Determine injector size from known physical parameters

NOTE! Incredibly bold assumptions – teee heee! HEY, you gotta start somewhere!!

 Assumptions:

 

      Highest fuel consumption will occur at peak torque

      Peak torque occurs at 3700 rev/min

      Peak fuel demand is 18 mi/gal or 6.4 Km/litre

      Differential ratio is 3.44:1

      4^th gear is 1:1

      Tyre size is 145x80x10 which is 1060 rev/mi or 660 rev/Km

      Maximum duty cycle available for balanced running is 58% of 1 revolution

 

6.4 Km	*	1000 m	*	litre	=	6.4 m	=	0.156 ml
litre		Km		1000 ml		ml		m

 

3700 engine rev	*	1 wheel rev	=	1075 wheel rev	*	Km	=	1.630 Km
min		3.44 engine rev		min		660 wheel rev		min

 

1.630 Km	*	1000 m	=	1630 m
min		Km		min

 

0.156 ml	*	1630 m	=	255 ml
m		min		min

So if the injector is running at 58% open time its total capacity will then be

255 ml	*	1	=	440 ml
min		0.58		min

 

Since those 440 cc are being fed to the engine by 2 injectors, each will have a capacity of 220 cc/min. Using those same parameters in a spreadsheet we get:

 

CHECK: The 5.7 L V8 Corvette uses eight 219-230 cc/min injectors. That’s an equivalent of a 2.8 L 4 cylinder roughly double our 1.3 L . Since we are using 2 injectors to feed the whole engine instead of 4, that’s about the same as a 1.4 litre. QED