From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Ed Anderson
Sent: Thursday, February 11, 2010 3:12 PM
To: Rotary motors in aircraft
Subject: [FlyRotary] Air Pump

I believe Al is right to be skeptical, but I agree with Al and hope it works out.

 

Here is where I believe you may find a problem.  There are two sort of opposing/conflicting parameters in the approach.  The first is your pressure increase and the second is your volume flow.  Generally, it’s fairly easy to get either by itself.  The problem is getting both at the same time.  For example if you put the end of your leaf blower a box and turn it on there is no question you are going to get a considerable increase in pressure in the box – but the air flow rate is going to be low.  Take the nozzle out of the box and the pressure at the nozzle will drop dramatically but the flow rate will increase greatly.

 

Most equations I have seen for calculating the output of a centrifugal impeller assumes they are exhausting into a static environment.  In your case you are going to exhausting into a huge air pump inlet.  This air pump (engine) is going to be trying to “suck’ the manifold void of air at the same time your impeller is going to be trying pump sufficient air to not only meet its ambient pressure air needs but actually provide it MORE than the engine can suck out - so that the air density will actually increase in the manifold – a tough order of business.

 

Air mass Flow = Volume of engine displacement * RPM * manifold air density /1728 = V * rpm *p/1728

 

1728 is just a conversion factor so we can ignore it for this discussion.  But what this formula shows is that there are only two variables in the equation - your rpm and your manifold density (p).  At a constant rpm (say 6000) then the only way to increase air mass flow through the engine is to increase the air density in the intake manifold.

 

The way a turbo/super charger works, of course, is not by increasing the air volume flow through the engine but by increasing the air density. The 100,000 rpm impeller accelerates the air velocity inside the compressor vanes and then using the old Diffuser principal, slows this air down at the compressor exit and converts the increased dynamic energy of the accelerated air stream into a static pressure increase reflecting the increased air density produced.

 

 So for our 13B at 6000 rpm it would still be displacing the same 277 CFM through the engine itself - but the air density per cubic foot would now be higher than ambient.  So while the volume flow is constant (at that rpm), the air mass flow is increased because the density is increased.  Yes, to increase this air density the air flow into the inlet to the turbocharger does have to be greater than 277 CFM, but it’s still 277 CFM (because air is compressible) through the engine itself – just at a higher air density.

 

There have been many “electric” turbo chargers for automobiles advertised on e bay and else where advertising up to 2 psi for prices ranging from $69 to $600.  Tests that have been done with them show that with some of the “best” ones, you can get around 2” psi with the flow nearly blocked off or you can get several hundred CFM air flow with no obstruction, but not both at the same time.  I think this is the essence of the challenge faced.

 

If you think about it - you are trying to design a blower that is competing with “suction” power of your engine.  There you have a 1300 CC engine that is pumping (“sucking”) like crazy at 6000 rpm drying to pump the manifold volume void of air.  The atmosphere is trying to push air into the manifold at 14.7 psi at a flow rate at this rpm of around 277 CFM.  To get any assist from a blower, it has to be able to not only provide for a flow rate the atmosphere pressure is pushing through it but then enhance that flow rate even further so that the air density in the intake manifold starts to go above ambient air density.

 

Now if you can get a 2 psi increase in the manifold with the engine running – then indeed you will get a very useful power increase.  If the equations say the 11” flywheel blower will give you 2lbs of pressure - do they also indicate what volumetric flow you will get at that pressure.  If they say you will get 2 psi at a flow rate of approx 300 - 500 CFM then you will definitely have something.

 

 I’m a rooting for you Earnest, but please expedite the rate of progress else I’m going to be gone before we get the answer {:>)

 

Ed