Question

A mixture of saturated hydrocarbon and N2 us burned in excess air supplied at 25°C, 740 torrs with 90% RH. An orsat analysis of the stack gas shows 7.6% CO2, 2.28% CO, 1.14% H2, 6.03% O2 and 82.95% N2. With a dew point of 53.46°C. The stack gases leave at 300°C, 765 mmHg with a volume ratio of 2.049 m3 wet stack gas/m3 wet air.

a. The formula of hydrocarbon is?

b. Mole % analysis of the fuel

c. % excess air

Answer 1

Saturates hydrocarbon and N2 is burned with air at 25°C and 740 torr

1 torr = 1 mmHg

760 mmHg = 1 atm

P = 740 mmHg = 0.9736 atm

R. H = 90%

R. H = PA/Pas

PA- partial pressure

Pas - saturation pressure

At T = 25°C saturation pressure of water from handbook = 0.0313 atm

0.90= PA/0.0313

PA =0.02817 atm

Mole fraction of water vapor in air = PA/P

= (0.02817/0.9736) = 0.02893

Normal air Contains 21% oxygen and 79% nitrogen

Mole fraction of oxygen in air = (1-0.02894) (0.21) = 0.2039

Mole fraction of nitrogen in air = (1-0.02894) (0.79) = 0.7671

The wet stack gas volume of 2.049 m3

at T = 300°C = 573 K

P = 765 mmHg

P = 765/760 = 1.0065 atm = 1.0196 ×105 Pa

n = PV/RT

n = (1.0196×105) (2.049) /(8.314×573)

n = 43.853 mol

Moles of wet stack gas = 43.853 mol

Dew point of stack gas = 53.46°C

At dew point saturation pressure = partial pressure of water vapor

At T = 53.46°C , partial pressure of water vapor =

0.144 atm

Mole fraction of water vapor =

Partial pressure /P = (0.144/1.0065) = 0.1430

Amount of water vapor present = 0.1430(43.853) = 6.2709 moles

Volume of wet air = 1 m3

T = 25°C = 298 K

P = 0.9736 atm = 0.98625×105 Pa

n = PV/RT

n = (0.98625×105) (1) /(8.314×298)

n = 39.8071 mol wet air

Mole fraction of water vapor in air = 0.02893

Amount of water vapor in air = 0.02893(39.8071) = 1.1516 moles

Amount of water vapor produced due to reaction = 6.2709-1.1516 = 5.1193 moles

The reaction occuring is

According to Stiochiometry amount of H reacted to form 5.1193 moles water vapor = 5.1193(1) = 5.1193 moles

Amount of dry gas present = total stack gas - amount of water vapor in stack

= 43.853-6.2709 = 37.5821 moles

From orsat analysis mole % of H2 = 1.14%

Moles of H2 in stack gas = 37.5821(0.0114) = 0.4284 moles

Moles of H present = 2(0.4284) =0.85687 moles

Total moles of H present = 0.85687+5.1193 = 5.97617

Since hydrocarbon is saturated only single bonds are present so

Each carbon attaches to 3 hydrogen atoms

Carbon atoms present = 5.97617/3 = 1.992 moles

The chemical formula of hydrocarbon

=

~

B)

Mol% analysis of fuel

Component	moles	mol%
C	1.992	25
H	5.976	75
Total	7.968	100

C)

Dry gas analysis

Dry gas present = 37.5821

Component	mol%	moles
N2	82.95	31.174
O2	6.03	2.266
CO	2.28	0.856
CO2	7.6	2.8562
H2	1.14	0.5219
Total	100	37.5821

The reactions occuring are

According to Stiochiometry amount of O2 needed for 2.8562 mole CO2 = 2.8562(1) = 2.8562 mole

According to Stiochiometry amount of O2

Needed for 0.856 mole CO = 0.856/2

= 0.428 mole

Total H present = 5.97619 moles

According to Stiochiometry amount of O2 needs

need 5.97619 moles = 5.97619(0.5) /2 = 1.4940 moles

Total O2 required theoretically = 1.4940+0.428+2.8562 = 4.7782 moles

Amount of O2 in stack gas = 2.266 moles

Excess % = (excess oxygen/theroetical O2)(100)

= (2.266/4.7782)= 0.47423 = 47.423%

Exces oxygen% = excess air% = 47.423%

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