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DISTILLATION SEPARATION FACTOR
STUDIES FOR IDEAL AND NON-IDEAL
SYSTEMS
Zawar Ahmad Khan
Supervisor
Dr. K. M. Bukhari
Co-Supervisor
Dr. Muhammad Aslam
Center for Nuclear Studies
Quaid-i-Azam University, Islamabad, Pakistan
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OBJECTIVE
To study the Distillation Separation Factor.
For theoretical study H2O-HDO-D2O system (ideal system) was selected, becauseexperimental study of this system requires sophisticated and costly equipment, like
mass spectrometer etc.
For experimental study H2O-C2H5OH system (non-ideal system) was selected as
theoretical study of this system is quite complicated.
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Theoretical Analysis
3.1 DERIVATION OF RELATIONSHIP (K = Ae-B/T)
This relationship is derived for exchange reaction H2O + D2O 2HDO in vapor phase .
For a typical reaction: A + B C + D (3.1)
the equilibrium constant K can be determined from the expression
K = [C] . [D] / [A]. [B] (3.2)
and RT lnK = -G (3.3)
In the case of isotopic reaction, the G quantities are very small and it is not possible to calculate them from Eqn. (3.3),
since the uncertainties with which the molar free energies of the various isotopic species are known, are generally greater
than G. Furthermore, in statistical thermodynamics one can demonstrate that, for one mole of perfect gas
G = -RT ln(Z/N) (3.9)
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Where Z is the partition function for the system under consideration.
Comparing Eqn. (3.9) with Eqn. (3.3) we have
K = ZC.ZD/ Z
A.Z
B (3.10)
Statistical mechanics thus permits the calculation of equilibrium constants
through the partition function Z, relevant to the reaction of various molecular
species. Since the total energy for each level can be broken down into theterms relating to the different degrees of freedom, we find that
Z = Ztrans. Zrot. Zvibr. Zel. Znucl. e-Eo/kT (3.11)
In which the first five factors refer in order to translational, rotational,
vibrational, electronic and nuclear degrees of freedom, and the final termexpresses the zero-point energy Eo.
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For isotopic exchange reactions of the general type:
aAXb + bB*Xa aA*Xb + bBXa (3.12)
In which X and *X are two isotopes of the same element and a and b are the number of moles
of different species
The equilibrium constant is
K = (*Z/Z)a AXb/ (*Z/Z)b
BXa (3.13)
Introducing the notation u = h/kT we can write the *Z/Z ratio that appears in Eqn. (3.13) as
*Z/Z = (*Z/Z)trans.(*Z/Z)rot.(*Z/Z)vibre()(*u-u) (3.15)
Where Ztrans= const(M3/2. T5/2)/P (3.16)
And Zrot= const. IT/s (3.17)For diatomic or linear polyatomic molecules (where M is the molecular weight, P is the
pressure, I is the moment of inertia and `s` is the molecule's symmetry number), or
Zrot= const.(Ix Iy Iz)1/2.T/s (3.18)
For non linear polyatomic molecules Ix, Iy and Izare the principal moments of inertia. The
partition function for the vibrational degree of freedom is given by :
Zvibr= [1/(1-e-u)] (3.19)
In which the product is extended to all the fundamental frequencies of the molecule. For
diatomic molecules, with only a single fundamental frequency, we obtain from Eqns. (3.16),
(3.17) and (3.19).
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*Z/Z = s/*s(*M/M)3/2.(*I/I).[e-*u/2/(1-e-*u) ].[(1-e-u)/e-u/2] (3.20)
Where as for nonlinear polyatomic molecules, we obtain from Eqns. (3.16),(3.18) and (3.19)
*Z/Z = [(s/*s)(*M/M)3/2.(*Ix*Iy*Iz /IxIyIz)1/2]{[e-*u/2/(1-e-*u)].[(1-e-u)/e-u/2]} (3.21)For polyatomic molecules the difficulties can be overcome in many cases bymaking use of the rule of the product of frequencies, stated by O. Redlich[86], which gives
(*M/M)3/2.(*Ix*Iy*Iz /IxIyIz)1/2= (*u/u) (3.22)
Eqn. (3.21) then becomes
*Z/Z = (s/*s){(*u/u).[e-*u/2/(1-e-*u)].[(1-e-u)/e-u/2]} (3.23)This equation is also applicable to diatomic molecules, yielding results veryclose to those obtained with Eqn.(3.21). Eqn. (3.23), developed in powersseries of of u = *u-u and approximated except for terms of a higher orderthan the first, becomes:
*Z/Z = (s/*s)eG(u).u (3.24)
With G(u) = 1/2 - (1/u) + 1/(eu-1).
Tabulation of the G(u) function [35], which varies from 0 to 1/2 for uvariable between 0 and , makes it far simpler to calculate the equilibriumconstants [87].
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APPLICATION TO EXCHANGE REACTION
H2O + D
2O 2HDO
H2O + D2O 2HDO or
OHH + ODD OHD + ODH (3.25)
Let a = b = 1; A= OH, B = OD, X = H and *X = D Using Eqs.3.12 and 3.25,
Eqn 3.13 becomesKvap=(*Z/Z) / (*Z/Z) HDO = (*Z) . (3.26)
Kvap= (Z)2
HDO/{(Z) .(Z) } (3.28)
Specifying the quantities of Eqn. (3.28) we have
Kvap=(Z)HDO.(Z)HDO /{(Z) .(Z) .}= ( . /sHDO.sHDO)
(uHDO.uHDO/ u .u ){exp[-(uHDO+ uHDO- - )]}f(u) (3.29)
where f(u) = [1-exp(- )] [1-exp(- )]/[1-exp(-uHDO)]2
or neglecting the factor f(u) which at 1000 K differ from unity by less than 4%
we have:
OH2
OH2 HDOZ)( }).(*)/{( 2 HDOOH ZZ
OH2 OD2
OH2 OD2 ODOH SS 22 .
OH2 OD2OH2u OD
2u
OD2u OH2u
Kvap= ( . / s2
HDO )(u2HDO/ . ).{exp[-(2uHDO- )]} (3.30)OH2S OD2S. OH2u OD2u
ODOH 22 uu
0
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Sample Calculations
Consider the exchange reaction:
(H2O)vap+ (D2O)vap (2HDO)vapFor the calculation of its equilibrium constant Table 3.1 below showsthe numerical values of the quantities required for the calculations.These calculations are made by neglecting anharmonocity correctionsand by using Eqn. (3.30).
Table 3.1. Molecular constants of H2O, D2O and HDO [20].Constant H2O D2O HDO
M(a.m.u) 18 20 19
(o/c)(cm-1) 4671 3430 4077
uT (K) 6785 4938 5871
s 2 2 1
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Plugging in the values of s's and u's in Eqn. (3.30) from Table 3.1, we haveKvap= [2x2/(1)
2]x[(5871)2]/[6785x4938)]x{exp[-(1/2T)(2x5871-6785-4938)]}
Kvap= 4.115 exp(-10/T)
Kvap(at 273 K) = 3.967
Kvap(at 298 K) = 3.979
Kvap(at 348 K) = 3.998 Relationship for Kvapis also determined by using Eqn. (3.23) and using the three
fundamental frequencies for every three isotopic molecules of water and the
relationship found is:
Kvap= 4 (0.753) (1..043) (1.3248) Exp (-19/T)
Kvap= 4.1732 Exp (-19/T)Kvap(at 273 K) = 3.893
Kvap(at 298 K) = 3.915
Kvap(at 348 K) = 3.951
The corresponding error found for the three temperature are 2.0%, 1.53%, and 1.27%.
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Table 3.25. Review of liquid and vapor phase equilibriumconstant data for the exchange reaction H
2O + D
2O
2HDO as a function of temperature by different workers.
Referenc
es
Temperat
ure C
Urey
1947
(data of
Libby and
Herzberg
1943)
K(HM)
[28]
Kirshenb
aum
1951
[35]
Bigelesi
n 1955
(Theo.)
[36]*
Pyper R. S
Newbury
and G. W.
Barton Jr
(1967) Kl
(Exptl.)
[37] A. J.
Kresgi
Chiang
(Substrate
Method)
1968 Kl
(Exptl.)
[38] J.
R.
Holston
(Calc.)
1969
[39] Wayne
C. Duer
and Gray L.
Bertand
(1970) K
(Exptl.)
[41] J. W.
Pyper, R.
J.Dupzyk, R. D.
Friesen, S. L.
Bernasek et.el
(1977) Kv
This
wor
k
199
8
0 3. 94 3.76 - - - 3.69 3.74 0.07 - 3.967
25 3.96 3.80 3.96 3.75 0.07 3.85 0.03 3.72 3.76 0.02 3.81 0.09 3.97
9
75 3.98 3.85 - - - 3.78 - - 3.99
8
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ConclusionsTheoretical values of gas- phase equilibrium constant are calculated for
different spectroscopic data reported by different authors, from 1933 to 1997
using Eqn.(3.30).
1. Generally theoretical calculations of liquid-phase equilibrium constant are
quite difficult due to the involvement of intermolecular forces. But this is not
true in case of isotopic species as intermolecular forces are isotope
independent and cancel out in ratios. Also the required data is not frequently
available.
2. Harmonic data gives values more close to the classical value of 4.0 i.e. as
high as 3.98.
3. Anharmonic data gives values quite smaller than the classical value of 4.0 i.e.
as low as 3.20.
4. Semi empirical values of Kvand Klshow that at low temperatures i.e., up to
298 oK, Kv Kl, but at higher temperatures KlKv .
5. Values of K increases with the increase in temperature for this exchange
reaction.
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SEPARATION FACTOR FOR H2O-HDO-D2O
SYSTEM
A theoretical approach is applied to develop a relationship between
the separation factor (
) for nearly ideal System i.e., H2O - HDO - D2O,
and the equilibrium constant for the exchange reaction H2O + D2OHDO
in liquid and vapor phase.
The resulting equation is of the type (distillation) = f(Kl,Kv) where Kland
Kv are equilibrium constants in liquid and vapor, where K = Ae-B/T
DEVELOPMENT OF SEMIEMPIRICAL RELATIONSHIP FOR
DISTILLATION SEPARATION FACTOR OF H2O-HDO-D2O SYSTEM
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DERIVATION OF RELATIONSHIP
[ (distillation) = f(Kl, Kv)]
Considering the exchange reaction in liquid phase:
H2O (liq) + D2O (liq) 2HDO (liq) ( 4.18)
The equilibrium constant for the reaction (4.18) is
Kl= [HDO]2liq /[H2O]liq [D2O]liq ( 4.19)
Applying Daltons Law of partial pressure and Roaults Law of ideal solution of mixtures
We arrive at Kl/ Kv= (4.23)
As HDO cannot exist in the absence of H2O and D2O its vapor pressure cannot be determined
directly and we have to make certain assumptions.
Assumption-Iis the geometric mean of and i.e.,
(4.24)
Therefore Eqn. (4.23) gives Kl/ Kv= 1 or Kl= Kv (4.25)
For separation to be possible KlKv, therefore this assumption is not valid [91].
2000 )/(22 HDOODOH PPP
0
2OHP
0
2ODP
)( 000 22 ODOHHDO PPP
0
HDOP
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Assumption-II
is the arithmetic mean of and i.e.,
= ( + )/2 (4.26)
Subsituting Eqn. (4.26) in Eqn. (4.23) we get
Kl/ K
v= 4 /( + )2 (4.27)
Kl/ Kv= ( / ) + ( / ) +
But / 2, from Eqn. (2.16); is used as it is assumed that GM rule is notexactly obyed.
Therefore Kv/ Kl(2/4) + (1/42) + (1/2) (4.28)
The operational range of is 1.07 at 298K and 1.02 at 373K.
Let = 1.07 then Eqn. (4.28) gives Kv/ Kl1.0046Therefore Kv1.0046Kl, i.e., Kv> Kl
Also for = 1.02; Kv1.0004 Klor Kv> Kl
This result is contradicting the experimental data in this temperature range
[28]. Hence this possibility is also not valid.
0
HDOP
0
2OHP 0
2ODP
0
HDOP
0
2OHP
0
2ODP
0
2OHP
0
2ODP
0
2ODP
0
2OH
P0
2OH
P0
2ODP
0
2ODP
0
2OHP
0
2OHP
0
2ODP
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We can assume: (i) when is slightly greater than 1 then log = (
-1) and
(ii) (+1)/2 1
Therefore Eqn. (4.31) becomes
Exp(
(Kl/ Kv- 1))(4.32)
If C is the correction factor, then above Eqn. (4.32) becomes
= C Exp((Kl/ Kv- 1))
(4.33)
As decreases and the ratio Kl / K
v increases with the increase in
temperature as shown in Table 4.16 the negative sign of exponentialpower will lead to correct solution.
EXPANSION OF RELATIONSHIP
Taking log of both sides of above Eqn. and approximating:
Exp{(Bv-B
l) /T} = 1+(B
v- B
l)/T, we get:
log = log Cavg- [ (Al/Av) { 1 + (Bv- Bl)/T }- 1 ]1/2
(4.34)
If K A e , K A e and C C , then Eqn. (4.33) becomes:
= C e
l l
B /T
v v
B /T
avg
avg
-A
Ae 1
l v
l
v
Bv Bl
T
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CALCULATION OF CORRECTION
FACTOR (C)
Firstly (Kl, Kv) data quoted in [30] is used for the
calculation of factor
Exp((Kl / Kv - 1) ).
Secondly is calculated using empirical Eqn. quotedin [63 ] i.e;
log = log ( / ) = (019645.4/T ) -(48.5026/T) +
0.013615
Then expression = C Exp((Kl / Kv - 1) ) is used for
the calculation of correction factor C.
Values of these factors are reported in Table 4.16.
PH O2
PHDO
21
21
2
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Temp. K Kl[28] Kv[28] Kl / KvExp (-(Kl / Kv- 1)
) C
270 3.42 3.448 0.99188 - - -
280 3.46 3.475 0.99570 - - -
290 3.50 3.497 1.00086 0.9711 1.083 1.11523
300 3.53 3.517 1.00370 0.9410 1.073 1.14030
310 3.55 3.533 1.00481 0.9331 1.064 1.14029
320 3.57 3.551 1.00540 0.9292 1.0554 1.13582
330 3.59 3.568 1.00620 0.9243 1.048 1.13383
340 3.61 3.583 1.00754 0.9170 1.042 1.13650
350 3.63 3.598 1.00890 0.9100 1.036 1.13864
360 3.64 3.621 1.00780 0.9155 1.031 1.12620
370 3.66 3.625 1.00970 0.9062 1.026 1.13220
380 3.67 3.638 1.00880 0.9105 1.0223 1.12280
Table 4.16. Calculated values of C and data used in its calculations
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Table 4.17 Data for the development of empirical equations and
evaluation of constants Al, Av, and Bl, Bv.
Temp.
K
X-axis
(1000/T)
Y-axis
(ln Kl)
Y-axis
(ln Kv)
270 3.704 1.22964 1.2378
280 3.571 1.24130 1.2456
290 3.448 1.25280 1.2519
300 3.333 1.26130 1.2576
310 3.226 1.26700 1.2622
320 3.125 1.27260 1.2672
330 3.030 1.27820 1.2720
340 2.941 1.28370 1.2762
350 2.857 1.28920 1.2804
360 2.778 1.29200 1.2843
370 2.703 1.29750 1.2879
380 2.632 1.30020 1.2914
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RESULTS
Two curves are are plotted in Figures 4.1 and 4.2 using data given in
Table 4.17. Empirical equations and values of constants evaluated are
given below
4.2.4.1 Statistics and fitted empirical equation for Kl(i) Statistics:
log Kl= -57.91/T + 1.4536 (4.35)
Regression sum of squares = 0.00226535
Residual sum of squares = 4.77678 E-006
Co-efficient of determination R2= 0.997896
Residual mean squares = 5.97097 E-007(ii) Empirical equation developed is of the type:
Kl= 4.2785 e-57.91/ T
4.2.1.1 Statistics and fitted empirical equation for Kv
(i) Statistics:
log Kv= - 48.463/ T + 1.41885 (4.37)
Regression sum of squares = 0.00158652Residual sum of squares = 2.6971 E-007
Co-efficient of determination R2= 0.99983
Residual mean squares = 3.3713-008
(ii) Empirical equation developed is of the type:
Kv= 4.132 e-48.463/T
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SAMPLE CALCULATIONSEmpirical equations obtained are given below:
Kl= 4.2785 e- 57.91/T
and Kv= 4.132 e- 48.463/T
The values of empirical constants obtained from these equations and
Table 4.16 are listed below:
log Cavg= 0.124155; Al= 4.2785, Bl= 57.91; Av= 4.132, Bv= 48.463
Plugging these values in Eqn. (4.34); we get
log = 0.124155 - [0.0355 - (9.782/T)]1/2 (4.39)
This equation is valid for the temperature range of 283 K to 370 K.
(i) Now at 300 K, log = 0.124155 - 0.0538 = 0.0704,
therefore = 1.073
(ii) And at 373 K, log = 0.124155 - 0.096306 = 0.02785,therefore = 1.028
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COMPARISON OF RESULTS
Table 4.18. Comparison of results of separation factor for H2O-HDO-D2O system of various workers as a function of temperature.
Re
f.N
o.
Temp. K 273 283 293 298 303 313 323 333 343 348 353 363 373
[6
3]
Lewis and
McDonald
1933(Calc;)
1.079 1.069
3
1.060
8
1.05
33
1.04
7
1.0406 1.0353 1.031 1.02
67
[2
1]
B. Topely
H. Eyring
1933
(Theoretical)
1.092 1.083 1.07
1
1.067 1.060 1.05
3
1.03
8
1.02
5
[5
6]
Riesenfeld
and Chang
1936 (Calc;)
1.082
2
1.071
4
1.062
9
1.056 1.04
94
1.04
37
1.0386 1.035 1.0298 1.02
58
[5
5]
Miles and
Menzies
1936
(Empirical)
1.093
6
1.079
5
1.069
8
1.061
3
1.05
34
1.04
65
1.0404 1.0348 1.030 1.02
58
[2
8]
Kirshenbau
m
1951
(Calculated)
1.12 1.087 1.07 1.066 1.059 1.05
2
1.02
6
[6
1]
Combs
Googin and
Smith 1954
(Calculated)
1.093
6
1.082
2
1.071
4
1.063 1.05
45
[6
1]
Combs
Googin and
Smith 1954
(Exptl.)
1.100
3
1.087
3
1.08
2
1.075 1.063 1.05
1
[5
9]
E. Whalley
1957
(Empirical)
1.109 1.07
2
1.05
0
1.03
5
1.02
5
[6
4]
W. M. Jones
1968(Calc.)
1.088
2
1.078 1.068
2
1.060 1.05
3
1.04
62
1.0403 1.0351 1.0304 1.02
61
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Table 4.18 continued........
Ref.
No.
Temp. K 273 283 293 298 303 313 323 333 343 348 353 363 373
[65] W. Alexander
Van Hook
1968(Empirical)
1.105 1.075 1.054 1.038 1.028
[66] Jovan Pupezine
et. al 1972
(Empirical)
1.1076 1.0932 1.081 1.0703 1.0612 1.0533 1.0464 1.0404 1.0352 1.0306 1.0267
[66] From salt
solutiondata (+0.3)
(Empirical)
1.1074 1.093 1.0807 1.0701 1.061 1.0531 1.0462 1.0403 1.0352 1.031 1.027
[38] J. H. Rolston
1976 (Empirical)
1.113 1.098 1.080 1.074 1.065 1.056 1.040 1.027
[69] Gy Jakli and
G. Jancso` 1980
(Empirical)
1.107 1.093 1.076 1.071 1.062 1.054 1.038
[72] Rong Sen Zhang
1988 (Empirical)
1.108 1.0932 1.081 1.075 1.070 1.061 1.053 1.0463 1.0404 1.038 1.0353 1.031 1.027
This work 1998
(S. Empirical)
Eqn. (4.39).
1.098 1.0813 1.0751 1.070 1.061 1.0533 1.047 1.0414 1.039 1.03654 1.032 1.0282
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Discussion
Previously it was thought that H2O-HDO-D2O system is a perfectlyideal mixture, and it was also assumed that vapor pressure of HDO is
the geometric mean (GM) of the vapor pressure of H2O and D2O, asHDO exists only in the presence of H2O and D2O, and cannot beisolated [21, 52, 53, 35]. But later on several workers have reportedthat this system deviates from perfect ideality and GM rule is notexactly valid [30, 65, 66, 67, 69, 90]. The main problem in water is to
be able to distinguish between the contributions arising from deviation
of the(i) GM rule for the vapor pressure of HDO (ii) that arising from thenon-ideality of liquid mixture (iii) and non ideality of the vapor, in thesense of deviation from perfect gas equation. We have attributed the alleffect to non-validity of GM rule.
Our semi-empirical model is based on the assumption that vaporpressure of HDO is the harmonic mean (HM) of vapor pressure of H2Oand D2O, and a comparison of our results with those of other workersis given in Table 4.18 and Figure 4.3.
122
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273.00 293.00 313.00 333.00 353.00 373.00283.00 303.00 323.00 343.00 363.00
Temperature [K]
1.02
1.04
1.06
1.09
1.11
1.03
1.05
1.08
1.10
1.12
Separa
tionFactor
Figure 4.3. Separation Factor vs Temperature curves representing data of differentworkers.
Separation Factorvs Temperature Plotsof different Workers
Ref. 61
Ref. 19
Ref. 54
Ref. 53
Ref. 26
Ref. 59[Calc;]
Ref. 59[Exptl;]
Ref. 57
Ref. 62
Ref. 63
Ref. 64[Empirical]
Ref. 64[salt solu; data]
Ref. 36Ref. 67
Ref. 70
This work 1998
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In[63] it is stated by Lewis and R. T. McDonald that [H(HDO)]liq[H(HDO)]vapdue to intermolecular forces in liquid phase.
It is also well known that [S(HDO)]liq[S(HDO)]vap, and as
G = H - S. Therefore [G(HDO)]liq[G(HDO)]vap.
Also it is known fact that
- G = RT ln K or K = Exp (- G/RT ) from where it is evident that KvKl, which proves that HM rule is applicable instead of GM or AMrule for
In [30] it is proved by Alfred Narten that geometric mean rule is not
obyed.
In [65] the non-ideality of H2O-HDO-D2O system is proved by
Alexander Van Hook.
In [67] the deviation from geometric mean rule is shown to be from0.02 to 0.08 by Alexander Van Hook.
PHDO
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In [68] and [69] Gy. J. Akli et. al also show the deviation from ideality
of H2O - HDO - D2O system and we have proved the validity of HM
assumption from their data and
KlKvindicates that separation is possible.
Kl- Kv) decreases with increase in temperature which explains the fact
that decreases with increase in temperature. Values of calculated from our semiempirical model match well with
the theoretical and empirical values of many other workers. The
values found by us are also in good agreement with the experimental
values quoted in litrature and it is safer to use our relationship in the
temperature range 295 K to 345 K.
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EXPERIMENTAL STUDY OF H2
O-C2
H5
OH
SYSTEM
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Experiments were performed to determine the rates of evaporation in ethanol-watermixtures under different conditions. The objectives of these experiments are to studythe separation of water from ethanol by using evaporation with forced circulation of airat ambient temperature.
APPARATUS
The major apparatus used along with their functions are given below :
i) Solution density measurement apparatus DMA-48 of Anton Parr was used todetermine the Vol% Ethanol in Ethanol-Water mixture.
ii) Air Drying unit (AD) which contains CaCl2. Air is sucked through this unit tomake it as dry as possible.
iii) Air flow rotameter (AFM) to measure the flow rate of air (range 0-3000 lt/hr).iv) Liquid flow rotameter (LFM) to measure flow rate of liquid feed mixture
(range 0-60 lt/hr).
v) Thermometer (T) of thermovalve type to record the temperature of the process.
EVAPORATION OF ETHANOL-WATER MIXTURE-I
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vi) Vacuumgauge (VG) with range 760 to 0 mm of Hg.vii) Peri-Staltic type variable flow liquid pump (LP) having range up to 100 lt/hr.
viii) Oil diffusion vacuum pump (VP). A vacuum of 650 mm of Hg can be
attained in the apparatus.
Diameter of column 80 mm (3 in)
Packing nominal diameter 8 mm (0.375 in)
Packed height 1.5 m (5 ft)
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PROCEDURE
1. A calibration curve was plotted between LFM readings and liquid flow rate (lt/hr),as
shown in Figure 5.2.
2. Air Flow meter was calibrated using a pre-calibrated flow-meter and confirmed by
the collection of air in a polyethylene bag for a fixed time. A calibration curve was
drawn between air-flow rate and vacuum gauge readings, as shown in Figure 5.3.
3. To measure liquid mixture composition a calibration curve was drawn between liquid
mixture density (gm/cc) and Vol% Ethanol using Anton Parr DMA-48 fluid density
measurement apparatus, as shown in Figure 5.4.
4. 1200 ml of mixture of a fixed composition, say 42% Ethanol by volume was taken in
each experiment with different air flow rates (420 lt/hr to 1220 lt/hr).
5. In each experiment the plant was operated for one hour with liquid flow rate of 6
lt/hr.
6 The experiments were repeated for 30.5, 26.5, 20.5, and 16.5 Vol% Ethanol in
mixtures.
7 In each experiment the column was first dried by passing dry air for 1/2 hr. The feed
liquid was then run through the column for 15 min. Then liquid is allowed to
accumulate in the calibrated liquid leg for 5 min and the liquid level is recorded and
its composition is measured by taking a small sample.
8 The change in liquid level giving the volume evaporated of liquid mixture and
change in composition was recorded after 1hr operation.
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1.00
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0.00 20.00 40.00 60.00 80.00 100.00
10.00 30.00 50.00 70.00 90.00
VOL% ETHANOL
0.80
0.84
0.88
0.92
0.96
0.82
0.86
0.90
0.94
0.98
DENSITY(gm/cc)
Figure 5.4. Density of ethanol-water mixture vs Vol% ethanol in mixture.
T bl 5 1 D t f th ti f i t f i l th l d t
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Table 5.1. Data for the evaporation of mixture of commercial ethanol and waterAir Flow Rate
(lt/hr)
Volume
of Mixture
Evaporated(ml/hr)
Vol% Ethanol
Initial
Vol% Ethanol
after 60 min
Vol% of Mixture
after 60 min
Process Temperature
oC
420 84 42.0 36.0 6.0 27.5
680 105 42.0 35.5 6.5 27.0
940 120 42.0 35.0 7.0 29.0
1220 140 42.0 34.0 8.0 27.5
420 65 30.5 27.0 3.5 27.0
680 85 30.5 25.5 5.0 28.5
940 90 30.5 25.5 5.0 27.0
1220 100 30.5 25.0 5.5 27.0
420 60 26.5 23.5 3.0 28.0
680 70 26.5 23.0 3.5 27.0
940 90 26.5 22.5 4.0 29.0
1220 90 26.5 21.5 5.0 27.5
420 55 20.5 18.5 2.0 29.0
680 65 20.5 17.0 3.5 27.5
940 80 20.5 17.0 3.5 27.0
1220 85 20.5 15.0 5.5 28.0
420 50 16.5 14.7 1.8 27.0
680 55 16.5 14.5 2.0 26.0
940 60 16.5 14.5 2.0 24.0
1220 60 16.5 14.0 2.5 24.0
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MATHEMATICAL MODELING ( An Analytical Approach)
To develop different mathematical models a kinetic approach is
applied to the rate of evaporation of volatile liquids [73]. Due to
universal nature of the Arrhenius Law as it is applied to biological and
medical fields we have used it to study the rate of evaporation and
separation of mixture of volatile liquids. Different models have been
made by using different sets of assumptions. In each model a
mathematical equation is made that contains a parameter K (samesymbol K and k are used for different models, they should be
considered as different for every model). The value of this parameter is
then found by testing it against the experimental data. If the parameter
has the same value for each data, then we say that the fit is good, and
the particular model is also good. However, if different experimentalresults give different values of K, then the model is not considered to
be good.
KINETIC MODEL -IA
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This model is made with the following assumptions [74]:
i) Due to the lower volatility of water as compared to ethanol,evaporation of water is neglected .
ii) Rate of evaporation of ethanol is proportional to itsconcentration, i.e. exponential decrease in concentrationof ethanol as a function of time.
This results in the equation : C(t) = C(0) , where K [1/hr] is thevelocity constant.
or (5.1a)
Values of 'K' are calculated for different air flow rates and feedconcentrations and as there is variation in the values of 'K',
therefore an average value is reported for different air flow ratesin Table 5.2 for ethanol water mixture. The values of Kavgincreases from 0.122 to 0.219 for air flow rate 420 to 1220 lt/hrfor Model-IA. Experimental and calculated values of vol%ethanol in product are reported in Table 5.3.
Kt
e
Kt
C
C t
1 0log
( )
( )
Table 5.2. Values of average rate constants Kavgfor different air flow rates and vol%
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g
ethanol in feed for Model-IA.
Air flow rate
(lt/hr)
420 680 940 1220
Kavg 0.122 0.161 0.168 0.219
Table 5.3. Experimental and calculated values of vol% ethanol in product(Model-IA).
Air Flow Rate
(lit/hr)
Vol% Ethanol
in feed C(0)
Vol% Ethanol
after 60 min
C(t)expt
Vol% Ethanol
after 60 min
C(t)cal
% error
420 42.0 36.0 37.176 -3.270
680 42.0 35.5 35.754 -0.716
940 42.0 35.0 35.505 -1.443
1220 42.0 34.0 33.740 +0.765
420 30.5 27.0 26.997 +0.011
680 30.5 25.5 25.964 -1.820
940 30.5 25.5 25.783 -1.110
1220 30.5 25.0 24.501 +1.996
420 26.5 23.5 23.456 +0.187
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Maximum error in this model is 9.787% which is much below the allowable value of 25%
[76]. Graphical represention of experimental and calculated vol% ethanol in product as a
function of vol% ethanol in feed and air flow rate are shown in Figures 5.5 and 5.6.
Table 5.3 continued
680 26.5 23.0 22.559 +1.917
940 26.5 22.5 22.402 +0.436
1220 26.5 21.5 21.288 +0.986
420 20.5 18.5 18.146 +1.914
680 20.5 17.0 17.452 -2.659
940 20.5 17.0 17.330 -1.941
1220 20.5 15.0 16.468 -9.787
420 16.5 14.75 14.620 +0.881
680 16.5 14.5 14.050 +3.103
940 16.5 14.5 13.950 +3.793
1220 16.5 14.0 13.250 +5.357
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16.00 26.00 36.0021.00 31.00 41.00
Vol% Ethanol in Feed
13.00
18.00
23.00
28.00
33.00
38.00
15.50
20.50
25.50
30.50
35.50
Vol%E
thanolinP
roduct
Air flow rate (lt/hr)
1220 lt/hr
1220 lt/hr
940 lt/hr
940 lt/hr
680 lt/hr
680 lt/hr
420 lt/hr
420 lt/hr
Markers represent Exptl. data.
Curves represent calc. values.
Figure 5.5. Experimental and calculated values of Vol% ethanol in product vs Vol%ethanol in feed for Model-IA.
140
400.00 600.00 800.00 1000.00 1200.00500.00 700.00 900.00 1100.00
Air Flow Rate (lt/hr)
13.00
18.00
23.00
28.00
33.00
38.00
15.50
20.50
25.50
30.50
35.50
Vol%EthanolinProduct
Figure 5.6. Experimental and calculated values of Vol% ethanol in productvs Air flow rate for Model-IA.
Markers represent Exptl. data.Curves represent Calc. values.
141
Vol% ethanol in feed
42%
42%
30.5%
26.5%
26.5%
20.5%
20.5%
16.5%
16.5%
KINETIC MODEL IB
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KINETIC MODEL-IB
i) Due to the lower volatility of water as compared to ethanol,
evaporation of water is neglected.
ii) Linear decrease in concentration of ethanol as a function of timeis assumed i.e.,
C(t) = C(0) - K t
or (5.1b)
where K [hr-1] is a velocity constant. Values of 'K' are calculated fordifferent air flow rates and feed concentration and average values of
'K' are calculated for every air flow rate.
The values of Kavg increases from 3.25 to 5.3 when air flow rate
increases from 420 to 1220 lt/hr for Model-IB.
Maximum error in this model is 20% which is below the allowable
value of 25%
KC(0) C(t)
t
KINETIC MODEL II
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KINETIC MODEL-II
The following four assumptions are used :
i) Due to the lower volatility of water as compared to ethanolevaporation of water is negligible i.e., Vw(t) = Vw(0) = Vw =
constant, where Vwis the volume of water in mixture.ii) Volume of ethanol in mixture is a function of time, i.e. Va(t).
iii) Volume fraction of ethanol (5.2)
iv) From Arrhenius Law, rate of decrease in volume of mixture isproportional to ethanol concentration.Therefore: (5.3)
(5.10)
Again values of 'K' are calculated for different air flow rates and feed
concentration, and as there is variations in the values of 'K' an averagevalue is calculated for every air flow rate, as reported in Table 5.6. Thevalue of Kavg increase from 264 to 405 for the Model-II when air flowrate increases from 420 to 1220 lt/hr.
Maximum error in this model is 17.65% which is below the allowablevalue of 25% [76]. A large positive error due to the reason that aconstant volume of water (VW) is assumed.
Ca tVa t
Vw Va t( )
( )
( )
dVa t
dtKC
at K
Va t
Vw Va t
KVa loss
t
V
t1
Va loss
Va
w log
( ) ( ) ( )
( )
0
0
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KINETIC MODEL -III
This model is based on the following two assumptions.
i) Due to the lower volatility of water as compared to ethanol evaporation ofwater is neglected.
ii) dV/dt = -k [C(t)] is assumed, where dV/dt = volumetric rate of
evaporation of mixture, C(t) = C(0) - Kt, where K [1/hr] is rate constant, n is order of
reaction and k = proportionality constant which is a function of air flow rate,
temperature, liquid flow rate, interfacial area and nature of the liquid.Therefore (5.11)
(5.16)
Where postscript I and J represents two different cases of volume evaporated (V),
vol% ethanol in feed (F), vol%ethanol in product (P) and rate constant (K).
The average values of ns varies from 0.41 to 0.87 for air flow rates from 420 to
1220 lt/hr.
dV
dtk C(t) k [C(0) -Kt]n n [ ]
VI
VJ
KJ
KI
FI PI
FJ PJ
n 1 n 1
n 1 n 1
n
MODELS IVA AND IVB
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MODELS IVA AND IVB
These models are based on the following assumptions :
i) Due to the lower volatility of water as compared to ethanol evaporation of wateris neglected.
ii) (5.17)
where, dV/dt is volumetric rate of evaporation, and n is order of reaction.
C(t) = C(0) , K is rate or velocity constant, and k is proportionality constantas defined in model - III.
To use computer for iterative calculations the above equation is written as under:
MODEL-IVA
Where postscript I and J represents two different cases of ethanol concentration in feed(C), rate constant (K) and volume evaporated (V).
As nK 0 e-nK = 1-nK
Eqn. (5.20) gives:
(5.21)
For iterative computer calculations the above equation is written as under:
MODEL-IVB
For different air flow rates anavgvaries from 0.396 to 0.920 andbnavgvaries from 0.404
to 0.829.
dV
dtk C t k [C(0) en
-Kt n [ ( )] ]
Vk C
nK1 1 nK k [C 0
nn
[ ( )][ ] ( )]
0
VI
VJ
CI
CJ
KJ
KI
1 e
1 e
n
n
nKI
nKJ
[ ]
[ ]* *
[ ]
[ ]
VI
VJ
CI
CJ
n
n
Kt
e
DISCUSSION
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DISCUSSION
The maximum error for Model-IA is 9.8%, which is reasonable for such experimentaldata.
The maximum error for Model-IB is 20.0% at low concentration of ethanol i.e., 16.5%.It is clear from Figure 5.8, that this model gives good results from 42 to 20% ethanol infeed.
The maximum error for Model-II is 17.65%. This negative error is due to the additionof a constant quantity Vwin the denominator of Eqn. (5.2). This assumption have moreeffect due to the more evaporation of water at higher air flow rates.
Therefore analytical Model-IA is more appropriate for this process conditions due tolesser error as compared to other two models. Furthermore by looking at Tables 5.8, 5.9and 5.10 it is clear that order of reaction i.e., values of nsincreases with air flow rate inModel-III, IVA and IVB show same behavior. The values of navg for these modelsmatch each other with maximum error of 4%.
The mathematical models developed by analytical techniques do not give a constantvalue of the parameter 'K' and n for different cases due to the following reasons.
i) Error in the measurements of 'volume evaporated' due to large hold up
of column andsmall volume change.ii) There is also temperature variation (i.e., 24 to 29C) which is not taken into
account.
iii) Error due to the Humidity of input air.
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EVAPORATION OF ETHANOL-WATER
MIXTURE-II
INTRODUCTION
There are two general approaches for curve fittings that are distinguishedfrom each other on the basis of the amount of error associated with the data.
First, when the data exhibits a significant degree of error or 'noise' the
strategy is to derive a single curve that represents the general trend of the
data. Approach of this nature is called Least Square Regression.Second, when the data is known to be very precise the basic approach is
to fit a curve or a series of curves that pass directly through each of the
points. This estimation of values between well-known discrete points iscalled Interpolation.
MATHEMATICAL MODELING (An Empirical Approach)
As our data is of first kind, so it is preferred to apply least square regressiontechnique to see the variation of 'volume evaporated' first as a function of
normalized air flow rate; secondly air flow rate as function of feedconcentration; and thirdly changes in ethanol concentration as a function of
air flow rate. The details of these curves which are made by using computersoftware and experimental data of Table 5.1 are given below.
2.80
153
2.80
154
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0.80 1.20 1.60 2.00 2.40 2.80 3.201.00 1.40 1.80 2.20 2.60 3.00
Air Flow Rate (Normalized)
1.00
1.40
1.80
2.20
2.60
1.20
1.60
2.00
2.40
VolumeEva
porated(Normalized)
Vol% ethanol in feed42%
42%
30.5%
30.5%
26.5%
26.5%
20.5%
20.5%
16.5%
16.5%
Figure 6.1. Shows Normalized (Volume of mixture evaporated vs Air flow rate)using straight line fit for different initial concentration of ethanol.
Markers represent data points.Curves represents straight line fit.
0.80 1.20 1.60 2.00 2.40 2.80 3.20
1.00 1.40 1.80 2.20 2.60 3.00
Air Flow Rate (Normalized)
1.00
1.40
1.80
2.20
2.60
1.20
1.60
2.00
2.40
VolumeEvap
orated(Normalized)
Vol% ethanol in feed42%
42%
30.5%
30.5%
26.5%
26.5%
20.5%
20.5%
16.5%
16.5%
Figure 6.2. Shows Normalized (Volume of mixture evaporated vs Air flow rate) using
Log regression curve fitting for different initial concentration of ethanol.
Markers represent Exptl.data.Curves represent log regression fit.
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Figure 6.6. Shows Volume of mixture evaporated vs Vol% ethanol in feed using Log
regression curve fitting for ethanol-water mixture for different air flow rates.
Markers represent Expt.; data.Curves represent Log regression fits.
16.00 26.00 36.0021.00 31.00 41.00
Vol% ethanol in feed
50.00
70.00
90.00
110.00
130.00
60.00
80.00
100.00
120.00
140.00
Volumeofmixtu
reevaporated(ml/hr)
Air Flow Rates420 lt/hr
420 lt/hr
680 lt/hr
680 lt/hr
940 lt/hr
940 lt/hr
1220 lt/hr1220 lt/hr
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EMPIRICAL EQUATIONS OF TYPE ( V = a + b Log F + e Log C )
From the above figures it is clear that 'volume evaporated' shows logarithmic relationshipwith 'air flow rate' and 'feed concentration of mixture'. On this basis the following empiricalequation is derived.
(6.1)where F is air flow rate, V1volume evaporated, a' and b are constants.
Similarly (6.2)
Where C is the ethanol concentration , V2is volume evaporated, d and e are empiricalconstants.
Combining Eqns. (6.1) and (6.2) we get:
(6.3)
Let a' + d = aThen (6.4)
The values of 'a', 'b', and 'e' are determined by 'Multiple Linear Regression' technique,for which computer program is attached at Appendix-4. The values determined for, a, b, eand V are given as output. These results indicate that more than 90% of the originaluncertainty has been explained by this model i.e. R 0.90 [75].
Forcommercial ethanol empirical equations for volume evaporated of ethanol andethanol-water mixture as a function of 'air flow rate' and 'feed concentration' are givenbelow:
V(ethanol) = -361.431 + 26.7 Log F + 77.3 Log C (6.5)
V(mixture) = -336.7 + 33.95 Log F + 59.23 Log C (6.6)
The three dimensional graphical representation for Eqns. (6.5) and (6.6) are given in Figures6.11 and 6.12 respectively.
V a b Log F1
/
V V V a d b Log F e Log C1 2 ( )/
V a b Log F + e Log C
2
V d e Log C2
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The maximum error is 16.67% which is below the allowable value of 25%
[76]. Graphical representation of experimental and calculated volume of
mixture evaporated (ml/hr) as a function of log of air flow rate and vol%
ethanol in feed arerepresented in Figures 6.13 and 6.14 respectively. Plot
of calculated vs experimental values of volume of mixture evaporated for
this model are shown in Figure 6.15, the fitted line have equation Y =
1.012*X with R = 0.995736 (which means more than 99.58% of
uncertainity has been explained by this model [75]).
2
Fi 6 11
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Fig 6.11
Figure 6.11. Three dimensional graphical representation for Volume of ethanol evaporated as a function of
Air flow rate and Vol% ethanol in feed for empirical Model-I [Eqn. (6.5)].
ETHA
NOLVOLUMEEV
APORATED(ml/hr).
AIR FLOW RATE(lt/hr). VOL% ETHANOL IN FEED
Fi 6 12
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Fig 6.12
MIX
TUREVOLUMEE
VAPORATED(ml/hr)
AIR FLOW RATE (lt/hr)
Figure 6.12. Three dimensional graphical representation for Volume of mixture evaporated as a function of
Air flow rate and Vol% ethanol in feed for empirical Model-I [Eqn. (6.6)].
165 166
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6.00 6.40 6.80 7.206.20 6.60 7.00
Log of air flow rate (lt/hr)
34.00
54.00
74.00
94.00
114.00
44.00
64.00
84.00
104.00
124.00
Volumeofmixtu
reevaporated(ml/hr)
Vol% ethanol in feed42%
42%
30.5%
30.5%
26.5%
26.5%
20.5%
20.5%
16.5%
16.5%
Figure 6.13. Shows Experimental and calculated values of Volume of mixture evaporatedvs Log of air flow rate for straight line fits for different Vol% ethanol in feed
for empirical Model-I.
Markers represent Exptl. data.Curves represent Calc. values
2.80 3.00 3.20 3.40 3.60 3.802.90 3.10 3.30 3.50 3.70
Log of vol% ethanol in feed
34.00
54.00
74.00
94.00
114.00
44.00
64.00
84.00
104.00
124.00
Volumeofmixtu
reevaporated(ml/hr)
Air flow rates420 lt/hr
420 lt/hr
680 lt/hr
680 lt/hr
940 lt/hr
940 lt/hr
1220 lt/hr
1220 lt/hr
Figure 6.14. Experimental and calculated values of volume of mixture evaporated vsLog of Vol% ethanol in feed for straight line fit for different Air flow
rates for empirical Model-I.
Markers represent Exptl. data.Curves represent Calc. values.
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45.00 65.00 85.00 105.00
55.00 75.00 95.00 115.00
Experimental volume of mixture evaporated (ml/hr)
45.00
65.00
85.00
105.00
55.00
75.00
95.00
115.00
Calcu
latedvolumeofm
ixtureevaporated(ml/hr)
Markers represent Exptl. data.Curves represent Calc. values.
Figure 6.15. Calculated values of volume of mixture evaporated vs Experimental valuesof volume of mixture evaporated for empirical Model-I [Eqn. (6.6)].
V(Calc.) = 1.012* V(Exptl.) and R**2= 0.995736for empirical Model-I [Eqn. (6.6)].
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EMPIRICAL EQUATIONS OF TYPE (Y = a XbZc)
We have done some further analysis of data of Table 6.1 using another technique given in
referance [76]. We have used the computer software and following procedure to find the
empirical equations of the type Y = a Xb
Zc
.
(i) Plotted experimental values of volume of mixture evaporated (VE) vs air flow rate (AF).
This will give us four curves one for each ethanol concentration in feed. Fitted the power
exponential equation: VE= a (AF)band found the value of aand bfor each concentration.
An average value of 0.42 is taken for btaken from Figure 6.16.
(ii) Plotted VE / (AF)0.42vs initial concentration (CF). Fitted the power exponential equation:
VE / (AF)0.42 = c (CF)
dagain. The value of c = 0.681, d = 0.643, and
R2(coefficient of determination) = 0.92 was obtained. Which means more than 92% of
uncertainty has been explained by this model [75] taken from Figure 6.17.
(iii) Plotted (VE/ (CF)0.643) vs AF. Fitted the power equation: VE/ (CF)
0.643= e (AF)f. Then
we obtained e = 0.572, f = 0.427, and R2= 0.921, taken from Figure 6.18.
(iv) The equation VC= 0.572 (AF)0.427 (CF)
0.643was obtained.
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In Table 6.2 are shown the initial concentrations of ethanol, and the
volume of mixture evaporated (experimental and calculated by Model-
II) for different air flow rates. As a first approximation since thetemperature are all within five degree Celsius of each other, we assume
that the effect of temperature may be neglected. The volume
evaporated has been related to the air flow rates and initial
concentration, and it has been found that best fit for the data is made
by the equation:
VC = 0.572 (AF)0.427 (CF)
0.643 (6.7)
Where VC= Total Volume Evaporated (ml)
AF = Air flow rate (lit/hr)
CF= Initial ethanol concentration in mixture (Vol %)
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20.00 25.00 30.00 35.00 40.00
22.50 27.50 32.50 37.50
Vol% ethanol in feed
4.00
5.00
6.00
7.00
8.00
4.50
5.50
6.50
7.50
VolumeEvaporated/(AirFlowRate)**0.42
Solid lines represent equations.Markers represents data.
Y=0.681*X**0.643R**2 = 0.916673
Figure 6.17. Plotted Volume evaporated / (Air flow rate)**0.42 vs Vol% ethanol in feed, using Power law fit for the development of empirical Model-II.
169
420.00 620.00 820.00 1020.00 1220.00520.00 720.00 920.00 1120.00
Air Flow Rate (lt/hr)
65.00
85.00
105.00
125.00
145.00
75.00
95.00
115.00
135.00
155.00
Volumeofmixtureevaporated(ml/hr)
Vol% ethanol in feed42%
42%
30.5%
30.5%
26.5%
26.5%
20.5%
20.5%
Solid lines represent equation.Markers represent exptl. data.
Figure 6.16. Experimental Volume of mixture evapora ted vs Air flow rate for differentVol% ethanol in feed using Power law fit for the development of empirical
Model-II.
168
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420.00 620.00 820.00 1020.00 1220.00520.00 720.00 920.00 1120.00
Air flow rate (lt/hr)
7.00
9.00
11.00
13.00
8.00
10.00
12.00
14.00
VolumeEvaporated/(Vol%
ethanolinfeed)**0.
641
Solid lines represent equation.Markers represernt Exptl. data.
Y=0.572*X**0.427
R**2=0.921
Figure 6.18. Plotted Volume of mixture evaporated / (Vol% ethanol in feed)**0.643 vsAir flow rate using Power law fit for the development of empiricalModel-II.
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The calculated values of the volume evaporated (VC) are given in
Table 6.2 and is plotted against the experimentally observed value of
volume evaporated (VE
) in Figure 6.19. The markers representexperimental values and continuous curve represents calculated values
of V. For ideal data, all points should lie on the 45 line. As seen in
Figure 6.19, most of the data lies about this line, represented by
equation VC= 0.999583 VEwith R2= 0.99738. The percent error in the
calculation of VCis found by:
Percent error = 100*( VE- VC)/ VE (6.8)
which is also given in Table 6.2. It can be seen that the error in Model-
II is always less than about 8.3 %, which shows that the fit is quite
good.
0
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55.00 75.00 95.00 115.00 135.0065.00 85.00 105.00 125.00
Experimental Volume Evaporated (ml/hr)
52.00
72.00
92.00
112.00
132.00
62.00
82.00
102.00
122.00
CalculatedVo
lumeEvaporated(ml/hr)
45 LineMarkers represent Exptl. data.Solid line represents fitted line.
Figure 6.19. Calculated values of volume evaporated vs Experimental values of Volumeevaporated for empirical Model-II.
Y=0.999583*XR**2=0.99738
DISCUSSION
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DISCUSSION
The error observed in calculated values of two models may be due to the following
reasons:
(i) Error in the measurements of "Volume evaporated " is due to large hold up ofcolumn and small volume change.
(ii) There is also temperature variation (i.e., 24 to 290 C) which is not taken into
account.
(iii) Humidity of input air may also be contributing some error.
We feel that the rate of separation is quite small, which can be improved by increasing
the amount of reflux, which can be done by increasing air flow rate, and by having
temperature difference between the liquids and air for condensation. Also, the inlet air
should be bubbled through the product. Air flow rate can be increased manifold by
having co-current flow, but how this will affect the separation factor, remains to beseen. As the maximum error in Model-II is 8.3%, while the error in Model-I is 16.67%.
Therefore empirical Model-II is better than the empirical Model-I.
o
SEPERATION FACTOR OF ETHONOL-WATER
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SEPERATION FACTOR OF ETHONOL-WATER
MIXTURECALCULATION OF SEPERATION FACTOR (semi-empirical) FOR
ETHONOL-WATER SYSTEM
INTRODUCTION
In this chapter we have discussed the calculations of distillation separation factor of
ethanol-water mixture by analytical (thermodynamic), semi-empirical, and
empirical techniques.
Distillation is the combination of two operations evaporation and condensation. For
the calculation of separation factor, only the experimental data of evaporation is
sufficient. Different setups of apparatus Figure 7.3 to Figure 7.18 and process
conditions, such as degree of vacuum (VM), process temperature (PT), room
temperature (RT), column top temperature (CTT), air flow rate (AFR), density of
feed (DF), density of product (DP), volume of mixture evaporated (VW), volume of
product (VP), volume of feed (VF), calculations factors (A1 and A2) and separation
factor (SF) were recorded in Tables 7.3 to 7.18. To decide about the best setup,
separation factor were calculated for different setups and process conditions.
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MATHEMATICAL
The simplest example of batch distillation is a single stage differential
distillation, starting with a still pot, initially full, heated at a constant rate. In thisprocess the vapor formed on boiling the liquid is removed at once from the
system. This vapor is richer in the more volatile component, with the result that
the composition of the liquid product progressively alters. Thus, whilst the
vapor formed over a short period is in equilibrium with the liquid, the total
vapor formed is not in equilibrium with the residual liquid. At the end of the
process the liquid which has not been vaporized is removed as the bottom
product. The analysis of this process was first proposed by Reyleigh [57].
Let M be the number of moles of material in the still and X be the mole
fraction of component A. Suppose an amount dM, containing a mole fraction Y
of A, be vaporized. Then a material balance on component A gives:
Y dM = d(MX) = MdX + XdM
If MP, MF are the number of moles and XP, XF are volume fraction of product
and feed, then after rearrangement and integration of above equation we get:
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(7.41)
The integral on the right-hand side can be solved graphically if the
equilibrium relationship between Y and X is available. In some cases a
direct integration is possible. Thus if over the range concerned the
equilibrium relationship is a straight line of the form Y= mX+C
where m is the slope of the straight line and C is the intercept,
then:
log {MP/MF} = [1/(m-1)] log {[(m-1)XP + C]/ [(m-1)XF + C]}
or MP/MF = [(Y-X)/(YF-XF)] (m-1) (7.42)
logMP
MF
dX
Y XXF
XP
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From this equation the amount of liquid to be distilled in order to
obtain a liquid of given concentration in the still may be calculated and
from this the average composition of the distillate can be found by amass balance.
Alternatively, if the relative volatility may be assumed constant
over the range concerned, then Y = X/[ 1 + (-1)X] can be
substituted in Eqn. (7.42). This leads to the solution:
log{MP/MF}=[1/(-1)]log{XP(1-XF)/XF(1 - XP)}+log {(1- XF)/(1 -XP)} (7.43)
As this process consists of only a single stage, a complete
separation is impossible unless the relative volatility is infinite.
Application is restricted to conditions where a preliminary separation
is to be followed by a more rigorous distillation, where high purities
are not required, or where the mixture is very easily separated.
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STEP-IIIIn this step overall and component volume balance
equations were written.
If VF, VP and VW are the volume of feed, product, and waste
respectively, and
VF = 1200 ml ; then ( 7.48 )
VP = 1200 - VW ( 7.49 )
Now as XF, XP and XW are the volume fraction ethanol in feed,
product and waste respectively and MF, MP and MW are the moles of
ethanol in feed, product and waste, then we can write:
MF = VF * XF * 0.018333 ( 7.50)
and
MP = VP * XP * 0.018333 ( 7.51 )
STEP-IV Writing formula for single stage batch distillation i.e., differential
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STEP IV Writing formula for single stage batch distillation i.e., differentialdistillation [10] we get:
Log (MP/MF) = (1/A) Log {[XP(1 - XF)]/ [XF(1 - XP)]}
+ Log{(1 - XF)/(1 - XP)} ( 7.52 )
rearranging
A = 1+ Log {[XP(1 - XF)]/[XF(1 - XP)]}(Log(MP/MF)
- Log {(1 - XF)/(1 - XP)} (7.53)
Where A is the separation factor:
Separation factor of ethanol-water system has been calculated fordifferent setups of apparatus. The input and output data are attached in Tables7.3 to 7.18. The symbols used in Tables, are not properly defined above are
given as under:VM = Vacuum (mm of Hg)
AFR = Air flow rate (lt/hr)
DD = DP - DF
PT = Process Temperature
RT = Room Temperature
CTT = Column Top TemperatureA1 = Log{ XP(1 - XF) / (XF(1 - XP))}
A2 = Log{ MP(1 - XP) / ( MF(1 - XF))}
A = 1 + ( A1/A2)
SF = A = Separation factor
100.00
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0.80 0.84 0.88 0.92 0.96 1.00
Density of ethanol (gm/cc)
0.00
20.00
40.00
60.00
80.00
Vol
%e
thanol
Figure 7.19. Vol% ethanol in ethanol-water mixture vs density of ethanol.
Markers represents Expt. data.Curve represents fitted polynomial.
V= -2050.4 D**2 + 3148.0 D - 1100
FIG7 3
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FIG7.3
Figure 7.3 Experimental setup for the separation factor study of ethanol-water mixture using packed column,recycling of liquid feed and dried air flow from the liquid surface at room temperature.
FIG7.5
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FIG7.5
Figure 7.5 Experimental setup for the separation factor study of ethanol-water mixture using, slightly heated
(liquid feed and dried air flow) with air flow from the liquid surface.
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TABLE 7.3 INPUT AND OUTPUT DATA SETUP OF
FIGURE 7.3
S.No VM AFR DF DP DD VW PT RT CTT VP XF XP MF MP A1 A2 SF
1 300 420 0.9508 0.9566 0.0058 84 27.5 30 28 1116 0.3951 0.3509 8.6930 7.1792 -0.1894 -0.1207 2.5685
2 200 680 0.9508 0.9576 0.0068 105 27 30 28 1095 0.3951 0.3431 8.6930 6.8881 -0.2237 -0.1502 2.4889
3 100 940 0.9508 0.9581 0.0073 120 29 30 28 1080 0.3951 0.3392 8.6930 6.7166 -0.2410 -0.1695 2.4217
4 0 1220 0.9508 0.9596 0.0088 140 27.5 30 28 1060 0.3951 0.3275 8.6930 6.3636 -0.2939 -0.2059 2.4278
5 300 420 0.9646 0.9680 0.0034 65 27 30 28 1135 0.2876 0.2599 6.3270 5.4080 -0.1394 -0.1188 2.1733
6 200 680 0.9646 0.9695 0.0049 85 28.5 30 28 1115 0.2876 0.2475 6.3270 5.0598 -0.2047 -0.1688 2.2131
7 100 940 0.9646 0.9696 0.0050 90 27 30 28 1110 0.2876 0.2467 6.3270 5.0203 -0.2092 -0.1755 2.1918
8 0 1220 0.9646 0.9703 0.0057 100 27 30 28 1100 0.2876 0.2409 6.3270 4.8580 -0.2407 -0.2007 2.1992
9 300 420 0.9689 0.9721 0.0032 60 28 30 28 1140 0.2525 0.2259 5.5547 4.7206 -0.1464 -0.1277 2.1463
10 200 680 0.9689 0.9725 0.0036 70 27 30 28 1130 0.2525 0.2225 5.5547 4.6097 -0.1657 -0.1472 2.1259
11 100 940 0.9689 0.9732 0.0043 90 29 30 28 1110 0.2525 0.2166 5.5547 4.4082 -0.2001 -0.1843 2.0856
12 0 1220 0.9689 0.9742 0.0053 90 27.5 30 28 1110 0.2525 0.2082 5.5547 4.2363 -0.2506 -0.2134 2.1745
13 300 420 0.9757 0.9779 0.0022 55 29 30 28 1145 0.1954 0.1766 4.2992 3.7061 -0.1247 -0.1253 1.9956
14 200 680 0.9757 0.9789 0.0032 65 27.5 30 28 1135 0.1954 0.1679 4.2992 3.4939 -0.1853 -0.1738 2.0665
15 100 940 0.9757 0.9790 0.0033 80 27 30 28 1120 0.1954 0.1670 4.2992 3.4299 -0.1916 -0.1912 2.0017
16 0 1220 0.9757 0.9817 0.0060 85 28 30 28 1115 0.1954 0.1435 4.2992 2.9332 -0.3714 -0.3198 2.1614
17 300 420 0.9800 0.9819 0.0019 50 27 30 28 1150 0.1584 0.1417 3.4838 2.9883 -0.1304 -0.1339 1.9742
18 200 680 0.9800 0.9824 0.0024 55 24 30 28 1145 0.1584 0.1373 3.4838 2.8830 -0.1671 -0.1646 2.0147
19 100 940 0.9800 0.9824 0.0024 60 26 30 28 1140 0.1584 0.1373 3.4838 2.8704 -0.1671 -0.1690 1.9884
20 0 1220 0.9800 0.9827 0.0027 60 24 30 28 1140 0.1584 0.1347 3.4838 2.8151 -0.1896 -0.1854 2.0224
TABLE 7 5 Input and Output Data For Setup of Figure 7 5
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TABLE 7.5 Input and Output Data For Setup of Figure 7.5
S.No VM AFR DF DP DD VW PT RT CTT VP XF XP MF MP A1 A2 SF
1 400 145 0.9529 0.9552 0.0023 20 30 12 23 1180 0.3793 0.3617 8.3441 7.8247 -0.0754 -0.0363 3.0746
2 350 270 0.9529 0.9564 0.0035 60 30 12 23 1140 0.3793 0.3524 8.3441 7.3659 -0.1157 -0.0824 2.4052
3 300 420 0.9529 0.9576 0.0047 75 30 12 23 1125 0.3793 0.3431 8.3441 7.0768 -0.1568 -0.1081 2.4504
4 200 680 0.9529 0.9593 0.0064 95 30 12 23 1105 0.3793 0.3298 8.3441 6.6816 -0.2164 -0.1455 2.4868
5 100 940 0.9529 0.9614 0.0085 100 30 12 23 1100 0.3793 0.3132 8.3441 6.3167 -0.2925 -0.1772 2.6502
6 350 270 0.9605 0.9657 0.0052 55 30 12 23 1145 0.3204 0.2787 7.0479 5.8500 -0.1989 -0.1268 2.5689
7 300 420 0.9605 0.9662 0.0057 80 35 12 23 1120 0.3204 0.2746 7.0479 5.6388 -0.2192 -0.1579 2.3880
8 250 540 0.9605 0.9654 0.0049 70 25 12 23 1130 0.3204 0.2811 7.0479 5.8238 -0.1868 -0.1346 2.3874
9 200 680 0.9605 0.9664 0.0059 90 30 12 23 1110 0.3204 0.2730 7.0479 5.5553 -0.2274 -0.1706 2.3329
10 150 810 0.9605 0.9678 0.0073 100 30 12 23 1100 0.3204 0.2615 7.0479 5.2743 -0.2859 -0.2069 2.381911 350 270 0.9687 0.9723 0.0036 50 35 12 23 1150 0.2541 0.2242 5.5910 4.7267 -0.1647 -0.1286 2.2813
12 300 420 0.9687 0.9743 0.0056 90 40 12 23 1110 0.2541 0.2073 5.5910 4.2190 -0.2645 -0.2207 2.1984
13 250 540 0.9687 0.9745 0.0058 80 35 12 23 1120 0.2541 0.2056 5.5910 4.2222 -0.2748 -0.2178 2.2618
14 200 680 0.9687 0.9746 0.0059 85 30 12 23 1115 0.2541 0.2048 5.5910 4.1860 -0.2800 -0.2253 2.2427
15 150 810 0.9687 0.9743 0.0056 85 25 12 23 1115 0.2541 0.2073 5.5910 4.2380 -0.2645 -0.2162 2.2233
16 350 270 0.9746 0.9783 0.0037 60 30 12 23 1140 0.2048 0.1731 4.5051 3.6178 -0.2071 -0.1803 2.1489
17 300 420 0.9746 0.9792 0.0046 70 35 12 23 1130 0.2048 0.1653 4.5051 3.4246 -0.2626 -0.2258 2.1629
18 250 540 0.9746 0.9798 0.0052 70 30 12 23 1130 0.2048 0.1601 4.5051 3.3167 -0.3008 -0.2516 2.1957
19 200 680 0.9746 0.9790 0.0044 85 30 12 23 1115 0.2048 0.1670 4.5051 3.4146 -0.2500 -0.2308 2.0835
20 150 810 0.9746 0.9797 0.0051 95 35 12 23 1105 0.2048 0.1610 4.5051 3.2609 -0.2944 -0.2696 2.0920
21 350 270 0.9795 0.9832 0.0037 60 40 12 23 1140 0.1627 0.1303 3.5795 2.7229 -0.2602 -0.2355 2.1048
22 300 420 0.9795 0.9836 0.0041 65 30 12 23 1135 0.1627 0.1267 3.5795 2.6373 -0.2918 -0.2634 2.1079
23 250 540 0.9795 0.9840 0.0045 70 35 12 23 1130 0.1627 0.1232 3.5795 2.5522 -0.3243 -0.2921 2.1099
24 200 680 0.9795 0.9845 0.0050 75 30 12 23 1125 0.1627 0.1188 3.5795 2.4493 -0.3660 -0.3282 2.1151
25 150 810 0.9795 0.9850 0.0055 80 25 12 23 1120 0.1627 0.1143 3.5795 2.3471 -0.4093 -0.3659 2.1186
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DEVELOPMENT OF EMPIRICAL RELATIONSHIP FOR
SEPERATION FACTOR
DEVELOPMENT OF COMPUTER PROGRAM
Computers programs have been written in FORTRAN to use the data obtained in the
experimental results reported in Tables 7.3 to 7.18. Separation factors (empirical) are
calculated by using the models of Eqns. (7.54), (7.55), (7.56) and (7.57) by using
computer programs mod-5 and mod-6 (modified form of mod-4) developed insection 6.2.1.
Mod-5 calculates the empirical constants of Eqns. (7.54), (7.55) and (7.56) with
two independent variables, where as mod-6 calculates the empirical constants of Eqn.
(7.57) with three independent variables.
Different combinations of last three columns of master input file are used taking
two at a time along with first two columns. For example column 3 and 4 are used forEqn. (7.54), column 3 and 5 for Eqn (7.55) and column 3 and 4 for ARF=0 for Eqn.
(7.56) for vacuum distillation. For Eqn. (7.57) all the three columns of input file are
used. The steps involved in this are given in the next slide.
STEP I Input data from Tables 7 3 to 7 18 are merged to a single file and an
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STEP-I Input data from Tables 7.3 to 7.18 are merged to a single file and an
master input file is prepared for 371 points, each data point having been taken
after 1.5 hr of operation time.
STEP-II A computer code called MOD-5 has been developed to calculate thevalues of constants of the following empirical equations:
SF = a + b Log (DF) + e Log (T) (7.54)
SF = a + b Log (DF) + e Log (AFR) (7.55)
SF(VD) = a + b Log (DF) + e Log (T) (7.56)
Where AFR is the air flow rate in lt/hr, and T is the temperature in K. Eqn.(7.56) is for vacuum distillation , in this case AFR = 0.
STEP-III A computer code called MOD-6 has been developed to
calculate the values of constants for empirical equation of the type:
SF = a + b Log (DF) + e Log (AFR) + d Log (T) (7.57)
The computer programs, Input and Output files for these models are
attached at Appendix-5 and 6.
RESULTS
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Resulting empirical equations for different cases are given below:
CASE-I SF = f (DF, T)
SF = 0.539 - 10.403 Log (DF) + 0.236 Log (T) (7.58)
DF(0.9492 gm/cc to 0.9969 gm/cc), T(294.5 K to 350 K)
CASE-II SF = f (DF, AFR)
SF = 1.797 - 10.466 Log(DF) +0.014 Log (AFR) (7.59)
DF(0.9492 gm/cc to 0.9969 gm/cc), AFR(25 lt/hr to 1220 lt/hr)
CASE-III SF (VD) = f (DF, T)
SF = -12.746 - 20.494 Log (DF) + 2.506 Log (T) (7.60)
DF(0.9810 gm/cc to 0.9944 gm/cc), T(314 K to 350 K)
CASE-IV SF = f (DF, AFR, T)
SF = 3.008 - 10.258 Log (DF) + 0.016 Log (AFR) - 0.214 Log(T) (7.61)
DF(0.9492 gm/cc to 0.9969 gm/cc), AFR(25 lt/hr to 1220 lt/hr), T(294 K to 335 K)
(i) Values of SF (Experimental) given in Tables 7.3 to 7.18 and SF (Calculated) by this program
are plotted in Figures 7.20 to 7.23. Fitted straight lines through origin have slopes almost equal to
1 i.e., 450 showing that the calculated values are in good agreement with the experimental values.
(ii) Surface plots of Eqns. (7.58), (7.59) and (7.60) are also shown in Figures 7.24, 7.25 and 7.26.
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1.80 2.00 2.20 2.40 2.601.90 2.10 2.30 2.50
Separation Factor (experimental)
1.80
2.00
2.20
2.40
2.60
Separation
Factor(calculated)
SF = f (DF, PT)
Y = 0.9995*X
R**2 = 0.9995
Figure 7.20. Calculated vs Experimental separation factor as a function of density of feedand process temperature.
Markers represents Calculated (SF) vs Experimental (SF).Straight line represents fitted straight line through origin.
1.80 2.00 2.20 2.40 2.601.90 2.10 2.30 2.50
Separation Factor (experimental)
1.80
2.00
2.20
2.40
2.60
1.90
2.10
2.30
2.50
SeparationFactor(calculated)
R**2 = 0.9995
SF = f (DF, AFR)
Y = 0.999999*X
Figure 7.21. Calculated vs Experimental separation factor as a function of density of feedand air flow rate.
Markers represents Calculated (SF] vs Experimental (SF).Line represents fitted straight line through origin.
197
199
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Markers represents Calculated (SF) vs Experimental (SF).Line represents fitted straight line through origin.
SF = f (DF, AFR, PT)
Number of data points = 293
Y = 0.999552*X
R**2 = 0.999539
Figure 7.23. Calculated vs Experimental separation factor as a function of density of feed, air flow rate and process temperature.
199
1.80 2.00 2.20 2.40
Separation Factor (experimental)
1.80
2.00
2.20
2.40
SeparationFa
ctor(calculated)
SF(VD) = f (DF, PT)
Y = 0.999961*X
R**2 = 0.999812
Markers represents Calculated vs Experimental Separation factor.Straight line represents straight line fit through origin.
Figure 7.22. Calculated vs Experimental separation factor as a function of density offeed and process temperature for vacuum distillation i.e. no air flow.
1.95 2.00 2.05 2.10 2.15 2.20 2.25
Separation Factor (experimental)
1.90
2.00
2.10
2.20
2.30
SeparationF
actor(calculated)
Fig 7.24
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Fig 7.24
0.95
0.96
0.97
0.98
0.99300
320
340
2
2.2
2.4
0.95
0.96
0.97
0.98
0.99
Figure 7.24. Surface plot of separation factor as a function of ethanol-water mixture feed density and
process temperature for empirical model MOD-5.
Density of Mixture (gm/cc)
Process Temperature ( K)Separation
Facto
r(calculated)
0
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FIG 7.25
0.95
0.96
0.97
0.98
0.99
250
500
750
10002
2.2
2.4
0.95
0.96
0.97
0.98
0.99
Figure 7.25. Surface plot of separation factor as a function of ethanol-water
mixture feed density and air flow rate for empirical model MOD-5.
Density of Mixture (gm/cc)Air Flow Rate (lt/hr)
Separation
Factor
(calculated)
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CONCLUSION AND SUGGESTION
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CONCLUSIONS
Separation of ideal (isotopic) liquid mixture e.g. H2O-HDO-D2O is
possible only due to mass difference (which gives rise to zero point
energy difference) at low temperature, because the Van der Waals fields
are almost equal in these systems. As the zero point energy (vibrational
energy) of lighter molecule is more as compared to the heavier molecule,
therefore lighter molecule needs less energy for detaching it from other
liquid molecules and this proves that normal effect prevails at low
temperature. Now as the temperature is raised the infrared frequency of
lighter molecule is absorbed more strongly by heavier molecule and as a
result it needs lesser energy for evaporation and becomes more volatile,
therefore inverseeffect prevails at higher temperature e.g.,
in case of H2O-HDO-D2O system this inversion temperature is 230C.
Th th ti l t d hi h i l d th i f lit t f 1933
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The theoretical study which includes the review of literature from 1933
to 1998 regarding isotopic exchange reaction H2O + D2O 2HDO
reveals that theoretical calculations of equilibrium constant for thisreaction are easier in vapor phase than in liquid phase due to the
involvement of Van der Waals forces in liquid phase in general cases,
but cancel out in case of isotopes. We also note that:
(i) Kl Kv at lower temperatures and Kl Kv at higher
temperatures.
(ii) In the theoretical derivation of relationship between K and T it is
seen that if harmonic molecular spectroscopic data is used the value ofK is more near to the classical value of 4.0, and if anharmonic data is
used the values are much smaller than 4.0.
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As the theoretical study of separation factor of ethanol water mixture
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As the theoretical study of separation factor of ethanol-water mixture
is very difficult due to non-ideality of this system therefore an
experimental study was conducted. Firstly the evaporation of ethanol-
water mixture was studied. In the first step four mathematical
(analytical) models were developed to calculate volume evaporated
and their validity was checked by fitting experimental data. In case of
analytical Model-IA the maximum error in the calculated volume %
ethanol in product is approximately 10%, in analytical Model-IB the
maximum error in the calculated volume % ethanol in product is 20%
and in case of analytical Model-II the maximum error in calculated
volume evaporated is 17.65%. Although these errors are less than the
acceptable value of 25% [74], still we have tested two more empirical
models.
Two empirical models were also developed and tested for this data in
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Two empirical models were also developed and tested for this data, in
case of empirical Model-I the maximum error in calculated volume
evaporated is 16.67%, while in case of empirical Model-II themaximum error is only 8.3%, which is lowest of all the tested models.
For the experimental separation factor studies of ethanol-water system,
sixteen different setups of apparatus and different process conditions
are used to collect the separation factor data. Approximately 370 data
points are collected. These data points values are firstly used in
Relayigh`s analytical equation for the calculation of separation factor
for each data point [55]. The separation factor values for ethanol-water
mixture for these process conditions are very much near to the
experimental value of 2.5.
Four different empirical relationships are also developed by using the above
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p p p y g
separation factor (experimental) values and two and three different independent
variables. The first model SF = f (DF, T) has goodness of fit value = 0.864, the
second model SF = f (DF, AFR) has goodness of fit value = 0.875, the third
model for vacuum distillation SF (VD) = f (DF, T) has goodness of fit value =
0.883, the fourth model SF = f (DF, AFR, T) has goodness of fit value = 0.880.
Although the value of co-efficient of determination is slightly less in case of
fourth model as compared to third model, but the number of data points are muchmore in case of 4th model, so the best model which has three independent
variables is the fourth model and this model is valid for the following ranges of
three different variables, DF (0.9492 gm/cc to 0.9969gm/cc), AFR (25 lt/hr to
1220 lt/hr), T (294 to 335 K). When calculated vs experimental separation factor
values of this model are plotted we get a straight line with slope nearly equal to 1
with goodness of fit value = 0.99954.
0
SUGGESTIONS FOR FUTURE WORK
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Reasons for the following experimental facts should be investigated.
(i) Why at low temperatures i.e. up to 25C, Kl Kv and at higher
temperatures i.e. 25C, Kl Kv.
(ii) Why K (HM) K (ANH).
(iii) As we also know that Kl Kv, does it indicate
any relevance between (i) and (ii).
(iv) Why heavy water distillation plants are not operated at higher
temperature to have more D2O in the vapor phase as at higher temperaturesinverseeffect prevails. A study should be conducted for the separation of
heavy water at higher pressure and temperatures (more than 230C).
(v) As and ( ) is maximum at 170C, a study should be
conducted at this temperature for the separation of heavy water from light
water.
P
P
H O
D O
2
2
P PH O D O2 2
Experimental results regarding ethanol-water separation can be improved by
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taking the following measures.
(i) Measurement of volume evaporated can be made more exact byreducing the quantity of hold-up volume.
(ii) Process temperature variations should be made minimum.(iii) Humidity control of evaporating air should be improved.
The rate of separation can be improved by increasing the amount of reflux,
which can be achieved by increasing temperature difference between theprocess temperature and cooling water temperature also the inlet air should be
bubbled through the liquid. Air flow rate can be increased manifold by having
concurrent flow, but how this will affect the separation factor remains to beseen.
A power law fit be done for four models by relating Log (SF) to Log (AFR),Log (DF) and Log (T). Although the value of goodness of fit is maximum for
VD, but the air flow is zero in this case. The introduction of air is giving somenegative effect on the separation of mixture due to the formation of a thin layer
of air between vapor and liquid films. So this process can be used for theliquid mixtures separation at low temperature if we have some way of
separating air-water mixture after helping the first evaporation of the liquid
mixture.
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