Gas Absorption with Plate Absorbers

Gas Absorption

One man’s magic is another man’s engineering. Lazarus Long Absorption represents a vapor recovery operation where the desired component is transferred from a gas to the liquid phase through countercurrent mass transfer.

This unit operation is modeled as a cascade of equilibrium stages

The assumption of equilibrium is weak and is corrected by the use of tray efficiencies Absorption represents a vapor recovery operation where the desired component is transferred from a gas to the liquid phase through countercurrent mass transfer.

This unit operation is modeled as a cascade of equilibrium stages

The assumption of equilibrium is weak and is corrected by the use of tray efficiencies Absorption represents a vapor recovery operation where the desired component is transferred from a gas to the liquid phase through countercurrent mass transfer.

This unit operation is modeled as a cascade of equilibrium stages

The assumption of equilibrium is weak and is corrected by the use of tray efficiencies Given a vapor feed, there are 4 degrees of freedom:

1. Pressure (P)

2. Number of equilibrium stages (N) for a desired recovery of key component (or vice versa)

3. Temperature (T_{0}) of the absorbing liquid stream 4. Flowrate (L_{0}) of the absorbing liquid stream Given a vapor feed, there are 4 degrees of freedom:

1. Pressure (P)

2. Number of equilibrium stages (N) for a desired recovery of key component (or vice versa)

3. Temperature (T_{0}) of the absorbing liquid stream 4. Flowrate (L_{0}) of the absorbing liquid stream Given a vapor feed, there are 4 degrees of freedom:

1. Pressure (P)

2. Number of equilibrium stages (N) for a desired recovery of key component (or vice versa)

3. Temperature (T_{0}) of the absorbing liquid stream 4. Flowrate (L_{0}) of the absorbing liquid stream Given a vapor feed, there are 4 degrees of freedom:

1. Pressure (P)

2. Number of equilibrium stages (N) for a desired recovery of key component (or vice versa)

3. Temperature (T_{0}) of the absorbing liquid stream 4. Flowrate (L_{0}) of the absorbing liquid stream Given a vapor feed, there are 4 degrees of freedom:

1. Pressure (P)

2. Number of equilibrium stages (N) for a desired recovery of key component (or vice versa)

3. Temperature (T_{0}) of the absorbing liquid stream 4. Flowrate (L_{0}) of the absorbing liquid stream Given these 4 specifications, we can derive the mass balance relations The liquid and vapor streams leaving a stage are at equilibrium. Thus

\frac{v_{i}}{V_{i}} = K_{i} \frac{l_{i}}{L_{i}}

If we define an absorption factor A_{i} = \frac{L_{i}}{K_{i}V_{i}}, we can write

the above equation as

l_{i} = A_{i}v_{i}

For each stage, the absorption factor A_{i} will be different. If we assume a constant absorption factor A_{E} for all the stages, we get simplified mass balance equations.

Gas Absorption with Plate Absorbers – p.5/13

Given these 4 specifications, we can derive the mass balance relations The liquid and vapor streams leaving a stage are at equilibrium. Thus

\frac{v_{i}}{V_{i}} = K_{i} \frac{l_{i}}{L_{i}}

If we define an absorption factor A_{i} = \frac{L_{i}}{K_{i}V_{i}}, we can write

the above equation as

l_{i} = A_{i}v_{i}

For each stage, the absorption factor A_{i} will be different. If we assume a constant absorption factor A_{E} for all the stages, we get simplified mass balance equations.

Gas Absorption with Plate Absorbers – p.5/13

Given these 4 specifications, we can derive the mass balance relations The liquid and vapor streams leaving a stage are at equilibrium. Thus

\frac{v_{i}}{V_{i}} = K_{i} \frac{l_{i}}{L_{i}}

If we define an absorption factor A_{i} = \frac{L_{i}}{K_{i}V_{i}}, we can write

the above equation as

l_{i} = A_{i}v_{i}

For each stage, the absorption factor A_{i} will be different. If we assume a constant absorption factor A_{E} for all the stages, we get simplified mass balance equations.

Gas Absorption with Plate Absorbers – p.5/13

Given these 4 specifications, we can derive the mass balance relations The liquid and vapor streams leaving a stage are at equilibrium. Thus

\frac{v_{i}}{V_{i}} = K_{i} \frac{l_{i}}{L_{i}}

If we define an absorption factor A_{i} = \frac{L_{i}}{K_{i}V_{i}}, we can write

the above equation as

l_{i} = A_{i}v_{i}

For each stage, the absorption factor A_{i} will be different. If we assume a constant absorption factor A_{E} for all the stages, we get simplified mass balance equations.

Gas Absorption with Plate Absorbers – p.5/13

Given these 4 specifications, we can derive the mass balance relations The liquid and vapor streams leaving a stage are at equilibrium. Thus

\frac{v_{i}}{V_{i}} = K_{i} \frac{l_{i}}{L_{i}}

If we define an absorption factor A_{i} = \frac{L_{i}}{K_{i}V_{i}}, we can write

the above equation as

l_{i} = A_{i}v_{i}

For each stage, the absorption factor A_{i} will be different. If we assume a constant absorption factor A_{E} for all the stages, we get simplified mass balance equations.

Gas Absorption with Plate Absorbers – p.5/13

Given these 4 specifications, we can derive the mass balance relations The liquid and vapor streams leaving a stage are at equilibrium. Thus

\frac{v_{i}}{V_{i}} = K_{i} \frac{l_{i}}{L_{i}}

If we define an absorption factor A_{i} = \frac{L_{i}}{K_{i}V_{i}}, we can write

the above equation as

l_{i} = A_{i}v_{i}

For each stage, the absorption factor A_{i} will be different. If we assume a constant absorption factor A_{E} for all the stages, we get simplified mass balance equations.

Gas Absorption with Plate Absorbers – p.5/13

For the first stage, we have:

\begin{aligned}& l_{1} + v_{1} & = & l_{0} + v_{2} \\ or & v_{2} & = & (A_{E} + 1)v_{1} − l_{0} \end{aligned}

For each subsequent stage, we have:

\begin{aligned}v_{i+1} & = & l_{i} + v_{i} − l_{i−1} \\ v_{i+1} & = & (A_{E} + 1)v_{i} − A_{E}v_{i−1} \end{aligned}

Writing the above balance from tray 1 to tray N relates the vapor flow at the bottom, v_{N+1} to the vapor flow at the top, v_{1} and the liquid flow at the top, l_{0}.

v_{N+1} = ∑_{i=0}^{N} (A_{E})^{i} v_{1} − ∑_{i=0}^{N−1} (A_{E})^{i} l_{0}

\begin{aligned}& l_{1} + v_{1} & = & l_{0} + v_{2} \\ or & v_{2} & = & (A_{E} + 1)v_{1} − l_{0} \end{aligned}

For each subsequent stage, we have:

\begin{aligned}v_{i+1} & = & l_{i} + v_{i} − l_{i−1} \\ v_{i+1} & = & (A_{E} + 1)v_{i} − A_{E}v_{i−1} \end{aligned}

Writing the above balance from tray 1 to tray N relates the vapor flow at the bottom, v_{N+1} to the vapor flow at the top, v_{1} and the liquid flow at the top, l_{0}.

v_{N+1} = ∑_{i=0}^{N} (A_{E})^{i} v_{1} − ∑_{i=0}^{N−1} (A_{E})^{i} l_{0}

\begin{aligned}& l_{1} + v_{1} & = & l_{0} + v_{2} \\ or & v_{2} & = & (A_{E} + 1)v_{1} − l_{0} \end{aligned}

For each subsequent stage, we have:

\begin{aligned}v_{i+1} & = & l_{i} + v_{i} − l_{i−1} \\ v_{i+1} & = & (A_{E} + 1)v_{i} − A_{E}v_{i−1} \end{aligned}

Writing the above balance from tray 1 to tray N relates the vapor flow at the bottom, v_{N+1} to the vapor flow at the top, v_{1} and the liquid flow at the top, l_{0}.

v_{N+1} = ∑_{i=0}^{N} (A_{E})^{i} v_{1} − ∑_{i=0}^{N−1} (A_{E})^{i} l_{0}

\begin{aligned}& l_{1} + v_{1} & = & l_{0} + v_{2} \\ or & v_{2} & = & (A_{E} + 1)v_{1} − l_{0} \end{aligned}

For each subsequent stage, we have:

\begin{aligned}v_{i+1} & = & l_{i} + v_{i} − l_{i−1} \\ v_{i+1} & = & (A_{E} + 1)v_{i} − A_{E}v_{i−1} \end{aligned}

Writing the above balance from tray 1 to tray N relates the vapor flow at the bottom, v_{N+1} to the vapor flow at the top, v_{1} and the liquid flow at the top, l_{0}.

v_{N+1} = ∑_{i=0}^{N} (A_{E})^{i} v_{1} − ∑_{i=0}^{N−1} (A_{E})^{i} l_{0}

Define

β_{N} = ∑_{i=0}^{N} (A_{E})^{i}

Multiplying both sides of the above equation by (1 − A_{E}) results in:

(1 − A_{E})β_{N} = ∑_{i=0}^{N} (A_{E})^{i} − ∑_{i=1}^{N+1} (A_{E})^{i}

which implies

β_{N} = \frac{1 − A_{E}^{N+1}}{1 − A_{E}}

which simplifies the relationship for v_{N+1} to:

β_{N} = ∑_{i=0}^{N} (A_{E})^{i}

Multiplying both sides of the above equation by (1 − A_{E}) results in:

(1 − A_{E})β_{N} = ∑_{i=0}^{N} (A_{E})^{i} − ∑_{i=1}^{N+1} (A_{E})^{i}

which implies

β_{N} = \frac{1 − A_{E}^{N+1}}{1 − A_{E}}

which simplifies the relationship for v_{N+1} to:

β_{N} = ∑_{i=0}^{N} (A_{E})^{i}

Multiplying both sides of the above equation by (1 − A_{E}) results in:

(1 − A_{E})β_{N} = ∑_{i=0}^{N} (A_{E})^{i} − ∑_{i=1}^{N+1} (A_{E})^{i}

which implies

β_{N} = \frac{1 − A_{E}^{N+1}}{1 − A_{E}}

which simplifies the relationship for v_{N+1} to:

β_{N} = ∑_{i=0}^{N} (A_{E})^{i}

Multiplying both sides of the above equation by (1 − A_{E}) results in:

(1 − A_{E})β_{N} = ∑_{i=0}^{N} (A_{E})^{i} − ∑_{i=1}^{N+1} (A_{E})^{i}

which implies

β_{N} = \frac{1 − A_{E}^{N+1}}{1 − A_{E}}

which simplifies the relationship for v_{N+1} to:

v_{N+1} = β_{N}v_{1} − β_{N−1}l_{0}

The liquid flow rate at the bottom can be calculated from an overall mass balance to yield:

l_{N} = v_{N+1} + l_{0} − v_{1}

The above equations are valid for any component k in the column.

Gas Absorption with Plate Absorbers – p.8/13

The liquid flow rate at the bottom can be calculated from an overall mass balance to yield:

l_{N} = v_{N+1} + l_{0} − v_{1}

The above equations are valid for any component k in the column.

Gas Absorption with Plate Absorbers – p.8/13

The liquid flow rate at the bottom can be calculated from an overall mass balance to yield:

l_{N} = v_{N+1} + l_{0} − v_{1}

The above equations are valid for any component k in the column.

Gas Absorption with Plate Absorbers – p.8/13

Calculation of Number of Stages

We define a recovery fraction, r, for the key component n as follows:

v_{1}^{n} = (1 − r)v_{N+1}^{n}

Substituting this in the mass balance equations in the previous slide, we get

v_{N+1}^{n} = β_{N}(1 − r)v_{N+1}^{n} − β_{N−1}l_{0}^{n}

Substituting for β_{N} in terms of A_{E}, and solving for N results in the Kremser equation:

N = \frac{ln \{\frac{l_{0}^{n} + (r − A_{E})v_{N+1}^{n}}{l_{0}^{n} − A_{E}(1 − r)v_{N+1}^{n}} \}}{ln \{A_{E}\}}

We define a recovery fraction, r, for the key component n as follows:

v_{1}^{n} = (1 − r)v_{N+1}^{n}

Substituting this in the mass balance equations in the previous slide, we get

v_{N+1}^{n} = β_{N}(1 − r)v_{N+1}^{n} − β_{N−1}l_{0}^{n}

Substituting for β_{N} in terms of A_{E}, and solving for N results in the Kremser equation:

N = \frac{ln \{\frac{l_{0}^{n} + (r − A_{E})v_{N+1}^{n}}{l_{0}^{n} − A_{E}(1 − r)v_{N+1}^{n}} \}}{ln \{A_{E}\}}

We define a recovery fraction, r, for the key component n as follows:

v_{1}^{n} = (1 − r)v_{N+1}^{n}

Substituting this in the mass balance equations in the previous slide, we get

v_{N+1}^{n} = β_{N}(1 − r)v_{N+1}^{n} − β_{N−1}l_{0}^{n}

Substituting for β_{N} in terms of A_{E}, and solving for N results in the Kremser equation:

N = \frac{ln \{\frac{l_{0}^{n} + (r − A_{E})v_{N+1}^{n}}{l_{0}^{n} − A_{E}(1 − r)v_{N+1}^{n}} \}}{ln \{A_{E}\}}

We define a recovery fraction, r, for the key component n as follows:

v_{1}^{n} = (1 − r)v_{N+1}^{n}

Substituting this in the mass balance equations in the previous slide, we get

v_{N+1}^{n} = β_{N}(1 − r)v_{N+1}^{n} − β_{N−1}l_{0}^{n}

Substituting for β_{N} in terms of A_{E}, and solving for N results in the Kremser equation:

N = \frac{ln \{\frac{l_{0}^{n} + (r − A_{E})v_{N+1}^{n}}{l_{0}^{n} − A_{E}(1 − r)v_{N+1}^{n}} \}}{ln \{A_{E}\}}

If none of the key component appears in the liquid feed (l_{0}^{n} = 0), the above equation simplifies to

N = \frac{ln \{\frac{(r − A_{E})}{A_{E}(r − 1)}\}}{ln \{A_{E}\}}

We choose 4 degrees of freedom:

r, the recovery of the key component, n Overhead column pressure Solvent temperature (assume isothermal operation) The absorption factor A_{E} = 1.4 (choose this as initial guess) which fixes the liquid flow rate

Gas Absorption with Plate Absorbers – p.10/13

N = \frac{ln \{\frac{(r − A_{E})}{A_{E}(r − 1)}\}}{ln \{A_{E}\}}

We choose 4 degrees of freedom:

r, the recovery of the key component, n Overhead column pressure Solvent temperature (assume isothermal operation) The absorption factor A_{E} = 1.4 (choose this as initial guess) which fixes the liquid flow rate

Gas Absorption with Plate Absorbers – p.10/13

N = \frac{ln \{\frac{(r − A_{E})}{A_{E}(r − 1)}\}}{ln \{A_{E}\}}

We choose 4 degrees of freedom:

r, the recovery of the key component, n Overhead column pressure Solvent temperature (assume isothermal operation) The absorption factor A_{E} = 1.4 (choose this as initial guess) which fixes the liquid flow rate

Gas Absorption with Plate Absorbers – p.10/13

N = \frac{ln \{\frac{(r − A_{E})}{A_{E}(r − 1)}\}}{ln \{A_{E}\}}

We choose 4 degrees of freedom:

r, the recovery of the key component, n Overhead column pressure Solvent temperature (assume isothermal operation) The absorption factor A_{E} = 1.4 (choose this as initial guess) which fixes the liquid flow rate

Gas Absorption with Plate Absorbers – p.10/13

N = \frac{ln \{\frac{(r − A_{E})}{A_{E}(r − 1)}\}}{ln \{A_{E}\}}

We choose 4 degrees of freedom:

r, the recovery of the key component, n Overhead column pressure Solvent temperature (assume isothermal operation) The absorption factor A_{E} = 1.4 (choose this as initial guess) which fixes the liquid flow rate

Gas Absorption with Plate Absorbers – p.10/13

N = \frac{ln \{\frac{(r − A_{E})}{A_{E}(r − 1)}\}}{ln \{A_{E}\}}

We choose 4 degrees of freedom:

r, the recovery of the key component, n Overhead column pressure Solvent temperature (assume isothermal operation) The absorption factor A_{E} = 1.4 (choose this as initial guess) which fixes the liquid flow rate

Gas Absorption with Plate Absorbers – p.10/13

N = \frac{ln \{\frac{(r − A_{E})}{A_{E}(r − 1)}\}}{ln \{A_{E}\}}

We choose 4 degrees of freedom:

r, the recovery of the key component, n Overhead column pressure Solvent temperature (assume isothermal operation) The absorption factor A_{E} = 1.4 (choose this as initial guess) which fixes the liquid flow rate

Gas Absorption with Plate Absorbers – p.10/13

Absorption Algorithm

Select key component n, fix recovery (typically r = 0.99), fix P and solvent temperature.

Calculate L_{0} from

\begin{aligned}A_{E} & = & \frac{L_{0}}{V_{N+1}K_{n}} & = & 1.4 \\ L_{0} & = & 1.4V_{N+1} \frac{P_{N}^{0}(T)}{P} \end{aligned}

Calculate the number of stages from the Kremser equation

N = \frac{ln \{\frac{l_{0}^{n} + (r − A_{E})v_{N+1}^{n}}{l_{0}^{n} − A_{E}(1 − r)v_{N+1}^{n}} \}}{ln \{A_{E}\}} Gas Absorption with Plate Absorbers – p.11/13

Select key component n, fix recovery (typically r = 0.99), fix P and solvent temperature.

Calculate L_{0} from

\begin{aligned}A_{E} & = & \frac{L_{0}}{V_{N+1}K_{n}} & = & 1.4 \\ L_{0} & = & 1.4V_{N+1} \frac{P_{N}^{0}(T)}{P} \end{aligned}

Calculate the number of stages from the Kremser equation

N = \frac{ln \{\frac{l_{0}^{n} + (r − A_{E})v_{N+1}^{n}}{l_{0}^{n} − A_{E}(1 − r)v_{N+1}^{n}} \}}{ln \{A_{E}\}} Gas Absorption with Plate Absorbers – p.11/13

Select key component n, fix recovery (typically r = 0.99), fix P and solvent temperature.

Calculate L_{0} from

\begin{aligned}A_{E} & = & \frac{L_{0}}{V_{N+1}K_{n}} & = & 1.4 \\ L_{0} & = & 1.4V_{N+1} \frac{P_{N}^{0}(T)}{P} \end{aligned}

Calculate the number of stages from the Kremser equation

N = \frac{ln \{\frac{l_{0}^{n} + (r − A_{E})v_{N+1}^{n}}{l_{0}^{n} − A_{E}(1 − r)v_{N+1}^{n}} \}}{ln \{A_{E}\}} Gas Absorption with Plate Absorbers – p.11/13

Select key component n, fix recovery (typically r = 0.99), fix P and solvent temperature.

Calculate L_{0} from

\begin{aligned}A_{E} & = & \frac{L_{0}}{V_{N+1}K_{n}} & = & 1.4 \\ L_{0} & = & 1.4V_{N+1} \frac{P_{N}^{0}(T)}{P} \end{aligned}

Calculate the number of stages from the Kremser equation

N = \frac{ln \{\frac{l_{0}^{n} + (r − A_{E})v_{N+1}^{n}}{l_{0}^{n} − A_{E}(1 − r)v_{N+1}^{n}} \}}{ln \{A_{E}\}}

Gas Absorption with Plate Absorbers – p.11/13

Calculate absorption factors for non-key components

\begin{aligned}& A^{k} & = & \frac{L_{0}}{V_{N+1}} \frac{P}{P_{k}^{0}(T)} & k ≠ n \\ or & A^{k} & = & \frac{1.4}{α_{k/n}} \end{aligned}

Calculate aggregate terms (β values) for non-key components

\begin{aligned}β_{N}^{k} & = & \frac{[1 − (A^{k})^{N+1}]}{1 − A^{k}} \\ β_{N−1}^{k} & = & \frac{[1 − (A^{k})^{N}]}{1 − A^{k}} \end{aligned}

\begin{aligned}& A^{k} & = & \frac{L_{0}}{V_{N+1}} \frac{P}{P_{k}^{0}(T)} & k ≠ n \\ or & A^{k} & = & \frac{1.4}{α_{k/n}} \end{aligned}

Calculate aggregate terms (β values) for non-key components

\begin{aligned}β_{N}^{k} & = & \frac{[1 − (A^{k})^{N+1}]}{1 − A^{k}} \\ β_{N−1}^{k} & = & \frac{[1 − (A^{k})^{N}]}{1 − A^{k}} \end{aligned}

Complete the mass balance for all components

\begin{aligned}v_{1}^{k} & = & \frac{v_{N+1}^{k}}{β_{N}^{k}} + \frac{β_{N−1}^{k}}{β_{N}^{k}} l_{0}^{k} \\ l_{N}^{k} & = & (1 − \frac{β_{N−1}^{k}}{β_{N}^{k}} ) l_{0}^{k} + (1 − \frac{1}{β_{N}^{k}} ) v_{N+1}^{k} \end{aligned}

If necessary, readjust P and T and return to step 1 under the following conditions

1. If the temperature of l_{N} is too high, increase L_{0} (check with bubble point equation).

2. If too much solvent vaporizes in v_{1}, increase P or decrease T.

3. If too many undesirable components are absorbed, increase T, decrease P, or select a more suitable solvent for absorption.

Gas Absorption with Plate Absorbers – p.13/13

\begin{aligned}v_{1}^{k} & = & \frac{v_{N+1}^{k}}{β_{N}^{k}} + \frac{β_{N−1}^{k}}{β_{N}^{k}} l_{0}^{k} \\ l_{N}^{k} & = & (1 − \frac{β_{N−1}^{k}}{β_{N}^{k}} ) l_{0}^{k} + (1 − \frac{1}{β_{N}^{k}} ) v_{N+1}^{k} \end{aligned}

If necessary, readjust P and T and return to step 1 under the following conditions

1. If the temperature of l_{N} is too high, increase L_{0} (check with bubble point equation).

2. If too much solvent vaporizes in v_{1}, increase P or decrease T.

3. If too many undesirable components are absorbed, increase T, decrease P, or select a more suitable solvent for absorption.

\begin{aligned}v_{1}^{k} & = & \frac{v_{N+1}^{k}}{β_{N}^{k}} + \frac{β_{N−1}^{k}}{β_{N}^{k}} l_{0}^{k} \\ l_{N}^{k} & = & (1 − \frac{β_{N−1}^{k}}{β_{N}^{k}} ) l_{0}^{k} + (1 − \frac{1}{β_{N}^{k}} ) v_{N+1}^{k} \end{aligned}

If necessary, readjust P and T and return to step 1 under the following conditions

1. If the temperature of l_{N} is too high, increase L_{0} (check with bubble point equation).

2. If too much solvent vaporizes in v_{1}, increase P or decrease T.

3. If too many undesirable components are absorbed, increase T, decrease P, or select a more suitable solvent for absorption.

\begin{aligned}v_{1}^{k} & = & \frac{v_{N+1}^{k}}{β_{N}^{k}} + \frac{β_{N−1}^{k}}{β_{N}^{k}} l_{0}^{k} \\ l_{N}^{k} & = & (1 − \frac{β_{N−1}^{k}}{β_{N}^{k}} ) l_{0}^{k} + (1 − \frac{1}{β_{N}^{k}} ) v_{N+1}^{k} \end{aligned}

If necessary, readjust P and T and return to step 1 under the following conditions

1. If the temperature of l_{N} is too high, increase L_{0} (check with bubble point equation).

2. If too much solvent vaporizes in v_{1}, increase P or decrease T.

3. If too many undesirable components are absorbed, increase T, decrease P, or select a more suitable solvent for absorption.

\begin{aligned}v_{1}^{k} & = & \frac{v_{N+1}^{k}}{β_{N}^{k}} + \frac{β_{N−1}^{k}}{β_{N}^{k}} l_{0}^{k} \\ l_{N}^{k} & = & (1 − \frac{β_{N−1}^{k}}{β_{N}^{k}} ) l_{0}^{k} + (1 − \frac{1}{β_{N}^{k}} ) v_{N+1}^{k} \end{aligned}

If necessary, readjust P and T and return to step 1 under the following conditions

1. If the temperature of l_{N} is too high, increase L_{0} (check with bubble point equation).

2. If too much solvent vaporizes in v_{1}, increase P or decrease T.

3. If too many undesirable components are absorbed, increase T, decrease P, or select a more suitable solvent for absorption.