Injector Math

[andy quaas]

Oh my god.  You must keep us up to date on your progress.  You are
going to use the eec-4, no?

Andy


---Frederic Breitwieser <frederic.breitwieser at xephic.dynip.com> wrote:
>
> Greetings, as most of you know by now, I'm building an
> efi-based mopar stroker engine using a turned down 440 crank
> combined with a .030 over 383 block, for a total
> displacement of 431 cubes.  With the below math, I'm trying
> to caluclate several things.
> 
> Approximate HP/Torque to be expected from the motor, both
> N/A as well as boosted (10 lbs of boost max), the
> appropriate injector size for a sequential-based injection
> system, as well as cfm of air required at 5000 RPM naturally
> aspirated as well as with 10lbs of boost.  Yes, I'm having
> fun.  Anyhoo:
> 
> 
> Using several different sources (boogs, mags, websites) I
> once again found this formula:
> 
>         fuel flow (lb/hr) = HP x BSFC
>                             ---------
>                               cyls
> 
> BSFC would be .45 for a N/A engine, .50 for a Nitrous
> engine, and .55 for a forced induction motor, and 1.1 for an
> alcohol motor, and 1.76 for a nitromethane motor.
> 
> Cyls is easy, I have eight.
> 
> So I have flow (lb/hr) = HP x .55
>                          --------
>                             8
> 
> Now, I have two variables left.  Flow, which I'm trying to
> calculate (gives me injector size), and HP.  How does one
> calculate the math for the HP without testing the engine?
> 
> The engine has these stats:  1969 383 cid stroked to 431 cid
> using a 1970 forged hipo crank turned down for the
> appropriate journal sizes of the 383 block.
> 
> Bore: 4.25" + .030 overbore (4.28" bore)
> Crank Stroke: 3.75" (1970 forged 440 crank)
> Rods: 6.358" (B rods) or 6.768" (RB rods)
> 
> I have chevy rods but I'm prolly going with Mpar 440YJ rods
> and having Ross or Wiseco make pistons with the pin higher
> into the head of the piston, with a final compression ratio
> targeted at 9.0:1 before turbocharging, which I think I have
> sized appropriately for 10lbs of boost as a maximum with a
> water-based intercooler system.
> 
> Also, the cam will be an RV cam once I find something
> appropriate, as this engine will be in a low RPM Torque
> motor for my Dodge truck, for towing purposes.  The goal is
> to make all the power on the low end for towing in the
> 1500-3000 RPM range, with a 5000 RPM maximum redline
> determined arbitrarily.
> 
> As far as the intake goes, I am using an aluminum Edelbrock
> 383 Streetmaster intake, which I have modified by welding on
> injector bungs so the fuel injectors are flush with the end
> of the runners right at the heads, and where the carb
> normally would have gone, I've welded together a bolt-on
> plenum that is 10" wide, 15" long, and 12" high, minus a 3"
> transmission cooler core configured to become a water-based
> intercooler.  The volume of the plenum therefore is 1575
> cubic inches including the volume of space taken up by the
> intercooler, and should be more than adequate.  The front of
> the plenum has two Ford throttle bodies that I have to
> interconnect mechanically so they are effectively ganged,
> each connected to a turbo. TE-34 would be my first guess for
> a turbo.
> 
> For horse power calculations, I used Corky Bell's PLAN
> formula, which is Power = bmep x stroke x bore area x
> putts.  Breaking this down further, I ended up with this:
> 
> power = bmep x stroke x area x putts (cyl * maxrpm/2)
> power = bmep x 3.75" x 57.5489 x (8 cyl * 2500)
> power = bmep x 4316167.5
> 
> How do I resolve this with two variables?  Same problem as
> before.  I could merge the above two formulas, however then
> I'd end up with this:
> 
> So I have flow (lb/hr) = HP x .55   and HP = bmep x
> 4316167.5
>                          --------
>                             8
> 
> Merged I have flow = bmep x 4216167.5 * .55
>                      ----------------------
>                               8
> 
> flow = bmep x 2318892.125
>        ------------------
>                8
> 
> Still, I have two variables, and can't resolve either of
> them.  So, I ventured onto more simplistic formulas.  Corky
> has a range formula like this:
> 
> powerlow = .052 x 431cid x (10lbs of boost + 14.7) =
> 553.5764 hp
> powerhigh = .077 x 431cid x (10lbs of boost + 14.7) =
> 819.7189 hp
> 
> Then, I opted to extrapolate the same formulas to work
> without boost, and ended up with the following:
> 
> powerNAlow = .052 x 431cid x 14.7 = 329.4564 hp
> powerNAhigh = .077 x 431cid x 14.7 = 487.8489 hp
> 
> Now I am truly confused.  However, I continued the math for
> giggles, and took the original formula using the highest
> value I ended up with (820 hp):
> 
>         fuel flow (lb/hr) = HP x BSFC
>                             ---------
>                               cyls
> 
> and substituted seemingly random numbers like so:
> 
>         fuel flow (lb/hr) = 819.82 HP x .55
>                             ---------------
>                                    8
> 
>         fuel flow (lb/hr) = 56.362625 lb/hr
> 
> Bendix/Siemans has injectors that flow 55 lbs/hr as well as
> 62 lbs/hr.  MSD has 50 lb/hrs (2012) that work with stock
> ECMs (high impedence) and fuel rail pressure can be
> increased to compensate I would imagine so there is more
> flow.  Injectors are rated based on certain fuel rail
> pressures, so increasing the rail pressure would increase
> the flow proportionally, within reason.  50 lb/hr injectors
> when I need 56.x lb/hr injectors could be close enough if
> the rail pressure goes up 13%, which I can do easily enough.
> 
> 
> How off am I in my math?  Any ideas?
> -- 
> Frederic Breitwieser
> Bridgeport, CT 06606
> 
> http://www.xephic.dynip.com
> 1993 Superchaged Lincoln Continental
> 1989 500cid Turbocharged HWMMV
> 1975 Dodge D200 Club Cab (soon to be twin turbo 440)
> 2000 Buick GTP (twin turbo V6)
> 

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