A. if the solvent is not soluble in the gas phase, what are the components of th
ID: 894650 • Letter: A
Question
A. if the solvent is not soluble in the gas phase, what are the components of the carbon dioxide lean stream?
B. Perform a degree of freedom analysis.
C. Find the flow rates of streams D and E in Kg/s.
D. Determine the composition of stream D. 3. A liquid solvent (ethanolamine) is used to remove carbon dioxide from the fl gas of a power plant in an absorption column as shown in the diagram below. in the diagram below. CO2 Lean Gas, D Pure liquid solvent, S = 1000 kg/s CO2 Rich Solvent, E Composition 83.5 % Solvent 15% CO2 1% SO2 0.5% NO2 Flue Gas, F = 100 kmol/s Composition 78 % N2 12%C02 5%2 190 CO 2% SO2 190 NO2 a. If the solvent is not soluble in the gas phase, what are the componer b. c. d. the CO2 lean stream? Perform a degrees of freedom analysis. Find the flow rates of streams D and E in kg/s. Determine the composition of stream D.
Explanation / Answer
A. if the solvent is not soluble in the gas phase, what are the components of the carbon dioxide lean
stream?
N2, CO2, O2, CO, SO2, NO2
B. Perform a degree of freedom analysis.
Inlet Streams:
(i)Pure solvent, S = 1000 kg/s
(ii)Flue Gas, F = 100 kmol/s
Mole percentage composition:
N2 = 78
CO2 = 12
O2 = 05
CO = 1
SO2 = 2
NO2 = 1
Average molar mass of flue gas = MWF = 0.78*28+0.12*44+0.05*32+0.01*28+0.02*64+0.01*46= 30.74
Outlet streams:
(iii)CO2, rich solvent = E kg/s
Mass percentage composition:
solvent, S = 83.5
CO2 = 15
SO2 = 1
NO2 = 0.5
(iv)Lean gas, CO2 = D kg/s
Average molar mass of lean gas = MWD
Mass fraction composition:
N2 = Wdn2
CO2 = Wdco2
O2 = Wdo2
CO = Wdco
SO2 = Wdso2
NO2 = Wdno2
Unknowns: D, E, Wdn2, Wdco2, Wdo2, Wdco, Wdso2, Wdno2
No of unknowns = 8
No of equations:
(i)solvent mass balance:
S = 0.835*E
(ii)Total mass balance:
S + F*MWF = D + E
(iii)CO2 mass balance:
0.12*F*44 = 0.15*E + Wdco2*D
(iv)SO2 mass balance:
0.02*F*64 = 0.01*E + Wdso2*D
(v) NO2 mass balance:
0.01*F*46 = 0.005*E + Wdno2*D
(vi) CO mass balance:
0.01*F*28 = Wdco*D
(vii) O2 mass balance:
0.05*F*32 = Wdo2*D
(viii) N2 mass balance:
0.78*F*28 = Wdn2*D
No of equations = 8
Degree of freedom = no of unknowns - no of independet equation = 8-8=0
C. Find the flow rates of streams D and E in Kg/s.
Basis: 1 sec
solvent mass balance:
S = 0.835*E
1000 = 0.835*E
E = 1000/.835=1197.6 kg
Total mass balance:
S + F*MWF = D + E
1000 + 100*30.74 = D + 1197.6
D = 1000 + 100*30.74 - 1197.6 = 2876.4 kg
D. Determine the composition of stream D.
CO2 mass balance:
0.12*F*44 = 0.15*E + Wdco2*D
0.12*100*44 = 0.15*1197.6 + Wdco2*2876.4
CO2 mass fraction in stream D, Wdco2 = 0.12
SO2 mass balance:
0.02*F*64 = 0.01*E + Wdso2*D
0.02*100*64 = 0.01*1197.6 + Wdso2*2876.4
SO2 mass fraction in stream D, Wdso2 = 0.04
NO2 mass balance:
0.01*F*46 = 0.005*E + Wdno2*D
0.01*100*46 = 0.005*1197.6 + Wdno2*2876.4
NO2 mass fraction in stream D, Wdno2 = 0.014
CO mass balance:
0.01*F*28 = Wdco*D
Wdco = 0.01*100*28/2876.4 = 0.01
O2 mass balance:
0.05*F*32 = Wdo2*D
Wdo2 = 0.05*100*32/2876.4 = 0.056
N2 mass balance:
0.78*F*28 = Wdn2*D
Wdn2 = 0.78*100*28/2876.4 = 0.76
Mass percentage
N2 = Wdn2*100 = 76%
CO2 = Wdco2*100 = 12%
O2 = Wdo2*100 = 5.6%
CO = Wdco*100 = 1%
SO2 = Wdso2*100 = 4%
NO2 = Wdno2*100 = 1.4 %
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