Answer:
8 kmol/s
Explanation:
From the given information:
The combustion reaction equation for Octane in a stoichiometric condition can be expressed as:
[tex]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)[/tex]
[tex]C_{8}H_{18} +12.5(O_2 + 3.76N_2) \to 9H_2O +8CO_2 + 12.5(3.76 \ N_2)[/tex]
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:
[tex]C_{8}H_{18} +20(O_2 + 3.76N_2) \to 9H_2O +8CO_2 + 7.5 \ O_2+ 75.2 \ N_2[/tex] 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
