Author

Dry-bulb temperature, °F

FIGURE 1.1

Moisture in air due to relative humidity.

Hence, M = 0.622 × 0.0231/(1.033 – 0.0231) = 0.0142 kg/kg air or 0.0142 lb/lb air. If 1000 kg of dry air is required for combustion in a boiler, the actual wet air that should be sent to the burner will be 1014.2 kg, and the boiler fan has to be sized for the volume of this amount of wet air.

Combustion Calculations

Knowing the fuel analysis, excess air, and ambient conditions, one may perform combustion calculations as shown in the following text.

Example 1.1

Natural gas having CH4 = 83.4%, C2H6 = 15.8%, and N2 = 0.8 by volume is fired in a boiler using 15% excess air. Ambient temperature is 20°C and relative humidity is 80%. Perform combustion calculations and determine the flue gas analysis.

Solution

From steam tables, the saturated vapor pressure at 20°C (68°F) is 0.34 psia = 0.0239 kg/cm2 a. At 80% relative humidity, Pw = 0.8 × 0.0239 = 0.0191 kg/cm2 a

Molecular weight of the fuel = (83.4 × 16 + 15.8 × 30 + 0.8 × 28)/100 = 18.3. wda = kg dry air/kg fuel = 1217 × 28.84/1830 = 19.18.

wwa = kg wet air/kg fuel = 19.18 + 23.5 × 18/1830 = 19.41.

wdg = kg dry flue gas/kg fuel = (115 × 44+33.3 × 32 + 961 × 28)/1830 = 18. wwg = kg wet flue gas/kg fuel = (115 × 44+33.3 × 32 + 961.8 × 28 + 237.7

× 18)/1830 = 20.4.

Water Formed per kg of Fuel

This is an important piece of information as it gives an idea of how much water can be condensed in a condensing economizer if used. In the earlier example, the MW of flue gas = 27.7. The % weight of H2O in the flue gas = 17.64 × 18/27.7 = 11.46. It was shown earlier that wwg = amount of wet flue gas formed per kg of fuel = 20.4. Hence, the amount of water vapor formed per kg fuel fired = 0.1146 × 20.4 = 2.34 kg. Typically, for natural gas, it varies from 2.15 to 2.4. Similar calculations may be carried out for fuel oil. For no. 2 fuel oil at 15% excess air, it can be shown that % volume CO2 = 11.57, H2O = 12.29, N2 = 73.63, O2 = 2.51. MW of flue gas = 11.57 × 44 + 12.29 × 18 + 73.63 × 28 + 2.51 × 32 = 28.72.% weight of H2O in flue gas = 12.29 × 18/28.72 = 7.7%. The amount of flue gas produced per kg fuel is about 18. Hence, 0.077 × 18 = 1.39 kg of water vapor is produced per kg of fuel fired at 15% excess air. On an average, it varies from 1.3 to 1.5 kg/kg fuel, much smaller than that for natural gas. Hence, more energy can be recovered in latent heat form with flue gas from combustion of natural gas than with fuel oils.

Excess Air from Flue Gas Analysis

In operating plants, data on flue gas analysis will be available using which the plant engineer may arrive at the excess air used. This will help the plant engineer evaluate the boiler efficiency, and air and flue gas quantities. A formula that is widely used to obtain excess air E in % is [2]

where O2, CO, and N2 are % volume of oxygen, carbon monoxide, and nitrogen on dry flue gas basis. Another formula is used when an Orsat-type analyzer is used for analyzing the flue gases; SO2 will be absorbed with CO2. The oxygen is on dry volumetric basis.

where C, H2, N2, S, and O2 are fraction by weight of carbon, hydrogen, nitrogen, sulfur, and oxygen in fuel.

Let us check the value of constant K1 for Example 1.1.

%weight of CH4 in fuel = 83.4 × 16/(83.4 × 16 + 15.8 × 30 + 0.8 × 28) = 0.729.

%weight of C2H6 in fuel = 15.8 × 30/(83.4 × 16 + 15.8 × 30 + 0.8 × 28) = 0.259

%N2 by weight in fuel = 0.012

Carbon C in CH4 = 0.75 × 0.729 = 0.5467 and fraction hydrogen = 0.1823 Carbon in C2H6 = (24/30) × 0.259 = 0.2072 and fraction hydrogen = 0.0518

Total C fraction by weight = 0.5467 + 0.2072 = 0.7539, and total hydrogen by weight = 0.0518 + 0.1823 = 0.2341. Hence,

K1 = [100 × 0.7539 + 237 × 0.2341 + 9 × 0.012]/[0.7539 + 3 × 0.2341] = 89.9

Let us compute the heating value of the fuel from the weight fractions. Lower heating value (LHV) = 0.729 × 21,529 + 0.259 × 20,432 = 20,980 Btu/lb = 11,655 kcal/kg = 48,800 kJ/kg (where 21,529 and 20,432 in Btu/lb are the LHV of methane and ethane from Tables 1.1 and 1.2).

HHV = 0.729 × 23,879 + 0.259 × 22,320 = 23,188 Btu/lb = 12,882 kcal/kg = 53,940 kJ/kg = 53.93 × 106 J/kg.

K1 may also be obtained from Table 1.4 for various fuels. If oxygen is measured on true volume basis (wet basis), then one uses constant K2 for excess air evaluation as shown:

Many modern analytical techniques, such as those employing infrared or paramagnetic principles, also measure on a dry gas basis because they require moisture-free samples to avoid damage to the detection cells. These analyzers are set up with a sample conditioning system that removes moisture from the gas sample. However, some analyzers, such as in situ oxygen detectors employing a zirconium oxide cell, measure on the wet gas basis. Results from such equipment need to be corrected to a dry gas basis before they are used in the ASME equations.

These values of K1 and K2 have been arrived at after performing calculations on several fuels with different fuel analysis and hence give a good working average value.

TABLE 1.4

Constants K1 and K2 for Excess Air Evaluation

Using Table 1.4, for methane (natural gas), K1 = 90. O2 on dry flue gas basis is 3%. Hence, excess air = 90 × 3/(21 – 3) = 15%. If wet basis is used, then K2 = 110.5. Then, excess air = 110.5 × 2.47/(21 – 2.47) = 14.7—close to 15%. These constants are good estimates for a type of fuel, and so some minor variations may be expected depending on actual fuel analysis. There is another approximate method to get the excess air from CO2 values, but the accuracy is not good; see Table 1.5.

E = max CO2 on dry flue gas basis/%CO2 measured. In our example, from Table 1.5, max CO2 = 12%, while actual is 10.3%. Hence, E = 12/10.3 = 1.165 or 16.5% excess air. This is only an estimate. The O2 basis is more accurate.

Simplified Combustion Calculations

One may develop a suitable computer code for performing detailed combustion calculations using the procedure given earlier. However, when one is interested only in an estimate of air or flue gas quantities generated in a boiler or heater, the simplified combustion calculation procedure described in the following text may be used. Each fuel such as natural gas, coal, or oil requires a certain amount of dry stoichiometric air for combustion per million joules of energy fired on HHV basis (or per MM Btu fired). This quantity does not vary much with fuel analysis for a type of fuel and therefore becomes a valuable method of estimating air for combustion when fuel analysis is not readily available.

For solid fuels and oil, the dry air wda in kg/kg fuel can be obtained from fundamentals:

(CO, H2, and CH4 are in fraction by weight. Only some constituents are shown here; if there are other combustibles, one can include those by using values from Tables 1.1 and 1.2. For example, CO requires 2.47 kg air/kg fuel and H2 requires 34.34 kg air/kg fuel.

Estimation of Heating Values

When the ultimate analysis of fuels is known, the heating values can be estimated using

Some texts use 9200 instead of 9720. Difference in LHV is small.

Where W is the weight fraction of moisture in fuel, and C, H2, O2, and S are the fractions by weight of carbon, hydrogen, oxygen, and sulfur. LHV and HHV are in Btu/lb. Multiply by 2.326 to obtain values in kJ/kg.

If fuel oil has 87.5% by weight carbon and 12.5% by weight hydrogen, HHV = 14,500 × 0.875 + 62,000 × 0.125 = 20,437 Btu/lb = 47,538 kJ/kg = 11,353 kcal/kg.

LHV = 20,427 – 9,200 × 0.125 = 19,277 Btu lb = 44,838 kJ/kg = 10,709 kcal/kg.

If ultimate analysis is not known, one may estimate the heating values of fuel oils using [3]

F = 25.2 for 20 <= °API <= 30

F = 25.45 <= °API <= 40

Burner Selection

The burner is an important component of any boiler. Burner designs have undergone several iterations during the last decade. Burner suppliers are offering burners with singledigit NOx emissions with low CO levels competing with the selective catalytic reduction (SCR) system. However, these burners require a large amount of FGR on the order of 35%, and flame stability is a concern at low loads. Fuel or air staging or steam injection are the other methods used to limit NOx. Today, single burners are used for capacities up to 300–350 MM Btu/h (316.5–369 GJ/h). Figure 1.2 shows a Natcom burner firing natural gas. It is a maintenance-free (refractory-less) burner. Burners are also available for 9 ppm NOx on natural gas. More on emissions and NOx control is discussed in Chapter 3. Table 1.8 shows typical burner emissions.

FIGURE 1.2

Low NOx burner. (Courtesy of Natcom Burner, Division of Cleaver Brooks, Thomasville, GA.)

Example 1.5

Let us determine the combustion temperature for the fuel gas in Example 1.1. The LHV is 11,655 kcal/kg (20,980 Btu/lb or 48,800 kJ/kg). Fraction methane by weight = 0.729 and that of ethane = 0.259.

Solution

Amount of flue gas produced = 20.4 kg/kg fuel. Assume air is not heated and is at ambient conditions. Hence, the enthalpy of the products of combustion = 48,800/20.4 = 2392.1 kJ/kg = 572.2 kcal/kg (dividing LHV of fuel by the amount of flue gas produced). From Table F.10 showing enthalpy of products of combustion, Tad = 1807°C (3285°F). This is the adiabatic combustion temperature, and the actual flame temperature will be slightly less due to dissociation losses of about 1%. Hence, 1788°C (3250°F) is more likely.

Let us say 15% FGR at 150°C from economizer exit is introduced into the boiler. From Example 1.1, we see that the ratio of flue gas to fuel is about 20.4 at 15% excess for natural gas. The LHV of the fuel is 11,655 kcal/kg as shown earlier.

FGR flow = 0.15 × 20.4 = 3.06 kg flue gas. At 150°C, the enthalpy is 36 kcal/kg (reference 15°C). Air at 27°C has enthalpy = 2.7 kcal/kg.

Hence, enthalpy of air, fuel, and flue gas mixture = (19.4 × 2.7 + 11,655 + 3.06 × 36)/ 20.4/1.15 = 503.7 kcal/kg or combustion temperature = 1600°C (2912°F); enthalpy obtained from Table F.10 by interpolation using the flue gas analysis shown in Example 1.1.

Simplified Procedure for Estimating Combustion Temperatures

An estimate of adiabatic combustion temperature may be made using the following formula. The air and flue gas produced are estimated using the simplified combustion calculation procedure.

Tad = [LHV + Aα × HHV × Cpa × (ta – tamb)/106]/[(1 + Aα × HHV/106) × Cpg] (1.13)

Basically, we estimate the air and flue gas produced using the A values discussed earlier and then use the specific heats of air and flue gas to obtain Tad. is the excess air factor. If 20% excess air is used, then = 1.2. Tad, ta, and tamb are the adiabatic combustion temperature, combustion air temperature, and ambient temperature respectively.

Example 1.6

Fuel oil is fired at 15% excess air in a boiler; estimate the combustion temperature.

Solution

LHV = 18,430 Btu/lb = 42,872 kJ/kg, HHV = 19,636 Btu/lb = 45,673 kJ/kg. There is no air heater. Hence ta = tamb. Hence, 1 million kJ on HHV basis has 106/45,673 = 21.89 kg oil.

Theoretical air required for 106 kJ = 320 kg from Table 1.5; hence, actual air = 1.15 × 320 = 368 kg; flue gas produced = (368 +2 1.89) = 390 kg or ratio of flue gas to fuel = (390/21.89) = 17.8.

Tad = 42,872/(17.8 × 1.328) = 1813°C (3295°F). (1.328 is the flue gas specific heat in kJ/kg °K.)

Effect of FGR on Combustion Temperature

Introducing flue gas at the point of combustion reduces the flame temperature significantly, which in turn lowers NOx levels.

FGR may be introduced at the FD fan suction as shown in Figure 3.8, or it may be introduced at the burner using a separate FGR fan. The suction system or induced system does not require an additional fan, but the FD fan must be sized to handle the higher flow at higher temperature of the flue gas and air mixture. Table 1.9 shows the effect of FGR on combustion temperatures.

(Natural gas: 97% methane, 2% ethane, and 1% propane. Flue gas analysis% volume CO2 = 8.29, H2O = 18.17, N2 = 71.07, O2 = 2.46. LHV = 49,864 (21,437 Btu/lb), HHV = 55,272 kJ/kg (23,762 Btu/lb). Fuel oil: carbon = 87.5% by weight, hydrogen = 12.5%,°API = 32. Flue gas analysis: % volume CO2 = 11.57, H2O = 12.29, N2 = 73.63, O2 = 2.51. LHV = 43,059 kJ/kg (18,512 Btu/lb), HHV = 45,885 kJ/kg (19,726 Btu/lb)).

Relating FGR and Oxygen in Windbox

FGR affects the oxygen in the windbox by diluting it. One may measure the oxygen values in the windbox and relate it to the FGR rate used. This may be used for relating actual FGR rates used with NOx emissions.

Example 1.7

A boiler firing natural gas at 15% excess air uses 54,118 kg/h of combustion air, and about 6,352 kg/h of flue gases is recirculated. Flue gas analysis is as follows: % volume CO2 = 8.29, H2O = 18.17, N2 = 71.07, O2 = 2.47. Determine the oxygen levels in the windbox. Assume air is dry and has 77% nitrogen and 23% oxygen by weight.

Solution

The amount of nitrogen in air = 0.77 × 54,118 = 41,671 kg/h and oxygen = (54,118 – 41,671) = 12,447 kg/h. MW of flue gas = (8.29 × 44 + 18.17 × 18 + 71.07 × 28 + 2.47 × 32)/100 = 27.61.

% CO2 by weight in flue gas is 8.29 × 44/27.61 = 13.21. Similarly, % weight of H2O = 11.84, N2 = 72.07, O2 = 2.88. The mixture has a flow of 6352 + 54,118 = 60,470 kg/h

CO2 in the mixture = 0.1321 × 6352 = 839 kg/h

H2O = 0.1184 × 6352 = 750 kg/h (neglect air moisture) N2 = 0.7207 × 6352 + 41,671 = 46,249 kg/h

O2 = 12,447 + 0.0288 × 6,352 = 12,630 kg/h Converting this to% volume, we have

CO2 = (839/44)/[(839/44) + (750/18) + (46,249/28) + (12,630/32)] = 100 × (19.07/2107) = 0.9%

Similarly, H2O = 1.98%, N2 = 78.37%, and O2 = 18.75%. We see that the % volume of oxygen has come down to 18.75 due to FGR. For finding out the % FGR for a desired NOx level, this reading of oxygen may be used as a reference.

Gas Turbine Exhaust Combustion Calculations

Supplementary firing or furnace firing is often done in HRSGs to increase the HRSG efficiency while generating additional steam on the order of 200%–400%. HRSGs that recover energy from gas turbine exhaust are often fired using distillate fuel or natural gas to raise the gas temperature entering the HRSG. Heavy oils are rarely used as emissions of NOx, and particulates and oxides of sulfur formed could affect the finned surfaces of the HRSG and the SCR catalyst if used. Depending upon the firing temperature, the HRSGs may be called supplementary-fired or furnace-fired HRSGs. This is discussed in Chapter 4. Here we will see how the combustion calculations are performed and how these calculations are different from those for a steam generator. Typically, turbine exhaust contains over 13%–15% by volume of oxygen in the exhaust, and one can increase the firing temperature by addition of fuel alone till an oxygen level of about 3%–3.5% by volume is reached without using additional air. Only fuel is introduced in the burner, and the oxygen available in the exhaust is consumed. A typical duct burner is shown in Figure 1.4. Arrangement of an HRSG with duct burner is shown in Figure 1.5.

Relating Oxygen and Energy Input in Turbine Exhaust Gases

Gas turbine exhaust gases typically contain 13%–16% oxygen by volume compared to 21% in atmospheric air. If steam is injected in the gas turbine for NOx control, the oxygen content will be further reduced. Still there is enough oxygen to raise the exhaust gases to about 1600°C (see Figure 1.6). Sometimes, augmenting air is introduced at the burner to ensure a stable combustion process.

The energy Q in GJ/h required to raise the temperature of exhaust gases from t1 to t2 °C is given by an energy balance around the burner, but approximately it is Q = 10–6 × Wg × (h2 – h1)

FIGURE 1.4

Burner for turbine exhaust. (Courtesy of Natcom Burner, Division of Cleaver Brooks, Thomasville, GA.)