Custom EFI Project

[Martin Evans]

Ric Rainbolt wrote:

>I will be using eight Bosch fuel injectors (19lb/hr @ 3 bar).  Fuel >pressure
will be 3-6 bar, controlled by the EFI.

Do you know the flow rate at 3.6bar? do you anticipate requiring fuel in excess
of 19lb/hr?

My initial thoughts follow. Feel free to criticise as i'm learning too!

If you are trying to implement a crank and idle strategy for yor engine I think
it will be necessary to look at TPS (to start with a switch contact indicating
closed throttle will do), TDC and RPM. 

SYSV will have an impact but you will need to evaluate the change in your
specific injectors flow rate with varying voltage.

MAP should be used to compensate for atmospheric pressure before start
commences. If you don't anticipate operating at altitude it makes life easier to
forget it.

Engine coolant temperature should, I think be pulled forwards from stage 4 of
your implementation as the mixture requirements do vary wildly with temperature.
My own vehicle indicates a change from approximately 0.8:1 at -40 deg C to
16.8:1 at 104 deg C. At 0 deg C the ratio is approximately 9.2:1.

Initially ignoring the O2 sensor would make sense as it would not give a useful
output  until the sensor gets warm. Are you intending to close loop control with
a 3 way catalyst eventually?

Calculation time!

For basic fuelling lambda will need to be controlled around 1.0 which implies
14.7:1 air to fuel ratio. 
Assuming 100% volumetric efficiency:
Calculate for one cylinder
Each cylinder measures 0.375 litre
In one cycle it will consume 0.375 litre of air
One cycle is 2 engine revolutions
800 revolutions per minute is 400 cycles
400 cycles is 400 x 0.375 litre = 150 litres/minute
The density of air at 15 deg C is 1.225kg / m-3
1 m3 is 1000 litres therefore 150 litres is:
150 x 1.225/1000 = 0.18375 kg/minute
For 14.7:1 air to fuel ratio the injector would need to supply per hour:
0.18375kg x 60min/hr /14.7 = 0.75kg
The injectors flow 19lb/hr = 8.6kg/hr
0.75kg assuming linear flow/time characteristic is:
0.75/8.6kg is 8.7% on time

The injector on time will therefore need controlling to give:

at 0 deg C 14.7/9.2 x 8.7% = 14% on time
at 104 deg C 14.7/16.8 x 8.7% = 7.6% on time

Obviously we cannot expect 100% volumetric efficiency so mixtures will tend to
be slightly richer than predicted so shorter times would be required.

Martin Evans

100341.377 at compuserve.com