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 w da in kg/kg fuel can be obtained from fundamentals:

(1.6) where C, H 2 , and S are carbon, hydrogen, and sulfur in the fuel by fraction weight.

w da = 11.53C + 34.34 × (H 2 –O 2 /8) + 4.29S

For gaseous fuels, w da = 2.47 × CO + 34.34 × H 2 + 17.27 × CH 4 + 13.3 × C 2 H 2 + 14.81 × C 2 H 4 + 16.12 × C 2 H 6 – 4.32O 2 (1.7)

(CO, H 2 , and CH 4 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 H 2 requires 34.34 kg air/kg fuel.

Combustion Calculations

Example 1.2

Estimate the amount of air required per million kJ of fuel oil fired. C = 0.875, H 2 = 0.125, and °API = 28.

Solution

The HHV of fuel oils may be written as

(1.8a) or

HHV = 17,887 + 57.5°API-102.2%S (Btu/lb)

(1.8b) Using this, HHV = 45,350 kJ/kg.

HHV = 41,605 + 133.74°API-237.7%S (kJ/kg)

Using (1.6), the amount of dry stoichiometric air required = 11.53 × 0.875 + 34.34 × 0.125 = 14.38 kg/kg fuel. Here, 1 million kJ of energy on HHV basis has 10 6 /45,350 = 22.05 kg fuel; hence, 22.05 kg fuel requires 14.38 × 22.05 = 317 kg dry air. (1 MM Btu of energy has 10 6 /19,500 = 51.3 lb fuel. The HHV, in Btu/lb, is 19,500. Hence, 1 MM Btu fuel requires 51.3 × 14.38 = 738 lb of dry air.)

Example 1.3

Check the combustion air required for million kJ of natural gas having methane = 83.4%, ethane = 15.8%, and nitrogen = 0.8% by volume.

Solution

MW of the fuel = 0.834 × 16 + 0.158 × 30 + 0.008 × 28 = 18.3. % weight of methane = 83.4 × 16/18.3 = 72.9; % weight of ethane = 15.8 × 30/18.3 = 25.9

w da = 17.27 × 0.729 + 16.12 × 0.259 = 16.76 kg/kg fuel (17.27 and 16.12 were taken from Tables 1.1 and 1.2).

HHV = (0.729 × 23,876 + 0.256 × 22,320) × 2.326 = 53,776 kJ/kg (convert from Btu/lb to

kJ/kg using a factor of 2.326). Here, 1 million kJ has 10 6 /53,776 = 18.60 kg fuel; hence, air required for firing 1 million kJ = 18.6 × 16.76 = 312 kg. Let us take 100% methane. HHV = 23,879 × 2.326 = 55,543 kJ/kg; methane requires 17.265 kg air/kg (from

Tables 1.1 and 1.2). Hence, 1 million kJ of energy has 10 6 /55,543 = 18 kg, and this requires 18 × 17.265 =

310.8 kg dry stoichiometric air. Thus, for a variety of fuels, we can perform this calculation and show that within a small % variation, the dry stoichiometric air required for combustion of a type of fuel is a constant (see Table 1.5).

Example 1.4

A fired heater is firing natural gas at 15% excess air in a furnace; energy input is 50 million kJ/h on an HHV basis. Assume HHV = 52,500 kJ/kg. Determine the air and flue gas produced.

Solution

We don’t have the fuel analysis and hence use the simplified procedure. Air required =

50 × 313 × 1.15 × 1.013 = 18,231 kg/h (1.013 refers to the moisture in air at the ambient temperature of 27°C and relative humidity of 60%. Factor A = 313). Fuel required = 50 ×

10 6 /52,500 = 952 kg/h; hence flue gas produced = 18,231 + 952 = 19,183 kg/h. Working the same problem in British units, (50 × 10 6 kJ/h = 50 × 10 9 /1055 Btu/h = 47.39 MM Btu/h. 1 MM Btu requires 728 lb dry theoretical amount of air. Hence, actual air required = 47.39 × 728 × 1.15 × 1.013 = 40,190 lb/h).

Gaseous fuels are often fired in steam generators, and Table 1.6 gives the fuel gas analysis for a few typical gaseous fuels, and Table 1.7 shows the ultimate analysis for typical fuel oils.

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

TABLE 1.6

Analysis of Some Gaseous Fuels

Gas

Coke Oven

Blast Furnace

Natural Gas

Natural Gas Refinery Gas

% Volume hydrogen 47.9 2.4 45 Methane

33.9 0.1 83.4 97 12.5 Ethylene

Ethane 15.8 1 7 Propane

23 Butane

11 Carbon monoxide

Carbon dioxide 2.6 14.4 0.5 Nitrogen

Water vapor

Hydrogen sulfide 1.5 Specific gravity

0.569 0.76 (relative to air)

HHV, MJ/m 3 (Btu/ft 3 )

Analysis of Typical Fuel Oils

Grade or No. 1 2 4 5 6

86.5–89.2 86.5–89.2 Hydrogen

Carbon, % weight

0–0.1 0.01–0.5 HHV, kJ/kg

43,170 42,333 HHV, Btu/lb (avg)

18,560 18,200 HHV, kcal/kg

Estimation of Heating Values

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

HHV = 14,500C + 62,000(H 2 –O 2 /8) + 4000S

(1.9a)

(1.9b) Some texts use 9200 instead of 9720. Difference in LHV is small.

LHV = HHV – 9720H 2 – 1110W

Where W is the weight fraction of moisture in fuel, and C, H 2 ,O 2 , 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.

Combustion Calculations

If ultimate analysis is not known, one may estimate the heating values of fuel oils using [3] HHV = 17,887 + 57.5 × °API – 102.2S

(1.10a) LHV = HHV – 92 × H 2 (HHV and LHV in Btu/lb) (1.10b)

(1.11) where F = 24.5 for 0 <= °API <= 9

%H 2 = F – 2122.5/(°API + 131.5)

F = 25 for 9 <= °API <= 20

F = 25.2 for 20 <= °API <= 30

F = 25.45 <= °API <= 40

Simplified Combustion Calculations