Answer to Question #93137 in Chemical Engineering for Nduduzo

Question #93137
A gas consists of 70% propane (C3H8) and 30% butane (C4H10) by volume. Calculate: 4.1. The stoichiometric amount of air required to burn 100 kmol/hr of feed if air is assumed to be 21% oxygen. 4.2. The percentage excess air present if a dry analysis of the combustion products shows 9% CO2 assuming complete combustion of propane and butane (hint - calculate the nitrogen that is left over from the reached oxygen first and assume the remainder of the exit stream is air). Reactions: C3H8 + O2 = CO2 + H2O C4H10 + O2 = CO2 + H2O
Expert's answer

A gas consists of 70% propane (C3H8) and 30% butane (C4H10) by volume. Calculate:

4.1. The stoichiometric amount of air required to burn 100 kmol/hr of feed if air is assumed to be 21% oxygen.

4.2. The percentage excess air present if a dry analysis of the combustion products shows 9% CO2 assuming complete combustion of propane and butane (hint - calculate the nitrogen that is left over from the reached oxygen first and assume the remainder of the exit stream is air).

Reactions:

C3H8 + O2 = CO2 + H2O

C4H10 + O2 = CO2 + H2O

Solution:

4.1. The combustion reactions for propane and butane are:

C3H8 + 5 O2 + 18.8 N2 = 3CO2 + 4H2O + 18.8 N2

C4H10 + 6.5 O2 + 24.5 N2 = 4CO2 + 5 H2O + 24.5 N2

toichiometric Air Requirement

On the basis of 1 volume of the fuel gas,

the propane content requires 0.7×(5 + 18.8) = 16.7 vols air

and the butane requires 0.3×(6.5 + 24.5) = 6.3 vols air

Hence the stoichiometric air-to-fuel ratio is 23:1.

4.2. Excess Air

The combustion products (dry) will contain

(0.7×3) + (0.3×4) = 3.3 vols CO2

 (0.7× 18.8) + (0.3×24.5) = 20.5 vols N2

plus υvolumes excess air, giving a total volume of products of(23.8 + υ ).

Given that the measured CO2 in the products is 9%, we can write:

9/100 = 3.3/(23.8 + υ )

hence υ = 12.87 vols

The stoichiometric air requirement is 23 vols so the percentage excess air is

12.87/23*100% = 55.9%

Answer:

4.1. 23:1

4.2. 55.9%

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