Solving Flowsheet Equations

Solving Flowsheet Equations – p.1/8

Why, a four-year-old child could understand this. Someone get me a four-year-old child

Groucho Marx

Solving Linear Mass Balance Equations

All units except the reactor have independent split fractions for each component.

The reactor mass balance relates flows to limiting components in the reaction.

If we solve for the limiting components first, we can solve for all other flows.

\underline{Option 1}: Solve simultaneously, the linear mass balances for the flowsheet.

\underline{Option 2}: Solve sequentially, using a tearing algorithm.

1. Choose tear streams that break all recycle loops in flowsheet (typically at the reactor inlet).

2. Trace path backwards from reactor until all loops are covered (end up at the reactor inlet again).

3. Fill all streams by using split fractions and moving

forward from the reactor feed.

Solving Flowsheet Equations – p.3/8

Solving Linear Mass Balance Equations

All units except the reactor have independent split fractions for each component.

The reactor mass balance relates flows to limiting components in the reaction.

If we solve for the limiting components first, we can solve for all other flows.

\underline{Option 1}: Solve simultaneously, the linear mass balances for the flowsheet.

\underline{Option 2}: Solve sequentially, using a tearing algorithm.

1. Choose tear streams that break all recycle loops in flowsheet (typically at the reactor inlet).

2. Trace path backwards from reactor until all loops are covered (end up at the reactor inlet again).

3. Fill all streams by using split fractions and moving

forward from the reactor feed.

Solving Flowsheet Equations – p.3/8

Solving Linear Mass Balance Equations

All units except the reactor have independent split fractions for each component.

The reactor mass balance relates flows to limiting components in the reaction.

If we solve for the limiting components first, we can solve for all other flows.

\underline{Option 1}: Solve simultaneously, the linear mass balances for the flowsheet.

\underline{Option 2}: Solve sequentially, using a tearing algorithm.

1. Choose tear streams that break all recycle loops in flowsheet (typically at the reactor inlet).

2. Trace path backwards from reactor until all loops are covered (end up at the reactor inlet again).

3. Fill all streams by using split fractions and moving

forward from the reactor feed.

Solving Flowsheet Equations – p.3/8

Solving Linear Mass Balance Equations

All units except the reactor have independent split fractions for each component.

The reactor mass balance relates flows to limiting components in the reaction.

If we solve for the limiting components first, we can solve for all other flows.

\underline{Option 1}: Solve simultaneously, the linear mass balances for the flowsheet.

\underline{Option 2}: Solve sequentially, using a tearing algorithm.

1. Choose tear streams that break all recycle loops in flowsheet (typically at the reactor inlet).

2. Trace path backwards from reactor until all loops are covered (end up at the reactor inlet again).

3. Fill all streams by using split fractions and moving

forward from the reactor feed.

Solving Flowsheet Equations – p.3/8

Solving Linear Mass Balance Equations

All units except the reactor have independent split fractions for each component.

The reactor mass balance relates flows to limiting components in the reaction.

If we solve for the limiting components first, we can solve for all other flows.

\underline{Option 1}: Solve simultaneously, the linear mass balances for the flowsheet.

\underline{Option 2}: Solve sequentially, using a tearing algorithm.

1. Choose tear streams that break all recycle loops in flowsheet (typically at the reactor inlet).

2. Trace path backwards from reactor until all loops are covered (end up at the reactor inlet again).

3. Fill all streams by using split fractions and moving

forward from the reactor feed.

Solving Flowsheet Equations – p.3/8

Solving Linear Mass Balance Equations

All units except the reactor have independent split fractions for each component.

The reactor mass balance relates flows to limiting components in the reaction.

If we solve for the limiting components first, we can solve for all other flows.

\underline{Option 1}: Solve simultaneously, the linear mass balances for the flowsheet.

\underline{Option 2}: Solve sequentially, using a tearing algorithm.

1. Choose tear streams that break all recycle loops in flowsheet (typically at the reactor inlet).

2. Trace path backwards from reactor until all loops are covered (end up at the reactor inlet again).

3. Fill all streams by using split fractions and moving

forward from the reactor feed.

Solving Flowsheet Equations – p.3/8

Solving Linear Mass Balance Equations

All units except the reactor have independent split fractions for each component.

The reactor mass balance relates flows to limiting components in the reaction.

If we solve for the limiting components first, we can solve for all other flows.

\underline{Option 1}: Solve simultaneously, the linear mass balances for the flowsheet.

\underline{Option 2}: Solve sequentially, using a tearing algorithm.

1. Choose tear streams that break all recycle loops in flowsheet (typically at the reactor inlet).

2. Trace path backwards from reactor until all loops are covered (end up at the reactor inlet again).

3. Fill all streams by using split fractions and moving

forward from the reactor feed.

Solving Flowsheet Equations – p.3/8

Solving Linear Mass Balance Equations

All units except the reactor have independent split fractions for each component.

The reactor mass balance relates flows to limiting components in the reaction.

If we solve for the limiting components first, we can solve for all other flows.

\underline{Option 1}: Solve simultaneously, the linear mass balances for the flowsheet.

\underline{Option 2}: Solve sequentially, using a tearing algorithm.

1. Choose tear streams that break all recycle loops in flowsheet (typically at the reactor inlet).

2. Trace path backwards from reactor until all loops are covered (end up at the reactor inlet again).

3. Fill all streams by using split fractions and moving

forward from the reactor feed.

Solving Flowsheet Equations – p.3/8

Ethanol BFD

Solving Flowsheet Equations – p.4/8

Balance for EL and P L

μ_{1}(EL) = μ_{02}(EL) + μ_{51}(EL) + μ_{81}(EL)
= μ_{02}(EL) + (0.995)(0.979)(0.985)(0.93)μ_{1}(EL)+

(1)(1){0.021(0.985) + 0.015}(0.93)μ_{1}(EL)

= 96 + 0.9255μ_{1}(EL)

Solving for μ_{1}(EL), we get: μ_{1}(EL) = 1289 gmol/s Similarly, for P L:

μ_{1}(P L) = μ_{02}(P L) + μ_{51}(P L) + μ_{81}(P L)
= 3 + (0.995)(0.901)(0.932)(0.993)μ_{1}(P L)+

(1)(1){0.099(0.932) + 0.068}(0.993)μ_{1}(P L)

Solving for μ_{1}(P L), we get: μ_{1}(P L) = 268.6 gmol/s

μ_{1}(EL) = μ_{02}(EL) + μ_{51}(EL) + μ_{81}(EL)
= μ_{02}(EL) + (0.995)(0.979)(0.985)(0.93)μ_{1}(EL)+

(1)(1){0.021(0.985) + 0.015}(0.93)μ_{1}(EL)

= 96 + 0.9255μ_{1}(EL)

Solving for μ_{1}(EL), we get: μ_{1}(EL) = 1289 gmol/s Similarly, for P L:

μ_{1}(P L) = μ_{02}(P L) + μ_{51}(P L) + μ_{81}(P L)
= 3 + (0.995)(0.901)(0.932)(0.993)μ_{1}(P L)+

(1)(1){0.099(0.932) + 0.068}(0.993)μ_{1}(P L)

Solving for μ_{1}(P L), we get: μ_{1}(P L) = 268.6 gmol/s

μ_{1}(EL) = μ_{02}(EL) + μ_{51}(EL) + μ_{81}(EL)
= μ_{02}(EL) + (0.995)(0.979)(0.985)(0.93)μ_{1}(EL)+

(1)(1){0.021(0.985) + 0.015}(0.93)μ_{1}(EL)

= 96 + 0.9255μ_{1}(EL)

Solving for μ_{1}(EL), we get: μ_{1}(EL) = 1289 gmol/s Similarly, for P L:

μ_{1}(P L) = μ_{02}(P L) + μ_{51}(P L) + μ_{81}(P L)
= 3 + (0.995)(0.901)(0.932)(0.993)μ_{1}(P L)+

(1)(1){0.099(0.932) + 0.068}(0.993)μ_{1}(P L)

Solving for μ_{1}(P L), we get: μ_{1}(P L) = 268.6 gmol/s

μ_{1}(EL) = μ_{02}(EL) + μ_{51}(EL) + μ_{81}(EL)
= μ_{02}(EL) + (0.995)(0.979)(0.985)(0.93)μ_{1}(EL)+

(1)(1){0.021(0.985) + 0.015}(0.93)μ_{1}(EL)

= 96 + 0.9255μ_{1}(EL)

Solving for μ_{1}(EL), we get: μ_{1}(EL) = 1289 gmol/s Similarly, for P L:

μ_{1}(P L) = μ_{02}(P L) + μ_{51}(P L) + μ_{81}(P L)
= 3 + (0.995)(0.901)(0.932)(0.993)μ_{1}(P L)+

(1)(1){0.099(0.932) + 0.068}(0.993)μ_{1}(P L)

Solving for μ_{1}(P L), we get: μ_{1}(P L) = 268.6 gmol/s

μ_{1}(EL) = μ_{02}(EL) + μ_{51}(EL) + μ_{81}(EL)
= μ_{02}(EL) + (0.995)(0.979)(0.985)(0.93)μ_{1}(EL)+

(1)(1){0.021(0.985) + 0.015}(0.93)μ_{1}(EL)

= 96 + 0.9255μ_{1}(EL)

Solving for μ_{1}(EL), we get: μ_{1}(EL) = 1289 gmol/s Similarly, for P L:

μ_{1}(P L) = μ_{02}(P L) + μ_{51}(P L) + μ_{81}(P L)
= 3 + (0.995)(0.901)(0.932)(0.993)μ_{1}(P L)+

(1)(1){0.099(0.932) + 0.068}(0.993)μ_{1}(P L)

Solving for μ_{1}(P L), we get: μ_{1}(P L) = 268.6 gmol/s

μ_{1}(EL) = μ_{02}(EL) + μ_{51}(EL) + μ_{81}(EL)
= μ_{02}(EL) + (0.995)(0.979)(0.985)(0.93)μ_{1}(EL)+

(1)(1){0.021(0.985) + 0.015}(0.93)μ_{1}(EL)

= 96 + 0.9255μ_{1}(EL)

Solving for μ_{1}(EL), we get: μ_{1}(EL) = 1289 gmol/s Similarly, for P L:

μ_{1}(P L) = μ_{02}(P L) + μ_{51}(P L) + μ_{81}(P L)
= 3 + (0.995)(0.901)(0.932)(0.993)μ_{1}(P L)+

(1)(1){0.099(0.932) + 0.068}(0.993)μ_{1}(P L)

Solving for μ_{1}(P L), we get: μ_{1}(P L) = 268.6 gmol/s

μ_{1}(EL) = μ_{02}(EL) + μ_{51}(EL) + μ_{81}(EL)
= μ_{02}(EL) + (0.995)(0.979)(0.985)(0.93)μ_{1}(EL)+

(1)(1){0.021(0.985) + 0.015}(0.93)μ_{1}(EL)

= 96 + 0.9255μ_{1}(EL)

Solving for μ_{1}(EL), we get: μ_{1}(EL) = 1289 gmol/s Similarly, for P L:

μ_{1}(P L) = μ_{02}(P L) + μ_{51}(P L) + μ_{81}(P L)
= 3 + (0.995)(0.901)(0.932)(0.993)μ_{1}(P L)+

(1)(1){0.099(0.932) + 0.068}(0.993)μ_{1}(P L)

Solving for μ_{1}(P L), we get: μ_{1}(P L) = 268.6 gmol/s

μ_{1}(EL) = μ_{02}(EL) + μ_{51}(EL) + μ_{81}(EL)
= μ_{02}(EL) + (0.995)(0.979)(0.985)(0.93)μ_{1}(EL)+

(1)(1){0.021(0.985) + 0.015}(0.93)μ_{1}(EL)

= 96 + 0.9255μ_{1}(EL)

Solving for μ_{1}(EL), we get: μ_{1}(EL) = 1289 gmol/s Similarly, for P L:

μ_{1}(P L) = μ_{02}(P L) + μ_{51}(P L) + μ_{81}(P L)
= 3 + (0.995)(0.901)(0.932)(0.993)μ_{1}(P L)+

(1)(1){0.099(0.932) + 0.068}(0.993)μ_{1}(P L)

Solving for μ_{1}(P L), we get: μ_{1}(P L) = 268.6 gmol/s

μ_{1}(EL) = μ_{02}(EL) + μ_{51}(EL) + μ_{81}(EL)
= μ_{02}(EL) + (0.995)(0.979)(0.985)(0.93)μ_{1}(EL)+

(1)(1){0.021(0.985) + 0.015}(0.93)μ_{1}(EL)

= 96 + 0.9255μ_{1}(EL)

Solving for μ_{1}(EL), we get: μ_{1}(EL) = 1289 gmol/s Similarly, for P L:

μ_{1}(P L) = μ_{02}(P L) + μ_{51}(P L) + μ_{81}(P L)
= 3 + (0.995)(0.901)(0.932)(0.993)μ_{1}(P L)+

(1)(1){0.099(0.932) + 0.068}(0.993)μ_{1}(P L)

Solving for μ_{1}(P L), we get: μ_{1}(P L) = 268.6 gmol/s

μ_{1}(EL) = μ_{02}(EL) + μ_{51}(EL) + μ_{81}(EL)
= μ_{02}(EL) + (0.995)(0.979)(0.985)(0.93)μ_{1}(EL)+

(1)(1){0.021(0.985) + 0.015}(0.93)μ_{1}(EL)

= 96 + 0.9255μ_{1}(EL)

Solving for μ_{1}(EL), we get: μ_{1}(EL) = 1289 gmol/s Similarly, for P L:

μ_{1}(P L) = μ_{02}(P L) + μ_{51}(P L) + μ_{81}(P L)
= 3 + (0.995)(0.901)(0.932)(0.993)μ_{1}(P L)+

(1)(1){0.099(0.932) + 0.068}(0.993)μ_{1}(P L)

Once we have the limiting reactant flow rate, we can determine other component flow rates as well. Balance for EA

Solving for μ_{1}(P L), we get: μ_{1}(P L) = 268.6 gmol/s μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) (No EA in feed)

μ_{51}(EA) = (1 − ξ)μ_{41}(EA) = 0.995μ_{41}(EA) μ_{41}(EA) = ξ_{41}μ_{31}(EA) = 0.01μ_{31}(EA) μ_{31}(EA) = 0.121μ_{2}(EA) μ_{2}(EA) = η_{1}μ_{1}(EL) + μ_{1}(EA)

= 0.07 × 1289 + μ_{1}(EA)

Thus: μ_{51}(EA) = (0.995)(0.01)(0.121){90.23 + μ_{1}(EA)} Similarly: μ_{81}(EA) = (0.005)(0.995){879 + 0.121(0.99)}{90.23 + μ_{1}(EA)} Substituting in μ_{1}(EA) = μ_{51}(EA) + μ_{81}(EA) and solving for μ_{1}(EA), we get: μ_{1}(EA) = 0.56 gmol/s

Solving Flowsheet Equations – p.6/8

Once μ_{1}(EA) is known, we can calculate the flow rates μ_{2}(EA), μ_{31}(EA), ...

μ_{31}(EA) = 10.99 gmol/s μ_{32}(EA) = 79.81 gmol/s μ_{41}(EA) = 0.11 gmol/s μ_{42}(EA) = 10.88 gmol/s μ_{51}(EA) = 0.1093 gmol/s μ_{52}(EA) = 0.0005 gmol/s μ_{6}(EA) = 90.68 gmol/s μ_{71}(EA) = 90.73 gmol/s μ_{72}(EA) = 0.45 gmol/s μ_{81}(EA) = 0.45 gmol/s μ_{82}(EA) = 89.77 gmol/s

Similarly, molar flow rates for all components can be calculated.

μ_{82} = μ_{82}(M) + μ_{82}(EL) + μ_{82}(P L)+
μ_{82}(DEE) + μ_{82}(EA) + μ_{82}(IP A) + μ_{82}(W)
= 0 + 0 + 0 + 0.1065 + 89.775 + 1.804 + 71.686
= 163.77 gmol/s

Solving Flowsheet Equations – p.7/8

μ_{31}(EA) = 10.99 gmol/s μ_{32}(EA) = 79.81 gmol/s μ_{41}(EA) = 0.11 gmol/s μ_{42}(EA) = 10.88 gmol/s μ_{51}(EA) = 0.1093 gmol/s μ_{52}(EA) = 0.0005 gmol/s μ_{6}(EA) = 90.68 gmol/s μ_{71}(EA) = 90.73 gmol/s μ_{72}(EA) = 0.45 gmol/s μ_{81}(EA) = 0.45 gmol/s μ_{82}(EA) = 89.77 gmol/s

Similarly, molar flow rates for all components can be calculated.

μ_{82} = μ_{82}(M) + μ_{82}(EL) + μ_{82}(P L)+
μ_{82}(DEE) + μ_{82}(EA) + μ_{82}(IP A) + μ_{82}(W)
= 0 + 0 + 0 + 0.1065 + 89.775 + 1.804 + 71.686
= 163.77 gmol/s

Solving Flowsheet Equations – p.7/8

μ_{31}(EA) = 10.99 gmol/s μ_{32}(EA) = 79.81 gmol/s μ_{41}(EA) = 0.11 gmol/s μ_{42}(EA) = 10.88 gmol/s μ_{51}(EA) = 0.1093 gmol/s μ_{52}(EA) = 0.0005 gmol/s μ_{6}(EA) = 90.68 gmol/s μ_{71}(EA) = 90.73 gmol/s μ_{72}(EA) = 0.45 gmol/s μ_{81}(EA) = 0.45 gmol/s μ_{82}(EA) = 89.77 gmol/s

Similarly, molar flow rates for all components can be calculated.

μ_{82} = μ_{82}(M) + μ_{82}(EL) + μ_{82}(P L)+
μ_{82}(DEE) + μ_{82}(EA) + μ_{82}(IP A) + μ_{82}(W)
= 0 + 0 + 0 + 0.1065 + 89.775 + 1.804 + 71.686
= 163.77 gmol/s

Solving Flowsheet Equations – p.7/8

μ_{31}(EA) = 10.99 gmol/s μ_{32}(EA) = 79.81 gmol/s μ_{41}(EA) = 0.11 gmol/s μ_{42}(EA) = 10.88 gmol/s μ_{51}(EA) = 0.1093 gmol/s μ_{52}(EA) = 0.0005 gmol/s μ_{6}(EA) = 90.68 gmol/s μ_{71}(EA) = 90.73 gmol/s μ_{72}(EA) = 0.45 gmol/s μ_{81}(EA) = 0.45 gmol/s μ_{82}(EA) = 89.77 gmol/s

Similarly, molar flow rates for all components can be calculated.

μ_{82} = μ_{82}(M) + μ_{82}(EL) + μ_{82}(P L)+
μ_{82}(DEE) + μ_{82}(EA) + μ_{82}(IP A) + μ_{82}(W)
= 0 + 0 + 0 + 0.1065 + 89.775 + 1.804 + 71.686
= 163.77 gmol/s

Solving Flowsheet Equations – p.7/8

μ_{31}(EA) = 10.99 gmol/s μ_{32}(EA) = 79.81 gmol/s μ_{41}(EA) = 0.11 gmol/s μ_{42}(EA) = 10.88 gmol/s μ_{51}(EA) = 0.1093 gmol/s μ_{52}(EA) = 0.0005 gmol/s μ_{6}(EA) = 90.68 gmol/s μ_{71}(EA) = 90.73 gmol/s μ_{72}(EA) = 0.45 gmol/s μ_{81}(EA) = 0.45 gmol/s μ_{82}(EA) = 89.77 gmol/s

Similarly, molar flow rates for all components can be calculated.

μ_{82} = μ_{82}(M) + μ_{82}(EL) + μ_{82}(P L)+
μ_{82}(DEE) + μ_{82}(EA) + μ_{82}(IP A) + μ_{82}(W)
= 0 + 0 + 0 + 0.1065 + 89.775 + 1.804 + 71.686
= 163.77 gmol/s

Solving Flowsheet Equations – p.7/8

μ_{31}(EA) = 10.99 gmol/s μ_{32}(EA) = 79.81 gmol/s μ_{41}(EA) = 0.11 gmol/s μ_{42}(EA) = 10.88 gmol/s μ_{51}(EA) = 0.1093 gmol/s μ_{52}(EA) = 0.0005 gmol/s μ_{6}(EA) = 90.68 gmol/s μ_{71}(EA) = 90.73 gmol/s μ_{72}(EA) = 0.45 gmol/s μ_{81}(EA) = 0.45 gmol/s μ_{82}(EA) = 89.77 gmol/s

Similarly, molar flow rates for all components can be calculated.

μ_{82} = μ_{82}(M) + μ_{82}(EL) + μ_{82}(P L)+
μ_{82}(DEE) + μ_{82}(EA) + μ_{82}(IP A) + μ_{82}(W)
= 0 + 0 + 0 + 0.1065 + 89.775 + 1.804 + 71.686
= 163.77 gmol/s

Solving Flowsheet Equations – p.7/8

μ_{31}(EA) = 10.99 gmol/s μ_{32}(EA) = 79.81 gmol/s μ_{41}(EA) = 0.11 gmol/s μ_{42}(EA) = 10.88 gmol/s μ_{51}(EA) = 0.1093 gmol/s μ_{52}(EA) = 0.0005 gmol/s μ_{6}(EA) = 90.68 gmol/s μ_{71}(EA) = 90.73 gmol/s μ_{72}(EA) = 0.45 gmol/s μ_{81}(EA) = 0.45 gmol/s μ_{82}(EA) = 89.77 gmol/s

Similarly, molar flow rates for all components can be calculated.

μ_{82} = μ_{82}(M) + μ_{82}(EL) + μ_{82}(P L)+
μ_{82}(DEE) + μ_{82}(EA) + μ_{82}(IP A) + μ_{82}(W)
= 0 + 0 + 0 + 0.1065 + 89.775 + 1.804 + 71.686
= 163.77 gmol/s

Solving Flowsheet Equations – p.7/8

μ_{31}(EA) = 10.99 gmol/s μ_{32}(EA) = 79.81 gmol/s μ_{41}(EA) = 0.11 gmol/s μ_{42}(EA) = 10.88 gmol/s μ_{51}(EA) = 0.1093 gmol/s μ_{52}(EA) = 0.0005 gmol/s μ_{6}(EA) = 90.68 gmol/s μ_{71}(EA) = 90.73 gmol/s μ_{72}(EA) = 0.45 gmol/s μ_{81}(EA) = 0.45 gmol/s μ_{82}(EA) = 89.77 gmol/s

Similarly, molar flow rates for all components can be calculated.

μ_{82} = μ_{82}(M) + μ_{82}(EL) + μ_{82}(P L)+
μ_{82}(DEE) + μ_{82}(EA) + μ_{82}(IP A) + μ_{82}(W)
= 0 + 0 + 0 + 0.1065 + 89.775 + 1.804 + 71.686
= 163.77 gmol/s

Solving Flowsheet Equations – p.7/8

μ_{31}(EA) = 10.99 gmol/s μ_{32}(EA) = 79.81 gmol/s μ_{41}(EA) = 0.11 gmol/s μ_{42}(EA) = 10.88 gmol/s μ_{51}(EA) = 0.1093 gmol/s μ_{52}(EA) = 0.0005 gmol/s μ_{6}(EA) = 90.68 gmol/s μ_{71}(EA) = 90.73 gmol/s μ_{72}(EA) = 0.45 gmol/s μ_{81}(EA) = 0.45 gmol/s μ_{82}(EA) = 89.77 gmol/s

Similarly, molar flow rates for all components can be calculated.

μ_{82} = μ_{82}(M) + μ_{82}(EL) + μ_{82}(P L)+
μ_{82}(DEE) + μ_{82}(EA) + μ_{82}(IP A) + μ_{82}(W)
= 0 + 0 + 0 + 0.1065 + 89.775 + 1.804 + 71.686
= 163.77 gmol/s

Solving Flowsheet Equations – p.7/8

μ_{31}(EA) = 10.99 gmol/s μ_{32}(EA) = 79.81 gmol/s μ_{41}(EA) = 0.11 gmol/s μ_{42}(EA) = 10.88 gmol/s μ_{51}(EA) = 0.1093 gmol/s μ_{52}(EA) = 0.0005 gmol/s μ_{6}(EA) = 90.68 gmol/s μ_{71}(EA) = 90.73 gmol/s μ_{72}(EA) = 0.45 gmol/s μ_{81}(EA) = 0.45 gmol/s μ_{82}(EA) = 89.77 gmol/s

Similarly, molar flow rates for all components can be calculated.

μ_{82} = μ_{82}(M) + μ_{82}(EL) + μ_{82}(P L)+
μ_{82}(DEE) + μ_{82}(EA) + μ_{82}(IP A) + μ_{82}(W)
= 0 + 0 + 0 + 0.1065 + 89.775 + 1.804 + 71.686
= 163.77 gmol/s

Solving Flowsheet Equations – p.7/8

μ_{31}(EA) = 10.99 gmol/s μ_{32}(EA) = 79.81 gmol/s μ_{41}(EA) = 0.11 gmol/s μ_{42}(EA) = 10.88 gmol/s μ_{51}(EA) = 0.1093 gmol/s μ_{52}(EA) = 0.0005 gmol/s μ_{6}(EA) = 90.68 gmol/s μ_{71}(EA) = 90.73 gmol/s μ_{72}(EA) = 0.45 gmol/s μ_{81}(EA) = 0.45 gmol/s μ_{82}(EA) = 89.77 gmol/s

Similarly, molar flow rates for all components can be calculated.

μ_{82} = μ_{82}(M) + μ_{82}(EL) + μ_{82}(P L)+
μ_{82}(DEE) + μ_{82}(EA) + μ_{82}(IP A) + μ_{82}(W)
= 0 + 0 + 0 + 0.1065 + 89.775 + 1.804 + 71.686
= 163.77 gmol/s

Solving Flowsheet Equations – p.7/8

μ_{31}(EA) = 10.99 gmol/s μ_{32}(EA) = 79.81 gmol/s μ_{41}(EA) = 0.11 gmol/s μ_{42}(EA) = 10.88 gmol/s μ_{51}(EA) = 0.1093 gmol/s μ_{52}(EA) = 0.0005 gmol/s μ_{6}(EA) = 90.68 gmol/s μ_{71}(EA) = 90.73 gmol/s μ_{72}(EA) = 0.45 gmol/s μ_{81}(EA) = 0.45 gmol/s μ_{82}(EA) = 89.77 gmol/s

Similarly, molar flow rates for all components can be calculated.

μ_{82} = μ_{82}(M) + μ_{82}(EL) + μ_{82}(P L)+
μ_{82}(DEE) + μ_{82}(EA) + μ_{82}(IP A) + μ_{82}(W)
= 0 + 0 + 0 + 0.1065 + 89.775 + 1.804 + 71.686
= 163.77 gmol/s

Solving Flowsheet Equations – p.7/8

Final Azeotropic Separation

We want to recover 99.5% of azeotropic composition ((85.5% EA and 14.5% W).

⟹ 99.5% EA + necessary water
⟹ μ_{91}(EA) = 0.995 × 89.775 = 89.33 gmol/s

μ_{91}(W) = \frac{14.5}{85.5} × 89.33 = 15.17 gmol/s

μ_{91}(DEE) = 0.01065 gmol/s (lighter than azeotrope)

We want IP A to be less than 0.1% in the top product. Thus:

\frac{μ_{91}(IP A)}{μ_{91}(IP A) + μ_{91}(DEE) + μ_{91}(W) + μ_{91}(EA)} = 0.001

We want to recover 99.5% of azeotropic composition ((85.5% EA and 14.5% W).

⟹ 99.5% EA + necessary water
⟹ μ_{91}(EA) = 0.995 × 89.775 = 89.33 gmol/s

μ_{91}(W) = \frac{14.5}{85.5} × 89.33 = 15.17 gmol/s

μ_{91}(DEE) = 0.01065 gmol/s (lighter than azeotrope)

We want IP A to be less than 0.1% in the top product. Thus:

\frac{μ_{91}(IP A)}{μ_{91}(IP A) + μ_{91}(DEE) + μ_{91}(W) + μ_{91}(EA)} = 0.001

We want to recover 99.5% of azeotropic composition ((85.5% EA and 14.5% W).

⟹ 99.5% EA + necessary water
⟹ μ_{91}(EA) = 0.995 × 89.775 = 89.33 gmol/s

μ_{91}(W) = \frac{14.5}{85.5} × 89.33 = 15.17 gmol/s

μ_{91}(DEE) = 0.01065 gmol/s (lighter than azeotrope)

We want IP A to be less than 0.1% in the top product. Thus:

\frac{μ_{91}(IP A)}{μ_{91}(IP A) + μ_{91}(DEE) + μ_{91}(W) + μ_{91}(EA)} = 0.001

We want to recover 99.5% of azeotropic composition ((85.5% EA and 14.5% W).

⟹ 99.5% EA + necessary water
⟹ μ_{91}(EA) = 0.995 × 89.775 = 89.33 gmol/s

μ_{91}(W) = \frac{14.5}{85.5} × 89.33 = 15.17 gmol/s

μ_{91}(DEE) = 0.01065 gmol/s (lighter than azeotrope)

We want IP A to be less than 0.1% in the top product. Thus:

\frac{μ_{91}(IP A)}{μ_{91}(IP A) + μ_{91}(DEE) + μ_{91}(W) + μ_{91}(EA)} = 0.001

We want to recover 99.5% of azeotropic composition ((85.5% EA and 14.5% W).

⟹ 99.5% EA + necessary water
⟹ μ_{91}(EA) = 0.995 × 89.775 = 89.33 gmol/s

μ_{91}(W) = \frac{14.5}{85.5} × 89.33 = 15.17 gmol/s

μ_{91}(DEE) = 0.01065 gmol/s (lighter than azeotrope)

We want IP A to be less than 0.1% in the top product. Thus:

\frac{μ_{91}(IP A)}{μ_{91}(IP A) + μ_{91}(DEE) + μ_{91}(W) + μ_{91}(EA)} = 0.001

We want to recover 99.5% of azeotropic composition ((85.5% EA and 14.5% W).

⟹ 99.5% EA + necessary water
⟹ μ_{91}(EA) = 0.995 × 89.775 = 89.33 gmol/s

μ_{91}(W) = \frac{14.5}{85.5} × 89.33 = 15.17 gmol/s

μ_{91}(DEE) = 0.01065 gmol/s (lighter than azeotrope)

We want IP A to be less than 0.1% in the top product. Thus:

\frac{μ_{91}(IP A)}{μ_{91}(IP A) + μ_{91}(DEE) + μ_{91}(W) + μ_{91}(EA)} = 0.001

We want to recover 99.5% of azeotropic composition ((85.5% EA and 14.5% W).

⟹ 99.5% EA + necessary water
⟹ μ_{91}(EA) = 0.995 × 89.775 = 89.33 gmol/s

μ_{91}(W) = \frac{14.5}{85.5} × 89.33 = 15.17 gmol/s

μ_{91}(DEE) = 0.01065 gmol/s (lighter than azeotrope)

We want IP A to be less than 0.1% in the top product. Thus:

\frac{μ_{91}(IP A)}{μ_{91}(IP A) + μ_{91}(DEE) + μ_{91}(W) + μ_{91}(EA)} = 0.001

We want to recover 99.5% of azeotropic composition ((85.5% EA and 14.5% W).

⟹ 99.5% EA + necessary water
⟹ μ_{91}(EA) = 0.995 × 89.775 = 89.33 gmol/s

μ_{91}(W) = \frac{14.5}{85.5} × 89.33 = 15.17 gmol/s

μ_{91}(DEE) = 0.01065 gmol/s (lighter than azeotrope)

We want IP A to be less than 0.1% in the top product. Thus:

\frac{μ_{91}(IP A)}{μ_{91}(IP A) + μ_{91}(DEE) + μ_{91}(W) + μ_{91}(EA)} = 0.001

We want to recover 99.5% of azeotropic composition ((85.5% EA and 14.5% W).

⟹ 99.5% EA + necessary water
⟹ μ_{91}(EA) = 0.995 × 89.775 = 89.33 gmol/s

μ_{91}(W) = \frac{14.5}{85.5} × 89.33 = 15.17 gmol/s

μ_{91}(DEE) = 0.01065 gmol/s (lighter than azeotrope)

We want IP A to be less than 0.1% in the top product. Thus:

\frac{μ_{91}(IP A)}{μ_{91}(IP A) + μ_{91}(DEE) + μ_{91}(W) + μ_{91}(EA)} = 0.001