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.