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01 POWER ISLAND / 01 CCPP / V. Ganapathy-Industrial Boilers and HRSG-Design (2003)

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and inlet gas temperature can vary significantly with load depending on the type of process or application.

6.25a

Q:

Discuss the importance of dew point corrosion in boilers and heaters fired with fuels containing sulfur.

A:

During the process of combustion, sulfur in fuels such as coal, oil, and gas is converted to sulfur dioxide. Some portion of it (1–5%) is converted to sulfur trioxide, which can combine with water vapor in the flue gas to form gaseous sulfuric acid. If the surface in contact with the gas is cooler than the acid dew point, sulfuric acid can condense on it, causing corrosion. ADP (acid dew point) is dependent on several factors, such as excess air, percent sulfur in fuel, percent conversion of SO2 to SO3, and partial pressure of water vapor in the flue gas. Manufacturers of economizers and air heaters suggest minimum cold-end temperatures that are required to avoid corrosion. Figures 6.1 and 6.2 are typical. Sometimes the minimum fluid temperature, which affects the tube metal temperature, is suggested. The following equation gives a conservative estimate of the acid dew point [8]:

Tdp

¼ 1:7842 þ 0:0269 log pw 0:129 log pSO3

ð18aÞ

 

þ 0:329 log pw log pSO3

 

 

where

 

 

Tdp ¼ acid dew point, K

 

pw ¼ partial pressure of water vapor, atm

 

pSO3

¼ partial pressure of sulfur trioxide, atm

 

Table 6.8 gives typical pSO3 values for various fuels and excess air. Q6.18c shows how ppm SO3 can be computed from which pSO3 is obtained.

A practical way to determine Tdp is to use a dew point meter. An estimation of the cold-end metal temperature can give an indication of possible corrosion.

6.25b

Q:

How is the dew point of an acid gas computed?

A:

Table 6.9 shows the dew point correlations for various acid gases [9,11].

Copyright © 2003 Marcel Dekker, Inc.

FIGURE 6.1 The relationship between SO3 and ADT. (Courtesy of Land Combustion Inc.)

Flue gas from an incinerator has the following analysis (vol%): H2O ¼ 12, SO2 ¼ 0.02, HCl ¼ 0.0015 and the rest oxygen and nitrogen. Gas pressure ¼ 10 in. wg. Compute the dew points of sulfuric and hydrochloric acids given that 2% of SO2 converts to SO3. In order to use the correlations, the gas pressures must be converted to mmHg. Atmospheric pressure ¼ 10 in. wg ¼ 10=407 ¼ 0.02457 atmg or 1.02457 atm abs.

pH2O ¼ 0:12 1:02457 760 ¼ 93:44 mmHg ln PH2O ¼ 4:537

PHCl ¼ 0:0015 1:0245 760 ¼ 0:1168 mmHg ln pHCl ¼ 2:1473

Partial pressures of sulfuric acid and SO3 are equal. Hence

PSO3 ¼ 0:02 0:0002 760 1:0245 ¼ 0:0031 mmHg ln PSO3 ¼ 5:7716

Substituting into the equations, we obtain the following.

Copyright © 2003 Marcel Dekker, Inc.

FIGURE 6.2 Limiting tube metal temperatures to avoid external corrosion in economizers and air heaters when burning fuels containing sulfur. (From Ref. 13, with permission.)

For hydrochloric acid:

1000 ¼ 3:7368 0:1591 4:537 þ 0:0326 2:1473

Tdp

0:00269 4:537 2:1473 ¼ 3:0588

or

 

Tdp ¼ 327 K ¼ 54 C ¼ 129 F

 

 

 

 

TABLE 6.8 SO3 in Flue Gas (ppm)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Sulfur (%)

 

 

 

 

 

 

 

 

 

 

 

Fuel

Excess air (%)

0.5

1.0

2.0

3.0

4.0

5.0

 

 

 

 

 

 

 

 

Oil

5

2

3

3

4

5

6

 

11

6

7

8

10

12

14

Coal

25

3–7

7–14

14–28

20–40

27–54

33–66

 

 

 

 

 

 

 

 

 

Copyright © 2003 Marcel Dekker, Inc.

TABLE 6.9 Dew Points of Acid Gasesa

Hydrobromic acid

1000=Tdp ¼ 3.5639 7 0.1350 ln PH2O 7 0.0398 ln PHBr þ 0.00235 ln PH2O ln PHBr

Hydrochloric acid

1000=Tdp ¼ 3.7368 7 0.1591 ln PH2O 7 0.0326 ln PHCl þ 0.00269 ln PH2O ln PHCl

Nitric acid

1000=Tdp ¼ 3.6614 7 0.1446 ln PH2O 7 0.0827 ln PHNO3 þ 0.00756 ln PH2O ln PHNO3

Sulfurous acid

1000=Tdp ¼ 3.9526 7 0.1863 ln PH3O þ 0.000867 ln PSO2 7 0.000913 ln PH2O ln PSO2

Sulfuric acid

1000=Tdp ¼ 2.276 7 0.0294 ln PH2O 7 0.0858 ln PH3SO4 þ 0.0062 ln PH2O ln PH2SO4

aTdp is dew point temperature (K), and P is partial pressure (mmHg). Compared with published data, the predicted dew points are within about 6 K of actual values except for H2SO4, which is within about 9 K. Source: HCl, HBr, HNO3 and SO2 correlations were derived from vapor–liquid equilibrium data. The H2SO4 correlation is from Ref. 5.

For sulfuric acid:

1000 ¼ 2:276 0:0294 4:537 þ 0:0858 5:7716

Tdp

0:0062 4:537 5:7716 ¼ 2:4755

or

Tdp ¼ 404 K ¼ 131 C ¼ 268 F

The dew points of other gases can be obtained in a similar manner.

6.25c

Q:

Does the potential for acid dew point corrosion decrease if the gas temperature at the economizer is increased?

Copyright © 2003 Marcel Dekker, Inc.

A:

Acid dew points were computed in Q6.25a. If the tube wall temperatures can be maintained above the dew point, then condensation of vapors is unlikely. However, the tube wall temperature in a gas-to-liquid heat exchanger such as the economizer is governed by the gas film heat transfer coefficient rather than the tube-side water coefficient, which is very high.

It can be shown by using the electrical analogy and neglecting the effects of fouling that [9]

tm ¼ to ðto tiÞ hi hi þ ho

where

tm ¼ tube wall temperature

to ¼ gasand tube-side fluid temperature hi ¼ tube-side heat transfer coefficient ho ¼ gas-side heat transfer coefficient

In an economizer, hi is typically about 1000 Btu=ft2 h F and h0 is about 15 Btu=ft2 h F.

Let us assume that water temperature ti ¼ 250 F and compute the wall temperature tm for two gas temperatures, 350 F and 750 F.

tm1

¼ 350 ð350 250Þ

1000

¼ 252 F

 

1015

tm2

¼ 750 ð750 250Þ

1000

¼ 258 F

 

1015

Hence for a variation of 400 F in gas temperature, the tube wall temperature changes by only 6 F because the gas film heat transfer coefficient is so low compared to the water-side coefficient. Even with finned tubes the difference would be marginal.

We see that if we specify a higher stack gas temperature when selecting or designing an economizer we cannot avoid corrosion concerns if the water temperature is low or close to the acid dew point. A better way is to increase the water temperature entering the economizer by raising the deaerator pressure or by using a heat exchanger to preheat the water.

6.25d

Q:

Using the correlation given below, evaluate the sulfuric acid dew point.

Tdp ¼ 203:25 þ 27:6 log PH2O þ 10:83 log PSO3 þ 1:06 ðlog PSO3 þ 8Þ2:19 ð18bÞ

Copyright © 2003 Marcel Dekker, Inc.

The partial pressures are in atmospheres and dew point is in degrees Celsius.

A:

Using the data from Q6.25b [14],

PSO3 ¼ 0:0031 mmHg ¼ 4:1 10 6 atm

log PSO3 ¼ 5:3872

PH2O ¼ 93:44 mmHg ¼ 0:1229 atm

log PH2O ¼ 0:9104

Tdp ¼ 203:25 27:6 0:9104 10:83 5:3872 þ 1:06 ð2:6128Þ2:19

¼ 128:4 C, or 263 F

which agrees with the other correlation. However, it should be mentioned that these calculations have some uncertainty, and experience should be taken as the guide.

6.26a

Q:

How do you convert pollutants such as NOx and CO from gas turbine exhaust gases from mass units such as lb=h to ppm?

A:

With strict emission regulations, plant engineers and consultants often find it necessary to relate mass and volumetric units of pollutants such as NOx and CO. In gas turbine cogeneration and combined cycle plants, in addition to the pollutants from the gas turbine itself, one has to consider the contributions from duct burners or auxiliary burners that are added to increase the steam generation from the HRSGs (heat recovery steam generators).

One can easily obtain the total lb=h of NOx or CO in the exhaust gas. However, regulations refer to NOx and CO in ppmvd (parts per million volume dry) referred to 15% oxygen in the gas. The conversion can be done as follows.

If w lb=h is the flow rate of NOx (usually reported as NO2) in a turbine exhaust flow of W lb=h, the following expression gives NOx in volumetric units on dry basis [9].

V

¼

100

 

ðw=46Þ=ðW =MWÞ

ð

19

Þ

100 %H2O

 

 

 

where

%H2O ¼ volume of water vapor

MW ¼ molecular weight of the exhaust gases

Copyright © 2003 Marcel Dekker, Inc.

ð20Þ
¼ V F

The value of V obtained with Eq. (19) must be converted to 15% oxygen on dry basis to give ppmvd of NOx:

V ð21 15Þ 106

Vn ¼ 21 100 %O2=ð100 %H2

where %O2 is the oxygen present in the wet exhaust gases and factor F converts V to 15% oxygen basis, which is the usual basis of reporting emissions. Similarly, CO emission in ppmvd can be obtained as

Vc ¼ 1:642 Vn (for the same w lb=h rateÞ

because the ratio of the molecular weights of NO2 and CO is 1.642.

Example

Determine the NOx and CO concentrations in ppmvd, 15% oxygen dry basis if 25 lb=h of NOx and 15 lb=h of CO are present in 550,000 lb=h of turbine exhaust gas that has the following analysis by volume percent (usually argon is added to the nitrogen content):

CO2 ¼ 3:5;

H2O ¼ 10;

 

N2 ¼ 75;

 

O2 ¼ 11:5

Solution.

First,

 

 

 

 

 

 

 

 

 

MW ¼ ð3:5 44 þ 10 18 þ 75 28 þ 11:5 32Þ=100 ¼ 28

Let us compute NOx on dry basis in the exhaust.

 

V

¼

 

100 ð25=46Þ

 

¼

0:00003074

 

ð550,000=28Þ=ð100 10Þ

 

 

 

 

 

 

F

¼

 

½

106 ð21 15Þ

11:5 ¼

0:73

 

106

 

21

ð

100

 

10

Þ&

 

 

 

 

100=

 

 

 

 

Hence

Vn ¼ 0:00003074 0:73 106 ¼ 22:4 ppmvd

Similarly, Vc ¼ (15=25) 1.642 22.4 ¼ 22.0 ppmvd.

6.26b

Q:

How can the emissions due to NOx and CO in fired boilers be converted from ppm to lb=MM Btu or vice versa [10]?

A:

Packaged steam generators firing gas or oil must limit emissions of pollutants in order to meet state and federal regulations. Criteria on emissions of common

Copyright © 2003 Marcel Dekker, Inc.

pollutants such as carbon monoxide (CO) and oxides of nitrogen (NOx) are often specified in parts per million volume dry (ppmvd) at 3% oxygen. On the other hand, burner and boiler suppliers often cite or guarantee values in pounds per million Btu fired.

Table 6.10 demonstrates a simple method for calculating the conversion. It should be noted that excess air has little effect on the conversion factor.

Table 6.10 shows the results of combustion calculations for natural gas and No. 2 oil at various excess air levels. The table shows the flue gas analysis, molecular weight, and amount of flue gas produced per million Btu fired on higher heating value (HHV) basis. Using these, we will arrive at the relationship between ppmvd values of NOx or CO and the corresponding values in lb=MM Btu fired.

Calculations for Natural Gas

 

 

 

 

From simple mass-to-mole conversions we have

 

 

 

V

106

 

Y

 

N

 

MW

21 3

ð

21

Þ

 

46

Wgm

21 O2 Y

n ¼

 

 

 

where

 

 

 

 

 

 

 

 

 

 

 

MW ¼ molecular weight of wet flue gases

 

 

 

N ¼ pounds of NOx per million Btu fired

 

 

 

O2 ¼ vol% oxygen in wet flue gases

 

 

 

Vn ¼ parts per million volume dry NOx

 

 

 

Wgm

¼ flue gas produced per MM Btu fired, lb

 

 

 

Y¼ 100=(100 7 %H2O), where H2O is the volume of water vapor in wet flue gases

TABLE 6.10 Results of Combustion Calculations (Analysis in vol%)

 

 

 

 

 

Percent excess air

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

10

20

30

0

10

20

30

Component

 

 

Natural gasa

 

 

No. 2 Oilb

 

 

 

 

 

 

 

 

 

 

CO2

9.47

8.68

8.02

7.45

13.49

12.33

11.35

10.51

H2O

19.91

18.38

17.08

15.96

12.88

11.90

11.07

10.36

N2

70.62

71.22

71.73

72.16

73.63

74.02

74.34

74.62

O2

0

1.72

3.18

4.43

0

1.76

3.24

4.50

MW

27.52

27.62

27.68

27.77

28.87

28.85

28.84

28.82

Wgm

768

841

914

966

790

864

938

1011

aNatural gas analysis assumed: C1 ¼ 97, C2 ¼ 2, C3 ¼ 1 vol%. (HHV and LLV ¼ 23,759 and 21,462 Btu=lb, respectively.)

bNo. 2 oil analysis assumed: C ¼ 87.5%, H2 ¼ 12.5%; API ¼ 32. (HHV and LLV ¼ 19,727 and 18,512 Btu=lb, respectively.)

Copyright © 2003 Marcel Dekker, Inc.

From Table 6.10; for zero excess air:

 

 

 

 

 

Wgm ¼ ð106=23,789Þ 18:3 ¼ 769

 

 

 

 

 

Y ¼ 100=ð100 19:91Þ ¼ 1:248

 

 

 

 

 

MW ¼ 27:53; O2 ¼ 0

 

 

 

 

 

Substituting these into Eq. (21)

we have

 

 

 

 

 

Vn ¼ 106 1:248 N

27:52

 

 

18

 

 

¼ 832 N

 

 

 

 

 

46

 

769

 

21

 

 

 

 

 

 

Similarly, to obtain ppmvd CO (parts per million volume dry CO), one would use 28 instead of 46 in the denominator. Thus the molecular weight of NOx would be 46 and the calculated molecular weight of CO would be 28.

Ve ¼ 1367 CO

where CO is the pounds of CO per MM Btu fired on higher heating value (HHV) basis.

Now repeat the calculations for 30% excess air:

Wgm ¼ 986:6;

Y ¼

 

 

 

100

 

¼ 1:189

 

 

 

 

 

 

100

 

15:96

MW ¼ 27:77;

 

 

 

 

 

 

 

 

O2 ¼ 4:43

 

 

 

Vn ¼ 106

1:189

N

27:77

 

 

 

 

 

 

 

 

46

986:6

 

 

 

 

18

 

 

 

¼ 832N

 

 

 

21 ð4:43 1:189Þ

 

Thus, independent of excess air, we obtain 832 as the conversion factor for NOx and 1367 for CO.

Similarly, for No. 2 oil and using values from Table 6.10,

Vn ¼ 783N and Vc ¼ 1286 CO

Example

If a natural gas burner generates 0.1 lb of NOx per MM Btu fired, then the equivalent would equal 832 0.1 ¼ 83 ppmvd.

6.26c

Q:

How can the emissions of unburned hydrocarbons (UHCs) be converted from lb=MM Btu to ppmv basis?

Copyright © 2003 Marcel Dekker, Inc.

A:

Refer to Table 6.10, which shows the results of combustion calculations for oil and gaseous fuels at various excess air levels. We can obtain UHC emissions on ppmv basis if lb=MM Btu values are known.

Let us assume that U is the emission of UHC (treated as methane) in lb=MM Btu in flue gases of natural gas at 20% excess air. Using Eq. (21) for converting from mass to volume units,

106 Y MW ð21 3Þ

Vu ¼ 16 Wgm ð21 O2 Y Þ

MW ¼ 16 for UHC and 27.68 for flue gases, water vapor in flue gases ¼ 17.08 vol% at 20% excess air for natural gas, Wgm ¼ 914 lb=MM Btu, and % oxygen wet ¼ 3.18. Hence,

Vu ¼ U 106

 

100

 

 

 

 

 

 

 

82:92

 

 

 

 

 

27:68 18

¼

2394U ppmvd

16 914 ð21 3:18 100=82:92Þ

 

For excess air at 10% excess air, MW ¼ 27.62 for flue gases, water vapor ¼ 18.38 vol%, oxygen wet ¼ 1.72 vol% Wgm ¼ 841.

V

U 106

 

100

 

 

 

 

27:62 18

 

 

82:62 16

 

 

ð

 

 

Þ

u ¼

 

841

21

 

1:72

 

100=82:62

 

 

 

 

 

¼

2365U ppmvd

 

 

 

 

 

 

 

 

 

Hence, if the UHC value is 0.1 lb=MM Btu for natural gas, it is equivalent to about 237 ppmv.

For No. 2 oil at 20%

excess air, Wgm ¼ 938, oxygen ¼ 3.24, MW flue

gases ¼ 28.84, water vapor ¼ 11.07 vol%.

 

 

 

 

 

 

V

U 106

100

16

 

 

 

28:84 18

 

 

 

 

 

 

ð

 

 

Þ

u ¼

 

88:93

938

21

 

3:24

 

100=88:93

 

 

 

 

 

¼ 2240U ppmvd

6.26d

Q:

Convert SOx values from lb=MM Btu to ppmvd.

Copyright © 2003 Marcel Dekker, Inc.