A mixture of propylene and propane is burned with excess oxygen in a furnace. Th

Question

A mixture of propylene and propane is burned with excess oxygen in a furnace. The combustion products contain 46% H_2O. After all the water is removed, the gas stream contains 65% CO_2 by mole and the rest O_2. Use an elemental balance to determine the ratio of propane to propylene (and/or the fraction of each) in the feed and the percent excess O_2. A fuel stream containing 70 mole% butane and 30 mole% hydrogen are is burned with 25% excess air. Fractional conversions of 85% of the butane and 90% of the hydrogen are achieved. Of the butane that reacts, 95% of it goes to form CO_2 and the balance to form CO. Calculate the composition of the effluent stream.

Explanation / Answer

First calculate the % of CO2 as follows:

(100-46H2O)% = 54%= 65% CO2 +35% O2

54 *65/100= CO2 %= 35.1%

Assuming that all of th CO2 and H2O in the stack gases derives from the fuel, then 100 mol of stack gas contains 35.1 mol of C atoms and 2x46 =92 mol of H atoms.

The ratio of H to C in the stack gases is 92/35.1=2.6

Suppose 1 mol of fuel mixture contains n mol of propane C3H8 and 1-n mol of C3H6. The mole amounts of C and H present in 1 mol of fuel are following:

3n + (1-n)3 = 3n+3-3n= 3

8n +(1-n)6=8n +6-6n =2n+6

The ratio of H to C should be the same in the fuel mixture as in the stack gases, ie

2n+6 / 3= 2.6

2n +6= 7.8

2n=1.8

n= 0.9

Therefore the fuel mixture is 0.9 mol% propane and 99.1% propylene.

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