A fuel composed, by weight, of 70% carbon, 28% hydrogen, and 2% sulfur is burned

Question

A fuel composed, by weight, of 70% carbon, 28% hydrogen, and 2% sulfur is burned with 10% excess air For an 800 MWe plant with a 40% thermal efficiency:

3. In the Combustion Example of Lecture 7, a "divide and conquer" approach was taken to solve the problem. Repeat that exercise except (1) perform all the calculations (for example, the fuel input mass flow rate) exclusively using SI units (i.e., kJ/kg and kg/hr), and (2) do not divide and conuer,rather use the single long stoichiometric balance equation and explicitly calculate the conquer, rather use the single long stoichiometric balance equation and explicitly calculate the number of moles of the reactants and products to determine the flue gas emissions in kg/hr. (You do not have to compute the scrubber system input/output masses.) [20 pts]

Explanation / Answer

I think you wish to know the reactions of the given elements with oxygen, as nothing is mentioned what to determine.

C + O2 --> CO2

H2 + 1/2O2 --> H2O

S + O2 --> SO2

so if 1 kg of fuel is taken then

amount of carbon = 700g

amount of hydrogen = 280 grams

Amount of sulphur = 20 grams

As per the stoichiometry

For 1 mole (12gram) of carbon 1 mole of oxygen(32g) is required

so for 700 grams of carbon = 32 X 700 / 12 grams of oxygen will be required = 1866.7 grams of oxygen

For 280 grams = 16 X 280 / 2 = 2240 grams

For 20 grams of sulphur = 20 grams of oxygen will be required

So total oxygen required = 2240 + 20 + 1866.7 grams = 4126.7 grams

The percentage of oxygen in air per kg is 23 %

so for 23 grams 100 grams of air is needed

For 4126.7 grams of oxygen air required = 100 X 4126.7 / 23 = 17942 grams of air = 17.94 Kg

20% excess air = 20 X 17942 / 100 = 3.588 Kg

total air = 17.94 + 3.588 = 21.52 Kg for 1Kg of fuel

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