LIQUID-LIQUID EXTRACTION
FORMAL REPORT
THOMAS SALERNO
GROUP MEMBERS:
GREGORY ROTHSCHING
AN DU
i
ABSTRACT
In order to determine the correlation of the mass transfer rate, overall mass
transfer coefficient, and terminal velocity with diameter during single drop
microextraction of acetic acid from toluene by water, this experiment was
designed to educate engineers in this practice. It is also the intention of this paper
to reveal the accuracy of predictive equations used to design extraction columns.
To achieve these objectives, my fellow teammates and I placed our toluene phase,
of .776 M acetic acid, into a large graduated cylinder. From a syringe suspended
above the column, spherical drops of our aqueous phase, made up of .01M NaOH
and phenolphthalein, were released at diameters of 3.5, 4.0, and 5.1mm.
By recording the time for each drop to traverse .213m of the column, we
determined that its terminal velocity increased with increasing diameter with
values of 9.5, 10.3, 11.9 cm/sec. Furthermore, the time recorded for each drop to
turn clear (i.e. the entire concentration of NaOH neutralized), we were able to
conclude the mass transfer rate increased with increasing diameters with values of
6.0, 7.9, 12.9 e-8 mole/sec. By mixing equal amounts of toluene and water with
acetic acid and titrating each phase, we ascertained that the equilibrium
distribution coefficient of our solution is 10.63±1. Thereby, we were able to
declare the equilibrium concentration of aqueous acetic acid at the surface of our
drop to be 8.25M. This data was inputted into a mass transfer form of Newton’s
Law of Cooling, N HAc  K * A *  C *  CHAc ,Wat  , to calculate the overall mass
transfer coefficient to decrease with our increasing diameters with values of 1.95,
1.92, 1.87 e-7 m/sec.
Because the terminal velocities measured were an average of 22% below those
predicted for rigid spheres, and the drops were observed to fall in a helix pattern,
we determined that the drops experienced oscillations on its track down the
column. By using the model of Handlos and Baron, the overall mass transfer
coefficients could be predicted within errors of only 1.5%. with an experimentally
determined instability factor of 113.3 as suggested by Henschke and Pfenning.
In summary, this report has established that an increase in diameter of a fallen
drop will cause an increase in its terminal velocity, a decrease in its overall mass
transfer coefficient (increase in inside mass transfer coefficient but overriding
decrease in outside mass transfer coefficient), and an increase in its mass transfer
rate. This trend indicates that the toluene (outside) phase is the controlling
resistance to mass transfer, and that the surface area of the drop is the controlling
factor for the mass transfer rate. We also verified that the trends of our state
variables were predicted, but that experimental measurements were required for
prediction of absolute numbers. When designing an extraction column, these
results are most significant for the chemical engineer.
ii
TABLE OF CONTENTS
ABSTRACT ................................................................................................................................................. II
TABLE OF CONTENTS ........................................................................................................................... III
INTRODUCTION: ....................................................................................................................................... 1
THEORY: ..................................................................................................................................................... 5
EXPERIMENTAL: ....................................................................................................................................... 5
Introduction: ........................................................................................................................................ 5
Figure 1: Addition of water phase into column as spherical drops. .............................................................. 5
Figure 2: Extraction of HAc from toluene phase by water phase. ................................................................ 6
Terminal Velocity:................................................................................................................................ 6
Figure 3: Schematic of forces acting on drop. ............................................................................................... 6
Molar Rate of Transfer: ....................................................................................................................... 7
Equilibrium Distribution Coefficient: ................................................................................................. 9
Overall Mass Transfer Coefficient: ..................................................................................................... 9
THEORETICAL:.........................................................................................................................................11
Introduction: .......................................................................................................................................11
Figure 4: Concentration Profile of out two phase liquid-liquid system. ...................................................... 12
Dimensional Analysis .........................................................................................................................13
Table 1: Important variables in our liquid-liquid extraction system. .......................................................... 13
Terminal Velocity:...............................................................................................................................15
Figure 3: Schematic of forces acting on drop. ............................................................................................. 15
Mass Transfer Coefficient Outside of Drop: ......................................................................................17
Mass Transfer Coefficient Inside of Drop..........................................................................................19
Figure 4: Circulation patterns in the drop. .................................................................................................. 20
Figure 5: Drop with tori superimposed on the right. ................................................................................... 21
Mass Transfer Coefficient Inside and Outside of Drop (Fudge Factor)...........................................26
Overall Mass Transfer Coefficient: ....................................................................................................27
Figure 6: Graphical representation of concentration difference driving force. .......................................... 28
EQUIPMENT AND PROCEDURE: .........................................................................................................31
PART 1: CALCULATING VELOCITY AND MASS TRANSFER RATE ...........................................................31
Figure 7: Schematic of graduate cylinder used to hold toluene phase. ....................................................... 31
Figure 8: Schematic of syringe, plunger, sample pot, and syringe pump Model 341. ................................ 32
PART 2: TITRATION .................................................................................................................................34
Figure 9: Schematic of titration equipment. ................................................................................................ 34
RESULTS AND DISCUSSIONS: ..............................................................................................................38
TERMINAL VELOCITY: ............................................................................................................................38
Table 2: Average times recorded for drop to fall .213m in the column. ...................................................... 39
Figure 10: Terminal velocity of the drop (Experimental and Predicted) as a function of drop diameter .. 39
Table 3: Measured and Predicted Terminal Velocities of Drop ................................................................... 40
CONCENTRATION OF ACETIC ACID IN TOLUENE: ..................................................................................41
Table 4: Concentration of HAc in the toluene phase. .................................................................................. 42
DISTRIBUTION COEFFICIENT: .................................................................................................................42
Table 5: Experimentally determined distribution coefficient. ...................................................................... 43
Table 6: Experimental and Predicted distribution coefficients. ................................................................... 44
OVERALL MASS TRANSFER RATE: .........................................................................................................44
Figure 11: Molar Transfer Rate as a function of drop diameter. ................................................................. 46
OVERALL MASS TRANSFER COEFFICIENT: ..............................................................................................46
Figure 12: Overall Mass Transfer Coefficient as a function of drop diameter ........................................... 49
Figure 13: Graph of Kpred versus Kexp to determine CIP ........................................................................... 52
Figure 14: Fudge Factor predicted and experimental overall mass transfer coefficient as a function of
drop diameter.................................................................................................................................................. 53
INDIVIDUAL MASS TRANSFER COEFFICIENTS: .........................................................................................53
iii
Figure15: Individual mass transfer coefficients as a function of drop diameter. ....................................... 54
REVISITING THE OVERALL MASS TRANSFER RATE: .............................................................................55
Table 7: Percentage Growth of the drop’s surface area and overall mass transfer coefficient. .................. 56
CONCLUSIONS: ........................................................................................................................................57
RECOMMENDATIONS: ...........................................................................................................................63
NOMENCLATURE: ...................................................................................................................................64
ABBREVIATIONS:......................................................................................................................................64
DIMENSIONLESS NUMBERS: ....................................................................................................................64
VARIABLES:..............................................................................................................................................64
REFERENCES: ...........................................................................................................................................66
APPENDIX: .................................................................................................................................................67
RAW DATA: ..............................................................................................................................................67
Table 8: Raw Data, Week 1, Large Syringe .................................................................................................. 68
Table 9: Raw Data, Week 1, Medium Syringe .............................................................................................. 69
Table 10: Raw Data, Week 1, Small Syringe ................................................................................................ 70
Table 11: Raw Data, Week 2, Titrations ....................................................................................................... 71
SAMPLE CALCULATIONS: ........................................................................................................................72
Experimental: ......................................................................................................................................72
Theoretical: .........................................................................................................................................78
Figure 13: Graph of Kpred versus Kexp to determine CIP ........................................................................... 83
iv
INTRODUCTION:
A relatively cheap method to separate a two component mixture is through the
process of liquid-liquid extraction. This particular unit operation is of major
importance to the chemical engineer and is the subject of this report.
Every research project needs a starting point. So let’s begin by supposing a
specific liquid reactant, R, is required for your process. In order to obtain your
reactant, there are two possible paths that may be chosen. First, is to purchase this
product in its purest form which could prove to be quite expensive. Second, is to
derive it indirectly from your own reaction in a non-pure state. It follows that
there will be present some impurity, A, that must be removed in order for it to be
used in your process.
Based on the components themselves, there are a number of approaches to
accomplish the removal of impurities from a product. One method is by
distillation. However, this method requires both substances to be thermally stable
over a wide range of temperatures, and have a substantial difference in their
relative volatilities. But, even if these criteria are met, distillation, itself, incurs
substantial operating expenses because of the requisite heat exchangers required
for a condenser and a reboiler. A more attractive option is to separate the impurity
through liquid – liquid extraction (or gas-liquid extraction if more applicable).
This method involves using a solvent, E, which is immiscible with the R phase,
and has a greater affinity for the impurity A. Thus, if an equilibrium stage
extraction unit is set up, we can transfer the impurity A out of our desired reactant
R, and into our solvent E. The main advantage in using this technique is the lack
of an energy expense, because these phases do not have to be boiled nor
condensed. Also, the lack of heating allows us to perform this procedure on many
components even if they are temperature sensitive.
By examining current industry processes, it is quite evident that this method is the
most attractive by its use in a wide variety of applications. For example, liquid –
1
liquid extraction is used in the production of penicillin and for the separation of
aromatic products from other mixed hydrocarbons that come out of the efflux of
the reactor, etc. Another chic characteristic of this process is the generality of the
equations used to design the unit. For example, in this experiment, we will study
the extraction of acetic acid from toluene by water. The correlations and theories
developed from this experiment will be applicable to the above mentioned
extractions and many other extraction processes that may arise in the chemical
engineer’s course of action.
To reduce costs and increase accuracy in the designing of industrial extraction
columns, engineers often necessitate similar experiments to be carried out on a
laboratory-scale, rather than a pilot-plant scale. Therefore, it is increasingly
important to be able to develop theoretical models that transfer mass between
single drops and a stagnant continuous phase and enhance the capability to adapt
this to industrial scale columns. In order to achieve this goal, mathematical
models are needed for the description of mass transfer and drop motion for a
single drop, and the effect of these parameters on drop diameter. This report is
limited to satisfying the modeling of mass transfer from a single drop and
discussing the application of these models in industrial processes. Thus, leaving
the actual math involved in this last part as a recommendation for further analysis.
More specifically, this experiment will study chemical extraction (mass transfer
aided by the push of the reactants to react to form product), as opposed to
mechanical extraction (mass transfer driven solely by the impurity’s greater
affinity for the E phase relative to the R phase), because the water phase will
contain .01M NaOH as well as several drops of phenolphthalein. In order to make
experimental measurements easier and more accurate without requiring expensive
equipment, this basic concentration of the water phase was utilized. In our
particular setup, as the acetic acid diffuses into the water phase, it will react with
the NaOH and neutralize it, thus making the solution less basic and
consequentially less red. This, also, has the effect of keeping the concentration of
acetic acid in the water phase drop at zero for the entire time of experimental
measurement. Once the known amount of NaOH has been completely neutralized,
2
the drop will appear clear, and we will be able to calculate the exact amount of
time required to diffuse in a known amount of acetic acid.
Note, that chemical extraction can differ in its analysis from mechanical
extraction in three major areas. First, the concentration of A in the E phase will
always remain zero because A will react immediately to form product. Second,
the diffusion path of the impurity A will be reduced, however, as this report will
demonstrate, the diffusion path will be unimportant due to the predicted
oscillations occurring which help to keep a uniform mix in the drop. Finally,
chemical extraction will cause the mass transfer to be irreversible, due to the A
disappearing upon diffusing into the E phase. Even though chemical extraction is
modeled exactly like that of mechanical extraction, there is one difference being
the extra effectiveness of the driving force in driving mass transfer due to the
reactants wanting to form product. This extra effectiveness will be modeled in the
testing by a simple experimental constant, which will be absorbed into the
instability constant to be discussed later. This analysis by a constant is allowed
because of the oscillations occurring in the drop, a theory that will be discussed
later. Thus the procedures, equations, and theories from this type of extraction
presented in this report can be equally applicable to mechanical extraction.
Included in this procedure, will be the use of an aqueous solution (AQ) consisting
of .01M NaOH and phenolphthalein (pink solution), introduced as spherical drops
into an organic solution (ORG) of known molarity of HAc (to be determined
later). The main focus of this experiment is to determine the effect our
independent variable (drop diameter, D) will have on our dependent variables: the
drops terminal velocity (v), mass transfer rate (Na), and mass transfer coefficient
(Ki). Furthermore, as a result of this experiment, we will be able to determine the
distribution coefficient (m) for acetic acid between T and W and reveal the
accuracy of the theoretically developed predictive equations.
Throughout the first week, a glass cylinder was filled with an unknown molarity of
the ORG. With the use of a cut syringe, a large (roughly .5cm diameter) drop of the
AQ was inserted at the top of the cylinder. The molarity of the ORG was adjusted
3
such as to make the drop turn clear approximately ¾’s of the way down the column.
Next, we inserted varying diameters of the AQ, namely .5, .45, and .4cm. For each
diameter, 10 drops were used to determine the amount of time needed to travel a
certain distance (finding v); and another 10 drops were used to determine the time
needed to become clear (neutralized). With this data, my lab partners and I
calculated the mass transfer rate to be ( N a  Voldrop  CNaOH / Time ).
In the second week, NaOH was used to make my ORG from the cylinder basic,
thus, ensuring that all of the acetic acid present in the solution had been
neutralized, and back titrated with HCl to find the molarity of the toluene, CA,T.
Then, by mixing 50ml of both T and W with a given amount of HAc (calculated
to achieve the same CA,T as found in the column) and titrating each phase, we
found the distribution coefficient for HAc in T and W at this concentration,
m  C A,W / C A,T .
Upon completion of this phase in the experiment, we were able to determine the
effect diameter had on the overall mass transfer coefficient, the mass transfer rate,
and the terminal velocity of the drop. Also with this resulting data, the
equilibrium distribution coefficient of acetic acid in water and toluene was
calculated.
This data was then compared to developed models of Handlos and Baron, with
adjustments made by Henschke and Pfenning to prove the accuracy in predicting
not only the trends produced but also the absolute numbers.
4
THEORY:
Experimental:
Introduction:
The Falling Drop Experiment studies the microextraction of Acetic Acid (HAc)
from toluene (Tol) by water (Wat). The term microextraction signifies that this
experiment does not involve continuous streams of both phases. Rather, the
toluene (or organic) phase has a certain concentration of acetic acid in it, namely,
CHAc ,Tol , and it remains uniformly mixed and stagnant in a large graduated
cylinder. Wat is then inserted as spherical drops, one at a time into the cylinder as
shown in Figure 1.
Vo lu metric Flo w:
mL per Minute
Vo lu metric Flo w:
mL per Hour
Figure 1: Addition of water phase into column as spherical drops.
The extraction of HAc occurs via a transfer from the bulk (toluene) phase, into a
small sphere of water, as can be seen in Figure 2.
5
NaOH
Toluene Phase
Wate r Phase
Neutralization
Reaction
HAc
Mass Transfer
Product
Figure 2: Extraction of HAc from toluene phase by water phase.
Terminal Velocity:
Once the sphere of water enters the large volume of toluene, it will have three
forces acting on it. These are described as the force of gravity which pulls the
sphere vertically down the column, the buoyancy force which acts to push the
sphere up the column, and a drag force which acts in the direction opposite
velocity. This situation is illustrated in Figure 3.
FDrag
FBouyancy
FDrag
Velocity
FGravity
Figure 3: Schematic of forces acting on drop.
It is important to note from Figure 3 that we have drawn the drag force as pushing the
spherical drop up. This only happens when the velocity of the particle is down, which we
know to be the case. Our math, however, will take this unknown direction of drag force
into account.
6
We, then, apply Newton’s Law, which states that the sum of the forces is equal to
the mass of the particle, multiplied by its acceleration.
(1)
M*
C v v surrounding fluid Aprojected
dv
M
 M * agravity 
surrounding fluid agravity  D
dt
drop
2
In this equation, M is the mass of the drop in kilograms, v is the velocity of the
drop in meters per second, agravity is the acceleration due to gravity in meters per
second squared,  surrounding fluid and  drop are the densities of the bulk surrounding
fluid and of the drop respectively, both in kg per meter cubed, CD is the drag
coefficient for a sphere, and Aprojected is the projected area of the sphere which is
the area of a circle.
However, for all practical purposes, the drop, upon entering the toluene phase will
immediately reach a terminal velocity, vT, at which
dv
 0 , and thus will travel
dt
with a constant velocity the entire way down the column.
In order to experimentally obtain this terminal velocity, we have to measure the
time it takes for the particle to travel a certain distance. Subsequently, Equation 2
is used to obtain vT.
(2)
vT 
distance traveled by particle
.
time
Molar Rate of Transfer:
In order to measure the rate of extraction, we are going to perform an in-situ
titration. 1M NaOH and a sufficient amount of phenolphthalein are used in the
aqueous phase creating a dark red solution. Since the Wat phase has a greater
affinity for HAc than the Tol phase, as the Wat drop passes down the column it
will continuously be extracting HAc into it. However, because of the presence of
the NaOH in the Wat, the HAc will react according to:
7
(3)
CH 3CH 2COOH (aq)  NaOH (aq)  H 2O(l )  CH 3CH 2COO   Na (aq )
Because of the high equilibrium coefficient of this reaction, the HAc will
immediately reach and disappear. Thus, as long as there is still NaOH present in
the water, the concentration of HAc in the water, C HAc ,Wat , will be assumed zero.
As the NaOH is being neutralized by the diffusing HAc, the phenolphthalein will
have a less basic atmosphere to cause a red color, thus, the deep red color will
fade to pink and eventually become clear. With our current equipment, the rate of
this change is impossible to quantitatively measure. However, we can measure the
time it takes for all of the reddish/pinkish color to disappear, and, thus, be able to
measure the time it takes to neutralize all of the NaOH present. We are able to
measure the amount of NaOH which had to be neutralized from Equation 4
because we have set the original concentration of NaOH present in the water drop.
(4)
4 3
Vdrop   rdrop
3
nNaOH ,initial  Vdrop  CNaOH ,initial
By taking an average of the volume of 50 drops from the same syringe, we can
estimate Vdrop . Then, from Equation 1, we determine that one mole of HAc must
have been extracted for every mole of NaOH that has been neutralized, and we
can write,
(5)
nHAc ,extracted  nNaOH ,neutralized .
Next, we can measure the experimental molar transfer rate by dividing the amount
of HAc transferred by the time it took to fully neutralize the drop of water, as is
shown in Equation 6.
8
(6)
n

N HAc   HAc ,extracted 
 time 
Equilibrium Distribution Coefficient:
Finding the experimental distribution coefficient of HAc between Wat and Tol is
easily accomplished after examining what this term represents. The distribution
coefficient reveals how the acetic acid will transfer itself when it is present in a
mixture of toluene and water and is defined as,
(7)
C
m   HAc ,Wat
C
 HAc ,Tol

 .

Thus, we could mix an equal amount of Tol and Wat with a certain amount of
acetic acid. Because these two phases are immiscible, we can easily separate these
phases with a separatory funnel. To find out the concentration of HAc in each
phase, we would titrate each.. Then by using Equation 7, we will find the
distribution coefficient.
However, the distribution coefficient, m, is concentration dependant. Thus, we
need to perform this procedure such that the CHAc ,Tol established, matches that
found in the column. The formula for this is outlined in the procedure section of
this report.
Overall Mass Transfer Coefficient:
The next step in the analysis of this experiment is to fit the data into a simple rate
law. We can develop such a law by analyzing the physical situation occurring.
As described earlier, once the sphere has entered the toluene phase, it will travel
down at an approximately constant velocity of vT. The sphere’s presence will
introduce a concentration gradient into the system because the bulk atmosphere
has a concentration of HAc of CHAc ,Tol , while the inside of the sphere will initially
have a concentration of HAc of CHAc ,Wat  0 . Thus, there exists a driving force to
9
balance out this concentration gradient and cause transfer of HAc across the
interface of the bulk toluene phase and the drop.
Because the drop in the toluene is moving, relative to the drop, so too is the
interface for this transfer, therefore we have convective mass transfer taking
place. We can model this transfer by deriving an expression from Newton’s Law
of Cooling shown in Equation 8.
(8)
Rate of Transfer = Constant* Driving Force
The rate of transfer we are looking for is the molar rate of transfer of acetic acid.
We have previously defined the driving force for this transfer, as being the
concentration difference at the surface of the drop and the concentration at the
center of the drop. Subsequently, we can write Equation 8 in a more useful form
as,
(9)
N HAc  K * A *  C *  CHAc ,Wat 
In this equation, we have defined K as the overall mass transfer coefficient, A is
the surface are of the drop, C* as the concentration of HAc at the surface of the
drop and C HAc ,Wat as the concentration of HAc present in the drop. We can
simplify this expression by ,again, noting that because of the high equilibrium
constant of Equation 3, the concentration of HAc present in the drop will be zero
at all times before total neutralization of the NaOH present in the water. We can
write this as,
(10)
N HAc  K * A *  C * 
All that remains is to find K, the overall mass transfer coefficient, is an expression
to find C*. This can be accomplished by assuming the diffusion of HAc into the
drop occurs via diffusion at temporary stagnant points at the surface of the drop.
At this moment in time, the molecules of water at the surface will be in
10
equilibrium with the toluene phase. Thus, the concentration of HAc in these
molecules of water will be such that they are in equilibrium with the surrounding
molecules of toluene at a known concentration of CHAc ,Tol . Hence, we are able to
use Equation 7 to write,
(11)
K
N HAc
m * CHAc,Tol * A
We can find K through Equation 11 because all of the variables are known.
Theoretical:
Introduction:
From a purely mathematical perspective, it is our intention to derive an expression
for the rate of mass transfer. This expression would allow us to predict transfer
rates from theory without always having to run laboratory experiments. Because
of its purely mathematical basis, this development would be practical for
formulating immediate predictions and permitting us to see which factors have a
strong impact on the mass transfer by allowing us to see trends in the data.
As described by the derived Newton’s law for mass transfer, Equations 8 and 9,
the rate of mass transfer in an extraction process will depend on three things, the
area of contact, the effective driving force, and the transfer coefficient. As
explained previously, we can write this equation for our microextraction process
as Equation 10, reproduced below,
(10)
N HAc  K * A *  C * 
Where N HAc is the molar rate of transfer of HAc, K is the overall heat transfer
coefficient, A is the surface area of the drop, and C* is the concentration of HAc
11
at the surface of the drop. In this equation, we can easily calculate the surface area
of the drop through
(12)
2
A  4* * rdrop
And, as stated before, we can use Equation 11, reproduced below, to calculate C*,
(11)
K
N HAc
.
m * CHAc,Tol * A
Thus, in order to find the molar rate of transfer, we only need to develop a method
to determine K, the overall mass transfer coefficient, from theory. This process,
however, is not so straightforward.
The difficulty becomes clear when you analyze the interface of two phases in
equilibrium. For our situation, we have toluene and water striving to obtain
equilibrium, at the surface of the aqueous drop, in the transfer of acetic acid,
shown in Figure 4.
Figure 4: Concentration Profile of out two phase liquid-liquid system.
It is evident from Figure 4, that each phase will develop its own smooth
concentration gradient within itself; however, these two concentration gradients
will not be continuous at the surface. This is due to non-proportionate affinities
among the two phases for acetic acid. In our situation, the acetic acid has a greater
chemical affinity for the water and therefore in order to obtain chemical
12
equilibrium by equating the chemical potentials of the two phases, the
concentration of acetic acid in the water must be greater than that in the toluene.
Thus, at the surface of the drop, we will experience an extra resistance which
creates the discontinuity in our concentration gradients of acetic acid. In order to
include this resistance into our overall mass transfer coefficient, we will have to
consider the three controlling factors: the area of contact, the effective driving
force, and the transfer coefficient, separately. We will perform analyses to
develop relations for the coefficient of mass transfer inside the drop as well as
outside the drop. Then, relate these coefficients to determine the overall mass
transfer coefficient of our system.
Dimensional Analysis
The purpose of a dimensionless analysis is to predict the most important
dimensionless parameters which are instrumental in modeling your situation. The
situation that we will model is that of mass transfer between a liquid droplet
flowing through another immiscible liquid phase.
The important variables of our system are listed below in Table 1.
Table 1: Important variables for Buckingham method
Variable
Drop Diameter
Symbol
Dimensions
Fluid Density
d m
3
ρ kg/m
Fluid Viscosity
μ kg/(m*t)
Drop Velocity
Fluid Diffusivity
Mass Transfer Coefficient
v m/s
2
DAB L /t
K L/t
Table 1: Important variables in our liquid-liquid extraction system.
These terms are the parameters describing the geometry, velocity, and fluid
properties of our system. Also the most important parameter, the mass transfer
coefficient is also listed.
Using the Buckingham method, we determine that there are six variables which
span over three units, consequently our system will require three dimensionless
13
groups in order to model it. We define our core variables to be DAB ,  , and D ,
and can form our three dimensionless groups as the following
a
 b D c kc
(13a)  1  DAB
d
eD f v
(13b)  2  DAB
g
 h Di 
(13c)  3  DAB
Writing 1 in terms of its dimensions, we arrive at
a
(14)
b
 Length 2   Mass 
c  Length 
1  
Length  
 

3  
 Time 
 time   Length 
In order for 1 to be dimensionless, the exponents in the above equation must be
a=-1, b=0, c=1 and our first dimensionless group must be
(15)
 1  Sc 
kc L
DAB
and is called the Sherwood number or the mass transfer Nusselt number. In a
similar manner, we can solve the other two dimensionless groups and obtain
Dv
DAB
(16)
2 
(17)
 3  Sc 

 DAB
Here, we have just introduced the Schmit number which is a ratio of the
momentum diffusivity to the mass diffusivity, analogous to the Prandtl number
for heat transfer. The dimensionless parameter  2 doesn’t add much in the
modeling of our system, however if we divide  2 by  3 we obtain a familiar
parameter known as the Reynolds number.
14
(18)
 Dv   DAB   Dv
.
Re  


 DAB     
The final result of this dimensionless analysis suggests we will be able to
correlate the mass transfer coefficient in our situation using
(19)
Sh  function(Re, Sc)
This is similar to that which was obtained in heat transfer, according to Thomas
Salerno in his Convection Formal Report dated June 2006.
(20)
Nu  function(Re, Pr)
Terminal Velocity:
Our theoretical treatment will begin in the same way we began our experimental
treatment, by finishing the complete description of our situation. We are aware
that we are extracting HAc from toluene into a spherical drop of water, however,
we need a method to determine the speed that this drop is moving. To determine
this, we again analyze Figure 3.
FDrag
FBouyancy
FDrag
Velocity
FGravity
Figure 3: Schematic of forces acting on drop.
15
It is important to note from Figure 3 that we have drawn the drag force as pushing the
spherical drop up. This only happens when the velocity of the particle is down, which we
know to be the case. Our math, however, will take this unknown direction of drag force
into account.
Then, we can re-apply Newton’s Law, which states that the sum of the forces is
equal to the mass of the particle, multiplied by its acceleration.
(1)
M*
C v v surrounding fluid Aprojected
dv
M
 M * agravity 
surrounding fluid agravity  D
dt
drop
2
In this equation, M is the mass of the drop in kilograms, v is the velocity of the
drop in meters per second, agravity is the acceleration due to gravity in meters per
second squared,  surrounding fluid and  drop are the densities of the bulk surrounding
fluid and of the drop respectively, both in kg per meter cubed, CD is the drag
coefficient for a sphere, and Aprojected is the projected area of the sphere which is
the area of a circle. Because of the dilute concentrations present in both the
toluene and water phases, we can allow their pure density to represent the density
of each phase. Thus, we can simplify Equation 1 to,
(21)
2
dv agravity  Wat  Tol  CD v v Tol * rdrop


dt
Wat
2* M
We will assume that the drop reaches its terminal velocity immediately after
entering the toluene. Therefore, our drop will be traveling at a constant terminal
velocity vT, which is defined as,
(22)
2
dv agravity  Wat  Tol  CD v v Tol * rdrop


0
dt
Wat
2* M
In trying to solve Equation 22, we encounter an obstacle. The coefficient of drag
is dependent on the Reynolds number, which depends on the velocity of drop.
Also, we have different correlations between the coefficient of drag and the
Reynolds number depending upon the value of the Reynolds number, which we
16
cannot calculate because we don’t know the velocity. These correlations were
derived from an experiment and are given from Welty, Wicks, Wilson, and Rorrer
to be:
(23a)
CD 
(23b) CD 
(23c)
24
for Re  2
Re
18.5
for 2  Re  500
Re.6
CD  .44 for Re  500 .
So in order to solve this equation, we have to guess a terminal velocity, choose
the correct correlation for the coefficient of drag, insert this into Equation 22 and
solve for the terminal velocity. Then, we calculate our actual Reynolds number
and ensure we chose the right correlation, if not we have to select the correct
correlation and solve again.
Mass Transfer Coefficient Outside of Drop:
We will now consider the transfer of HAc from the continuous phase of toluene to
the surface of the spherical drop of water. As was done by Higbie, we can derive
an expression for this outside mass transfer coefficient analytically.
When the water drop descends down the column, through the toluene, a turbulent
boundary layer will form around the sphere. We can analyze this boundary layer
by using the typical boundary layer equations. Note the derivation of these
equations is outside the scope of this report. For an abbreviated derivation, the
reader should consult the Convection Experiment Formal Report written by
Thomas Salerno.
First, we perform a mass balance on the system and obtain continuity equation
(24)
vx v y

0
x y
17
Then, we can perform a momentum balance on the system and obtain the equation
of motion,
(25)
vx
vy   2vx
vx
 vy

x
y  y 2
Along with this momentum boundary layer, there will form a thermal boundary
layer, which can be described by performing an energy balance on the system,
(26)
T
T
 2T
vx
 vy
 2
x
y
y
Finally, we need an analogous differential equation for the mass transfer boundary
layer which can be obtained by comparing the similarities in Equations 16 and 17
for momentum and thermal boundary layers. Each equation has the same form on
the left hand side with the components of the velocities being multiplied by the
state variables gradient in that direction. The right hand sides are also very similar
with the derivative of the state variables gradient being multiplied by the
particular equations’ diffusivity. Thus, we can write our mass transfer boundary
layer equation as,
(27)
vx
C A
C A
 2C A
 vy
 DAB
x
y
y 2
Higbie, then, transformed Equations 24, 25, 26, and 27 into spherical coordinates.
Assumptions were made that mass transfer occurs when differential volume
elements of the continuous toluene phase comes into contact with the forward
stagnation point in the drop. It diffuses the acetic acid into the drop as it travels
around the drop. He formulated a conjecture following the work of Pigford, that
the radial velocity at any point during the transfer is negligible, and that the
tangential velocity is always equal to the terminal velocity of the falling drop.
Finally, he used West’s postulation that the time of contact between the toluene
phase and the spherical drop is simply calculated from,
18
(28)
time 
d
vT
Where d is the diameter of the drop and vT is the drop’s terminal velocity.
Note the mathematics in the current analysis to obtain Equation 29, although
extensive, are rather straightforward. The derivation follows the normal
calculation procedure for any boundary layer analysis. It will not be shown here
because it is not the purpose of this report to solve straightforward equations, but,
instead to show the techniques, theories, and assumptions used to arrive at these
relations. If the reader would like to see the typical steps needed in boundary
layer analysis, he or she should again consult the Convection Formal Report,
written by Thomas Salerno
After making these assumptions, the boundary layer equations can be solved to
produce a correlation for the outside mass transfer coefficient in terms of the
dimensionless variables we have previously defined, shown in Equation 29.
Shoutside 
koutside  2  rdrop
(29)
koutside  1.13
DHAc
1/ 2
 1.13Peoutside
 vT  2  rdrop
 1.13 
 D
HAc

1/ 2



1/ 2
vT1/ 2 DHAc
2  r 
1/ 2
drop
Mass Transfer Coefficient Inside of Drop
The simplest approach to model the mass transfer inside the drop is to assume the
drop is stagnant then apply and solve the ordinary differential equations presented
above. However, this method will generally lead to mass transfer coefficients that
are usually one or two order of magnitudes too low. This should be expected
because there is motion between the two phases creating a convective mass
transfer along with a diffusion mass transfer.
19
Because we currently don’t have an exact method to analyze this turbulence flow,
it is impossible to solve this situation accurately. Therefore, developing a
correlation for mass transfer inside the drop is much more difficult and will be
discussed in detail in this section.
According to Handlos and Baron, the circulation patterns in the drop can be
represented as shown in Figure 4.
Figure 4: Circulation patterns in the drop.
In this schematic, Handlos and Baron combined the tangential motion of the drop
with the (assumed) random radial motions due to vibrations in the evaluation of
eddy diffusivities within the drop.
The assumption of this circulation pattern is a valid one because experiments have
been done comparing the velocity of falling solid spheres and falling liquid drops.
That of spheres followed closely to its terminal velocity prediction (which
assumes a rigid surface with only vertical velocity), however, that of the liquid
drops shows large over predictions of the velocity. Also, observations of the drops
in this experiment show some spinning of the drops which can be caused from
this internal circulation in the drops. Finally, this circulation pattern can be given
a more physical interpretation. Garner, Skelland, and Hale have shown that the
transfer of solute from the continuous phase into the drop can cause such
circulation patterns.
20
Handlos and Baron then replaced, for mathematical purposes, the streamlines in
the drop by a system of tori shown in Figure 5.
Figure 5: Drop with tori superimposed on the right.
In this drop, particles are constantly being moved along streamlines due to the
random radial motion. If we assume the circulation patterns are giving complete
mixing inside the drop, then the probability that a particle is found between
positions p  and p  dp is the ratio of the differential volume element at p  to
the total volume of the torus. Thus, we can form an equation for the position of
any molecule as
(30)
P  p  dp 
32 p
dp
d2
We can put this equation into a more useful form by using the substitution
(31)
r
4p
d
where d is the drop diameter. Combining Equations 28 and 29 gives
(32)
P  r  dr   2r dr  .
21
We can develop a relation for the square of the displacement in our circular torus
as,
(33)
z2 
d2
2
 r  r  .
16
If we run this experiment multiple times in our drop, the mean square
displacement will simply be the expectation value of z 2 , which can be found via
1
(34)
z 2   z 2  r  P  r   dr  
0
1
d2
d2
2



r
r

r
dr

6r 2  8r  3 .




8 0
96
The average time required for this circulation was investigated and given by
Kronig and Brink to be,
(35)

16d  i 
1   ,
3V  o 
where V is the volume of the drop, i is the viscosity of the phase inside the drop,
and o is the viscosity of the phase outside the drop.
Proceeding to analyze the transfer process by eddy diffusion, we can represent our
mass transfer through use of the Einstein equation, which combines the mean
square displacement for a given time to the effective diffusivity. The result of this
analysis of our situation yields,
(36)
2
z2
dV  6r  8r  3
E (r ) 

.
4 2048 
i 
1  
 o 
22
We can, then, multiply and divide by the molecular diffusivity, and obtain, after
some manipulation,
(37)
E (r ) 
D  Pe
6 r 2  8r  3 

2048
where we have defined a new term, Pe , which is simply
(38)
Pe 
Pe
 i 
1  
 o 
which uses a dimensionless parameter known as the Peclet number which is given
by the multiplication of the Reynolds number and the Schmit number.
Next, we apply the equation of continuity to our drop. For our drop this equation
yields, in spherical terms,
(39)
c 16 1  
c 
 2
E r  .
t d r r 
r 
where c is the concentration of solute within the drop. We can rewrite this
equation into a more useful form by making the substitution r  1  y . We arrive
at, after plugging in Equation 37,
(40)
2048d 2 c  1 
c 

1  5 y  10 y 2  6 y 3   .

16 DPe t 1  y y
y 
The boundary conditions for this equation are
(41a)
c  co at t  0, 0  y  1
(41b) c  0 at y  0, t
0.
23
We can then solve Equation 40 and 41 by using the separations of variables
technique. In this technique we assume that the concentration function we seek is
made up of a function dependent only on time multiplied by a function dependent
only on the position term. In mathematical terms we define,
(42)
c  T (t )Y ( y )
We can then plug Equation 40 into Equation 38 and split this equation into a set
of two ordinary differential equations:
(43)
dT
16 DPe

T
dt
2048d 2
(44)
d
dY
1  5 y  10 y 2  6 y 3 
  1  y  Y  0

dy
dy
Where  represents one of the many Eigen values. Accordingly the general
solution of Equation 38 is
(45)

 16n DtPe 
c  c0  AnYn exp  
2 
 2048d 
1
Where co is the initial concentration of the solute in the drop, An are the constants
which need to be found from the boundary conditions, and Yn are the eigen
functions corresponding to  n . However, as is the case for most eigen-value
problems, each successive Eigen value is larger than the first. In our present case,
the second Eigen value is of sufficient magnitude that only the first term in
Equation 45 is needed to model our system. Thus, we can write the ratio of the
mass of solute in the drop at time t, to the amount present at time zero is,
(46)

M (t )
 16n DtPe 
 2 An 2 exp  
.
2 
M (0)
 2048d 
1
24
By performing a material balance on our system, we can go back and obtain a
relation for our mass transfer coefficient via another means. This yields,
(47)
d dc
 ki  c *  c 
6 dt
However, we can simplify this equation by noting the conditions we labeled in
Equation 41, in which we arrive at,
(48)
d dc
 ki c *
6 dt
From which, we can solve via the same method just described to yield a relation
of the form,
(49)
M (t )
 6k t 
 exp   i  .
M (0)
 d 
Thus, we can equate Equations 46 and 49 to obtain the mass transfer coefficient
expressed as a function of the first eigen value,
(50)
ki 
161 DPe
6 2048 d 2
As long as we can find the first eigen value of Equation 44, we have a relation for
the inside mass transfer coefficient. This can be solved using the Ritz method
which finds an approximate eigen function via,
j
(51)
Yn   c j y j
1
With j=5 we are able to determine that the lowest eigen value is 1  2.88 . Thus,
we can finalize Equation 50 into,
25
(52)
ki 
.00375vT
 i 
1  
 o 
We can rearrange this equation into the more general and useful form of,
(53)
Shi 
.00375Pe
 i 
1  
 o 
which is the relation, developed by Handlos and Baron, that we will use in our
experiment.
Mass Transfer Coefficient Inside and Outside of Drop (Fudge Factor)
Researchers Henschke and Pfenning studied the extraction from a falling drop
into a continuous phase. They propose a similar argument for the theory of a
falling drop which reinforces many of the ideas presented previously, introduced
by Handlos and Baron. In fact, in the article, they agreed with the derivations and
calculations made by Handlos and Baron. However, they did want to extend their
theory a bit further. Because the model of Handlos and Baron was developed to
be general for all liquid - liquid extraction systems, Henschke and Pfenning
noticed that using this relation for liquid-liquid extraction brought with it a
significant assumption, that the interface of the surface is ideal. They suggest that
the mass-transfer induced turbulence will distort the internal eddies in a manner
which is strongly dependent on the drop’s surface tension at its exterior. To
account for this in Henschke and Pfenning’s own system, they only needed to add
an extra term to Equation 53,
(54)
Shi 
.00375Pe
  
CIP 1  i 
 o 
26
The constant, CIP , introduced in this equation is termed the instability constant
and it characterizes the instabilities at the interface which is specific to the system.
As it is difficult to foretell the surface tension in any particular system, they have
not determined a method of predicting this constant. The exact compositions of
each phase, the temperature, the pressure, the Reynolds number, etc. would have
to be taken into account. Instead, they strongly recommended solving for it
experimentally. Because they have determined the instability constant to be a
substance-specific constant, we can calculate it after one measurement. Its value
should stay constant throughout the experiment, or any similar extractions with
the same materials at same temperature and pressure.
Of course, our system is different than their system. In their development of
Equation 54, they assumed the coefficient for mass transfer outside the drop is
negligible, i.e., the inside coefficient is the controlling resistance. However, as our
data will prove later, our system is such that the outside mass transfer is the
controlling resistance. Similarly, Henschke and Pfenning did discuss this situation
in their paper. They suggest, “if turbulence occurs simultaneously inside and
outside the drop, the mass transfer resistance outside will be reduced by roughly
the same factor as inside the drop. Thus we have to transform Equation 29 to
Shoutside 
(55)
koutside 
koutside  2  rdrop
DHAc
1/ 2
1.13Peoutside
1.13  vT  2  rdrop



CIP
CIP  DHAc
1/ 2



1/ 2
1.13 vT1/ 2 DHAc
CIP  2  r 1/ 2
drop
In their report, they suggested values from 100-2500 are reasonable for CIP, for
turbulent systems.
Overall Mass Transfer Coefficient:
Now, we are prepared to determine the overall mass transfer coefficient governing
our situation. This is accomplished by applying the two resistance theory. In this
27
theory, we need to determine separate relationships for mass transfer inside and
outside the drop, which depend on their own driving force as if the other phase
wasn’t present. These equations take the form,
(56)
N A  kin  y A,Wat  y Ai 
(57)
N A  kout  xAi  xA,Tol 
Evidently, these two equations are equal due to the principle of conservation of
mass, which states the moles leaving the outside of the drop has to equal the
moles entering the inside of the drop, hence, we can write,
(58)
N A  kin  y A,Wat  y Ai   kout  xAi  xA,Tol  .
Note, that we have already developed relations to determine the mass transfer
coefficient inside and outside the drop. The next step is to graphically relate the
driving forces for both the inside and outside of the drop. Recall the driving force
on the inside, is the concentration of solute on the inside less the concentration at
the interface. Also, you must remember the driving force on the outside is the
concentration at the interface less the bulk concentration in the continuous phase.
The graphical relation is shown in Figure 6.
y Ai  y *A
 m'
x Ai  x AL
y AG  y Ai
 m"
x*A  x Ai
Figure 6: Graphical representation of concentration difference driving force.
28
As exemplified in Figure 6, the driving force outside the drop can be modeled to a
straight line with slope m . Similarly, we can fit the driving force inside the drop
to a straight line with slope m . Consequently, we have defined the following
two relations from our equilibrium graph.
(59)
m 
(60)
m 
y Ai  y *A
x Ai  x A,Tol
y A,Wat  y Ai
x*A  xAi
Another important concept of note from Figure 6 and Equations 59 and 60, is that
we are now able to link the driving force inside the drop to that outside the drop.
This link was provided through the concentration of solute in each phase at the
interface which is related by equilibrium.
To create an equation for the overall mass transfer coefficient, we will utilize
Figure 6, the driving force for the overall equation should be the combination of
the driving forces for both inside and outside the drop. With this method, the
concentration at the interface drops out and we are left with  y A,Wat  y *A  . There
may be some debate on what our driving force should be that of the inside or that
of the outside. However, the choice of driving force doesn’t matter since we are
algebraically defining the overall equation and not theoretically deriving it. One
should realize that because of this flexibility, there will be two possibilities of the
overall coefficient both of which will be different. However, both are acceptable,
as long as they clearly stated which driving force coefficient they are using. For
our experiments, we will base the overall coefficient off of the driving force on
the inside of the drop, which we write as
(61)
N A  Kin  y A,Wat  y*A 
Now, we can combine, substitute, and algebraically manipulate Equations 56
through 61, as follows,
29
N A  K in  y A,Wat  y *A   kin  y A,Wat  y Ai   kout   x Ai  x A,Tol 
(62)
y
A,Wat
 y *A    y A,Wat  y *A   add and subtract : y Ai
y
A,Wat
 y *A    y A,Wat  y Ai    y Ai  y *A   substitute : m 
y
A,Wat
 y *A    y A,Wat  y Ai   m  x Ai  x A,Tol  
y Ai  y*A
x Ai  x A,Tol
substitute : N A  K in  y A,Wat  y *A   kin  y A,Wat  y Ai   kout   x Ai  x A,Tol 
N A N A N A m


K  kin
kout
Thus we have derived a relation for the overall mass transfer coefficient to be,
(63)
1
1
m


.
K  kin kout
30
EQUIPMENT AND PROCEDURE:
Part 1: Calculating Velocity and Mass Transfer Rate
For the first part of the experiment, I inserted a drop of .01M NaOH in water, into
a continuous stream of stagnant toluene which is at a previously unknown amount
of acetic acid.
The toluene phase was placed in a large 1-gallon graduated cylinder with
markings on the side. This would allow the drop to fall vertically downward
without any interruptions for at least .6 meters. In order to calculate the velocity
of the drop, the experimenter utilizes the equally spaced markings on the side
(.001775m per marking) of the graduated cylinder. At the bottom of the column,
there is an outlet valve from which the toluene phase can be extracted to take a
sample, and to facilitate the empting of the column when the experiment is done.
These features can be seen in Figure 7.
Vo lu metric Flo w:
mL per Minute
Vo lu metric Flo w:
mL per Hour
Figure 7: Schematic of graduate cylinder used to hold toluene phase.
In order to insert the drops at the same height and location in the column, a cut
syringe which hung from a circular clamp just above the graduated cylinder was
used. A cut syringe was preferred for this experiment as it was constructed with a
31
plastic tip. By simply cutting the plastic tip on the syringe to a required size, the
diameter of the drop leaving the syringe was at a constant diameter throughout
each portion of the experiment. Additionally, the syringe was fixed to a plunger
which laid on an syringe pump. To assist in the regulation of the amount of liquid
being pushed out of the plunger per unit of time from the syringe the syringe
pump Model 341 from Sage Instruments, a division of Orion Research Inc., was
required. This automated pump regulated the rate of compression on the plunger
to ensure the same pressure would be used to expel a drop from the syringe which
would in turn ensure constant velocity and diameter of the drop leaving the
syringe. The controller on this unit allowed for a number of settings ranging from
a fast to slow output. Thus, allowing all three members of my group to accurately
record the time for the drops to fall a certain distance. Finally, the plunger was
connected to a sample pot filled with the .01M NaOH solution, which was closed
to the environment, to guarantee proper loading of this sample without any
evaporation or air pockets which could distort either the sample itself or the drop
size. The features for this part of the apparatus are shown in Figure 8.
Vo lu metric Flo w:
mL per Minute
Vo lu metric Flo w:
mL per Hour
Syringe Pump
Model 341
Sage Instruments
Figure 8: Schematic of syringe, plunger, sample pot, and syringe pump Model 341.
Having described the equipment to be used during the procedure, I will begin to
explain the course of action for this part of the lab. The first variable that must be
32
measured is the volume of the drop at the current syringe size. To correctly
manage the volume of the drop, a sample was loaded into the plunger which was
connected to the automatic dispenser, and set on .54mL/min. A high setting is
practical at this point, as we had time restraints and a certain number of drops
needed to be calculated. We then proceeded to place 50 or 60 drops into a 10mL
graduated cylinder and recorded the volume of these drops. Then, we divided by
the number of drops which allowed us to solve for our volume per drop.
The next variable of interest is the drop’s terminal velocity. To make this
calculation easier, we made certain that the toluene phase started exactly at the
zero line on the graduated cylinder. We used a ruler to calculate the distance it
takes for 600 marks to be passed, namely, .213 meters. Once the drop enters the
toluene, at this zero line, we will start the first stopwatch. As it passed the 600
line, the first stopwatch was stopped and the second stopwatch started.
Sequentially, when the drop reached the 1200 line, the second stopwatch was
stopped and the third stopwatch started. Till finally, when the drop reaches the
1800 line the third stopwatch is stopped. All of these time recordings measure the
time it takes for this drop to pass .213 meters, but in different locations of the
tube. To find the terminal velocity of the drop, these values can be averaged. The
experimenter should also note the average time in each section of the cylinder in
order to reveal if the particle is at a terminal velocity or if its velocity is changing
throughout the cylinder. Because of the necessity of accuracy, it is recommended
to take 10 trials of this procedure. For both the determination of the drop’s
terminal velocity and the determination of the mass transfer rate, it is
recommended to set the syringe pump at .12mL/hr, thus allowing time to take
measurements and record them before the next drop falls.
The next part of the experiment requires the experimenter to ascertain the time it
takes this particle to turn clear. This method is similar to that used to find the
drop’s terminal velocity. Simultaneously, all three experimenters will start their
stopwatches when the drop enters the toluene phase. Of special note, is that the
particle must be falling when it reaches the toluene phase, if it is still hanging
from the syringe but not falling, as it sits in the toluene phase it will be
33
transferring mass of the acetic acid before you can begin your timed experiment.
One must ensure the syringe is far enough above the toluene phase and that the
drop is completely in free fall before it enters the stagnant continuous phase. The
drop will be monitored by the three evaluators as it travels downward in the
column. Each of the three participants will then bring to a halt their stopwatches
at the moment they deem the drop has turned completely clear. Again, the
experimenter should note that the pink will disappear from the bottom of the drop
first but still remain in the top of the drop, do not stop the stopwatches until the
drop has completely turned clear. It is recommended to execute 10 trials of this
part of the procedure to gain proficiency at documenting a clear drop versus a
slightly pink one. After each trial record the time, as this will be used to calculate
the mass transfer rate.
Part 2: Titration
For the second part of this lab, standard titration equipment was needed. This
included a 5 ml pipette , a 1 ml pipette, two 100mL titrating graduated cylinders
with control valves at the bottom secured to a rotating stand, a 250 ml beaker, a
250 ml separatory funnel, and a magnetic stirrer.
These materials will be arranged as seen in Figure 9
Figure 9: Schematic of titration equipment.
34
The concentration of acetic acid present in our toluene phase is the first variable
we want to calculate in this stage of the lab.. In order to do this, place a magnetic
stirrer into a 250 ml beaker. Then, place 5ml of your toluene phase solution from
your column into this beaker using the 5ml pipette. Next, place 3 drops of
phenolphthalein solution into the beaker. Place the beaker on the electronic stirrer
and mix the solution. After that, add pure NaOH from one of your titrating
cylinders slowly into your solution until you achieve a light pink color. Record
the amount of NaOH needed for this. Rotate your HCl titrating cylinder to be over
your solution. Add HCl very slowly into your mixture to reach a fading pink
endpoint, the point where your solution is only slightly pink and about to turn
clear. Record the amount of HCl needed. From this data you should be able to
calculate the amount of acetic acid present in your toluene phase from the
following calculation,
mmole HAc   mL NaOH  mL HCl  
(64)
CHAc , Tol 
.1 mmole
mL
mmole HAc
5 mL sample
Next, the experimenter needs to calculate the distribution coefficient of acetic acid
in water and toluene. To accomplish this, we need to mix a given amount of
water, toluene, and acetic acid, and measure the concentration of acetic acid
present in each of the water and toluene phase after mixing. However, we should
note that this value is concentration dependent. Thus, we must calculate the
amount relative proportions of each component that we need to add such that the
concentration of acetic acid present in the toluene phase is equal to that we have
found in our column.
To do this calculation, we first assume a distribution coefficient, usually a value
of 10 is a good estimate. Then, we approximate the amount of acetic acid we
would need to add in order to obtain the same concentration of acetic acid in our
toluene phase that we recorded from our column. This calculation procedure is as
follows,
35
mmole
mL
mmole
mmole HAc in Wat  50mL Wat *10*.8
mL
mmole HAc  mmole HAc in Tol  mmole HAc in Wat
mmole HAc in Tol  50 mL Tol * .8
(65)
mL HAc  mmole HAc *
mole
60.05 g 1mL
*
*
1000mmole mole 1.05 g
After this is completed, mix equal amounts of pure water and pure toluene (in the
amounts used for the calculation in Equation 65 and add the appropriate amount
of acetic acid, the amount calculated from Equation 65. Mix this solution for
approximately 30 minutes to ensure there is good intermingling between these
two immiscible phases which will guarantee the acetic acid will separate
according to equilibrium. Next, place the mixture into a separatory funnel and let
it settle for about a minute. Take a 5ml pipette and extract the toluene (top) phase
from the funnel itself. Continue to titrate with this via the same method described
previously. The concentration of acetic acid in the toluene phase can be calculated
through the use of Equation 64. Remove a small amount from the bottom of the
separatory funnel and place into a small jar. Use a 1ml marked pipette to transfer
.5ml of this into a beaker. Add water from a squirt bottle around the sides to make
sure the entire .5ml is in solution. Then, add three drops of phenolphthalein, and
titrate with HCl until the solution reaches its faded pink endpoint. Using similar
equations to Equation 64, the experimenter should be able to calculate the amount
of acetic acid present in the water phase via,
mmole HAc   mL NaOH  
(66)
CHAc , Wat 
.1 mmole
mL
mmole HAc
.5 mL sample
With this data, the experimenter can use Equation 7, and find the distribution
coefficient as,
36
(7)
C
m   HAc ,Wat
C
 HAc ,Tol

 .

This will complete the calculations and experimental procedure for this lab. The
experimenter would now follow the theory section of this report to calculate all of
the parameters needed to satisfy the objectives of this experiment.
37
RESULTS AND DISCUSSIONS:
Terminal Velocity:
In analyzing our data, our first step was to determine the terminal velocity of our
falling drop. As described previously in our theory section, we have three forces
acting on our particle: Gravity, which pulls the molecule down; Buoyancy which
pushes the molecule up; and Drag, which acts to retard the molecules velocity, in
our case the Drag will always be pushing up since our drop is always traveling
downward. These forces will act on the particle via Newton’s Second law which
allows us to model the drops velocity as,
(21)
2
dv agravity  Wat  Tol  CD v v Wat * rdrop


dt
Wat
2* M
To determine this value theoretically, we simply have to solve Equation 21 for v,
while setting
dv
 0 , which is the definition of terminal velocity. This maybe a
dt
bit tedious because the correct correlation for CD depends on the value of the
terminal velocity as given by Equations 23a,b, and c. However, a simple guess
and check procedure was used and worked satisfactory in obtaining the required
value.
To determine the experimental value, we has to assume that the particle reaches
terminal velocity almost immediately after entering the toluene phase. Thus, we
released a drop in the toluene phase, recorded the time it needed to vertically
travel down the column’s first .213 meter section, then recorded the time for the
drop to travel down the column’s second .213 meters, and finally the time needed
for it travel the column’s third .213 meters. The averages for these times are
shown in Table 2.
38
Table 2: Average Times (sec) for Drop to Fall .213m Increments
Division of Column
Size of Drop:
First .213m
Second .213m
Third .213m
Small
2.19
2.33
2.24
Medium
1.98
2.20
2.05
Large
1.68
1.88
1.82
Table 2: Average times recorded for drop to fall .213m in the column.
From this table, it is evident that the first .213m generally traveled in the shortest
time. However, its values are in close approximation to the other average times.
This validates our assumption about the molecule reaching its terminal velocity
immediately and is acceptable to give us the same results we would have obtained
if we didn’t make this assumption.
Subsequently, we can now calculate the experimental terminal velocity by
dividing the average time it took each drop to travel the specified .213 meters.
The results of this calculation as well as the results of theoretical calculations are
shown together in Figure 10.
Terminal Velocity of Drop as a Function of Drop Diameter
0.18
y = 1.8303x - 0.0327
R2 = 0.9754
0.16
Terminal Velocity (m/s)
0.14
0.12
0.1
Experimental Velocity
Predicted Velocity
Linear (Predicted Velocity)
0.08
0.06
0.04
0.02
0
0.07
0.075
0.08
0.085
0.09
0.095
0.1
0.105
0.11
Square Root of Drop Diameter (m^.5)
Figure 10: Terminal velocity of the drop (Experimental and Predicted) as a function of drop
diameter
39
It should be noted from this graph that as the diameter of the particle rises, the
terminal velocity of the particle rises. The reader should also observe that at all
diameters our theoretical predictions are consistently overestimating the velocity
by 21%. The data for this is shown in Table 3.
Table 3: Measured and Predicted Terminal Velocities of Drop
Diameter (m)
Measured (m/s)
Error
Predicted (m/s)
Error
Experimental Error
0.00346
0.095
5.0%
0.117
3.1%
-19.5%
0.00398
0.103
6.4%
0.134
3.2%
-23.1%
0.00516
0.119
6.1%
0.152
3.5%
-22.0%
Table 3: Measured and Predicted Terminal Velocities of Drop
The result of the terminal velocity increasing as the diameter increases is logical
from a simple thought analysis prospective. As was previously mentioned, the
drop has three forces acting on it, a force from gravity always pushing the
molecule down, a buoyancy force from displacing the toluene which always
pushes the molecule up the column, and a drag force which always acts opposite
the molecule’s velocity. We know that the force of gravity is driven by the weight
of the particle, which is mostly water and that the force of buoyancy is driven by
the weight of the continuous phase which it has displaced, which is mostly
toluene. Since the drop is the same size, we can compare these forces by
comparing the densities of water to toluene. In view of the fact that the water is
denser than toluene the force of gravity will be greater and the particle will fall.
Therefore, if we increase the diameter of the particle, we also increase its volume
and since the densities remain constant, the force due to gravity has to grow
greater in proportion to the force due to buoyancy. Clearly, this implies that as the
diameter of the particle grows, the greater the force pulling the particle down is,
thus. the velocity of the particle must increase, which is exactly the results we see
here. (Note we have neglected the drag force in this analysis, but this force is
dependent mainly on the velocity of the particle and less on its size for the range
of sizes with which we have experimented. For that reason, its effect on causing
the magnitude of the velocity change is substantial, but its effect on causing the
velocity to either increase or decrease is negligible for our purposes.)
40
To explain the reason for the substantial over prediction of the terminal velocity
of our experiments, we need to consider the assumptions being made. In order to
develop our theoretical equation for describing the velocity of the falling particle,
we had to assume that the particle remains a perfect rigid sphere. However, as the
analysis of Handlos and Baron have proven, the molecule is continuously
oscillating the material on the inside and also changing shape and oscillating the
material at the surface, due to the mass transfer and the lack of rigidity of liquids.
Thus, in our theoretical model, Equation 21, our assumptions predict a straight
vertical drop of our particle. Where, in reality, we have observed our particle to
fall in more of a helix pattern. This helix flow path is caused by the internal
circulations of the particle and has the effect of increasing the drag coefficient of
the particle as it falls. This result was proven by Bond and Newton. Hence, the
increased drag coefficient on our particle will cause the terminal velocity of our
particle to be reduced as compared to the value we have predicted.
As a result, the experimentally measured value of the terminal velocity is a more
accurate answer and will be used as the velocity in all of our calculations in this
experiment. The argument has been brought up that the experimentally measured
value is also off because the helix path of the particle is causing the actual
distance the particle is traveling to be more than the measure .213m of the
column. However, we take the classic physics assumptions that the velocity in the
vertical direction is independent of the velocity in the horizontal direction, then as
long as the forces on the particle are always acting vertically; our experimental
measurement will still be accurate.
Concentration of Acetic Acid in Toluene:
The next step of my analysis was to calculate the amount of acetic acid that was
present in my toluene phase in the column. This was accomplished through
titration.
I added 5 ml of my toluene phase in a beaker and than added an excess of NaOH
to ensure that all of the acetic acid had reacted. This was an important step in the
process as NaOH does not mix well with the toluene phase. Then, I back titrated
41
with HCl and calculated the amount of acetic acid from subtracting the amount of
NaOH needed, from the amount of HCl needed to reach the phenolphthalein
endpoint.
.776 mol/L of acetic acid in my toluene was my end result. This molarity was the
outcome of my adjusting the unknown molarity of the original toluene phase
present in the column, such that the large drop of NaOH will turn clear roughly
¾’s of the way down the column. Thus, there is no predicted value this
concentration should match. However, I did compare this with other values
recorded from other students who performed the same experiment and it was in
the same ballpark as their answers, in consequence, I have confidence this value is
accurate.
Due to the time restraint in running this lab, I was only able to titrate this mixture
twice, therefore my error is simply the standard deviation of two points. This
value is probably not as accurate as I would like but is all I can work with, only
given two – three hour labs. Thus, the error is .037 mol / L .
This data is shown in Table 4.
Table 4: Molarity of Toluene Phase
Current Expriment:
Concentration of HAc 1
Concentration of HAc 2
Average Conc HAc
Std Dev Conc HAc
Concentration of Hac:
0.75
0.802
0.776
0.037
mol/L
mol/L
mol/L
mol/L
Previous Experiments:
Concentration of HAc 1
Concentration of HAc 2
0.803 mol/L
0.813 mol/L
Table 4: Concentration of HAc in the toluene phase.
Distribution Coefficient:
The next step of the calculations was to find out the distribution coefficient for
our system. In other words, I need to determine how the acetic acid will distribute
itself when it is in equilibrium with water and toluene.
42
In order to determine this, I mixed 50 ml of both water and toluene with 27 ml of
acetic acid and stirred the mixture until it was uniform (about 30 minutes) and
allowed it to settle. I proceeded to titrate each phase of the two phase mixture to
find the concentration of acetic acid in each phase. From this, I was able to
calculate the distribution coefficient based of the concentration of acetic acid in
each phase following Equation 7.
(7)
C
m   HAc ,Wat
C
 HAc ,Tol



The results of this calculation are shown in Table 5.
Table 5: Experimental Distribution Coefficient
Phase:
Toluene
Water
Distribution Coeff - m
Assumed Error - m
Concentration of Hac:
0.803 mol/L
8.54 mol/L
10.635
1.01
Table 5: Experimentally determined distribution coefficient.
Because of the time constraint placed on my lab, I only had sufficient time to run
this portion of the experiment once. As a consequence, I was not able to calculate
my values through an average, nor was I able to find my experimental error
through standard deviation. However, examining this process in detail, I am able
to perceive that the only probable place of error could occur during my titration
calculations. Thus, it is a good assumption that my titrating ability is fairly
consistent and the error associated with each titration is equal in percent error
obtained during my previous titration calculation (4.8%). The error associated
with the distribution coefficient should be the sum of each of the errors of the two
titrations, and it is this value that appears in Table 5.
In an attempt to compare this with theory, I again encountered a problem. There is
currently no accurate way to predict the distribution coefficient of my system at
its exact concentration. However, once again, I can use previous experimental
43
results of this calculation for my theoretical prediction. Although this time, the
experiments were controlled accurately and recorded in literature as an average of
multiple runs, thus, my confidence in this predicted value will be higher than it
was for the predicted value of the molarity of the toluene phase.
According to Fuse and Iguchi, for my system of 8.0 mole percent acetic acid in
the toluene phase, the distribution ratio should be 2.32  .132 mole percent in
water phase / mole percent in toluene phase. Through some data, I can convert
their mole percent distribution coefficient can be converted to a value of
9.0  .512 .
This data can be combined in a more user friendly way by use of Table 6.
Table 6: Experimental and Predicted Distribution Coefficient
Value
Basis
Measured m
10.635 concentration
Error in Measured m
9.5% concentration
Predicted m
2.32 mole
Predicted m
9 concentration
Error in Pred m
5.7% concentration
Experimental Error
18.2%
Table 6: Experimental and Predicted distribution coefficients.
With the help of this table, we can see that our measured distribution coefficient is
close to the predicted value but still substantially far from it with 18% error
between the two values. However, this error is acceptable, because as stated
before, the data used to make our prediction was based on past experiments. Thus,
this isn’t an absolute calculated value, just another group of researchers’
experimental prediction for its value. The fact that our two distribution
coefficients doesn’t match, doesn’t imply any failure in this experiment, but the
fact that the two values are close in approximation afford me confidence in my
results.
Overall Mass Transfer Rate:
44
To satisfy our next objective, we need to calculate the mass transfer rate of acetic
acid entering the drop for each diameter.
To measure this value experimentally is relatively straightforward. We have
previously calculated the volume of each drop that we add to the system, and we
have set the concentration of NaOH in the drop equal to .01M. Thus, we can
calculate the amount of NaOH present in each drop by multiplying the two afore
mentioned quantities. This value will also be the amount of HAc needed to enter
the drop to completely neutralize the NaOH, because this reaction has a one to
one ratio of these components as given from Equation 3.
(3)
CH 3CH 2COOH (aq)  NaOH (aq)  H 2O(l )  CH 3CH 2COO   Na (aq )
Next, we can measure the time needed for the drop to be completely neutralized,
by calculating the time needed for the previously calculated amount of acetic acid
to enter the drop. Dividing these two values will give the molar rate of transfer as
evidenced by Equation 6.
(6)
n

N HAc   HAc ,extracted 
 time 
The possibility of error can penetrate this procedure in one of two areas. First, the
error in calculating the volume is derived from the error in reading the volume of
a certain number of drops, in a 10 ml graduated cylinder, which will be .1 mL .
The second area of possible error will be in recording the time needed for
complete neutralization. This value can be calculated from the standard deviation
of all of the measurements taken to record this time and is ±.22 seconds. It should
be noted that this error is probably much lower than the actual error, because each
of the participants have a different perspective of the exact point the drop turned
clear.
This data is compiled into Figure 11.
45
Molar Transfer Rate as a function of Drop Diameter
15
Molar Transfer Rate (mol per second *10^-8)
14
13
y = 4058.2x - 8.0895
R2 = 0.9984
12
11
Molar Transfer Rate
Linear (Molar Transfer Rate)
10
9
8
7
6
5
0.003
0.0035
0.004
0.0045
0.005
0.0055
Diameter of Drop (m)
Figure 11: Molar Transfer Rate as a function of drop diameter.
The most important aspect of this graph is the functionality of the molar transfer
rate with drop diameter. It seems the molar transfer rate increases almost linearly
as the diameter increases. It is impossible for me to be fully confident that the
relationship is purely linear because of the few number of data points, however
the strong correlation coefficient does suggest it is most likely linear.
The result of the molar transfer rate increasing with increasing diameter is hard to
prove logically at this point. We know that the molar transfer rate, according to
Newton’s Laws, should be a function of how the surface area depends on
diameter and how the overall mass transfer coefficient depends on diameter. As
for the latter, we can be sure the surface area increases with increasing diameter
simply from the formula for surface area. However, we have yet to determine the
relationship for mass transfer coefficient. Thus, we will return to this topic of
intellectually proving or disproving this result later in the report.
Overall mass transfer coefficient:
46
To determine the overall mass transfer coefficient experimentally, is again, a
relatively straightforward task. We have previously calculated the molar transfer
rate. From Equation 11 all we need to calculate the overall mass transfer
coefficient (based on the inside mass transfer) is the equilibrium concentration of
acetic aid at the surface in the water phase.
(11)
K
N HAc
m * CHAc,Tol * A
The C* can be easily calculated from the definition of the distribution coefficient.
The distribution coefficient, that we have previously calculated to be 10.64, is the
concentration of acetic acid in water divided by the concentration in acetic acid in
toluene which would be achieved when the phases are in equilibrium. Since, we
have already calculated the concentration of acetic acid in the toluene, the
concentration of acetic acid in the water at the surface of the drop (which, from
our assumptions, must be in equilibrium with the toluene) can easily be calculated
from
(7)
C
m   HAc ,Wat
C
 HAc ,Tol

 .

The error associate with this calculation is easily calculated from the error
associated with the distribution coefficient (previously found) and the error
associated with the calculation of the concentration of acetic acid in the toluene
(previously calculated).
To determine the theoretical value for the overall mass transfer coefficient is
much more complicated. It involves using many assumptions and calculations
introduced by Handlos and Baron, which were explained in detail in the theory
section of this report. The equations derived from this rigorous analysis are
47
Shoutside 
koutside  2  rdrop
(29)
koutside  1.13
DHAc
 1.13Pe
1/ 2
outside
 vT  2  rdrop
 1.13 
 D
HAc

1/ 2



1/ 2
vT1/ 2 DHAc
2  r 
1/ 2
drop
(53)
Shi 
.00375Pe
 i 
1  
 o 
The error introduced by these calculations can be easily seen from the variables
involved in these calculations, the error in velocity (mainly associated with the
standard deviation of the multiple time recordings), the error in recorded
diffusivity (can be assumed to be low), the error in recorded viscosities (can be
assumed to match the same percentage as recorded density), the error in the radius
(derived from the error in reading the volume of 50 drops).
To satisfy our objective of determining how the overall mass transfer coefficient
depends on diameter, I have graphed the data of in Figure 12.
48
Overall Mass Transfer Coefficient as a Function of Drop Diameter
295
Overall Mass Transfer Coefficient (10^-7 m/s))
245
195
K Exp
K Pred
Linear (K Pred)
145
95
45
-5
0.003
0.0035
0.004
0.0045
0.005
0.0055
0.006
Diameter of Drop (m)
Figure 12: Overall Mass Transfer Coefficient as a function of drop diameter
The most important thing to note from Figure 12 is that the overall mass transfer
coefficient drops as diameter increases. However, it is also key to notice the
significant error obtained in our prediction of the overall mass transfer coefficient
using Handlos and Baron’s equations.
The overall mass transfer coefficient is a proportionality factor added to the rate
equation. Its purpose is to measure the effectiveness of the surface area at
achieving mass transfer. In other words, it measures the resistance of the drop at
transferring the acetic acid across its surface. The result of decreasing overall
mass transfer coefficient with increasing diameter is realistic from a logical point
of analysis. If we view this system from outside of the drop’s perspective, as we
have previously discussed in the theory section of this report, there is convective
mass transfer (at turbulent levels) due to the drop moving through the continuous
phase at its terminal velocity. However, as was mentioned in the theory of
Handlos and Baron, there is also a considerable amount of turbulence induced by
the transfer of mass occurring at the interface. This mass transfer acted to cause
49
turbulence due to radial motion of particles inside the drop which formed large
eddies currents, which aided mass transfer. These eddies distortions are based off
of two characteristics of the system. First, the rate of mass transfer at particular
points along the surface, and secondly, the volume of the particle because as
represented from Figure 5, after the particles are induced to move radially by
mass transfer they must travel a path around half the volume of the molecule until
they are forced back to the surface. Therefore as the diameter of the molecule
increases, so does its volume, and the path that these molecules must traverse also
becomes larger. But since the surface area has grown less, as compared to the
volume, the added velocity of these particles is not enough to overcome the extra
distance and consequently, the eddie currents must move slower. These slower
oscillations should cause less of a mass transfer induced turbulence and result in
the area to be less effective at mass transfer as the diameter increases.
However, if we analyze this from inside the drop’s perspective, the result of the
overall mass transfer coefficient decreasing with increasing diameter can seem
inaccurate. Because the volume of the drop is increasing faster than the
corresponding surface area, we see that the ratio of molecules further away from
the acetic acid is growing relative to the molecules in equilibrium with the acetic
acid at the drops surface. Thus, we have a higher proportion of molecules that will
have reacted with the NaOH and have zero concentration of acetic acid. Despite
the slower eddies, the increase in the molecules with no acetic acid should have
an overall effect of increasing the average driving force among all of the
molecules present. Therefore, the effective driving force among all molecules
would have increased and we would expect the overall mass transfer coefficient to
increase with increasing drop diameter.
As a result, we now have a quandary. Why is the first explanation correct and the
second explanation wrong? This question will be answered later in my report.
The result of our model over predicting the mass transfer coefficient is not
surprising. As was discussed in the Fudge Factor section of the Theory in this
report, many researches have felt that the model produced by Handlos and Baron
50
was very accurate in its analysis; however, they assumed ideality in their mass
transfer system. This is because Handlos and Baron, developed their model to be
general and applicable for all liquid - liquid extraction systems. On the other
hand, as pointed out by Henschke and Pfenning each liquid – liquid extraction
system comes with certain non negligible instabilities at the surface such as
surface tension and coalescence, which will cause the mass transfer coefficients to
be less. They suggest only the addition of a single constant to adjust for this,
though, they do point out this constant, although relatively consistent for every
similar system, must be measured experimentally.
With this understanding, I decided to model my system using their adjusted
equations,
(54)
Shi 
.00375Pe
  
CIP 1  i 
 o 
Shoutside 
(55)
koutside
koutside  2  rdrop
DHAc
1/ 2
1.13Peoutside
1.13  vT  2  rdrop



CIP
CIP  DHAc
1/ 2
1.13 vT1/ 2 DHAc

CIP  2  r 1/ 2
1/ 2



.
drop
As you can see from Equations 54 and 55, the instability constant can be easily
factored out to give K pred  CIP K exp . We are able to find this value by graphing
the originally predicted overall mass transfer coefficients versus
the
experimentally determined mass transfer coefficients, in which case the slope of
the graph should be CIP. This can be seen in Figure 13.
51
Predicting the Instability Constant
224
K pred orig (10^-7 m/s)
222
220
y = 113.28x + 2.0928
R2 = 0.9953
218
216
214
212
1.86
1.87
1.88
1.89
1.9
1.91
1.92
1.93
1.94
1.95
K exp (10^-7 m/s)
Figure 13: Graph of Kpred versus Kexp to determine CIP
A statistical regression of this data in Excel provided me with CIP  113.28  7.28 .
After adding this instability constant, my predicted and experimental results
showed good conformity as evidenced by Figure 14.
52
1.96
Overall Mass Transfer Coefficient as a Function of Drop Diameter
Experimental versus Predicted with Fudge Factor
2.5
Overall Mass Transfer Coefficient (10^-7 m/s))
2.4
2.3
2.2
2.1
K Exp
K Pred FF
Linear (K Pred FF)
2
1.9
1.8
1.7
1.6
1.5
0.003
0.0035
0.004
0.0045
0.005
0.0055
Diameter of Drop (m)
Figure 14: Fudge Factor predicted and experimental overall mass transfer coefficient as a
function of drop diameter
The most significant aspect of Figure 14 is that with the addition of a simple
instability constant, the predicted values are within 1.5% error of the experimental
values and follow the same functionality with diameter, each averaging a 2% drop
in the value of the overall mass transfer coefficient with a 1mm increase in
diameter.
Individual mass transfer coefficients:
We now return to the question of what is causing the overall mass transfer
coefficient to decrease with increasing diameter. The answer lies in the
functionality of the individual mass transfer coefficients with increasing diameter.
Analysis of the individual mass transfer coefficient reveals that they follow
different trends with diameter. As shown in Figure 15, the mass transfer
coefficient outside the drop decreases with increasing diameter, where as the mass
transfer coefficient inside the drop increases with increasing diameter.
53
Individual Mass Transfer Coefficient as a Function of Drop Diameter
3
Individual Mass Transfer Coefficient (10^-4 m/s))
2.8
2.6
2.4
2.2
K inside drop
K outside drop
Linear (K outside drop)
Linear (K inside drop)
2
1.8
1.6
1.4
1.2
1
0.003
0.0035
0.004
0.0045
0.005
0.0055
Diameter of Drop (m)
Figure15: Individual mass transfer coefficients as a function of drop diameter.
The most vital aspect the reader should take away from Figure 15, is that the
outside mass transfer coefficient follows the same trend as the overall mass
transfer coefficient, where as the inside mass transfer coefficient increases with
increasing diameter.
Therefore our explanation from outside of the drop’s perspective as to why the
mass transfer coefficient decreases with increasing diameter is true and this is
reflected by the trend of the outside mass transfer coefficient. Also, our
explanation from inside of the drop’s perspective as to why the mass transfer
coefficient increases with increasing diameter is true and is reflected by the trend
of the inside mass transfer coefficient.
It is with confidence in our thorough analysis that we can conclude that the
overall mass transfer coefficient follows the trend of decreasing with increasing
diameter because the outside mass transfer coefficient is the controlling resistance
54
in our system. This is because its value of resistance is larger and is combined
with the distribution coefficient which represents the discontinuity resistance of
the acetic acid concentration at the surface.
Revisiting the Overall Mass Transfer Rate:
Now, let us revisit the trend we observed in the overall mass transfer rate as we
have accumulated pertinent data and information to process this information.
Recall from earlier in this report, we showed that the mass transfer rate increases
with increasing diameter. This result agrees with what we would expect if we
performed a rational analysis of this situation.
Starting form Equation 9
(9)
N HAc  K * A *  C *  CHAc ,Wat 
We can see that the overall mass transfer rate depends on the overall mass transfer
coefficient, the surface area, and the driving force. If we change the diameter of
our particle, there should be no effect on our driving force, although the total
number of moles of acetic acid transferred would be different, the driving force
which is based on the concentration would stay the same. This means the
functionality of N with diameter must be based on only K and A. The overall
mass transfer coefficient we have previously found to fall with increasing
diameter, while the surface area is easily proven to rise with increasing diameter.
Hence, it depends on which of these terms rises faster relative to the other, which
will control the direction the overall mass transfer rate will follow. There is no
definite theory that one of these terms always has to grow faster than the other,
however, since we have already proven that the mass transfer rate increases with
increasing diameter, we should expect that the surface area is growing relatively
faster, and is the controlling factor in determining N.
This can be proved through the use of Table 7
55
Table 7: Percent Growth of S.A. and K
Surface Area (sq m)
Percent Change
K overall (m/s)
Percent Change
Diameter
0.00346
3.76E-05
Initial
1.95E-07
Initial
0.00398
4.98E-05
32.4%
1.92E-07
-1.5%
0.00516
8.37E-05
68.2%
1.87E-07
-2.9%
Table 7: Percentage Growth of the drop’s surface area and overall mass transfer coefficient.
This table shows that the percentage increase in surface area as diameter increases
is much greater than the percentage decrease in the overall mass transfer
coefficient. Accordingly, our observation of the mass transfer rate increasing with
increasing diameter is logical in conclusion.
56
CONCLUSIONS:
The purpose of this experiment was to provide a complete description of the
extraction of acetic acid from toluene into water. This unit operation was studied
through micro extraction, measuring the molar transfer rate, overall mass transfer
coefficient, and the equilibrium driving force of the transfer of acetic acid in a
stagnant continuous toluene phase into a single droplet of water (which also
contained NaOH and phenolphthalein).
In this experiment we discovered the equilibrium distribution coefficient for
acetic acid in water and toluene is 10.63. This result was 18% higher than the
value obtained by Fuse and Iguchi, in their experiments. The percent error
obtained in my experiment was found to be 9.5%. The percent error in Fuse and
Iguchi’s experiment was estimated to be 5.7% based on the scatter in their data.
However, it is not unreasonable to assume that there may be other errors present
which is not represented by the scatter in the data. Thus, it is not unreasonable to
believe their total experimental error, including the scatter and any uncertainty in
the values of the physical parameters they must have used in their calculations, is
also close to 9.5% in this range of concentration of acetic acid in toluene.
Therefore, the two values for the distribution coefficient are in agreement. Table 6
summarizes the data from this portion of the experiment.
As a result, we can conclude from this experimental analysis that the distribution
coefficient for an industrial liquid-liquid extraction system involving these
components is 10.63. This value can be used in making the mass transfer
calculations around this unit operation to ensure we set the correct parameters to
achieve the desired separation. Of course, in order to use this value, the engineer
will have to assume that the system is able to obtain equilibrium at the interface of
the two immiscible liquids.
The next objective of this experiment was to determine how the terminal velocity
of the falling drop depends on diameter. It was easy to predict, before performing
this experiment, that as the diameter of the particle increased, the terminal
velocity would also increase. This was easily accomplished by looking at the
57
forces acting on the falling drop and using Newton’s second law to relate the sum
of the forces. This prediction should be simple for any certified engineer.
However, it wasn’t the trend that was the most important part of this objective,
instead it was the ability to determine an experimental correlation and see if it
differs from that predicted by theory.
As can be seen from Figure 10 and Table 3 the predicted trend of increasing
terminal velocity with increasing diameter is proven. However more importantly,
notice the large amount of error between the predicted values and the
experimentally measured values. This 21% average percent error reveals that this
system does not travel in the ideal fashion predicted by our theoretical equations.
This proves that the drag coefficient of our drop is significantly large than that
which would be expected by a solid rigid sphere. Allowing us to conclude that our
drop doesn’t travel down the column as a sphere, but instead will distort itself.
Also, this distortion will cause oscillations or eddies in the drop. These eddies will
act to slow down the particle having it result in much lower velocities as we see
here. But more importantly, the major significance of proving the larger error in
the terminal velocity prediction is the proof of the formation of these eddies. As
described by Handlos and Baron and outlined in the Theory section of this report,
the formation of eddies is key to model the turbulence involved in the mass
transfer in this system.
Additionally, this data provides a correlation for engineers to predict how the
oscillating particle’s terminal velocity truly depends on drop diameter. This
functionality will be useful in the engineer’s design of an extraction column.
The next objective was to determine how the overall mass transfer coefficient
depends on diameter. To completely answer this question, it was easier to see how
the individual mass transfer coefficients depended on diameter. From Figure 15,
the reader should observe that the outside mass transfer coefficient decreases as
the diameter increases. It was my conclusion that the increase in the volume to
surface area ratio caused the internal oscillations (eddies) to become relatively
slower. Hence less turbulence induced mass transfer was occurring and causing
58
the additional surface area to become less effective. Also from Figure 15, the
reader should take note that the inside mass transfer coefficient increased with
increasing diameter. This led me to the conclusions that the increase in volume,
allowed for a higher percentage of molecules to be present with zero
concentration of acetic acid (i.e. away from the surface), as compared with those
molecules in equilibrium with the acetic acid (i.e. at the surface). Thus, the overall
average driving force will be higher and make the surface area more effective for
mass transfer.
By looking at Figure 12, the evidence reveals that the overall mass transfer
coefficient was proven to fall with increasing diameter. With this data, I was able
to conclude that the outside mass transfer coefficient is the controlling resistance,
because it caused the overall coefficient to follow its trend. This conclusion
became evident because the coefficients for the outside mass transfer were greater
in value, and because this term included the effect of the resistance induced by the
discontinuity of the concentration of acetic acid among the two phases at the
interface.
The significance of this result is three-fold. First, the engineer can plan on his
diameter size by realizing its effect on the overall mass transfer coefficient. This
will help with his mass transfer equations in the design of the extraction unit
operation. Second, and more importantly, this allowed us to see that the outside
mass transfer coefficient is the controlling resistance. Thus, if the engineer has the
ability to change any physical parameters (or maybe even materials) in his process
to increase the mass transfer, he should ensure that the variables he is changing
have the major effect on the outside mass transfer coefficient since it is this term
that controls the process. Finally, realizing that the outside mass transfer
coefficient is the controlling resistance, his time can wisely be spent modeling this
factor accurately, and using this functionality as the basis for his scale-up
calculations.
Figure 12 is, also, the basis of answering our other objective concerning the
overall mass transfer coefficient, namely, how well can we predict this value. It
59
was suggested by researchers that the model of Handlos and Baron, developed for
general liquid-liquid extraction of a falling drop, grossly over predicts the overall
mass transfer coefficient by 2 orders of magnitude. The reasoning for this was the
instabilities at the surface of liquid-liquid extraction, such as surface tension, are
significant, but not included in the original model for gas-liquid adsorption.
However, as was pointed out by Henschke and Pfenning, the model can be easily
adjusted to account for this extra instability by adding a constant instability factor
in the denominator of both individual mass transfer coefficients. This value
should remain constant for similar systems, though, it must be determined
experimentally.
For our system, I have determined the instability constant to be 113.28 with a
possible error of 7%, which is in the range predicted for turbulent mass transfer
systems. Addition of this factor improved our prediction to only 1.5% error. Thus,
I have concluded that the model developed by Handlos and Baron with the
adjustment by Henschke and Pfenning will work very well for predicting the trend
of liquid-liquid extraction systems. This model can also be used with confidence
for predicting absolute values. The only other requirement would be that the
engineers run a small scale experiment in order to obtain a few data points to
determine the instability factor. It is this value that will remain constant for the
industrial extraction unit.
My final objective was to determine the effect that diameter has on the overall
mass transfer rate. This functionality can be predicted by looking at the factors
controlling this transfer rate. From Newton’s relation, we know the factors of
interest are the driving force, the surface area, and the overall mass transfer
coefficient. We can immediately ignore the driving force because this value will
not change with the diameter of the particle. Next, it should be fairly obvious that
the surface area will increase with increasing diameter. Finally, we have already
shown that the overall mass transfer coefficient decreases with increasing
diameter. Thus we can say the functional relationship of the mass transfer rate
will depend on which of these two factors is the controlling variable for our
process. Table 7 reveals to us, from previous calculations that the surface area is
60
growing faster, relative to the decreasing of the overall mass transfer coefficient.
From this information, we should expect the surface area to be the controlling
factor and our mass transfer rate should increase.
As evidenced by Figure 11, this dependency is noted. In this figure, we see the
molar transfer rate increases almost linearly with an increase in diameter. Once
again, I need to point to the significance of this result for engineering an
extraction process. The large the engineer makes the diameter, the faster the
particle will transfer mass. However, it is also important for him to note our
previous functional relationship we found for the same particle’s terminal
velocity. The particle is transferring mass at a higher rate, but it is also traveling
much faster and has more mass to transfer before reaching an equilibrium state.
Therefore to achieve equilibrium for larger drops, greater amounts of travel time
and distance in the column is needed. These factors are essential and should be
included in the design of the extraction column.
In summary, when the engineer is designing an extraction column, he should take
into consideration the following trends. First, that the terminal velocity of the
particle and the mass transfer rate will both increase with an increase in diameter,
while the overall mass transfer coefficient will decrease with an increase in
diameter. The engineer should also note the following characteristics. One, the
overriding individual mass transfer coefficient is that of the outside. Two, the
overriding factor influencing the mass transfer rate is the surface area. Next, the
engineer should realize that the distribution coefficient should lie in the vicinity of
10.63 (concentration of HAc in water to that in toluene), but that this value is
concentration dependent. Also, he should note that the terminal velocity must be
measured experimentally as the oscillations in the drop cause it to be far off the
theoretical predictions but the theoretical trend predicted for the terminal velocity
is accurate on a percentage basis. Finally, the model of Handlos and Baron,
provides and excellent description of the trend of the overall mass transfer
coefficient, but an experimentally determined instability factor is needed for
accurate predictions of the absolute numbers. This factor should stay constant for
all similar situations involving these materials.
61
The usefulness of these conclusions is evident in the design of extraction
columns. The trends observed can be used in the rough sketch of the unit
operation, while the predictive equations developed can be used to predict the
column concentration profile and exit concentrations. Of course, to apply this, we
have to change our assumption that the surface concentration and the internal drop
concentration are both constant, as these values will vary with position in the
column. Thus, an analytical solution would not be easy, an a finite difference
algorithm is suggested, using the equations developed in this paper.
62
RECOMMENDATIONS:
In this experiment, I determined, as did Henschke and Pfenning, that an experimentally
determined instability constant allowed the modeling of the overall mass transfer
coefficient to within 2% of the actual values. However, it was hard to determine for
certain how accurate I was able to predict the value of this instability constant with only
three data points.
I recommend that the experimenter make up a known molarity of acetic acid in their
toluene phase since we know that a molarity of .8 moles per liter acetic acid provides
good results in this experiment, thus, freeing up some time in the second week for more
runs to be made at different diameters.
The benefits of this are that more data points can be run. This will better determine the
accuracy of the instability constant by allowing it to be calculated through regression
which allows the scatter to represent the error more accurately.
The only negative concept that I encountered is that the titration of the toluene phase of
the column is not required. In the process of determining the distribution coefficient, it is
necessary to titrate a toluene phase and as a result the experience of doing technique is
not lost. In fact, it eliminates the overkill of having to titrate this phase twice.
63
NOMENCLATURE:
Abbreviations:
A  impurity component
AQ  aqueous phase
E  extract solvent
HAc  acetic acid
ORG  organic phase
R  raffinate phase
T or Tol  toluene or toluene phase
W or Wat  water or water phase
Dimensionless Numbers:
Nu  Nusselt Number
Pe  Peclet Number
Pr  Prandtl Number
Re  Reynolds Number
Sc  Schmit Number
Sh  Sherwood Number
 #  Dimensionless Group
Variables:
 
 Projected Area  m 
A  surface area of drop m 2
AProj
2
 m2 
ag  acceleration due to gravity 

 sec 
 mole 
Ccomponent i  concentration of component i  3 
 m 
 mole 
C *  equilibrium concentration of HAc at surface  3 
 m 
CD  drag coefficient
CIP  instability constant
64
D or d  drop diameter  m 
 m2 
DAB or DHAc  fluid diffusivity 

 sec 
F  Force  N 
m
K  overall mass transfer coefficient based on inside  
s
m
kin  inside mass transfer coefficient  
s
m
kout  outside mass transfer coefficient  
s
M  mass of drop  kg 
MW  molecular weight  kg / kmole 
m  distribution coefficient
 mole 
N  mass transfer rate 

 sec 
ni  moles of component i  mole 
p  position in drop  m 
r  radius of drop  m 
t  time  sec 
Vol or V  volume  m3 
 m 
vT  terminal veloctiy 

 sec 
z 2  expectation value of displacement
 m2 

 sec 
  thermal diffusivity 
  error
  eigen value
 kg 

 m  sec 
 kg 
  density  3 
m 
  viscosity 
65
REFERENCES:
1.
Bond and Newton. Philadelphia Magazine. 5,No. 7. Page 794. 1928
2.
Fuse and Iguchi. Kagaku Kogaku. Experimental Tie Lines in Mole Percent
of Acetic Acid in Toluene and Water. 35(1971)107.
3.
Handlos and Baron. Mass and Heat Transfer from Drops in Liquid Liquid Extraction. AICHE Journal. Volume 3, No. 1. Pages 127-136.
March 1957.
4.
Henschke and Pfenning. Mass-Transfer Enhancement in Single Drop
Extraction Experiments. AICHE Journal. Volume 45, No. 10. Pages 20792086. October 1999.
5.
Higbie. Trans American Institute of Chemical Engineers. 31, 365 (1935).
6.
Johnstone and Pigford and Chapin. Trans American Institute of Chemical
Engineers. 37, 95 (1941).
7.
Kronig and Brink. Applied Science Research. A2, 142 (1950).
8.
Salerno, Tom. Convection Formal Report. June 2006.
9.
Welty and Wicks and Wilson and Rorrer. Fundamentals of Momentum,
Heat, and Mass Transfer – 4th Ed. John Wiley & Sons, Inc. New York.
(2001)
10.
West and Herrman and Chong and Thomas. Ind. Eng. Chem. 44, 621
(1952)
66
APPENDIX:
Raw Data:
During week one of this experiment, measurements of time were taken of the drop
to travel certain distances in the column. This documented data was used to
calculate the terminal velocity of the drop. The procedure and corresponding time
measurements were done for each syringe size. Additionally during the first week,
my group and I took measurements of the time needed for the drop to turn clear,
aiding us in our calculation of the molar transfer rate. This data was also recorded
for each corresponding syringe size. The raw numbers from these measurements
are shown below in Tables 8, 9, and 10.
67
Large Syringe
Calculating Volume
Number of Drops
Total Volume
Volume per Drop
Surface Area per Drop
Radius of Drop
50
0.0000036
0.000000072
8.36962E-05
0.002580762
(quantity)
(m^3)
(m^3)
(m^2)
(m)
Calculating Terminal Velocity
Trial #
0-600
1
2
3
4
5
6
7
8
9
10
1.72
1.72
1.71
1.57
1.59
1.68
1.7
1.73
1.72
1.64
1.678
Average Time (s)
Length per 100 marks
Length per 600 marks
Average Velocity
600-1200 1200-1800
1.94
1.84
2.08
1.8
1.85
1.74
1.86
1.82
1.8
1.81
1.92
1.84
1.73
1.83
1.86
1.85
1.97
1.86
1.77
1.84
1.878
1.823
0.0355 (m)
0.213 (m)
0.118795315 (m/s)
0.001775
Calculating Mass Transfer Rate
Trial #
1
2
3
4
5
6
7
8
9
10
11
Average time (s)
Concentration of NaOH
Moles of NaOH in drop
Moles of HAc transferred
N - transfer rate
K overall - exp
Distance 1
5.36
5.37
5.33
5.52
5.65
5.43
5.6
5.38
5.74
5.74
5.89
5.578636364
0.01
0.00000072
0.00000072
1.29064E-07
Distance 2
5.76
5.43
5.44
5.64
5.05
5.83
5.53
5.81
5.74
5.57
5.92
(m)
M
moles
moles
mole/s
1.86851E-07
Table 8: Raw Data, Week 1, Large Syringe
68
Medium Syringe
Calculating Volume
Number of Drops
Total Volume
Volume per Drop
Surface Area per Drop
Radius of Drop
50
0.00000165
0.000000033
4.97539E-05
0.001989795
(quantity)
(m^3)
(m^3)
(m^2)
(m)
Calculating Terminal Velocity
Trial #
0-600
1
2
3
4
5
6
7
8
9
10
Average Time (s)
Length per 100 marks
Length per 600 marks
Average Velocity
2.01
1.97
1.97
1.9
1.95
1.95
1.95
1.96
2.01
2.08
1.975
600-1200 1200-1800
2.01
2.16
1.92
2.01
2.27
2.07
2.15
2.09
2.33
1.93
2.33
2.05
2.39
2.01
2.11
2.02
2.24
2.07
2.21
2.06
2.196
2.047
0.0355 (m)
0.213 (m)
0.102766163 (m/s)
Calculating Mass Transfer Rate
Trial #
1
2
3
4
5
6
7
8
9
10
11
Average time (s)
Concentration of NaOH
Moles of NaOH in drop
Moles of HAc transferred
N - transfer rate
K overall - exp
Distance 1
4.34
3.91
4.3
4.18
4.48
3.98
4.07
4.01
4.01
3.96
4.49
4.177272727
0.01
0.00000033
0.00000033
7.89989E-08
Distance 2
4.41
4.14
4.05
3.98
4.56
4.06
4.06
4.06
4.11
4.54
4.2
(m)
M
moles
moles
mole/s
1.92393E-07
Table 9: Raw Data, Week 1, Medium Syringe
69
Small Syringe
Calculating Volume
Number of Drops
Total Volume
Volume per Drop
Surface Area per Drop
Radius of Drop
60
0.0000013
2.16667E-08
3.75848E-05
0.001729423
(quantity)
(m^3)
(m^3)
(m^2)
(m)
Calculating Terminal Velocity
Trial #
0-600
1
2
3
4
5
6
7
8
9
10
Average Time (s)
Length per 100 marks
Length per 600 marks
Average Velocity
2.06
2.26
2.12
2.34
2.12
2.24
2.18
2.25
2.1
2.22
2.189
600-1200 1200-1800
2.38
2.5
2.34
2.23
2.41
2.13
2.42
2.17
2.28
2.24
2.32
2.12
2.37
2.27
2.13
2.3
2.39
2.1
2.26
2.3
2.33
2.236
0.0355 (m)
0.213 (m)
0.094596595 (m/s)
Calculating Mass Transfer Rate
Trial #
Distance 1
1
2
3
4
5
6
7
8
9
10
11
3.7
3.67
3.61
3.46
3.56
3.73
4.12
3.52
3.53
3.24
Average time (s)
Concentration of NaOH
Moles of NaOH in drop
Moles of HAc transferred
N - transfer rate
3.576
0.01
2.16667E-07
2.16667E-07
6.05891E-08
K overall - exp
1.95334E-07
Distance 2
3.61
3.71
3.61
3.6
3.3
3.6
3.63
3.67
3.21
3.44
(s)
M
moles
moles
mole/s
Table 10: Raw Data, Week 1, Small Syringe
70
In week two of this experiment, the necessary titrations were performed by the
group. This data allowed us to immediately calculate the concentration of acetic
acid in the toluene phase and the equilibrium distribution coefficient at this
concentration of acetic acid in the toluene phase. This data is shown below in
Table 11.
Titration 1: Calculating HAc in Toluene
Amount of T solution added
Amount of .1M NaOH added
Amount of .1M HCl added
Amount of HAc present
Concentration of HAc in bulk
0.005
0.0506
0.0131
0.00375
0.75
(L)
(L)
(L)
moles
mol/L
0.005
0.05
0.0099
0.00401
0.802
(L)
(L)
(L)
moles
mol/L
should make a strong pink color
back titrated to light pink fading endpoint
Titration 2: Calculating HAc in Toluene
Amount of T solution added
Amount of .1M NaOH added
Amount of .1M HCl added
Amount of HAc present
Concentration of HAc in bulk
should make a strong pink color
back titrated to light pink fading endpoint
Titration Avg: Calculating HAc in Toluene
should make a strong pink color
back titrated to light pink fading endpoint
Concentration of HAc in bulk
Stdev Concentration of HAc in bulk
0.776 mol/L
0.036769553 mol/L
Calculating Distribution Coefficient
Amount of pure T added
Amount of pure W added
Estimated amount of Hac needed:
0.05 L
0.05 L
Concentration of HAc needed in T
Concentration of HAc in T
Moles HAc need in T
Volume of HAc in T
Est. Distribution Coeff
Concentration of HAc needed in W
Concentration of HAc in W
Moles HAc need in W
Volume of HAc in W
0.03677
0.800085
0.041923
0.002398
10
8.000854
8.000198
0.737393
0.042172
mol/L
mol/L
mol
L
Total moles HAc needed
Total volume HAc needed
0.779316 mol
0.044569 L
Amount of T soln added
Amount of .1M NaOH added
Amount of .1M HCl added
Amount of HAc present
Concentration of HAc in T Phase
0.005
0.0402
0.00005
0.004015
0.803
L
L
L
moles
mol/L
Amount of W soln added
Amount of .1M NaOH added
Amount of .1M HCl added
Amount of HAc present
Concentration of HAc in T Phase
0.0005
0.0427
0
0.00427
8.54
L
L
L
moles
mol/L
mol/L
mol/L
mol
L
Titrating Toluene Phase:
Titrating Water Phase:
m - Distribution Coefficient orig est
m - Distribution Coefficient
10
10.63511831
C* - conc of HAc at interface
C* - conc of HAc at interface
8.252851806 mol/L
8252.851806 mol/m^3
K - overall mass trans coeff
1.86851E-07
Table 11: Raw Data, Week 2, Titrations
71
Sample Calculations:
Experimental:
During the first week of the experiment, I calculated the terminal velocity of my
falling drop. As described previously in my report, this involved taking numerous
measurements of the time needed for the drop to pass through .213 meter sections
of the column.
Starting with the large drop, in order to find the average velocity, I first had to
calculate the average time for the particle to pass each .213 meter section. This
was accomplished by adding up each measurement unit and dividing by the total
number of units. The data received from this part is displayed in Table 8.
Mathematically this is calculated as,
30
Average time 
(67)
Average time 
 time
i 1
i
30
1.62  1.72  1.71  1.67  1.59  1.68  1.7  1.73  1.72  1.64 
1.84  2.08  1.85  1.86  1.8  1.92  1.73  1.86  1.97  1.87 


 1.84  1.8  1.78  1.82  1.81  1.84  1.83  1.85  1.82  1.84 
30
Average time  1.793 sec
Next, I calculated the average velocity by dividing the .213 meter distance that
each time was recorded with by the average time, this gave me the following
result,
(68)
v
Distance Traveled
.213 m

 .1188 m / sec
Average Time
1.793 sec
The error measurement in this calculation is derived from the error in measuring
the distance and the time. Since I used a ruler that had markings every millimeter,
it can be assumed that the error in measuring the distance to be half a millimeter..
The error in the time can be assumed to be the scatter in the time measurements,
72
or more specifically the standard deviation of the time measurements, which is
calculated by,
30
(69)
std dev 
 Time measurment
i
i 1
 Average Time 
2
30
 .105 sec
Thus we can calculate the error in our measured velocity as,
error in distance std dev in time

distance
time
.0005m .105sec
v 

 6.1%
.213m 1.793sec
v 
(70)
After this was completed, I needed to find the molar transfer rate from the large
drop. In order to do this, as described previously, I needed to take time
measurements starting from the time the drop entered the toluene phase, and
ending when the drop turned clear in the toluene phase. This data allowed me to
ascertain the time needed for the acetic acid to diffuse into the drop and neutralize
the entire amount of NaOH present. Thus, the first step in finding the value of this
variable was to find out how much NaOH was present. To do this I needed two
important statistics, the concentration of NaOH in the water phase, which is
known to be .01M and the volume of the drop of the water phase from the large
syringe.
To calculate the volume of each drop my group and I measured the volume of 50
of drops, and then divided the total volume by this amount. My group and I
proceeded to drop 50 drops of our water phase solution into a 10ml graduated
cylinder, and measured its volume to be 0.0000036m3. We can calculate the
volume per drop from,
(71)
Vdrop 
Total Volume 0.0000036m3

 7.2e8 m3 / drop
# of drops
50 drops
73
This data allowed us to calculate the average amount of NaOH present in each
drop from,
(72)
nNaOH , Wat  Vdrop * CNaOH , Wat  7.2e 8 m3 *
1000 L
mole
*.01
 7.2e 7 mole NaOH
m3
L
The next step was to calculate how much acetic acid needed to enter the drop to
neutralize the 7.2e7 mole NaOH present. To attain this, we need to look at the
neutralization reaction between these compounds given to us by Equation 3,
(3)
CH 3CH 2COOH (aq)  NaOH (aq)  H 2O(l )  CH 3CH 2COO   Na (aq )
From this equation, you can see that there is a 1:1 relationship between NaOH and
acetic acid in the neutralization reaction. Hence, the amount of acetic acid that
gets transferred into the water phase while the drop is still colored is equal to the
amount of NaOH originally present which we calculated from Equation 70.
Now that we know how much was entered, we need to know the time required for
this phase in order to find the mass transfer rate. This requires us to calculate the
average of the measurements of the time it took for this drop to turn clear. This
can be done by,
22
taverage to turn clear 
(74)
taverage to turn clear
taverage to turn clear
 time measurement
i 1
i
22
5.36  5.37  5.33  5.52  5.65  5.43  5.6  5.38  5.74  5.74  5.89
5.76  5.43  5.44  5.64  5.05  5.83  5.53  5.81  5.74  5.57  5.92

22
 5.579sec
Now, we can calculate the molar transfer rate as,
74
N
nNaOH , Wat
amount HAc diffused

average time
taverage to turn clear
N
7.2e7 mole
 1.29e7 mole / sec
5.58 sec
(75)
Next, we can calculate the error involved in the experimental measurement of this
value. This error is derived from the error in predicting the amount of NaOH in
the drop, which comes from the error in finding the volume, the error in the
molarity, and the standard deviation in the time measurements. The error in the
volume depended largely upon the ability of our group to read the volume of the
50 drops, which can be assumed to be .1ml since the graduated cylinder had
markings every .2mL. The error in the molarity of the solution must be estimated,
since this molarity is very small, a 5% error isn’t unlikely. The standard deviation
in the time can be estimated as shown in Equation 69 and is found to be .217
seconds. Thus, we can calculate the error in the molar transfer rate as follows,
error in volume error in molarity std dev in time


volume
molarity
time
1
.1mL 
.217 sec
N  50
 5% 
 11.7%
1
5.579sec
 3.6mL 
50
N 
(76)
In week two, my group and I titrated the toluene phase. This involved taking 5ml
of my toluene phase and placing it in a stirred beaker. An excess of NaOH was
added to ensure that all of the acetic acid present was neutralized. It was essential
to add excess because the two phases are immiscible. Then, we had to back titrate
with HCl to reach the phenolphthalein endpoint. As described in the procedure
section of this report, the calculation procedure to determine the amount of acetic
acid present in the toluene phase is,
75
mmole HAc   mL NaOH  mL HCl  
.1 mmole
mL
mmole HAc  .0506mL NaOH  .0311mL HCl  
.1 mmole
mL
mmole HAc  3.75
mmole HAc
CHAc , Tol 
5 mL sample
3.75
CHAc , Tol 
 .75mol / L HAc
5
(77)
This procedure was then repeated and resulted in CHAc , Tol  .802mol / L HAc .
Thus, the average value was taken as our actual concentration from,
CHAc , Tol 
(78)
.75  .802  mol / L HAc  .776mol / L HAc
2
The error involved in this measurement can be derived from a number of sources.
First, we have to estimate the amount of NaOH and HCl in the titrating columns
both before titration and after titration. There is an error in estimating the 5ml of
sample that needed to be taken from my toluene phase. Finally, there is an error in
judging when the solution has reached its phenolphthalein endpoint. Because of
the numerous possible errors and the uncertainty in estimating the latter
mentioned error source, the measurement error in the value of Equation 78 was
calculated from the standard deviation of the two data points taken. Which is
written,
.75  .776   .802  .776 
2
std dev, CHAc , Tol 
(79)
CHAc , Tol 
std dev, CHAc , Tol
CHAc , Tol
2
.0368

 4.7%
.776
2
 .0368
.
My group and I recognize that using the standard deviation of only two points
isn’t a true measure of the error involved; however, our answer of 4% error with
76
the number of sources involved seems to be in the range of percent error we
would have expected. Therefore, we accept this as our error.
Next, we had to determine the distribution coefficient of acetic acid in toluene and
water at .776mol / L HAc in Tol . Following the calculation steps described in the
procedure section of this report, we have to estimate the amount of acetic acid to
add in order to give us this concentration of acetic acid in our toluene phase. This
is performed,
mmole
 40
mL
mmole
mmole HAc in Wat  50mL Wat *10*.8
 400
mL
mmole HAc  mmole HAc in Tol  mmole HAc in Wat  440
mole
60.05 g 1mL
mL HAc  mmole HAc *
*
*
1000mmole mole 1.05 g
mole
60.05 g 1mL
mL HAc  440 mmole HAc *
*
*
 25.2mL HAc
1000mmole mole 1.05 g
mmole HAc in Tol  50 mL Tol * .8
(80)
Approximately 50ml of both toluene and water and approximately 26 ml of acetic
acid were added into a separatory funnel and mixed. We titrated each via the
procedure already described, and performed the calculations by the procedure
already described to find that
(81)
m
CHAc , Wat
CHAc , Tol

8.54mol / L
 10.64
.803mol / L
The error in this calculation is derived again by the error involved in making the
titrations, which we have defined to comprise numerous sources. We would have
liked to take the standard deviation of a number of measurements of this value,
however, time did not allow us to obtain another measurement of this value. Thus,
we had to assume that our titration ability is relatively consistent, such that we
would incur the same error in each of these titrations. The error in our distribution
coefficient would then be,
77
2
(82)
m   Percent Error in Titration i  2* 4.7%
i 1
m  9.4%
Theoretical:
Our next task was to calculate predictions of these experimental values based
from theory.
From our force balance, performed in the theory section of this report, we find our
velocity depends on time according to the following differential equation, which
has been set equal to zero since we are finding the terminal velocity, i.e., the
velocity at which the velocity no longer depends on time.
(20)
2
dv agravity  Wat  Tol  CD v v Tol * rdrop


0
dt
Wat
2* M
Our first step in solving Equation 20 is to ensure the only variable in the equation
is the unknown velocity. In order to replace the drag coefficient with a function
that depends on velocity alone, we have the following correlations to choose
from:
(21a)
CD 
(21b) CD 
(21c)
24
for Re  2
Re
18.5
for 2  Re  500
Re.6
CD  .44 for Re  500 .
For our large particle, we will assume that the velocity will be large enough to
give us a Reynolds number above 500. The major reasoning of this prediction was
so that as our first guess, we will simply have to plug in a constant for C D into our
equation instead of a function form of the Reynolds number which depends on
velocity. Performing this we arrive at
78
(83)
v
agravity  Wat  Tol 
2* M
Wat
2
.44* Tol * rdrop
Next we can replace the unknown mass term with the volume of the drop times
the density, which we both know. We now have
(84)
4 3
agravity  Wat  Tol  2* Wat  3  rdrop
.
v

2
Wat
.44* Tol * rdrop
Plug in all of the known values into this equation and solve for the terminal
velocity.
(85)
4 3
8
1
agravity  Wat  Tol  2* Wat  3  rdrop
agravity  Wat  Tol  3 Wat rdrop
v



2
Wat
.44* Tol * rdrop
Wat
.44 Tol
2
v
8
3
 867 kg / m3  3 *998kg / m * .00258m 

 .152m / sec
998kg / m3
.44*867 kg / m3
 9.81m / sec 998kg / m
3
Our next step will be to determine the error in this calculation. Since all terms are
multiplied, we can set up our error equation as
(86)
v 1 a 1   Wat  Tol  1 Wat 1 rdrop
.




v 2 a 2  Wat  Tol  2 Wat 2 rdrop
In this equation, we can assume the error due to the acceleration due to gravity is
only 1%, because we had numerous measurements and calculations of this value
in literature. The error in the density can be gathered by averaging all of the
values listed for the densities in many literature sources and finding the standard
deviation of this error. The results of this allowed us to find the error due to the
density was approximately 1.06% . Finally, the error in the radius of the drop has
79
to be calculated. We calculated the radius of the drop by measuring the volume of
50 drops, calculating the volume of one drop and finally the average radius of
each drop. The error in this calculation is
(87)
rdrop
rdrop
1
.1mL 
 50
 2.78%
1
 3.6mL 
50
Thus, we can calculate our error in predicted velocity to be
v 1 a 1   Wat  Tol  1 Wat 1 rdrop




v
2 a 2  Wat  Tol  2 Wat 2 rdrop
(88)
v 1
1
1
1
 *1%  * 2*1.06%  *1.06%  * 2.8%
v
2
2
2
2
v
 3.5%
v
The experimental error can now be calculated by the difference of the terminal
velocity measured in week 1 and the predicted value calculated in Equation 85.
Exp Error 
(89)
Exp Error 
vT , exp  vT , pred
vT , pred
*100%
.119  .152
*100%  22.0%
.152
To determine the predicted value of the distribution coefficient, my group and I
used recorded experimental data given to us by Fuse and Iguchi. However, their
data is in mole percent, whereas, our experimental data is in weight percent, thus
it became necessary to translate my data into mole percent to compare.
This task can be accomplished by first assuming we have 1000mL of both toluene
and water, of which we have determined their concentration to be .803 and 8.54
80
mol per liter respectively. Thus to calculate the mole percent of HAc in the
toluene phase we have,
mL Tol phase  1000mL
mole HAc in Tol  1L *.803mol / L  .803mole
.803mol *60.01g / mole
mL HAc in Tol  1L 
 45.89mL
1.05 g / L
mL Tol  1L  45.89mL  954.11mL
(90)
g
954.11mL *.867  
954.11mL * Tol
 L   8.98mole
mole Tol 

MWTol
 g 
92.14 

 mol 
.803
mole% HAc in Tol 
 .0821
.803  8.98
Similarly we can calculate the mole percent of HAc in the water phase to be,
mL Wat phase  1000mL
mole HAc in Wat  1L *8.54mol / L  8.54mole
8.54mol *60.01g / mole
mL HAc in Wat  1L 
 488mL
1.05 g / L
mL Wat  1L  488mL  512mL
(91)
g
512mL *.999  
512mL * Wat
 L   28.416mole
mole Wat 

g
MWWat


18 

 mol 
8.54
mole% HAc in Wat 
 .231
8.54  28.416
Thus our experimental distribution coefficient in mole percent turns out to be
(92)
mole percent HAc in Wat phase .231

mole percent HAc in Tol phase .0821
 2.82
mmole 
mmole
81
Which is close (only 18.6% difference) to 2.32, the experimental found by Fuse
and Iguchi in their research.
Next, my group and I had to calculate the predicted values for the individual mass
transfer coefficients. As noted earlier, the velocity used for these equations is the
experimental velocity since this was determined to be a better estimate of the true
velocity. The calculation of these are done by,
koutside  1.13
1/ 2
vT1/ 2 DHAc
2  r 
1/ 2
drop
(93)
.119   2.26e9 
koutside  1.13
1/ 2
 2  .00258
1/ 2
1/ 2
 2.57e 4
.00375vT

i 
1  
 o 
.00375 .119 
ki 
 1.8e 4
.001 

1 

 .00068 
ki 
(94)
Next we had to calculate the overall mass transfer coefficient. This was achieved
by graphing the equilibrium between the two phases and plotting the driving
forces. The calculation procedure is as follows,
1
1
m


K  kin kout
(95)
1
1
10.635


4
2.57e 4
K  1.8e
K   2.14e 5
It was then determined that the equations presented by Handlos and Baron
required an experimentally determined instability factor to be added. The
predictive equations took the form,
82
(54)
Shi 
.00375Pe
  
CIP 1  i 
 o 
Shoutside 
(55)
koutside 
koutside  2  rdrop
DHAc
1/ 2
1.13Peoutside
1.13  vT  2  rdrop



CIP
CIP  DHAc
1/ 2



1/ 2
1.13 vT1/ 2 DHAc
CIP  2  r 1/ 2
drop
In order to determine this coefficient, we need only graph the K from experiment
versus that from predicted. The slope of this graph gave us CIP.
Predicting the Instability Constant
224
K pred orig (10^-7 m/s)
222
220
218
y=111.11*Kexp+6.07
216
214
212
1.86
1.87
1.88
1.89
1.9
1.91
1.92
1.93
1.94
1.95
K exp (10^-7 m/s)
Figure 13: Graph of Kpred versus Kexp to determine CIP
From Figure 13, we have estimated our CIP constant to be 113.28. The error
involved from this calculation can again be associated to the scatter in the data,
which is
83
1.96
(96)
CIP  7.0% .
The experimental error can be calculated as done before in Equation 89.
This concludes a sample of each type of calculation for the liquid-liquid
extraction, falling drop experiment.
THESE CALCULATIONS HAVE BEEN REVIEWED AND APPROVED BY:
THOMAS SALERNO:
_______________________
GREGORY ROTHSCHING: _______________________
AN DU:
_______________________
84