Problem 2. Air Pollution Control Systems (15 points) Combusting coal at a power

Question

Problem 2. Air Pollution Control Systems (15 points) Combusting coal at a power plant releases 12,800 lbs of SO2 per hour. In order to comply with Clean Air Act regulations, the power plant is only allowed to emit 640 lbs of SO2 per hour. a. How much SO2 will need to be captured per hour (in kg/hr)? (2 points) In order to capture this SO2, the power plant will use a slurry of either calcium carbonate (CaCO3) or hydrated lime (Ca(OH)2). The stoichiometry of these reactions are below: The process will use a 10% (by wt%) slurry b. What is the mass low rate (in kg/hr) of CaCO3 required to comply with the SO2 standards? (3 points) c. What is the volumetric flow rate (in m'/hr) of CaCO3 slurry? Assume a water density of 1000 kg/m3. (3 points) d. CaCO3 will cost S120/short ton. How much will the plant pay for CaCO3 each hour if it chooses to use it as its reagent? (2 points) e, What is the mass flow rate (in kg/hr) of Ca(OH)2 required to comply with the SO2 standards? (3 points) f. What is the volumetric flow rate (in m/hr) of Ca(OH)2 slurry? Assume a water density of 1000 kg/m3. (3 points) g. Ca(OH)2 will cost $170/short ton. How much will the plant pay for Ca(OH)2 each hour if it chooses to use it as its reagent? (2 points) h. Considering only the costs for the CaCOs or Ca(OH)2 reagent, which should the power plant select? (2 points)

Explanation / Answer

Answer a)

SO2 released = 12800lb/h = 5443.108 kg /h

SO2 limit = 640lb/h = 640*0.454 kg/h = 290.56 kg/h

Thus SO2 needed to be captured = 5443.108 - 290.56= 5152.548 kg/h

Thus moles of SO2 captured = 5152548/64 mol/h = 80508.5625 mol/h

Mass flow rate of CaCO3 = 80508.5625 *0.1 kg/h = 8050.85625 kg/h

Answer d)

Cost = $ 120*80.5085625 = $ 9661.0275 per hour

Answer e)

Assuming that 10% of total CaCO3 required above is hydrated to form a slurry then mass flow rate of hydrated lime = 805.085 kg/h

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