Question

Dry coke composed of 4% inert solids (ash), 90% carbon, and 6% hydrogen is burned in a furnace with dry air. The solid refuse left after combustion contains 10% carbon and 90% inert ash (and no hydrogen). The inert ash content does not enter into the reaction. The orsat analysis of the flue gas gives 13.9% co2, 0.8% co, 4.3% o2, and 81.0% n2. Calculate the percent of excess air based on complete combustion of the coke

Answer 1

Answer:

                            Mass fraction         mass fraction

$O_{2}$         0.21          C            0.10

$N_{2}$         0.79       inert         0.90

                                 1.00                        1.00

                              mole%              C

$CO_{2}$     13.9               13.9

CO                            0.8               0.8

$O_{2}$         4.3

$N_{2}$       81.0

Basis is 100 kg mol P

Here unknowns are  F,R,A,W

Probably of but check for redundancy

Balances (in moles):

      C:       $\frac{F(0.90)}{12} =P(0.139+0.008)+\frac{R(0.10)}{12} +W(0)$

      H:       $\frac{F(0.06)}{1.005} =P(0)+R(0)+W(2)$

    2N:       $A(0.79)=P(0.81)+R(0)+W(0)$

    2O:       $A(0.21)=P(0.139+\frac{0.008}{2} +0.43)+\frac{W}{2}$

    Inert(mass):  F(0.04)=R(0.90)

   From 2N,      $A=\frac{100(0.81)}{0.79} =102.53kgmol$

   From 2O,      W=5.863kgmol

   From H,         F=196.98kg

   From inert     R=8.7548kg

            H in:$\frac{196.98(0.06)}{2} =5.94$

            O2 in:$\frac{5.94}{2} =2.96$

            C in:$\frac{196.98(0.90)}{12} =14.77$

check via C:  196.98(0.90)/12=100(0.147)+(8.7548(0.10)/12)

Required O2: 14.7+(5.863/2)=17.63 kg mol

                       Total O2 in=102.53(0.21)=21.53

Excess(21.53-17.63)/17.63=0.221 or 22%