SpudFiles

Hey I can't figure out the theoretical flow rate of Isooctane fuel through a small engine. I keep getting numbers on the order of 4 L/min. It's ridiculously high and I don't know how I keep getting it. I'm basing the ratios off of the stoichiometric ratio of fuel to air. 12.5O2+C8H18->9H20+8CO2. Attached I have an excel sheet containing all of the calculations and figures I've used to get my answer. Maybe one of you can help me?

Must be using the new version of Office. I can't open the file. Can you save it in an older format such as office 97 or 2K?

Technician1002 wrote:Must be using the new version of Office. I can't open the file. Can you save it in an older format such as office 97 or 2K?

Here ya go

You are calculating it by multiplying the intake consumption with the volumetric fraction (why are you dividing it by two?!), that seems okay.
This indeed yields 4,3 (or 8,7 when not divided by two).

This is the volume of gaseous fuel that passes the engine in one minute.
I don't get why you are dividing by two so I'll work with 8,7 from now on.

Divide the volume by the molar gas constant:
8,781295781/24,12941111=0,363924993 mol
This is the number of mol of fuel that passes in one minute.

Multiply this by the molar mass:
0,363924993*114,2656=41,58410773 g
This is the mass of the amount of fuel that passes in one minute.

Density of liquid: 688 kg/m^3
Divide by 1000 and then by the density.
(41,58410773/1000)/688=6,0442E-05 m^3
This is the volume of the liquid fuel that passes in one minute.

Multiply by 10^6 to convert m^3 into cm^3:
60,44201705 cm^3 (or mL, or cc) per minute!

So it consumes about 1 cc of liquid fuel per second.
Feel free correct me if I'm wrong, but I think this is it.

psycix wrote:You are calculating it by multiplying the intake consumption with the volumetric fraction (why are you dividing it by two?!), that seems okay.
This indeed yields 4,3 (or 8,7 when not divided by two).

This is the volume of gaseous fuel that passes the engine in one minute.
I don't get why you are dividing by two so I'll work with 8,7 from now on.

Divide the volume by the molar gas constant:
8,781295781/24,12941111=0,363924993 mol
This is the number of mol of fuel that passes in one minute.

Multiply this by the molar mass:
0,363924993*114,2656=41,58410773 g
This is the mass of the amount of fuel that passes in one minute.

Density of liquid: 688 kg/m^3
Divide by 1000 and then by the density.
(41,58410773/1000)/688=6,0442E-05 m^3
This is the volume of the liquid fuel that passes in one minute.

Multiply by 10^6 to convert m^3 into cm^3:
60,44201705 cm^3 (or mL, or cc) per minute!

So it consumes about 1 cc of liquid fuel per second.
Feel free correct me if I'm wrong, but I think this is it.

You sound right. Thank you. Oh and by the way it is divided by two because it is a 4 stroke engine. This means in only takes fuel in every other down stroke