Octane C 8 H 18 is burned with theoretical amount of air at a pressure of 500 kP

Question

Octane C8H18 is burned with theoretical amount of air at a pressure of 500 kPa. Determine:  

a.       The chemical reaction equation

b.      The air fuel ratio on a mole basis

c.       The air fuel ratio on a mass basis

d.      If the products are cooled at a constant pressure of 100 kPa, at what temperature will dew start to form?

e.       Suppose the products are cooled way below the dew point temperature so that all the water from the products is removed. Neglect any vapor which may still be present at very low amount. What would be the volume fraction of CO2in the dry products?

f.       If you have a smog test report of the gasoline burning car at hand, compare the CO2 level of your analysis with the actual smog test result and discuss your finding.

Explanation / Answer

a. C8H18 + O2 = CO2 + H2O

b. defining air as 3.76 molecules of N2 for every O2 (Since, we dont use pure air, rather oxygen)

a(C8H18) + b(O2 + 3.76 N2) c(CO2) + d(H2O) +b(3.76 N2)

balancing the equation gives,

2(C8H18) + 25(O2 + 3.76 N2) 16(CO2) + 18(H2O) +50(3.76 N2)

Now air to fuel ratio (AFR) = mair/mfuel

As we need 25 mol of air for 2 mol of C8H18, air/fuel ratio = 25/2 = 12.5 (AFR on mole basis)

c. To convert into mass basis, we need to know how many grams in each mole?

1mol C8H18 = 8(12)+18(1) kg = 114 g
1mol air = 2(16)+3.76*2*(14) = 137.28 g

Since we require 12.5 mol of air for each mol of C8H18, air/fuel ratio = 12.5(137.28)/114 = 15.05

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