I think that if we talk about IQ, we increase in percentage, the formula I think is valid for the rail pressure
I think that if we talk about IQ, we increase in percentage, the formula I think is valid for the rail pressure
I know that the increments the data from the home are not a fixed percentage, but the formula above gives us always the same percentage
why? prediamo of the case values just to make an example
For example, I have that:
PRAIL 1400bar
IQ 70mm3
Tinj 1000us
K= (√Prail * Tinj)/ IQ
K=534,42
the calculated K-factor for that curve ? 534,42
Now, to calculate the new tinj, we use the formula of mario
Suppose that you insert 70 + 10%= 77mm3 in bp
Tinj=(IQ*(K)/√Prail = (77*534,42)/sqrt1000=37352/37,41= 1100 cio? 1000+ 10%= 1100
ANOTHER EXAMPLE example I:
PRAIL 2000bar
IQ 90mm3
Tinj 1000us
K= (√Prail * Tinj)/ IQ
K=496,90
the calculated K-factor for that curve ? 496,90
Now, to calculate the new tinj, we use the formula of mario
Suppose that you insert 90 + 20%= 108mm3 in bp
Tinj=(IQ*(K)/√Prail = (108*496,90)/sqrt2000=53665,62/44,72= 1200 to cio? 1000+ 20%= 1200
If you then changed the rail pressure then the thing remains the same, but with rail unchanged, the formula gives you a result that is always the same percentage that increased Your IQ
Check a little bit with different values? But I think I have reason.
Hello to all
If you increase the rail pressure then, but apressione rail invarita not need
If you can be useful in annex I place a tiny little program to calculate the time of injection, I don't know if when andrette to ziparlo there will still be the formulas. Let me know?
Hello to all
Give us a look and let me know, why? if all goes well as soon as I have time I improved it better.
Hello
Last edited by msport (exil77grande); 24-02-2014 to 13:43