Question #97451

Solid fuel 70% carbon, 9% hydrogen, 14% Sulphur, 3% oxygen, 2% nitrogen and inorganic ash.

Theoretical amount of air required to combust 400kg of this fuel.

Theoretical amount of air required to combust 400kg of this fuel.

Expert's answer

First, we should calculate the mass of each element in this amount of the fuel:

"m(C)=M_f*\\frac{w(C)}{100\\%}=400kg*\\frac{70\\%}{100\\%}=280kg;"

"m(H)=M_f*\\frac{w(H)}{100\\%}=400kg*\\frac{9\\%}{100\\%}=36kg;"

"m(S)=M_f*\\frac{w(S)}{100\\%}=400kg*\\frac{14\\%}{100\\%}=56kg;"

"m(N)=M_f*\\frac{w(N)}{100\\%}=400kg*\\frac{2\\%}{100\\%}=8kg;"

"m(O)=M_f*\\frac{w(O)}{100\\%}=400kg*\\frac{3\\%}{100\\%}=12kg."

Three elements will be oxidized through the combustion process (carbon, hydrogen and sulfur):

"C+O_2 \\rightarrow CO_2;"

"S+O_2 \\rightarrow SO_2;"

"H+\\frac{1}{4}O_2 \\rightarrow \\frac{1}{2}H_2O."

Nitrogen will be oxidized only under conditions close to those in the internal combustion engine.

Now, we must calculate the amount in moles of carbon, hydrogen, sulfur and oxygen in the fuel:

"n(C)=\\frac{m(C)}{M(C)}=\\frac{280kg}{12*10^{-3}kg\/mol}=23333mol;"

"n(S)=\\frac{m(S)}{M(S)}=\\frac{56kg}{32*10^{-3}kg\/mol}=1750mol;"

"n(H)=\\frac{m(H)}{M(H)}=\\frac{36kg}{1*10^{-3}kg\/mol}=36000mol;"

"n(O)=\\frac{m(O)}{M(O)}=\\frac{12kg}{16*10^{-3}kg\/mol}=750mol."

Lets calculate the total amount in moles of the oxygen required for the oxidation of the fuel. We also should not forget about oxygen in the fuel, which is also involved into the oxidation process ("2O \\rightarrow O_2"):

"n_{total}(O_2)=n(C)+n(S)+\\frac{1}{4}n(H)-\\frac{1}{2}n(O)_{fuel}=23333mol+1750mol+\\frac{1}{4}*36000mol-\\frac{1}{2}*750mol=33708mol."

Now, we have to found the equivalent volume of pure oxygen measured under normal conditions:

"V_{O_2}=n_{total}(O_2)*V_m=33708mol*22.4L\/mol=755059L(\\approx755m^3)."

Finally, we all know - the air contains approximately 1/5 volume part of pure oxygen:

"V_{air}=5*V_{O_2}=5*755059L=3775295L(\\approx3775m^3)."

Thus, 400 kg of this fuel requires approx. 3775 m^{3} of air and this is the theoretical minimum, which does not involve oxidation of nitrogen etc.

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