**Batch Distillation: Thermodynamic Efficiency**

José C. Zavala-Loría\* and Asteria Narváez-García

*Universidad Autónoma del Carmen (UNACAR), DES-DAIT: Facultad de Química, Cd. del Carmen, Cam. México* 

### **1. Introduction**

90 Distillation – Advances from Modeling to Applications

[18] Zhang Junliang, Wang Feng, Peng Weicai, Xiao Fukui, Wei Wei, Sun Yuhan. Process

39(6): 646-650

Simulation for Separation of Dimethyl Carbonate and Methanol Through Atmospheric-Pressurized Rectification. PETROCHEMICAL TECHNOLOGY, 2010,

> The batch distillation is a separation processes that requires great amounts of energy. Due to high energy costs, the study of energy consumption is of great interest in this process (Zavala et al. 2007). According to Luyben (1990), the energy consumption increases when operation occurs in a batch manner. Determining how efficient is the heat transfer under specific conditions and to modify them in order to find how efficient the heat is used, is an important task.

> The analysis of thermodynamic efficiency in a batch distillation column has been presented by Kim and Diwekar (2000), Zavala-Loría (2004), Zavala et al. (2007), Zavala & Coronado (2008) and Zavala et al. (2011). The first work only developed expressions to calculate thermodynamic efficiency while the rest of the works developed expressions to calculate thermodynamic efficiency and have applied them to a new problem of optimal control: Maximum thermodynamic efficiency.

#### **2. Description of the process**

For this study we used a conventional batch distillation column consisting of:


Figure 1 Shows a conventional batch column like the one used for this study.

Mass balances resulting from the process shown in Figure 1, allow us to obtain the mathematical model in Table 1. The model considers the following elements:


<sup>\*</sup> Corresponding Author

Batch Distillation: Thermodynamic Efficiency 93

Considering ideal mixtures, the vapor-liquid equilibrium (VLE) can be obtained from

*<sup>x</sup> <sup>P</sup>* ; *i* = 1, 2,…,*n* (6)

; *i* = 1, 2,…,*n* (7)

*<sup>i</sup>* represents the activity coefficient, *Psat* is the saturation

; *i* = 1, 2,…,*n* (8)

*<sup>t</sup>*) of a process based on availability or exergy

(9)

 *<sup>i</sup> sat <sup>i</sup> <sup>i</sup> i <sup>y</sup> <sup>P</sup> <sup>K</sup>*

*<sup>y</sup> <sup>P</sup> <sup>K</sup> x P* 

 *<sup>i</sup> sat <sup>i</sup> i i i*

obtained with an equation of state (Redlich-Kwong, Soave, Peng-Robinson, etc.)

ln *sat <sup>i</sup> i i*

Coefficients *Ai*, *Bi* and *Ci* appear in literature related to this area of study.

*t*

where *W* is the work and *LW* is the total loss of work.

*<sup>B</sup> P A*

*i*

Where *K*(*<sup>i</sup>*) is the constant of the VLE, and *y*(*<sup>i</sup>*) is the mole fraction of the vapor phase, *x*(*i*) is the

represents the system pressure. Activity coefficients can be obtained with a solution model (Wilson, NRTL, etc.) or the Chao-Seader correlation. The fugacity coefficients can be

*<sup>i</sup>* denotes the partial fugacity coefficient, all referring to component *i*, and *P*

*sat* can be obtained with Antoine's equation.

*i*

*T C*

min min

Since the minimum work is determined by changes in exergy, they can be determined using the First and Second Law of Thermodynamics. Figure 2 shows the control volume used to

Figure 2 shows that the process can exchange energy with the environment but does not perform any mechanical work. The energy balance (enthalpy) given by the First Law of Thermodynamics, considering the reboiler, the column of trays and the condenser-reflux

> 1 *<sup>n</sup> sist <sup>B</sup> t t <sup>i</sup> D rt*

*<sup>m</sup>* (10)

*i d I d BI d HI dHI dt dt dt dt*

obtain the equations that represent thermodynamic efficiency in batch distillation.

*W W W W LW*

min

*total sep sep*

For non-ideal mixtures, we can use the following equation:

Raoult's law equation:

mole fraction of the liquid phase,

At standard or low pressures *Pi*

**3. Thermodynamic efficiency** 

By definition, the thermodynamic efficiency (

pressure and

is defined as:

drum, is:

where:

Fig. 1. Conventional batch distillation column.

	- Vapor
	- Liquid.
	- Stages
	- Condenser-Reflux drum.

$$\frac{dB}{dt} = -D\_\iota = -\frac{V}{R\_\iota + 1} \tag{1}$$

$$\frac{d\mathbf{x}\_{B}^{(\ell)}}{dt} = \left(\frac{V}{B}\right) \left\{ \mathbf{x}\_{B}^{(\ell)} - \mathbf{y}\_{B}^{(\ell)} + \left(\frac{L}{V}\right) \left[\mathbf{x}\_{1}^{(\ell)} - \mathbf{x}\_{B}^{(\ell)}\right] \right\} \tag{2}$$

$$\frac{d\mathbf{x}\_j^{(i)}}{dt} = \left(\frac{V}{H\_j}\right) \left[\mathbf{y}\_{j-1}^{(i)} - \mathbf{y}\_j^{(i)} + \left(\frac{L}{V}\right) \left[\mathbf{x}\_{j+1}^{(i)} - \mathbf{x}\_j^{(i)}\right]\right] \tag{3}$$

$$\vdots \quad i = 1, \dots, n \quad \text{ } j = 1, \dots, N$$

$$\frac{d\mathbf{x}\_D^{(i)}}{dt} = \left(\frac{V}{H\_D}\right) \left[\mathbf{y}\_n^{(i)} - \mathbf{x}\_D^{(i)}\right] \tag{4}$$
 
$$\vdots \text{ } i = 1, \dots, n$$

$$\sum\_{i=1}^{n} \mathbf{y}\_{j}^{(i)} = \sum\_{i=1}^{n} \mathbf{K}\_{j}^{(i)} \mathbf{x}\_{j}^{(i)} = 1 \quad ; \; i = 1, \ldots, n \; ; \; j = 1, \ldots, N \tag{5}$$

Table 1. Mathematical model of a batch distillation column (*n*=number of components and *N*=number of equilibrium stages).

1 *<sup>t</sup> t*

1

; *i n* 1, ,

; *i n* 1, , ; *<sup>j</sup>* 1, , *<sup>N</sup>*

; *i n* 1, ,

*y Kx i n j N*

1 ; 1, , ; 1, ,

Table 1. Mathematical model of a batch distillation column (*n*=number of components and

*n D*

*dt R* (1)

(2)

(3)

(4)

(5)

*dB <sup>V</sup> <sup>D</sup>*

*B B B*

*B ii ii*

 1 1

> *<sup>i</sup> D i i*

> > *D*

*y x dt H*

*jj jj*

*xy xx dt B <sup>V</sup>*

*dx V L*

*j ii ii*

*yy xx dt H <sup>V</sup>*

1 1

*i i*

*N*=number of equilibrium stages).

*dx V*

*n n i ii j jj*

Fig. 1. Conventional batch distillation column.

 Constant flows: Vapor Liquid.

Stages

*i*

Adiabatic column

Constant liquid holdup:

*i*

*j*

*dx V L*

Condenser-Reflux drum.

Considering ideal mixtures, the vapor-liquid equilibrium (VLE) can be obtained from Raoult's law equation:

$$K^{(i)} = \frac{y^{(i)}}{x^{(i)}} = \frac{P\_i^{sat}}{P} \; ; \; i = 1, 2, \dots, m \tag{6}$$

For non-ideal mixtures, we can use the following equation:

$$K^{(i)} = \frac{y^{(i)}}{\mathbf{x}^{(i)}} = \frac{\mathcal{Y}\_i P\_i^{sat}}{\widehat{\phi\_i} P} \; ; \; i = 1, 2, \dots, n \tag{7}$$

Where *K*(*<sup>i</sup>*) is the constant of the VLE, and *y*(*<sup>i</sup>*) is the mole fraction of the vapor phase, *x*(*i*) is the mole fraction of the liquid phase, *<sup>i</sup>* represents the activity coefficient, *Psat* is the saturation pressure and *<sup>i</sup>* denotes the partial fugacity coefficient, all referring to component *i*, and *P* represents the system pressure. Activity coefficients can be obtained with a solution model (Wilson, NRTL, etc.) or the Chao-Seader correlation. The fugacity coefficients can be obtained with an equation of state (Redlich-Kwong, Soave, Peng-Robinson, etc.)

At standard or low pressures *Pi sat* can be obtained with Antoine's equation.

$$\ln\left[P\_i^{\text{sat}}\right] = A\_i - \frac{B\_i}{T + C\_i} \; ; i = 1 \; \text{2} \; ... \; \text{m} \tag{8}$$

Coefficients *Ai*, *Bi* and *Ci* appear in literature related to this area of study.

#### **3. Thermodynamic efficiency**

By definition, the thermodynamic efficiency (*<sup>t</sup>*) of a process based on availability or exergy is defined as:

$$\eta\_t = \left(\frac{\mathcal{W}\_{\text{min}}}{\mathcal{W}\_{\text{total}}}\right)\_{\text{sep}} = \left(\frac{\mathcal{W}\_{\text{min}}}{\mathcal{W}\_{\text{min}} + L\mathcal{W}}\right)\_{\text{sep}}\tag{9}$$

where *W* is the work and *LW* is the total loss of work.

Since the minimum work is determined by changes in exergy, they can be determined using the First and Second Law of Thermodynamics. Figure 2 shows the control volume used to obtain the equations that represent thermodynamic efficiency in batch distillation.

Figure 2 shows that the process can exchange energy with the environment but does not perform any mechanical work. The energy balance (enthalpy) given by the First Law of Thermodynamics, considering the reboiler, the column of trays and the condenser-reflux drum, is:

$$\frac{d\left(mI\right)\_{\text{sist}}}{dt} = \frac{d\left(BI\_B\right)}{dt} + \sum\_{i=1}^{n} \frac{d\left(H\_I I\_t\right)\_i}{dt} + \frac{d\left(H\_D I\_{rt}\right)}{dt} \tag{10}$$

where:

Batch Distillation: Thermodynamic Efficiency 95

*B C d m I TS T T Q Q D I TS Td S dt <sup>T</sup> <sup>T</sup>* 

the availability is represented as 0 *A I TS* , then:

*mA*

Another way to estimate the value of *sist d m*

*mA*

 *sist d m dt*

Equation (10):

component ( *Achem* ), i.e.:

<sup>0</sup> 0 0

 0 0 <sup>0</sup> 1 1 *sist*

*B C <sup>d</sup> T T <sup>Q</sup> Q D LW dt T <sup>T</sup>* 

Where the loss of work is defined as *LW T dS* <sup>0</sup> *irr* . The terms of the right hand side of Equation (16) represent the difference between the exergy or the availability of flows entering and leaving the system. The term on the left hand side is the change of exergy in

> 0 0 1 1 *sist B CD*

*T T d m*

*<sup>A</sup>* can be calculated, if discretised, by multiplying Equation (12) by *T*0 minus

 1

*B t t D rt i*

*B*

*B*

*<sup>A</sup> A A <sup>A</sup>* (19)

*dt dt A A <sup>A</sup>* (20)

*AA A phis chem* (21)

*B HH*

*n*

 *mA A AA*

*dt*

*B B sist*

*B D*

*d m dB d dB B dt dt dt dt* 

*B B*

According to Kim and Diwekar (2000) and Zavala-Loría and Coronado-Velasco (2008), the total exergy can be calculated from its physical component ( *Aphis* ) and its chemical

*dB d*

*B C <sup>d</sup> T T Q Q D Td S dt T <sup>T</sup>* 

0 0 1 1 *sist*

*B C*

*sist i*

bottom of the column. Applying an exergy balance in the reboiler, we have:

Introducing Equation (1) from Table (1) in Equation (19), we obtain:

*t t*

*LW Q Q D <sup>T</sup> <sup>T</sup> dt* 

the system. Finding the values in Equation (16), the work loss is:

0 0 1 1 *<sup>B</sup> C DD irr*

*B C D irr*

*B CD*

(14)

*A* (15)

*A* (16)

*A* (17)

(18)

*A*

*<sup>A</sup>* is by applying an exergy balance at the

$$\frac{d\left(mI\right)\_{sist}}{dt} = Q\_0 + Q\_\mathcal{B} - Q\_\mathcal{C} - DI\_\mathcal{D} \tag{11}$$

According to the Second Law of Thermodynamics, the entropy balance is:

$$\frac{d\left(mS\right)\_{\text{sist}}}{dt} = \frac{d\left(BS\_B\right)}{dt} + \sum\_{i=1}^{n} \frac{d\left(H\_t S\_t\right)\_i}{dt} + \frac{d\left(H\_D S\_{rt}\right)}{dt} \tag{12}$$

where:

$$\frac{d\left(mS\right)\_{sist}}{dt} = \frac{Q\_0}{T\_0} + \frac{Q\_B}{T\_B} - \frac{Q\_C}{T\_C} - DS\_D + d\left(S\_{irr}\right) \tag{13}$$

Fig. 2. Control volume for batch process.

*I* represents enthalpy, *S* is entropy, *Q* is the amount of heat, *C* represents the condenser, *D* represents the distillate, *B* represents the bottoms, *m* is the system mass, 0 is the reference state, *t* is the stage or tray and *T* is the temperature. The available work can be obtained if we combine equations (11) and (13). To do so, equation (13) must be multiplied by the temperature of the reference state *T*0 (it is considered that the reference state is liquid at 25 °C and 1.0 atm):

$$\frac{d\left(\pi T\_0 S\right)\_{\text{sist}}}{dt} = Q\_0 + \frac{T\_0}{T\_B} Q\_B - \frac{T\_0}{T\_C} Q\_C - DT\_0 S\_D + T\_0 d\left(S\_{irr}\right)$$

$$\frac{d\left(\pi I\right)\_{\text{sist}}}{dt} - \frac{d\left(\pi T\_0 S\right)\_{\text{sist}}}{dt} = Q\_0 + Q\_B - Q\_C - DI\_D - Q\_0 - \frac{T\_0}{T\_B} Q\_B + \frac{T\_0}{T\_C} Q\_C + DT\_0 S\_D - T\_0 d\left(S\_{irr}\right)$$

*B C*

$$\frac{d\left[m\left(I - T\_0 S\right)\right]}{dt} = \left(1 - \frac{T\_0}{T\_B}\right)Q\_B - \left(1 - \frac{T\_0}{T\_C}\right)Q\_C - D\left(I\_D - T\_0 S\_D\right) - T\_0 d\left(S\_{irr}\right) \tag{14}$$

the availability is represented as 0 *A I TS* , then:

$$\frac{d\left(m\mathcal{A}\right)\_{\text{sist}}}{dt} = \left(1 - \frac{T\_0}{T\_B}\right)Q\_B - \left(1 - \frac{T\_0}{T\_C}\right)Q\_C - D\mathcal{A}\_D - T\_0 d\left(S\_{irr}\right) \tag{15}$$

$$\frac{d\left(m\mathcal{A}\right)\_{sist}}{dt} = \left(1 - \frac{T\_0}{T\_B}\right)Q\_B - \left(1 - \frac{T\_0}{T\_C}\right)Q\_C - D\mathcal{A}\_D - L\mathcal{W} \tag{16}$$

Where the loss of work is defined as *LW T dS* <sup>0</sup> *irr* . The terms of the right hand side of Equation (16) represent the difference between the exergy or the availability of flows entering and leaving the system. The term on the left hand side is the change of exergy in the system. Finding the values in Equation (16), the work loss is:

$$L\mathcal{W} = \left(1 - \frac{T\_0}{T\_B}\right) Q\_B - \left(1 - \frac{T\_0}{T\_C}\right) Q\_C - D\mathcal{A}\_D - \frac{d\left(m\mathcal{A}\right)\_{ist}}{dt} \tag{17}$$

*sist d m*

94 Distillation – Advances from Modeling to Applications

*Q Q Q DI dt*

 1 *<sup>n</sup> sist <sup>B</sup> t t <sup>i</sup> D rt*

<sup>0</sup>

*B C d S Q Q <sup>Q</sup> DS d S*

*I* represents enthalpy, *S* is entropy, *Q* is the amount of heat, *C* represents the condenser, *D* represents the distillate, *B* represents the bottoms, *m* is the system mass, 0 is the reference state, *t* is the stage or tray and *T* is the temperature. The available work can be obtained if we combine equations (11) and (13). To do so, equation (13) must be multiplied by the temperature of the reference state *T*0 (it is considered that the reference state is liquid at 25

> <sup>0</sup> 0 0 0 0 0

<sup>0</sup> 0 0

*B C d TS T T Q Q Q DT S T d S dt T T <sup>m</sup>*

*d I d TS T T Q Q Q DI Q Q Q DT S T d S dt dt T T <sup>m</sup> <sup>m</sup>*

*B C D irr*

0 0 0 0

*BC D B C D irr B C*

*i d S d BS d HS dHS dt dt dt dt*

*BC D*

*<sup>m</sup>* (11)

*<sup>m</sup>* (12)

*D irr*

*<sup>m</sup>* (13)

0

*d I*

where:

Fig. 2. Control volume for batch process.

*sist sist*

*sist*

°C and 1.0 atm):

*sist*

According to the Second Law of Thermodynamics, the entropy balance is:

0 *sist B C*

*dt T T T*

*dt <sup>A</sup>* can be calculated, if discretised, by multiplying Equation (12) by *T*0 minus Equation (10):

$$\frac{\Delta\left(m\mathcal{A}\right)\_{\text{sist}}}{\Delta t} = \frac{\Delta\left(B\mathcal{A}\_{\text{B}}\right) + \sum\_{i=1}^{n} \Delta\left(H\_{t}\mathcal{A}\_{t}\right)\_{i} + \Delta\left(H\_{D}\mathcal{A}\_{rt}\right)}{\Delta t} \tag{18}$$

Another way to estimate the value of *sist d m dt <sup>A</sup>* is by applying an exergy balance at the bottom of the column. Applying an exergy balance in the reboiler, we have:

$$\frac{d\left(m\mathcal{A}\right)\_{\text{sist}}}{dt} = \frac{d\left(B\mathcal{A}\_{\text{B}}\right)}{dt} = B\frac{d\left(\mathcal{A}\_{\text{B}}\right)}{dt} + \mathcal{A}\_{\text{B}}\frac{d\left(B\right)}{dt} \tag{19}$$

Introducing Equation (1) from Table (1) in Equation (19), we obtain:

$$\frac{d\left(B\mathcal{A}\_{\mathcal{B}}\right)}{dt} = B\frac{d\left(\mathcal{A}\_{\mathcal{B}}\right)}{dt} - D\mathcal{A}\_{\mathcal{B}}\tag{20}$$

According to Kim and Diwekar (2000) and Zavala-Loría and Coronado-Velasco (2008), the total exergy can be calculated from its physical component ( *Aphis* ) and its chemical component ( *Achem* ), i.e.:

$$\mathcal{A} = \mathcal{A}\_{\text{phys}} + \mathcal{A}\_{\text{chem}} \tag{21}$$

Batch Distillation: Thermodynamic Efficiency 97

,

*V L*

*i B B i i B*

*x*

,

Substituting Equation (28) on the right hand side of Equation (20) for an ideal mixture, and

*d B <sup>L</sup> VRT x y x x <sup>x</sup>*

*B ii ii i*

*B ii ii i*

*A*

0 , 1

*A*

, 0 ln *i i D D phis D D i*

,0 , ln *i i D D phis D D i D i RT x x* 

The exergy transfer associated with the transmission of energy as heat in the process can be calculated by the energy balances in the reboiler and condenser. Considering that *Hvap* is the same for each component and is not related to the temperature of the process, the Clausius-Clapeyron equation can be used to calculate *Hvap*, and the first two terms on the

right hand side of Equation (16) can be calculated by the following equation;

The exergy of the current production (dome) that will be used in Equation (6) for an ideal

*dt*

 

Equation (29) on the right hand side of Equation (20) for a non-ideal mixture yields:

0 1

*d B <sup>L</sup> VRT x y x x <sup>x</sup>*

*x d*

*i i B i B*

1 ln

*dt dt dt*

*B B i B i i i i B B i B*

*i*

*i*

*dt V*

*dt V*

,

*i i B*

*dx x d*

*xy xx x B V*

, 0

*B B B i B B*

*D RT x x*

*B phis B B i*

0

 

*RT x x A A* (32)

*A A* (33)

,0 ,

*t B phis B B i B i*

*D RT x x*

*BRT*

*B B B B*

*ii ii i B B B i B B*

,

,

1 ,

*x d*

*i B i B*

1 ln

,

*i i B*

ln

,

*dt*

*i i*

*i i*

ln

1 ln

1 ln

(29)

(30)

(31)

*A*

*A*

*i i*

ln

*dx x*

**Ideal Mixture:** 

**Non-ideal Mixture:** 

mixture can be calculated as:

for a non-ideal mixture:

,

where *Aphis* considers the physical processes that involve thermal interactions with the surroundings and *Achem* considers the mass and heat transfer with the surroundings. The main contribution to this energy is due to mixing effects and can be estimated from the chemical potential at low pressures (Kim and Diwekar, 2000). *Aphis* is relatively lower than *Achem* . Thus, the physical component can be regarded as constant for all chemical species

and the derivative of this term is eliminated. The chemical component of exergy for an ideal mixture can be expressed as:

$$\mathcal{A}\_{\text{chem}} = RT\_0 \sum\_{i} \mathbf{x}^{(i)} \ln \left[ \mathbf{x}^{(i)} \right] \tag{22}$$

whereas a non-ideal mixture can be calculated as:

$$\mathcal{A}\_{\text{chem}} = RT\_0 \sum\_{i} \mathbf{x}^{(i)} \ln \left[ \mathbf{y}\_i \mathbf{x}^{(i)} \right] \tag{23}$$

and the exergy exchange in the reboiler for an ideal mixture can be calculated as:

$$\mathcal{A}\_{\rm B} = \mathcal{A}\_{\rm B\_{\rm ph}} + \mathcal{A}\_{\rm B\_{\rm chw}} = \mathcal{A}\_{\rm B\_{\rm ph}} + RT\_0 \sum\_{i} \mathbf{x}\_{\rm B}^{(i)} \ln \left\lfloor \mathbf{x}\_{\rm B}^{(i)} \right\rfloor \tag{24}$$

for a non-ideal mixture:

$$\mathcal{A}\_{B} = \mathcal{A}\_{B\_{\text{phys}}} + \mathcal{A}\_{B\_{\text{cluu}}} = \mathcal{A}\_{B\_{\text{phys}}} + RT\_0 \sum\_{i} \mathbf{x}\_{B}^{(i)} \ln \left[ \mathcal{V}\_{B,i} \mathbf{x}\_{B}^{(i)} \right] \tag{25}$$

Taking the derivative of Equations (24) and (25) yields:

$$\frac{d\left(\mathcal{A}\_{B}\right)}{dt} = \frac{d\left(\mathcal{A}\_{B\_{\text{dom}}}\right)}{dt} = RT\_0 \frac{d\left\{\sum\_{i} \mathbf{x}\_{B}^{(i)} \ln\left[\mathbf{x}\_{B}^{(i)}\right]\right\}}{dt} \tag{26}$$

$$\frac{d\left(\mathcal{A}\_{\rm B}\right)}{dt} = \frac{d\left(\mathcal{A}\_{\rm B\_{\rm chem}}\right)}{dt} = RT\_0 \frac{d\left\{\sum\_{i} \mathbf{x}\_{\rm B}^{(i)} \ln\left[\mathcal{I}\_{\rm B,i} \mathbf{x}\_{\rm B}^{(i)}\right]\right\}}{dt} \tag{27}$$

The derivative of the term on the right hand side of Equations (26) and (27) can be represented in terms of the mathematical model of the column; thus, substituting Equation (2) of Table 1 yields:

$$\begin{split} \frac{d\left\{\sum\_{i} \mathbf{x}\_{B}^{(i)} \ln \left[\mathbf{x}\_{B}^{(i)}\right] \right\}}{dt} &= \sum\_{i} \left\{ \left[1 + \ln \mathbf{x}\_{B}^{(i)}\right] \frac{d\mathbf{x}\_{B}^{(i)}}{dt} \right\} \\ = \left(\frac{V}{B}\right) \sum\_{i} \left\{ \mathbf{x}\_{B}^{(i)} - y\_{B}^{(i)} + \left(\frac{L}{V}\right) \left[\mathbf{x}\_{1}^{(i)} - \mathbf{x}\_{B}^{(i)}\right] \right\} \end{split} \tag{28}$$

$$\begin{split}d\left\{\frac{\sum\nolimits^{(i)}\ln\left[\mathcal{Y}\_{i,B}\mathbf{x}\_{B}^{(i)}\right]}{dt}\right\} &=\sum\_{i}\left\{\left[1+\ln\mathcal{Y}\_{i,B}\mathbf{x}\_{B}^{(i)}\right]\frac{d\mathbf{x}\_{B}^{(i)}}{dt}+\frac{\mathbf{x}\_{B}^{(i)}}{\mathcal{Y}\_{i,B}}\frac{d\mathcal{Y}\_{i,B}}{dt}\right\} \\ &=\left(\frac{V}{B}\right)\sum\_{i}\left\{\mathbf{x}\_{B}^{(i)}-\mathbf{y}\_{B}^{(i)}+\left(\frac{L}{V}\right)\left[\mathbf{x}\_{1}^{(i)}-\mathbf{x}\_{B}^{(i)}\right]\right\}\left[1+\ln\mathcal{Y}\_{i,B}\mathbf{x}\_{B}^{(i)}\right] \\ &+\sum\_{i}\left[\frac{\mathbf{x}\_{B}^{(i)}}{\mathcal{Y}\_{i,B}}\frac{d\mathbf{y}\_{i,B}}{dt}\right] \end{split} \tag{29}$$

Substituting Equation (28) on the right hand side of Equation (20) for an ideal mixture, and Equation (29) on the right hand side of Equation (20) for a non-ideal mixture yields:

#### **Ideal Mixture:**

96 Distillation – Advances from Modeling to Applications

where *Aphis* considers the physical processes that involve thermal interactions with the surroundings and *Achem* considers the mass and heat transfer with the surroundings. The main contribution to this energy is due to mixing effects and can be estimated from the chemical potential at low pressures (Kim and Diwekar, 2000). *Aphis* is relatively lower than *Achem* . Thus, the physical component can be regarded as constant for all chemical species and the derivative of this term is eliminated. The chemical component of exergy for an ideal

*i*

*chem i i RT x x*

*BB B B B B*

*BB B B B B B i*

 0

 , 0

The derivative of the term on the right hand side of Equations (26) and (27) can be represented in terms of the mathematical model of the column; thus, substituting Equation

*ii ii B B B*

*B B i i i B*

*dt dt*

*i*

*xy xx B V*

1

1 ln

*RT dt dt dt*

*RT dt dt dt*

and the exergy exchange in the reboiler for an ideal mixture can be calculated as:

*phis chem phis*

*phis chem phis*

*chem*

*chem*

*V L*

*i i*

ln

*dx x*

*i*

*B B i*

*B B i*

*chem*

 <sup>0</sup> ln *i i*

 <sup>0</sup> ln *i i*

*RT x x <sup>A</sup>* (22)

*<sup>A</sup>* (23)

*i i*

 0 , ln

*i i*

<sup>0</sup> ln

*AA A A* (25)

ln

*dx x*

ln

*dx x*

*i i B B*

*i i B B B i*

*dx*

(28)

*B*

*x*

*<sup>A</sup> <sup>A</sup>* (27)

*<sup>A</sup> <sup>A</sup>* (26)

*i RT x x AA A A* (24)

> *i RT x x*

mixture can be expressed as:

for a non-ideal mixture:

(2) of Table 1 yields:

whereas a non-ideal mixture can be calculated as:

Taking the derivative of Equations (24) and (25) yields:

*d d*

*d d*

$$\begin{split} \frac{d\left(\boldsymbol{B}\,\boldsymbol{\mathcal{A}}\_{B}\right)}{dt} = \boldsymbol{V}\boldsymbol{R}\boldsymbol{T}\_{0} \sum\_{i} \left\{ \mathbf{x}\_{B}^{(i)} - \mathbf{y}\_{B}^{(i)} + \left(\frac{L}{V}\right) \left[\mathbf{x}\_{1}^{(i)} - \mathbf{x}\_{B}^{(i)}\right] \right\} \left\{ \mathbf{1} + \ln \left[\mathbf{x}\_{B}^{(i)}\right] \right\} \\ - \boldsymbol{D} \left\{ \boldsymbol{\mathcal{A}}\_{B,plis} + \boldsymbol{R}\boldsymbol{T}\_{0} \sum\_{i} \mathbf{x}\_{B}^{(i)} \ln \left[\mathbf{x}\_{B}^{(i)}\right] \right\} \end{split} \tag{30}$$

**Non-ideal Mixture:** 

$$\begin{split} \frac{d\left(\mathcal{B}\mathcal{A}\_{\mathcal{B}}\right)}{dt} = VRT\_{0}\sum\_{i} \left[\mathbf{x}\_{\mathcal{B}}^{(i)} - \mathbf{y}\_{\mathcal{B}}^{(i)} + \left(\frac{L}{V}\right) \left[\mathbf{x}\_{1}^{(i)} - \mathbf{x}\_{\mathcal{B}}^{(i)}\right] \right] \left[1 + \ln \mathcal{Y}\_{i,\mathcal{B}} \mathbf{x}\_{\mathcal{B}}^{(i)}\right] \\ &+ BRT\_{0} \sum\_{i} \left[\frac{\mathbf{x}\_{\mathcal{B}}^{(i)}}{\mathcal{Y}\_{i,\mathcal{B}}} \frac{d\mathcal{Y}\_{i,\mathcal{B}}}{dt} \right] \\ &- D\_{t} \left[\mathcal{A}\_{\mathcal{B},\mathrm{phys}} + RT\_{0} \sum\_{i} \mathbf{x}\_{\mathcal{B}}^{(i)} \ln \mathcal{Y}\_{i,\mathcal{B}} \mathbf{x}\_{\mathcal{B}}^{(i)}\right] \end{split} \tag{31}$$

The exergy of the current production (dome) that will be used in Equation (6) for an ideal mixture can be calculated as:

$$\mathcal{A}\_{\rm D} = \mathcal{A}\_{\rm D, plus} + RT\_0 \sum\_{i} \mathbf{x}\_{\rm D}^{(i)} \ln \left[ \mathbf{x}\_{\rm D}^{(i)} \right] \tag{32}$$

for a non-ideal mixture:

$$\mathcal{A}\_{\rm D} = \mathcal{A}\_{\rm D, phys} + RT\_0 \sum\_{i} \mathbf{x}\_{\rm D}^{(i)} \ln \left[ \mathcal{Y}\_{i,\rm D} \mathbf{x}\_{\rm D}^{(i)} \right] \tag{33}$$

The exergy transfer associated with the transmission of energy as heat in the process can be calculated by the energy balances in the reboiler and condenser. Considering that *Hvap* is the same for each component and is not related to the temperature of the process, the Clausius-Clapeyron equation can be used to calculate *Hvap*, and the first two terms on the right hand side of Equation (16) can be calculated by the following equation;

Batch Distillation: Thermodynamic Efficiency 99

*LW VRT*

*<sup>L</sup> VRT x y x x x <sup>V</sup>*

0 0, ,

*i i*

*W DRT x x x x*

*<sup>L</sup> VRT x y x x <sup>x</sup> V*

ln ln ln

*B DD i i*

0 1 , 0

*B B B i B B*

*VRT x y x x x BRT*

*LW VRT DRT x x x x*

0

0 1

*i*

*K*

*K*

min 0

*i*

0

and the thermodynamic efficiency [Equation (9)] results in:

*BRT*

*i*

0 1

min 0 , ,

*W DRT x x x x*

*i i*

,

*dt*

*<sup>x</sup> <sup>R</sup>*

*x d*

*i B i B*

,

*i i B*

1, 1, 1,

the minimum work can be calculated as:

1, 1, 1,

 

1 1 ln

1

1 ln

*B BD D i B i D*

1 ln

1 ln

1 1

(40)

,

(42)

(43)

(44)

(41)

,

*i*

 

ln ln

0 1

1 1

*D B ii ii B B DD*

*x*

*x*

*i i i B*

*D BB i ii i*

1 ln

*ii ii i B i B*

*ii ii DD B B*

ln ln

*ii ii i B B B B*

*i ii i D DB B i D i B*

ln ln

0 , 1

*i i i i ii ii i D DB B B BB B*

ln ln 1 ln

*x x x x Rx y x y x*

*D B*

*x*

1 1 1 ln

 

 

1 1

*<sup>L</sup> VRT x y x x <sup>x</sup> V*

 

*ii ii i B B B i B B*

*L x d*

*V dt*

*i i ii ii i B B B B*

*DRT x x x x*

**Ideal Mixture:** 

**Non-ideal Mixture:** 

**Ideal Mixture:** 

**Non-ideal Mixture:** 

**Ideal Mixture:** 

*<sup>i</sup> <sup>t</sup>*

**Ideal Mixture:** 

$$\left(1 - \frac{T\_0}{T\_B}\right) \mathbf{Q}\_B - \left(\mathbf{1} - \frac{T\_0}{T\_D}\right) \mathbf{Q}\_D = V \Delta H^{vap} T\_0 \left(\frac{\mathbf{1}}{T\_C} - \frac{\mathbf{1}}{T\_B}\right) \tag{34}$$

and considering constant relative volatility:

$$\left(1-\frac{T\_0}{T\_B}\right)Q\_B-\left(1-\frac{T\_0}{T\_D}\right)Q\_D=VRT\_0\ln\left|\frac{x\_D^{(1)}(a\_1-1)+1}{x\_B^{(1)}(a\_1-1)+1}\right|\tag{35}$$

#### **Non-ideal Mixture:**

$$\left(1-\frac{T\_0}{T\_B}\right)Q\_B - \left(1-\frac{T\_0}{T\_D}\right)Q\_D = VRT\_0\ln\left(\frac{\mathcal{Y}\_{1,D}\Phi\_{1,B}K\_{1,B}}{\mathcal{Y}\_{1,B}\Phi\_{1,D}K\_{1,D}}\right) \tag{36}$$

where, K is the liquid-vapor equilibrium constant and Φ is defined as:

$$\Phi\_k = \frac{\widehat{\phi\_k}}{\phi\_k^{\rm sat}} \exp\left[-\frac{V\_k \left(P - P\_k^{\rm sat}\right)}{RT}\right]; k = 1, 2, \dots, n \tag{37}$$

Therefore, the term for exergy loss or work loss, LW in Equation (17) for an ideal mixture, can be calculated with Equation (35) and for a non-ideal mixture with Equation (36):

#### **Ideal Mixture:**

$$LPV = VRT\_0 \ln\left[\frac{\mathbf{x}\_D^{(1)}(a\_1 - 1) + 1}{\mathbf{x}\_B^{(1)}(a\_1 - 1) + 1}\right] + DRT\_0 \left\{\sum\_i \mathbf{x}\_B^{(i)} \ln\left[\mathbf{x}\_B^{(i)}\right] - \sum\_i \mathbf{x}\_D^{(i)} \left[\ln \mathbf{x}\_D^{(i)}\right]\right\}$$

$$+ D\left(\mathcal{A}\_{\mathcal{B}, \eta \text{bias}} - \mathcal{A}\_{\mathcal{D}, \eta \text{bias}}\right)$$

$$-VRT\_0 \sum\_i \left\{\mathbf{x}\_B^{(i)} - \mathbf{y}\_B^{(i)} + \left(\frac{L}{V}\right) \left[\mathbf{x}\_1^{(i)} - \mathbf{x}\_B^{(i)}\right]\right\} \left[1 + \ln \mathbf{x}\_B^{(i)}\right]$$

**Non-ideal Mixture:** 

$$LPV = VRT\_0 \ln\left(\frac{\mathcal{I}\_{1,D}\Phi\_{1,B}K\_{1,B}}{\mathcal{I}\_{1,B}\Phi\_{1,D}K\_{1,D}}\right) + D\_tRT\_0 \left\{\sum\_i \mathbf{x}\_B^{(i)} \ln\left[\mathcal{I}\_{i,B}\mathbf{x}\_B^{(i)}\right] - \sum\_i \mathbf{x}\_D^{(i)} \ln\left[\mathcal{I}\_{i,D}\mathbf{x}\_D^{(i)}\right]\right\}
$$

$$+ D\_t \left(\mathcal{A}\_{B,\text{plis}} - \mathcal{A}\_{D,\text{plis}}\right)$$

$$-VRT\_0 \sum\_i \left\{\mathbf{x}\_B^{(i)} - \mathbf{y}\_B^{(i)} + \left(\frac{L}{V}\right) \left[\mathbf{x}\_1^{(i)} - \mathbf{x}\_B^{(i)}\right]\right\} \left\{1 + \ln\left[\mathcal{I}\_{i,B}\mathbf{x}\_B^{(i)}\right]\right\}\tag{39}$$

$$-BRT\_0 \sum\_i \left[\frac{\mathbf{x}\_B^{(i)}}{\mathcal{I}\_{i,B}} \frac{d\mathcal{I}\_{i,B}}{dt}\right]$$

Considering that , , 0 *A A B phis D phis* , then:

**Ideal Mixture:** 

98 Distillation – Advances from Modeling to Applications

1 1 1 1 *vap*

0 0 1

 

0 0 1, 1, 1,

*B D B DD T T K <sup>Q</sup> Q VRT T T <sup>K</sup>*

*sat*

0 1

0 0, ,

ln ln ln

*LW VRT D RT x x x x*

*B DD i i*

1 1 ln ln ln

*B i i*

*D ii ii*

*<sup>L</sup> VRT x y x x x <sup>V</sup>*

*D BB i ii i*

0 , 1

*<sup>L</sup> VRT x y x x <sup>x</sup> V*

1 1 ln *D BB*

*B D B T T x <sup>Q</sup> Q VRT T T <sup>x</sup>*

exp

can be calculated with Equation (35) and for a non-ideal mixture with Equation (36):

*<sup>x</sup> LW VRT DRT x x x x*

*VPP RT*

Therefore, the term for exergy loss or work loss, LW in Equation (17) for an ideal mixture,

*k k <sup>k</sup>*

*B D C B T T Q Q VH T T T T T* 

0

1

*D*

(36)

0 1

0

 

1 1

1 1

; *k* = 1, 2, …,*n* (37)

*B phis D phis*

*A A* (38)

1 ln

, ,

*t B phis D phis*

*A A*

 

*x d*

*i B i B i B*

,

 

0

*i*

*BRT*

1 ln

,

(39)

*dt*

, ,

1

1, 1, 1,

*B B DD*

*D*

*ii ii i B B B i B B*

*t B BD D i B i D*

*D*

*ii ii i B B B B*

(34)

(35)

0 0

and considering constant relative volatility:

*B D*

1 1 ln

*B D*

where, K is the liquid-vapor equilibrium constant and Φ is defined as:

*k sat k*

 

1, 1, 1,

Considering that , , 0 *A A B phis D phis* , then:

*K*

*K*

1, 1, 1,

1 1

*i*

*i*

1 1 0 0 1 1

*x*

*B D*

**Ideal Mixture:** 

**Non-ideal Mixture:** 

**Ideal Mixture:** 

**Non-ideal Mixture:** 

$$L\mathcal{W} = VRT\_0 \ln\left[\frac{\mathbf{x}\_D^{(1)}(\alpha\_1 - 1) + 1}{\mathbf{x}\_B^{(1)}(\alpha\_1 - 1) + 1}\right]$$

$$+ DRT\_0 \left\{\sum\_i \mathbf{x}\_B^{(i)} \ln\left[\mathbf{x}\_B^{(i)}\right] - \sum\_i \mathbf{x}\_D^{(i)} \left[\ln \mathbf{x}\_D^{(i)}\right]\right\}\tag{40}$$

$$- VRT\_0 \sum\_i \left\{\mathbf{x}\_B^{(i)} - \mathbf{y}\_B^{(i)} + \left(\frac{L}{V}\right) \left[\mathbf{x}\_1^{(i)} - \mathbf{x}\_B^{(i)}\right]\right\} \left[1 + \ln \mathbf{x}\_B^{(i)}\right]$$

**Non-ideal Mixture:** 

$$LIN = VRT\_0 \ln\left(\frac{\mathcal{Y}\_{1,D}\Phi\_{1,B}K\_{1,B}}{\mathcal{Y}\_{1,B}\Phi\_{1,D}K\_{1,D}}\right) + DRT\_0 \left\{\sum\_{i} \mathbf{x}\_B^{(i)} \ln\left[\mathcal{Y}\_{i,B}\mathbf{x}\_B^{(i)}\right] - \sum\_{i} \mathbf{x}\_D^{(i)} \ln\left[\mathcal{Y}\_{i,D}\mathbf{x}\_D^{(i)}\right]\right\}
$$

$$-VRT\_0 \sum\_{i} \left\{\mathbf{x}\_B^{(i)} - \mathbf{y}\_B^{(i)} + \left(\frac{L}{V}\right) \left[\mathbf{x}\_1^{(i)} - \mathbf{x}\_B^{(i)}\right]\right\} \left\{1 + \ln\left[\mathcal{Y}\_{i,B}\mathbf{x}\_B^{(i)}\right]\right\} - BRT\_0 \sum\_{i} \left[\frac{\mathbf{x}\_B^{(i)}}{\mathcal{Y}\_{i,B}} \frac{d\mathcal{Y}\_{i,B}}{dt}\right] \tag{41}$$

the minimum work can be calculated as:

#### **Ideal Mixture:**

$$\begin{split} \mathcal{W}\_{\text{min}} &= DRT\_0 \left\{ \sum\_{i} \mathbf{x}\_D^{(i)} \ln \left[ \mathbf{x}\_D^{(i)} \right] - \sum\_{i} \mathbf{x}\_B^{(i)} \ln \left[ \mathbf{x}\_B^{(i)} \right] \right\} \\ &+ VRT\_0 \sum\_{i} \left\{ \mathbf{x}\_B^{(i)} - \mathbf{y}\_B^{(i)} + \left( \frac{L}{V} \right) \left[ \mathbf{x}\_1^{(i)} - \mathbf{x}\_B^{(i)} \right] \right\} \left\{ 1 + \ln \left[ \mathbf{x}\_B^{(i)} \right] \right\} \end{split} \tag{42}$$

**Non-ideal Mixture:** 

$$\begin{split} IV\_{\text{min}} &= DRT\_0 \left\{ \sum\_{i} \mathbf{x}\_D^{(i)} \ln \left[ \boldsymbol{\mathcal{Y}}\_{i,D} \mathbf{x}\_D^{(i)} \right] - \sum\_{i} \mathbf{x}\_B^{(i)} \ln \left[ \boldsymbol{\mathcal{Y}}\_{i,B} \mathbf{x}\_B^{(i)} \right] \right\} \\ &+ VRT\_0 \sum\_{i} \left\{ \mathbf{x}\_B^{(i)} - \mathbf{y}\_B^{(i)} + \left( \frac{L}{V} \right) \left[ \mathbf{x}\_1^{(i)} - \mathbf{x}\_B^{(i)} \right] \right\} \left\{ 1 + \ln \left[ \boldsymbol{\mathcal{Y}}\_{i,B} \mathbf{x}\_B^{(i)} \right] \right\} \\ &+ BRT\_0 \sum\_{i} \left[ \frac{\mathbf{x}\_B^{(i)}}{\mathcal{Y}\_{i,B}} \frac{d\boldsymbol{\mathcal{Y}}\_{i,B}}{dt} \right] \end{split} \tag{43}$$

and the thermodynamic efficiency [Equation (9)] results in:

#### **Ideal Mixture:**

$$\eta\_{t} = \frac{\sum\_{i} \left\{ \mathbf{x}\_{D}^{(i)} \ln \left[ \mathbf{x}\_{D}^{(i)} \right] - \mathbf{x}\_{B}^{(i)} \ln \left[ \mathbf{x}\_{B}^{(i)} \right] \right\} + \left\{ R \left[ \mathbf{x}\_{1}^{(i)} - y\_{B}^{(i)} \right] + \mathbf{x}\_{B}^{(i)} - y\_{B}^{(i)} \right\} \left\{ 1 + \ln \left[ \mathbf{x}\_{B}^{(i)} \right] \right\}}{\left( R + 1 \right) \ln \left[ \frac{\mathbf{x}\_{D}^{(1)} \left( \alpha\_{1} - 1 \right) + 1}{\mathbf{x}\_{B}^{(1)} \left( \alpha\_{1} - 1 \right) + 1} \right]} \tag{44}$$

Batch Distillation: Thermodynamic Efficiency 101

The conditions and data used to solve the mathematical model of thermodynamic efficiency are given in Table 2. Relative volatilities were calculated with the Soave-Redlich-Kwong

Table 2. Data used to solve the mathematical model of thermodynamic efficiency

The first step for the resolution of the mathematical model was to reach the steady state to extract the product during one hour. Figure 3, shows the concentrations at the top of the

> Hexane Benzene

Clorobenzene

0 0.5 1 1.5 2 **Time, h**

**Variable/Parameter Amount Units** 

200.00 240.00 5.00 1.00 1.00 0.20 0.30 0.50 8.54 4.71 15 5

kmol kmol/h kmol kmol h

**Case of study 1: Hexane/Benzene/Chlorobenzene mixture** 

equation of state.

Condenser liquid holdup (*HD*) Tray liquid holdup (*HT*)

column (including the steady state).

Fig. 3. Concentrations at the top of the column.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Feed (*F*) Vapor flow (*V*)

Batch time *Z*Hexane *Z*Benzene *Z*Chlorobenzene *α*Hexane *α*Benzene

Number of trays Reflux ratio (Constant)

**Non-ideal Mixture:** 

$$\eta\_t = \frac{\sum\_{i} \left| \mathbf{x}\_D^{(i)} \ln \boldsymbol{\gamma}\_{i,D} \mathbf{x}\_D^{(i)} - \mathbf{x}\_B^{(i)} \ln \boldsymbol{\gamma}\_{i,B} \mathbf{x}\_B^{(i)} + \left( R \left[ \mathbf{x}\_1^{(i)} - \mathbf{y}\_B^{(i)} \right] + \mathbf{x}\_B^{(i)} - \mathbf{y}\_B^{(i)} \right) \left[ \mathbf{1} + \ln \boldsymbol{\gamma}\_{i,B} \mathbf{x}\_B^{(i)} \right] + \left( \frac{B}{V} \right) \left| \frac{\mathbf{x}\_B^{(i)}}{\boldsymbol{\gamma}\_{i,B}} \frac{d \boldsymbol{\gamma}\_{i,B}}{dt} \right|}{(R+1) \ln \left( \frac{\mathbf{K}\_{1,B}}{\mathbf{K}\_{1,D}} \frac{\boldsymbol{\Phi}\_{1,B}}{\boldsymbol{\Phi}\_{1,D}} \frac{\boldsymbol{\gamma}\_{1,D}}{\boldsymbol{\gamma}\_{i,B}} \right)} \tag{45}$$

Equation (45) can be reduced if one considers that the changes of activity coefficients in the reboiler are so small that they can be neglected. Then:

$$\eta\_{t} = \frac{\sum\_{i} \left\{ \mathbf{x}\_{B}^{(i)} + \mathbf{x}\_{D}^{(i)} \ln \gamma\_{i,D} \mathbf{x}\_{D}^{(i)} + \left[ R \left( \mathbf{x}\_{1}^{(i)} - \mathbf{y}\_{B}^{(i)} \right) - \mathbf{y}\_{B}^{(i)} \right] \left[ 1 + \ln \gamma\_{i,B} \mathbf{x}\_{B}^{(i)} \right] \right\}}{(R+1) \ln \left( \frac{K\_{1,B}}{K\_{1,D}} \frac{\Phi\_{1,B}}{\Phi\_{1,D}} \frac{\mathcal{V}\_{1,D}}{\mathcal{V}\_{1,B}} \right)} \tag{46}$$

If the gas phase is ideal or close to ideality, then Equation (46) can be reduced even more:

$$\eta\_t = \frac{\sum\_{i} \left\{ \mathbf{x}\_B^{(i)} + \mathbf{x}\_D^{(i)} \ln \gamma\_{i,D} \mathbf{x}\_D^{(i)} + \left[ R \left( \mathbf{x}\_1^{(i)} - \mathbf{y}\_B^{(i)} \right) - \mathbf{y}\_B^{(i)} \right] \left[ 1 + \ln \gamma\_{i,B} \mathbf{x}\_B^{(i)} \right] \right\}}{\left( R + 1 \right) \ln \left( \frac{K\_{1,B}}{K\_{1,D}} \frac{\mathcal{V}\_{1,D}}{\mathcal{V}\_{1,B}} \right)} \tag{47}$$

The solution of Equation (45) or its simplifications [Equation (46) and (47)] require the calculation of activity coefficients of the mixtures, which can be obtained with an appropriate solution model (Wilson, NRTL, UNIQUAC, UNIFAC, etc.). According to Equations (44), (45), (46) and (47), the thermodynamic efficiency not only depends on the reflux ratio but on the concentration of components in both phases and the kind of mixture. A convenient way to express thermodynamic efficiency is the average thermodynamic equation used by Zavala Loría (2004), Zavala et al. (2007), Zavala and Coronado (2008). and Zavala et al. (2011) to solve the maximum thermodynamic efficiency problem:

$$
\eta\_{\text{average}} = \frac{\int\_0^t \eta\_t dt}{t} \tag{48}
$$

#### **4. Results and discussion**

Using the First and Second Law of Thermodynamics, this work has developed an expression for calculating the thermodynamic efficiency of a batch distillation process [Equations (44) and (45)]. To solve the mathematical model, the feeding was introduced at the top of the column at boiling temperature, neglecting the accumulation of vapor in each stage. The process was performed with total condenser, atmospheric pressure and adiabatic column.

To solve the model, two mixtures were considered. The first one is an ideal mixture of Hexane/Benzene/Chlorobenzene (HBC); the second one is a non-ideal mixture of Ethanol-Water (EW).

*K*

 

*K*

*x x x x Rx y x y x V dt*

*i ii i ii ii i B i B*

Equation (45) can be reduced if one considers that the changes of activity coefficients in the

1, 1, 1, 1, 1, 1,

 

*K*

 

*K*

0 *t t*

Using the First and Second Law of Thermodynamics, this work has developed an expression for calculating the thermodynamic efficiency of a batch distillation process [Equations (44) and (45)]. To solve the mathematical model, the feeding was introduced at the top of the column at boiling temperature, neglecting the accumulation of vapor in each stage. The process was performed with total condenser, atmospheric pressure and adiabatic column. To solve the model, two mixtures were considered. The first one is an ideal mixture of Hexane/Benzene/Chlorobenzene (HBC); the second one is a non-ideal mixture of Ethanol-

*dt t* 

*x x x Rx y y x*

The solution of Equation (45) or its simplifications [Equation (46) and (47)] require the calculation of activity coefficients of the mixtures, which can be obtained with an appropriate solution model (Wilson, NRTL, UNIQUAC, UNIFAC, etc.). According to Equations (44), (45), (46) and (47), the thermodynamic efficiency not only depends on the reflux ratio but on the concentration of components in both phases and the kind of mixture. A convenient way to express thermodynamic efficiency is the average thermodynamic equation used by Zavala Loría (2004), Zavala et al. (2007), Zavala and Coronado (2008). and

*ii i ii i i BD D i D B B i B B*

ln 1 ln

*x x x Rx y y x*

*K*

*K*

If the gas phase is ideal or close to ideality, then Equation (46) can be reduced even more:

1 ln

*ii i ii i i BD D i D B B i B B*

ln 1 ln

*B BD D DB*

, , 1

1, 1, 1, 1, 1, 1,

, , 1

1, 1, 1, 1,

*B D D B*

*B BD D DB*

 

 

(48)

,

(46)

(47)

(45)

,

*i*

*B x d*

*R*

 

reboiler are so small that they can be neglected. Then:

*<sup>i</sup> <sup>t</sup>*

*<sup>i</sup> <sup>t</sup>*

**4. Results and discussion** 

Water (EW).

ln ln 1 ln

, , 1 ,

*D DB B i D i B B BB i B B i i B*

1 ln

*R*

1 ln

*R*

Zavala et al. (2011) to solve the maximum thermodynamic efficiency problem:

*average*

**Non-ideal Mixture:** 

*t*

#### **Case of study 1: Hexane/Benzene/Chlorobenzene mixture**

The conditions and data used to solve the mathematical model of thermodynamic efficiency are given in Table 2. Relative volatilities were calculated with the Soave-Redlich-Kwong equation of state.


Table 2. Data used to solve the mathematical model of thermodynamic efficiency

The first step for the resolution of the mathematical model was to reach the steady state to extract the product during one hour. Figure 3, shows the concentrations at the top of the column (including the steady state).

Fig. 3. Concentrations at the top of the column.

Batch Distillation: Thermodynamic Efficiency 103

The conditions used for solving the mathematical model of thermodynamic efficiency for a non ideal mixture (Ethanol/Water) are given in Table 4. Wilson's equation was used to obtain the activity coefficients; the vapor phase is considered to have an ideal behavior. The steady state was reached when Ethanol presented a purity level of 89.61% mol obtained

3.045 hours was the period of time needed to reach the steady state, we obtained an average thermodynamic efficiency of 36.96% and the product presented an average concentration of

Table 4. Data used to solve the mathematical model of thermodynamic efficiency.

012345 **Time, h**

Ethanol Water

Fig. 5. Concentration profiles at the top of the column.

**Variable/Parameter Amount Units** 

100.00 120.00 5.00 1.00 2.00 0.50 0.50 8.54 4.71 15 2

kmol kmol/h kmol kmol h

**Case of study 2: Ethanol/Water mixture** 

86.74% mol of the most volatile component.

using Wilson's equation.

Condenser liquid holdup (*HD*) Tray liquid holdup (*HT*)

Feed (*F*) Vapor flow (*V*)

Batch time *Z*Ethanol *Z*Water *α*Ethanol *α*Water

Number of trays Reflux ratio (Constant)

> 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

The average thermodynamic efficiency is 17.58% for this production time while the average concentration of the most volatile component at the top of the column is 88.55% mol. Table 3, shows the variations observed when the reflux ratio is changed.


Table 3. Thermodynamic efficiency behavior based on variations in the reflux ratio.

Figure 4 shows the thermodynamic efficiency behavior in the product obtaining.

Fig. 4. Thermodynamic efficiency behavior in the product obtaining.

Such variations make evident the influence of the reflux ratio over the thermodynamic efficiency; in other words, the thermodynamic efficiency of the process is smaller when the reflux ratio is higher. We also could observe that the reflux ratio affects the product concentration. If the reflux ratio increases the product concentration increases as well.

The average thermodynamic efficiency is 17.58% for this production time while the average concentration of the most volatile component at the top of the column is 88.55% mol. Table 3,

2 55.60 49.50

3 28.61 64.34

4 20.60 77.66

5 17.58 88.55

6 16.16 95.55

7 15.08 98.51

8 14.02 99.40

Table 3. Thermodynamic efficiency behavior based on variations in the reflux ratio.

Figure 4 shows the thermodynamic efficiency behavior in the product obtaining.

Fig. 4. Thermodynamic efficiency behavior in the product obtaining.

0

0.05

0.1

0.15

0.2

0.25

Such variations make evident the influence of the reflux ratio over the thermodynamic efficiency; in other words, the thermodynamic efficiency of the process is smaller when the reflux ratio is higher. We also could observe that the reflux ratio affects the product concentration. If the reflux ratio increases the product concentration increases as well.

0 0.2 0.4 0.6 0.8 1

**Time, h**

*D average* , *x* **(%)** 

**Reflux ratio average (%)**  <sup>1</sup>

shows the variations observed when the reflux ratio is changed.

#### **Case of study 2: Ethanol/Water mixture**

The conditions used for solving the mathematical model of thermodynamic efficiency for a non ideal mixture (Ethanol/Water) are given in Table 4. Wilson's equation was used to obtain the activity coefficients; the vapor phase is considered to have an ideal behavior. The steady state was reached when Ethanol presented a purity level of 89.61% mol obtained using Wilson's equation.

3.045 hours was the period of time needed to reach the steady state, we obtained an average thermodynamic efficiency of 36.96% and the product presented an average concentration of 86.74% mol of the most volatile component.


Table 4. Data used to solve the mathematical model of thermodynamic efficiency.

Fig. 5. Concentration profiles at the top of the column.

Batch Distillation: Thermodynamic Efficiency 105

process. The resulting equation was used to find the batch distillation thermodynamic efficiency for an ideal mixture and a non-ideal mixture. The equation obtained is a generalization of the equation developed by Zavala-Loría et al. (2007), Zavala and Coronado

The results obtained by solving the Equations, allowed us to observe the relationship between reflux and thermodynamic efficiency of the process. Furthermore, variables such as the product purity are affected by the reflux ratio, in other words, the purity of the product

requires a greater amount of reflux to obtain a higher concentration.

(2008) and Zavala et al. (2011).

*Hj* molar hold-up on tray *j* 

*n* number of components *N* number of trays in the column

*z* mole fraction in the feed

( )*<sup>i</sup> Kj* equilibrium constant

*<sup>t</sup>* punctual thermodynamic efficiency

*T* temperature *A* availability

Hvap vaporization heat

*B* boiler or reboiler

*I* enthalpy (J/mol) *S* entropy (J/mol K) *V* vapor flow rate (mol/h) *L* liquid flow rate (mol/h) *D* distillate flow rate (mol/h)

*HD* molar hold-up on the condenser

*B* amount of moles in the reboiler, mol

*<sup>j</sup> x* mole fraction in the liquid of component *i* at plate *j*

*<sup>j</sup> y* mole fraction in the vapor of component *i* at plate *j*

*Bx* mole fraction in the liquid of component *i* in the reboiler

*Dx* mole fraction in the vapor of component *i* in the distillate

**6. Nomenclature** 

*Rt* reflux ratio *t* time (h) *W* work (J) *LW* work loss (J)

( )*i*

( )*i*

( )*i*

( )*i*

**Subindex** 

0 reference *c* condenser f end *D* distillate

*t* time

Figure 5 shows the behavior of the concentrations at the top of the column, while the behavior of punctual thermodynamic efficiency is shown in Figure 6. Table 5 shows the variations observed when different reflux ratios are used.

Fig. 6. Thermodynamic efficiency profile (obtaining product).


Table 5. Thermodynamic efficiency behavior with regards to the reflux ratio variation.

As in the last case, we observe that when the reflux ratio increases, the efficiency decreases; however, the concentration of the most volatile component presents also an increment. We can formulate the following heuristic rule: If the variables of the process are maintained, the reflux ratio will have an inverse effect on the thermodynamic efficiency of the process.

### **5. Concluding remarks**

Using the First and Second Law of Thermodynamics (exergy concept), this work has developed an expression for calculating the thermodynamic efficiency of a batch distillation process. The resulting equation was used to find the batch distillation thermodynamic efficiency for an ideal mixture and a non-ideal mixture. The equation obtained is a generalization of the equation developed by Zavala-Loría et al. (2007), Zavala and Coronado (2008) and Zavala et al. (2011).

The results obtained by solving the Equations, allowed us to observe the relationship between reflux and thermodynamic efficiency of the process. Furthermore, variables such as the product purity are affected by the reflux ratio, in other words, the purity of the product requires a greater amount of reflux to obtain a higher concentration.

## **6. Nomenclature**

104 Distillation – Advances from Modeling to Applications

Figure 5 shows the behavior of the concentrations at the top of the column, while the behavior of punctual thermodynamic efficiency is shown in Figure 6. Table 5 shows the

variations observed when different reflux ratios are used.

Fig. 6. Thermodynamic efficiency profile (obtaining product).

**5. Concluding remarks** 

0

0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45

**Reflux ratio average (%)**  <sup>1</sup>

Table 5. Thermodynamic efficiency behavior with regards to the reflux ratio variation.

As in the last case, we observe that when the reflux ratio increases, the efficiency decreases; however, the concentration of the most volatile component presents also an increment. We can formulate the following heuristic rule: If the variables of the process are maintained, the reflux ratio will have an inverse effect on the thermodynamic efficiency of the process.

Using the First and Second Law of Thermodynamics (exergy concept), this work has developed an expression for calculating the thermodynamic efficiency of a batch distillation

5 36.96 86.74 6 32.75 86.96 7 29.24 87.12 8 26.38 87.24 9 24.00 87.32 10 22.02 87.39

0 0.511.5 2 **Time, h**

*D average* , *x* **(%)** 



**Part 2** 

**Food and Aroma Concentration** 

#### **7. References**

