Sizing Injectors

[Bruce Bowling]

There have beeen some questions about determining fuel
injector flow sizing.  The following is one way to determine
size based on knowing the engines volumetric efficiency
at peak torque (WOT).  This can be obtained from a dyno
run, or one can use 75 to 80% for street engine, 90-95%
for race motors, and above 100% for turbocharged engines.

The following determines the time required to hold the
injector open (in milliseconds) per "squirt" event.  Using
this info should help in determining injector size, and
help determine if sequential or batch firing is more suited
for an application.

To start, one must compute how much air is "sucked" into the
engine per completion of a "cycle" (two revolutions for a 4-cycle).
This is determined by using the volumetric efficiency and the
weight of air at standard pressure and temperature (60 deg F, 
29.92" Hg, dry air):

W(lbs air) = CID * VE * 4.428 x 10-5 (Lbs/CID)          (1)
where CID = Cubic Inch Displacement
      VE = Volumetric Efficiency

Now find the pounds of fuel per cylinder at some air/fuel ratio:

S(Lbs fuel/cyclinder) = W * (1/cyl) * (1/AF)            (2)
where cyl = number of cylinders
      AF = desired air/fuel ratio (like 14.7)

Modify S if more than one injector "squirt" per engine
cycle (ie. batch injection fires 2 times, sequential
fires only once)

S = S/(no of squirts)                                   (3)

Now find the injector flow in pounds/millisecond:

F(lbs/millisecond) = Injector(lbs/hr) * (1/3,600,000)   (4)
where Injector is the rated static flow

Finally, the injector fire time in milliseconds:

TIME(ms) = S/F                                          (5)


For example, a Chevy 350 with 0.8 VE wanting a AF of 13.0,
running true sequential using 30 Lb/hr injectors, works as:

W = 350 * 0.8 * 4.428 x 10-5  =  1.2398 x 10-2
S = W * (1/8) * (1/13.0) = 1.1921 x 10-4 
F = 30 * (1/3600000) = 8.333 x 10-6
TIME = 1.1921 x 10-4 / 8.333 x 10-6 = 14.3 milliseconds

One thing to watch out for is maximum squirt time, which is
RPM dependent.  If one assumes that the injector can be on
for one complete engine revolution (like TDC exhaust to TDC
power), one can compute how long the injector is open (ms)

1 rev/minute = 1.666 x 10-5 rev/milliseconds, 
or  1 minute/rev = 60,000 milliseconds/rev

So time(ms) = 60,000/RPM
 
If the Chevy has a maximum RPM of 7000, this represents
a time of (60,000/7000) = 8.6 milliseconds per rev.
If the injector is on all the time (static, and not
a condition), the time for a "cycle" is 2 * 8.6 = 17.2.
The number computed above (TIME = 14.3 ms) is very close
to the injector being on all the time, so watch out
when using the above.



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               Bruce A. Bowling
  Staff Scientist - Instrumentation and Controls
 The Continuous Electron Beam Accelerator Facility
    12000 Jefferson Ave - Newport News, VA 23602
                 (804) 249-7240
               bowling at cebaf.gov
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