Distillation

Distillation – p.1/19

Distillation

One man’s magic is another man’s engineering. Lazarus Long Distillation is a cascade of flash units. The art of distillation dates back to atleast the first century A.D. Currently, multistage distillation is by far the most widely used method for separating liquid mixtures of chemical components. Distillation is a highly energy intensive technique.

Distillation is a cascade of flash units. The art of distillation dates back to atleast the first century A.D. Currently, multistage distillation is by far the most widely used method for separating liquid mixtures of chemical components. Distillation is a highly energy intensive technique.

Distillation is a cascade of flash units. The art of distillation dates back to atleast the first century A.D. Currently, multistage distillation is by far the most widely used method for separating liquid mixtures of chemical components. Distillation is a highly energy intensive technique.

Distillation is a cascade of flash units. The art of distillation dates back to atleast the first century A.D. Currently, multistage distillation is by far the most widely used method for separating liquid mixtures of chemical components. Distillation is a highly energy intensive technique.

Distillation Modeling

Feed flowrate, F Vapor flowrate, V Distillate flowrate, D Bottoms flowrate, B Reflux ratio, R = \frac{L}{D}

Design Issues Number of trays Column Diameter Tray spacing Operation Issues T and P Reflux Ratio, R Feed tray location Product recoveries

Distillation – p.4/19

Feed flowrate, F Vapor flowrate, V Distillate flowrate, D Bottoms flowrate, B Reflux ratio, R = \frac{L}{D}

Design Issues Number of trays Column Diameter Tray spacing Operation Issues T and P Reflux Ratio, R Feed tray location Product recoveries

Distillation – p.4/19

Feed flowrate, F Vapor flowrate, V Distillate flowrate, D Bottoms flowrate, B Reflux ratio, R = \frac{L}{D}

Design Issues Number of trays Column Diameter Tray spacing Operation Issues T and P Reflux Ratio, R Feed tray location Product recoveries

Distillation – p.4/19

Feed flowrate, F Vapor flowrate, V Distillate flowrate, D Bottoms flowrate, B Reflux ratio, R = \frac{L}{D}

Design Issues Number of trays Column Diameter Tray spacing Operation Issues T and P Reflux Ratio, R Feed tray location Product recoveries

Distillation – p.4/19

Consider a distillation unit with N trays and a multicomponent feed F_{j} with N_{c} components that comes to the jth tray and liquid products P L_{j} and vapor P V_{j}. There could be multiple feed trays and multiple product trays.

Each tray has a liquid holdup M_{j} and a much smaller vapor holdup with liquid and vapor mole fractions x_{ij} and y_{ij} respectively. Tray j has liquid L_{j} and vapor V_{j} flowing out of the tray and is connected to streams above and below. A mass and energy balance could be written for each component on each tray.

Distillation – p.5/19

Consider a distillation unit with N trays and a multicomponent feed F_{j} with N_{c} components that comes to the jth tray and liquid products P L_{j} and vapor P V_{j}. There could be multiple feed trays and multiple product trays.

Each tray has a liquid holdup M_{j} and a much smaller vapor holdup with liquid and vapor mole fractions x_{ij} and y_{ij} respectively. Tray j has liquid L_{j} and vapor V_{j} flowing out of the tray and is connected to streams above and below. A mass and energy balance could be written for each component on each tray.

Distillation – p.5/19

Consider a distillation unit with N trays and a multicomponent feed F_{j} with N_{c} components that comes to the jth tray and liquid products P L_{j} and vapor P V_{j}. There could be multiple feed trays and multiple product trays.

Each tray has a liquid holdup M_{j} and a much smaller vapor holdup with liquid and vapor mole fractions x_{ij} and y_{ij} respectively. Tray j has liquid L_{j} and vapor V_{j} flowing out of the tray and is connected to streams above and below. A mass and energy balance could be written for each component on each tray.

Distillation – p.5/19

Consider a distillation unit with N trays and a multicomponent feed F_{j} with N_{c} components that comes to the jth tray and liquid products P L_{j} and vapor P V_{j}. There could be multiple feed trays and multiple product trays.

Each tray has a liquid holdup M_{j} and a much smaller vapor holdup with liquid and vapor mole fractions x_{ij} and y_{ij} respectively. Tray j has liquid L_{j} and vapor V_{j} flowing out of the tray and is connected to streams above and below. A mass and energy balance could be written for each component on each tray.

Distillation – p.5/19

Consider a distillation unit with N trays and a multicomponent feed F_{j} with N_{c} components that comes to the jth tray and liquid products P L_{j} and vapor P V_{j}. There could be multiple feed trays and multiple product trays.

Each tray has a liquid holdup M_{j} and a much smaller vapor holdup with liquid and vapor mole fractions x_{ij} and y_{ij} respectively. Tray j has liquid L_{j} and vapor V_{j} flowing out of the tray and is connected to streams above and below. A mass and energy balance could be written for each component on each tray.

Distillation – p.5/19

Mass Balance:

\begin{matrix}F_{j}z_{ij} + L_{j−1}x_{i,j−1} + V_{j+1}y_{i,j+1} − (P L_{j} + L_{j})x_{ij}− \\ (V_{j} + P V_{j})y_{ij} = 0 i = 1, 2, ..., N_{c}; j = 1, 2, ..., N \end{matrix}

Equilibrium Expressions:

\begin{matrix}y_{ij} = K_{ij}x_{ij} \\ K_{ij} = K(T_{j}, P_{j}, x_{ij}) \end{matrix}

Summation Equations:

∑_{i} x_{ij} = 1, ∑_{i} y_{ij} = 1, j = 1, 2, ..., N

Heat Balance:

\begin{matrix}F_{j}HF_{j} + L_{j−1}H_{l,j−1} + V_{j+1}H_{v,j+1} − (P L_{j} + L_{j})H_{lj}− \\ (V_{j} + P V_{j})H_{vj} + Q_{j} = 0 j = 1, 2, ..., N \end{matrix}

Distillation – p.6/19

Mass Balance:

\begin{matrix}F_{j}z_{ij} + L_{j−1}x_{i,j−1} + V_{j+1}y_{i,j+1} − (P L_{j} + L_{j})x_{ij}− \\ (V_{j} + P V_{j})y_{ij} = 0 i = 1, 2, ..., N_{c}; j = 1, 2, ..., N \end{matrix}

Equilibrium Expressions:

\begin{matrix}y_{ij} = K_{ij}x_{ij} \\ K_{ij} = K(T_{j}, P_{j}, x_{ij}) \end{matrix}

Summation Equations:

∑_{i} x_{ij} = 1, ∑_{i} y_{ij} = 1, j = 1, 2, ..., N

Heat Balance:

\begin{matrix}F_{j}HF_{j} + L_{j−1}H_{l,j−1} + V_{j+1}H_{v,j+1} − (P L_{j} + L_{j})H_{lj}− \\ (V_{j} + P V_{j})H_{vj} + Q_{j} = 0 j = 1, 2, ..., N \end{matrix} Distillation – p.6/19

Mass Balance:

\begin{matrix}F_{j}z_{ij} + L_{j−1}x_{i,j−1} + V_{j+1}y_{i,j+1} − (P L_{j} + L_{j})x_{ij}− \\ (V_{j} + P V_{j})y_{ij} = 0 i = 1, 2, ..., N_{c}; j = 1, 2, ..., N \end{matrix}

Equilibrium Expressions:

\begin{matrix}y_{ij} = K_{ij}x_{ij} \\ K_{ij} = K(T_{j}, P_{j}, x_{ij}) \end{matrix}

Summation Equations:

∑_{i} x_{ij} = 1, ∑_{i} y_{ij} = 1, j = 1, 2, ..., N

Heat Balance:

\begin{matrix}F_{j}HF_{j} + L_{j−1}H_{l,j−1} + V_{j+1}H_{v,j+1} − (P L_{j} + L_{j})H_{lj}− \\ (V_{j} + P V_{j})H_{vj} + Q_{j} = 0 j = 1, 2, ..., N \end{matrix} Distillation – p.6/19

Mass Balance:

\begin{matrix}F_{j}z_{ij} + L_{j−1}x_{i,j−1} + V_{j+1}y_{i,j+1} − (P L_{j} + L_{j})x_{ij}− \\ (V_{j} + P V_{j})y_{ij} = 0 i = 1, 2, ..., N_{c}; j = 1, 2, ..., N \end{matrix}

Equilibrium Expressions:

\begin{matrix}y_{ij} = K_{ij}x_{ij} \\ K_{ij} = K(T_{j}, P_{j}, x_{ij}) \end{matrix}

Summation Equations:

∑_{i} x_{ij} = 1, ∑_{i} y_{ij} = 1, j = 1, 2, ..., N

Heat Balance:

\begin{matrix}F_{j}HF_{j} + L_{j−1}H_{l,j−1} + V_{j+1}H_{v,j+1} − (P L_{j} + L_{j})H_{lj}− \\ (V_{j} + P V_{j})H_{vj} + Q_{j} = 0 j = 1, 2, ..., N \end{matrix}

Distillation – p.6/19

MESH Equations

The Mass, Equilibrium, Summation, and Heat (MESH) equations form the standard model for tray-by-tray distillation model.

Note that the thermodynamic properties (K values and specific enthalpies) are expressed as implicit functions that require physical property models.

These are a large number of equations that are not trivial to solve.

For this reason, short-cut methods are typically used for designing columns.

Once a column is designed, its performance is analyzed using a detailed model.

The Mass, Equilibrium, Summation, and Heat (MESH) equations form the standard model for tray-by-tray distillation model.

Note that the thermodynamic properties (K values and specific enthalpies) are expressed as implicit functions that require physical property models.

These are a large number of equations that are not trivial to solve.

For this reason, short-cut methods are typically used for designing columns.

Once a column is designed, its performance is analyzed using a detailed model.

The Mass, Equilibrium, Summation, and Heat (MESH) equations form the standard model for tray-by-tray distillation model.

Note that the thermodynamic properties (K values and specific enthalpies) are expressed as implicit functions that require physical property models.

These are a large number of equations that are not trivial to solve.

For this reason, short-cut methods are typically used for designing columns.

Once a column is designed, its performance is analyzed using a detailed model.

The Mass, Equilibrium, Summation, and Heat (MESH) equations form the standard model for tray-by-tray distillation model.

Note that the thermodynamic properties (K values and specific enthalpies) are expressed as implicit functions that require physical property models.

These are a large number of equations that are not trivial to solve.

For this reason, short-cut methods are typically used for designing columns.

Once a column is designed, its performance is analyzed using a detailed model.

The Mass, Equilibrium, Summation, and Heat (MESH) equations form the standard model for tray-by-tray distillation model.

Note that the thermodynamic properties (K values and specific enthalpies) are expressed as implicit functions that require physical property models.

These are a large number of equations that are not trivial to solve.

For this reason, short-cut methods are typically used for designing columns.

Once a column is designed, its performance is analyzed using a detailed model.

Short-cut Distillation Model

Light Key: The component that is present in significant quantity in the bottoms stream is called the light key component.

Components lighter than the light key can be assumed to come out completely in the distillate steam.

Heavy Key: The component that is present in significant quantity in the distillate stream is called the heavy key component.

Components heavier than the heavy key can be assumed to come out completely in the bottoms stream.

Degrees of Freedom: A column has three degrees of freedom and is fully specified if the following are known:

1. Overhead split fraction of light key , ξ_{lk}

2. Overhead split fraction of heavy key, ξ_{hk}

3. Column pressure or temperature

Distillation – p.8/19

Short-cut Distillation Model

Light Key: The component that is present in significant quantity in the bottoms stream is called the light key component.

Components lighter than the light key can be assumed to come out completely in the distillate steam.

Heavy Key: The component that is present in significant quantity in the distillate stream is called the heavy key component.

Components heavier than the heavy key can be assumed to come out completely in the bottoms stream.

Degrees of Freedom: A column has three degrees of freedom and is fully specified if the following are known:

1. Overhead split fraction of light key , ξ_{lk}

2. Overhead split fraction of heavy key, ξ_{hk}

3. Column pressure or temperature

Distillation – p.8/19

Short-cut Distillation Model

Light Key: The component that is present in significant quantity in the bottoms stream is called the light key component.

Components lighter than the light key can be assumed to come out completely in the distillate steam.

Heavy Key: The component that is present in significant quantity in the distillate stream is called the heavy key component.

Components heavier than the heavy key can be assumed to come out completely in the bottoms stream.

Degrees of Freedom: A column has three degrees of freedom and is fully specified if the following are known:

1. Overhead split fraction of light key , ξ_{lk}

2. Overhead split fraction of heavy key, ξ_{hk}

3. Column pressure or temperature

Distillation – p.8/19

Short-cut Distillation Model

Light Key: The component that is present in significant quantity in the bottoms stream is called the light key component.

Components lighter than the light key can be assumed to come out completely in the distillate steam.

Heavy Key: The component that is present in significant quantity in the distillate stream is called the heavy key component.

Components heavier than the heavy key can be assumed to come out completely in the bottoms stream.

Degrees of Freedom: A column has three degrees of freedom and is fully specified if the following are known:

1. Overhead split fraction of light key , ξ_{lk}

2. Overhead split fraction of heavy key, ξ_{hk}

3. Column pressure or temperature

Distillation – p.8/19

Short-cut Distillation Model

Light Key: The component that is present in significant quantity in the bottoms stream is called the light key component.

Components lighter than the light key can be assumed to come out completely in the distillate steam.

Heavy Key: The component that is present in significant quantity in the distillate stream is called the heavy key component.

Components heavier than the heavy key can be assumed to come out completely in the bottoms stream.

Degrees of Freedom: A column has three degrees of freedom and is fully specified if the following are known:

1. Overhead split fraction of light key , ξ_{lk}

2. Overhead split fraction of heavy key, ξ_{hk}

3. Column pressure or temperature

Distillation – p.8/19

Short-cut Distillation Model

Light Key: The component that is present in significant quantity in the bottoms stream is called the light key component.

Components lighter than the light key can be assumed to come out completely in the distillate steam.

Heavy Key: The component that is present in significant quantity in the distillate stream is called the heavy key component.

Components heavier than the heavy key can be assumed to come out completely in the bottoms stream.

Degrees of Freedom: A column has three degrees of freedom and is fully specified if the following are known:

1. Overhead split fraction of light key , ξ_{lk}

2. Overhead split fraction of heavy key, ξ_{hk}

3. Column pressure or temperature

Distillation – p.8/19

Short-cut Distillation Model

Light Key: The component that is present in significant quantity in the bottoms stream is called the light key component.

Components lighter than the light key can be assumed to come out completely in the distillate steam.

Heavy Key: The component that is present in significant quantity in the distillate stream is called the heavy key component.

Components heavier than the heavy key can be assumed to come out completely in the bottoms stream.

Degrees of Freedom: A column has three degrees of freedom and is fully specified if the following are known:

1. Overhead split fraction of light key , ξ_{lk}

2. Overhead split fraction of heavy key, ξ_{hk}

3. Column pressure or temperature

Distillation – p.8/19

Short-cut Distillation Model

Light Key: The component that is present in significant quantity in the bottoms stream is called the light key component.

Components lighter than the light key can be assumed to come out completely in the distillate steam.

Heavy Key: The component that is present in significant quantity in the distillate stream is called the heavy key component.

Components heavier than the heavy key can be assumed to come out completely in the bottoms stream.

Degrees of Freedom: A column has three degrees of freedom and is fully specified if the following are known:

1. Overhead split fraction of light key , ξ_{lk}

2. Overhead split fraction of heavy key, ξ_{hk}

3. Column pressure or temperature

Distillation – p.8/19

Short-cut Distillation Model

Assumptions: Vapor is an ideal gas and the liquid is an ideal solution.

Each stage is at equilibrium and is modeled as a flash unit. α_{k/n} is independent of temperature and pressure. The split fractions of the light and heavy keys are known (ξ_{lk} and ξ_{hk} are known). The column is under total reflux.

Assumptions: Vapor is an ideal gas and the liquid is an ideal solution. Each stage is at equilibrium and is modeled as a flash unit. α_{k/n} is independent of temperature and pressure. The split fractions of the light and heavy keys are known (ξ_{lk} and ξ_{hk} are known). The column is under total reflux.

Assumptions: Vapor is an ideal gas and the liquid is an ideal solution. Each stage is at equilibrium and is modeled as a flash unit. α_{k/n} is independent of temperature and pressure. The split fractions of the light and heavy keys are known (ξ_{lk} and ξ_{hk} are known). The column is under total reflux.

Assumptions: Vapor is an ideal gas and the liquid is an ideal solution. Each stage is at equilibrium and is modeled as a flash unit. α_{k/n} is independent of temperature and pressure. The split fractions of the light and heavy keys are known (ξ_{lk} and ξ_{hk} are known). The column is under total reflux.

Assumptions: Vapor is an ideal gas and the liquid is an ideal solution. Each stage is at equilibrium and is modeled as a flash unit. α_{k/n} is independent of temperature and pressure. The split fractions of the light and heavy keys are known (ξ_{lk} and ξ_{hk} are known). The column is under total reflux.

Fenske Equation

Under the above assumptions, the minimum number of stages is given by the Fenske equation:

N = \frac{ln [\frac{(ξ_{lk}(1 − ξ_{hk}))}{(ξ_{hk}(1 − ξ_{lk}))}]}{ln(α_{lk/hk})}

Once N is known, all other component split fractions can be obtained by substituting k for lk in the Fenske equation. With minor rearrangement:

ξ_{k} = \frac{α_{k}^{N}ξ_{hk}}{1 + (α_{k}^{N} − 1)ξ_{hk}}

Note that when N = 1, we get the expression for split fractions in a flash unit.

Distillation – p.10/19

Under the above assumptions, the minimum number of stages is given by the Fenske equation:

N = \frac{ln [\frac{(ξ_{lk}(1 − ξ_{hk}))}{(ξ_{hk}(1 − ξ_{lk}))}]}{ln(α_{lk/hk})}

Once N is known, all other component split fractions can be obtained by substituting k for lk in the Fenske equation. With minor rearrangement:

ξ_{k} = \frac{α_{k}^{N}ξ_{hk}}{1 + (α_{k}^{N} − 1)ξ_{hk}}

Note that when N = 1, we get the expression for split fractions in a flash unit.

Distillation – p.10/19

Under the above assumptions, the minimum number of stages is given by the Fenske equation:

N = \frac{ln [\frac{(ξ_{lk}(1 − ξ_{hk}))}{(ξ_{hk}(1 − ξ_{lk}))}]}{ln(α_{lk/hk})}

Once N is known, all other component split fractions can be obtained by substituting k for lk in the Fenske equation. With minor rearrangement:

ξ_{k} = \frac{α_{k}^{N}ξ_{hk}}{1 + (α_{k}^{N} − 1)ξ_{hk}}

Note that when N = 1, we get the expression for split fractions in a flash unit.

Distillation – p.10/19

Under the above assumptions, the minimum number of stages is given by the Fenske equation:

N = \frac{ln [\frac{(ξ_{lk}(1 − ξ_{hk}))}{(ξ_{hk}(1 − ξ_{lk}))}]}{ln(α_{lk/hk})}

Once N is known, all other component split fractions can be obtained by substituting k for lk in the Fenske equation. With minor rearrangement:

ξ_{k} = \frac{α_{k}^{N}ξ_{hk}}{1 + (α_{k}^{N} − 1)ξ_{hk}}

Note that when N = 1, we get the expression for split fractions in a flash unit.

Distillation – p.10/19

Selection of column P or T

• Column pressure is lower at the top than at the bottom • More volatile components are higher in concentration at the top

T_{cw} ≤ T_{bub,C} ≤ T_{dew,C} ≤ T_{bub,R} ≤ T_{dew,C} ≤ T_{st} • Column pressure is lower at the top than at the bottom • More volatile components are higher in concentration at the top

T_{cw} ≤ T_{bub,C} ≤ T_{dew,C} ≤ T_{bub,R} ≤ T_{dew,C} ≤ T_{st} • Column pressure is lower at the top than at the bottom • More volatile components are higher in concentration at the top

T_{cw} ≤ T_{bub,C} ≤ T_{dew,C} ≤ T_{bub,R} ≤ T_{dew,C} ≤ T_{st} • Column pressure is lower at the top than at the bottom • More volatile components are higher in concentration at the top

T_{cw} ≤ T_{bub,C} ≤ T_{dew,C} ≤ T_{bub,R} ≤ T_{dew,C} ≤ T_{st} • Column pressure is lower at the top than at the bottom • More volatile components are higher in concentration at the top

T_{cw} ≤ T_{bub,C} ≤ T_{dew,C} ≤ T_{bub,R} ≤ T_{dew,C} ≤ T_{st} • Column pressure is lower at the top than at the bottom • More volatile components are higher in concentration at the top

T_{cw} ≤ T_{bub,C} ≤ T_{dew,C} ≤ T_{bub,R} ≤ T_{dew,C} ≤ T_{st} • Column pressure is lower at the top than at the bottom • More volatile components are higher in concentration at the top

T_{cw} ≤ T_{bub,C} ≤ T_{dew,C} ≤ T_{bub,R} ≤ T_{dew,C} ≤ T_{st} • Column pressure is lower at the top than at the bottom • More volatile components are higher in concentration at the top

T_{cw} ≤ T_{bub,C} ≤ T_{dew,C} ≤ T_{bub,R} ≤ T_{dew,C} ≤ T_{st} Column pressure is selected so that the following constraints hold.

Select the condenser pressure so that T_{bub,C} ≥ 310 K. This ensures that T_{cw} ∼ 303 K.

Select condenser pressure so that all bubble point temperatures are below the critical temperature of the mixture. T_{bub} ≤ T_{cm} where T_{cm} = ∑ x_{k}T_{c}^{k}.

From the bubble point equation we note that T_{bub} increases with P and we prefer to choose P above 1 atm. Thus P = ᾱ_{n}P_{n}^{0}(T_{bub}) ≥ 1 atm Column pressure is selected so that the following constraints hold.

Select the condenser pressure so that T_{bub,C} ≥ 310 K. This ensures that T_{cw} ∼ 303 K.

Select condenser pressure so that all bubble point temperatures are below the critical temperature of the mixture. T_{bub} ≤ T_{cm} where T_{cm} = ∑ x_{k}T_{c}^{k}.

From the bubble point equation we note that T_{bub} increases with P and we prefer to choose P above 1 atm. Thus P = ᾱ_{n}P_{n}^{0}(T_{bub}) ≥ 1 atm Column pressure is selected so that the following constraints hold.

Select the condenser pressure so that T_{bub,C} ≥ 310 K. This ensures that T_{cw} ∼ 303 K.

Select condenser pressure so that all bubble point temperatures are below the critical temperature of the mixture. T_{bub} ≤ T_{cm} where T_{cm} = ∑ x_{k}T_{c}^{k}.

Partial Condenser and Reboiler

From the bubble point equation we note that T_{bub} increases with P and we prefer to choose P above 1 atm. Thus P = ᾱ_{n}P_{n}^{0}(T_{bub}) ≥ 1 atm The conditions given in the previous slide can be difficult to satisfy when we have non-condensible or non-volatile comonents.

If there are non-condensibles in the distillate, a partial condenser should be used.

If there are non-volatiles in the bottoms, a partial reboiler should be used.

Mass and energy balance calculations for partial condensers and reboilers are greatly simplified by noting that the product streams are saturated liquid and vapor as shown in the next slide.

The conditions given in the previous slide can be difficult to satisfy when we have non-condensible or non-volatile comonents.

If there are non-condensibles in the distillate, a partial condenser should be used.

If there are non-volatiles in the bottoms, a partial reboiler should be used.

Mass and energy balance calculations for partial condensers and reboilers are greatly simplified by noting that the product streams are saturated liquid and vapor as shown in the next slide.

The conditions given in the previous slide can be difficult to satisfy when we have non-condensible or non-volatile comonents.

If there are non-condensibles in the distillate, a partial condenser should be used.

If there are non-volatiles in the bottoms, a partial reboiler should be used.

Mass and energy balance calculations for partial condensers and reboilers are greatly simplified by noting that the product streams are saturated liquid and vapor as shown in the next slide.

The conditions given in the previous slide can be difficult to satisfy when we have non-condensible or non-volatile comonents.

If there are non-condensibles in the distillate, a partial condenser should be used.

If there are non-volatiles in the bottoms, a partial reboiler should be used.

Mass and energy balance calculations for partial condensers and reboilers are greatly simplified by noting that the product streams are saturated liquid and vapor as shown in the next slide.

Partial Condenser Calculations

Figure 1 is equivalent to Figure 2.

The calculation of flowrates in Figure 2 is a straightforward flash calculation. T_{cond} can be easily calculated as P is specified and \frac{D_{v}}{D} is

specified. D_{v} is the mole rate of non-condensibles and D is the total distillate flowrate.

Figure 1 is equivalent to Figure 2.

The calculation of flowrates in Figure 2 is a straightforward flash calculation. T_{cond} can be easily calculated as P is specified and \frac{D_{v}}{D} is

specified. D_{v} is the mole rate of non-condensibles and D is the total distillate flowrate.

Figure 1 is equivalent to Figure 2.

The calculation of flowrates in Figure 2 is a straightforward flash calculation. T_{cond} can be easily calculated as P is specified and \frac{D_{v}}{D} is

specified. D_{v} is the mole rate of non-condensibles and D is the total distillate flowrate.

Figure 1 is equivalent to Figure 2.

The calculation of flowrates in Figure 2 is a straightforward flash calculation. T_{cond} can be easily calculated as P is specified and \frac{D_{v}}{D} is

specified. D_{v} is the mole rate of non-condensibles and D is the total distillate flowrate.

Figure 1 is equivalent to Figure 2.

The calculation of flowrates in Figure 2 is a straightforward flash calculation. T_{cond} can be easily calculated as P is specified and \frac{D_{v}}{D} is

specified. D_{v} is the mole rate of non-condensibles and D is the total distillate flowrate.

Total vs. Partial Reboiler

The temperature of the stream exiting the reboiler is the highest temperature in the column.

To avoid excessively high temperatures, a partial reboiler effectively adds an extra equilibrium stage.

The temperature of the stream exiting the reboiler is the highest temperature in the column.

To avoid excessively high temperatures, a partial reboiler effectively adds an extra equilibrium stage.

The temperature of the stream exiting the reboiler is the highest temperature in the column.

To avoid excessively high temperatures, a partial reboiler effectively adds an extra equilibrium stage.

The temperature of the stream exiting the reboiler is the highest temperature in the column.

To avoid excessively high temperatures, a partial reboiler effectively adds an extra equilibrium stage.

Total Reboiler Calculations

Consider a mixture with three components, 1, 2 and 3 where 2 is the key component in the bottoms stream and 3 is a very high boiling component.

In a total reboiler, the y_{1} = x_{1}, y_{2} = x_{2} and y_{3} = x_{3}. The dew point temperature can be computed from

P_{2}^{0}(T_{dew}) = P (\frac{y_{1}}{α_{1/2}} + \frac{y_{2}}{α_{2/2}} + \frac{y_{3}}{α_{3/2}} ).

Since component 3 has a high boiling point, α_{3/2} is a very small number and so the dew temperature T_{dew} will be very high.

Distillation – p.16/19

Consider a mixture with three components, 1, 2 and 3 where 2 is the key component in the bottoms stream and 3 is a very high boiling component.

In a total reboiler, the y_{1} = x_{1}, y_{2} = x_{2} and y_{3} = x_{3}. The dew point temperature can be computed from

P_{2}^{0}(T_{dew}) = P (\frac{y_{1}}{α_{1/2}} + \frac{y_{2}}{α_{2/2}} + \frac{y_{3}}{α_{3/2}} ).

Since component 3 has a high boiling point, α_{3/2} is a very small number and so the dew temperature T_{dew} will be very high.

Distillation – p.16/19

Consider a mixture with three components, 1, 2 and 3 where 2 is the key component in the bottoms stream and 3 is a very high boiling component.

In a total reboiler, the y_{1} = x_{1}, y_{2} = x_{2} and y_{3} = x_{3}. The dew point temperature can be computed from

P_{2}^{0}(T_{dew}) = P (\frac{y_{1}}{α_{1/2}} + \frac{y_{2}}{α_{2/2}} + \frac{y_{3}}{α_{3/2}} ).

Since component 3 has a high boiling point, α_{3/2} is a very small number and so the dew temperature T_{dew} will be very high.

Distillation – p.16/19

Consider a mixture with three components, 1, 2 and 3 where 2 is the key component in the bottoms stream and 3 is a very high boiling component.

In a total reboiler, the y_{1} = x_{1}, y_{2} = x_{2} and y_{3} = x_{3}. The dew point temperature can be computed from

P_{2}^{0}(T_{dew}) = P (\frac{y_{1}}{α_{1/2}} + \frac{y_{2}}{α_{2/2}} + \frac{y_{3}}{α_{3/2}} ).

Since component 3 has a high boiling point, α_{3/2} is a very small number and so the dew temperature T_{dew} will be very high.

Distillation – p.16/19

Partial Reboiler Calculations

Consider the same mixture as in the previous slide. In a partial reboiler, y_{i} and x_{i} are in equilibrium since the reboiler is effectively a flash unit.

Since component 3 has a very high boiling point, its mole fraction in the vapor stage is expected to be almost zero at reasonable temperatures. The dew point temperature can be computed from

P_{2}^{0}(T_{dew}) = P (\frac{y_{1}}{α_{1/2}} + \frac{y_{2}}{α_{2/2}} ).

Since the term \frac{y_{3}}{α_{3/2}} is negligible, the dew temperature

T_{dew} will not be as high as in a total reboiler.

Distillation – p.17/19

Consider the same mixture as in the previous slide. In a partial reboiler, y_{i} and x_{i} are in equilibrium since the reboiler is effectively a flash unit.

Since component 3 has a very high boiling point, its mole fraction in the vapor stage is expected to be almost zero at reasonable temperatures. The dew point temperature can be computed from

P_{2}^{0}(T_{dew}) = P (\frac{y_{1}}{α_{1/2}} + \frac{y_{2}}{α_{2/2}} ).

Since the term \frac{y_{3}}{α_{3/2}} is negligible, the dew temperature

T_{dew} will not be as high as in a total reboiler.

Distillation – p.17/19

Consider the same mixture as in the previous slide. In a partial reboiler, y_{i} and x_{i} are in equilibrium since the reboiler is effectively a flash unit.

Since component 3 has a very high boiling point, its mole fraction in the vapor stage is expected to be almost zero at reasonable temperatures. The dew point temperature can be computed from

P_{2}^{0}(T_{dew}) = P (\frac{y_{1}}{α_{1/2}} + \frac{y_{2}}{α_{2/2}} ).

Since the term \frac{y_{3}}{α_{3/2}} is negligible, the dew temperature

T_{dew} will not be as high as in a total reboiler.

Distillation – p.17/19

Consider the same mixture as in the previous slide. In a partial reboiler, y_{i} and x_{i} are in equilibrium since the reboiler is effectively a flash unit.

Since component 3 has a very high boiling point, its mole fraction in the vapor stage is expected to be almost zero at reasonable temperatures. The dew point temperature can be computed from

P_{2}^{0}(T_{dew}) = P (\frac{y_{1}}{α_{1/2}} + \frac{y_{2}}{α_{2/2}} ).

Since the term \frac{y_{3}}{α_{3/2}} is negligible, the dew temperature

T_{dew} will not be as high as in a total reboiler.

Distillation – p.17/19

Reflux Ratio and Theoretical Stages

Calculate the actual theoretical stages and reflux ratio from the following equations (Westerberg correlations)

N_{i} = \frac{12.3}{(α_{lk/hk} − 1)^{2/3}(1 − β_{i})^{1/6}} i = lk, hk

where β_{lk} is the split fraction of the light key at the top (β_{lk} = ξ_{lk}) and β_{hk} is the split fraction of the heavy key at the bottom (β_{hk} = 1 − ξ_{hk}).

R_{i} = \frac{1.38}{(α_{lk/hk} − 1)^{0.9}(1 − β_{i})^{0.1}} i = lk, hk

From N_{lk} and N_{hk} calculate the actual theoretical stages as: N_{T} = 0.8max(N_{i}) + 0.2min(N_{i})

From R_{lk} and R_{hk} calculate the reflux ratio as: R = 0.8max(R_{i}) + 0.2min(R_{i})

Distillation – p.18/19

Calculate the actual theoretical stages and reflux ratio from the following equations (Westerberg correlations)

N_{i} = \frac{12.3}{(α_{lk/hk} − 1)^{2/3}(1 − β_{i})^{1/6}} i = lk, hk

where β_{lk} is the split fraction of the light key at the top (β_{lk} = ξ_{lk}) and β_{hk} is the split fraction of the heavy key at the bottom (β_{hk} = 1 − ξ_{hk}).

R_{i} = \frac{1.38}{(α_{lk/hk} − 1)^{0.9}(1 − β_{i})^{0.1}} i = lk, hk

From N_{lk} and N_{hk} calculate the actual theoretical stages as: N_{T} = 0.8max(N_{i}) + 0.2min(N_{i})

From R_{lk} and R_{hk} calculate the reflux ratio as: R = 0.8max(R_{i}) + 0.2min(R_{i})

Distillation – p.18/19

Calculate the actual theoretical stages and reflux ratio from the following equations (Westerberg correlations)

N_{i} = \frac{12.3}{(α_{lk/hk} − 1)^{2/3}(1 − β_{i})^{1/6}} i = lk, hk

where β_{lk} is the split fraction of the light key at the top (β_{lk} = ξ_{lk}) and β_{hk} is the split fraction of the heavy key at the bottom (β_{hk} = 1 − ξ_{hk}).

R_{i} = \frac{1.38}{(α_{lk/hk} − 1)^{0.9}(1 − β_{i})^{0.1}} i = lk, hk

From N_{lk} and N_{hk} calculate the actual theoretical stages as: N_{T} = 0.8max(N_{i}) + 0.2min(N_{i})

From R_{lk} and R_{hk} calculate the reflux ratio as: R = 0.8max(R_{i}) + 0.2min(R_{i})

Distillation – p.18/19

Calculate the actual theoretical stages and reflux ratio from the following equations (Westerberg correlations)

N_{i} = \frac{12.3}{(α_{lk/hk} − 1)^{2/3}(1 − β_{i})^{1/6}} i = lk, hk

where β_{lk} is the split fraction of the light key at the top (β_{lk} = ξ_{lk}) and β_{hk} is the split fraction of the heavy key at the bottom (β_{hk} = 1 − ξ_{hk}).

R_{i} = \frac{1.38}{(α_{lk/hk} − 1)^{0.9}(1 − β_{i})^{0.1}} i = lk, hk

From N_{lk} and N_{hk} calculate the actual theoretical stages as: N_{T} = 0.8max(N_{i}) + 0.2min(N_{i})

From R_{lk} and R_{hk} calculate the reflux ratio as: R = 0.8max(R_{i}) + 0.2min(R_{i})

Distillation – p.18/19

Calculate the actual theoretical stages and reflux ratio from the following equations (Westerberg correlations)

N_{i} = \frac{12.3}{(α_{lk/hk} − 1)^{2/3}(1 − β_{i})^{1/6}} i = lk, hk

where β_{lk} is the split fraction of the light key at the top (β_{lk} = ξ_{lk}) and β_{hk} is the split fraction of the heavy key at the bottom (β_{hk} = 1 − ξ_{hk}).

R_{i} = \frac{1.38}{(α_{lk/hk} − 1)^{0.9}(1 − β_{i})^{0.1}} i = lk, hk

From N_{lk} and N_{hk} calculate the actual theoretical stages as: N_{T} = 0.8max(N_{i}) + 0.2min(N_{i})

From R_{lk} and R_{hk} calculate the reflux ratio as: R = 0.8max(R_{i}) + 0.2min(R_{i})

Distillation – p.18/19

Design Procedure for Distillation

Distillation – p.19/19

Design Procedure for Distillation

Distillation – p.19/19