UNIVERSITY OF THE PHILIPPINES, DILIMANCollege of Engineering Department of Chemical Engineering CASE STUDY: The Use of Limestone Slurry Scrubbing to Remove Sulfur Dioxide from Power Plant Flue Gases In partial fulfillment of the requirements of ChE 132: Stagewise Operations Submitted by: Kimberly A. Gines SN 2011-47993 Submitted to Dr. Richard Chu Submitted on November 24, 2014, First Semester, AY 2014-2015 ChE 132 CASE STUDY - GINES 2014 PROBLEM STATEMENT Five hundred megawatts of electrical power is to be generated in the present facility. Coal with properties shown in Table 1 is fed at 25°C to a furnace where it is burned with 15% excess air. During combustion of the coal, sulfur reacts to form SO2 and a negligible amount of sulfur trioxide (SO3), while carbon and hydrogen are oxidized completely to CO2 and H2O. Essentially all of the nitrogen in the coal leaves the furnace as N2. The ash in the coal leaves the furnace in two streams: 80% leaves as fly ash in the furnace flue gas, and the remainder leaves the furnace as bottom ash at 900°C. Combustion air is brought into the process at 25°C and 50% relative humidity and sent to a heat exchanger, where its temperature is increased to 315°C by exchanging heat with the furnace flue gas. It is then fed to the boiler, where it reacts with coal. The flue gas leaves the furnace at 330°C, goes to an electrostatic precipitator where 99.9% of the particulate material is removed, and then to the air preheater where it exchanges heat with the combustion air. The flue gas leaves the air preheater and is split into two equal streams, with each being the feed stream to one of two identical scrubber trains. In each of the scrubber trains, the divided off-gas stream is fed to a scrubber, where it contacts aqueous slurry of limestone and undergoes adiabatic cooling to 53°C. Sulfur dioxide is absorbed in the slurry and reacts with the limestone: The solid-liquid limestone slurry enters the scrubber at 50°C; the liquid portion of the slurry slows at a rate of 15.2 kg liquid/kg inlet gas and the solid to liquid ratio in the slurry is 1:9 by weight. The cleaned glue gas meets the EPA standard on SO2 emissions; it leaves the scrubber saturated with water at 53°C, containing the carbon dioxide generated in the scrubbing but none of the entering fly ash. The fresh ground limestone is fed to the blending tank at a rate that is 5.2% in excess of that required to react with the sulfur dioxide absorbed from the flue gas. The limestone material fed consists of 92.1% CaCO3 and the remainder is inert insoluble material. The generation of steam and its utilization in the production of electricity in this facility are typical of many power cycles. The boiler used in the present situation generates steam at supercritical conditions: 540°C and 24.1 MPa absolute. The low pressure steam extracted from the power system contains 27.5% liquid water at 6.55 kPa absolute. Heat is removed from the wet low-pressure steam in a condenser by cooling water that enters the condenser at 25°C and leaves at 28°C. Saturated condensate at 38°C is produced by the condenser and pumped back into the boiler. 1. Construct a flowchart of the process and completely label the streams. Show the details of only one train in the SO2 scrubber operation. 2. Estimate the molar flow rate (kmol/min) of each element in the coal (other than those in the ash). 3. Determine the feed rate (kmol/min) of O2 required for complete combustion of the coal. Page 2 of 20 The system may be assumed to meet the standard of 90% removal of the SO 2 released upon combustion. If 15% excess O2 is fed to the combustion furnace. Determine the flow rate (kg/min) of slurry entering the scrubber. b. c. Speculate on the reason for choosing 15% excess air in the prescribed process by giving one possible reason for not using less air and one for not using more. standard cubic meters/min. Estimate this value for the two alternative air flow rates corresponding to 5% and 25% excess oxygen. From this efficiency and the specified power output of 500 MWe. and the molar flow rate (kmol/min) of water in the air stream. 11. kg/min). a. Estimate the flow rate (kg/min and kmol/min) of each component and the composition (mole fractions) of the furnace flue gas. The coal feed rate (kg/h). 8.ChE 132 CASE STUDY .. c. The oxygen and nitrogen feed rates (kmol/min).GINES 2014 4. the dew point and degrees of superheat of the wet air. d. 9. Estimate the feed rate (kg/min) of fresh ground limestone to the blending tank. The air feed rate (kmol/min. the average molecular weight.39 unit is converted to electrical energy. At what rate is heat removed from the furnace? Assuming that all of the heat removed from the furnace is used to generate steam (i. cubic meters/min). fly ash. that is. CaSO3. standard cubic meters/min. for each unit of heat released with the combustion of coal. Estimate the solid-to-liquid mass ratio in the slurry leaving the scrubber. Determine the temperature of the flue gas as it leaves the heat exchanger (air preheater) following the boiler. estimate the rate of steam generation in the power cycle. The rate of steam generation (kg/h). Estimate the rate (kg/min. Compare the release of SO2 in the scrubbed flue gas from Problem 6 with the EPA limit of no more than 520 nanograms SO2 per joule of heat input to the boiler. The mole fraction of water in the wet air. CaCO3. d. L/min) at which the filtrate is recycled to the blending tank. b. What are the flow rates (kg/min) of inerts. cubic meters/min). and water in the wet solids removed from the filter? What fractions of the CaSO3 and CaCO3 are dissolved in the liquid portion of the wet solids? f. Power plants of the type described here operate with an efficiency of about 39%. 0. Determine the effect of the percent excess air fed to the boiler furnace by calculating the rate of steam generation (kg/min) for air flow rates that are 5% and 25% in excess of that theoretically required. none is lost to the surroundings). The flow rate of each component in the gas leaving the furnace (kmol/min. 5. At what rate (kg/min.e. c. 10.) At what rate (kg/min) is fly ash removed from the flue gas by the electrostatic precipitator? 6. (Ignore the fly ash in calculating mole fractions. e. The air feed rate (kmol/min. b. determine the following: a. L/min) is makeup water added to the blending tank? 7. Page 3 of 20 . Determine the flow rate (kg/min and kmol/min) of each component in the cleaned flue gas leaving the scrubber. estimate the following: a. b. Page 4 of 20 . at what rate would methane would have to be burned to raise the stack-gas temperature to the desired value? c. the slurry flow rates entering and leaving the scrubber. scale the results from Problem 6 to determine for each scrubber train the actual flow rates (kg/h) of wet solids and filtrate from the filter.ChE 132 CASE STUDY . a. 13. The following are alternatives by which the scrubbed flue gas can be reheated: (1) bypassing the scrubber with a fraction of the glue gas leaving the air preheater and mixing this stream with cleaned flue gas. For the required coal feed rate. the flow rates of gas (kg/h and m3/min) entering and leaving the scrubber. If the combustion products are blended with the cleaned flue gas.GINES 2014 12. and the flow rates (kg/h) of fresh water and limestone fed to each blending tank. Give a reason for rejecting Alternative 1. How much more coal (kg dry coal/h) would have to be burned for Alternative 3 if the amount of heat released by burning coal (kJ/kg dry coal) is the same as determined in Problem 7? Suggest two process alternatives by which the heat can be transferred to the flue gas. (2) burning natural gas and blending the combustion products with the scrubbed flue gas. assume that the natural gas consists entirely of methane at 25°C and that it is burned adiabatically with 10% excess air that has the same conditions as the air fed to the furnace. Why is the scrubbed flue gas reheated before it is sent to the stack? 14. In evaluating Alternative 2. and (3) using steam from the power cycle either to heat air that is blended with the scrubbed flue gas or to heat the flue gas in a heat exchanger. 2011).0 Nitrogen 1.7 250 230.046 kJ/(kg °C) Ash: Cp = 0.6 Sulfur 3.98 ln (Psat) = A – [B/(T + C)] Psat in kPa and T in K Table 3.2 Hydrogen 5. Pressure. Composition (Ultimate Analysis) Component Dry weight % Carbon 75. Superheated steam table in kJ/kg (Felder. Average properties of coal. Antoine Equation and Constants for Water A 16.7 334 Table 4. bar 50°C 75°C 221.7 C -42.2 228.GINES 2014 GIVEN DATA Table 1.ChE 132 CASE STUDY .5 Ash 7. 2005). Heats of formation of components (Perry’s 8th edition) Component Temperature kJ/mol Component HsO(v) 25 54078 N2 53 -240885 80 -239969 330 54623 CO2 53 -392444 O2 80 -391396 330 12982 SO2 53 -392369 80 391252 Ash 330 13635 500°C 3210 3166 Temperature 25 53 80 330 25 53 80 330 330 900 550°C 3370 3337 kJ/mol 8571 815 1602 9024 8950 827 1633 9434 281 806 Page 5 of 20 . Antoine constants for water (SVA.5 Oxygen 7.2 331.780 kJ/kg dry coal Dry coal: Cp = 1.921 kJ/(kg °C) OTHER DATA AND SOURCE Table 2.80 kg/100 kg dry coal HHV: 30.3872 B 3885.2 Moisture: 4. Show the details of only one train in the SO2 scrubber operation. A schematic diagram of the problem statement (Felder. 2005). A similar figure can be found in Felder (2005) where in this particular case study was obtained. Figure 1 shows the schematic diagram of the problem statement.GINES 2014 PROBLEM 1: Construct a flowchart of the process and completely label the streams. Figure 1.ChE 132 CASE STUDY . Page 6 of 20 . 0 18. S.26637070 Table 6. and O of dry coal.0 kmol/min 6.0 0.46875000 (-) (-) Wet Coal kg/min kmol/min 75.2 6.6 3.26144879 5. along with the given ultimate analysis (Table 1) and the components’ corresponding molecular weights were used to obtain the molar flow rates of each component in the coal feed.0 Weight % 75. A basis of 100 kg dry coal/min.5 7.0 100.26144879 H 4.5 0.10913626 7.0 14.11420414 S 0.10913626 0. Table 5.95049505 N 0.0 100.0 1.0 4. Analysis of coal feed.ChE 132 CASE STUDY .5 0. The molar flow rates of the base components in the coal feed.2 (-) 4.11420414 0.95049505 1. Component Molar Flow Rate (kmol/min) C 6.46875000 Total 11. H.0 Dry Coal kg/min 75.11420414 3. N. as well as that of wet coal (which includes the moisture content).5 7. Table 5 shows the molar flow rates of C.95049505 0.GINES 2014 PROBLEM 2: Estimate the molar flow rate (kmol/min) of each element in the coal (other than those in the ash).2 0.0 32. Component Molecular Weight C H N S O Ash Water Total 12. The oxygen required for complete combustion of coal is determined by using the molar flow rates obtained in Problem 2 and the following reactions: Page 7 of 20 .6 3.2 0.0 1.5 7.1 16.2 5.5 7.46875000 7.0 1.10913626 O 0.90403425 PROBLEM 3: Determine the feed rate (kmol/min) of O2 required for complete combustion of the coal.6 0.8 0.26144879 4.2 5. b.1) The mole fraction of water in the wet air can be calculated using the definition for the mole fraction in terms of pressures: The partial pressure of water in the given system can be obtained using the definition of the relative humidity (RH): Page 8 of 20 .485 kmol O2/min amount of N2 fed = (79 kmol N2 / 21 kmol O2)*(amount O2 fed) = (79kmol N2/21 kmol O2)*(8. then: amount of O2 fed = (1 + 0. The mole fraction of water in the wet air. c.378 kmol O2/min) amount of O2 fed = 8.15) * (amount O2 required) = (1.46875000 1. Reaction coefficients for oxygen. The air feed rate (kmol/min.46875 kmol O/min) *(1 kmol O2 / 2 kmol O) O2 required = 7. Component Molar Flow Rate (kmol/min) Stoichiometric Coefficient to Oxygen C H S O 6.485 kmol O2/min) amount of N2 fed = 31.95049505 0. The oxygen and nitrogen feed rates (kmol/min).5 kmol O2/ 1 kmol H2) + (0.00 0.15)*(7.00 0.2614 kmol C/min)*(1 kmol O2/ 1 kmol C) + (4. a) If 15% excess O2 is fed.109136 kmol S/min)*(1 kmol O2/ 1 kmol S) – (0. the average molecular weight.ChE 132 CASE STUDY .378 kmol O2/min PROBLEM 4: If 15% excess O2 is fed to the combustion furnace. cubic meters/min). the dew point and degrees of superheat of the wet air. standard cubic meters/min.10913626 0.25 1.918 kmol N2/min b.GINES 2014 The stoichiometric amount of oxygen to each component is as follows: Table 6.950495 kmol H/min)*(1 kmol H2/ 2 kmol H)*(0.50 The amount of oxygen required is given by the following equation: O2 required = (6.26144879 4. and the molar flow rate (kmol/min) of water in the air stream. estimate the following: a. the dry basis of air is first used and then corrected to include the amount of water in the wet air.015730282 From the table.89°C or 287.01573.015730282 Mole fraction (wet) 0.325 kPa RH 50 percentage Psat.7778)(28) + (0.672 kg/kmol.15 K 1 atm Pressure 101.2) To get the average molecular weight of the air feed. Component Molecular Weight N2 O2 Water Total 28. Summary of values for the average molecular weight.3872 – [3885. Psat = PH2O. 25 Celcius Temperature 298. Tdew = 13. This can be calculated by using the Antoine relation but this time solving for the temperature (T) when Psat = 1.672 kg/kmol Therefore.00 18.79 0. Average Molecular Weight = 28.187741617 kPa PH2O 1.00 32. b.21 0. b.7/ (Tdew – 42.02 Mole fraction (dry) 0. That is. Table 8.593870809 kPa yH2O 0.040774 K b.777765529 0. the mole fraction of water in the wet air is 0.59387 (See Table 7): ln (1.015486672 1 The average molecular weight is computed using the following equation: Average MW = (0.98)] Solving for T gives.ChE 132 CASE STUDY .3) The dew point is the temperature at which water has a vapor pressure equal to its partial pressure. The table below summarizes the vales calculated using the equations aforementioned: Table 7.01549)(18.015730282 1. Therefore: Page 9 of 20 .206747799 0. H2O 3.GINES 2014 The vapor pressure of water (Psat) can be calculated using the Antoine relation given in Table 2.4) The degrees of superheat is the number of temperature degrees through which air has been heated above the saturation temperature.02) = 28.2067)(32) + (0. Summary of values to obtain the mole fraction of water in the wet air.59387081 kPa) = 16. 015486672/0.95771171 0.05020922 N2 31. Flow rates 41.992360461 O2 1.Tdew = 25°C – 13.02 56.02 895. Analysis of air fed into the system.479908892 Water 0.) At what rate (kg/min) is fly ash removed from the flue gas by the electrostatic precipitator? The flow rates of each component and the composition of the furnace flue gas are shown in the following tables: Table 11.002565207 64. Component kmol/min Mole Fraction CO2 6.109226 °C b.ChE 132 CASE STUDY .106075073 0.751154065 28.206747799) Amount of water in air = 0. Estimate the flow rate (kg/min and kmol/min) of each component and the composition (mole fractions) of the furnace flue gas.01 kg/min 275. Component Flow rate (kmol/min) N2 31.GINES 2014 Degrees of Superheat = Tfeed .073109834 18.O2 required 1269.01571547 Table 10.07 6.5) The molar flow rate (kmol/min) of water in the air stream is calculated using the amount of oxygen fed and the (wet) molar fractions of water and oxygen from Table 8: Amount of water in air = (amount O2 fed)*(mole fraction of H2O/mole fraction of O2) = (8.9560138 scfm 922.110444463 0. Table 9.63519694 kmol H2O/min c) The air rate is simply the sum of the flow rates of nitrogen.47990899 kmol O2/min)*(0. Table 9 summarizes these values. Summary of feed air flow rates.261448793 0. oxygen and water fed into the system.635196938 Total 41.0796639 m3/min PROBLEM 5.109136264 0.4550822 SO2 0. (Ignore the fly ash in calculating mole fractions.147173013 MW 44.01571547 kmol/min 855.5663614 H2O 3. Analysis of the flue gas.90060964 O2 8.458416 Page 10 of 20 .025997881 32 35.89°C Degrees of Superheat = 11.39440233 Total 42.54481631 1 Composition C in coal H2O in air + H in coal N2 in air + N in coal S in coal O2 fed . 5663614 H2O 3.054568132 0. e.05020922 N2 31. CaSO3.002565207 6.109136264 0. What are the flow rates (kg/min) of inerts. b. Table 13.073109834 0.147173013 kg/min 275. Analysis of the gas entering the scrubber.754 kg/min. and water in the wet solids removed from the filter? What fractions of the CaSO3 and CaCO3 are dissolved in the liquid portion of the wet solids? f.999)*(5.073109834 56. Mole kmol/min Fraction 6.025997881 35.025997881 1 MW kg/min 44.27240815 Mole Fraction 0. CaCO3.555222231 15. c.76 kg/min Fly ash removed by EP = (0.2 kg/min) = 5.783181 28.110444463 0.458416 The fly ash removed by the electrostatic precipitator (EP) is determined by: Fly ash before EP = (0.727541 3.0251046 447. Compositionof the Flue Gas.39440233 Total 42.49618023 17. At what rate (kg/min.732088 Page 11 of 20 . Determine the flow rate (kg/min and kmol/min) of each component in the cleaned flue gas leaving the scrubber.002565207 0. a.95771171 0. Table 13 summarizes the analysis of the gas entering the scrubbers.00288 634.01 18.97885586 0. Estimate the rate (kg/min.54481631 1 1269.106075073 0.147173013 0. L/min) is makeup water added to the blending tank? The feed to each scrubber (which comes from the electrostatic precipitator) is half of the total flue gas flow rate since they are equally divided into the scrubber.02 64.07 32 - 137.751154065 0. L/min) at which the filtrate is recycled to the blending tank.751154065 895.4550822 SO2 0. d. Determine the flow rate (kg/min) of slurry entering the scrubber. PROBLEM 6: The system may be assumed to meet the standard of 90% removal of the SO2 released upon combustion.8)*(7. Estimate the feed rate (kg/min) of fresh ground limestone to the blending tank.130724396 1. Component kmol/min CO2 H2O N2 SO2 O2 Fly Ash Total 3. fly ash.261448793 0.GINES 2014 Component CO2 Table 12.6972012 0.992360461 O2 1.76 kg/min) = 5.75424 kg/min Therefore the fly ash removed by the electrostatic precipitator is 5.02 28.ChE 132 CASE STUDY . Estimate the solid-to-liquid mass ratio in the slurry leaving the scrubber.553037536 21. 179835715 3. To get the solid-to-liquid mass ratio of the slurry leaving the scrubber: Table 15.049111319 64.141808336 0.005456813 0.73378358 Mole Fraction 0.139872701 0.702868302 0.02 28.647.003 kg/100 kg water The water in the liquid is determined using the following equation (x = amount of water): Page 12 of 20 .07 3.34961802 17. Analysis of the gas leaving the scrubber. Meanwhile.647.GINES 2014 Among the components in the flue gas leaving the scrubber.14656221 CaSO3 formed 0.94457 54.732088 kg gas/min) = 9. only nitrogen and oxygen will remain constant.01 18.92 kg/min. the total amount of slurry in the scrubber is 10.93 kg/min CaCO3 in liquid = 0.93 kg/min)*(1 solid/9 liquid in slurry) = 1.99 kg/min Total amount of slurry = 9.049111319 120.719. Sulfur dioxide and Carbon dioxide are both consumed and used in another reaction or generated.07802 The amount of slurry fed into the scrubber is calculated as follows: Amount of liquid in slurry = (15.727541 0.93 kg/min Amount of solids in the slurry = (9.719.01659766 15.92 kg/min Therefore.07 32 (-) 139. Analysis of the Reaction of Sulfur dioxide and Calcium Sulfite.024326683 0 MW kg/min 44.071.647.2 kg liquid/kg inlet gas)*(634.99 kg/min = 10.986155285 Amount of liquid in the slurry = 9.97885586 0.17 5.000240031 0. All of the fly ash is also removed here.553037536 0 22.02 64.91555192 Added solids 0.6972012 0 660.647. 90% of the sulfur dioxide is absorbed and reacted with calcium carbonate forming the added carbon dioxide (1:1 ratio). kmol/min MW kg/min SO2 absorbed 0. the amount of water in the stream leaving the scrubber is adjusted so that the water mole fraction is the one calculated from the vapor pressure.3590898 447. Component kmol/min CO2 H2O N2 SO2 O2 Fly Ash Total 3.002 kg/100 kg water CaSO3 in liquid = 0.09 4.93 kg/min + 1. Table 14.9017072 CaCO3 consumed 0.071.ChE 132 CASE STUDY .049111319 100. 52/100 = 0. Analysis of wet solids obtained from the filter. Mass Component kg/min Fraction Inerts 0.052 kmol CaCO3/ 1 kmol SO2) *(1 kg limestone / 0. the amount of inerts is 0.52/100 = 0.4816104 kg inerts/min Therefore.2556087 0. The mass of liquid in that slurry is: Mass of liquid in slurry to filter = 50.517624 kg/min Therefore.56147238 kg Therefore.482 kg/min.2324 0. Since inerts leave the system only in the wet solids: Mass of inerts in the wet solids = 5.93 Solving for X: X = 9647.111207765. the amount of added liquid due to calcium carbonate and calcium sulfite are: Amount of CaCO3 dissolved in liquid = 0.0001 8. the feed rate of freshly ground limestone is given by: Amount of limestone = 0.GINES 2014 X + 0.040558352 CaCO3 solids 0. Since there is 5.289425529) = 0.00288 0.002*9647.6147 kg/min.003X/100 = 9647.901707203 0. the amount of limestone feed is 5.8745063 1 From the solid-to-liquid ratio calculated earlier.497006532 CaSO3 liquid 0. the slurry leaving the absorber is 11 wt% solids.19295903652 + 0. Table 16.000242536 H2O 5. the total solids sent to the filter is given by Table 16.2% excess CaCO3 required to react with the absorbed SO2.986155)/(9647.921 kg CaCO3) =0.0546 kmol SO2)&(1. Therefore.111207765 The solid-to-liquid mass ratio of the slurry leaving the scrubber is 0.021525838 CaCO3 liquid 0.09 kg CaCO3/mol) = 0.ChE 132 CASE STUDY .9 kg/min.99197 + 0.0002 1.440641477 Total 11. Page 13 of 20 .481610402 0.289425529 kg/min Therefore.003*9647.517624 + 0.6147 kg limestone/min*(0.002X/100 + 0.921 kg CaCO3)*(100.68428E-05 Fly Ash 0.079 kg inerts/0.4214E-06 CaSO3 solids 5. the mass ratio of solid-to-liquid slurry leaving the scrubber is: Solid-to-Liquid Ratio = (1071.1929503652 kg/min Amount of CaSO3 dissolved in liquid = 0.9*(0. Using the energy balance around the furnace and assuming that no heat is lost to the surroundings.01659766 0 31. using the energy balance equation.. Table 17 shows the analysis of the steam coming in and out of the furnace.9785 From here.097 kJ/min. Kmol or kg H (kJ/min) Coal 100 30780 In 41. the rate at which heat is removed can be determined. IN Component kmol/min kJ/mol kJ/min Coal 100 0 0 CO2 0 0 0 H2O vapor 0.52 L/min PROBLEM 7: At what rate is heat removed from the furnace? Assuming that all of the heat removed from the furnace is used to generate steam (i.44 OUT kJ/mol 0 12982 54623 0 9024 13635 9434 281 806 kJ/min 0 82561. whose values are used for the terms in the energy balance.1252 SO2 0 0 0 O2 8.279069832 0 0 N2 31. using the energy balance: Table 18.010913626 1.95771171 0.90060964 8571 273420.614 0 288386.64 549085. estimate the rate of steam generation in the power cycle. Mass of recycle = 45.106075073 5.ChE 132 CASE STUDY . Analysis of steam in and out of the furnace.7122 1618.479908892 8950 75895. Q = -4.e.18459 Ash (-) (-) (-) Bottom Ash (-) (-) (-) Total 383665.39 148.387.635196938 54078 34350. none is lost to the surroundings).52 kg/min Volume of make-up water = 33.359671431 3.GINES 2014 The recycle flow rate is the flow rate of liquid in the slurry minus the flow rate of liquid in the wet solids. Page 14 of 20 .807297 10434. Table 17.4898 kmol/min 0 6.6 kg/min Volume of recycle = 45.76 1. Summary for the rate of heat removal.848.4898 Out 42.1 L/min The make-up water is computed as follows: Mass of make-up water = liquid water leaving with wet solids + water leaving as CaSO3 hemihydrate + water leaving with cleaned off gas – water entering absorber Mass of make-up water = 33.18001 H2O liquid 0.2947853 383665.2545 164775.4509695 549085.56 1160.979 From this. 2 The mass of steam generated is calculated using the following equation: From Table 20. Data for interpolation. T (°C) P (bar) H (kJ/kg) 540 241 3313.7 75 331.60 kg/min.8 550 3370 3337 Table 20.4948 Out In 50 228.2 230.116 241 bar 179.928 250 bar 181.6 kJ/kg = 1401.ChE 132 CASE STUDY .60 kg/min.097 kJ/min / 3130. T (°C) 38 221.6 kJ/kg Therefore. the mass of steam generated = 4387848.GINES 2014 To get the mass of steam generated: Table 19.2 kJ/kg = 3130.8 38 241 183.8 kJ/kg – 183. Page 15 of 20 . Summary of interpolation. The mass of steam generated is 1401. ΔHsteam = 3313.8 3313.2 bar 175.7 334 500 3210 3166 540 3338 3302. 64 517097.2733 Out 38.841. as done in Problem 7.76 281 Bottom Ash (-) (-) (-) 1.56 1160.217292274 8950 82494.807297 17391.010913626 13635 O2 9.010913626 13635 O2 7. Kmol H Coal 100 30780 In 37. For 5% and 25% excess air.2545 164775.ChE 132 CASE STUDY .12664359 8571 249644.843458455 9434 Ash (-) (-) (-) 5.257.15218 3.1871 1618. Table 23. the same calculations were done until solving for the heat evolved.44 806 Total 417027.723 Page 16 of 20 .359671431 12982 H2O vapor 0.01659766 54623 H2O liquid 0.76 281 Bottom Ash (-) (-) (-) 1. the Q here is equal to -3. Summary of HHV values and the amount of heat evolved (5% excess air). IN OUT Component kmol/min kJ/mol kJ/min kmol/min kJ/mol Coal 100 0 0 0 0 CO2 0 0 0 6.64 581074.01659766 54623 H2O liquid 0.66 148.72820135 350303. Analysis of steam coming in and out of the furnace (5% excess air).56 1160.44 806 Total 350303.73167777 9024 SO2 0 0 0 0.359671431 12982 H2O vapor 0. IN OUT Component kmol/min kJ/mol kJ/min kmol/min kJ/mol Coal 100 0 0 0 0 CO2 0 0 0 6.93962006 517097.279069832 0 0 0 0 N2 34.23741 1618.18374566 9024 SO2 0 0 0 0.40 kJ/min.279069832 0 0 0 0 N2 29.121 148. Analysis of steam coming in and out of the furnace (25% excess air).614 0 263354.2733 kJ/min 0 82561.807297 3478.76586 1.234 Table 22.7064 kJ/min 0 82561.7883 34.4622 29. Table 21.234 Using the same equation in solving for Q.74252551 8950 69295.GINES 2014 PROBLEM 8: Determine the effect of the percent excess air fed to the boiler furnace by calculating the rate of steam generation (kg/min) for air flow rates that are 5% and 25% in excess of that theoretically required.20783 3.60332 0.2545 164775.6745757 8571 297195.614 0 313418.579962422 54078 31363.368691691 9434 Ash (-) (-) (-) 5. The effect of the percent excess air fed to the boiler furnace is determined by calculating the rate of steam generation for air flow rates that are 5% and 25% in excess. Speculate on the reason for choosing 15% excess air in the prescribed process by giving one possible reason for not using less air and one for not using more.690431454 54078 37337. Summary of HHV values and the amount of heat evolved (25% excess air). PROBLEM 10: Compare the release of SO2 in the scrubbed flue gas from Problem 6 with the EPA limit of no more than 520 nanograms SO2 per joule of heat input to the boiler.86136926 417027.097 -4921107. The table below summarizes the heat evolved as well as the mass of steam generated for each percentage of excess air.3573959 kJ/min ng/kJ ng/J From the table.823 kJ/min. the flue gas for this system has met the standard EPA limit. Q (kJ/min) -3841257.36 ng SO2/J in the scrubber flue gases.823 5% excess air 15% excess air 25% excess air Mass of steam generated (kg/min) 1227. SO2 Scrubber 1 0.005456813 kmol/min Scrubber 2 0.399 -4387848.003577 1401. Summary of conversions for SO2 in the scrubbers. Q = -4. Table 25.921.96231894 581074. Summary of comparison.699236046 kg/min 6.GINES 2014 Table 24.107.010913626 kmol/min Total 0.005456813 kmol/min 0. Page 17 of 20 . PROBLEM 9: Determine the temperature of the flue gas as it leaves the heat exchanger (air preheater) following the boiler. which should be avoided. The release of SO2 in the scrubbed flue has can be compared to the EPA standard by simply converting the values obtained in Problem 6 into the desired units (ng/J): Table 26.097 159357.723 From the values in the table above.599724 1571. Estimate this value for the two alternative air flow rates corresponding to 5% and 25% excess oxygen.99236E+11 ng/min Q SO2 from scrubbers/Q 4387848. Kmol H Coal 100 30780 In 44.7064 Out 45.3959 159. it can be seen that there is 159.937591 The 15% excess air is chosen in the prescribed process because the excess air affects the amount of unburned carbon or soot that is formed in the process. Since it is less than 520 ng SO2/J.ChE 132 CASE STUDY . cubic meters/min).8556 735.62892 900342.30957 1337494.01571547 Air 855.47956412 1163.9560138 922.010913626 1.885 Page 18 of 20 .690912 1337494.6061 110234.01659766 31. The air feed rate (kmol/min. The flow rate of each component in the gas leaving the furnace (kmol/min.5152 43142. kg/min).097 4615384615 1051.95771171 0. c.39 unit is converted to electrical energy.0833 44652.93811 941889.599724 *Flue Gas Composition Component CO2 H2O N2 SO2 O2 Ash Total MW 44. With an efficiency of about 39% and a specified power output of 500 MWe.02 28.088 57177.02 64. Table 28.45894 3173.76 42.ChE 132 CASE STUDY .30957 kg/min 294403.8 41. 0.4509695 kg/hr kg/hr kmol/hr scfh m3/hr kmol/hr kg/hr kg/hr Actual kmol/hr 6689. The rate of steam generation (kg/h).026532 33614. Summary of the Actual Flow Rates for the System.521 969895. the scale factor is first determined in order to solve for the desired feed rates: Table 27.359671431 3. for each unit of heat released with the combustion of coal.885 1474281.GINES 2014 PROBLEM 11: Power plants of the type described here operate with an efficiency of about 39%. The coal feed rate (kg/h).91276 11.856061 kJ/min kJ/hr min/hr Using this value of the scale factor. From this efficiency and the specified power output of 500 MWe. all computed flow rates are then scaled up into the actual flow rates for the system. Determining the scale factor for the scale-up calculations.431769 44652. d.39 Q Qactual Scale Factor 4387848. determine the following: a. Power output 500 (MWe) Efficiency 0.106075073 5. that is.0796639 42.55695 Steam 1401.07 32 - Actual 105185.4509695 Flue gas* 1271. standard cubic meters/min.165 Basis kmol/min 6.01 18.81662 6058. b.4956732 37229. Flow Rates Basis Dry Coal 100 Wet Coal 104. 891 61637.582 1135564.72947 0.48918 0. the flow rates of gas (kg/h and m3/min) entering and leaving the scrubber.126 kg/min 147428.07 O2 581.02934546 665721.9691 470944. scale the results from Problem 6 to determine for each scrubber train the actual flow rates (kg/h) of wet solids and filtrate from the filter.717 Actual 10719.02 N2 16807.15478 1 Component kmol/min Mole Fraction MW CO2 3349.215 470944.45638 0.928 367.79 kg/hr 5.45638 0. The table below summarizes the actual flow rates of the gas entering and leaving the scrubbers. the slurry flow rates entering and leaving the scrubber.7158847 0.02 N2 16807.138653856 44.07 O2 581.ChE 132 CASE STUDY .024077514 32 Fly Ash 0 0 (-) Total 24160.573978206 2.7747837 18614.73978206 0.01 H2O 3420.026055355 32 Fly Ash 0 0 (-) Total 22326.544 28588.071060748 18.27240815 22375.6956691 28. Table 29.52 kg/hr 10731.5623 m3/hr Flue Gas exit from each scrubber 22. Component kmol/min Mole Fraction MW CO2 3344.51145 kmol/h 935.02 Gas entering the scrubber SO2 5.614723797 5905. Actual Flow Rates for the Flue Gases entering and leaving the scrubbers.7158847 0.9083 3. The calculations for Problem 6 were used again using the values obtained in Problem 11.GINES 2014 PROBLEM 12: For the required coal feed rate.73378358 23912.141575773 18.79421 11288302.513266 0.37572E-05 64.752814649 28.9197 11275812.1307 1 0 Basis Slurry feed to each scrubber Slurry exit from each scrubber Limestone to blending tank kg/min 147201.87 m3/hr PROBLEM 13: Why is the scrubbed flue gas reheated before it is sent to the stack? Heating the scrubbed gas before sending it to the stack allows it density to reduce allowing the scrubbed gas to rise through the stack more easily and consequently out into the atmosphere.02 Gas leaving the scrubber SO2 0.01 H2O 1586.3557 983859. Page 19 of 20 .928 36.9083 0 698662. and the flow rates (kg/h) of fresh water and limestone fed to each blending tank.895274 0.66805 kmol/h 1079.747837 18614.14981216 44.000257088 64.881257 kg/hr Flue Gas feed to each scrubber 21. Analysis of the scrubbed flue gas and stack gas.48918 16807. the heat evolved is much greater than the value calculated beforehand.7158847 24160. an additional 461.1307 -2124534278 20238215.121 949942. If the combustion products are blended with the cleaned flue gas.573978206 -392369 -225211.48918 -240855 -823841921.238.97 kmol/h 3349. a.95 0.895274 3420. In evaluating Alternative 2.12 -224570.895274 -392444 -1314646301 3420. Therefore.7158847 827 481079. The heat from the scrubbed flue gas and the stack gas are determined using the actual flow rates from Problem 12 and the heats of formation at 53°C for the flue gas and 80°C for the stack gas. assume that the natural gas consists entirely of methane at 25°C and that it is burned adiabatically with 10% excess air that has the same conditions as the air fed to the furnace.2547 581. Therefore. Moreover. Page 20 of 20 . an added amount of coal is needed to reach this amount of heat.GINES 2014 PROBLEM 14: The following are alternatives by which the scrubbed flue gas can be reheated: (1) bypassing the scrubber with a fraction of the glue gas leaving the air preheater and mixing this stream with cleaned flue gas. The first alternative to reheat the scrubbed flue gas makes controlling the temperature of the bypassed gas more difficult.481 kg coal/hr) = 461.ChE 132 CASE STUDY .97 kJ/hr) / (43.233 kg of coal per hour is needed to satisfy the heat evolution solved.0366 24160. Table 30. at what rate would methane would have to be burned to raise the stack-gas temperature to the desired value? c.45638 815 13698076. Component CO2 H2O N2 SO2 O2 Total Q Scrubbed Flue Gas kmol/hr kJ/mol kJ/h 3349.233 kg/hr Therefore. b. (2) burning natural gas and blending the combustion products with the scrubbed flue gas.878.45638 0. Mass of additional coal = (20. and (3) using steam from the power cycle either to heat air that is blended with the scrubbed flue gas or to heat the flue gas in a heat exchanger.215. this requires a more accurate analysis of sulfur dioxide which is not found in this study. Give a reason for rejecting Alternative 1. How much more coal (kg dry coal/h) would have to be burned for Alternative 3 if the amount of heat released by burning coal (kJ/kg dry coal) is the same as determined in Problem 7? Suggest two process alternatives by which the heat can be transferred to the flue gas. the first alternative is rejected.1307 Stack Gas kJ/mol -391396 -239969 1602 -391252 1633 kJ/hr -1311135611 -820811368 26925545.573978206 581. It can also directly release particulates as well as sulfur dioxide which is being prevented.0397 -2104296062 kJ/hr From the calculations.4 16807.