3rd Class Edition 3 Part A2 Fuels, Combustion, and Flue Gas Analysis PDF

?& Chapter 3 • Fuels, Combustion, and Flue Gas Analysis OBJECTIVE 5 Given the ultimate analysis of a fuel, use Dulong's formula to calculate the heating value of fuel. DULONG'S FORMULA The heating value of a fuel can be calculated using the results of the ultimate analysis of the fuel. From the f...

?& Chapter 3 • Fuels, Combustion, and Flue Gas Analysis OBJECTIVE 5 Given the ultimate analysis of a fuel, use Dulong's formula to calculate the heating value of fuel. DULONG'S FORMULA The heating value of a fuel can be calculated using the results of the ultimate analysis of the fuel. From the fuel analysis, the percentages of the combustibles (carbon, hydrogen, and sulfur) are known. Since the heat of combustion for these elements is known, the calorific value of a fuel can be calculated using Dulongs formula, which is: AS Higher heating value = 33.7 C + 144 ( W.2-— ) +9.3 S 8 Where C, H2, and S represent the mass of carbon, hydrogen, and sulfur per kilogram of fuel. Dulongs formula is also found in the PanGlobal Academic Supplement. The percentages of each substance are entered into the formula as decimals. Note: When using Dulong's formula, the result is stated in MJ per kg of fuel, since: 33.7 MJ = heat produced by combustion of 1 kg of C 144 MJ = heat produced by combustion of 1 kg of H^ 9.3 MJ = heat produced by combustion of 1 kg of S 02N The equation [H^-— } represents available hydrogen. If the fuel contains 0^ it is assumed that the oxygen is contained in moisture, in the form ofH^O. Therefore, the amount of 0 '2 . hydrogen available for combustion is reduced because an amount of-^- is locked up as water. 8 Example 12 Calculate the higher heating value of a fuel in units ofkj/kg, given the following ultimate analysis: Carbon 65.0% (0.65 kg/kg of fuel) Hydrogen 4.7% Sulfur 0.5% Oxygen 9.8% Nitrogen 18.2°/o Ash 1.8% Solution 12 Higher heating value = 33.7 C + 144 ( V.2--1) +9.3 S = (33.7 x 0.65) + 144 (0.047 - 0:09S-} + (9.3 x 0.005) = 21.905 + 144 (0.047 - 0.01225) + 0.0465 21.905 + 5.004 + 0.0465 = 26.9555 MJ/kJ = 26 956 kj/kg (Ans.) 132 3rd Class Edition 3 • Part A2 Fuels, Combustion, and Flue Gas Analysis • Chapter 3 Example 13 A sample of coal has the following analysis by mass: carbon 82%, hydrogen 8%, sulfur 2%, oxygen 4%, and ash 4%. Calculate the higher heating value of the fuel using Dulongs formula. Solution 13 Use Dulongs formula and remember to insert the numbers as decimal fractions. Higher heating value = 33.7 C + 144 (^—1) +9.3 S 8 = 33.7 x 0.82 + 144 fo.08-a04^ + 9.3 x 0.02 = 27.634+144 (0.08-0.005)+0.186 = 27.634 + 144 (0.075) + 0.186 27.634+10.8+0.186 = 38.62 MJ/kg (Ans.) Example 14 Given the following compositions by mass, calculate the higher and lower heating values: carbon 85.5%, hydrogen 12.5%, and oxygen 2%. Take the heating value of carbon as 33.7 MJ/m3 and hydrogen as 144 MJ/m3. Take the enthalpy of evaporation of steam to be 2257 kj/kg at atnospheric pressure. Solution 14 Use Dulongs formula to calculate the higher heating value (HHV) and remember to insert the numbers as decimal fractions. 0, Higher heating value = 33.7 C + 144 ( V.2-— ) + 9.3 S = 33.7 x 0.855 + 144 f 0.125-ao2^+ 9.3x0 = 28.8135 + 144 (0.125 - 0.0025) 28.8135 + 144 x 0.1225 28.8135 + 17.64 = 46.45 MJ/kg (Ans.) Based on 1 kg of fuel, calculate the enthalpy of evaporation of the water content. Start by writing out the combustion equation for hydrogen on a mass basis. 2H2 + Os -> 2H20 2 kmol + 1 kmol 2 kmol 4kg Hz + 32 kg 02 -> 36 kg HzO IkgHz + 8kg02 ^ 9kgH20 From the analysis by mass, 1 kg of fuel has 0.125 kg of N2. Multiply the last equation by 0.125 to calculate the moisture formed during the combustion of 1 kg of fuel. 0.125 kg Hz + 1kg 02 -> 1.125 kg HzO Enthalpy of evaporation of 1.125 kg H^O = 1.125 x 2.257 MJ/kg = 2.54 MJ/kg LHV = HHV - enthalpy of evaporation = 46.45 - 2.54 = 43.91 MJ/kg (Ans.) 3rd Class Edition 3 • Part A2 133