Question 18.5: A mixture of 50 mole percent benzene and toluene is to be se......

x_{F}=0.50\qquad x_{D}=0.98\qquad x_{B}=0.02

From Eq. (18.5)

\frac{D}{F}=\frac{x_{F}-x_{B}}{x_{D}-x_{B}}               (18.5)

{\frac{D}{F}}={\frac{x_{F}-x_{B}}{x_{D}-x_{B}}}={\frac{0.5-0.02}{0.98-0.02}}=0.50

Basis:                F= 100 mol               D = 50 mol             B = 50 mol

From Eq. (18.3)

Total-material balance              F = D + B  (18.3)

R_{D m}={\frac{x_{D}-y^{\prime}}{y^{\prime}-x^{\prime}}}\qquad{\mathrm{based~on~}}{\frac{L}{V}}=\mathrm{const}

For this feed, q = 1.0 and x^{\prime}=x_{F}=0.50
From the equilibrium curve, y’ = 0.72, and

R_{D m}={\frac{0.98-0.72}{0.72-0.50}}=1.18

R_{D}=1.2(1.18)=1.42

R=1.42(50)=71\;\mathrm{mol}

At top of column

V_{1}=R+D=71+50=121\;\mathrm{mol}

Enthalpy balance Choose 80°C as a reference temperature so that the reflux and
distillate product at 80°C have zero enthalpy. For benzene vapor, the enthalpy is the heat of vaporization at 80°C plus the sensible heat of the vapor:

H_{y}=7360+23(T-80){\mathrm{cal/mol}}

For toluene vapor, the enthalpy of vaporization at 80°C is calculated from the value at the boiling point:

\Delta H_{\mathrm{v}}=\Delta H_{\mathrm{v,b}}+(C_{p,l}-C_{p,v})(T_{b}-T)

For toluene at T=80^{\circ}{\mathrm{C}},\,\Delta H_{\mathrm{v}}=7960+(40-33)(30.6)=8174\;{\mathrm{cal/mol}}. For toluene at T^{\circ}{\mathbf C},\,H_{y}=8174+33(T-80).

From Eq. (18.47) with H_{D}=0{\mathrm{~and~}}V_{a}=V_{1}

V_{n+1}H_{y,n+1}=L_{n}H_{x,n}+V_{1}H_{y,1}

Assume the top plate temperature is about 80°C to evaluate H_{y,1}. Since y_{1}= x_{0}=0.98,

H_{y}=0.98(7360)+0.02(8174)=7376\;\mathrm{cal/mol}

{\mathrm{Pick~}}x_{n}=0.5{\mathrm{~and~}}\mathrm{get~}T_{b}=92{}^{\circ}{\mathrm{C~from~Fig.~18.3}}.{\mathrm{~Then}}

H_{x,n}=[(0.5\times33)+(0.5\times40)](92-80)

= 438 cal/mol

To estimate y_{n+1}, use the operating line for constant molal overflow (dashed line in Fig. 18.25):

y_{n+1}\approx0.70\qquad T_{d}\approx93^{\circ}{\mathrm{C~from~Fig.~18.3}}

H_{y,n+1}=0.7[7360+(23\times13)]+0.3[8174+(33\times13)]

= 7942 cal/mol

Since L_{n}=V_{n+1}-D, from Eq. (18.47)

V_{n+1}H_{y,n+1}=L_{n}H_{x,n}+V_{a}H_{y,a}-R H_{D}               (18.47)

V_{n+1}(7942)=(V_{n+1}-50)(438)+121(7376)

V_{n+1}={\frac{870.596}{7504}}=116.0\,{\mathrm{mol}}\qquad L_{n}=66.0\,{\mathrm{mol}}

From Eq. (18.48)

y_{n+1}={\frac{L_{n}x_{n}}{V_{n+1}}}+{\frac{D x_{D}}{V_{n+1}}}.                     (18.48)

y_{n+1}={\frac{66}{116}}(0.50)+{\frac{50(0.98)}{116}}=0.707

This is close enough to 0.70.
A similar calculation for x_{n}=0.7  gives

V_{n+1}=118\qquad L_{n}=68\qquad y_{n+1}=0.818

The operating line, shown in Fig. 18.25 as a solid line, is almost straight but lies above the operating line based on constant molal overflow (dashed line).
To get the vapor rate at the reboiler, an overall balance is made to get q_{r}{\mathrm{:}}

F H_{F}+q_{r}=D H_{D}+B H_{B}+q_{c}

For feed at 92°C,

H_{F}=[(0.5\times33)+(0.5\times40)](92-80)

= 438 cal/mol

For bottoms at 111 °C,

H_{B}=[(0.02\times33)+(0.98\times40)](111-80)

={1236\ {\mathrm{cal/mol}}}

q_{c}=121\times7376=892.496\,\mathrm{cal}

q_{r}=(50\times0)+(50\times1236)+892,496-(100\times438)

= 910,496 cal

An enthalpy balance around the reboiler is then made:

q_{r}+L_{b}H_{x,b}=V_{b}H_{y,b}+B H_{b}

The vapor from the reboiler is about 5 percent benzene at 111 °C, and

H_{y,b}=0.05[7360+(23\times31)]+0.95[8174+(33\times31)]

= 9141 cal/mol

The liquid to the reboiler is about 4 percent benzene at 110°C, and

H_{x,b}=0.04(33\times30)+0.96(40\times30)=1192\;\mathrm{cal/moI}

Since L_{b}=V_{b}+50,\;\mathrm{and}\;H_{y,b}-H_{x,b}=9141-1192=7949\;\mathrm{cal/mol},

V_{b}={\frac{910.496+50(1192)-50(1236)}{7949}}=114.3

L_{b}=114.3+50=164.3

Approximately the same value for {\mathit{V}}_{b} could have been obtained from q_{r} and the heat of vaporization of toluene:

V_{b}\approx{\frac{q_{r}}{\Delta H_{v}}}={\frac{910,496}{7960}}=114.4

Use Eq. (18.50) to get an intermediate value of V_{m+1}:

V_{m+1}H_{y,m+1}=L_{m}H_{x,m}+q_{r}-B H_{B}            (18.50)

For x_{m}=0.4,y_{m+1}=0.55 (from the operating line in Fig. 18.25); also

T_{m}=95^{\circ}C\qquad T_{m+1}=97^{\circ}C

H_{y,m+1}=0.55[7360+23(97-80)]+0.45[8174+33(17)]=8194\ \mathrm{cal/mol}

H_{x,m}=[(0.4\times33)+(0.6\times40)](95-80)

= 558 cal/mol

L_{m}=V_{m+1}+50

From Eq. (18.50),

8194V_{m+1}=558(V_{m+1}+50)+910,496-(1236\times50)

V_{m+1}=114.8\;\mathrm{mol}

L_{m}=164.8\;\mathrm{mol}

Note that in this case there is almost no change in L and V in the stripping section, in contrast to the 7 percent decrease in L in the rectifying section. The lower operating line can be drawn as a straight line to the intersection of the upper operating line and the q line.
Counting steps, about 27 ideal stages are required for this separation, compared to 21 based on the assumption of constant molal overflow. The difference would be smaller if a higher reflux ratio were used. The calculations were based on 1.2 times the nominal value of R_{\mathrm{min}}, but this really corresponds to about 1.1 times the true minimum reflux, as can be seen from Fig. 18.25.