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

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 (C_{3}H_{8}) and 30% butane (C_{4}H_{10}) 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:

C_{3}H_{8} + O_{2} = CO_{2} + H_{2}O

C_{4}H_{10} + O_{2} = CO_{2} + H_{2}O

**Solution:**

4.1. The combustion reactions for propane and butane are:

C_{3}H_{8} + 5 O_{2} + 18.8 N_{2} = 3CO_{2 }+ 4H_{2}O + 18.8 N_{2}

C_{4}H_{10} + 6.5 O_{2} + 24.5 N_{2 }= 4CO_{2} + 5 H_{2}O + 24.5 N_{2}

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 CO_{2}

(0.7× 18.8) + (0.3×24.5) = 20.5 vols N_{2}

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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