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I've got a 2.5 L engine here for which I do not know the performance specs (yet), because it's customized.

I need to select injectors for this build, but there seems to be only one equation on the interweb to get the injector size, and this takes the engine's power in account, which I don't have.

My idea was to calculate the maximum mass airflow that the engine will displace, and then calculate how much fuel gets mixed with that at the richest AFR I will ever maintain (12.05). Then dividing by max. duty cycle should give me the injector size. I took 100% VE at WOT.

This is the result:

Specific air mass:  1.27 kg/m3  
Specific fuel mass: 0.75 kg/L
Volume air flow:    6000rpm * 2.5L * 1/2
                  = 7500 L/min = 7.5 m3/min
                  = 0.125 m3/sec (because 4 stroke)  
Mass air flow:      0.125 * 1.27 = 0.16 kg/sec  
Mass fuel flow:     0.16 / 12.05 = 0.013 kg/sec  
Volume fuelflow:    0.013 / 0.75 = 0.018 L/sec = 1062 cc/min  
Duty-Cycle 0.8:     1062 / 0.8 = 1328 cc/min

This looks a bit too high to me... VE and AFR could be a bit lower but it'd still be too large.

Is there any other method to determine injector size?

Cheers

Zaid
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Bart
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2 Answers2

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Assuming this is for a naturally-aspirated application, your calculations are reasonable.

I think you just missed dividing the value obtained by the number of cylinders.

Usually 2.5 L engines have 4 cylinders and (subsequently) 4 injectors.

So

1328 cc/min / 4 = 332 cc/min

You would select the next biggest injector size available (although 330 cc/min injectors would work just fine here)

Zaid
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2

Looks good except the Number of Cylinders (or more specifically, injectors) part.

In the case of a EJ257 Subaru motor (ironically 2.5 liters, 4 cylinders) a set of Deatschwerks 750cc units will get you well into 500+ horsepower with room to spare (90% max IDC).

And keep in mind this is a turbo setup, with probably a large turbo, and a VE over 100%.

I ilke your math and perservation of units. One thing that is very educational about this kind of exercise is to truly realize what IDC "Injector Duty Cycle" truly means. In your example, a very conservative max IDC of .8 is used. That means the injector is injecting 80% of the time. Of ALL time.

So? There's a common misconception that injectors only inject when the intake valve is open.

For "fun" (yes, I realize I will never date a girl or reproduce... count your blessings) take a nice streetable intake cam with a 270 degree duration, and determine the timing window you can inject in with intake valve open (ok to assume 270) and what flow you would need to accomplish this at say, 7500rpm. Remember it's still an Otto cycle.

Wasn't that fun?? [cough]

Injectors on performance engines, especially turbo cars, are on almost continuously at high loads. Doesn't seem intuitive to me, but it's the truth. With the right tuning and intake airflow, the fuel doesn't even condense out on cold valves like it used to with CIS and bank injection systems. The details of the flow and aerosol physics invovled at that level are far beyond my comprehension.

On edit:

It appears my ramblings may have lead the OP astray. The screenshot below is for a turbo car with a target desired 500hp at flywheel. It also describes the most extreme WOT condition. However, the link is useful as it does the math the OP is already quite adept at, in an easy plug-and-play. Note that "Normally Aspirated" (non-turbo) is a button selection. I picked what was appropriate for a Subaru EJ257 (which I know and love) but I did not mean to imply the screenshot was an answer to the original question. By all means, knowing the underlying math is a far greater skill than depending on an online calculator.

enter image description here

SteveRacer
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