Question

3. A six cylinder, four-stroke cycle SI engine with multipoint fuel injection has a displacement of 2.4 liters (0.0024 m). The engine operates on a hydrocarbon fuel with H/C = 1.9 and MW =105 kg/kmol, at an equivalence ratio of 1.05. Each cylinder has one fuel injector fired once per cycle. The flow rate for the injector when continuously open is 0.003 kg/s. Consider the wide-open throttle, peak power operating point where the design volumetric efficiency is 90% and the speed is 5000 rpm. Assume the volumetric efficiency is based on an inlet air density of 1.18 kg/m a) Calculate the pulse width required for the injector to meeting the requirements of this operating point. (ms) b) Would the flow rate of this injector (i.e. 0.003 kg/s) be acceptable for this engine at this given operating point? Why or why not?

Answer 1

dispalcement for 1 cylinder = 0.0024/6 = 0.0004 m3

mass flow rate = 0.003 kg/sec

vol. eff. = 0.9

speed = 5000 rpm

density of air = 1.18 kg/m3

a) no. of cycles for 2 stroke engine = 5000/2 = 2500 rpm

injector is open in each cycle, so in one sec it open for 2500/60 = 42 times.

so, duration of opening of injector for one time is 1/42 = 0.0238 s

pulse width is 23.8 ms

b) mass of air sucked during injector opening = mass flow rate* pulse width

= 0.003 * 0.023

= 0.0000714 kg

vol. of this mass = mass sucked/density of air

= 0.0000714/1.18

= 0.0000605 m3

volumetric efficiency with this mass = 0.0000605/ vol. of cylinder

= 0.0000605/0.0004 = 0.151 = 15.1%

this efficiency is less than the required eficiency, so given flow rate is not acceptable.