Problem 4.95 Combustion of Propane and Butane Mixture A mixture of propane and b
ID: 701361 • Letter: P
Question
Problem 4.95 Combustion of Propane and Butane Mixture A mixture of propane and butane is burned with pure oxygen. The combustion products contain 47.6 mole% H2O. After all the water is removed from the products, the residual gas contains 68.6 mole% CO2 and the balance 02 a. What is the mole percent of propane in the fuel? b. It now turns out that the fuel mixture may contain not only propane and butane but also other hydrocarbons. The fuel does not contain oxygen. However, the dry combustion gases still contain 68.6% carbon dioxide we wish to determine the elementa composition carbon and hydrogen molar percentage ofte i (Hint: Calculate the elemental compositions on an oxygen free basis), what is the mole% of carbon in the fuel? Click if you would like to Show Work for this question: Open Show WorkExplanation / Answer
Basis - 100 moles of residual gas (wet basis)
Dry basis
Mol% of CO2 = 68.6%
Mol% of O2 = 100-68.6 = 31.4%
Moles H2O = 47.6 mol
Moles of dry gas = 100 - 47.6 = 52.4 moles
Mol fraction of CO2 = 0.686 x 0.524 = 0.359
Mol fraction of O2 = 0.524- 0.359 = 0.165
Now do the atomic balance of O
Total O2 in = O2 (in H2O) out + O2 (in CO2) out + O2 out
= 100*0.476 + 100*2*0.359 + 100*2*0.165
= 76.2
C balance
3 x C in C3H8 + 4 x C in C4H10 = 35.9
3n1 + 4n2 = 35.9
0.75 n1 + n2 = 8.975
H balance
8n1 + 10 n2 = 95.2
0.8 n1 + n2 = 9.52
Solve these above equations simultaneously
0.05 n1 = 0.545
n1 = 0.545/0.05 = 10.9 mol C3H8
n2 = 9.52 - 0.8*10.9 = 0.8
Mol% C3H8 = 10.9/(10.9+0.8)
= 0.9316
Part b
In the product gas
Mol of C = 100 mol x (0.359 mol CO2/mol) x (1 mol C/mol CO2)
= 35.9 mol C
Mol of H = 100 mol x (0.476 mol H2O/mol) x 2 mol H/mol H2O)
= 95.2 mol H
Mol% C = 35.9*100/(35.9 + 95.2) = 27.38%