Question

Calculate air to fuel ratio by mass, for a four-cylinder SI engine running on i) gasoline and ii) a mixture of 20% methanol and 80% gasoline both at stoichiometric conditions. Assume gasoline is represented with C8H18 . Calculate any loss of power when gasoline-methanol mixture is used and discuss your results.

Answer 1

a)

gasoline is typically octane = C8H18

C8H18 + O2 = CO2 + H2O

balance

C8H18 + 25/2O2 = 8CO2 + 9H2O

so... for 1 mol of octane, we need 25/2 mol fo O2

mol of air = 25/2 / 0.21 = 59.523 mol of air

ratio in mass

mass of octane ( 1mol ) = 1 * 114.23 = 114.23 g of octane

mass of air = 59.52 mol of air = 59.52*29 = 1726.08 g of air

so..

ratio

1726.08/114.23 = 15.110 ratio

b)

for

20% methanol and 80% gasoline

assume this is per mass

so..

20 g of methanol. 80 g of octane

mol of methanol = mass/MW = 20/32 = 0.625

mol of octane = mass/MW = 80/114 = 0.7017

ratio of methanol:

2CH3OH(l) + 3O2(g) arrow 2CO2(g) + 4H2O(l)

so ratio is

3/2 mol of O2 per 1 mol of methanol --> 1.5/0.21 = 7.142 mol of air required

so...

0.625 mol of methanol --> 0.625 *7.142 = 4.46375 mol of air for methanol

0.7017 mol of gasoline --> 0.7017*59.523 = 41.767 mol of air for gasoline

total = 4.46375 +41.767 = 45.46375 mol of air

mass of air = mol*MW = 45.46375*29 = 1318.448 g of air

g of mix = 100 g

ratio

1318.448/100 = 13.18

15.110 ratio vs. 13.18

clearly, we require less air for the 20/80 mix