Unformatted text preview: d reflux and distillate.
Property constants in Example 2.3.
Assumptions: Ideal gas and ideal gas and liquid solutions.
Find: Condenser duty in kJ/h.
Analysis: For the thermodynamic path, cool the overhead from 331 K to 325 K. Then
condense at 325 K. Because this temperature change is small, compute vapor specific
heats at the two temperatures and take the arithmetic average. From Eq. (235) and the
constants in Example 2.3, for vapor heat capacity in J/kmolK and temperature in K,
o
CPEB = −43098.9 + 707.151T − 0.481063T 2 + 130084 × 10 −4 T 3
.
o
CPS = −28248.3 + 615.878T − 0.40231T 2 + 0.993528 × 10 −4 T 3 Solving, Comp.
EB
S kg/h
77,500
2,500 MW
106.17
104.15 kmol/h
729.975
24.003 Vapor heat capacity, J/kmolK
331 K
325 K
Avg.
142,980
140,380
141,680
132,130
132,830
133,980 The sensible vapor enthalpy change, Qsensible from 331 K to 325 K is, Qsensible =
i o
ni CPi ∆T = i o
ni CPi (331 − 325) = 729.975(141,680)(331 − 325) + 24.003(133,980)(331 − 325)
= 620,560,000 J / h = 620,560 kJ / h
For the latent heat of condensation, use Eq. (241) to estimate the molar heats of
vaporization of at 325 K for the two components, using the vapor pressure constants
given in Example 2.3. Exercise 2.11 (continued) vap
∆HEB = 8314 325 2 − −7440.61
325 2 + 0.00623121 + −9.87052
5
+ 6(4.13065 × 10−18 ) 325
325 = 40,740,000 J / kmol = 40,740 kJ / kmol
∆HSvap = 8314 325 2 − −914107
.
325 2 + 0.0143369 + −17.0918
5
+ 6(18375 × 10−18 ) 325
.
325 = 42,440,000 J / kmol = 42,440 kJ / kmol
The latent heat of condensation, Qlatent , in kJ/kmol is, Qlatent = ni ∆Hivap = 729.975(40,740) + 24.003(42,440)
i = 30,760,000 kJ / h
The total condenser duty = Qsensible + Q latent = 620,000 + 30,760,000 = 31,380,000 kJ/h Exercise 2.12
Subject: Thermodynamic properties of a benzene (B) toluene (T) feed to a distillation
column.
Given: Temperature, pressure, and component flow rates for the column feed. Property
constants, critical temperature
Assumptions: Phase condition is liquid (needs to be verified). Ideal gas and liquid
solution.
Find: Molar volume, density, enthalpy, and entropy of the liquid feed.
Analysis: The feed is at 100oF and 20 psia. From Fig. 2.4, since the vapor pressures of
benzene and toluene are 3.3 and 0.95 psia, respectively, the feed is a subcooled liquid. T
= 311 K.
From Eqs. (4) , Table 2.4 and (238), υL =
i Mi ρL i υL = = Mi
− 1− 2/ 7 AB
78.11 311
− 1−
562 B υL = T
Tc 2 /7 304.1(0.269)
92.14 T 290.6(0.265)
Total flow rate = 415 + 131 = 546 kmol/h
Benzene mole fraction = 415/546 = 0.760 311
− 1−
593.1 2/ 7 = 0.0905 m3 / kmol
= 01083 m3 / kmol
. Toluene mole fraction = 10.76 = 0.240 From Eq. (4), Table 2.4 for a mixture (additive volumes),
υL = 0.76(0.0905) + 0.24(0.1083) = 0.0948 m3/kmol = 1.52 ft3/lbmol
Mixture molecular weight = M = 0.76(78.11) + 0.24(92.14) = 81.48
From Eq. (4), Table 2.4 for conversion to density,
ρL = M/υL = 81.48/1.52 = 53.6 lb/ft3 Exercise 2.13
Subject: Liquid density of the bottoms from a distillation column.
Given: Temperature, pressure, component flow rates
Assumptions: Ideal liquid solution so th...
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 Spring '11
 Levicky
 The Land

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