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Old Feb 6th 2014, 01:59 AM   #1
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Internal Combustion Engine (Injectors)

I need a solution to this problem.

An automobile has a 3.2-litter, five-cylinder, four-stroke cycle diesel engine [8]
operating at 2400 RPM. Fuel injection occurs from 20 bTDC to 5 aTDC. The
engine has a volumetric efficiency of 0.95 and operates with fuel equivalence ratio of
0.80. Light diesel fuel is' used. Calculate:
Time for one injection,
Fuel flow rate through an injector.



If there are any assumptions made please let me know.
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Old Feb 7th 2014, 09:19 AM   #2
MBW
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The first part is relatively easy:

20 bTDC to 5 aTDC indicates fuel injection occurs over 25 of the cycle
i.e 25/360 of one cycle of the cylinder
2400 rpm = 40 cycles per second = 0.025 seconds per cycle
from this the time for one injection is readily determined.

The second part requires an understanding of the terms:
volumetric efficiency and fuel equivalence ratio
which I don't have.
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Old Feb 7th 2014, 10:28 AM   #3
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Thank You! I have figured them out right before 1 hour :P

I thought it was more complicated than that!

The second part:
volumetric efficiency = ( mass of actual air) / (mass of ideal air)

equivalence ratio = [(mass of fuel)/( mass of air)/ (mass of fuel)/( mass of air) in stoichiometric proportions]
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