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

We know that out of the given elements in the coal carbon , hydrogen and sulphur will undergo combustion

I. The reactions will be

C(s) + O2 --> CO2(g)

H2 + 1/2O2 --> H2O

S + O2 --> SO2

II. The mass of fuel is = 200Kg

So if the composition given by mass then ,

(a) Mass of carbon = 39.25 X 200Kg / 100 = 78.5 Kg

So moles of carbon = 78.5 Kg / 12 g/mole = 6.54 X 10^3 moles

(b) Mass of hydrogen = 6.93 X 200 / 100 = 13.86 Kg

So moles of hydrogen = Mass / molecular weight = 13.86 X 1000 / 2 = 6.93 X 10^3 moles

(c) Mass of oxygen = 41.11 X 200 / 100 = 82.22 Kg

So moles = 82.22 X 1000 / 16 = 5.14 X 10^3 moles

(d) Mass of sulphur = 0.79 X 200 / 100 = 1.58 Kg

So moles = 1.58 / 32 = 49.375 moles

III. Air fuel ratio will be

For combustion of each

C(s) + O2 --> CO2(g)

For one mole of C we need 1 mole of O2

So 6.54 X 10^3 moles of C will need = 6.54 X 10^3 moles of Oxygen

H2 + 1/2O2 --> H2O

For one mole of H2, 0.5 moles of O2

So for 6.93 X 10^3 of H2 will need = 3.47 X 10^ 3 moles of O2

S + O2 --> SO2

One mole requires one mole of O2

so 49.375 moles will need 49.375 moles of O2

Total moles of O2 required = 49.375 + 3.47 X 10^ 3 + 6.54 X 10^3 = 10.059 X 10^3 moles

Oxygen already present = 5.14 X 10^3 moles

so moles of oxygen required = 4.91 X 10^3 moles

For each mole of oxygen the air required =4.762 moles

So air required = 4.762 X 4.91 X 10^3 moles

Moles of fuel undergoing combustion = 49.375 + 6.93 X 10^ 3 + 6.54 X 10^3 = 13.519 X 10^3 moles

Ratio of fuel / air = 13.519 / 23.381 = 0.578 : 1

IV )

The heat will be releaes due to following combustion reactions

(a) C(s) + O2 --> CO2(g)

Heat of reaction = Enthalpy of formation of CO2 =393.520 KJ / mole

So heat released due to combustion of 6.54 X 10^3 moles will = -248.82 X 6.54 KJ = -2573.62X 10^3 KJ

H2 + 1/2O2 --> H2O

HEat of reaction = Heat of formation of water

So heat of reaction = -241.820 KJ / mole

Mole of water formed = 6.93 X 10^3 moles

So heat released will be = 6.93 X 10^3 X -241.52 = 1673.73 X 10^3 KJ

S + O2 --> SO2

Heat of reaction = Heat of formation of SO2 = -297.1 KJ / mole

So heat of formatio of 49.375 moles = 49.375 X -297.1 X 10^3 KJ = 14.669 X 10^ KJ

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