Boiler duty and efficiency calculations

Combination firing—two fuels fired PROJECT sample units—British

Boiler Parameters

1. Boiler duty—MM Btu/h = 109.93

1. Steam press—psig = 400

2. Excess air—% = 18

2. Steam temp.—°F = 750

3. Amb. temp.—°F = 80

3. Steam flow—lb/h = 100,000

4. Exit gas temp.—°F = 404

4. Feed water temp.—°F = 320

5. Ref. humidity—% = 60

5. Blowdown—% = 1

Flue Gas Analysis—% vol (wet/dry)

6. Process steam—Lb/h =

6. CO 2 8.99/10.73 (gas flow, analysis, air flow, efficiencies shown here pertain 2 O 16.19 7. H

to effective average conditions if multiple fuels are fired.) 2 8. N 71.92/85.81 2 9. O 2.9/3.46

10. SO 2 — 11. Total air—lb/h = 115,402 12. Total flue gas—lb/h = 121,312 13. Efficiency—% HHV = 82.7 14. Efficiency—% LHV = 82.7 15. Adiab. comb temp.—°F = 3207

Air heated by boiler flue gas or no air heater Fuel no. 1 gas—% volume [% duty = 71]

Methane = 97, ethane = 2, propane = 1 Fuel and air input HHV—MM Btu/h = 95.54 fuel fred—lb/h = 4021 ad. comb temp.—°F = 3164 Air/fuel ratio = 20.5 gas/fuel = 21.5 eff-lhv—% = 90.55 eff-hhv—% = 81.69 Losses: dry gas—% = 6.22 air moisture = 0.17 fuel moisture = 11.12 radiation = 0.3 unaccounted = 0.5

Flue gas analysis:% vol CO 2 = 8.1 H 2 O = 17.8 N 2 = 71.22 O 2 = 2.89 SO 2 =.

LHV—Btu/lb = 21,439 HHV—Btu/lb = 23,764 fuel temp. °F = 80 air temp. °F = 80 NO x —ppmv/0.1 lb per MM Btu HHV = 83.1 CO—ppmv/0.1 lb CO per MM Btu HHV = 136.6

Fuel no. 2 oil—% weight [% duty = 29] Carbon = 87. Hydrogen = 13. Sulfur =. Nitrogen =. Oxygen =. °API = 32 moisture =. Fuel AND air input HHV—MM Btu/h = 37.38 fuel fired—Lb/h = 1895 ad comb temp.—°F = 3288 Air/fuel ratio = 17.41 gas/fuel = 18.41 eff-lhv—% = 91.12 eff-hhv—% = 85.28 Losses: dry gas—% = 6.71 air moisture = 0.17 fuel moisture = 7.04 radiation = 0.3 unaccounted = 0.5

Flue gas analysis:% vol CO 2 = 11.29 H 2 O = 12.04 N 2 = 73.73 O 2 = 2.94 SO 2 =.

LHV—Btu/lb = 18,463 HHV—Btu/lb = 19,727 fuel temp. °F = 80 air temp. F = 80 NO x —ppmv/0.1 lb per MM Btu HHV = 77.5 CO—ppmv/0.1 lb CO per MM Btu HHV = 127.4

FIGURE 1.8

Results from combination firing of fuels.

Emissions of NO x in mass units such as g/GJ may be converted to ppmvd from the fol- lowing equation:

V n = 10 6 × Y × (N/46) × (MW/W gm ) × (21 – 3)/(21 – O 2 × Y) (1.17) where

MW = molecular weight of wet flue gases

V n is the amount of NO x equivalent of 1 g/GJ in ppmvd O 2 = % volume of oxygen in wet flue gases Y = 100/(100 – %H 2 O) where %H 2 O is the volume of water vapor in wet flue gases N = NO x produced in g/GJ (we may replace this by CO or SO x as shown later) W gm = flue gas produced per GJ fuel fired, g, obtained as shown later (lb/MM Btu values

shown in brackets)

Combustion Calculations

TABLE 1.15

Conversion from g/GJ to ppmvd of NO x , CO, UHC

Excess Air, % 0 10 20 30 0 10 20 30

Fuel Gas

Oil Oil Oil CO 2 9.47 8.68 8.02 7.45 13.49 12.33 11.35 10.51

H 2 O 19.91 18.38 17.08 15.96 12.88 11.90 11.07 10.36 N 2 70.62 71.22 71.73 72.16 73.63 74.02 74.34 74.62 O 2 0 1.72 3.18 4.43 0 1.76 3.24 4.5

MW 27.52 27.62 27.68 27.77 28.87 28.85 28.84 28.82 W gm (g/GJ)

371,400 403,300 435,000 W gm (lb/MM Btu)

864 938 1011 V n , ppmvd-g/GJ

1.816 1.813 1.810 V n , ppmvd-lb/MM Btu

782 782 782 V c , ppmvd-g/GJ

2.986 2.978 2.974 V c , ppmvd-lb/MM Btu

1284 1281 1279 V h , ppmvd-g/GJ

5.226 5.212 5.210 V h , ppmvd-lb/Mm Btu

2247 2241 2240 V s , ppmvd-g/GJ

1.305 1.303 1.300 V s , ppmvd-lb/MM Btu

Let us see how V n is obtained. For the case of zero excess air, air required per GJ is obtained using the million GJ method discussed earlier or from detailed combustion calculations. From the A values in Table 1.5 and the HHV of fuel for the natural gas case.

Flue gas per GJ = 314 + 10 6 /55,263 = 332 kg/GJ = 332,000 g/GJ. It may also be computed more accurately from a computer program using the fuel analysis, results of which are shown in Table 1.15. Fuel consumed per GJ = 10 6 /55,263 = 18.1 kg/GJ. MW = 27.53; Y = 100/

(100 – 19.91) = 1.2486; O 2 =0

Let us convert 1 g/GJ to obtain V n , the NO x in ppmvd.

V n = 10 6 × Y × (N/46) × (MW/W gm ) × (21 – 3)/(21 – O 2 × Y) = 10 6 × 1.2486 × (1/46) × (27.53/331,300) × 18/(21 – 0) = 1.933 ppmvd (1 lb/MM Btu = 430 g/GJ = 831 ppmvd).

Let us check at 30% excess air for natural gas: Flue gas per GJ = 314 × 1.3 + 10 6 /55,263 = 426.3 kg/GJ = 426,300 g/GJ

approximately or more accurately 425,200 from Table 1.15.

Y = 100/(100 – 15.96) = 1.19.

V n = 10 6 × 1.19 × (1/46) × (27.77/425,200) × 18/(21 – 4.43 × 1.19) = 1.9335 ppmvd Hence 1 g/GJ of NO x = 1.933 ppmvd or 1 lb/MM Btu NO x = (453.7/1.05485) × 1.933 = 831

ppmvd. (453.7 is the number of g in 1 lb and 1.05485 is the conversion from MM Btu to GJ). Similarly, for CO, one would use 28 in the denominator of this equation (1.17) instead of

32 Steam Generators and Waste Heat Boilers: For Process and Plant Engineers

46 and obtain 1 g/GJ = 3.176 ppmvd (1 lb/MM Btu = 1362 ppmvd) (to convert to lb/MM Btu from g/GJ multiply by 430). For UHC, insert MW of 16 in the earlier equation (UHC is considered as methane, which has 16 as its MW). Then, V u = 5.543U, where U is the UHC in g/GJ and V u is the ppmvd of UHC. 1 lb/MM Btu UHC = 5.543 × 430 = 2383 ppmvd.

For SO x , using an MW of 64, we have V s = 1.389S, where S is the emissions in g/GJ and V s is in ppmvd or 1lb/MM Btu of SO x = 1.389 × 430 = 597 ppmvd. For no. 2 oil, using values from Table 1.15, for the case of zero excess air, for 1 g/GJ NO x , W gm = 320 + 10 6 /45,885 = 341.79 kg/GJ = 341,790 g/GJ or, more accurately, from combustion calculations, 339,700. Y = 100/(100 – 12.88) = 1.1478

V n = 10 6 × 1.1478 × (1/46) × (28.87/339,700) × 18/(21 – 0) = 1.818 ppmvd or 1 lb/MM Btu NO x = 1.818 × 430 = 781 ppmvd. It can be shown that at 30% excess also, the value in ppmvd is the same as shown in Table 1.15.

Similarly, 1 g/GJ of CO is equivalent to V c = 2.987 ppmvd. Or 1 lb/MM Btu CO = 1285 ppmvd. 1 g/GJ of UHC or V u = 5.22 ppmvd and 1 lb/MM Btu UHC = 2245 ppmvd.

1 g/GJ of SO x = 1.303 ppmvd and 1 lb/MM Btu = 1.303 × 430 = 560 ppmvd

Converting ppmvd of NO x to mg/Nm 3

Let 1 N m 3 of flue gas contain 1 mg of NO x . Hence, volume of NO x = 22.4 × (1/46) × 10 –6 =

0.487 ppmvd. Hence 1 ppmvd = 1/0.487 = 2.053 mg/N m 3 of NO x .

Similarly, 1 ppmvd of CO = 1.25 mg/N m 3 (as MW of CO = 28 versus 46 for NO x ) and

(22.4/64) mg/N m 3 of SO x = 1 ppmvd or 2.857 mg/N m 3 .

From Table 1.16, one can convert emissions of pollutants to any unit. For example, 1 lb/MM Btu of NO x in oil firing is 1/0.000128 = 781 ppmvd. CO of 1 g/GJ in gas firing will be 1/0.315 = 3.174 ppmvd.

Example 1.15

Fuel oil containing 1.5% sulfur is fired in a furnace. What is the emission of SO 2 in g/GJ

if the HHV = 46,000 kJ/kg (19,776 Btu/lb)? What is the ppmw if 15% excess air is used for firing this oil in a boiler?

TABLE 1.16

Summary of Emission Conversions

Item

ppmvd

mg/Nm 3 g/GJ

lb/MM Btu

0.00120 NO x oil

0.00128 CO gas

0.000732 CO oil

0.000786 UHC gas

0.000419 UHC oil

0.000445 SO x gas

0.001675 SO x oil

Combustion Calculations

1 g of S gives rise to 2 g of SO 2 . Hence, SO 2 emission = 0.015 × 2 × 1,000 × 10 6 /46,000 = 652 g/GJ (1.516 lb/MM Btu)

Fuel of 1 GJ requires 320 kg air for combustion (Table 1.5). With 15% excess air, flue gas generated = 1.15 × 320 = 10 6 /46,000 = 389.7 kg/GJ

Fuel input of 1GJ generates 0.03 × 21.7 = 0.651 kg SO 2 . Hence, % volume wet of SO 2 in

flue gas = (0.651/64)/(389.7/28.5) = 744 ppmv wet., where 28.5 is the assumed MW of flue

gas and 64 that of SO 2 . If 3% of SO 2 converts to SO 3 , the amount of SO 3 in ppmv in flue

gas = 744 × 0.03 × 64/80 = 17.8 ppmv (80 is the MW of SO 3 ) These conversion calculations are important when the acid dew point needs to be estimated.

Example 1.16

Natural gas with C 1 = 97% volume, C 2 = 2%, C 3 = 1% is fired with 15% excess air in a boiler. If fuel contains 700 ppm H 2 S, what is the ppmv of SO 2 formed? If 2% of this converts to SO 3 , what is the ppmv SO 3 formed? Let ambient temperature = 27°C and relative humidity = 60% so that the amount of moisture in air = 0.013 kg/kg. One comes across this type of problem often in a boiler containing H 2 S. It is neces- sary to find out the SO 2 and then the SO 3 formed so that the acid dew point may be computed.

Amount of theoretical dry air required for combustion of 100 moles of fuel = 97 × 9.528 + 2 × 16.675 + 1 × 23.821 = 981.4 mol. Hence, actual air = 1.15 × 981.4 = 1128.6 mol. Moisture in air = 1128.6 × 0.013 × 27.5/18 = 22.4 mol.

CO 2 = 97 + 3.5 × 2 + 5 × 1 = 109 mol H 2 O = 2 × 97 + 3 × 2 + 4 × 1 + 22.4 = 226.4 mol N 2 = 0.79 × 1128.6 = 891.6 mol O 2 = (1128.6 – 981.4) × 0.21 = 30.9 mol

Total = 109 + 226.4 + 891.6 + 30.9 = 1257.9 mol % volume CO 2 = 100 × 109/1257.9 = 8.66. H 2 O = 100 × (226.4/1257.9) = 18.00% N 2 = 100 ×

(891.6/1257.9) = 0.70.88% and O 2 = 100 × 30.9/1257.9 = 2.46%

MW flue gas = 8.66 × 44+18x18 + 70.88 × 28 + 2.46 × 32 = 27.68 700 ppm H 2 S in fuel gives rise to 700/12.579 = 55.6 ppmv SO 2 in flue gas Assuming 3% conversion of SO 2 to SO 3 , the amount of SO 3 in flue gas = 0.03 × 55.6 × 64/80 = 1.33 ppmv SO 3 . (First convert the SO 2 to weight basis and then to SO 3 on volume basis. This value is used in acid dew point estimation.)

As discussed earlier, higher combustion temperatures will result in higher NO x . When fuels containing hydrogen are fired, the NO x levels will be higher. Table 1.17 gives an idea of the NO x and CO with various fuels.

TABLE 1.17

Typical Emissions from Various Fuels

Gas

NO x , lb/MM Btu (g/GJ)

CO (lb/MM Btu) (g/GJ)

Natural gas

Hydrogen gas

Refinery gas

0.03–0.08 (12.9–34.4) Blast furnace gas

Producer gas

34 Steam Generators and Waste Heat Boilers: For Process and Plant Engineers

Boiler duty and efficiency calculations