COURSE NUMBER ChE 302 TITLE STUDY OF DIFFUSION EXPERIMENT NO 02 DATE OF PERFORMANCE 15/01/2014 DATE OF SUBMISSION 22/01/2014 SUBMITTED TO DR.MD.EASIR ARAFAT KHAN ASSISTANT PROFESSOR DEPARTMENT OF CHEMICAL ENGINEERING BUET,DHAKA-1000,BANGLADESH SUBMITTED BY KAMONASHIS SARKAR (TENDUL) STUDENT ID 1002008(L-3/T-1) GROUP NUMBER 02 (A1) PARTNERS STUDENT ID 1002006,1002007,1002009,1002010 CHEMICAL ENGINEERING DEPARTMENT BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY
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COURSE NUMBER ChE 302
TITLE STUDY OF DIFFUSION
EXPERIMENT NO 02
DATE OF PERFORMANCE 15/01/2014
DATE OF SUBMISSION 22/01/2014
SUBMITTED TO
DR.MD.EASIR ARAFAT KHAN
ASSISTANT PROFESSOR
DEPARTMENT OF CHEMICAL ENGINEERING
BUET,DHAKA-1000,BANGLADESH
SUBMITTED BY
KAMONASHIS SARKAR (TENDUL)
STUDENT ID 1002008(L-3/T-1)
GROUP NUMBER 02 (A1)
PARTNERS STUDENT ID 1002006,1002007,1002009,1002010
CHEMICAL ENGINEERING DEPARTMENT
BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY
Summary
The main objective of this experiment is to determine the liquid diffusion co-efficient of NaCl solution in
distilled water by using the concept of mass transfer.Mass transfer is the net movement of mass from one
location, usually meaning a stream, phase, fraction or component, to another. Mass transfer occurs in
many processes, such as absorption, evaporation, adsorption, drying, precipitation, membrane filtration,
and distillation.The phrase is commonly used in engineering for physical processes that involve diffusive
and convective transport of chemical species within physical systems. Mass transfer takes place in either a
gas phase or a liquid phase or in both simultaneously. Mass transfer is a result of diffusion. Diffusion is
the phenomenon of material transport by atomic motion. The driving force of diffusion process is the
concentration gradient of the component. Diffusivity is one of the most important fluid properties.
Diffusion is of great importance in Fluid flow and mass transfer operations such as chemical process plant
design, catalyst design in chemical industry, formation of alloys etc.
Experimental setup
Figure1 Schematic diagram for liquid diffusivity measurement
Diffuser
Conductivity
meter
Capillaries
Conductivity
Probe
Magnetic
stirrer
Stirrer bar
Observed data
Volume of water, V = 1.1 Liter
Length of capillaries, x = 0.5 cm
Diameter of capillaries, d = 0.1 cm
Numbers of capillaries, N = 97
Table1 Observed data for determining diffusivity of NaCl solution.
Observation
number
Time (seconds)
1M NaCl solution 4M NaCl solution
Conductivity, k (µS) Conductivity, k (µS)
01 0 0.70 11.29
02 300 83.42 641.60
03 600 100.60 758.90
04 900 114.20 845.60
05 1200 124.20 892.50
06 1500 133.20 940.50
07 1800 141.20 980.30
08 2100 148.60 1013.00
09 2400 155.60 1045.00
Calculated data
Observation
number
Concentration
of NaCl
solution
,M(mol/L)
Rate of conductivity change
over
time(slope),dk/dt(µs/sec)
Electrical conductivity
change per unit molar
concentration
change,Cm(µs/mol/L)
Diffusivity
,D(cm2/sec)
01 1 0.0331 85543.96 2.79×10-4
02 4 0.1804 85543.96 3.81×10-4
Graphs
y = 0.0331x + 80.381 for 1M NaCl solution
0
20
40
60
80
100
120
140
160
180
0 500 1000 1500 2000 2500 3000
Con
du
ctiv
ity,µ
s
Time,sec
Graph-1(Conductivity vs time)
y = 0.1804x + 646.12 for 4M NaCl solution
0
200
400
600
800
1000
1200
0 500 1000 1500 2000 2500 3000
Con
du
ctiv
ity,µ
s
Time,sec
Graph2(Conductivity vs time)
Graph3 Determination of conductivity change over time
y = 0.0331x + 80.381
for 1M solution
y = 0.1804x + 646.12 for 4M solution
0
200
400
600
800
1000
1200
0 500 1000 1500 2000 2500 3000
Co
nd
uct
ivit
y,µ
s
Time,sec
Graph3(Conductivity vs time)
Sample calculation
Determination of electrical conductivity change per unit molar concntration change
Water was 500ml and 2ml 1M NaCl solution was added,then we can write
V1S1=V2S2 V1=Volume of NaCl solution=2ml
S2=(V1S1/V2) S1=Concentration of NaCl solution
=1M
S2=(2×1)/502 V2=Volume of solution=502ml
S2=3.98×10-3M S2=Concentration of solution=?
Water was 500ml and 4ml 1M NaCl solution was added,then we can write
V1S1=V3S3 V1=Volume of NaCl solution=2ml
S3=(V1S1/V3) S1=Concentration of NaCl solution
=4M
S3=(4×1)/504 V3=Volume of solution=502ml
S3=7.94×10-3M S3=Concentration of solution=?
Water was 500ml and 6ml 1M NaCl solution was added,then we can write
V1S1=V4S4 V1=Volume of NaCl solution=2ml
S4=(V1S1/V4) S1=Concentration of NaCl solution
=1M
S4=(6×1)/506 V4=Volume of solution=502ml
S4=11.86×10-3M S4=Concentration of solution=?
From conductivity meter we have measured conductivity(K).
when S2=3.98×10-3
M then K2=400.9 µs
when S3=7.94×10-3
M then K3=741 µs
when S4=11.86×10-3
M then K4=1075 µs
Now electrical conductivity change per unit molar conductivity change(C)