che132 - case study writeup

UNIVERSITY OF THE PHILIPPINES, DILIMAN

College of EngineeringDepartment of Chemical Engineering

CASE STUDY:The Use of Limestone Slurry Scrubbing to Remove Sulfur Dioxide from Power Plant Flue GasesIn partial fulfillment of the requirements of ChE 132: Stagewise Operations

Submitted by: Kimberly A. GinesSN 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.

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4. If 15% excess O2 is fed to the combustion furnace, estimate the following:a. The oxygen and nitrogen feed rates (kmol/min).b. The mole fraction of water in the wet air, the average molecular weight, the dew

point and degrees of superheat of the wet air, and the molar flow rate (kmol/min) of water in the air stream.

c. The air feed rate (kmol/min, standard cubic meters/min, cubic meters/min).

5. Estimate the flow rate (kg/min and kmol/min) of each component and the composition (mole fractions) of the furnace flue gas. (Ignore the fly ash in calculating mole fractions.) At what rate (kg/min) is fly ash removed from the flue gas by the electrostatic precipitator?

6. The system may be assumed to meet the standard of 90% removal of the SO 2 released upon combustion.

a. Determine the flow rate (kg/min and kmol/min) of each component in the cleaned flue gas leaving the scrubber.

b. Determine the flow rate (kg/min) of slurry entering the scrubber.c. Estimate the solid-to-liquid mass ratio in the slurry leaving the scrubber.d. Estimate the feed rate (kg/min) of fresh ground limestone to the blending tank.e. What are the flow rates (kg/min) of inerts, CaSO3, CaCO3, fly ash, 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. Estimate the rate (kg/min, L/min) at which the filtrate is recycled to the blending tank. At what rate (kg/min, L/min) is makeup water added to the blending tank?

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.e., none is lost to the surroundings), estimate the rate of steam generation in the power cycle.

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. 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.

9. Determine the temperature of the flue gas as it leaves the heat exchanger (air preheater) following the boiler. Estimate this value for the two alternative air flow rates corresponding to 5% and 25% excess oxygen.

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.

11. Power plants of the type described here operate with an efficiency of about 39%; that is, for each unit of heat released with the combustion of coal, 0.39 unit is converted to electrical energy. From this efficiency and the specified power output of 500 MWe, determine the following:

a. The coal feed rate (kg/h).b. The air feed rate (kmol/min, standard cubic meters/min, cubic meters/min).c. The flow rate of each component in the gas leaving the furnace (kmol/min, kg/min).d. The rate of steam generation (kg/h).

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12. For the required coal feed rate, 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, the slurry flow rates entering and leaving the scrubber, the flow rates of gas (kg/h and m3/min) entering and leaving the scrubber, and the flow rates (kg/h) of fresh water and limestone fed to each blending tank.

13. Why is the scrubbed flue gas reheated before it is sent to the stack?

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; (2) burning natural gas and blending the combustion products with the scrubbed flue gas; 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.

a. Give a reason for rejecting Alternative 1.b. In evaluating Alternative 2, 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. If the combustion products are blended with the cleaned flue gas, at what rate would methane would have to be burned to raise the stack-gas temperature to the desired value?

c. 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.

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GIVEN DATA

Table 1. Average properties of coal. Composition (Ultimate Analysis)

Component Dry weight %Carbon 75.2

Hydrogen 5.0Nitrogen 1.6

Sulfur 3.5Oxygen 7.5

Ash 7.2Moisture: 4.80 kg/100 kg dry coal

HHV: 30,780 kJ/kg dry coalDry coal: Cp = 1.046 kJ/(kg °C)

Ash: Cp = 0.921 kJ/(kg °C)

OTHER DATA AND SOURCE

Table 2. Antoine constants for water (SVA, 2011).Antoine Equation and Constants for Water

A 16.3872B 3885.7C -42.98

ln (Psat) = A – [B/(T + C)]Psat in kPa and T in K

Table 3. Superheated steam table in kJ/kg (Felder, 2005).Pressure, bar 50°C 75°C 500°C 550°C

221.2 228.2 331.7 3210 3370250 230.7 334 3166 3337

Table 4. Heats of formation of components (Perry’s 8th edition)Component Temperature kJ/mol Component Temperature kJ/mol

HsO(v) 25 54078 N2 25 857153 -240885 53 81580 -239969 80 1602

330 54623 330 9024CO2 53 -392444 O2 25 8950

80 -391396 53 827330 12982 80 1633

SO2 53 -392369 330 943480 391252 Ash 330 281

330 13635 900 806

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PROBLEM 1: Construct a flowchart of the process and completely label the streams. Show the details of only one train in the SO2 scrubber operation.

Figure 1. A schematic diagram of the problem statement (Felder, 2005).

Figure 1 shows the schematic diagram of the problem statement. A similar figure can be found in Felder (2005) where in this particular case study was obtained.

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PROBLEM 2: Estimate the molar flow rate (kmol/min) of each element in the coal (other than those in the ash).

A basis of 100 kg dry coal/min, 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. Table 5 shows the molar flow rates of C, H, N, S, and O of dry coal, as well as that of wet coal (which includes the moisture content).

Table 5. Analysis of coal feed. Component Molecular Weight Dry Coal Wet Coal

Weight % kg/min kmol/min kg/min kmol/minC 12.0 75.2 75.2 6.26144879 75.2 6.26144879H 1.0 5.0 5.0 4.95049505 5.0 4.95049505N 14.0 1.6 1.6 0.11420414 1.6 0.11420414S 32.1 3.5 3.5 0.10913626 3.5 0.10913626O 16.0 7.5 7.5 0.46875000 7.5 0.46875000

Ash 0.0 7.2 7.2 (-) 7.2 (-)Water 18.0 0.0 0.0 (-) 4.8 0.26637070Total 100.0 100.0

Table 6. The molar flow rates of the base components in the coal feed.

Component Molar Flow Rate (kmol/min)C 6.26144879H 4.95049505N 0.11420414S 0.10913626O 0.46875000

Total 11.90403425

PROBLEM 3: Determine the feed rate (kmol/min) of O2 required for complete combustion of the coal.

The oxygen required for complete combustion of coal is determined by using the molar flow rates obtained in Problem 2 and the following reactions:

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The stoichiometric amount of oxygen to each component is as follows:

Table 6. Reaction coefficients for oxygen.

Component Molar Flow Rate (kmol/min)Stoichiometric Coefficient to

Oxygen

C 6.26144879 1.00H 4.95049505 0.25S 0.10913626 1.00O 0.46875000 0.50

The amount of oxygen required is given by the following equation:

O2 required = (6.2614 kmol C/min)*(1 kmol O2/ 1 kmol C) + (4.950495 kmol H/min)*(1 kmol H2/ 2 kmol H)*(0.5 kmol O2/ 1 kmol H2) + (0.109136 kmol S/min)*(1 kmol O2/ 1 kmol S)– (0.46875 kmol O/min) *(1 kmol O2 / 2 kmol O)

O2 required = 7.378 kmol O2/min

PROBLEM 4: If 15% excess O2 is fed to the combustion furnace, estimate the following:a. The oxygen and nitrogen feed rates (kmol/min).b. The mole fraction of water in the wet air, the average molecular weight, the dew point

and degrees of superheat of the wet air, and the molar flow rate (kmol/min) of water in the air stream.

c. The air feed rate (kmol/min, standard cubic meters/min, cubic meters/min).

a) If 15% excess O2 is fed, then:

amount of O2 fed = (1 + 0.15) * (amount O2 required) = (1.15)*(7.378 kmol O2/min)

amount of O2 fed = 8.485 kmol O2/min

amount of N2 fed = (79 kmol N2 / 21 kmol O2)*(amount O2 fed) = (79kmol N2/21 kmol O2)*(8.485 kmol O2/min)

amount of N2 fed = 31.918 kmol N2/min

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):

The vapor pressure of water (Psat) can be calculated using the Antoine relation given in Table 2. The table below summarizes the vales calculated using the equations aforementioned:

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Table 7. Summary of values to obtain the mole fraction of water in the wet air.

Temperature25 Celcius

298.15 K

Pressure1 atm

101.325 kPaRH 50 percentage

Psat, H2O 3.187741617 kPaPH2O 1.593870809 kPayH2O 0.015730282

From the table, the mole fraction of water in the wet air is 0.01573.

b.2) To get the average molecular weight of the air feed, the dry basis of air is first used and then corrected to include the amount of water in the wet air.

Table 8. Summary of values for the average molecular weight.

Component Molecular WeightMole fraction

(dry)

Mole fraction

(wet)N2 28.00 0.79 0.777765529O2 32.00 0.21 0.206747799

Water 18.02 0.015730282 0.015486672Total 1.015730282 1

The average molecular weight is computed using the following equation:

Average MW = (0.7778)(28) + (0.2067)(32) + (0.01549)(18.02) = 28.672 kg/kmol

Therefore, Average Molecular Weight = 28.672 kg/kmol.

b.3) The dew point is the temperature at which water has a vapor pressure equal to its partial pressure. That is, Psat = PH2O. This can be calculated by using the Antoine relation but this time solving for the temperature (T) when Psat = 1.59387 (See Table 7):

ln (1.59387081 kPa) = 16.3872 – [3885.7/ (Tdew – 42.98)]

Solving for T gives, Tdew = 13.89°C or 287.040774 K

b.4) The degrees of superheat is the number of temperature degrees through which air has been heated above the saturation temperature. Therefore:

Degrees of Superheat = Tfeed - Tdew

= 25°C – 13.89°CDegrees of Superheat = 11.109226 °C

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b.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) = (8.47990899 kmol O2/min)*(0.015486672/0.206747799)

Amount of water in air = 0.63519694 kmol H2O/min

c) The air rate is simply the sum of the flow rates of nitrogen, oxygen and water fed into the system. Table 9 summarizes these values.

Table 9. Analysis of air fed into the system.Component Flow rate (kmol/min)

N2 31.90060964O2 8.479908892

Water 0.635196938Total 41.01571547

Table 10. Summary of feed air flow rates.Flow rates

41.01571547 kmol/min855.9560138 scfm922.0796639 m3/min

PROBLEM 5. Estimate the flow rate (kg/min and kmol/min) of each component and the composition (mole fractions) of the furnace flue gas. (Ignore the fly ash in calculating mole fractions.) 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. Analysis of the flue gas.Componen

tkmol/min Mole Fraction MW kg/min Composition

CO2 6.261448793 0.147173013 44.01 275.5663614 C in coal

H2O 3.110444463 0.073109834 18.02 56.05020922H2O in air + H in

coal

N2 31.95771171 0.751154065 28.02 895.4550822N2 in air + N in

coalSO2 0.109136264 0.002565207 64.07 6.992360461 S in coal

O2 1.106075073 0.025997881 32 35.39440233O2 fed - O2

requiredTotal 42.54481631 1 1269.458416

Table 12. Compositionof the Flue Gas.

Component kmol/minMole

Fraction kg/min

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CO2 6.261448793 0.147173013 275.5663614

H2O 3.110444463 0.073109834 56.05020922

N2 31.95771171 0.751154065 895.4550822

SO2 0.109136264 0.002565207 6.992360461

O2 1.106075073 0.025997881 35.39440233

Total 42.54481631 1 1269.458416

The fly ash removed by the electrostatic precipitator (EP) is determined by:

Fly ash before EP = (0.8)*(7.2 kg/min) = 5.76 kg/minFly ash removed by EP = (0.999)*(5.76 kg/min) = 5.75424 kg/min

Therefore the fly ash removed by the electrostatic precipitator is 5.754 kg/min.

PROBLEM 6: The system may be assumed to meet the standard of 90% removal of the SO2

released upon combustion.a. Determine the flow rate (kg/min and kmol/min) of each component in the cleaned flue gas

leaving the scrubber.b. Determine the flow rate (kg/min) of slurry entering the scrubber.c. Estimate the solid-to-liquid mass ratio in the slurry leaving the scrubber.d. Estimate the feed rate (kg/min) of fresh ground limestone to the blending tank.e. What are the flow rates (kg/min) of inerts, CaSO3, CaCO3, fly ash, 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. Estimate the rate (kg/min, L/min) at which the filtrate is recycled to the blending tank. At what rate (kg/min, 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. Table 13 summarizes the analysis of the gas entering the scrubbers.

Table 13. Analysis of the gas entering the scrubber.

Component kmol/min Mole Fraction

MW kg/min

CO2 3.1307243960.14717301

344.01

137.783181

H2O 1.5552222310.07310983

418.02

28.0251046

N2 15.978855860.75115406

528.02

447.727541

SO2 0.0545681320.00256520

764.07

3.49618023

O2 0.5530375360.02599788

132

17.6972012

Fly Ash - - - 0.00288

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Total 21.27240815 1634.73208

8Among the components in the flue gas leaving the scrubber, only nitrogen and oxygen will remain constant. Sulfur dioxide and Carbon dioxide are both consumed and used in another reaction or generated. 90% of the sulfur dioxide is absorbed and reacted with calcium carbonate forming the added carbon dioxide (1:1 ratio). All of the fly ash is also removed here. Meanwhile, 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.

Table 14. Analysis of the gas leaving the scrubber.

Component kmol/minMole

Fraction MW kg/min

CO2 3.1798357150.13987270

144.01 139.94457

H2O 3.016597660.14180833

618.02

54.3590898

N2 15.978855860.70286830

228.02

447.727541

SO2 0.0054568130.00024003

164.07

0.34961802

O2 0.5530375360.02432668

332

17.6972012

Fly Ash 0 0 (-) 0Total 22.73378358 660.07802

The amount of slurry fed into the scrubber is calculated as follows:

Amount of liquid in slurry = (15.2 kg liquid/kg inlet gas)*(634.732088 kg gas/min) = 9,647.93 kg/min

Amount of solids in the slurry = (9,647.93 kg/min)*(1 solid/9 liquid in slurry) = 1,071.99 kg/min

Total amount of slurry = 9,647.93 kg/min + 1,071.99 kg/min= 10,719.92 kg/min

Therefore, the total amount of slurry in the scrubber is 10,719.92 kg/min.

To get the solid-to-liquid mass ratio of the slurry leaving the scrubber:

Table 15. Analysis of the Reaction of Sulfur dioxide and Calcium Sulfite.kmol/min MW kg/min

SO2 absorbed 0.049111319 64.07 3.14656221CaSO3 formed 0.049111319 120.17 5.9017072

CaCO3

consumed 0.049111319 100.09 4.91555192

Added solids 0.986155285

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Amount of liquid in the slurry = 9,647.93 kg/minCaCO3 in liquid = 0.002 kg/100 kg waterCaSO3 in liquid = 0.003 kg/100 kg water

The water in the liquid is determined using the following equation (x = amount of water):X + 0.002X/100 + 0.003X/100 = 9647.93

Solving for X: X = 9647.517624 kg/min

Therefore, the amount of added liquid due to calcium carbonate and calcium sulfite are:

Amount of CaCO3 dissolved in liquid = 0.002*9647.52/100 = 0.1929503652 kg/minAmount of CaSO3 dissolved in liquid = 0.003*9647.52/100 = 0.289425529 kg/min

Therefore, the mass ratio of solid-to-liquid slurry leaving the scrubber is:

Solid-to-Liquid Ratio = (1071.99197 + 0.986155)/(9647.517624 + 0.19295903652 + 0.289425529) = 0.111207765

The solid-to-liquid mass ratio of the slurry leaving the scrubber is 0.111207765.

Since there is 5.2% excess CaCO3 required to react with the absorbed SO2, the feed rate of freshly ground limestone is given by:

Amount of limestone = 0.9*(0.0546 kmol SO2)&(1.052 kmol CaCO3/ 1 kmol SO2)*(1 kg limestone / 0.921 kg CaCO3)*(100.09 kg CaCO3/mol)

= 0.56147238 kg

Therefore, the amount of limestone feed is 5.6147 kg/min.

Since inerts leave the system only in the wet solids:

Mass of inerts in the wet solids = 5.6147 kg limestone/min*(0.079 kg inerts/0.921 kg CaCO3)=0.4816104 kg inerts/min

Therefore, the amount of inerts is 0.482 kg/min.

Table 16. Analysis of wet solids obtained from the filter.

Component kg/minMass

FractionInerts 0.481610402 0.040558352

CaCO3 solids 0.2556087 0.021525838CaCO3 liquid 0.0001 8.4214E-06CaSO3 solids 5.901707203 0.497006532CaSO3 liquid 0.0002 1.68428E-05

Fly Ash 0.00288 0.000242536H2O 5.2324 0.440641477

Total 11.8745063 1

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From the solid-to-liquid ratio calculated earlier, the slurry leaving the absorber is 11 wt% solids. Therefore, the total solids sent to the filter is given by Table 16. The mass of liquid in that slurry is:

Mass of liquid in slurry to filter = 50.9 kg/min.The recycle flow rate is the flow rate of liquid in the slurry minus the flow rate of liquid in the wet solids.

Mass of recycle = 45.6 kg/minVolume of recycle = 45.1 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 = 33.52 kg/minVolume of make-up water = 33.52 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.e., none is lost to the surroundings), estimate the rate of steam generation in the power cycle.

Using the energy balance around the furnace and assuming that no heat is lost to the surroundings, the rate at which heat is removed can be determined. Table 17 shows the analysis of the steam coming in and out of the furnace, whose values are used for the terms in the energy balance.

Table 17. Analysis of steam in and out of the furnace.IN OUT

Component kmol/min kJ/mol kJ/min kmol/min kJ/mol kJ/minCoal 100 0 0 0 0 0CO2 0 0 0 6.359671431 12982 82561.2545

H2O vapor 0.635196938 54078 34350.18001 3.01659766 54623 164775.614H2O liquid 0.279069832 0 0 0 0 0

N2 31.90060964 8571 273420.1252 31.95771171 9024 288386.39SO2 0 0 0 0.010913626 13635 148.807297O2 8.479908892 8950 75895.18459 1.106075073 9434 10434.7122Ash (-) (-) (-) 5.76 281 1618.56

Bottom Ash (-) (-) (-) 1.44 806 1160.64Total 383665.4898 549085.979

From this, using the energy balance:

Table 18. Summary for the rate of heat removal.Kmol or kg H (kJ/min)

Coal 100 30780

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In 41.2947853 383665.4898Out 42.4509695 549085.9785

From here, using the energy balance equation, Q = -4,387,848.097 kJ/min.To get the mass of steam generated:

Table 19. Data for interpolation.T (°C) 38 50 75 500 540 550221.2 bar 175.928 228.2 331.7 3210 3338 3370250 bar 181.116 230.7 334 3166 3302.8 3337241 bar 179.4948 3313.8

Table 20. Summary of interpolation.T (°C) P (bar) H (kJ/kg)

Out 540 241 3313.8In 38 241 183.2

The mass of steam generated is calculated using the following equation:

From Table 20, HΔ steam = 3313.8 kJ/kg – 183.2 kJ/kg = 3130.6 kJ/kg

Therefore, the mass of steam generated = 4387848.097 kJ/min / 3130.6 kJ/kg = 1401.60 kg/min.

The mass of steam generated is 1401.60 kg/min.

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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. 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.

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, as done in Problem 7.

For 5% and 25% excess air, the same calculations were done until solving for the heat evolved.

Table 21. Analysis of steam coming in and out of the furnace (5% excess air).IN OUT

Component kmol/minkJ/mol

kJ/min kmol/min kJ/mol kJ/min

Coal 100 0 0 0 0 0

CO2 0 0 06.35967143

112982 82561.2545

H2O vapor0.57996242

254078

31363.20783

3.01659766 54623 164775.614

H2O liquid0.27906983

20 0 0 0 0

N229.1266435

98571

249644.4622

29.18374566

9024 263354.121

SO2 0 0 00.01091362

613635 148.807297

O2 7.74252551 895069295.6033

20.36869169

19434 3478.23741

Ash (-) (-) (-) 5.76 281 1618.56Bottom Ash (-) (-) (-) 1.44 806 1160.64

Total350303.273

3517097.234

Table 22. Summary of HHV values and the amount of heat evolved (5% excess air).Kmol H

Coal 100 30780In 37.72820135 350303.2733

Out 38.93962006 517097.234

Using the same equation in solving for Q, the Q here is equal to -3,841,257.40 kJ/min.

Table 23. Analysis of steam coming in and out of the furnace (25% excess air).IN OUT

Component kmol/min kJ/mol kJ/min kmol/min kJ/mol kJ/minCoal 100 0 0 0 0 0CO2 0 0 0 6.35967143 12982 82561.2545

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1

H2O vapor0.69043145

454078

37337.15218

3.01659766 54623 164775.614

H2O liquid0.27906983

20 0 0 0 0

N2 34.6745757 8571297195.788

334.7316777

79024 313418.66

SO2 0 0 00.01091362

613635 148.807297

O29.21729227

48950

82494.76586

1.843458455

9434 17391.1871

Ash (-) (-) (-) 5.76 281 1618.56Bottom Ash (-) (-) (-) 1.44 806 1160.64

Total417027.706

4581074.723

Table 24. Summary of HHV values and the amount of heat evolved (25% excess air).Kmol H

Coal 100 30780In 44.86136926 417027.7064

Out 45.96231894 581074.723

From the values in the table above, Q = -4,921,107.823 kJ/min.

The table below summarizes the heat evolved as well as the mass of steam generated for each percentage of excess air.

Table 25. Summary of comparison.Q (kJ/min) Mass of steam generated (kg/min)

5% excess air -3841257.399 1227.00357715% excess air -4387848.097 1401.59972425% excess air -4921107.823 1571.937591

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, which should be avoided.

PROBLEM 9: Determine the temperature of the flue gas as it leaves the heat exchanger (air preheater) following the boiler. Estimate this value for the two alternative air flow rates corresponding to 5% and 25% excess oxygen.

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.

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. Summary of conversions for SO2 in the scrubbers.SO2

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Scrubber 1 0.005456813 kmol/minScrubber 2 0.005456813 kmol/min

Total0.010913626 kmol/min0.699236046 kg/min6.99236E+11 ng/min

Q 4387848.097 kJ/min

SO2 from scrubbers/Q159357.3959 ng/kJ159.3573959 ng/J

From the table, it can be seen that there is 159.36 ng SO2/J in the scrubber flue gases. Since it is less than 520 ng SO2/J, the flue gas for this system has met the standard EPA limit.PROBLEM 11: Power plants of the type described here operate with an efficiency of about 39%; that is, for each unit of heat released with the combustion of coal, 0.39 unit is converted to electrical energy. From this efficiency and the specified power output of 500 MWe, determine the following:

a. The coal feed rate (kg/h).b. The air feed rate (kmol/min, standard cubic meters/min, cubic meters/min).c. The flow rate of each component in the gas leaving the furnace (kmol/min, kg/min).d. The rate of steam generation (kg/h).

With an efficiency of about 39% and a specified power output of 500 MWe, the scale factor is first determined in order to solve for the desired feed rates:

Table 27. Determining the scale factor for the scale-up calculations.Power output

(MWe)500

Efficiency 0.39

Q 4387848.097 kJ/minQactual 4615384615 kJ/hr

Scale Factor 1051.856061 min/hr

Using this value of the scale factor, all computed flow rates are then scaled up into the actual flow rates for the system.

Table 28. Summary of the Actual Flow Rates for the System.Flow Rates Basis Actual

Dry Coal 100 105185.6061 kg/hrWet Coal 104.8 110234.5152 kg/hr

Air 41.01571547 43142.62892 kmol/hr855.9560138 900342.521 scfh922.0796639 969895.0833 m3/hr

Flue gas* 42.4509695 44652.30957 kmol/hr1271.55695 1337494.885 kg/hr

Steam 1401.599724 1474281.165 kg/hr

*Flue Gas Composition Basis Actual

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Component MW kmol/min kmol/hr kg/minCO2 44.01 6.359671431 6689.45894 294403.088H2O 18.02 3.01659766 3173.026532 57177.93811N2 28.02 31.95771171 33614.91276 941889.8556

SO2 64.07 0.010913626 11.47956412 735.4956732O2 32 1.106075073 1163.431769 37229.81662

Ash - 5.76 6058.690912Total 42.4509695 44652.30957 1337494.885

PROBLEM 12: For the required coal feed rate, 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, the slurry flow rates entering and leaving the scrubber, the flow rates of gas (kg/h and m3/min) entering and leaving the scrubber, and the flow rates (kg/h) of fresh water and limestone fed to each blending tank.

The calculations for Problem 6 were used again using the values obtained in Problem 11. The table below summarizes the actual flow rates of the gas entering and leaving the scrubbers.

Table 29. Actual Flow Rates for the Flue Gases entering and leaving the scrubbers.

Gas entering the scrubber

Component kmol/min Mole Fraction MW kg/minCO2 3344.72947 0.14981216 44.01 147201.544H2O 1586.513266 0.071060748 18.02 28588.9691N2 16807.45638 0.752814649 28.02 470944.928

SO2 5.73978206 0.000257088 64.07 367.747837O2 581.7158847 0.026055355 32 18614.9083

Fly Ash 0 0 (-) 3.02934546Total 22326.15478 1 665721.126

Gas leaving the scrubber

Component kmol/min Mole Fraction MW kg/minCO2 3349.895274 0.138653856 44.01 147428.891H2O 3420.48918 0.141575773 18.02 61637.215N2 16807.45638 0.6956691 28.02 470944.928

SO2 0.573978206 2.37572E-05 64.07 36.7747837O2 581.7158847 0.024077514 32 18614.9083

Fly Ash 0 0 (-) 0Total 24160.1307 1 0 698662.717

Basis ActualSlurry feed to each scrubber

10719.9197 11275812.52 kg/hr

Slurry exit from each scrubber

10731.79421 11288302.79 kg/hr

Limestone to blending tank

5.614723797 5905.881257 kg/hr

Flue Gas feed to each scrubber

21.27240815 22375.51145 kmol/h

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935.3557 983859.5623 m3/hr

Flue Gas exit from each scrubber

22.73378358 23912.66805 kmol/h

1079.582 1135564.87 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.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; (2) burning natural gas and blending the combustion products with the scrubbed flue gas; 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.

a. Give a reason for rejecting Alternative 1.b. In evaluating Alternative 2, 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. If the combustion products are blended with the cleaned flue gas, at what rate would methane would have to be burned to raise the stack-gas temperature to the desired value?

c. 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 to reheat the scrubbed flue gas makes controlling the temperature of the bypassed gas more difficult. It can also directly release particulates as well as sulfur dioxide which is being prevented. Moreover, this requires a more accurate analysis of sulfur dioxide which is not found in this study. Therefore, the first alternative is rejected.

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.

Table 30. Analysis of the scrubbed flue gas and stack gas.Scrubbed Flue Gas Stack Gas

Component kmol/hr kJ/mol kJ/h kmol/h kJ/mol kJ/hrCO2 3349.895274 -392444 -1314646301 3349.895274 -391396 -1311135611H2O 3420.48918 -240855 -823841921.4 3420.48918 -239969 -820811368N2 16807.45638 815 13698076.95 16807.45638 1602 26925545.12

SO2 0.573978206 -392369 -225211.2547 0.573978206 -391252 -224570.121O2 581.7158847 827 481079.0366 581.7158847 1633 949942.0397

Total 24160.1307 -2124534278 24160.1307 -2104296062

Q 20238215.97 kJ/hr

From the calculations, the heat evolved is much greater than the value calculated beforehand. Therefore, an added amount of coal is needed to reach this amount of heat.

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Mass of additional coal = (20,238,215.97 kJ/hr) / (43,878.481 kg coal/hr) = 461.233 kg/hr

Therefore, an additional 461.233 kg of coal per hour is needed to satisfy the heat evolution solved.

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che132 - case study writeup

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