Post on 15-Oct-2014
1
Multicomponent Flash Distillation
If there are more than two components, an analytical procedure is needed. The equation used are equilibrium, mass and energy balances, and stoichiometric relations. The mass and energy balances are very similar to those used in the binary case, but the equilibrium equations are usually written of K values. iii xKy where in general idrumdrumii xallpTKK ,,
These two equations are written once for each component.
2
Fortunately, for many systems the K values are approximately independent of composition. thus,
๐พ=๐พ (๐ ,๐ )
For light hydrocarbons, the approximate K values can be determined from the monographs prepared by DePriester
3The DePriester Chart
4
5
McWilliams, M.L., โAn Equation to Relate K-factors to Pressure and Temperature,โ Chem. Eng., 80(25), 138, 1973.
The DePriester charts have been fit to the following equation (McWilliams, 1973):
Note that T is in oR and p is in psia. The constants are given in the following table. This equation is valid from โ70o C (365.7o R) to 200o C (851.7o R) for pressure from 101.3 kPa (14.69 psia) to 6,000 kPa (870.1 psia)
p
a
p
apaa
T
a
T
aK pp
PTTT 3
2
213
221 lnln
See Table 2-3 p. 33 for constants
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C is the number of components.
0.11
C
iiy 0.1
1
C
iix
The K values are used along with the stoichiometric equations which the mole fraction in liquid and vapor phases must equal to 1.0
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For ideal system Raoultโs law holds
๐๐ด=๐ฅ๐ด (๐๐ )๐ด
By Daltonโs law of partial pressure
๐ฆ ๐ด=๐๐ด/๐
Combining these equations
๐ฆ ๐ด= (๐๐ )๐ด ๐ฅ๐ด /๐
iii xKy Comparing with
๐พ ๐ด=(๐๐ )๐ด /๐
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The mass balances,
and the energy balance,
iii VyLxFz
VLF
LVflashF LhVHQFh
Multicomponent Flash Distillation
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Usually the feed, F , and the feed mole fractions z1 for C-1 of the components will be specified. If pdrum and Tdrum or one liquid or vapor composition are also specified, then a sequential procedure can be used.Suppose we have 10 components (C = 10). Then we must find 10 Kโs, 10 xโs, 10 yโs one L and one V or 32 variables. To do this we must solve 32 equations [10 from 10 from 2 from , and 10 independence mass balances]
iii xKy idrumdrumii xallpTKK ,, 0.1
1
C
iiy
0.11
C
iix
10
How does one solve 32 simultaneous equations? However we will restrict ourselves to ideal solution where
drumdrumii pTKK ,
Since and are known, the 10 Ki can be determined easily (say, from the DePriester charts or using the McWilliams relation). Now there are only 22 equations to solve simultaneously.
drumT drump
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To simplify the solution procedure, we first use equilibrium,
to remove ii xKy iy
iiii xVKLxFz
Substituting for L
i
ii VKVF
Fzx
Then upon rearranging we have
FV
K
zx
i
ii
11 Ci ,1 (2.38
)
i
ii VKL
Fzx
Solving for xi, we have
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From iii xKy
and
0.1
1111
C
ii
iiC
ii
FV
K
zKy **
Then *
0.11111
C
ii
iC
ii
FV
K
zx
We obtain
FV
K
zKy
i
iii
11 (2.39
)
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If C is greater than 3, a trial-and-error procedure or root finding technique must be used to find . Although the equations * and ** are both valid, they do not have the good convergence properties. That is, if the wrong is chosen, the that is chosen next may not be better.
FV
FVFV
Subtracting * from **
0
1111 11
C
ii
iC
ii
ii
FV
K
z
FV
K
zK
Subtracting the sum term by term, we have
011
1
1
C
ii
ii
FV
K
zK
F
Vf
14
011
1
1
C
ii
ii
FV
K
zK
F
Vf
This equation, which is known as the Rachford-Rice equation, has excellence convergence properties.
FV
Since the feed composition, are specified and can be calculated when and are given, the only variable is the fraction vaporize . This equation can be solved by many different convergence procedures. For instance, the Secant method can be used by selecting two values of and calculating the values of the summation (it will be zero only at the correct value of ).
iziK
drumTdrump
FV
FV
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The Newton convergence procedure will
converge faster. Since is a function of
that should have a zero value, the equation
for the Newton convergence is
F
Vf FV
FVFVd
dfff k
kk 1
where is the value of the function for trial
k and is the value of the
derivative of the function for trial k. We desire
to have equal to zero, so we set
and solve for
kf
FVddfk
1kf
01 kf FV
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FVddf
fFVFVFV
k
kkk 1
This equation gives us the best next guess for the fraction vaporized. To use it, however, we need equations for both the function and derivative. For fk, use the Rachford-Rice equation. Then the derivative is
C
i i
iik
FVK
zK
FVd
df
12
2
11
1(2-45)
(2-44)
Substituting the Rachford-Rice equation and (2-45) into (2-44)
C
i i
ii
C
i i
ii
kk
FVK
zK
FVK
zK
F
V
F
V
12
2
1
1
11
1
11
1
(2-46)
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Once is calculated the value of the Rachford-
Rice equation can be determined. If it is close enough
to zero, the calculation is finished, otherwise, repeat
the Newton convergence for the next trial.
1kFV
Once has been found, and are calculated
from (2-38) and (2-39). L and V are determined
from the overall mass balances.
FV ix iy
The enthalpy and can now be calculated.For ideal solutions the enthalpies can be determined from the sum of the pure component enthalpies multiplied by the corresponding mole fractions:
Lh VH
1
,~
idrumdrumViV pTHyH
i(2-47a)
where and are enthalpies of the pure component. iVH
~iLh
~
(2-47b)
1
,~
idrumdrumiLiL pThxh
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Example 2-2 Multicomponent Flash Distillation
A flash chamber operating at 50 oC and 200 kPa is separating 1,000 kg moles/hr of a feed that is 30 mole % propane, 10 mole % n-butane, 15 mole % n-pentane and 45 mole % n-hexane. Find the product compositions and flow rates.
F = 1,000 kg mole/hrz1 = .30z2 = .10z3 = .15z4 = .45
CT odrum 50
kPapdrum 200
?,, iyV
?,, ixL
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Since and are given, a sequential solution can be used. We can use the Rachford-Rice equation to solve for and then find
drumT drumpFV
VLyx ii ,,,
Find from the DePriester charts (or use McWilliams equation)
iK
From the DePriester charts at 50 oC and 200 kPa we find
0.71 K
4.22 K
8.03 K
30.04 K
C3
n-C4
n-C5
n-C6
20
8785.03387.00306.01228.0125.1
Since f(0.1) is positive, a higher value for is required. Note that only term in the denominator of each term changes. Thus we can set up the equation so that only will change. Then equals
FV
FVFV
Calculate from the Rachford-Rice equation FVf
4
1 11
1
ii
ii
FV
K
zK
F
Vf
FV
FV
FV
FV
f13.01
45.013.0
18.01
15.018.0
14.21
1.014.2
10.71
3.010.71.0
Pick = 0.1 as first guess.
1.013.01
45.013.0
1.018.01
15.018.0
1.014.21
1.014.2
1.010.71
3.010.71.0
f
FV
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The first derivative from (2-45)
2
2
22
22
1
12
1
1 11
1
11
1
FVK
zK
FVK
zK
FVd
df
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42
42
3
32
3
11
1
11
1
FVK
zK
FVK
zK
22 4.11
196.0
0.61
8.10
FVFV
22 7.01
2205.0
2.01
006.0
FVFV
With = 0.1 this derivative is -4.631. From (2-46) the next guess for is
FVFV
29.0631.4
8785.01.0
2
F
V
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Calculating the value of the Rachford-Rice equation, we have . This is still positive and V/F is still too low.
329.029.0 f
2nd Trial
891.12
FVd
df
46.0891.1
329.029.0
3
F
V
This is closer, but V/F is still too low. 066.046.0 f
3rd Trial
32.13
FVd
df51.0
32.1
066.046.0
3
F
V 00173.051.0 f
Thus 51.0F
V
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Now we calculate from (2-38) and from ix iy iii xKy
0739.051.010.71
30.0
11 1
11
FV
K
zx
By similar calculations,
1400.0,0583.0 22 yx
1336.0,1670.0 33 yx
2099.0,6998.0 44 yx
Since F = 1,000 and = 0.51 FV
and kmol/h
51051.0 FV 4905101000 VFL
5172.00739.00.7111 xKy
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Check and ix iy
999.04
1
i
ix 0007.14
1
i
iy
These are close enough.
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Simultaneous Multicomponent Convergence
If the feed rate F, the feed composition consisting of (C-1)zi values, the flash drum pressure pdrum , and the feed temperature TF are specified, the hot liquid will vaporized when its pressure is dropped. This โflashingโ cools the liquid to provide energy to vaporize some of the liquid. The result Tdrum is unknown; thus we must use a simultaneous solution procedure.
1
,i
c
F i F F Fi
h z h T p
For wide-boiling point feed, Tdrum cannot have much effect on V/F
(2-50)
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Estimate Tdrum
Finished
Yes
No
Calculate ,i drum drumK T p
Estimate V/F
Solve Rachford-Rice eq.
( / ) 0f V F
Calculate xi, yi, L, V, H and h
Energy balance satisfied
Yes
NoEstimate Tdrum
Tbp < Tdrum < Tdp
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Arranging the energy balance into the functional form
LVflashF LhVHQFh
( ) 0k drum V L F flashE T VH Lh Fh Q (2-51)
The Newtonian procedure estimates EK+1(Tdrum) from the derivative,
(2-52)๐ธ๐+1โ๐ธ๐=๐๐ธ๐
๐๐ ๐๐๐ข๐( โ๐ ๐๐๐ข๐ )
- (2-53)
,
k v LPV PL
drum k drum drum
dE dH dhV L VC LC
dT dT dT (2-54)
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Set Ek+1 =0
โ ๐๐๐๐ข๐=โ๐ธ๐ (๐ ๐๐๐ข๐ ,๐ )
๐๐ธ๐
๐๐ ๐๐๐ข๐ .๐
(2-55)
,, 1 ,
,
( )k drum kdrum k drum k
drum k
E TT T
dE
dT
Convergence when
= 0.2 or 0.01oC
(2-56a)
|โ๐ ๐๐๐ข๐|<๐
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Example We have a liquid feed that is 20 mole % methane, 45 mole % n-pentane and 35 mole % n-hexane. Feed rate is 1500 kmol/h and feed temperature is 45oC and pressure is 100 psia. The flash drum operates at 30 psia and is adiabatic. Find Tdrum, V/F, xi, yi, L, V. Give for Tdrum = 0.02 and for V/F = 0.005
Component (kcal/gmol) Tbp, normal, oC CPL, cal/gmoloCMethane 1.955 -161.48 11.0 n-pentane 6.160 36.08 39.66 n-hexane 6.896 68.75 45.58
Vapor heat capacities in cal/gmoloC; T in oC:7 2 9 3
, 8.20 0.01307 8.75*10 2.63*10PV MC T T T 5 2 9 3
, 27.45 0.08148 4.538*10 10.1*10PV PC T T T 5 2 9 3
, 32.85 0.09763 5.716*10 13.78*10PV HC T T T
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Answer: Tdrum = 23.892, V/F = 0.2549
F = 1,500 kg mole/hrzM = .20zP = .45zH = .35TF = 45oCPF=100 psia
?odrumT C
30drump psia
?,, iyV
?,, ixL
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IterationNo.
1 15.000 16 0.2361 27.903
2 27.903 13 0.2677 21.385
3 21.385 14 0.2496 25.149
4 25.149 7 0.2567 23.277
5 23.277 3 0.2551 24.128
6 24.128 2 0.2551 23.786
7 23.786 2 0.2550 23.930
8 23.930 2 0.2550 23.875
9 23.875 2 0.2549 23.900
10 23.900 2 0.2549 23.892
drumTFVFV
drumTInitial
Trials tofind
Calc