200 kg of coal is burned steadily in a power plant boiler. This coal has an ulti

Question

200 kg of coal is burned steadily in a power plant boiler. This coal has an ultimate analysis by mass of consists 39.25% Carbon, 6.93 % Hydrogen, 41.11% Oxygen, 0.72% Nitrogen, 0.79% Sulfur and 11.20% ash (non-combustible). This coal is burned with 25% excess air in the boiler. The coal and air enter this boiler at standard conditions (1 atmospheric pressure and 298K) and the products of combustion are released through the chimney at 400 K. Utilizing the conservation of mass principle, derive the chemical reaction equation for this combustion process by balancing the no. of atoms of each element. Determine the mole number of each component in the products. Determine the Air-fuel ratio for this combustion process and the dew point temperature of the products. Calculate the amount of heat released in KJ/kmol of fuel for this combustion process. Use the mini property table below for the enthalpy of substance at different states:

Explanation / Answer

Let the basis of calculation be 100 kg of coal

2

Based on the above table, the answers for above questions are given below.

1. The balaned eqautions based on conservation mass principle are

39.25 kg of C + 104.666 kg of O2 = 143.916 kg of CO2

6.93 kg of H2 + 55.44 kg of O2 = 62.37 kg of H2O

0.79 kg of S + 0.79 kg of O2 =1.58 kg of SO2

There is no combustion for N2, O2 and Ash.

2.

3.27 K mols of C + 3.27 K mols of O2 = 3.27 K mols of CO2

3.465 K mols of H2 + 1.7325 K mols of O2 = 3.465 K mols ofof H2O

0.024 K mols of S + 0.024 K mols of O2 =0.024 K mols of SO2

There is no combustion for N2, O2 and Ash.

3. the theoritical quantity of O2 required for the combustion of 100 kg of coal =  119.786 kg

For 200 kg = 2x92.066 = 239.572 kg

Wt. of air required for the complete combustion for 200 kg of coal (25 % excess)

= 239.572 x (100/23) x (125/100)

= 239.572 x 4.3478 x 1.25

= 1302.01 kg

Therefore Air:Fuel is 1302.01:1

Constituent % by Wt. Mol. Wt No. of Kmole Chemical reaction Wt. of O2 required in kg Kmols of O2 required for combustion C 39.25 12 39.25/12 = 3.27 C+O2 = CO2 (39.25x32)/12 = 104.666 3.27x1 = 3.27 H2 6.93 2 6.93/2 = 3.465 H2+0.5O2 = H2O (6.93x0.5x32)/2 =55.44 3.465x0.5 = 1.7325 O2 41.11 32 41.11/32 =1.284 - -41.11(Because Oxygen is present in coal, , So No need of Extra Oxygen) -1.284 (Because Oxygen is present in coal, So No need of Extra Oxygen) N2 0.72 28 0.72/28 = 0.0257 - - - S 0.79 32 0.79/32 = 0.024 S+O2 = SO2 (0.79x32)/32 = 0.79 0.024x1 = 0.024 Ash 11.20 - - - - Total wt.(in kg) Of O2 required per 100 kg of coal is (104.666+ 55.44+0.79)-(41.11) =119.786 kg Total K mols Of O2 required per 100 kg of coal is (3.27+ 1.7325+ 0.024)-(1.284) = 3.7425

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