Chapter 10.7, Problem 28P

A separation stream off the main reactor effluent contains almost exclusively ethyl benzene, benzene, and toluene at 1 bar and 100°C. You determine that the stream flow rate is made up of 34 kg/s of benzene, 10 kg/s of toluene, and 57.75 kg/s of the other component. You send this mixture into a flash distillation unit operating at 0.6 bar and 100°C.

1. A. Estimate if this mixture flashes.
2. B. If the mixture flashes, determine the composition and amount of the equilibrium liquid and vapor.
3. C. You send the liquid exiting the flash distillation unit into another flash distillation unit operating at 1.5 bar and 140°C. Determine if this mixture flashes. If so, determine the composition and amounts of the equilibrium phases.
4. D. What percentage of the original benzene that left the reactor is now a vapor (you have to consider both flash units).

Interpretation Introduction

Interpretation:

Estimate if this mixture flashes.

Concept Introduction:

The expression to obtain the mole fraction for component 1 is,

$x_1=\frac{{\dot{n}}_1}{{\dot{n}}_1+{\dot{n}}_2+{\dot{n}}_3}$

Here, molar flow rate of components 1, 2, and 3 is ${\dot{n}}_1,{\dot{n}}_2,\text{and}{\dot{n}}_3$ respectively.

The expression to obtain the mole fraction for component 2 is,

$x_2=\frac{{\dot{n}}_2}{{\dot{n}}_1+{\dot{n}}_2+{\dot{n}}_3}$

The expression to obtain the mole fraction for component 3 is,

$x_3=\frac{{\dot{n}}_3}{{\dot{n}}_1+{\dot{n}}_2+{\dot{n}}_3}$

The Antoine equation for the components 1 and 2 is,

${\log }_{10}\left(P^{sat}\right)=A-\frac{B}{T+C}$

Here, Antoine constants or coefficients are A, B, and C, vapor pressure is $P^{sat}$, and temperature is T.

The expression of bubble point pressure using Raoult’s Law is,

$P=x_1{P}_1^{sat}+x_2{P}_2^{sat}+x_3{P}_3^{sat}$

Here, liquid phase mole fraction for components 1, 2, and 3 is $x_1$, $x_2$, and $x_3$ respectively.

The expression to obtain the dew point pressure is,

$P=\frac{1}{\frac{y_1}{P_1^{sat}}+\frac{y_2}{P_2^{sat}}+\frac{y_3}{P_3^{sat}}}$

Here, vapor phase composition for the components 1, 2, and 3 is $y_1,y_2,\text{and}{y}_3$ respectively.

Explanation of Solution

Convert the mass flow rates into molar flow rates to obtain the composition of the incoming stream by using the molecular weights of the three substances as in Table (1).

Substances	Mass flow rate $\left(\text{kg}/\text{s}\right)$	Molecular weight $\left(\text{kg}/\text{mol}\right)$	Molar flow rate $\left(\text{mol}/\text{s}\right)$
Ethyl benzene (1)	57.75	0.10617	6.13
Benzene (2)	34	0.07811	2.66
Toluene (3)	10	0.09214	0.92

Write the expression to obtain the mole fraction for component 1.

$x_1=\frac{{\dot{n}}_1}{{\dot{n}}_1+{\dot{n}}_2+{\dot{n}}_3}$        (1)

Substitute $6.13\text{mol}/\text{s}$ for ${\dot{n}}_1$, $2.66\text{mol}/\text{s}$ for ${\dot{n}}_2$, and $0.92\text{mol}/\text{s}$ for ${\dot{n}}_3$ in Equation (1).

$\begin{array}{c}{x}_1=\frac{6.13\text{mol}/\text{s}}{6.13\text{mol}/\text{s}+2.66\text{mol}/\text{s}+0.92\text{mol}/\text{s}}\\ =0.632\end{array}$

Write the expression to obtain the mole fraction for component 2.

$x_2=\frac{{\dot{n}}_2}{{\dot{n}}_1+{\dot{n}}_2+{\dot{n}}_3}$        (2)

Substitute $6.13\text{mol}/\text{s}$ for ${\dot{n}}_1$, $2.66\text{mol}/\text{s}$ for ${\dot{n}}_2$, and $0.92\text{mol}/\text{s}$ for ${\dot{n}}_3$ in Equation (2).

$\begin{array}{c}{x}_2=\frac{2.66\text{mol}/\text{s}}{6.13\text{mol}/\text{s}+2.66\text{mol}/\text{s}+0.92\text{mol}/\text{s}}\\ =0.274\end{array}$

Write the expression to obtain the mole fraction for component 3.

$x_3=\frac{{\dot{n}}_3}{{\dot{n}}_1+{\dot{n}}_2+{\dot{n}}_3}$        (3)

Substitute $6.13\text{mol}/\text{s}$ for ${\dot{n}}_1$, $2.66\text{mol}/\text{s}$ for ${\dot{n}}_2$, and $0.92\text{mol}/\text{s}$ for ${\dot{n}}_3$ in Equation (3).

$\begin{array}{c}{x}_3=\frac{0.92\text{mol}/\text{s}}{6.13\text{mol}/\text{s}+2.66\text{mol}/\text{s}+0.92\text{mol}/\text{s}}\\ =0.094\end{array}$

Write the Antoine equation for the components 1 and 2.

${\log }_{10}\left(P^{sat}\right)=A-\frac{B}{T+C}$        (4)

Refer Appendix E, “Antoine Coefficients”; obtain the constants A, B, and C for components as in Table (1).

Constants	Ethyl benzene (1)	benzene (2)	toluene (3)
A	6.95719	6.90565	6.95464
B	1,424.255	1,211.033	1,344.800
C	213.21	220.790	219.48

Substitute $100°\text{C}$ for T, 6.95719 for A, 1,424.255 for B, and 213.21 for C for component 1 in Equation (4).

$\begin{array}{l}{\log }_{10}\left(P_1{}^{sat}\right)=6.95719-\frac{1,424.255}{100°\text{C}+213.21}\\ P_1^{sat}={10}^{2.4099}\\ P_1^{sat}=256.98\text{mmHg}\end{array}$

Substitute $100°\text{C}$ for T, 6.90565 for A, 1,211.033 for B, and 220.790 for C for component 2 in Equation (4).

$\begin{array}{l}{\log }_{10}\left(P_2{}^{sat}\right)=6.90565-\frac{1,211.033}{100°\text{C}+220.790}\\ P_2^{sat}={10}^{3.130492}\\ P_2^{sat}=1,350.49\text{mmHg}\end{array}$

Substitute $100°\text{C}$ for T, 6.95464 for A, 1,344.800 for B, and 219.48 for C for component 3 in Equation (4).

$\begin{array}{l}{\log }_{10}\left(P_3{}^{sat}\right)=6.95464-\frac{1,344.800}{100°\text{C}+219.48}\\ P_3^{sat}={10}^{2.7453}\\ P_3^{sat}=556.29\text{mmHg}\end{array}$

Write the expression of bubble point pressure using Raoult’s Law.

$P=x_1{P}_1^{sat}+x_2{P}_2^{sat}+x_3{P}_3^{sat}$        (5)

Substitute 0.632 for $x_1$, 0.274 for $x_2$, 0.094 for $x_3$, 256.98 mmHg for $P_1^{sat}$, 1,350.49 mmHg for $P_2^{sat}$, and 556.29 mmHg for $P_3^{sat}$ in Equation (5).

$\begin{array}{c}P=\left(0.632\right)\left(256.98\text{mmHg}\right)+0.274\left(1,350.49\text{mmHg}\right)+0.094\left(556.29\text{mmHg}\right)\\ =584.52\text{mmHg}\end{array}$

Write the expression to obtain the dew point pressure.

$P=\frac{1}{\frac{y_1}{P_1^{sat}}+\frac{y_2}{P_2^{sat}}+\frac{y_3}{P_3^{sat}}}$        (6)

Substitute 0.632 for $y_1$, 0.274 for $y_2$, 0.094 for $y_3$, 256.98 mmHg for $P_1^{sat}$, 1,350.49 mmHg for $P_2^{sat}$, and 556.29 mmHg for $P_3^{sat}$ in Equation (6).

$\begin{array}{c}P=\frac{1}{\frac{0.632}{256.98\text{mmHg}}+\frac{0.274}{1,350.49\text{mmHg}}+\frac{0.094}{556.29\text{mmHg}}}\\ =353.27\text{mmHg}\end{array}$

Since the working pressure (0.6 bar = 450.04 mmHg) is between the bubble (584.52 mmHg) and dew points (353.27 mmHg) pressure, it will flash at this temperature.

Interpretation Introduction

Interpretation:

If the mixture flashes, determine the composition and amount of the equilibrium liquid and vapor.

Concept Introduction:

The expression of overall material balance is,

$\dot{N}=\dot{L}+\dot{V}$

Here, molar flow rate of the mixture is $\dot{L}$ and $\dot{V}$ respectively.

The expression of material balance on ethyl benzene is,

$w_1\dot{N}=x_1\dot{L}+y_1\dot{V}$

Here, the composition of the inlet stream 1 is $w_1$.

The expression of material balance on benzene is,

$w_2\dot{N}=x_2\dot{L}+y_2\dot{V}$

Here, the composition of the inlet stream 2 is $w_2$.

The Raoult’s law for ethyl benzene is,

$x_1{P}_1^{sat}=y_1{P}_2^{sat}$

Here, vapor pressure for component 1 is $P_1^{sat}$ and vapor phase composition for component 1 is $y_1$.

The Raoult’s law for benzene is,

$x_2{P}_2^{sat}=y_2{P}_3^{sat}$

Here, vapor pressure for component 2 is $P_2^{sat}$ and vapor phase composition for component 2 is $y_2$.

The Raoult’s law for toluene is,

$\left(1-x_1-x_2\right)P_3^{sat}=\left(1-y_1-y_2\right)P_1^{sat}$

Explanation of Solution

Write the expression of overall material balance.

$\dot{N}=\dot{L}+\dot{V}$        (7)

Substitute $9.708\frac{\text{mol}}{\sec }$ for $\dot{N}$ in Equation (7).

$9.708\frac{\text{mol}}{\sec }=\dot{L}+\dot{V}$        (8)

Write the expression of material balance on ethyl benzene.

$w_1\dot{N}=x_1\dot{L}+y_1\dot{V}$        (9)

Substitute 0.632 for $w_1$ and $9.708\frac{\text{mol}}{\sec }$ for $\dot{N}$ in Equation (9).

$\left(0.632\right)\left(9.708\frac{\text{mol}}{\sec }\right)=x_1\dot{L}+y_1\dot{V}$        (10)

Write the expression of material balance on benzene.

$w_2\dot{N}=x_2\dot{L}+y_2\dot{V}$        (11)

Substitute 0.274 for $w_2$ and $9.708\frac{\text{mol}}{\sec }$ for $\dot{N}$ in Equation (11).

$\left(0.274\right)\left(9.708\frac{\text{mol}}{\sec }\right)=x_2\dot{L}+y_2\dot{V}$        (12)

Write the Raoult’s law for ethyl benzene.

$x_1{P}_1^{sat}=y_1{P}_2^{sat}$        (13)

Substitute 256.98 mmHg for $P_1^{sat}$ and 1,350.49 mmHg for $P_2^{sat}$ in Equation (13).

$x_1\left(256.98\text{mmHg}\right)=y_1\left(1,350.49\text{mmHg}\right)$        (14)

Write the Raoult’s law for benzene.

$x_2{P}_2^{sat}=y_2{P}_3^{sat}$        (15)

Substitute 1350.49 mmHg for $P_2^{sat}$ and 556.29 mmHg for $P_3^{sat}$ in Equation (15).

$x_2\left(1,350.49\text{mmHg}\right)=y_2\left(556.29\text{mmHg}\right)$        (16)

Write the Raoult’s law for toluene.

$\left(1-x_1-x_2\right)P_3^{sat}=\left(1-y_1-y_2\right)P_1^{sat}$        (17)

Substitute 256.98 mmHg for $P_1^{sat}$ and 556.29 mmHg for $P_3^{sat}$ in Equation (17).

$\left(1-x_1-x_2\right)556.29\text{mmHg}=\left(1-y_1-y_2\right)256.98\text{mmHg}$        (18)

Solve equations (14), (16), and (18) and obtain the values of $x_1,x_2,y_1,\text{and}{y}_2$ as $0.760$, $0.153$, $0.434$, and $0.458$ respectively.

Substitute 0.760 for $x_1$, 0.153 for $x_2$, 0.434 for $y_1$, and 0.458 for $y_2$ in Equations (8), (10) and (12) to obtain the values of $\dot{L}$ and $\dot{V}$ as $5.873\frac{\text{mol}}{\sec }$ and $3.835\frac{\text{mol}}{\sec }$ respectively.

Interpretation Introduction

Interpretation:

Determine if this mixture flashes. If so, determine the composition and amounts of the equilibrium phases.

Concept Introduction:

The expression of overall material balance is,

$\dot{N}=\dot{L}+\dot{V}$

The expression of material balance on ethyl benzene is,

$w_1\dot{N}=x_1\dot{L}+y_1\dot{V}$

Here, the composition of the inlet stream 1 is $w_1$.

The expression of material balance on benzene is,

$w_2\dot{N}=x_2\dot{L}+y_2\dot{V}$

Here, the composition of the inlet stream 2 is $w_2$.

The Raoult’s law for ethyl benzene is,

$x_1{P}_1^{sat}=y_1{P}_2^{sat}$

The Raoult’s law for benzene is,

$x_2{P}_2^{sat}=y_2{P}_3^{sat}$

The Raoult’s law for toluene is,

$\left(1-x_1-x_2\right)P_3^{sat}=\left(1-y_1-y_2\right)P_1^{sat}$

Explanation of Solution

The bubble point pressure is 450.33 mmHg, while the dew point pressure is 309.88 mmHg. The system pressure is 0.5 bar (375.03 mmHg) and hence the system will flush one more time.

Write the expression of overall material balance.

$\dot{N}=\dot{L}+\dot{V}$        (19)

Substitute $5.873\frac{\text{mol}}{\sec }$ for $\dot{N}$ in Equation (19).

$9.708\frac{\text{mol}}{\sec }=\dot{L}+\dot{V}$        (20)

Write the expression of material balance on ethyl benzene.

$w_1\dot{N}=x_1\dot{L}+y_1\dot{V}$        (21)

Substitute 0.760 for $w_1$ and $5.873\frac{\text{mol}}{\sec }$ for $\dot{N}$ in Equation (21).

$\left(0.760\right)\left(5.873\frac{\text{mol}}{\sec }\right)=x_1\dot{L}+y_1\dot{V}$        (22)

Write the expression of material balance on benzene.

$w_2\dot{N}=x_2\dot{L}+y_2\dot{V}$        (23)

Substitute 0.153 for $w_2$ and $5.873\frac{\text{mol}}{\sec }$ for $\dot{N}$ in Equation (23).

$\left(0.153\right)\left(5.873\frac{\text{mol}}{\sec }\right)=x_2\dot{L}+y_2\dot{V}$        (24)

Write the Raoult’s law for ethyl benzene.

$x_1{P}_1^{sat}=y_1{P}_2^{sat}$        (25)

Substitute 256.98 mmHg for $P_1^{sat}$ and 1,350.49 mmHg for $P_2^{sat}$ in Equation (25).

$x_1\left(256.98\text{mmHg}\right)=y_1\left(1,350.49\text{mmHg}\right)$        (26)

Write the Raoult’s law for benzene.

$x_2{P}_2^{sat}=y_2{P}_3^{sat}$        (27)

Substitute 1,350.49 mmHg for $P_2^{sat}$ and 556.29 mmHg for $P_3^{sat}$ in Equation (27).

$x_2\left(1,350.49\text{mmHg}\right)=y_2\left(556.29\text{mmHg}\right)$        (28)

Write the Raoult’s law for toluene.

$\left(1-x_1-x_2\right)P_3^{sat}=\left(1-y_1-y_2\right)P_1^{sat}$        (29)

Substitute 256.98 mmHg for $P_1^{sat}$ and 556.29 mmHg for $P_3^{sat}$ in Equation (29).

$\left(1-x_1-x_2\right)556.29\text{mmHg}=\left(1-y_1-y_2\right)256.98\text{mmHg}$        (30)

Solve equations (26), (28), and (30) and obtain the values of $x_1,x_2,y_1,\text{and}{y}_2$ as $0.837$, $0.087$, $0.573$, and $0.314$ respectively.

Substitute 0.837 for $x_1$, 0.087 for $x_2$, 0.573 for $y_1$, and 0.314 for $y_2$ in Equations (20), (22), and (24) and obtain the values of $\dot{L}$ and $\dot{V}$ as $4.164\frac{\text{mol}}{\sec }$ and $1.709\frac{\text{mol}}{\sec }$ respectively.

Interpretation Introduction

Interpretation:

What percentage of the original benzene that left the reactor is now a vapor?

Concept Introduction:

The expression to obtain the amount of benzene in the vapor exiting the 1^st flash distillation unit is,

$\text{Benzene}\text{amount}=\dot{V}\times y_2$

The expression to obtain the amount of benzene in the vapor exiting the 2^nd flash distillation unit is,

$\text{Benzene}\text{amount}=\dot{V}\times y_2$

The expression to obtain the percentage of benzene in vapor is,

$\%\text{benzene}\text{in}\text{vapor}=\left[\frac{\text{Benzene}\text{}\text{amount for 1st + 2nd flash distillation}}{\text{Original}\text{benzene}}\right]\times 100\%$

Explanation of Solution

The original benzene that left the reactor is $2.66\frac{\text{mol}}{\sec }$.

Write the expression to obtain the amount of benzene in the vapor exiting the 1^st flash distillation unit.

$\text{Benzene}\text{amount}=\dot{V}\times y_2$        (31)

Substitute $3.835\text{mol}/\sec$ for $\dot{V}$ and 0.458 for $y_2$ in Equation (31).

$\begin{array}{c}\text{Benzene}\text{amount}=3.835\text{mol}/\sec \times 0.458\\ =1.758\text{mol}/\sec \end{array}$

Write the expression to obtain the amount of benzene in the vapor exiting the 2^nd flash distillation unit.

$\text{Benzene}\text{amount}=\dot{V}\times y_2$        (32)

Substitute $1.709\text{mol}/\sec$ for $\dot{V}$ and 0.314 for $y_2$ in Equation (32).

$\begin{array}{c}\text{Benzene}\text{amount}=1.709\text{mol}/\sec \times 0.314\\ =0.536\text{mol}/\sec \end{array}$

Write the expression to obtain the percentage of benzene in vapor.

$\%\text{benzene}\text{in}\text{vapor}=\left[\frac{\text{Benzene}\text{}\text{amount for 1st + 2nd flash distillation}}{\text{Original}\text{benzene}}\right]\times 100\%$        (33)

Substitute $1.758\text{mol}/\sec$ for $\text{Benzene}\text{}\text{amount for 1st}$ flash distillation unit, $0.536\text{mol}/\sec$ for 2^nd flash distillation unit, and $2.66\text{mol}/\text{s}$ for $\text{Original}\text{benzene}$ in Equation (33).

$\begin{array}{c}\%\text{benzene}\text{in}\text{vapor}=\left[\frac{1.758\text{mol}/\sec \text{ + }0.536\text{mol}/\sec }{2.66\text{mol}/\text{s}}\right]\times 100\%\\ =86.2\%\end{array}$

Thus, the percentage of benzene in vapor is $86.2\%$.