La venta termina hoyObtenga un 30% de descuento en cualquier curso (excepto paquetes)
Termina en --- --- ---
Publicaciones del foro
Cursos
Blog
Artículos técnicos
Hello friends ,
I have a 4-cylinder diesel engine, i would like to know if the formula described in the courses applies to the 4-cylinder engine?
For example :
1.7ms duration time injector
120 / 1600= 75ms 720 /75 = 9.6. 9.6 /1.7 = 16.3
50% before 50% after = 16.3x0.5=8.5
Many thanks
Yes, you're correct. the formula works for any 4 stroke diesel engine irrespective of number of cylinders.
Big thanks André !!! 😁💪
On the mathematics you did above, I have a confusion.
you have 9.6 degrees and 1.7ms
Here you divide 9.6 /1.7 = 16.3, but the result is multiplication, I mean 9.6/1.7 = 5.647 (division) / multiplication is 9.6*1.7 = 16.3
The result is ms or degrees before and after TDC??
For example :
1.7ms duration time injector
120 / 1600= 75ms 720 /75 = 9.6. 9.6 /1.7 = 16.3
50% before 50% after = 16.3x0.5=8.5
Many thanks
The 9.6 is degrees per MS.
His example is at 1600rpm. So to change that it rev's a second would be /60, but to make that cycle time (two revolutions on a 4 stroke) you can divide by 120 which equals 75
as there are 720 degrees in a cycle, we then divide this by the time, so 720/75 is 9.6degrees per ms.
9.6 times 1.7ms means the injection period is 16.3 degrees (yes, it should have been multiplication instead of division)
He has then gone on a 50% burn after TDC it has been split, for an injection timing of 8.5 degrees BTDC
