Question

A 5,000 mol/h fuel is fed to a furnace and burned with excess air. The fuel contains 80.0 mole % carbon (C), 15.0 mole % hydrogen (H) and the remainder sulphur (S). The ratio of feed fuel to feed air is 1:7. The selectivity of carbon dioxide (CO2) to carbon monoxide (CO) production is 12. The conversion of fuel in this process is only 80%. One of the components in the flue gas is sulphur dioxide (SO2).

(a) Write all equations that involved in this combustion process.
(b) Draw and label the process completely.
(c) Using Atomic Species Balance method, determine the molar flow rates (mol/h) of all components in the product. stream.                                                                             
(d) Determine the percentage of excess air in this process.

Answer 1

Fuel flowrate = 5000 mol/h

The fuel composition

80 mol% C

15 mol% H

5 mol% S

The ratio of fuel fed to air = 1: 7

Molar flowrate of air = 7(5000) = 35000 mol/h

A)

The equations involving are

B)

C)

The conversion of fuel is only 80%

Fuel reacted = 5000(0.80) =4000 mol/h

Unreacted fuel = 1000 mol/h

Selectivity of CO₂ to CO = 12

Amount of C reacted = 5000(0.80) (0.80) =3200 mol/h

Moles of C undergoing reaction 1 = 3200(12/13) = 2953.846 mol/h

Moles of C undergoing reaction 2 = 3200-2953.846 = 246.153 mol/h

Moles of H reacted = 5000(0.15) (0.80)

= 600 mol/h

Moles of S reacted = 5000(0.05) (0.80)

= 200 mol/h

According to stiochiometry O₂ reacted = 2953.846(1) + 246.153(0.5) + 600(0.25) + 200(1) = 3426.9225 mol/h

Air contains 79% N2 and 21 % O2

O₂ in feed = 35000(0.21) = 7350 mol/h

Product analysis based on stiochiometry

Component	mol/h	mol%
Unreacted fuel	1000	2.83
CO2	2953.846(1)= 2953.846	8.37
CO	256.153	0.725
H2O	300	0.8502
SO2	200	0.566
N2	35000(0.79) = 27650	78.366
O2	7350- 3426.9225= 3923.0775	11.11
Total	35283.076	100

D)

% excess air = % excess O2

O2 needed according to stiochiometry if all reactants reacted = (5000×0.8×(12/13) ×1)+ (5000×0.8×(1/13) ×0.5) + (5000×0.15×0.25) +(5000×0.05×1) = 4283.653 mol/h

O2 supplied = 7350 mol/h

% excess =( (7350-4283.653) /4283.653) (100) =

71.582 %

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