Question

1. Approximately 250,000 kg of coal are burned each day at the Iowa State University Power Plant. The coal is barged up from Kentucky to Davenport and trucked to Ames in semi=s and is a mixture of Kentucky, Illinois and Colorado coal with an average sulfur content of 2.55%. What would be the average daily output of sulfur dioxide (SO2) assuming that 5% of the sulfur content of the coal ends up in the ash (i.e. 5% of S is unreacted) and the rest is released in the stack gas?

2. Assuming that the remainder of the coal is essentially carbon (i.e., the coal contains only carbon and sulfur), how much oxygen would be consumed per day in the reactions with coal? Express your results in both mass (kg/d) and volume (m3 /d) units.

3. As a rough approximation, the average content of dry air is about 21% by volume. If 10% excess air is used in the combustion process, how much air will be required each day?

4. What would be the concentration of SO2 in the stack gas? Calculate your result on both a mass (g/ m3) and volume (ppm) basis.

5. What would be the concentration of CO2 in the stack gas? Calculate your result on both a mass (g/ m3 ) and volume (ppm) basis.

6. Using an emission factor of 8.3x10-5 lb mercury/ton coal burned, what would be the annual mercury emission from the facility in lb/y and kg/y. (1 ton = 2000 lb)

Answer 1

Ans. #1. Total mass of sulfur in coal (per day basis) = 2.55 % of 250000 kg coal

                                                                                    = 6375.00 kg

% of S emitted in stack gas = 100 % - % of S ending up in ash = 100 % - 5% = 95 %

Mass of S in stack gas = 95 % of total S in coal

                                    = 95 % of 6375.00 kg

                                    = 6056.25 kg

SO2 (molar mass =64.0648 g/mol) gas consists of 1 molecule of S-atom (atomic mass = 32.066 g/mol). So, 32.066 g sulfur is equivalent to 64.0648 g SO₂ gas.

Or,

            32.066 kg S is equivalent to 64.0648 kg SO₂

            Or, 1.0 kg S    -           -           (64.0648/ 32.066) kg SO₂

            Or, 6056.25 kg S       -           (64.0648/ 32.066) x 6056.25 kg SO₂

                                                            = 12099.81 kg

Therefore, mass of SO2 gas released /day = 12099.81 kg

#2. Mass of Carbon in coal = Total mass – Mass of S = 250000 kg – 6375.00 kg

= 243625.00 kg

Balanced reaction:    C + O₂ ------> CO₂

Stoichiometry: 1 mol C ( = 12.0 g) consumes 1 mol O₂ (= 32.0 g) for complete combustion.

Or,

                  12.0 g C consumes 32.0 g O₂

            Or, 12.0 kg C consumes 32.0 kg O₂

            Or, 1.0 kg C    -           -    (32.0 / 12.0) kg O₂

            Or, 243625.00 kg      -    (32.0 / 12.0) x (32.0 / 12.0) kg O₂

                                                            = 649666.666667 kg         

Thus, mass of O2 required = 649666.67 kg/day = 649666666.67 g /day

= 20302083.33 mol /day

Assuming ideal and STP conditions (because temperature in not specified), 1 mol of O2 gas occupies a volume of 22.414 L.

So, volume of O2 required = Ideal molar volume at STP x Moles of O2 consumed

                                                = (22.414 L/mol) x 20302083.33 mol

                                                = 455050895.83 L

                                                = 455050.9 m³                                 ; [1000 L = 1 m³]

Therefore, volume of O2 required = 455050.89 m³/day

#3. Total volume of O2 supplied = Actual O2 consumed + 10 % excess

                                                = 455050.9 m³ + 10% of 455050.9 m³

                                                = 500556.0 m³

Let the total volume of air required/day = X m³

Now,

            21 % of X m³ = 500556.0 m³

            Or, 0.21 X = 500556.0

            Or, X = 2383600.0

Therefore, volume of air required = 2383600.0 m³/day

#4. Part A. 1 mol C (12.0 g) produces 1 mol CO2 (44.0 g)

Total mass of CO2 produced = (44.0 / 12.0) x 243625.00 kg = 893291.67 kg

Total mass of stack gas = Mass of CO2 + Mass of SO2

                                                = 893291.67 kg + 12099.81 kg

                                                = 905391.47 kg

Now,

            [SO₂] ppm = mg of SO₂ / Mass of stack gas in Kg

                                    = (12099.81 x 10⁶) mg SO₂ / 905391.47 kg stack gas

                                    = 133641.73 mg/ kg

                                    = 133641.73 ppm

Part B: Volume of CO2 emitted at STP = Molar volume x (Mass of CO2 in g / Molar mass)

                                                            = (22.414 L/mol) x [893291670 g / 44.0 g mol^-1]

                                                            = 455050896 L

                                                            = 455050.896 m³

Volume of SO₂ emitted at STP = Molar volume x (Mass of SO₂ in g / Molar mass)

                                                            = (22.414 L/mol) x [12099810 g / 64.0 g mol^-1]

                                                            = 4237579.7 L

                                                            = 4237.579 m³

Total volume of stack gas = Volume of OC2 + Volume of SO2

                                                = 455050.896 m³ + 4237.579 m³

                                                = 459288.48 m³

Now,

            [SO2] = Mass of SO2 in gram / Volume of stack gas in m³      

                        = 12099810 g / 459288.48 m³

                        = 26.34 g/ m³

#5. [CO₂] ppm = mg of CO₂ / Mass of stack gas in Kg

                                    = (893291.67 x 10⁶) mg CO₂ / 905391.47 kg stack gas

                                    = 9866358.48 mg/ kg

                                    = 9866358.48 ppm

Now,

            [CO2] = Mass of CO2 in gram / Volume of stack gas in m³     

                        = 893291670 g / 459288.48 m³

                        = 194495 g/ m³

#6. 2000 lb = 907.1847 kg                                      ; [1 lb = 0.453592 kg]

            1 ton = 2000 lb

            So, 1 ton = 907.1847 kg

Mass of Coal used per day = 250000 kg / (907.1847 kg / ton) = 275.58 ton / day

Mass of coal consumed in 1 yr = (275.58 ton / day) x (365 days/ year)

                                                = 100585.91 ton/ yr

Emission factor of Hg = 8.3 x 10^-5 lb Hg/ ton coal

Now,

            Annual emission of mercury = (100585.91 ton coal / yr) x (8.3 x 10^-5 lb Hg/ ton coal)

                                                            = 8.35 lb Hg/ yr

                                                            = 3.79 kg Hg/yr                  ; [1 lb = 0.453592 kg]

                                                            = 0.0092 ton Hg/yr            ; [1 ton = 907.1847 kg]