Question

A mixture of octane, C8H18, and air flowing into a combustor has 60% excess air and 1 kmol/s of octane. What is the mole flow rate (kmol/s) of CO2 in the product stream?

Answer 1

Answer:

8 kmol/s

Explanation:

From the given information:

The combustion reaction equation for Octane in a stoichiometric condition can be expressed as:

$C_{8}H_{18} +12.5(O_2 + \dfrac{79}{21} N_2) \to 9H_2O +8CO_2 + 12.5(\dfrac{79}{21}N_2)$

$C_{8}H_{18} +12.5(O_2 + 3.76N_2) \to 9H_2O +8CO_2 + 12.5(3.76 \ N_2)$

In the combustor, it is said that 60% of excess air and 1 mole of Octane is present.

Thus;

the air supplied = 1.6  × 12.5 = 20

The equation can now be re-written as:

$C_{8}H_{18} +20(O_2 + 3.76N_2) \to 9H_2O +8CO_2 + 7.5 \ O_2+ 75.2 \ N_2$ because for 1 mole of Octane, 8 moles of CO2 can be found in the combustion product.

Thus, for 1 kmol/s of Octane also produce 8 kmol/s of CO2.

∴

The mole flow rate in Kmol/s of CO2 in the product stream = 8 kmol/s