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Unformatted text preview: mixture.
Given: Feed mixture, of composition below, at 250oF and 500 psia. Pressure exiting valve =
300 psia.
Find: (a)
(b)
(c)
(d) Phase condition of feed.
Temperature downstream of valve.
Mole fraction vaporized across valve.
Mole fraction compositions of vapor and liquid phases downstream of valve. Analysis: Use CHEMCAD process simulator with SRK method for Kvalues and enthalpies.
Results are as follows:
Stream
Temperature, oF
Pressure, psia
Phase condition
Mole fraction of feed
Mole fractions:
Ethylene
Ethane
Propylene
Propane
Isobutane
nButane Feed
250
500
Liquid
1.00 Vapor from valve
207.6
300
Vapor
0.4043 Liquid from valve
207.6
300
Liquid
0.5957 0.02
0.03
0.05
0.10
0.20
0.60 0.0348
0.0491
0.0658
0.1260
0.1953
0.5290 0.0099
0.0171
0.0393
0.0823
0.2032
0.6482 Exercise 4.43
Subject and to Find: Algorithm for flash calculation when Ψ = V/F and P are specified.
Given: Isothermal flash algorithm of Fig. 419a, and equations of Table 4.4.
Analysis: Specify feed rate and composition, and values of Ψ = V/F and P. Use the isothermal
flash algorithm of Fig. 419a as an inner loop. Guess the flash temperature and enter the inner
loop. If the calculated Ψ = V/F is not the specified value, guess a new value of T = say 1.05
times the initial guess of T, and repeat the inner loop. For the next and subsequent iterations, k,
apply the false position method to provide a new guess of T:
T k + 2 = T k +1 + Ψspec − Ψ k +1 T k +1 − T k / Ψ k +1 − Ψ k
This assumes that T is a linear function of Ψ = V/F. Iterate until the computed Ψ = V/F is within
say 0.1% of the specified value. Exercise 4.44
Subject and to Find: Algorithms for flash calculations with 6 different sets of specified
variables given in the table below.
Given: Isothermal flash algorithm of Fig. 419a, and equations of Table 4.4.
Assumption: All flashes are adiabatic.
Analysis: The equations to be solved for each algorithm are those for the standard adiabatic
flash procedure, where the specifications are outlet P and Q = 0.
RachfordRice Eq. (3), Table 4.4: f1 = zi 1 − Ki
=0
i =1 1 + Ψ Ki − 1
C Adiabatic energy balance, Eq. (419): f 2 = ΨhV + (1 − Ψ )hL − hF
1,000 (1) (2) For each of the 6 algorithms, we must choose a tear variable, the output variable for f1, and the
output variable for f2. If Kvalues are compositiondependent, then outer loop iterations with f1
are necessary as in Fig. 4.19. Note that the specification of hF is equivalent to specifying TF or
Q = 0. In some cases, it may be necessary to solve f1 and f2 simultaneously. Case Specifications Find 1
2
3
4
5
6 hF , P
hF , T
hF , Ψ
Ψ, T
Ψ, P
T, P Ψ, Τ
Ψ, P
T, P
hF , P
hF , T
hF , Ψ Output Variable
in
Tear
f1
f2
Variable
TV
TV
Ψ
PV
Ψ
Ψ
TV
PV
TV
h F{T F}
PV
hF
h F{T F}
TV
hF
h F{T F}
hF
Ψ As an example of one of the algorithms, consider Case 1, which is equivalent to the standard
adiabatic flash specification. The algorithm is shown in diagram form on the following page. Exercise 4.44 (continued)
Analysis: (c...
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 Spring '11
 Levicky
 The Land

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