Unformatted text preview: ontinued) Analysis: (c) (continued)
For this condition, the partial pressure of water is 0.2(3.32) = 0.664 psia (less than 1.79 psia).
The secondary dew point is much more difficult to determine. We now know that
as the pressure is increased, the hydrocarbons condense first. At the primary dew point, the
partial pressure of W is 0.664 psia. W will not condense until its partial pressure increases to its
vapor pressure of 1.79 psia. As the pressure is increased above the primary dew point pressure
of 3.32 psia, the hydrocarbons will condense, reducing their mole fractions in the vapor phase
and causing the mole fraction of water to increase in the vapor phase. This plus the increased
total pressure will eventually cause the water partial pressure to reach its vapor pressure and
begin condensing. We know the secondary dew point pressure must lie between the bubble point
pressure of 5.29 psia and the primary dew point pressure of 3.32 psia.
The equations that apply at the secondary dew point, where all pressures are in psia, are:
Dalton's law: P = pB + pT + 1.79 = yBP + yTP + 1.79
Raoult's law: pB = xB PBs = xB (5.22) (2) (3) pT = xT P = xT (178)
.
(4)
Substituting Eqs. (3) and (4) into (2),
P = 5.22xB +1.78xT + 1.79
(5)
But in the hydrocarbonrich liquid phase, xT = 1  xB . Therefore, Eq. (5) becomes:
(6)
P = 5.22xB +1.78(1  xB) + 1.79 = 3.44 xB + 3.57
s
T Material balance equations for a basis of 100 moles, with 40 moles of B, 40 moles of T, and 20
moles of W:
At the secondary dew point, only HCs have condensed. Therefore, the total moles of vapor is:
V = 20P/1.79 = 11.17P
(7)
The moles of liquid is:
L = 100  V
(8)
A benzene material balance is:
40 = yBV + xBL
(9)
Substituting Eqs. (7) and (8) into (9), We can also write from Eq. (3), 40 = 11.17 yBP + 100xB 11.17xBP
yB = 5.22xB/P (10) (11) Combining Eqs. (6), (10), and (11) to eliminate xB and yB ,
3.247P2 57.61P + 204.29 = 0 (12) Solving for the root between the bubble point and the primary dew point, the secondary dew
point pressure = P = 4.90 psia. Exercise 4.72
Subject: Bubble and dew points of a mixture of benzene, toluene, and water.
Given: Pressure of 2 atm.
(a) 50 mol% benzene and 50 mol% water.
(b) 50 mol% toluene and 50 mol% water.
(c) 40 mol% benzene, 40 mol% toluene, and 20 mol% water.
Assumptions: hydrocarbons and water immiscible.
Find: For each mixture, bubblepoint, primary dewpoint, and secondary dewpoint
temperatures.
Analysis: Use the CHEMCAD simulator with APISRK for Kvalues, with the immiscible
option and the threephase flash model. The following results are obtained:
Bubble point
Secondary dew point
Primary dew point T for Case (a), oC
89.4
89.4
100.0 T for Case (b), oC
104.3
104.3
110.3 T for Case (c), oC
93.9
98.0
99.7 Exercise 4.73
Subject: Bubblepoint of a mixture of toluene, ethyl benzene, and water.
Given: Mixture of 30 mol% toluene, 40 mol% ethylbenzene, and 30 mol% water at 0.5 atm.
Assumptions: Hydrocarbons and water are immiscible.
Find: Bubblepoint temperature and vapor composition.
Analysis: : Use the CHEMCAD simulato...
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 Spring '11
 Levicky
 The Land

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