Unformatted text preview: from condensation of acetone with ∆Hvap = 237 Btu/lb and a
molecular weight of 58.08 = 14.9(58.08)(237) = 205,000 Btu/h
Energy available from the cooling of evaporated water from 90oF to 80oF =
22(18)(0.44)(9080) = 2,000 Btu/h
Total available energy = 205,000 + 2,000 = 207,000 Btu/h
Energy required  Energy available = 423,000  207,000 = 216,000 Btu/h
This energy must come from cooling of the water absorbent from 90oF to T , and
condensed acetone from 78oF to T.
Therefore, using a CP of 0.53 for liquid acetone,
216,000 = 14.9(58.08)(78T) + 1,722(18)(1.0)(90T)
Solving, T = temperature of exiting liquid = 83oF.
The mol% acetone in the entering gas = 15/702 x 100% = 2.14 %.
This is outside of the explosive limits range of 2.5 to 13 mol%. Exercise 2.17
Subject: Volumetric flow rates of entering and exiting streams of an adsorber for
removing nitrogen from subquality natural gas.
Given: Temperature and pressure of feed gas and two product gases. Composition of
the feed gas. Specification of 90% removal of nitrogen and a 97% methane natural gas
product.
Assumptions: Applicability of the RedlichKwong equation of state.
Find: Volumetric flow rates of the entering and exiting gas streams in actual ft3/h.
Analysis: Removal of nitrogen = 0.9(176) = 158.4 lbmol/h
Nitrogen left in natural gas = 176  158.4 = 17.6 lbmol/h
Methane in natural gas product = 17.6(97/3) = 569.1 lbmol/h
Material balance summary:
Component
Nitrogen
Methane
Totals
Temperature, oF
Pressure, psia lbmol/h:
Feed gas
176
704
880
70
800 Waste gas
158.4
134.9
293.3
70
280 Natural gas
17.6
569.1
586.7
100
790 Using the ChemCAD simulation program, the following volumetric flow rates are
computed using the RedlichKwong equation of state: Stream
Feed gas
Waste gas
Natural gas Actual volumetric
flow rate, ft3/h
5,844
5,876
4,162 Exercise 2.18
Subject: Estimation of partial fugacity coefficients of propane and benzene using the
RK equation of state.
Given: From Example 2.5, a vapor mixture of 39.49 mol% propane and 60.51 mol%
benzene at 400oF and 410.3 psia.
Assumptions: Applicability of the RedlichKwong equation of state.
Find: Partial fugacity coefficients
Analysis: From Example 2.5, the following conditions and constants apply, where the
RedlichKwong constants, A and B, for each component are computed from Eqs. (247)
and (248) respectively.
T = 477.6 K
P = 2829 kPa ZV = 0.7314
R = 8.314 kPam3/kmolK A = 0.2724
B = 0.05326 From Eqs. (247) and (248),
aP
ai (2819)
a
Ai = 2i 2 =
=i
2
2
RT
(8.314) (477.6)
5593
bP
bi (2819)
Bi = i =
= 0.7099bi
RT (8.314)(477.6) Component, i
Propane
Benzene a, kPam6/kmol2
836.7
2,072 b, m3/kmol
0.06268
0.08263 Ai
0.1496
0.3705 Bi
0.04450
0.05866 The RedlichKwong equation for partial vapor fugacity coefficients is given by Eq. (256). Applying it to propane (P) and benzene (B), using the above values for constants
and conditions,
φPV = exp (0.7314 − 1) 0.04450
0.2724
0.1496 0.04450
0.05326
− ln ( 0.7314 − 0.05326 ) −
2
−
ln 1 +
0.05326
0.05326
0.2...
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 Spring '11
 Levicky
 The Land

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