3.0 Results and Calculations
Result for calibration curve:
Sample Table For Preparation of Refractive Index Versus Composition:
Molar volume of ethanol= 17.13 mol/L Molar volume of water = 55.49 mol/L
Table 3.1
No of mol of ethanol
No of mol of water
Total mole mole fraction ethanol
mole fraction water
refractive index
0 0.554938957 0.55493896 0 1 1.332990.01712622 0.499445061 0.51657128 0.033153644 0.966846356 1.337190.03425244 0.443951165 0.47820361 0.071627315 0.928372685 1.34328
0.051378661 0.38845727 0.43983593 0.116813241 0.883186759 1.349190.068504881 0.332963374 0.40146826 0.17063586 0.82936414 1.354230.085631101 0.277469478 0.36310058 0.235833006 0.764166994 1.358450.102757321 0.221975583 0.3247329 0.316436432 0.683563568 1.360960.119883542 0.166481687 0.28636523 0.418638611 0.581361389 1.363040.137009762 0.110987791 0.24799755 0.552464168 0.447535832 1.363580.145572872 0.083240844 0.22881372 0.636206932 0.363793068 1.363250.154135982 0.055493896 0.20962988 0.735276783 0.264723217 1.362730.162699092 0.027746948 0.19044604 0.854305462 0.145694538 1.361160.171262202 0 0.1712622 1 0 1.36096
Xd = 0.3675
Xb = 0.1706
0 0.2 0.4 0.6 0.8 1 1.20
2
4
6
8
10
12
f(x) = NaN x + NaNR² = 0 refractive index vs. mole fraction of
ethanol
y- dataLinear (y- data)
Graph 3.1 : refractive index against mole fraction of ethanol (calibration curve)
1) Calculation Molar Volume of Ethanol
From appendix E in manual,
Specific gravity of ethanol = 0.789
Specific gravity of water = 1.000.
Density of water = 1.000 g/mL.
Density of ethanol = 789 kg/m3
= 0.789 g/mL
Molecular weight (MW) of ethanol= 46.07g/mol
Molar volume of ethanol =1mol46.07 g X 0.789g
1mL X 1000mL1 L =17.13 mol/L
2) Calculation Molar Volume of Water
Density of water = 1000kg/m3
= 1 g/mL
Molecular weight (MW) of water= 18.02 g/mol
Molar volume of water=1mol18.02g X 1 g
1mL X 1000mL1 L =55.49 mol/L
3) Calculation of Mole Fraction
For this sample, second set of data is used.
For second set of data,
Volume of Ethanol = 1.0mL or 0.001 L
Volume of water = 9.0mL or 0.009 L
Mole of ethanol = Molar Volume X Volume
=17.13mol
LX 0.001 L
= 0.01712622 mole
Mole of water = 55.49mol
LX 0.009 L
= 0.499445061 mole
Mole fraction of ethanol = Mole of ethanol
Mole of ethanol+Mole of water
=0.01712
0.01712+0.49941
= 0.033153644
Mole fraction of water = Mole of water
Mole of ethanol+Mole of water
= 0.49941
0.01712+0.49941
= 0.966846356
Calculation is repeated for all set of data.
Experiment 1 : Batch Distillation at Total Reflux
Heater Distillate Bottom HETP
Temp Refractive index
Mole frac (Xd)
Temp Refractive index
Mole frac (Xb)
100 74.5 1.3562 0.20107 80.1 1.35585 0.19566 1120
80 79.8 1.35446 0.17419 73.3 1.36191 0.14306 1120
60 79.7 1.35356 0.16348 72.9 1.36225 0.08101 1120
40 79.8 1.35259 0.15312 72.5 1.36206 0.11568 1120
20 79.7 1.35187 0.14543 72.4 1.36229 0.07371 1120Table 3.2
Taking mole fraction of distillate = 0.400394, mole fraction of bottom product = 0.108481, the no. of theoretical plates can be determined from the X-Y equilibrium diagram for ethanol-water system at 1atm from APPENDIX A.2.
10 20 30 40 50 60 70 80 90 100 110800
850
900
950
1000
1050
1100
1150
HETP vs Power Supply
HETP
Power Supply (%)
HETP
(mm
)
Graph 3.2 – HETP Vs Power Supply
0.15 0.16 0.17 0.18 0.19 0.2 0.211.35
1.351
1.352
1.353
1.354
1.355
1.356
1.357
Refractive index vs Mole Fraction
Distillate
Mole Fraction,X
Refr
activ
e in
dex
Graph 3.3 Refractive index Vs Mole Fraction
0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.221.3521.3531.3541.3551.3561.3571.3581.359
1.361.3611.3621.363
Refractive index vs Mole Fraction
Bottom
Mole fraction,x
Refr
activ
e in
dex
Graph 3.4 - Refractive index vs Mole Fraction
Sample Calculation For experiment 1:
To calculate HETP:
For this sample calculation, data for 100% heater power is used.
Height of column = 1120mm
HETP= Height of columnNumber of stages
HETP=11201
HETP=1120mm
#Repeat the calculations for all the data respective to the percentage heater power.
Experiment 2 : Batch Distillation at Constant RefluxResults:
time temp Refractive index
volume Mole frac (Xd)
Flow rate temp Refractive
indexMole frac
(Xb)volume HETP 1/y-x
0 76.1 1.36094 11 0.31579450/170s 79.9 1.36224 0.082831 6.5 1120 4.29253
10 75.3 1.35923 11 0.26088150/214s 80 1.36216 0.097432 8.75 1120 6.1181
20 75.3 1.35937 10.75 0.26537750/219s 80.1 1.36208 0.112032 10.5 1120 6.52125
30 75.3 1.35916 11 0.25863360/264s 80.1 1.36205 0.117507 9.5 1120 7.08587
40 75.3 1.35937 12.5 0.26537750/217s 80.1 1.362 0.126632 7.2 1120 7.20749
50 75.5 1.35991 12.75 0.28271850/221s 80.2 1.36199 0.128457 10.5 1120 6.48254
60 75.5 1.3601 11.5 0.28881950/243s 80.2 1.36186 0.152183 13 1120 7.31869
Calculation for Experiment 2:
To calculate XD /(R+1) :
In this sample, data for time interval=0 is used
For Y-intercept value,( XD /(R+1) )
t=00
X D=0.315794
R=1
X D ÷(R+1)
¿0.315794÷ (1+1 )
¿0.315794÷2
¿0.157897
Then, to calculate HETP at time 0,
t=0
Based on graph given in the appendices, number of stages=1
Height of column= 1120mm
HETP=Height of co lumnNumber of stages
HETP=1120mm1
HETP=1120mm
# Repeat the calculation for others data using this step to calculate HETP
0 10 20 30 40 50 60 700
200
400
600
800
1000
1200
HETP
HETP
Time
HETP
(mm
)
Graph 3.5 - graph number of theoretical stage, HETP against time interval
For Mass Balance Calculations:
Initial Conditions:
Reboiler volume = 15.0 L
Reboiler composition = 0.082831 (value taken from XB at t=0min table 3.3)
Final Conditions:
Reboiler volume = 13.0L
Reboiler composition = 0.152183 ( value taken from XB at t=60min Table 3.3)
Distillate volume = 1.5 L
Distillate composition = 0.288819 (value taken from XD at t=60 min Table 3.3)
Around 0.14645 L of sample from the reboiler has been withdraw in order to test the refractive index.
Final reboiler volume
= initial reboiler volume – final distillate volume – loss of volume due to sample testing
= 15.0L – 1.5L – 0.14645L
= 13.353 L
0.25 0.26 0.27 0.28 0.29 0.3 0.31 0.321.358
1.3585
1.359
1.3595
1.36
1.3605
1.361
1.3615
Refractive index vs Mole Fraction
Distillate
Mole Fraction,Xd
Refr
activ
e in
dex
Graph 3.6 – refractive vs mole fraction
0.05 0.07 0.09 0.11 0.13 0.15 0.171.3616
1.3617
1.3618
1.3619
1.362
1.3621
1.3622
1.3623
Refractive index vs Mole Fraction
Bottom
Mole Fraction,Xb
Refr
activ
e in
dex
Graph 3.7 – Refractive index vs Mole fraction
0.001 0.021 0.041 0.061 0.081 0.101 0.121 0.141 0.161 0.1810
1
2
3
4
5
6
7
8
1/(Y-X) vs Xb
Xb
1/(y
-X)
Graph 3.8 : Graph 1
y−x versus XB for Different Time Interval to Determine the Area under the Curve
To calculate Area Under the Curve:
Area of a Complete big box= 0.02 X 1 =0.02
Approximate Number of the boxes under the curve= 20.5
Total area under the curve= 20.5 (0.02) = 0.41
According to Rayleigh’s equation as in the manual:
∫n1
n0 dnn
=∫x 1
x 0 dxy−x
=ln n0n1
n1= n0exp ¿¿
Where t0= initial condition,t1= final condition, x= bottom composition, y= top composition, n= total number of moles of ethanol in the reboiler.
Calculation of Theoretical Value of Amount of Ethanol Left in the Reboiler
lnn0n1
=∫ dxy−x
# lnn0n1
=area under thecurve of Graph3.8¿Graph 1
y−x versus XB for Different Time Interval)
n1=n0
exp (area under the graph)
n0ethanol=17.12mol ethanolL
(0.3)(15.0L mixture)=77.04mol ethanol
n0water=55.49mol waterL
(0.7 ) (15.0 L mixture )= 582.645 mol water
n1=77.04mol ethanolexp (0.174)
=64.74mol
0 10 20 30 40 50 60 700
0.020.040.060.08
0.10.120.140.16
Mole Fraction of ethanol VS Time
Xb
Time
Xb
Graph 3.9 : Graph mole fraction of ethanol left in reboiler against time
Performing a Mass Balance Calculation on the Distillation Column and Compare with the Experimental Results
i) Total mass balance formula
Fo = D1 + B1
F0 = total mole in feed
D1 = total mole in distillate
B1 = total mole in reboiler
ii) Component mass balance formula
F0 x0 = D1 xD1 + B1 xB1
x0 = mole fraction of ethanol in feed
xD1 = mole fraction of ethanol in distillate
xB1 = mole fraction of ethanol in reboiler
Initial Conditions:
Reboiler volume = 15.0 L
Reboiler composition = 0.082831
Final Conditions:
Reboiler volume = 13.0L
Reboiler composition = 0.152183
Distillate volume = 1.5L
Distillate composition = 0.288819
Substituting the final value of the composition of the distillate and bottom produce into the mass balance equation,
15(xo)=13.0(0.152183)+1.5(0.288819)
X0¿13.0 (0.152183 )+1.5 (0.288819 )
15
= 0.161 (Theoretical value for the initial value of mole fraction of ethanol)
Experimental value for the initial value of mole fraction of ethanol= 0.126
Percentage error =Experimental Value−Theoretical Value
Theoretical ValueX 100%
¿|0.152183−0.1610.161 |X 100%
=5.47%