Question

A coal from Texas which has an ultimate analysis (by mass \()\) as 39.25 percent \(C, 6.93\) percent \(H_{2}, 41.11\) percent \(O_{2}\) 0.72 percent \(\mathrm{N}_{2}, 0.79\) percent \(\mathrm{S},\) and 11.20 percent ash (non combustibles) is burned steadily with 40 percent excess air in a power plant boiler. The coal and air enter this boiler at standard conditions and the products of combustion in the smokestack are at \(127^{\circ} \mathrm{C}\). Calculate the heat transfer, in \(\mathrm{kJ} / \mathrm{kg}\) fuel, in this boiler. Include the effect of the sulfur in the energy analysis by noting that sulfur dioxide has an enthalpy of formation of \(-297,100 \mathrm{kJ} / \mathrm{kmol}\) and an average specific heat at constant pressure of \(c_{p}=41.7 \mathrm{kJ} / \mathrm{kmol} \cdot \mathrm{K}\).

Short answer

Answer: The heat transfer in the boiler is -21535.27 kJ/kg of fuel.

Step-by-step solution

Find the number of moles of each element in the coal

First, convert the given mass percentages of each element into their respective number of moles. The molar mass of C is 12 g/mol, H2 is 2 g/mol, O2 is 32 g/mol, N2 is 28 g/mol, and S is 32 g/mol. Assume a 100 g mass for simplicity. C: (39.25 g C) / (12 g/mol) = 3.2708 mol C H2: (6.93 g H2) / (2 g/mol) = 3.4650 mol H2 O2: (41.11 g O2) / (32 g/mol) = 1.2847 mol O2 N2: (0.72 g N2) / (28 g/mol) = 0.0257 mol N2 S: (0.79 g S) / (32 g/mol) = 0.0247 mol S

Determine the moles of excess air

We are given that there is 40% excess air used in the combustion process. Since we need 1 mol O2 for each mole of C to form CO2, and 0.5 moles O2 for each mole of H2 to form H2O, we can determine the required moles of O2 for complete combustion and then find the amount of excess air. Required O2 for C: 3.2708 mol C Required O2 for H2: 0.5 * 3.4650 mol H2 = 1.7325 mol O2 Total required O2: 3.2708 + 1.7325 = 5.0033 mol O2 40% excess air means 1.4 times the required O2: Actual moles of O2 in air: 5.0033 * 1.4 = 7.0046 mol O2 Now, find the moles of N2 present in the air (assuming atmospheric air contains 21% O2 and 79% N2): Moles of N2 in air: 7.0046 * (79/21) = 26.4054 mol N2

Determine the moles of products after combustion

Determine the moles of each product formed after combustion: CO2: 3.2708 mol (all C combusts) H2O: 3.4650 mol (all H2 combusts) N2: 0.0257 (in coal) + 26.4054 (in air) = 26.4311 mol SO2: 0.0247 mol (all S combusts)

Calculate the change in enthalpy for the reactants and products

Using the enthalpy of formation for each species, and given the enthalpy of formation for SO2, find the sum of the enthalpies of formation for reactants and products: ΔH_reactants = ΔH_C + ΔH_H2 + ΔH_O2 + ΔH_N2 + ΔH_S ΔH_products = ΔH_CO2 + ΔH_H2O + ΔH_N2 + ΔH_SO2 Using standard enthalpy of formation values (ΔHf°) and given average specific heat capacity (cp) for SO2, we get: ΔH_reactants = (3.2708 mol * 0) + (3.4650 mol * 0) + (1.2847 mol * 0) + (0.0257 mol * 0) + (0.0247 mol * 0) = 0 kJ ΔH_products = (3.2708 mol * -393,520 kJ/kmol) + (3.4650 mol * -241,820 kJ/kmol) + (26.4311 mol * 0) + (0.0247 mol * -297,100 kJ/kmol) + temperature_difference * (0.0247 mol * 41.7 kJ/kmol*K) Where the temperature difference is (127 - 25) = 102°C, which is the difference in temperature between the product and the reactants (assuming the reactants are at standard conditions of 25°C). ΔH_products = -2153630.96 kJ + (0.0247 mol * 41.7 * 102) = -2153630.96 + 103.876 kJ = -2153527.084 kJ

Calculate the heat transfer in the boiler

Now, we can calculate the heat transfer per unit mass of fuel (Q) by finding the difference between the enthalpy of the products and the reactants: Q = ΔH_products - ΔH_reactants Q = -2153527.084 kJ - 0 kJ Q = -2153527.084 kJ Since we assumed 100 g of fuel for simplicity, to find the heat transfer per kg of fuel, we need to divide this value by 0.1 kg (100 g): Q = (-2153527.084 kJ) / (0.1 kg) = -21535.27084 kJ/kg fuel Therefore, the heat transfer in the boiler is -21535.27 kJ/kg of fuel, considering the given composition of coal and the effect of sulfur in the energy analysis.