# Example 2.8: Energy balance in a column feed

An adiabatic distillation column is used to separate a mixture of n-butane (1) and n-heptane (2). The liquid feed is introduced directly on the third stage as shown on figure 1.

The column operates at an isobaric pressure of 2.26 atm. Some additional pieces of information concerning the characteristics of the feed and the surrounding stages are also given as follows:

Table 1: Data of the example
Position Feed Stage 2 Stage 4
Temperature (°C) 47.5
Vapour molar flow (mol/s) 0 43.5
Liquid molar flow (mol/s) 100 0.46 118.2
Vapour molar composition y1 0.5 0.565 0.990
Liquid molar composition x1 0.969

For the simplicity of the model, some basic expressions have been selected for enthalpy calculation and for distribution coefficient predictions: the expressions are (with T in °C, and h in cal/mol) for both phases and (with T in K). These equations apply to each component and the values of the coefficients are found below:

Table 2: Parameters for enthalpies and equilibrium constants
Vapour enthalpy Liquid enthalpy Equilibrium constant
Component
n-Butane 23.3 5470 34 0 -2530.4 8.5426
n-Heptane 39.7 9128 54 0 -4124.6 10.412
• What is the temperature of stage number 2?
• What is the temperature of stage number 4?
• What are the temperature and compositions of stage number 3?

## Analysis:

The properties to be evaluated are the temperature of the various equilibria on the 3rd stage around the feed. Pressure is fixed and part of the compositions is given.

• a) For the 2nd stage the liquid and vapour composition are known, so the distribution equilibrium are directly obtained. The temperature can be calculated using the relationship between Ki and T.
• b) For the 4th stage, composition of liquid is known meaning that a bubble temperature point has to be determined.
• c) On the 3rd stage, all incoming flows are known, so the overall composition is known as the total enthalpy may be calculated. A PH flash must be solved using the set of following equations:

The fluids are regular light hydrocarbons. All properties are well known. Particular expressions are recommended and no binary interaction is considered (ideal mixture, i.e. ).

All fluids will be found in both phases as vapour or liquid due to the under-critical pressure zone.

## Solution:

### See complete results in file (xls):

Some help on nomenclature and tips to use this file can be found here.

Plate number 2 is defined in both phases. So the distribution coefficients of each component can be calculated (it is a binary mixture!).

and

Normally, knowledge of T on the 2nd stage will give both and , so there are two equations for one unknown. Both equations lead to:

and

.

The average is 316.08.

For the 4th stage, the bubble calculation has to be undertaken. The iterative procedure must start with an initial value (perhaps the temperature of stage number 2, or any other chosen by the user). If the temperature of stage 2 is used, equation yields

The temperature must be lowered to reduce the sum. After a few iterations, a value of 298.66 K is found for plate number 4.

Material and energy balances around the 3rd stage must be made.

This implies that enthalpies of each flow must be calculated. For example, the enthalpy of the vapour leaving stage 2, it is expressed by:

Using this technique, the material and enthalpy content of each of the streams entering the 3rd plate can be calculated and summed as shown in table 3.

Table 3: Material and enthalpy content of the stream entering plate 3
Component
n-Butane (mol/s) 50 42.15 109.93 202.08
n-Heptane (mol/s) 50 1.35 8.27 59.62
Total nC4+nC7 100 43.5 118.2 261.7
h (cal/mol) 2090 6605 903 2303
H (cal/s) 209000 287336 106740 603077

Introducing the equilibrium condition in the above material balances (2 first equations), they can be treated as mass balance leading to the Rachford Rice equation. For a binary mixture, the vapour fraction can be calculated directly:

We are now left with two equations (the enthalpy balance and the above Rachford Rice equation) with two unknowns (θ and T). The complete algorithm becomes very simple:

• Assume a temperature
• Calculate K1 and K2
• Solve the Rachford Rice equation to find θ
• Calculate the enthalpies of streams V3 and L3
• Compare the sum of these two outlet enthalpies with the inlet enthalpy calculated in table 3.

A complete iteration is shown in table 4.

Table 4: Results of one iteration, using an assumed temperature of 40°C, yielding a vapour fraction θ=0.4366 and
a total enthalpy of 989.444 kcal/s
Temperature (°C) = 40 0.4366
Component
n-Butane (mol/s) 202.08 0.772 1.587 0.615 0.976 90.62 111.46 202.08
n-Heptane (mol/s) 59.62 0.228 0.063 0.385 0.024 56.83 2.79 59.62
Total nC4+nC7 261.7 1 1 1 147.45 114.25 261.7
h (cal/mol) 1668 6507 3781
H (cal/s) 603077 246002 743442 989444

As observed in the results above, calculated at a temperature of 40 °C, the vapour fraction of the flash is 0.4366 and the total enthalpy of is 989444, greater than the value of 603077 for . The temperature must therefore be less than 40 °C. A few additional iterations lead to a final value of 34.29 °C with the corresponding compositions and vapour fraction as seen below:

Table 5: Result of the enthalpy balance
Temperature (°C) = 34.29 0.1904
Component
n-Butane (mol/s) 202.08 0.772 1.367 0.722 0.986 152.93 49.15 202.08
n-Heptane (mol/s) 59.62 0.228 0.050 0.278 0.014 58.94 0.69 59.62
Total nC4+nC7 261.7 1 1 1 211.87 49.84 261.7
h (cal/mol) 1357 6327 2303
H (cal/s) 603077 287442 315209 603077