Question

A mixture of propane and butane is burned with air. Partial analysis of the stack gas produces the following dry-basis volume percentages: 0.0527% C3H8, 0.0527% C4H10, 1.48% CO, and 7.12% CO2. The stack gas is at an absolute pressure of 780 mmHg and the dew point of the gas is 46.5 ˚C. Calculate the molar composition of the fuel.

Answer 1

Pressure = 780 mmHg

Dew point temperature = 46.5 oC

Both the gases burned with air as follows:

C3H8 + 5 O2       3 CO2    +     4 H2O

C4H10   + 13/2 O2       4 CO2    +     5 H2O

At dew point (46.5 oC):

y_w P = P^ow          (at 46.5 oC, P^ow = 77.6 mmHg)

y_w * 780 = 77.6

y_w = 0.0995 mol H2O / mol

y_w = n_w / (100 + n_w)

0.0995 = n_w / (100 + n_w)

n_w = 11.0 mol H2O

Balancing carbons:

3 np + 4 nb = 100 [0.000527 * 3 + 0.000527 * 4 + 0.0148 + 0.712]

3 np + 4 nb = 100 [0.08969]

3 np + 4 nb = 8.969           ...........(1)

Balancing Hydrogen:

8 np + 10 nb = 100 [0.000527 * 8 + 0.000527 * 10] + 11.0 * 2

8 np + 10 nb = 23.149           ...........(2)

Solving (1) and (2):

np = 1.454 mol C3H8

nB = 1.152 mol C4H10

Hence, the fuel contains:

C3H8 = 56%

C4H10 = 44%