Patent Abstract:
Optimizing fly ash resistivity by controlling concentration of sulfur trioxide (SO 3 ) in flue gas by the use of an algorithm.

Full Description:
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 SO 2  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 SO 3  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 SO 3  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 SO 3  concentration needed to produce optimum fly ash resistivity. These correlations can be combined (as described below) to produce a superior SO 3  conditioning system control algorithm. 
     Current SO 3  conditioning systems use a preset rate of SO 3  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 SO 3  addition when coal and ash properties are changing. 
     The primary application will be at utility plants that use SO 3  conditioning to improve ESP performance. These plants are located both here and abroad. In addition, SO 3  is used in some industrial applications, and the new SO 3  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 (SO 3 ) 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 SO 3  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 SO 3  background level in the flue gas using correlation relating flue gas SO 3  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 SO 3  concentration to determine the flue gas SO 3  concentration needed to produce the optimum fly ash resistivity. 
     Step 6. Subtract the background SO 3  concentration from the needed SO 3  concentration from the needed SO 3  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 SO 3  conditioning system. 
     Another method encompassed by the invention involves determining a most effective injection rate for SO 3  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 SO 3  background level in the flue gas using correlation relating flue gas SO 3  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 SO 3  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 SO 3  background level in the flue gas using correlation relating flue gas SO 3  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 SO 3  concentration to fly ash resistivity to determine the flue gas SO 3  concentration to needed to produce the optimum fly ash resistivity, 
     Step 7. Subtract the background SO 3  from the needed SO 3  concentration from Step 6 to determine the amount of SO 3  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 SO 3  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 SO 3  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 SO 3  concentration measured at the ESP inlet—0 to 4 ppm. SO 3 —example number=3.5. 
     b. Moderate to high fly ash resistivity—8×10 10  ohm-cm to 5×10 12  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/cm 2  and in the outlet field, it might be 15.19 na/cm 2 . These current densities correspond to a fly ash resistivity of 1.0×10 11  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 SO 3  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—8×10 9  ohm-cm to 4×10 10  ohm-cm, depending on ESP collection and reentrainment characteristics—example number 1×10 10  ohm-cm. 
     c. High ESP power levels as indicated by current density levels. 
     For example, when the correct level of SO 2  has been added to the flue gas, the average current densities in the ESP would increase to 27.67 na/cm 2  in the inlet field, 33.50 na/cm 2  in the middle field and 39.50 na/cm 2  in the outlet field. The current densities correspond to a fly ash resistivity of 1×10 10  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/cm 2   
                 na/cm 2   
                 na/cm 2   
                 na/cm 2   
                 na/cm 2   
               
               
                   
               
             
          
           
               
                 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 SO 3  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 
                 Na 2 O 
                 0.96 
               
               
                 Oxygen 
                 6.00 
                 K 2 O 
                 2.43 
               
               
                 Nitogen 
                 1.00 
                 MgO 
                 0.78 
               
               
                 Sulfur 
                 2.20 
                 CaO 
                 2.62 
               
               
                 Moisture 
                 3.60 
                 Fe 2 O 3   
                 7.76 
               
               
                 Ash 
                 16.34 
                 Al 2 O 3   
                 17.85 
               
               
                 SUM 
                 100.00 
                 SiO 2   
                 61.00 
               
               
                   
                   
                 TiO 2   
                 0.62 
               
               
                   
                   
                 P 2 O 5   
                 0.55 
               
               
                   
                   
                 SO 3   
                 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 SO 3  background level in the flue gas using correlation relating flue gas SO 3  to coal type and coal sulfur content. The SO 3  concentration is calculated as a percentage of SO 2  in the flue gas which can be determined from a combustion calculation using the coal analysis and flue gas O 2  or CO 2  or if the flue gas SO 2  is available from plant instruments, this number can be used in the SO 3  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 O 2  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 
                   
                 Multipliers 1   
                 O 2   
                 Dry Air 
               
               
                   
               
             
          
           
               
                 C 
                 68.00 
                 ÷ 
                 12.01 
                 = 
                 5.662 
                 × 
                    1.0 and 4.76 
                 5.662 
                 26.951 
               
               
                 H 2   
                 3.86 
                 ÷ 
                 2.02 
                 = 
                 1.911 
                 × 
                   0.50 and 2.38 
                 0.956 
                 4.548 
               
               
                 O 2   
                 6.00 
                 ÷ 
                 32.00 
                 = 
                 0.188 
                 × 
                 −1.00 and −4.76 
                 −0.188 
                 −0.895 
               
               
                 N 2   
                 1.00 
                 ÷ 
                 28.01 
                 = 
                 0.036 
               
               
                 S 
                 1.20 
                 ÷ 
                 32.06 
                 = 
                 0.037 
                 × 
                   1.00 and 4.76 
                 0.037 
                 0.176 
               
               
                 H 2 O 
                 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 O 2  for 30% Excess Air (Typical Excess Air Level). 
     
       
         
               
               
               
               
             
               
               
               
               
             
               
               
               
               
             
           
               
                   
                   
               
               
                   
                   
                 Required for 
                   
               
               
                   
                   
                 Combustion 
               
               
                   
                   
                 moles/100 lb fuel 
               
               
                   
                   
                 at 30% excess air 
               
             
          
           
               
                   
                   
                 O 2   
                 Dry air 
               
               
                   
                   
               
             
          
           
               
                   
                 O 2  and air × 130/100 total 
                 8.407 
                 40.014 
               
               
                   
                 Excess air = 40.014 − 30.780 
                 — 
                 9.234 
               
               
                   
                 Excess O 2  = 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 
               
               
                   
               
             
          
           
               
                 CO 2   
                 5.662 
                 = 
                 5.662 
                 13.406 
                 14.412 
               
               
                 H 2 O 
                 1.911 + 0.200 + 0.838 a   
                 = 
                 2.949 
                 6.983 
                 — 
               
               
                 SO 2   
                 0.037 
                 = 
                 0.037 
                 0.088 
                 0.094 
               
               
                 N 2   
                 0.036 + 31.611 b   
                 = 
                 31.647 
                 74.931 
                 80.555 
               
               
                 O 2   
                 1.940 
                 = 
                 1.940 
                 4.593 
                 4.938 
               
               
                 Sum wet 
                   
                   
                 42.235 
               
               
                 Sum dry = 
                 42.235 − 2.949 
                   
                 39.286 
               
               
                   
               
               
                   a Moles H 2 O in air = (40.014 × 29 × 0.013) ÷ 18 = 0.838 
               
               
                   b Moles N 2  in air = (40.014 × 0.79) = 31.611 
               
             
          
         
       
     
     The critical numbers from these calculations are the SO 2  concentrations:
         0.088%, wet basis, and   094%, dry basis.
 
The moisture concentration, 6.98% is also critical Once these numbers are known, the native SO 3  concentration in the flue gas can be calculated as follows:
       

     The SO 2  concentration dry (the resistivity concentration in this example uses the equivalent SO 3  concentration of “dry” flue gas) is equal to 0.094%. the appropriate SO 2  to SO 3  conversion factor for this coal is 0.4% so the approximate SO 3  concentration is:
 
0.00094×0.004=3.76 PPM (dry basis)
 
As an alternative, the flue gas SO 2  concentration can be obtained from the plant&#39;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 SO 3  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 SO 3  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 
               
               
                   
               
             
             
               
                   
               
             
          
           
               
                 Li 2 O 
                 0.01 
                 0.01 
                 29.88 
                 0.00034 
                 0.024 
                 0.67 
                 0.016 
               
               
                 Na 2 O 
                 0.96 
                 0.99 
                 61.98 
                 0.01600 
                 1.116 
                 0.67 
                 0.744 
               
               
                 K 2 O 
                 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 
               
               
                 Fe 2 O 3   
                 7.76 
                 8.00 
                 159.70 
                 0.05009 
                 3.500 
                 0.40 
                 1.400 
               
               
                 Al 2 O 3   
                 17.85 
                 18.40 
                 101.96 
                 0.18046 
                 12.608 
                 0.40 
                 5.043 
               
               
                 SiO 2   
                 61.00 
                 62.89 
                 60.09 
                 1.04660 
                 73.123 
                 0.33 
                 24.368 
               
               
                 TiO 2   
                 0.62 
                 0.64 
                 79.90 
                 0.00801 
                 0.560 
                 0.33 
                 0.186 
               
               
                 P 2 O 5   
                 0.55 
                 0.57 
                 141.94 
                 0.00402 
                 0.281 
                 0.29 
                 0.080 
               
               
                 SO 3   
                 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, ρ v , the surface resistivity, ρ s , and the acid resistivity, ρ a . 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 
       ρ v =exp[−1.8916 ln  X −0.9696 ln  Y +1.234 ln  Z +3.62876−(0.069078) E +9980.58 /T]   ρ s =exp[27.59774−2.233348 ln  X−( 0.00176) W− (0.069078) E− (0.00073895)( W )exp(2303.3 /T )] ρ a =exp[85.1405−(0.708046)CSO 3 −23267.2 /T− (0.069078) E] , for  z &lt;3.5% or  K &gt;1.0% ρ a =exp[59.0677−(0.854721)CSO 3 −13049.47 /T− (0.069078) E] , for  z &lt;3.5% or  K &gt;1.0% 1/ρ vs =1/ρ v +1/ρ s   1/ρ vsa =1/ρ vs +1/ρ a   
     ρ v =volume resistivity (ohm-cm) 
     ρ s =surface resistivity (ohm-cm) 
     ρ a =adsorbed acid resistivity (ohm-cm) 
     ρ vs =volume and surface resistivity (ohm-cm) 
     ρ vsa =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 %) 
     CSO 3 =concentration of SO 3  (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 
     CSO 3 =(from Calculation 2)3.76 ppm, dry 
     E=10(typical electric field value)
 
ρ v =exp[−1.8916 ln(0.76)−0.9696 ln(1.40)+1.237 ln(2.376)+3.62876−(0.069078)(10)+9980.58/417]
 
=1.636×10 12  ohm-cm
 
ρ s =exp[27.59774−2.23348 ln(0.76)−(0.00176)(6.983)−(0.069078)(10)−(0.00073895)(6.983)exp(2303.0/417)]
 
=2.392×10 11  ohm-cm
 
ρ a =exp[85.1405−(0.708046)(3.76)−23267.2/417−(0.069078)(10)]
 
=1.939×10 11  ohm-cm
 
1/ρ vs =1/1.636×10 12 +1/2.392×10 11 =4.792×10 −12 
 
ρ vs =2.1×10 11  ohm-cm
 
1/ρ vsa =1/4.792×10 −12 +1/1.939×10 11 =9.949×10 −12 
 
ρ vsa =1.0 ×10 11  ohm-cm
 
     In this example, the calculated resistivity is found to be 1.0×10 11  ohm-cm, which is too high for optimum ESP performance, so additional SO 3  must be added to the flue gas. 
     Step 5. Use a correlation relating the base fly ash resistivity and flue gas SO 3  concentration to determine the flue gas SO 3  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 1.0×10 10  ohm-cm. Hence, the calculation proceeds as follows:
 
1/ρ vsa =1/ρ vs +1/ρ a 
 
where ρ vsa =1×10 10  ohm-cm(the desirable ρ)
 
ρ vs =2.1×10 11  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                   ρ va =1.05×10 10  ohm-cm 
     also from the preceding calculation,
 
ρ a =exp[85.1405−(0.708046)CSO 3 −23267.2 /T− (0.069078) E] 
 
where T=417 (from preceding calculation)
 
     E=10 (from preceding calculation) 
     hence
 
1.05×10 10 =exp[85.1405−(0.708046)CSO 3 −23267.2/417−(0.069078)(10)]
 
ln(1.05×10 10 )=85.1405−(0.708046)CSO 3 −55.79664−0.69078
 
23.07464109=28.652−(0.708046)CSO 3 
 
(0.708046)CSO 3 =5.578
 
CSO 3 =7.878
 
Correcting for wet conditions
 
SO 3  needed=7.878×(39.286/42.235)=7.33 ppm
 
     Step 6. Subtract the background SO 3  concentration from the needed SO 3  concentration from the needed SO 3  that must be added to the flue gas to produce the optimum fly ash resistivity. 
     The combustion calculation results and the background SO 3  calculation in step 2, the SO 3  concentration is estimated to be 0.00088×0.004=3.52 ppm (wet basis). From the desired level calculation above, the desirable SO 3  level=7.33 ppm, hence the difference=7.33−3.52=3.81 ppm. Hence, 3.81 ppm, SO 3  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 SO 3  conditioning system. In this case, the signal should be sent that will cause the SO 3  conditioning system to add 3.8 ppm SO 3  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 SO 3  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 SO 3  concentration measured at the ESP inlet—0 to 4 ppm: SO 3  —example number=3.5. 
     b. Moderate to high fly ash resistivity—8×10 10  ohm-cm to 5×10 12  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 2 , in the middle field it might be 12.41 na/cm 2  and the outlet field might be 15.19 na/cm 2 . These current densities correspond to a fly ash resistivity of 1.0×10 11  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/cm 2   
                 na/cm 2   
                 na/cm 2   
                 na/cm 2   
                 na/cm 2   
               
               
                   
               
             
          
           
               
                 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 SO 3  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—8×10 9  ohm-cm to 4×10 10  ohm-cm, depending on ESP collection and reentrainment characteristics—example number 1×10 10  ohm-cm. 
     c. High ESP power levels as indicated by current density levels. 
     For example, when the correct level of SO 3  has been added to the flue gas, the average current densities in the ESP would increase to 27.67 nA/cm 2  in the inlet field, 33.50 na/cm 2  in the middle field, and 39.08 na/cm 2  in the outlet field. The current densities correspond to a fly ash resistivity of 1×10 10  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 SO 3 : 
     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 
                 Na 2 O 
                 0.96 
               
               
                 Oxygen 
                 6.00 
                 K 2 O 
                 2.43 
               
               
                 Nitogen 
                 1.00 
                 MgO 
                 0.78 
               
               
                 Sulfur 
                 2.20 
                 CaO 
                 2.62 
               
               
                 Moisture 
                 3.60 
                 Fe 2 O 3   
                 7.76 
               
               
                 Ash 
                 16.34 
                 Al 2 O 3   
                 17.85 
               
               
                 SUM 
                 100.00 
                 SiO 2   
                 61.00 
               
               
                   
                   
                 TiO 2   
                 0.62 
               
               
                   
                   
                 P 2 O 5   
                 0.55 
               
               
                   
                   
                 SO 3   
                 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 SO 3  background level in the flue gas using correlation relating flue gas SO 3  to coal type and coal sulfur content. The SO 3  concentration is calculated as a percentage of SO 2  in flue gas which can be determined from a combustion calculation using the coal analysis and flue gas O 2  or CO 2  or if the flue gas SO 2  is available from plant instruments, this number can be used in the SO 3  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 O 2  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 
                   
                 Multipliers 1   
                 O 2   
                 Dry Air 
               
               
                   
               
             
          
           
               
                 C 
                 68.00 
                 ÷ 
                 12.01 
                 = 
                 5.662 
                 × 
                    1.0 and 4.76 
                 5.662 
                 26.951 
               
               
                 H 2   
                 3.86 
                 ÷ 
                 2.02 
                 = 
                 1.911 
                 × 
                   0.50 and 2.38 
                 0.956 
                 4.548 
               
               
                 O 2   
                 6.00 
                 ÷ 
                 32.00 
                 = 
                 0.188 
                 × 
                 −1.00 and −4.76 
                 −0.188 
                 −0.895 
               
               
                 N 2   
                 1.00 
                 ÷ 
                 28.01 
                 = 
                 0.036 
               
               
                 S 
                 1.20 
                 ÷ 
                 32.06 
                 = 
                 0.037 
                 × 
                   1.00 and 4.76 
                 0.037 
                 0.176 
               
               
                 H 2 O 
                 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 O 2  for 30% Excess Air (Typical Excess Air Level). 
     
       
         
               
               
               
               
             
               
               
               
               
             
               
               
               
               
             
           
               
                   
                   
               
               
                   
                   
                 Required for 
                   
               
               
                   
                   
                 Combustion 
               
               
                   
                   
                 moles/100 lb fuel 
               
               
                   
                   
                 at 30% excess air 
               
             
          
           
               
                   
                   
                 O 2   
                 Dry air 
               
               
                   
                   
               
             
          
           
               
                   
                 O 2  and air × 130/100 total 
                 8.407 
                 40.014 
               
               
                   
                 Excess air = 40.014 − 30.780 
                 — 
                 9.234 
               
               
                   
                 Excess O 2  = 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 
               
               
                   
               
             
          
           
               
                 CO 2   
                 5.662 
                 = 
                 5.662 
                 13.406 
                 14.412 
               
               
                 H 2 O 
                 1.911 + 0.200 + 0.838 a   
                 = 
                 2.949 
                 6.983 
                 — 
               
               
                 SO 2   
                 0.037 
                 = 
                 0.037 
                 0.088 
                 0.094 
               
               
                 N 2   
                 0.036 + 31.611 b   
                 = 
                 31.647 
                 74.931 
                 80.555 
               
               
                 O 2   
                 1.940 
                 = 
                 1.940 
                 4.593 
                 4.938 
               
               
                 Sum wet 
                   
                   
                 42.235 
               
               
                 Sum dry = 
                 42.235 − 2.949 
                   
                 39.286 
               
               
                   
               
               
                   a Moles H 2 O in air = (40.014 × 29 × 0.013) ÷ 18 = 0.838 
               
               
                   b Moles N 2  in air = (40.014 × 0.79) = 31.611 
               
             
          
         
       
     
     The critical numbers from these calculations are the SO 2  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 SO 3  concentration in the flue gas can be calculated as follows: 
     The SO 2  concentration dry (the resistivity concentration in this example uses the equivalent SO 3  concentration “dry” flue gas) is equal to 0.094%. The appropriate SO 2  to SO 3  conversion factor for this coal is 0.4% so the approximate SO 3  concentration is:
 
0.00094×0.004=3.76 PPM (dry basis).
 
As an alternative, the flue gas SO 2  concentration can be obtained from the plant&#39;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 SO 3  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 ft 2  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 ft 2 )=8.488×10 −6  a/ft 2 =8.488 μa/ft 2   8.488×10 −6 ×1.076=9.133 na/cm 2   
Note: 1.0 μa/ft 2 =1.076 na/cm 2    
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/cm 2    
     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   log 10  (J, nA/cm 2 ) = (6.455 ± 0.370) − 0.5013 log 10  (ρ, ohm-cm)       Field 2   log 10  (J, nA/cm 2 ) = (6.839 ± 0.360) − 0.5214 log 10  (ρ, ohm-cm)       Field 3   log 10  (J, nA/cm 2 ) = (5.497 ± 0.304) − 0.3905 log 10  (ρ, ohm-cm)       Field 4   log 10  (J, nA/cm 2 ) = (5.718 ± 0.327) − 0.4005 log 10  (ρ, ohm-cm)       Field 5   log 10  (J, nA/cm 2 ) = (3.328 ± 0.306) − 0.1736 log 10  (ρ, ohm-cm)                    
Where J is in nA/cm 2  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/cm 2             Field 2   12.41 na/cm 2             Field 3   15.19 na/cm 2             Field 4   18.73 na/cm 2                          
Using the ρ vs. J equation gives:
 
                                                       ρ                                        Field 1    9.1 × 10 10  ohm-cm           Field 2   10.4 × 10 10  ohm-cm           Field 3   11.3 × 10 10  ohm-cm           Field 4   16.2 × 10 10  ohm-cm           Average   11.8 × 10 10  ohm-cm                        
Note that the resistivity is much higher than the optimum value of 10 10  ohm-cm.
 
     Step 6. Use a correlation relating fly ash composition and flue gas temperature and SO 3  concentration to fly ash resistivity to determine the flue gas SO 3  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                                Li 2 O   0.01   0.01   29.88   0.00034   0.024   0.67   0.016       Na 2 O   0.96   0.99   61.98   0.01600   1.116   0.67   0.744       K 2 O   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       Fe 2 O 3     7.76   8.00   159.70   0.05009   3.500   0.40   1.400       Al 2 O 3     17.85   18.40   101.96   0.18046   12.608   0.40   5.043       SiO 2     61.00   62.89   60.09   1.04660   73.123   0.33   24.368       TiO 2     0.62   0.64   79.90   0.00801   0.560   0.33   0.186       P 2 O 5     0.55   0.57   141.94   0.00402   0.281   0.29   0.080       SO 3     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).
 ρ v =exp[−1.8916 ln  X− 0.9696 ln  Y+ 1.234 ln  Z+ 3.62876−(0.069078) E+ 9980.58 /T]   ρ s exp[27.59774−2.233348 ln  X−( 0.00176) W− (0.069078) E− (0.00073895)( W )exp(2303.3 /T )] ρ a exp[85.1405−(0.708046)CSO 3 −23267.2 /T− (0.069078) E] , for  z&lt; 3.5% or  K&gt; 1.0% ρ a =exp[59.0677−(0.854721)CSO 3 −13049.47 /T− (0.069078) E] , for  z&gt; 3.5% and  K&lt; 1.0% 1/ρ vs =1/ρ v +1/ρ s   1/ρ vsa =1/ρ vs +1/ρ a   
     ρ v =volume resistivity (ohm-cm) 
     ρ s =surface resistivity (ohm-cm) 
     ρ a =adsorbed acid resistivity (ohm-cm) 
     ρ vs =volume and surface resistivity (ohm-cm) 
     ρ vsa =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 %) 
     CSO 3 =concentration of SO 3  (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 
     CSO 3 =(from Calculation 2)3.76 ppm, dry 
     E=10(typical electric field value)
 
ρ v =exp[−1.8916 ln(0.76)−0.9696 ln(1.40)+1.237 ln(2.376)+3.62876−(0.069078)(10)+9980.58/417]
 
=1.636×10 12  ohm-cm
 
ρ s =exp[27.59774−2.23348 ln(0.76)−(0.00176)(6.983)−(0.069078)(10)−(0.00073895)(6.983)exp(2303.0/417)]
 
=2.392×10 11  ohm-cm
 
                     ρ   a     =     exp   ⁢           [     85.1405   -       (   0.708046   )     ⁢     (   3.76   )       -     23267.2   /   417     -       (   0.069078   )     ⁢     (   10   )         ]                 =       1.939   ×     10   11     ⁢   ohm     -   cm                                 1/ρ vs =1/1.636×10 12 +1/2.392×10 11 =4.792×10 −12   ρ vs =2.1 ×10 11  ohm-cm 1/ρ vsa =1/4.792×10 −12 +1/1.939×10 11 =9.949×10 −12   ρ vsa =1.0 ×10 11  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 SO 3  injection rate which will be refined in the following steps. The approximate level of SO 3  injection is calculated as follows: 
     From the preceding calculation,
 
1/ρ vsa =1/ρ vs +1/ρ a 
 
where ρ vsa =1×10 10  ohm-cm(the desirable ρ)
 
ρ vs =2.1×10 11  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                   ρ va =1.05×10 10  ohm-cm 
     also from Calculation 6,
 
ρ a =exp[85.1405−(0.708046)CSO 3 −23267.2 /T− (0.069078) E] 
 
where T=417 (from preceding calculation)
 
     E=10 (from preceding calculation) 
     hence
 
1.06×10 10 =exp[85.1405−(0.708046)CSO 3 −23267.2/417−(0.069078)(10)]
 
ln(1.05×10 10 )=85.1405−(0.708046)CSO 3 −55.79664−0.69078
 
23.07464109=28.652−(0.708046)CSO 3 
 
(0.708046)CSO 3 =5.578
 
CSO 3 =7.878
 
Correcting for wet conditions
 
Hence, the approximate total
 
SO 3  needed=7.878×(39.286/42.235)
 
=7.33 ppm
 
     Step 7. Subtract the background SO 3  from the needed SO 3  concentration from Step 6 to determine the amount of SO 3  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 SO 3  from combustion calculation and background calculation,
 
=0.00088×0.004=3.52 ppm (wet basis)
 
     From the calculation above, the approximate desirable SO 3  level=7.33 ppm.
 
Difference=7.33−3.52=3.81 ppm.
 
     This calculation shows that approximately 3.8 ppm of SO 3  should be added to the flue gas to produce an optimum level of fly ash resistivity. 
     Consequently, output to SO 3  control system a signal that will raise the SO 3  level in the flue gas by 3.8 ppm. 
     Step 8. Send rate of additional signal to the controls that operate the SO 3  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.

Technology Classification (CPC): 5