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%