1. Air and hydrogen sulphite mixture at 1 atm is fed continuously into a single-
ID: 905435 • Letter: 1
Question
1. Air and hydrogen sulphite mixture at 1 atm is fed continuously into a single-stage mixer with a pure water stream at 20 oC in the opposite direction. The exit gas and the exit liquid stream are in equilibrium. The inlet gas flow rate is 100 kmol/h containing 30% H2S. The liquid flow rate is 300 kmol/h. Estimate the amounts and compositions of the two exit streams, assuming water doesn’t evaporate to the gas phase. The Henry law constant (H) for H2S in a H2S-water system at 20oC is 4.83x102 atm/mol fraction. Note that the partial pressure of A in the gaseous phase is PA = H xA where xA is the mole fraction of A in liquid.
Explanation / Answer
Inlet:
Gasin:Qgin
Composition in mole fraction:
H2S: yH2Si = 0.3,
Air: yairi = 0.7
Liquid in:Qlin
composition in mole fraction: xwi = 1.0
Outlet:
Gas out: Qgout
Composition in mole fraction:
H2S mole fraction = yH2So
Air mole fraction = yairo = 1-yH2So
Liquid out:Qlout
Composition in mole fraction:
Henry's constant, H = 4.83E2 atm/mole fraction
Total pressure = P = 1 atm
partial pressure of H2S, pH2So = P*yH2So = yH2So
xH2So = pH2So/H = yH2So/483; Equation(0)
xwo = 1-xH2So = 1 - yH2So/483
Basis: 1 h of operation
Mass balance H2S: 100*0.3 = 100*yH2So/483 + Qgout*yH2So; Equation(1)
Mass balance air: 100*0.7 = Qgout*(1-yH2So); Equation(2)
Mass balance water: 300*1.0 = Qlout*xwo = Qlout*(1 - yH2So/483); Equation(3)
100*0.3 = 100*yH2So/483 + Qgout*yH2So;
From Equation(2), Qgout = 70/(1-yH2So) substitute in Equation(1)
30 = 100*yH2So/483 + 70*yH2So/(1-yH2So);
(1-yH2So) = (1-yH2So)*0.007*yH2So + 2.32*yH2So
0.007*yH2So^2 - 3.327*yH2So + 1 = 0
yH2So = 0.3
Exit gas composition
yH2So = 0.3
yairo = 1-yH2So = 0.7
Qgout = 70/(1-yH2So) = 100 kmol/h
Exit liquid composition:
xH2So = pH2So/H = yH2So/483 = 0.0006
xwo = 1 - xH2So = 0.9994
Equation(3) 300 = Qlout*xwo
Qlout = 300/xwo = 300.18 kmol/h
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