine's byproduct energy or
other engine energy into power and so far, turbo and
blower is the best that's out.

Here, read this: http://cgi.ebay.com/ebaymotors/ws/eBayISAPI.dll?ViewItem&item=2426797798&category=33742

and this: http://www.bamaclassifieds.com/ebay/warning.htm

And check this out: eBay search page

and

http://www.electricsupercharger.com (haha)

Those are all the current scams out because this topic of
electric superchargers is such an easily convincible
subject, because at first quick thought it
sounds 'feasible' but if it REALLY was then don't you
suppose the professional racers (with millions of dollars in
sponser money mind you) and the automakers would be in on this too? Trust
me, this is not just some well kept secret. There's just
too much efficiency loss, and high-powered electric motors
are really HEAVY. Why not just use a pulley/belt and a
shaft (ie. Vortec) to get the power directly and avoid all the heat losses from
electron movement (not to mention the weight from motors and batteries)
, or use the 'free' exhaust gases (heat energy) that the
motor just throws away anyways and convert that to usable
energy (ie. conventional methods)??
By the way, I had your same idea when I was like 18-19 years
old and it DOES sure sound like a good idea on paper, but it just can't
work with today's technology. Wouldn't you just rather work
extra hard for a summer and then just buy the complete turbo
kit? Sounds much easier to me than trying to defy the laws of physics


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