Observation and Calculatons:

a)Specific Gravity Calculations

Literature Value of 98%Ethanol at 25ᵒC = 0.79117

1.1−0.79117
Relative Error = ∗100
1.1

= 28.07%

Xf = 0.5

1.1−0.79
Now composition of water in top product = × 100=39.24 %
0.79

So ethanol in top product = 100-39.24 = 60.75%

This gives,

Xd = 0.61

Xw = 0.39

Mass Balance on Column

100 = D + W

Balance on Ethanol gives,

100 x 0.5 = 0.61 D +0.39 W

50 = 0.61 (100 – W) + 0.39 W

50 = 61 – 0.61 W + 0.39 W

11 = 0.22 W

W = 50

100 = D + 50
 D = 50

As by reference to example 11.7

Ln = 3D = 3 x 50 = 150

Vn = Ln + D = 150 + 50 = 200

Thus the top operating line from the equation 11.35

yn = 0.75 xn+1 + 0.1525

Since the feed is all liquid at boiling point, this will run down as increased flux to the plate
below.

Lm = Ln + F =150 +100 = 250

Vm = Lm – W =250 – 50 = 200

Using 11.37 equation for bottom plate

ym = 1.25 xm+1 – 0.0975

Part 2
 Plates calculation;Graph b/w X&Y
1.2

1 1
0.89
vapour composition(Y)

0.8 0.82
0.8
0.75
0.7 0.7
0.66
0.6 0.62 0.6
0.58
0.53 0.5
0.4 0.43 0.4
0.3
0.2 0.2
0.1
0 0
0 0.2 0.4 0.6 0.8 1 1.2

Liquid Composition(X)

y x y

Lewis Sorel Method:

For top fraction

Top operating line

ln D
Yn = * Xn+1 + * Xd
Vn Vn

150 50
Yn = * Xn+1 + * 0.61
200 200

Yn = 0.75 Xn+1 + 0.1525

Bottom operating line

Lm = Ln + F

Lm = 150 + 100

Lm = 250

Vm = Lm – W

Vm = 250 – 50
 Vm = 200

Now

Lm D
Ym = * Xm+1 - * Xw
Vm Vm

250 50
Ym = * Xm+1 – * 0.39
200 200

Ym = 1.25 Xm+1 – 0.0975

For calculation 0f composition and number of plates by lewis Sorel method

XD = yn = 0.61

See equilibrium curve for

Xm at Yn = 0.61

Xn = 0.388

Now

Yn = 0.75 Xn+1 + 0.1525

Yn-1 = 0.291 (0.388) + 0.1525

Yn-1 = 0.4435

Now at Yn-1 = 0.4435

X n-1 = 0.19

c)Plate Efficiency of the Column

N OG
ϵ=
N OL

Where NOG and NOL are given by

y e− y n
N OG=ln ( )
y e− y n−1

x n+1−x e
N OL=ln ( )
x n−x e
Where ye and xe are calculated by

ye = mxn and xe = myn

ye = 0.2067 and xe = 0.3233

So putting values in their respective formulas,

0.2067−0.89
N OG=ln ( )=0.10
0.2067−0.82

0.89−0.3233
N OL=ln ( )=0.17
0.8−0.3233

0.10
ϵ= =0.5882 ×100=58.82%
0.17