# Patent

## Title
Sulfur trioxide conditioning system control algorithm

## Abstract

Optimizing fly ash resistivity by controlling concentration of sulfur trioxide (SO3) in flue gas by the use of an algorithm.

## Background
RELATED APPLICATION
This application is related to provisional application 60/338,152, filed Dec. 6, 2001, the contents of which are herein incorporated by reference.


FIELD OF THE INVENTION
The herein disclosed invention finds applicability in producing optimum fly ash resistivity in flue gas.
BACKGROUND OF THE INVENTION
Many utilities now burn a variety of coals at their fossil fuel plants. This practice is growing for several reasons, including (1) the need to lower SO2 emissions by burning low-sulfur coals and (2) the need to reduce fuel costs to enhance their competitive position. Frequently, these coal changes have adverse affects on ESPs (ESP=electrostatic precipitator). Low-sulfur coals produce high resistivity ash that is difficult to collect in an electrostatic precipitator (the technology most commonly used to control particulate emissions from coal-fired power plants). Inexpensive coals are frequently variable in their properties and sometimes high in ash or low in sulfur. Conditioning the ash with SO3 before the ash enters a precipitator, can lower ash resistivity and improve ESP performance. In fact, this well established technology is used at several hundred plants both here and abroad to control fly ash resistivity in low-sulfur or variable-sulfur coals. While commercial conditioning systems are relatively reliable, the controls for these are not sophisticated, and this lack of sophistication can result in non-optimum ESP performance, and sometimes excess SO3 addition rates (and emissions).
The inventors have developed a correlation between certain ESP electrical operation parameters and fly ash resistivity. In particular, the inventors have found it to be possible to monitor the current density in an ESP electrical section, and using this number, estimate the resistivity of the fly ash in that electrical section. Using these correlations makes it possible to determine if the resistivity in this section is at an optimum level or not. Further, the inventors have also participated in the development of correlations between fly ash resistivity and the flue gas SO3 concentration needed to produce optimum fly ash resistivity. These correlations can be combined (as described below) to produce a superior SO3 conditioning system control algorithm.
Current SO3 conditioning systems use a preset rate of SO3 addition that is only adjusted for unit load. The invention, described herein, uses a unique combination of calculations to provide a rate of addition that is based on actual ESP operating data. These data, which can easily be obtained from modem ESP controls, are both real time and continuous. Hence, the new control algorithm is capable of producing an optimum rate of SO3 addition when coal and ash properties are changing.
The primary application will be at utility plants that use SO3 conditioning to improve ESP performance. These plants are located both here and abroad. In addition, SO3 is used in some industrial applications, and the new SO3 control algorithm could be used at those plants as well.
SUMMARY OF INVENTION
The herein disclosed invention is directed to a process for treating fly ash found in flue gas to produce effective fly ash electrical resistivity comprising employing an algorithm to determine the optimum amount of sulfur trioxide (SO3) to be added to the flue gas. The sulfur trioxide can result from the burning of coal, or the sulfur trioxide can result from the burning of coal plus the extrinsic addition of sulfur trioxide. The process of this invention involves an algorithm which takes into account 1) flue gas SO3 concentration, 2) initial fly ash resistivity, 3) electrostatic precipitator (ESP) current densities, 4) flue gas temperature and moisture and 5) fly ash composition.
Also, embraced by this invention is a process for treating fly ash found in flue gas to produce effective fly ash resistivity comprising the following steps:
Step 1. Obtain the proximate ultimate analyses of coal being burned in boiler and ash mineral analysis for this coal;
Step 2. Determine the average temperature of flue gas entering the electrostatic precipitator (ESP);
Step 3. Estimate SO3 background level in the flue gas using correlation relating flue gas SO3 to coal type and coal sulfur content.
Step 4. Calculate the base ash resistivity using empirical equations relating ash resistivity to ash composition, flue gas moisture and flue gas temperature.
Step 5. Use a correlation relating the base fly ash resistivity and flue gas SO3 concentration to determine the flue gas SO3 concentration needed to produce the optimum fly ash resistivity.
Step 6. Subtract the background SO3 concentration from the needed SO3 concentration from the needed SO3 that must be added to the flue gas to produce the optimum fly ash resistivity and
Step 7. Send rate of addition signal to the controls that operate the SO3 conditioning system.
Another method encompassed by the invention involves determining a most effective injection rate for SO3 into flue gas comprising the following steps:
Step 1. Obtain the proximate and ultimate analysis of the coal being burned in the boiler and the ash mineral analysis for the coal,
Step 2. Determine the average temperature of the flue gas entering the ESP from plant instrumentation,
Step 3. Estimate SO3 background level in the flue gas using correlation relating flue gas SO3 to coal type and coal sulfur content,
Step 4. The secondary current applied to the electrostatic precipitator is obtained from the controls for each transformer-rectifier set that is powering the precipitators,
Step 5. Determine effective fly ash resistivity level in the ESP using a correlation that relates fly ash resistivity to ESP current density for each electrical field, average the results to produce an effective resistivity for the ESP.
Step 6. a. If indicated ash resistivity is equal to or less than optimum resistivity, decrease rate of injection by x percent where x is between 5 and 25, or

    
    
        b. if indicated ash resistivity is greater than optimum resistivity, increase rate of injection by x percent where x is between 5 and 25.
    
    


Step 7. Repeat Step 6 until indicated fly ash resistivity passes through optimum resistivity point and then set rate of injection at a point in the range bounded by the levels calculated in the last two interactions, and then
Step 8. Every y minutes, where y is number between 5 and 30, restart the process beginning at Step 2.
A still further method involves a method for determining a most effective injection rate for SO3 into flue gas comprising the following steps,
Step 1. Obtain the proximate and ultimate analysis of the coal being burned in the boiler and the ash mineral analysis for the coal,
Step 2. Determine the average temperature of the flue gas entering the ESP from plant instrumentation,
Step 3. Estimate SO3 background level in the flue gas using correlation relating flue gas SO3 to coal type and coal sulfur content,
Step 4. The secondary current applied to the electrostatic precipitator is obtained from the controls for each transformer-rectifier set that is powering the precipitators,
Step 5. Determine effective fly ash resistivity level in the ESP using a correlation that relates fly ash resistivity to ESP current density for each electrical field. Average the results to produce an effective resistivity for the ESP. If this resistivity is not close to, or lower than, the optimum range, proceed with Step 6; otherwise, go to Step 10.
Step 6. Use a correlation relating fly ash composition and flue gas temperature and SO3 concentration to fly ash resistivity to determine the flue gas SO3 concentration to needed to produce the optimum fly ash resistivity,
Step 7. Subtract the background SO3 from the needed SO3 concentration from Step 6 to determine the amount of SO3 that must be added to the flue gas to produce the optimum fly ash resistivity.
Step 8. Send rate of additional signal to the controls that operate the SO3 conditioning system.
Step 9. Repeat Steps 4 and 5.
Step 10. a. If indicated ash resistivity is equal to or less than optimum resistivity, decrease rate of injection by x percent where x is between 5 and 25, or

    
    
        b. if indicated ash resistivity is greater than optimum resistivity, increase rate of injection by x percent where x is between 5 and 25.
    
    


Step 11. Repeat Step 10 until indicated fly ash resistivity passes through optimum resistivity point and then set rate of injection at a point in the range bounded by the levels calculated in the last two interactions, and then
Step 12. Every y minutes, where y is number between 5 and 30, restart the process beginning at Step 2.


DETAILED DESCRIPTION OF THE INVENTION
The herein disclosed invention involves completing a sequence of unique calculations that result in the estimation of the amount of SO3 that must be added to flue gas to produce optimum fly ash electrical resistivity. This sequence of steps is as follows:
“Typical” Starting Conditions:
a. Low flue gas SO3 concentration measured at the ESP inlet—0 to 4 ppm. SO3—example number=3.5.
b. Moderate to high fly ash resistivity—8×1010 ohm-cm to 5×1012 ohm-cm.
c. Low ESP power level characterized by low average current densities.
For example, in a three-field electrostatic precipitator the average current densities in the inlet field might be 9.13 na/cm, in the middle field it might be 12.41 na/cm2 and in the outlet field, it might be 15.19 na/cm2. These current densities correspond to a fly ash resistivity of 1.0×1011 ohm-cm and this level of resistivity is too high to allow optimum ESP performance (see Table 1).
Desired “End” Conditions:
a. Increased flue gas SO3 measured at ESP inlet—from 2 to 12 ppm, depending on flue gas temperature, flue gas moisture, and fly ash composition.
b. Optimum fly ash resistivity—8×109 ohm-cm to 4×1010 ohm-cm, depending on ESP collection and reentrainment characteristics—example number 1×1010 ohm-cm.
c. High ESP power levels as indicated by current density levels.
For example, when the correct level of SO2 has been added to the flue gas, the average current densities in the ESP would increase to 27.67 na/cm2 in the inlet field, 33.50 na/cm2 in the middle field and 39.50 na/cm2 in the outlet field. The current densities correspond to a fly ash resistivity of 1×1010 ohm-cm and this level of resistivity should produce optimum ESP performance (see Table 1).







TABLE 1







Typical Per-Field Current Densities for a Range of Resistivies













FIRST
SECOND
THIRD
FOURTH
FIFTH



FIELD 1
FIELD 2
FIELD 3
FIELD 4
FIELD 5


PARAMETER 1
6.255
5.839
5.697
5.018
4.718


PARAMETER 2
0.4813
0.4314
0.4105
0.3405
0.3036


RESISTIVITY
CURRENT
CURRENT
CURRENT
CURRENT
CURRENT


(ohm * cm)
na/cm2
na/cm2
na/cm2
na/cm2
na/cm2















1.00E+10
27.67
33.50
39.08
41.02
48.08


2.00E+10
19.82
24.84
29.41
32.40
38.96


4.00E+10
14.20
18.42
22.12
25.59
31.57


6.00E+10
11.68
15.46
18.73
22.29
27.91


8.00E+10
10.17
13.66
16.64
20.21
25.58


1.00E+11
9.13
12.41
15.19
18.73
23.90


2.00E+11
6.54
9.20
11.43
14.79
19.36


4.00E+11
4.69
6.82
8.60
11.68
15.69


6.00E+11
3.86
5.73
7.28
10.18
13.87


8.00E+11
3.36
5.06
6.47
9.23
12.71


1.00E+12
3.02
4.59
5.90
8.55
11.88


2.30E+12
2.02
3.21
4.19
6.44
9.23


4.00E+12
1.55
2.53
3.34
5.33
7.80


6.00E+12
1.27
2.12
2.83
4.65
6.90


8.00E+12
1.11
1.87
2.51
4.21
6.32


1.00E+13
1.00
1.70
2.29
3.90
5.90





Note:


Resistivities and current densities above the line are in the range that will produce optimum ESP performance. Resistivities and current densities below the line are in the range that will produce suboptimum ESP performance






This invention has several methods to determine the rate of SO3 addition that will produce the optimum level of fly ash resistivity and hence optimum ESP performance. The first method does not require data feed back from the ESP, while the second method does.
Method 1 is as follows:
Step 1. Obtain the proximate and ultimate analyses of the coal being burned in the boiler and the ash mineral analysis for this coal. Table 2 contains examples of typical analyses.










TABLE 2







Example Coal Composition











As Received
Example Fly Ash Composition


Ultimate Analysis
As Constituents


(%)
(%)













Carbon
68.00
LiO2
0.01


Hydrogen
3.86
Na2O
0.96


Oxygen
6.00
K2O
2.43


Nitogen
1.00
MgO
0.78


Sulfur
2.20
CaO
2.62


Moisture
3.60
Fe2O3
7.76


Ash
16.34
Al2O3
17.85


SUM
100.00
SiO2
61.00




TiO2
0.62




P2O5
0.55




SO3
2.43




SUM
97.01









Step 2. Determine the average temperature of the flue gas entering the ESP from plant instrumentation. For example, the instrumentation indicates the temperature of the flue gas entering the RSP is 291° F.
Step 3. Estimate SO3 background level in the flue gas using correlation relating flue gas SO3 to coal type and coal sulfur content. The SO3 concentration is calculated as a percentage of SO2 in the flue gas which can be determined from a combustion calculation using the coal analysis and flue gas O2 or CO2 or if the flue gas SO2 is available from plant instruments, this number can be used in the SO3 calculation. Using standard, well known chemical formulas and procedures, that calculation is as follows if the assumption for no excess air is used.
A. Calculation of Combustion Products, Air, and O2 for 100% Combustion.


























Required for










combustion



Ultimate






Moles/100 lb fuel


Coal
analysis

Molecular

Moles per


at 100% total air
















Constiuent
lb/100 lb fuel

weight

100 lb fuel

Multipliers1
O2
Dry Air



















C
68.00
÷
12.01
=
5.662
×
   1.0 and 4.76
5.662
26.951


H2
3.86
÷
2.02
=
1.911
×
  0.50 and 2.38
0.956
4.548


O2
6.00
÷
32.00
=
0.188
×
−1.00 and −4.76
−0.188
−0.895


N2
1.00
÷
28.01
=
0.036


S
1.20
÷
32.06
=
0.037
×
  1.00 and 4.76
0.037
0.176


H2O
3.60
÷
18.02
=
0.200


Ash
16.34

—

—



Sum
100.00



8.034


6.467
30.780









A correction for excess air, which is always added to the furnace to ensure complete combustion is next made as follows.
B. Calculation of Air and O2 for 30% Excess Air (Typical Excess Air Level).
















Required for





Combustion




moles/100 lb fuel




at 30% excess air












O2
Dry air















O2 and air × 130/100 total
8.407
40.014



Excess air = 40.014 − 30.780
—
9.234



Excess O2 = 8.407 − 6.467
1.940
—










Using the values from these two calculations, the final composition of the flue gas is calculated, again using established and well known formulas and procedures.
C. Calculation of Flue Gas Composition.










Products of Combustion















Total




Flue gas


moles/100
% by volume
% by volume


Constituent
Combustion/Fuel/Air

lb fuel
wet basis
dry basis















CO2
5.662
=
5.662
13.406
14.412


H2O
1.911 + 0.200 + 0.838a
=
2.949
6.983
—


SO2
0.037
=
0.037
0.088
0.094


N2
0.036 + 31.611b
=
31.647
74.931
80.555


O2
1.940
=
1.940
4.593
4.938


Sum wet


42.235


Sum dry =
42.235 − 2.949

39.286





aMoles H2O in air = (40.014 × 29 × 0.013) ÷ 18 = 0.838


bMoles N2 in air = (40.014 × 0.79) = 31.611






The critical numbers from these calculations are the SO2 concentrations:

    
    
        0.088%, wet basis, and
        094%, dry basis.

The moisture concentration, 6.98% is also critical Once these numbers are known, the native SO3 concentration in the flue gas can be calculated as follows:

    
    


The SO2 concentration dry (the resistivity concentration in this example uses the equivalent SO3 concentration of “dry” flue gas) is equal to 0.094%. the appropriate SO2 to SO3 conversion factor for this coal is 0.4% so the approximate SO3 concentration is:

0.00094×0.004=3.76 PPM (dry basis)

As an alternative, the flue gas SO2 concentration can be obtained from the plant's Continuous Emissions Monitoring (CEM) system, corrected for flue gas moisture concentration using factors from the combustion calculation and multiplied by the factor 0.004 to estimate to inherent or background SO3 concentration. For other coals, for example, western coals, the appropriate conversion factor is 0.001 and for Powder River Basin Coals, the conversion factor is 0.005 (as opposed to 0.004).

Step 4. Calculate the base ash resistivity using empirical equations relating ash resistivity to ash composition, flue gas moisture and flue gas temperature. The Bickelhaupt equations are an example of relationships that can be used for this calculation. This particular calculation is made using the ash mineral analysis from Table 2 and the moisture and SO3 calculations from step 2 using the following sequence of substeps:

    Substep 1: Normalize the weight percentages to sum 100% by dividing each specified percentage by the sum of the specified percentages.
    Substep 2: Divide each oxide percentage by the respective molecular weight to obtain the mole fractions.
    Substep 3: Divide each mole fraction by the sum of the mole fractions and multiply by 100 to obtain the molecular percentages as oxides.
    Substep 4: Multiply each molecular percentage by the decimal fraction of cations in the given oxide to obtain the atomic concentrations.

These substeps are illustrated for the example ash in the following table.



























Atomic



Specified
Normalized
Molecular
Mole
Molecular
Cationic
Concentration


Oxide
Weight %
Weight %
Weight
Fraction
Percentage
Fraction
Of Cation






















Li2O
0.01
0.01
29.88
0.00034
0.024
0.67
0.016


Na2O
0.96
0.99
61.98
0.01600
1.116
0.67
0.744


K2O
2.43
2.50
94.20
0.02654
1.854
0.67
1.236


MgO
0.78
0.80
40.31
0.01985
1.387
0.50
0.694


CaO
2.62
2.70
56.08
0.04815
3.364
0.50
1.682


Fe2O3
7.76
8.00
159.70
0.05009
3.500
0.40
1.400


Al2O3
17.85
18.40
101.96
0.18046
12.608
0.40
5.043


SiO2
61.00
62.89
60.09
1.04660
73.123
0.33
24.368


TiO2
0.62
0.64
79.90
0.00801
0.560
0.33
0.186


P2O5
0.55
0.57
141.94
0.00402
0.281
0.29
0.080


SO3
2.43
2.50
80.06
0.03123
2.183
0.25
0.546


Sum
97.01
100.00

1.43129
100.000









Now that the atomic concentrations of the critical ash mineral constituents are known, the rest of the calculation proceeds by calculating three separate resistivities, the volume resistivity, ρv, the surface resistivity, ρs, and the acid resistivity, ρa. These three resistivities are then combined to give the net resistivity of the ash using the parallel resistance formula. For the example coal, the calculation proceeds as follows using the following formulas and definitives.
Bickelhaupt Equations


ρv=exp[−1.8916 ln X−0.9696 ln Y+1.234 ln Z+3.62876−(0.069078)E+9980.58/T]

ρs=exp[27.59774−2.233348 ln X−(0.00176)W−(0.069078)E−(0.00073895)(W)exp(2303.3/T)]

ρa=exp[85.1405−(0.708046)CSO3−23267.2/T−(0.069078)E], for z<3.5% or K>1.0%

ρa=exp[59.0677−(0.854721)CSO3−13049.47/T−(0.069078)E], for z<3.5% or K>1.0%

1/ρvs=1/ρv+1/ρs

1/ρvsa=1/ρvs+1/ρa

ρv=volume resistivity (ohm-cm)
ρs=surface resistivity (ohm-cm)
ρa=adsorbed acid resistivity (ohm-cm)
ρvs=volume and surface resistivity (ohm-cm)
ρvsa=total resistivity (ohm-cm)
X=Li+Na percent atomic concentration
Y=Fe percent atomic concentration
Z=Mg+Ca percent atomic concentration
K=K percent atomic concentration
T=absolute temperature (K)
W=moisture in flue gas (volume %)
CSO3=concentration of SO3 (ppm, dry)
E=applied electric field (kV/cm)
Using the above definitions, equations and calculated values, the calculation proceeds for the example case as follows:
X=0.016+0.744=0.76
Y=1.40
Z=0.694+1.682=2.376
K=1.236
T=417 (Example gas temperature 291° F.)
W=6.983
CSO3=(from Calculation 2)3.76 ppm, dry
E=10(typical electric field value)

ρv=exp[−1.8916 ln(0.76)−0.9696 ln(1.40)+1.237 ln(2.376)+3.62876−(0.069078)(10)+9980.58/417]

=1.636×1012 ohm-cm

ρs=exp[27.59774−2.23348 ln(0.76)−(0.00176)(6.983)−(0.069078)(10)−(0.00073895)(6.983)exp(2303.0/417)]

=2.392×1011 ohm-cm

ρa=exp[85.1405−(0.708046)(3.76)−23267.2/417−(0.069078)(10)]

=1.939×1011 ohm-cm

1/ρvs=1/1.636×1012+1/2.392×1011=4.792×10−12

ρvs=2.1×1011 ohm-cm

1/ρvsa=1/4.792×10−12+1/1.939×1011=9.949×10−12

ρvsa=1.0 ×1011 ohm-cm

In this example, the calculated resistivity is found to be 1.0×1011 ohm-cm, which is too high for optimum ESP performance, so additional SO3 must be added to the flue gas.
Step 5. Use a correlation relating the base fly ash resistivity and flue gas SO3 concentration to determine the flue gas SO3 concentration needed to produce the optimum fly ash resistivity.
From the preceding step, the relationships between the acid resistivity, surface resistivity, volume resistivity and net ash resistivity are known. Further, it is known that the desirable level of resistivity is 1.0×1010 ohm-cm. Hence, the calculation proceeds as follows:

1/ρvsa=1/ρvs+1/ρa

where ρvsa=1×1010 ohm-cm(the desirable ρ)

ρvs=2.1×1011 ohm-cm(from preceding calculation)





  
    
      
        
          1
          /
          
            ρ
            a
          
        
        =
        
          
            1
            /
            
              ρ
              vsa
            
          
          -
          
            1
            /
            
              ρ
              vs
            
          
        
      
    
  
  
    
      
        =
        
          
            1.0
            ×
            
              10
              10
            
          
          -
          
            2.761905
            ×
            
              10
              12
            
          
        
      
    
  
  
    
      
        =
        
          9.5238
          ×
          
            10
            11
          
        
      
    
  




ρva=1.05×1010 ohm-cm

also from the preceding calculation,

ρa=exp[85.1405−(0.708046)CSO3−23267.2/T−(0.069078)E]

where T=417 (from preceding calculation)

E=10 (from preceding calculation)
hence

1.05×1010=exp[85.1405−(0.708046)CSO3−23267.2/417−(0.069078)(10)]

ln(1.05×1010)=85.1405−(0.708046)CSO3−55.79664−0.69078

23.07464109=28.652−(0.708046)CSO3

(0.708046)CSO3=5.578

CSO3=7.878

Correcting for wet conditions

SO3 needed=7.878×(39.286/42.235)=7.33 ppm

Step 6. Subtract the background SO3 concentration from the needed SO3 concentration from the needed SO3 that must be added to the flue gas to produce the optimum fly ash resistivity.
The combustion calculation results and the background SO3 calculation in step 2, the SO3 concentration is estimated to be 0.00088×0.004=3.52 ppm (wet basis). From the desired level calculation above, the desirable SO3 level=7.33 ppm, hence the difference=7.33−3.52=3.81 ppm. Hence, 3.81 ppm, SO3 must be added to the flue gas to produce the desired level of fly ash resistivity.
Step 7. Send rate of addition signal to the controls that operate the SO3 conditioning system. In this case, the signal should be sent that will cause the SO3 conditioning system to add 3.8 ppm SO3 to the flue gas.
Notice that this procedure uses the equations developed by Dr. Bickelhaupt to relate flue gas composition and ash mineral analysis in the calculations, but any set equations relating flue gas SO3 concentrations and ash mineral analysis to fly ash resistivity could be used. For example, the equations developed by Joe McCain and published in EPRI technical report 1004075 can be used.
This concludes Method 1.
Method 2 is as follows:
The example calculation for this method resumes the same starting conditions that were assumed for method 1. They are as follows:
a. Low flue gas SO3 concentration measured at the ESP inlet—0 to 4 ppm: SO3 —example number=3.5.
b. Moderate to high fly ash resistivity—8×1010 ohm-cm to 5×1012 ohm-cm.
c. Low ESP power level characterized by low average current densities.
For example, in a three-field electrostatic precipitator the average current densities in the inlet field might be 9.13 na/cm2, in the middle field it might be 12.41 na/cm2 and the outlet field might be 15.19 na/cm2. These current densities correspond to a fly ash resistivity of 1.0×1011 ohm-cm and this level of resistivity is too high to allow optimum ESP performance (see Table 1).










Typical Per-Field Current Densities for a Range of Resistivies













FIRST
SECOND
THIRD
FOURTH
FIFTH



FIELD 1
FIELD 2
FIELD 3
FIELD 4
FIELD 5


PARAMETER 1
6.255
5.839
5.697
5.018
4.718


PARAMETER 2
0.4813
0.4314
0.4105
0.3405
0.3036


RESISTIVITY
CURRENT
CURRENT
CURRENT
CURRENT
CURRENT


(ohm * cm)
na/cm2
na/cm2
na/cm2
na/cm2
na/cm2















1.00E+10
27.67
33.50
39.08
41.02
48.08


2.00E+10
19.82
24.84
29.41
32.40
38.96


4.00E+10
14.20
18.42
22.12
25.59
31.57


6.00E+10
11.68
15.46
18.73
22.29
27.91


8.00E+10
10.17
13.66
16.64
20.21
25.58


1.00E+11
9.13
12.41
15.19
18.73
23.90


2.00E+11
6.54
9.20
11.43
14.79
19.36


4.00E+11
4.69
6.82
8.60
11.68
15.69


6.00E+11
3.86
5.73
7.28
10.18
13.87


8.00E+11
3.36
5.06
6.47
9.23
12.71


1.00E+12
3.02
4.59
5.90
8.55
11.88


2.30E+12
2.02
3.21
4.19
6.44
9.23


4.00E+12
1.55
2.53
3.34
5.33
7.80


6.00E+12
1.27
2.12
2.83
4.65
6.90


8.00E+12
1.11
1.87
2.51
4.21
6.32


1.00E+13
1.00
1.70
2.29
3.90
5.90





Note:


Resistivities and current densities above the line are in the range that will produce optimum ESP performance. Resistivities and current densities below the line are in the range that will produce suboptimum ESP performance






As in the Method 1 example calculation, the desired end point is the same. It is described in the following paragraph:
Desired “End” Conditions:
a. Increased flue gas SO3 measured at ESP inlet—from 2 to 12 ppm, depending on flue gas temperature, flue gas moisture, and fly ash composition.
b. Optimum fly ash resistivity—8×109 ohm-cm to 4×1010 ohm-cm, depending on ESP collection and reentrainment characteristics—example number 1×1010 ohm-cm.
c. High ESP power levels as indicated by current density levels.
For example, when the correct level of SO3 has been added to the flue gas, the average current densities in the ESP would increase to 27.67 nA/cm2 in the inlet field, 33.50 na/cm2 in the middle field, and 39.08 na/cm2 in the outlet field. The current densities correspond to a fly ash resistivity of 1×1010 ohm-cm and this level of resistivity should produce optimum ESP performance (see Table 1).
Method 2 uses the following alternative sequence of steps to determine the optimum injection rate for SO3:
Step 1. Obtain the proximate and ultimate analysis of the coal being burned in the boiler and the ash mineral analysis for this coal. Table 2 contains examples of typical analysis.










TABLE 2







Example Coal Composition











As Received
Example Fly Ash Composition


Ultimate Analysis
As Constituents


(%)
(%)













Carbon
68.00
LiO2
0.01


Hydrogen
3.86
Na2O
0.96


Oxygen
6.00
K2O
2.43


Nitogen
1.00
MgO
0.78


Sulfur
2.20
CaO
2.62


Moisture
3.60
Fe2O3
7.76


Ash
16.34
Al2O3
17.85


SUM
100.00
SiO2
61.00




TiO2
0.62




P2O5
0.55




SO3
2.43




SUM
97.01









Step 2. Determine the average temperature of the flue gas entering the ESP from plant instrumentation. For example, the instrumentation indicates the temperature of the flue gas entering the ESP is 291° F.
Step 3. Estimate SO3 background level in the flue gas using correlation relating flue gas SO3 to coal type and coal sulfur content. The SO3 concentration is calculated as a percentage of SO2 in flue gas which can be determined from a combustion calculation using the coal analysis and flue gas O2 or CO2 or if the flue gas SO2 is available from plant instruments, this number can be used in the SO3 calculation. Using standard, well known chemical formulas and procedures, that calculation is as follows if the assumption for no excess air is used.
A. Calculation of Combustion Products, Air, and O2 for 100% Combustion.


























Required for










combustion



Ultimate






Moles/100 lb fuel


Coal
analysis

Molecular

Moles per


at 100% total air
















Constiuent
lb/100 lb fuel

weight

100 lb fuel

Multipliers1
O2
Dry Air



















C
68.00
÷
12.01
=
5.662
×
   1.0 and 4.76
5.662
26.951


H2
3.86
÷
2.02
=
1.911
×
  0.50 and 2.38
0.956
4.548


O2
6.00
÷
32.00
=
0.188
×
−1.00 and −4.76
−0.188
−0.895


N2
1.00
÷
28.01
=
0.036


S
1.20
÷
32.06
=
0.037
×
  1.00 and 4.76
0.037
0.176


H2O
3.60
÷
18.02
=
0.200


Ash
16.34

—

—




Sum
100.00



8.034


6.467
30.780









A correction for excess air, which is always added to the furnace to ensure complete combustion is next made as follows:
B. Calculation of Air and O2 for 30% Excess Air (Typical Excess Air Level).
















Required for





Combustion




moles/100 lb fuel




at 30% excess air












O2
Dry air















O2 and air × 130/100 total
8.407
40.014



Excess air = 40.014 − 30.780
—
9.234



Excess O2 = 8.407 − 6.467
1.940
—










Using the values from these two calculations, the final composition of the flue gas is calculated, again using established and well known formulas and procedures.
C. Calculation of Flue Gas Composition.










Products of Combustion















Total




Flue gas


moles/100
% by volume
% by volume


Constituent
Combustion/Fuel/Air

lb fuel
wet basis
dry basis















CO2
5.662
=
5.662
13.406
14.412


H2O
1.911 + 0.200 + 0.838a
=
2.949
6.983
—


SO2
0.037
=
0.037
0.088
0.094


N2
0.036 + 31.611b
=
31.647
74.931
80.555


O2
1.940
=
1.940
4.593
4.938


Sum wet


42.235


Sum dry =
42.235 − 2.949

39.286





aMoles H2O in air = (40.014 × 29 × 0.013) ÷ 18 = 0.838


bMoles N2 in air = (40.014 × 0.79) = 31.611






The critical numbers from these calculations are the SO2 concentrations:

    
    
        0.0870, wet basis, and
        0.0970 dry basis.
    
    


The moisture concentration 6.970 is also critical.
Once these numbers are known, the native SO3 concentration in the flue gas can be calculated as follows:
The SO2 concentration dry (the resistivity concentration in this example uses the equivalent SO3 concentration “dry” flue gas) is equal to 0.094%. The appropriate SO2 to SO3 conversion factor for this coal is 0.4% so the approximate SO3 concentration is:

0.00094×0.004=3.76 PPM (dry basis).

As an alternative, the flue gas SO2 concentration can be obtained from the plant's Continuous Emissions Monitoring (CEM) system, corrected for flue gas moisture concentration using factors from the combustion calculation and multiplied by the factor 0.004 to estimate inherent or background SO3 concentration. For other coals, for example, western coals, the approximate conversion factor is 0.001 and for Powder River Basin Coals, the conversion factor is 0.005 (as apposed to 0.004).

To this point, the calculations for Method 1 and Method 2 are the same, however, they are different from this point on.
Step 4. The secondary current applied to the electrostatic precipitator is obtained from the controls for each transformer-rectifier set that is powering the precipitator. These current numbers are translated into current densities by dividing the plate area powered by the transformer-rectifier set. In this example-case, the precipitator has four electrical fields in the direction of gas flow with four transformer-rectifier sets per field.
Readings from the transformer/rectifier sets are as follows:


















TR1
TR2
TR3
TR4









Field 1
165 ma
165 ma
165 ma
165 ma



Field 2
224 ma
224 ma
224 ma
224 ma



Field 3
274 ma
274 ma
274 ma
274 ma



Field 4
338 ma
338 ma
338 ma
338 ma










In this example-case, each transformer/rectifier set energized 19,440 ft2 of plate area.

For a typical field, these currents translate into current densities as follows:

165 ma×(1.0×10−3 ma/a)/(19,440 ft2)=8.488×10−6 a/ft2=8.488 μa/ft2

8.488×10−6×1.076=9.133 na/cm2

Note: 1.0 μa/ft2=1.076 na/cm2 

Similar calculations can be used to produce the following table



















TR1
TR2
TR3
TR4






















Field 1
9.13
9.13
9.13
9.13



Field 2
12.41
12.41
12.41
12.41



Field 3
15.19
15.19
15.19
15.19



Field 4
18.73
18.73
18.73
18.73










Where the units are nA/cm2 

Notice that in this example, all of the TR sets in the same field have been assumed to have the same operating point, i.e., the same voltage and current levels. If these numbers were different, an averaging process, in Step 5, would be used to deal with this more common situation.
Step 5. Determine effective fly ash resistivity level in the ESP using a correlation that relates fly ash resistivity to ESP current density for each electrical field. Average the results to produce an effective resistivity for the ESP. If this resistivity is close to, or lower than, the optimum range, go to Step 10, otherwise proceed to Step 6.
In this example-case, the correlations published in EPRI report CS-5040, table 3–4 are used.
These correlations, after amplification are as follows:













Field 1
log10 (J, nA/cm2) = (6.455 ± 0.370) − 0.5013 log10 (ρ, ohm-cm)


Field 2
log10 (J, nA/cm2) = (6.839 ± 0.360) − 0.5214 log10 (ρ, ohm-cm)


Field 3
log10 (J, nA/cm2) = (5.497 ± 0.304) − 0.3905 log10 (ρ, ohm-cm)


Field 4
log10 (J, nA/cm2) = (5.718 ± 0.327) − 0.4005 log10 (ρ, ohm-cm)


Field 5
log10 (J, nA/cm2) = (3.328 ± 0.306) − 0.1736 log10 (ρ, ohm-cm)









Where J is in nA/cm2 and ρ is in ohm-cm.

Since, log(e)=1/ln(10) substitution gives:

log(J)=log(e)ln(J)=ln(J)/ln(10)=ln(J)/2.302585 similarly,

log(ρ)=ln(ρ)/ln(10)=ln(ρ)/2.302585 and further substitution gives:

















Field 1
ln(J)/ln(10) = 6.455 − 0.5013 ln(ρ)/ln(10)



or
ln(J) = 2.302585 × 6.455 − 0.5013 ln(ρ)



Field 1
ln(J) = 14.8632 − 0.5013 ln(ρ)



similarly



Field 2
ln(J) = 15.74738 − 0.5214 ln(ρ)



Field 3
ln(J) = 12.65731 − 0.3905 ln(ρ)



Field 4
ln(J) = 13.16618 − 0.4005 ln(ρ)



Field 5
ln(J) = 7.66300 − 0.1736 ln(ρ)










These equations are inverted to give the following:

















Field 1
ln(ρ) = 29.64931 − 1.994813 ln(J)



Field 2
ln(ρ) = 30.20211 − 1.917913 ln(J)



Field 3
ln(ρ) = 32.41309 − 2.560819 ln(J)



Field 4
ln(ρ) = 32.87435 − 2.496879 ln(J)



Field 5
ln(ρ) = 44.14171 − 5.76037 ln(J)










From Calculation 3, we have the following:














J



















Field 1
 9.13 na/cm2



Field 2
12.41 na/cm2



Field 3
15.19 na/cm2



Field 4
18.73 na/cm2










Using the ρ vs. J equation gives:














ρ



















Field 1
 9.1 × 1010 ohm-cm



Field 2
10.4 × 1010 ohm-cm



Field 3
11.3 × 1010 ohm-cm



Field 4
16.2 × 1010 ohm-cm



Average
11.8 × 1010 ohm-cm










Note that the resistivity is much higher than the optimum value of 1010 ohm-cm.

Step 6. Use a correlation relating fly ash composition and flue gas temperature and SO3 concentration to fly ash resistivity to determine the flue gas SO3 concentration to needed to produce the optimum fly ash resistivity.
That calculation proceeds in a sequence of substeps as follows using the equations developed by Dr. Bickelhaupt and published in EPRI report C9-4145, Appendix A. Starting with the example ash composition in Table 2, complete substep as follows:

    Substep 1: Normalize the weight percentages to sum 100% by dividing each specified percentage by the sum of the specified percentages.
    Substep 2: Divide each oxide percentage by the respective molecular weight to obtain the mole fractions.
    Substep 3: Divide each mole fraction by the sum of the mole fractions and multiply by 100 to obtain the molecular percentages as oxides.
    Substep 4: Multiply each molecular percentage by the decimal fraction of cations in the given oxide to obtain the atomic concentrations.

All of these sub-steps are illustrated in the following table for the data in Table 2.



























Atomic



Specified
Normalized
Molecular
Mole
Molecular
Cationic
Concentration


Oxide
Weight %
Weight %
Weight
Fraction
Percentage
Fraction
Of Cation






















Li2O
0.01
0.01
29.88
0.00034
0.024
0.67
0.016


Na2O
0.96
0.99
61.98
0.01600
1.116
0.67
0.744


K2O
2.43
2.50
94.20
0.02654
1.854
0.67
1.236


MgO
0.78
0.80
40.31
0.01985
1.387
0.50
0.694


CaO
2.62
2.70
56.08
0.04815
3.364
0.50
1.682


Fe2O3
7.76
8.00
159.70
0.05009
3.500
0.40
1.400


Al2O3
17.85
18.40
101.96
0.18046
12.608
0.40
5.043


SiO2
61.00
62.89
60.09
1.04660
73.123
0.33
24.368


TiO2
0.62
0.64
79.90
0.00801
0.560
0.33
0.186


P2O5
0.55
0.57
141.94
0.00402
0.281
0.29
0.080


SO3
2.43
2.50
80.06
0.03123
2.183
0.25
0.546


Sum
97.01
100.00

1.43129
100.000









Using the % atomic concentrations from the above calculations, use the following equations for calculation of fly ash resistivity (Bickelhaupt equations).

ρv=exp[−1.8916 ln X−0.9696 ln Y+1.234 ln Z+3.62876−(0.069078)E+9980.58/T]

ρsexp[27.59774−2.233348 ln X−(0.00176)W−(0.069078)E−(0.00073895)(W)exp(2303.3/T)]

ρaexp[85.1405−(0.708046)CSO3−23267.2/T−(0.069078)E], for z<3.5% or K>1.0%

ρa=exp[59.0677−(0.854721)CSO3−13049.47/T−(0.069078)E], for z>3.5% and K<1.0%

1/ρvs=1/ρv+1/ρs

1/ρvsa=1/ρvs+1/ρa

ρv=volume resistivity (ohm-cm)
ρs=surface resistivity (ohm-cm)
ρa=adsorbed acid resistivity (ohm-cm)
ρvs=volume and surface resistivity (ohm-cm)
ρvsa=total resistivity (ohm-cm)
X=Li+Na percent atomic concentration
Y=Fe percent atomic concentration
Z=Mg+Ca percent atomic concentration
K=K percent atomic concentration
T=absolute temperature (K)
W=moisture in flue gas (volume %)
CSO3=concentration of SO3 (ppm, dry)
E=applied electric field (kV/cm)
For the example case,
X=0.016+0.744=0.76
Y=1.40
Z=0.694+1.682=2.376
K=1.236
T=417 (Example gas temperature 291° F.)
W=6.983
CSO3=(from Calculation 2)3.76 ppm, dry
E=10(typical electric field value)

ρv=exp[−1.8916 ln(0.76)−0.9696 ln(1.40)+1.237 ln(2.376)+3.62876−(0.069078)(10)+9980.58/417]

=1.636×1012 ohm-cm

ρs=exp[27.59774−2.23348 ln(0.76)−(0.00176)(6.983)−(0.069078)(10)−(0.00073895)(6.983)exp(2303.0/417)]

=2.392×1011 ohm-cm





  
    
      
        
          ρ
          a
        
        =
        
          exp
          ⁢
          
            
          
          [
          
            85.1405
            -
            
              
                (
                0.708046
                )
              
              ⁢
              
                (
                3.76
                )
              
            
            -
            
              23267.2
              /
              417
            
            -
            
              
                (
                0.069078
                )
              
              ⁢
              
                (
                10
                )
              
            
          
          ]
        
      
    
  
  
    
      
        =
        
          
            1.939
            ×
            
              10
              11
            
            ⁢
            ohm
          
          -
          cm
        
      
    
  
  
    
      
        
      
    
  




1/ρvs=1/1.636×1012+1/2.392×1011=4.792×10−12

ρvs=2.1 ×1011 ohm-cm

1/ρvsa=1/4.792×10−12+1/1.939×1011=9.949×10−12

ρvsa=1.0 ×1011 ohm-cm

This resistivity is consistent with the resistivity calculated from the precipitator current densities, but this consistency is not required for Method 2 since this calculation is being used to obtain an approximate SO3 injection rate which will be refined in the following steps. The approximate level of SO3 injection is calculated as follows:
From the preceding calculation,

1/ρvsa=1/ρvs+1/ρa

where ρvsa=1×1010 ohm-cm(the desirable ρ)

ρvs=2.1×1011 ohm-cm(from preceding calculation)





  
    
      
        
          1
          /
          
            ρ
            a
          
        
        =
        
          
            1
            /
            
              ρ
              vsa
            
          
          -
          
            1
            /
            
              ρ
              vs
            
          
        
      
    
  
  
    
      
        =
        
          
            1.0
            ×
            
              10
              10
            
          
          -
          
            2.761905
            ⁢
            
              
            
            ×
            
              10
              12
            
          
        
      
    
  
  
    
      
        =
        
          9.5238
          ⁢
          
            
          
          ×
          
            10
            11
          
        
      
    
  




ρva=1.05×1010 ohm-cm

also from Calculation 6,

ρa=exp[85.1405−(0.708046)CSO3−23267.2/T−(0.069078)E]

where T=417 (from preceding calculation)

E=10 (from preceding calculation)
hence

1.06×1010=exp[85.1405−(0.708046)CSO3−23267.2/417−(0.069078)(10)]

ln(1.05×1010)=85.1405−(0.708046)CSO3−55.79664−0.69078

23.07464109=28.652−(0.708046)CSO3

(0.708046)CSO3=5.578

CSO3=7.878

Correcting for wet conditions

Hence, the approximate total

SO3 needed=7.878×(39.286/42.235)

=7.33 ppm

Step 7. Subtract the background SO3 from the needed SO3 concentration from Step 6 to determine the amount of SO3 that must be added to the flue gas to produce the optimum fly ash resistivity. That calculation, for the example-case, proceeds as follows:
The SO3 from combustion calculation and background calculation,

=0.00088×0.004=3.52 ppm (wet basis)

From the calculation above, the approximate desirable SO3 level=7.33 ppm.

Difference=7.33−3.52=3.81 ppm.

This calculation shows that approximately 3.8 ppm of SO3 should be added to the flue gas to produce an optimum level of fly ash resistivity.
Consequently, output to SO3 control system a signal that will raise the SO3 level in the flue gas by 3.8 ppm.
Step 8. Send rate of additional signal to the controls that operate the SO3 conditioning system.
Step 9. Repeat Steps 4 and 5.
Step 10.
a. If indicated ash resistivity is equal to or less than optimum resistivity, decrease rate of injection by x percent where x is between 5 and 25.
Or
b. If indicated ash resistivity is greater than optimum resistivity, increase rate of injection by x percent where x is between 5 and 25.
Step 11. Repeat Step 10 until indicated fly ash resistivity passes through optimum resistivity point and then set rate of injection at a point in the range bounded by the levels calculated in the last two interactions; for example, at a point that is halfway between the two levels.
Step 12. Every y minutes, where y is number between 5 and 30, restart the process beginning at Step 2.
Obviously, many modifications may be made without departing from the basic spirit of the present invention.

## Claims

1. A method for determining a most effective injection rate for SO3 into a flue gas resulting from a coal burned in a boiler, the method comprising:
1, obtaining proximate and ultimate analyses of the coal being burned in the boiler and ash mineral analysis for the coal,
2, determining an average temperature of the flue gas entering an electrostatic precipitator (ESP) from plant instrumentation, wherein the ESP is powered with multiple electrical fields,
3, estimating SO3 background level in the flue gas using correlation relating flue gas SO3 to coal type and coal sulfur content,
4, obtaining a value of a current applied to the ESP from controls for each transformer-rectifier set that is powering the precipitator,
5, determining an effective fly ash resistivity level in the ESP using a correlation that relates fly ash resistivity to ESP current density for each electrical field, and averaging the results to produce the effective resistivity of the fly ash collected in the ESP,
6, a, if the indicated ash resistivity is equal to or less than an optimum resistivity, decreasing the rate of injection by x percent where x is between 5 and 25, or
b, if the indicated ash resistivity is greater than the optimum resistivity, increasing the rate of injection by x percent where x is between 5 and 25,
7, repeating step 6 until the indicated fly ash resistivity passes through an optimum resistivity point, setting the rate of injection at a point in the range bounded by the levels calculated in the last two interactions, and then
8, every y minutes, where y is number between 5 and 30, restarting the process beginning at step 2.

2. A method for determining a most effective injection rate for SO3 into a flue gas resulting from a coal burned in a boiler, the method comprising:
1, obtaining proximate and ultimate analyses of the coal being burned in the boiler and ash mineral analysis for the coal,
2, determining an average temperature of the flue gas entering an electrostatic precipitator (ESP) from plant instrumentation, wherein the ESP is powered with multiple electrical fields,
3, estimating SO3 background level in the flue gas using correlation relating flue gas SO3 to coal type and coal sulfur content,
4, obtaining a value of a current applied to the ESP from controls for each transformer-rectifier set that is powering the precipitator,
5, determining an effective fly ash resistivity level in the ESP using a correlation that relates fly ash resistivity to ESP current density for each electrical field, averaging the results to produce the effective resistivity of the fly ash collected in the ESP, and if this resistivity is not close to, or lower than, the optimum range, proceeding with step 6; otherwise, going to step 10,
6, using a correlation relating fly ash composition and flue gas temperature and SO3 concentration to the fly ash resistivity to determine a flue gas SO3 concentration needed to produce the optimum fly ash resistivity,
7, subtracting the background SO3 from the needed SO3 concentration from step 6 to determine the amount of SO3 that must be added to the flue gas to produce the optimum fly ash resistivity,
8, sending a signal to the controls that operate an SO3 conditioning system to increase the amount of SO3 injected into the flue gas by the additional amount determined in step 7,
9, repeating steps 4 and 5,
10, a, if the indicated ash resistivity is equal to or less than the optimum resistivity, decreasing the rate of injection by x percent where x is between 5 and 25, or
b, if the indicated ash resistivity is greater than the optimum resistivity, increasing the rate of injection by x percent where x is between 5 and 25,
11, repeating step 10 until the indicated fly ash resistivity passes through the optimum resistivity point, setting the rate of injection at a point in the range bounded by the levels calculated in the last two interactions, and then
12, every y minutes, where y is number between 5 and 30, restarting the process beginning at step 2.