Homework set 14 solutions:
Problem 1 (number 13.6 from the 3rd ed.)
First balance the equation that involves the complete reaction of the fuel
with 100% theoretical air.
.7CH4+.2CO+.05O2+.05N2+a(O2+3.76N2)->bCO2+cH2O+(3.76a+.05)N2
Solve for a=1.45, b=.9, and c=1.4
Now, for 120% theoretical air, we want to replace a with 1.2 a
1.2 times a = 1.74
Now the equation we want to balance is:
.7CH4+.2CO+.05O2+.05N2+1.75(O2+3.76N2)>bCO2+cH2O+(3.76*1.74+.05)N2 + dO2
Solve for b=.9, c=1.4, d=0.29
Because we had excess air, there will now be O2 in the products…
Problem 2: (number 13.12 from the 3rd ed.)
The relationship between mass and number of moles is m=nM
where m is mass, n is moles, and M is molar mass.
For 100Kg coal, we’d have 88kg C, 6 Kg H, 4 Kg O, 1 Kg of N and S.
The number of moles present for each of these is:
7.33 moles of C, 6 moles of H, 0.25 moles of O, .07 moles of N and .03
moles of S.
For a complete reaction with 100% air:
7.33C+6H+.25O+.07N+.03S+a(O2+3.76N2)------------ >bCo2+cH2O+dSo2+(3.76a+.07/2)N2
Solve for:
a=8.735, b=7.33, c=3, and d=.03
So, there are .03 kmoles of SO2 produced per 100kg of coal. The mass of
.03 kmole SO2 is 1.92 kg
So, there is 1.92 kg of SO2 produced per 100Kg coal, or 0.0192 kgSO2 per 1
kg coal.
Problem 3 (number 13.20 from the 3rd ed.)
The chemical equation for dry air would be:
CH4+3(O2+3.76N2)Co2+2H2O+O2+11.28N2
There is O2 because the reaction involves excess air.
Now lets add moisture:
Pv=phi*Pg=0.0237bars
also, Pv=nv/(na+nv)P
Na is the moles of dry air which in this case is 3+3*3.76=14.28
Solve the above equation for nv = 0.367 kmoles This is how much water
vapor is in the moist air.
Now the chemical equation is:
CH4+3(O2+3.76N2)+0.367H2OCo2+2.367H2O+O2+11.28N2
To get the dew point,
Pv=(nv/ntotal)P=2.367/(1+2.367+1+11.28)*(0.945)=.143 bars
Taable A-2 gives Tdp=53C