# fuel map discussion

Ciciora Steve sciciora at aztec.al.bldrdoc.gov
Thu Jul 7 21:28:58 GMT 1994

```Below is my first draft of a discussion of how I think fuel injection systems
work.  I faithfully submit it for peer review...
_______________________________________________________________________________
To: Ciciora Steve
From: CICIORA STEVEN JOSEP on Thu, 7 Jul 1994 3:26 PM
Subject: fuel map

Lets start off by assuming a few things before we get started.  For this
example, let's assume we have a 4 cyl 4 liter engine and we want an air/fuel
ratio of 14.7/1.  This means 14.7 grams of air for each gram of fuel. Also for
now, lets assume this engine is operating at an RPM that gives us 100%
volumetric effency.  Since our engine is 1 liter per cylinder, in order to
determine how much fuel to deliver to each cylinder, we need to know the mass
of the air in each cylinder.  Remember back in high school physics, PV=nRT?
P is the pressure in atmosphers, V is the volume in Liters, n is the number
of mols of air, R is the universal gas constant (8.206x10E-2 Liters *
atmosphers/mols*K), and T is the temperature in degrees Kelvin.  A little bit
of manipulation will give us P/(R*T)=n/V, where n/V is the density of air in
mols per liter.  This isn't very useful, so to convert mols to grams, we
multiply bolth sides of the equation by the molecular weight of air.  I know
that somewhere there must be a 'standard' average value for air, but I
couldn't find it.  I calculated the molecular weight of dry air to be
28.96475143.  This should be close.  So anyway, the result is:
density of air in grams per liter = (mw * P)/(R*T) where mw is the Molecular
Weight of air, P is the pressure of air in Atmospheres, R is the gas constant
and T is the Temperature in degrees Kelvin.  For example, let's plug in 1
atmosphere and 300 degrees Kelvin (24 degrees C, about room temperature).  At
this temperature and pressure, the density of air is 1.1765 grams per liter.
Dividing by the desired air/fuel ratio (14.7) tells us that we want 0.08019
grams of fuel for each liter of engine per cycle.  Using a lookup table
(generated using a fuel injector flow bench like the one described in
Performance Engineering Magazine) we would determine the desired amount of
time to hold the fuel injector open.
So by measuring the intake manifold pressure and the intake air temperature
(hopefully at the same spot that the pressure was measured) we can calculate
the amount of fuel needed for a given a/f ratio.  But wait!  This assumes a
constant 100% volumetric effency.  We all know that most of the time it is
less than 100%, and some times greater.  A good estimate is to assume
volumetric effency changes with engine RPM and to correct the mass of air
calculation with an RPM factor.  This is usually done with a 3D Fuel Map.  On
the X axis we would have the mass of air, the Y axis we would have the engine
RPM, and on the Z axis we output the amount of fuel needed.  This can be in
grams per cycle or if we include the fuel injector correction factor in the
fuel map, it can output pulse width.

(ranges of values for map/temp)
(corrections for cold start, acc, power mode, emmissions mode, fuel econ, etc)
(what do you think?)

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