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raniero
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A quantity of coal used in a boiler had the following analysis: 82% C; 5% H; 6% O; 2% N; 5% ash. The dry flue gas analysis showed 14% CO2 and some oxygen. Calculate the oxygen content of the dry flue gas.
I have first written the formula for combustion including the variables for balancing the equation:
Considering 1kg of coal:
0.82/12 C + 0.05 H + 0.06/1 O + 0.02/14 N + x(0.233/32 O2 + 0.767/28N2) → B(0.14 CO2 + a O2 + (0.86-a) N2)+ b H2O
When balancing the equation this yields:
Carbon Balance: B = 0.4881
Hydrogen Balance: 0.05 = 2b → b =0.025
The problem I am encountering arises in the oxygen balancing:
Oxygen balance: 0.06/16 + x(0.233/32)2 = (0.14)2B + 2aB + b
The book's answer states: 0.06/16 + x(0.233/32) = (0.14)B + aB + b/2
How come the book's answer doesn't add double the amount of substance (kmol) as I did for Oxygen molecules?
Other answers from the book compensate by multiplying by two when molecules such as O2 are present. Another disagreement I have with the solution is the 'b' divided by 2.
Thanks in advance
I have first written the formula for combustion including the variables for balancing the equation:
Considering 1kg of coal:
0.82/12 C + 0.05 H + 0.06/1 O + 0.02/14 N + x(0.233/32 O2 + 0.767/28N2) → B(0.14 CO2 + a O2 + (0.86-a) N2)+ b H2O
When balancing the equation this yields:
Carbon Balance: B = 0.4881
Hydrogen Balance: 0.05 = 2b → b =0.025
The problem I am encountering arises in the oxygen balancing:
Oxygen balance: 0.06/16 + x(0.233/32)2 = (0.14)2B + 2aB + b
The book's answer states: 0.06/16 + x(0.233/32) = (0.14)B + aB + b/2
How come the book's answer doesn't add double the amount of substance (kmol) as I did for Oxygen molecules?
Other answers from the book compensate by multiplying by two when molecules such as O2 are present. Another disagreement I have with the solution is the 'b' divided by 2.
Thanks in advance