Question

Question 1 Wash-oil containing 7.5 mol% benzene is fed to a 0.5m diameter stripper at 298 K and 1 atm. Benzene-free stearn flows to the tower countercurrently and is used to recover 80% of the benzene by using a steam rate 1.2 times the minimum steam rate. A solute-free wash-oil flow rate of 1850 g/s will be used. Determine the height of the stripper in ft. Under the given operating conditions, the overall mass transfer capacity coefficient is equal to 150 mol/m?.s.AXA. Molecular weight of the wash-oil is 0.26 kg/mol. The equilibrium data for the benzene-wash oil-steam system are as follows: х Y 0 0 0.02 0.073 0.04 0.147 0.06 0.222 0.08 0.326 0.1 0.424 0.12 0.533 0.14 0.667 31.493 ft 21.251 ft 8.602 ft 25.0785 ft 17.697 ft 4.949 ft 12.885 ft

Answer 1

Given the pure wash oil flow rate = L_S = 1850 g/s = 1.85 kg/s = 1.85 / 0.26 = 7.115 mol/s

mol fraction of benzene = 0.075

mol ratio of benzene = X₁ = 0.075 / ( 1 - 0.075 ) = 0.0811

molar flowrate of benzene entering = X₁ L_S = 0.0811 x 7.115 = 0.57689 mol/s

molar flowrate of benzene leaving the stripper = 0.2 x 0.57689 kmol/s = 0.115378 mol/s

mol ratio of benzene leaving the stripper to wash oil leaving X₂ = 0.115378 / 7.115 = 0.016216

mol fraction of benzene in the steam entering = Y₂ = 0

the equiliibrium curve is drawn.

100 0.7 0.6 0.5 0.4 0.3 0.2. 0.1 0 0 10.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16

an line is drawn from the point (X₂ , Y₂ ) i.e ( 0.01622 , 0 ) onto the intersection point of line X₁ = 0.0811 and equillibrium curve.

X = 0.0811 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 X2 = 0.01622 0 0.02 "0.04 0.06 0.08 0.1 0.12 0.14 0.16

the red line is the operating line corresponding to minimum steam rate.

the slope of the line gives minimum steam rate.

slope = L_s / G_smin = 5.1056

L_S = 7.115 mol/s

G_Smin = 7.115 / 5.1056 = 1.394 mol/s

actual steam rate = 1.394 x 1.2 = 1.6723 mol/s

slope of the actual operating line = L_S / G_S = 7.115 / 1.6723 = 4.2546

the actual operating line is drawn using the slope =4.2546 and point ( 0.01622 , 0)

100 0.7 0.6 0.5 0.4 0.3 0.2. 10.1 0 0 0.02 0.04 0.06 0.08. 0.1 0.12 0.14 0.16

the blue line is the equillibrium line and the red line is the actual operating line.
the height of the tower is obtained by using material balance i.e benzene lost from liquid phase = benzene transfered into the gas phase.

LdX = - NĄā Adh

A is the area of the tower = $\pi$ x D² / 4 = $\pi$ x 0.5² / 4 = 0.1963 m²

Nā= Kya (X-X)

L dx = -Kyā A( X - X*)dhe

by separating variables and integrating on both sides

Ls H = { dh = Kya -X1 dx (X-X) X2

X^* is obtained from graph by reading the equillibrium value of Y corresponding to X on the operating line.

the bulk mol fraction of benzene in gas and liquid phases is represented by operating line.

0.8 0.7 0.6 0.5 0.4 0.341 X1 = 0.0811 0.2 0.1 0 X2 = 0.01622 0.02 0.04 X1* = 0.0704 0.06 0.08 0 0.1 0.12 0.14 0.16

the whole interval of X₁ - X₂ is divided into 6 intervals.

the X, Y and the corresponding X^* are found

the integral value 1/(X - X^* ) is found and tablated.

X	Y	X*	(X-X) T
0.01622	0	0	61.65228
0.027033	0.045	0.01265	69.5265
0.037846	0.0899	0.024827	76.811
0.048659	0.13724	0.037506	89.662
0.059472	0.1819	0.04995	105.02
0.070285	0.2269	0.061394	112.4733
0.0811	0.27604	0.0704	93.46

by trapezoidal rule the area of the integral was found

the value of the integral = (0.010813 / 2 ) x ( 69.6523 + 2 ( 69.5265 + 76.811 + 89.662 + 105.02 + 112.4733 ) + 93.46 ) = 1070.6173

the area of the integral = no of transfer units = 5.78829 units

A is the area of the tower = $\pi$ x D² / 4 = $\pi$ x 0.5² / 4 = 0.1963 m²​​​​​​​​​​​​​​

the height of the transfer unit = L_S / K_Xa A = 7.115 / 150 x 0.1963 = 0.2415 m

height = 0.2415 x 5.78829 = 1.3983 m = 4.56 ft

nearer value is 4.949 ft. ( error is due to numerical inetgration)

height of the tower = 4.949 ft ( nearly)