[Diy_efi] Intercooled question

[elcamino73]

Scary, very scary


Blown 407 CID
73 ElCamino


----- Original Message -----
From: "Brian Michalk" <michalk at awpi.com>
To: <diy_efi at diy-efi.org>
Sent: Tuesday, June 25, 2002 9:56 AM
Subject: RE: [Diy_efi] Intercooled question


> What about a big can of propane.
>
> Put an expansion valve before the intercooler, then route the expanded
> propane into the throat of the intake manifold.
>
> > -----Original Message-----
> > From: diy_efi-admin at diy-efi.org [mailto:diy_efi-admin at diy-efi.org]On
> > Behalf Of Brian Dessent
> > Sent: Tuesday, June 25, 2002 4:01 AM
> > To: diy_efi at diy-efi.org
> > Subject: Re: [Diy_efi] Intercooled question
> >
> >
> > elcamino73 wrote:
> >
> > > Maybe if I give you some details you might make things clearer
> > (easier) for
> > > me to understand.
> > > I want to inject freon into an intercooler to bring air which will be
at
> > > approximately 120 degrees down (F) to 40 degrees(F). The max
> > pressure is 15
> > > psi, the max cfm is 1500. The freon will be pre-pressurized in a 5 to
10
> > > pound container which will be recovered when off boost. I would
> > be happy to
> > > be able to do this for thirty seconds but 60 seconds would be
> > preferable. I
> >
> > The way I would approach this is to think in terms of the mass flow rate
> > -- kg/s (I like to work in SI units.)  So the air entering the
> > intercooler has T = 322K, P = 205kPa, and volumetric flow rate =
> > 0.708m^3/s.  From the ideal gas equation (air can be treated as ideal),
> > PV = mRT, or m = PV/RT.  R for air is 0.287 kJ/(kg*K).
> >
> > So the mass flow rate (or "m dot", the dot meaning first derivative
> > w.r.t time) = 205 * 0.708 / 0.287 / 322 = 1.57kg/s.
> >
> > Since the air is free to contract as heat is removed, we want to use
> > C_p, the specific heat at constant pressure.  C_p is about 1000
> > J/(kg*K).  You desire to lower the temperature by 44.4K.  Therefore
> > m-dot * C_p * delta-T = Q-dot, which is the rate that energy (heat) must
> > be removed.  In this example it comes to 67.7 kJ/s.  1kJ = 0.948 BTU, so
> > that means 64.2 BTU/s.
> >
> > If one pound of R12 can remove 70 BTU, then you're looking at about 11
> > seconds worth of cooling for 10 lbs of R12.  Now, this is worst-case, in
> > that the pressure and flow rate that you mentioned are the maximums, so
> > if in reality they are less, then the R12 would last longer.
> >
> > (You said that you were planning on recovering all the R12, and since
> > it's pretty expensive these days and requires certification to acquire,
> > I would assume you know what you're doing.  It sure seems that R134a
> > would be a much easier and cheaper alternative though, and safer for the
> > environment in case something goes wrong...)
> >
> > Brian
> >
> > _______________________________________________
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> >
>
>
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