Temperature = 1200 F = 922.04 K Waste gas flow rate = 1000 scfm = 471.95 sLitre/
ID: 699707 • Letter: T
Question
Temperature = 1200 F = 922.04 K
Waste gas flow rate = 1000 scfm = 471.95 sLitre/s
= 471.95 * 922.04 / 273.15 L/s = 1593.1 L/s
At STP, 1 mol = 22.414 L
Waste gas flow rate = 471.95 sLitre/s / 22.414 L/mol
= 21.056 mol/s
Flow rate of benzene = 1 / 100 * 1000 = 10 scfm = 4.72 sL/s = 0.21 mol/s
Flow rate of toluene = 2 / 100 * 1000 = 20 scfm = 9.44 sL/s = 0.42 mol/s
Flow rate of ethyl mercaptan = 0.5 / 100 * 1000 = 5 scfm = 2.36 sL/s = 0.105 mol/s
Flow rate of nitrogen = 1000 – 10 – 20 – 5 = 965 scfm = 455.43 sL/s = 20.32 mol/s
C6H6 + 15/2 O2 --> 6 CO2 + 3 H2O : Benzene combustion
C7H8 + 9 O2 --> 7 CO2 + 4 H2O : Toluene combustion
C2H6S + 9/2 O2 --> 2 CO2 +3 H2O + SO2 : Ethyl mercaptan combustion
n = stoichiometric coefficient
a)
For benzene combustion:
Flow rate of O2 required = Flow rate of benzene * n * efficiency
= 0.21 mol/s * 15/2 * 99.9 / 100 = 1.578 mol/s
For toluene combustion:
Flow rate of O2 required = Flow rate of toluene * n * efficiency
= 0.42 mol/s * 9 * 99.9 / 100 = 3.786 mol/s
For ethyl mercaptan combustion:
Flow rate of O2 required = Flow rate of ethyl mercaptan * n * efficiency
= 0.105 mol/s * 9/2 * 99.99 / 100 = 0.474 mol/s
Total theoretical flow of O2 required = 1.578 mol/s + 3.786 mol/s + 0.474 mol/s
= 5.84 mol/s
Assuming air contains 21% O2 and 79 % N2,
Total theoretical flow of air required = 5.84 mol/s * 100 / 21 = 27.8 mol/s
Or
Total theoretical flow of air required = 27.8 mol/s * 22.414 sL/s
= 623 sL/s
b)
In the effluent gas,
Total flow of nitrogen = 20.32 mol/s + 27.8 mol/s - 5.84 mol/s = 42.28 mol/s
Total flow of O2 = 0 (Assuming no excess oxygen in air)
Total flow of benzene = 0.21 mol/s * (1 – 99.9/100) = 2.11 x 10-4 mol/s
Total flow of toluene = 0.42 mol/s * (1 – 99.9/100) = 4.21 x 10-4 mol/s
Total flow of C2H6S = 0.105 mol/s * (1 – 99.99/100) = 1.05 x 10-5 mol/s
Total flow of CO2 = Flow rate of reactants * n * efficiency
= (0.21 * 6 * 99.9/ 100) + (0.42 * 7 * 99.9/100) + (0.105 * 2 * 99.99/100)
= 4.42 mol/s
Total flow of H2O = Flow rate of reactants * n * efficiency
= (0.21 * 3 * 99.9/ 100) + (0.42 * 4 * 99.9/100) + (0.105 * 3 * 99.99/100)
= 2.63 mol/s
Total flow of SO2 = Flow rate of reactants * n * efficiency
= (0.105 * 1 * 99.99/100)
= 0.105 mol/s
Component
Mol/s
%
N2
42.279
85.53
O2
0
0.00
Benzene
2.11E-04
0.00
Toluene
4.21E-04
0.00
C2H6S
1.05E-05
0.00
CO2
4.418
8.94
H2O
2.630
5.32
SO2
0.105
0.21
c)
Rate of benzene reaction = A exp(-Ea/RT) [benzene]
1 / time1 = 7.43 x 2021 * exp(-95.9 * 1000 / (1.987 * 922.04)) = 1.72 x 105 s-1
Time1 = 5.82 x 10-6 s
Rate of toluene reaction = A exp(-Ea/RT) [toluene]
1 / time2 =2.28 x 1013 * exp(-56.6 * 1000 / (1.987 * 922.04)) = 0.873 s-1
Time2 = 1.14 s
Rate of C2H6S reaction = A exp(-Ea/RT) [C2H6S]
1 / time3 =5.2 x 105 * exp(-14.7 * 1000 / (1.987 * 922.04)) = 170 s-1
Time3 = 0.006 s
Residence time required = maximum of (time1, time2, time3)
= 1.14 s
d)
Inlet gas flow rate = 1593.1 L/s
Residence time required = 1.14 s
Size of chamber = 1593.1 L/s * 1.14 s
= 1825 L
Component
Mol/s
%
N2
42.279
85.53
O2
0
0.00
Benzene
2.11E-04
0.00
Toluene
4.21E-04
0.00
C2H6S
1.05E-05
0.00
CO2
4.418
8.94
H2O
2.630
5.32
SO2
0.105
0.21
Explanation / Answer
A waste gas containing by volume 1% benzene (C6, Hs), 2% toluene (C7 H8) and 0.5% ethyl mercaptan (C2H6S), with the rest nitrogen. The waste gas flow rate is 1000 scfm d the operating temperature is 12000F. It is desired to achieve 99.9% destruction efficiency for toluene and benzene, and 99.99% destruction efficiency for ethyl mercaptan, calculate the following: a)theoretical air requirement in molar and volumetric rates; b) the composition of the efficient gas; c) the required residence time of the waste gas in the treatment system; d) the size of the combustion chamber based on the inlet gas flow rate.
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