b)For water as solvent: Ms18.015 gm
mol
:=
10.36
Acetone: Psat1T() e
14.3145 2756.22
T
degC 228.060+
−
kPa⋅:=
Acetonitrile Psat2T() e
14.8950 3413.10
T
degC 250.523+
−
kPa⋅:=
a) Find BUBL P and DEW P values
T 50degC:= x10.5:= y10.5:=
10.35 a) The equation from NIST is: Mikiyi
⋅P⋅=Eq. (1)
The equation for Henry’s Law is:
xiHi
⋅yiP⋅=Eq. (2)
Solving to eliminate P gives: Hi
Mi
kixi
⋅
=Eq. (3)
By definition: Mi
ni
nsMs
⋅
=where M is the molar mass and the
subscript s refers to the solvent.
If xi is small, then xs is approximately equal to 1 and: Hi1
Mski
⋅
=Eq. (5)
333
10.37 Calculate x and y at T = 90 C and P = 75 kPa
Benzene: Psat1T() e
13.7819 2726.81
T
degC 217.572+
−
kPa⋅:=
Toluene: Psat2T() e
13.9320 3056.96
T
degC 217.625+
−
kPa⋅:=
a) Calculate the equilibrium composition of the liquid and vapor at the flash T and P
T 90degC:= P 75kPa:= Guess: x10.5:= y10.5:=
b)Find BUBL T and DEW T values
P 0.5atm:= x10.5:= y10.5:= Guess: T 50degC:=
Given x1Psat1T()⋅1x
1
−
()
Psat2T()⋅+ P=
334
10.38 yO210.0387:= yN210.7288:= yCO210.0775:= yH2O10.1550:=
ndot 10 kmol
hr
:= T1100degC:= T225degC:= P 1atm:=
PsatH2O T() e
16.3872 3885.70
T
degC 230.170+
−
kPa⋅:=
b) Assume that the measured values are correct. Since air will not dissolve
in the liquid to any significant extent, the mole fractions of toluene in the
liquid can be calculated.
x10.1604:= y10.2919:= x21x
1
−:= x20.8396=
Now calculate the composition of the vapor. y3 represents the mole
fraction of air in the vapor.
Guess: y20.5:= y31y
2
−y1
−:=
Given
335
ndotliq
ndotvap
⎛
⎜
⎜
⎞
⎟
yCO220.0775:=yN220.7288:=yO220.0387:=
ndotliq ndot
2
:=ndotvap ndot
2
:=
Guess:
Assume that two streams leave the process: a liquid water stream at rate
ndotliq and a vapor stream at rate ndotvap. Apply mole balances around
the cooler to calculate the exit composition of the vapor phase.
Calculate the mole fraction of water in the exit gas if the exit gas is
saturated with water.
336
Assume the liquid is stored at the bubble point at T = 40 F10.39
T2T2273.15K+:=T1T1273.15K+:=∆HlvH2O 40.66 kJ
mol
:=
Apply an energy balance around the cooler to calculate heat transfer rate.
337
b)Calculate the vapor stream molar flow rate using balance on SO2
10.40 H2S + 3/2 O2 -> H2O + SO2
By a stoichiometric balance, calculate the following total molar flow rates
ndotH2S 10 kmol
hr
:= ndotO2 3
2ndotH2S
:=
Feed:
Products ndotSO2 ndotH2S
:= ndotH2O ndotH2S
:=
Exit conditions:
P 1atm:= T270degC:= PsatH2O T() e
16.3872 3885.70
T
degC 230.170+
−
kPa⋅:=
a) Calculate the mole fraction of H2O and SO2 in the exiting vapor stream
assuming vapor is saturated with H2O
338
10.42 ndot150 kmol
hr
:= Tdp1 20degC:= Tdp2 10degC:= P 1atm:=
MH2O 18.01 gm
mol
:=
PsatH2O T() e
16.3872 3885.70
T
degC 230.170+
−
kPa⋅:=
y1PsatH2O Tdp1
()
P
:= y10.023=y2PsatH2O Tdp2
()
P
:= y20.012=
By a mole balances on the process
Guess: ndot2liq ndot1
:= ndot2vap ndot1
:=
10.41 NCL 0.01 kg
kg
:= MH2O 18.01 gm
mol
:= Mair 29 gm
mol
:=
a) YH2O NCL Mair
MH2O
⋅:= YH2O 0.0161=
339
Cyclohexane: A2 13.6568:= B2 2723.44:= C2 220.618:=
Psat1 T( ) exp A1 B1
T
degC C1+
−
⎛
⎜
⎜
⎝
⎞
⎠
kPa:=
Psat2 T( ) exp A2 B2
T
degC C2+
−
⎛
⎜
⎜
⎝
⎞
⎠
kPa:=
10.43 Benzene: A1 13.7819:= B1 2726.81:= C1 217.572:=
340