pintle vs. disc injector types

[Michael McBroom]

Jeffrey Engel wrote:
> 
> I don't think outside pressure would affect it.  The systems I am
> aware of reference the FPR to the intake manifold, so:
> 
> Fuel pressure is: P1 + Constant (about 30-41 psi) and P1 is the pressure
> inside the intake
> 
> the pressure at the end of the injector is: P1 (minus a bit due to 
> air flow resistance.  Maybe a max of 1-3inHg)

Well, yes.  It looks as if you're essentially agreeing with what I
wrote, though.  The "outside pressure" I was referring to IS P1.  In
that respect, you're also comparing different values than I was.  In the
original posting, I had taken the equation to mean that P2 was the "new"
fuel rail pressure and P1 was the "old" one, since the solution was for
Q2, a new flow rate.  In your example above, P1 is intake pressure,
where P2 (or P1 + Constant [aka fuel rail pressure]) is adjusted fuel
pressure, which is a different set of parameters.

Boy, did my equations get fouled up in the word-wrap process.  Let me
try this again.  For a turbocharged application in which a proportional
fuel pressure regulator is used, then it seems to me that one cannot use
the equation:

             P2
 Q2 = Q1 * (----)^0.5
             P1

to determine flow.  It seems that instead one must use the ratio of the
differences between fuel rail pressures and intake pressures in order to
determine new flow rates, since fuel rail pressure increases with
boost.  The way I see it, the following equation would have to be used:

             P2 - (atmospheric + boost)       
 Q2 = Q1 x (----------------------------)^0.5
                 (P1 - atmospheric)

where: P1 = fuel rail pressure + atmospheric pressure
       P2 = P1 + (boost * fpr ratio)

EG #1: using a 1:1 fuel pressure regulator

Given: 1:1 fpr
       0.8 bar boost
       Q1 = 300ml/min
       P1 = 3 bar + 1 bar (rail pressure + atmospheric pressure)
       P2 = P1 + (boost * fpr ratio)

               ([3 + 1] + [0.8 * 1]) - (1 + 0.8)
Q2  =  300 x (-----------------------------------)^0.5
                          ([3 + 1] - 1)

    = 300ml/min = Q1


EG #2: using a 2:1 rising rate fuel pressure regulator

Given: all of the above, but with a 2:1 rrfpr

               ([3 + 1] + [0.8 * 2]) - (1 + 0.8)
 Q2 = 300 x (------------------------------------)^0.5
                         ([3 + 1] - 1)

    = 338ml/min


This still seems to be the right approach to me.  If I've committed an
egregious error anywhere, though, please let me know.

-- 
Best,

Michael McBroom

'87 745T 123k w/APC                 Visit the Volvo Performance Site:
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Graduate Student, Linguistics                         Author of 
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