#2: Control of SOx.

(a) With 2% S in the coal and assuming a 90% capacity utilization factor, the (maximum) SO₂ emissions are 124,000 tons per year.

In NSPS units this is 3.3 lb SO₂/10⁶ BTU.

Therefore, the required SO₂ removal efficiency is 82%.

(b) From the relevant handout and/or report by the SOx group, K > 1 (deltaG_rxn < 0) at T~1200 K. Above 1200 K, conversion of SO₂ to SO₃ becomes less favorable.

(c) Follow the solution to P5.19 in Benítez (op. cit.), t = 0.66 s (with k = 5.95 s^-1).

#3: Control of NOx.

(a) With 18 lb NO₂ per ton of coal (typical emission factor rating), the uncontrolled NO₂ emissions are 2.15 lb NO₂/s or, in NSPS units, 0.75 lb/10⁶ BTU. This is larger than 0.6 lb/10⁶ BTU and this plant does need to install NOx control equipment.

(b) Follow Ex6.2 and the solution to P6.3 in Benítez (op. cit.) and note that the approach used does not work (very well) for a stoichiometric mixture. The equilibrium mole fractions of N₂, CO₂ and H₂O are 0.74, 0.13 and 0.13, respectively. Obviously, if there is no excess O₂, NO cannot be formed in the post-combustion zone. However, if N₂ can compete for oxygen with carbon and hydrogen, then some NO formation could be expected. As the equivalence ratio increases from 0.8 (with ~600 ppm NO formed) to 0.95 (with ~300 ppm formed) and finally to 1.0, the mole fraction of NO is probably (much?) less than 300 ppm.

#4: Control of COx.

(a) With 1 lb CO/ton of coal (typical emission factor rating), the uncontrolled CO emissions are 0.12 lb CO/s. The rest (i.e., the overwhelming majority) is CO₂, amounting to some 700 lb CO₂/s (assuming ~80%C in coal).

(b) Follow the solution to P1.7 in Benítez (op. cit.). At t = 0 (5% saturation), there are 0.0039 mL HbCO/mL blood. At t = x (50% saturation), there are 0.037 mL HbCO/mL blood. At 1.4 mL HbCO/min (assuming 350 ppm CO in air), these 166 mL HbCO will be delivered to the lungs in approx. 2 hours.

(c) At 108% of 1990 levels, Australia will have to reduce its 2010 CO₂ emissions by 124 million tons of CO₂ per year.

#5: Control of PM.

(a) At 16A lb PM/ton, the uncontrolled emissions are 10 lb PM/10⁶ BTU. This is much larger than the allowed 0.03 lb/10⁶ BTU and the plant needs a (very efficient!) PM control equipment.

(b) Follow the solution to P8.20 (de Nevers, op. cit.) to conclude that ~6% of the particles have diameter < 1 µm and that ~39% of the particles will be removed by this collector.

#6: Control of VOCs.

(a) See handout "On the roles of CO, NOx and VOCs in smog formation."

(b) Follow the solution to P3.9a (Benítez, op. cit.); with k = 8.35 s^-1, determine the temperature that results in this value of the rate constant.

(c) Toulene concentration is 2.43x10^-4 ppm, i.e., 0.000243 moles C7/mol air, which is 0.0017 ppmC. This is much less than the scale shown in Figure D.1.

lrr3@psu.edu (revised 11/25/98)