Question

A distillation column receives a feed that is 40 mole % n-pentane and 60 mole % n-hexane. The feed is saturated liquid at 2,500 lbmol/hr. The column is at 1 atm. A distillate that is 97 mole % n-pentane is desired. A total condenser is used. The reflux is a saturated liquid. The external reflux ratio is L₀/D = 3. The bottoms from the partial reboiler is 2 mole % n-pentane. Determine the intersection (x_q, y_q) of the q-line with the operating line, the number of equilibrium trays, and the minimum reflux ratio.

Data: Vapor pressure, P^sat, ln P^sat = A ? B/(T + C), where P^sat is in kPa and T is in K.

n-pentane: A = 13.9778, B = 2554.6, C = ?36.2529

n-hexane: A = 14.0568, B = 2825.42, C = ?42.7089

Answer 1

Given

x_F=0.4

x_D=0.97

x_B=0.02

as n pentane is light key component

taking overall material balance

D+B=2500

taking component balance for n -pentane

0.97*D+0.02*B=0.4*2500

0.97*D+0.02*B=1000

put D=2500-B

on solving it we get B=1500 lbmoles/h

& D = 1000 lbmoles/h

as we get distillate is almost pure pentane & bottom pure hexane .Hence

so temp of reflux stream & bottom are the boiling temp. of pentane & hexane resp.

T_D=309 K & T_B=342K

assuming ideal solution we can find equilibrium data can by found by using matlab codes

%

A=[13.9778 14.0568];B =[2554.6 2825.42];C=[-36.2529 -42.7089];

dT=.01;Tb=B./(A-pl)-C;

x=0:0.05:1;

for i=1:21;

xi=x(i);

% Assume a temperature for the buble point calculation

T=xi*Tb(1)+(1-xi)*Tb(2);

% Solve for the bubble point temperature using Newton's method for n=1:20;

f=xi*exp(A(1)-B(1)/(T+C(1)))+(1-xi)*exp(A(2)-B(2)/(T+C(2)))-P;

T1=T+dT;f1=xi*exp(A(1)-B(1)/(T1+C(1)))+(1-xi)*exp(A(2)-B(2)/(T1+C(2)))-P;

fp=(f1-f)/dT;eT=f/fp;T=T-eT;

if abs(eT)<0.001,break, end

end

% Solve for the mole fraction in the vapor phase using equilibrium relation for ideal system

yi=xi*exp(A(1)-B(1)/(T+C(1)))/P;

fprintf('x = %8.5f , y = %8.5f, T(K) = %8.2f\n',xi,yi,T)

end

--------------------------------------------------------------------------------

x	y	T (K)
0.0000	0.0000	342.06
0.05	0.12705	339.40
0.1	0.23699	336.91
0.15	0.33263	334.58
0.2	0.41626	332.39
0.25	0.48975	330.32
0.30	0.55462	328.38
0.35	0.61214	326.53
0.4	0.66335	324.79
0.45	0.70911	323.14
0.5	0.75016	321.56
0.55	0.78711	320.07
0.6	0.82048	318.64
0.65	0.8507	317.28
0.7	0.87816	315.97
0.75	0.90317	314.72
0.8	0.92601	313.53
0.85	0.94692	312.38
0.9	0.96610	311.28
0.95	0.98374	310.22
1	1	309.20

where x = mole fraction of n pentane in liquid

y= mole fraction of n pentane in vapor

Number of equilibrium stages

Making a balance over the top part of the tower

on n-pentane gives the operating line

y_n+1V_n+1 = x_nL_n + x_DD

assume, Vn+1 = V = constant and Ln = L = constant,

we have

y_n+1 = x_nL/V + x_DD/V

From V = L + D = D/V = 1 ? L/V.

In terms of the reflux ratio R = L/D

the operating line becomes

y_n+1=Rx_n/R+1 + x_D/R+1

since R=3

y_n+1=0.75x_n + 0.2425

feed is fed at 30^oC (303K) subcooled liquid

as we know slope of q-line = q/q-1

for subcooled liquid its vertical line

now we have operating line & q-line we can plot it on graph

q line starts from x_F,x_F (0.4,0.4 )

to plote graph first draw equilibrium curve & line x=y

then q-line starts from (0.4,0.4) vertical upward

then draw operating line with y intercept 0.2425 starting from point (0.97,0.97)

now operating line for stripping section can be drawn staring from (0.02,0.02) & passing through intersection of q-line & rectifying operating line.

now we just left with drawing stages

on drawing we get 9 equilibrium trays & 1 partial reboiler